Ivars Rankis, Janis Zakis, Anastasia Zhiravetska. Power Electronics. Riga, RTU Press, 2018. 288 p.

The textbook is developed for the students, who study power electronics and other close segments of science and technologies. It can be useful for the students and professionals who deal with semiconductor devices, DC-DC converters, rectifiers, inverters, and their application.

Reviewer prof. Dr. habil. sc. ing. L. Ribickis English proofreading Technical Editor Computer design Cover design

Cover picture from Shutterstock.com

Oksana Ivanova Irēna Skārda Baiba Puriņa Jekaterina Lukina

Published by RTU Press 1-422 Kalku Street, Riga, LV-1658 phone: +37167089441 e-mail: izdevnieciba@rtu.lv Printed by RTU Digital Center

This book has been completed with the financial support of Latvian Council of Science. The authors from Riga Technical University are responsible for the content of this document. This publication reflects the views of the authors only. The book is developed within the frames of the project «Research and Development of Impedance Source DC/DC Converters», Latvian Council of Science grant No. 416/2012 and «New single stage buck-boost multilevel inverters for renewable energy applications», Latvian Council of Science grant No. 673/2014.

ISBN 978-9934-22-068-5 (print) ISBN 978-9934-22-069-2 (epub)

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© Riga Technical University, 2018 © Ivars Rankis, Janis Zakis, Anastasia Zhiravetska, 2018

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Contents Preface.............................................................................................................. 5 Introduction..................................................................................................... 6 1. Elements of semiconductor converters and the basic conceptions........... 11

1.1. Fundamental electrical parameters..................................................................... 11 1.2. Properties of the passive reactive elements........................................................ 15 1.3. Properties of diode and its characteristics......................................................... 20 1.4. Bipolar transistor (BT) as an element of powerful converter........................... 29 1.5. Thyristors................................................................................................................ 35 1.6. Power MOSFETs.................................................................................................... 47 1.7. Insulated Gate Bipolar Transistors IGBT........................................................... 49 1.8. Thermal mode calculations of semiconductor elements................................... 52 1.9. Control and protection of transistors................................................................. 55 Review questions................................................................................................... 58 Tasks........................................................................................................................ 61

2. Line-frequency converters (LFC)................................................................ 65

2.1. Diode rectifiers....................................................................................................... 65

2.1.1. Rectifier circuits...................................................................................................... 65 2.1.2. Single-phase bridge rectifier.................................................................................. 67 2.1.3. Single-phase centre-tap rectifier with transformer........................................... 72 2.1.4. Three-phase half-cycle (centre-tap) rectifier....................................................... 74 2.1.5. Three-phase full-bridge rectifier.......................................................................... 78 2.1.6. Six-phase rectifier................................................................................................... 80 2.1.7. Twelve-pulse rectifier............................................................................................. 83 2.1.8. Ripple factor of the rectified voltage.................................................................... 86 2.1.9. Form of the network current................................................................................. 88 2.1.10. The process of current commutation................................................................... 90

Review questions................................................................................................... 96 Tasks........................................................................................................................ 99 2.2. Line-frequency phase controlled converters...................................................... 101

2.2.1. Control characteristics........................................................................................... 101 2.2.2. Inverting operation mode of the controlled rectifiers....................................... 105 2.2.3. Commutation processes in the controlled rectifiers......................................... 108 2.2.4. Power factor of the controlled rectifier................................................................ 114 2.2.5. Reversible rectifiers................................................................................................. 120

2.3. Cycloconverters..................................................................................................... 125 2.4. AC voltage controllers........................................................................................... 133 2.5. Control principles of the line-frequency converters......................................... 140 Review questions................................................................................................... 144 Tasks........................................................................................................................ 146

3. Switch-mode dc-dc converters.................................................................... 149 3.1. DC/DC converters................................................................................................. 149

3.1.1. Step-down (buck) converter.................................................................................. 149 3.1.2. Step-up (boost) converter...................................................................................... 152

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3.1.3. Improvement of the converter characteristics with autotransformer........... 154 3.1.4. Buck-boost converter............................................................................................. 155 3.1.4.1. Conventional topology........................................................................... 155 3.1.4.2. Cuk converter.......................................................................................... 157 3.1.4.3. Buck-boost converter of the load voltage of the same polarity........ 158 3.1.4.4. Buck-boost converter of commutated configuration......................... 159 3.1.5. Input and output filters.......................................................................................... 159 3.1.6. Multi-phase buck converters................................................................................. 165 3.1.7. Discontinuous load current operation mode of buck converter..................... 168 3.1.8. Full-bridge DC-DC converter............................................................................... 172 3.1.9. Implementation of qZ Source network................................................................ 177 3.1.10. DC-DC converters with thyristors....................................................................... 180 3.1.11. Operation of transistor switch in real circuit..................................................... 183 3.1.11.1. Process of transistor switching............................................................. 183 3.1.11.2. Safe operation area (SOA)...................................................................... 187 3.1.11.3. Improvement of the turn-off process................................................... 189 3.1.11.4. Improvement of the turn-on process................................................... 194 3.1.12. Isolated DC-DC power supply.............................................................................. 196 3.1.13. Calculation of the DC smoothing inductor........................................................ 201

Review questions................................................................................................... 205 Tasks........................................................................................................................ 208 3.2. Voltage and current source inventers (VSI and CSI)........................................ 212

3.2.1. Single-phase voltage source inverters (VSI)....................................................... 212 3.2.2. Features of bridge-type single-phase voltage source inverter.......................... 214 3.2.3. Features of neutral point transformer based single-phase inverter................ 218 3.2.4. Single-phase voltage source inverter control...................................................... 220 3.2.5. PWM-regulated single-phase full-bridge voltage source inverter.................. 225 3.2.6. Three-phase voltage source inverter operation principle................................. 229 3.2.7. Three-phase voltage source inverter control....................................................... 234 3.2.8. Multi-level voltage source inverters..................................................................... 240

3.3. Current source inverters....................................................................................... 242

3.3.1. Operation principle of current source inverter.................................................. 242 3.3.2. Three-phase current source inverters.................................................................. 248 3.3.3. Control of the voltage of current source inverter.............................................. 252

3.4. Resonance inverters.............................................................................................. 253 3.5. Commutated converter control systems............................................................. 257 Review questions................................................................................................... 262 Tasks........................................................................................................................ 265

4. Network power converters with free commutation capacities.................. 267 4.1. Rectifier — power conditioner.............................................................................. 267 4.2. Rectifier — network current waveform corrector.............................................. 279 Review questions................................................................................................... 285 Tasks........................................................................................................................ 287

References......................................................................................................... 288

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PREFACE This is an issue of the textbook in one of the modern study courses — Power Electronics. The course studies the application of the semiconductor elements — diodes, thyristors, transistors — for converting of the electric energy and control of its parameters. The tasks of alternative current converting into direct current, the inverting, control of AC without transformer and DC without rheostat are topical in both electrical and power engineering. Therefore nowadays the mastering of this course is very significant for the electrical engineering education. The students are expected to solve all the assignments related to each chapter for better acquisition of the material. Note that letter m in the assignments is the penult digit of the student’s ID card number, but n — the last one. The solutions of almost all the assignments can be examined by means of laboratory software VirtualLab available at home page www.etdv.rtu.lv . The virtual laboratories allow better understanding of the material, provide visual representation of the circuits as well as the technical solution and obtained results. The virtual laboratories have two versions v1.0 and v2.0; they can be easily downloaded into the students’ computers by means of a correspondent command and continuing with the commands run, next, next, next, install, finish, placing an icon of this software onto the desktop. Version v1.0 contains the models of single-phase and DC converters as well as some models of three-phase converters; version v2.0 contains mostly three-phase converters circuits and systems. All the converters operate in accordance with the particular algorithms correspondent to those considered in the textbook that gives an opportunity to know the influence of different parameters on the output of the converters. The step of the simulation is 10 µs, that restricts the use of very high modulation frequency, it cannot exceed 10 kHz that is not very high indicator for modern converters. All suggestions about content of the book as well as recommendations on improving of explanation can be sent to the Press of Riga Technical University.

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INTRODUCTION Economy requires such electrical energy supply sources (SS) the output parameters of which do not correspond to the ones traditionally used from the primary sources (most often Electrical Plants). Such alternative SS are denominated as the secondary ones. For example the primary SS produces three-phase alternating current AC but a technologic process demands regulated direct current DC. In this case the secondary SS provides convertion of AC voltage to DC and this operation is named a r ectification. If the primary SS, for instance, photovoltaic source, produces DC voltage, but this source has to be connected with the traditional AC grid, then the secondary SS has to provide conversion of DC to the grid’s AC. Such operation mode can be denominated as Grid inverting. Often it is necessary to change the number of phases, frequency and voltage of the standard AC supply for technology applications. Such operations also can be realized using the secondary SS. The tasks can be solved first converting AC voltage to DC and then inverting the obtained voltage with an autonomous inverter. In some cases variation of the primary AC source voltage can be obtained using simple devices — transformers. However, in most of the cases for solving the tasks special power electronic converters are to be used as more efficient. Still a more difficult task is connected with the change of the primary source DC voltage to DC of another voltage level. As direct electro-magnetic transformation of DC voltage is impossible a converter type secondary SS in this case can be applied; for instance, the first stage of it is an autonomous inverter then electro-magnetic transformation is realized and at last — rectification. Such conversions in contemporary technologies are provided by means of semiconductor (usually silicon) unidirectional current control elements — controlled and uncontrolled ones. Historically the first created uncontrolled elements were the diodes (Fig.1), which allow to develop uncontrolled rectifiers. a

b p

A

c

iV

n K

A

V

iV uV

K 0

uV

Fig.1. Bipolar structure of the diode (a), denomination in electrical circuits (b), simplified volt-amper (VA) characteristic (c) 6

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A diode comprises two layers — a semiconductor layer with high concentration of positive charges (p) and a layer with high concentration of negative charges (n). The electrode-terminal connected to the first layer is named as anode (A), the terminal of the second layer — cathode (K). Electrical current through the diode can pass only in the direction from anode A to the cathode K and voltage across current conducting diode uV between A and K is close to zero value (Fig.1,c). If real polarity of voltage across diode is opposite to the diode’s conductivity direction then the current of the diode is close to zero but its cathode is of positive polarity in respect to its anode, i.e., the diode is applied with a reverse voltage uV < 0 V which is equal to the voltage of supply circuit — the diode is reverse biased. In 1948 scientists of the USA invented a bipolar transistor (BT) — controlled unidirectional current element with certain continuous connection between controlled current and control current, that allows to develop contact-less high frequency switch for realization of switching converters. Really in one structure a BT comprises two contrary connected diodes (Fig.2), common connection point of which is applied as a control terminal — base B. a

b p

n

p

E

n C

VT C B

n

C

E B

B E

p

C

VT

E

B

Fig. 2. Substitution circuits of the p-n-p structure (a) and n-p-n (b) structure of the BT, three-layer semiconductor structures and BT denominations in electrical circuits

In such way p-n-p (a) and n-p-n (b) structures of the bipolar transistor can be developed. If the both end side terminals of transistor are switched to the AC current circuit then the current through transistor will be at zero level because each time one of the diodes is reverse biased. But if through one of the diodes from one side terminal E (emitter) to the terminal of the middle layer B (base) in p-n-p structure a DC current is passed (in n-p-n structure from B to E), then the middle n layer is supplied with p charges coming from emitter. Collector-base internal diode’s potential barrier is weakened and at connected to another side terminal C (collector) DC voltage of same polarity in respect to the emitter as the base has, a part of p charges emitted by emitter will move from middle layer to the collector layer and current will flow in the main emitter-collector circuit. The higher is the base current the higher are also emitter and 7

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collector currents. This transfer process of emitter charges to collector layer is named a transistor effect. In 1951 two different sort silicon BTs have been integrated in one structure (Fig.3) developing a semiconductor element turned-on with the help of short firing pulse — thyristor. Such element allowed to develop controlled converters operating directly with network AC voltage — rectifiers, AC controllers, network inverters, cycloconverters — dividers of network AC voltage frequency. Using forced commutation devices and circuits gave a possibility also to develop autonomous thyristor inverters and DC voltage choppers. Emitter of the p-n-p transistor of thyristor is connected to the anode terminal A, emitter of the n-p-n transistor — to the cathode terminal K (Fig.3), but collector of the p-n-p transistor and base of the n-p-n transistor both are connected to the firing (turn-on) terminal — gate G. Normally current between A and K can flow if potential of the ano de is more positive than that of the cathode and only after activating of control circuit between G and K with relative small DC current (about 0.5 A) short-time firing pulse. Main disadvantage of BT is a relatively small relation between controlled and control current (usually 10 till 100 - fold). Such characteristic of BT did not allow the development of easy controlled semiconductor switches for powerful applications, because then it is necessary to apply control current of high level. Improving the structures of transistors in 80-ties a powerful controlled with voltage level a field effect transistors MOSFET, i.e. metal-oxide-semiconductor-field-effect transistors, have been created. Application of the MOSFET transistors solved the problem with necessity of high control current and that allowed produce powerful switched converters. Really the MOSFET transistors have one essential drawback connected with its channel voltage drop which rises up when applying voltages of the level above some two hundred volts and therefore it limited application of MOSFET in large scale high voltage converters. a

b

A

p-n-p

n-p-n

G

A

K

VS G

K Fig.3. Substitution circuit of thyristor (a) and its denomination in electrical circuits (b) 8

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Efforts to develop in full extent controlled thyristors were crowned with elaboration of GTO element (GTO — gate turn off thyristor), which can be turned on as well as turned off using a bidirectional gate control signal. But practical implementation of such thyristors has been rather difficult and dynamic characteristics of GTO are not excellent thus limiting wide application of the elements. Integrating the structures of bipolar transistor, thyristors and field effect transistors in the 80-ies the most popular contemporary powerful transistors IGBT (Insulated Gate Bipolar Transistor) have been created. They can be switched with discrete changing of control voltage applying it to its input MOSFET, but in conductivity position IGBT has the properties similar to those of thyristors with relative small voltage drop in turn-on situation. Development of each version of powerful semiconductor device allowed create converters with new and improved characteristics as well as increase their power, quality and functionality. Nowadays a large number of converter circuits should be studied but all range of them is not possible to master in student courses. Therefore all the range of the converters in the book is divided into 3 parts: • line-frequency converters; • switching ones; • line-frequency with switching elements. For the first group all electric-magnetic processes and characteristics are directly dependent on the sine-form and frequency of network and the group comprises rectifiers, grid inverters, AC controllers and cycloconverters — dividers of network voltage frequency. Because of the dependence on the supply AC grid parameters such conver ters are named as line-frequency. The main semiconductor elements for implementation of such converters are diodes and thyristors, the latter for controlled applications. The second group includes the converters with DC supply — DC voltage choppers and autonomous inverters — autonomous converters of DC voltage to AC with an arbitrary frequency and phase number of output voltage. Converters of the group can be denominated as switching. The main semiconductor elements for implementation of such converters nowadays are powerful semiconductor transistors — IGBT and MOSFET. Operation frequency of the converters reaches-up to hundreds kilohertz. In the last time more and more line-frequency converter’s sorts are supplied with switching elements that allow obtaining the new properties at operation of line-frequency converters — improved quality and power factor of converters as well as bidirectional power flow of converter-grid circuits. In such combined converter systems the ordinary applied elements — diodes and thyristors — are complemented with high frequency switching transistor elements. The line-frequency switching converter group comprises AC voltage choppers, active rectifiers, active filters, direct matrix

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type frequency converters. It can be mentioned that converters of the third group now are at the stage of its rapid improvement for different applications. As it has been mentioned, in the student course it is not possible to discuss all the range of possible converter circuits and applications. Therefore here only the most popular circuits having wide application are discussed. The main method for examination of the circuits is connected with learning of electric-magnetic processes in the circuits and its elements — investigation of instantaneous meanings and time diagrams of voltages and currents, which allow to find connections between averaged and rms (effective) values of voltage and current parameters as well as their power at different stages and links of circuits. The main fundamental aspects are formulated in the first part of the book for starting a learning of the converter circuits. They have to be applied for investigation of characteristics of electric-magnetic processes, discussing of engineering approach to the applications of different semiconductor elements and their characteristics, examining of the main passive elements forming converter circuits. In the second part rectifiers — diode and controlled ones and their applications on the base of thyristors for grid inverting, output voltage reversion and dividing of grid voltage frequency, as well as AC voltage controllers are discussed. In the third part DC supply voltage converters — non-insulated and insulated in respect to load are discussed — DC choppers, reversible pulse converters, DC-AC autonomous inverters. The fourth part considers line frequency converters with switching elements in its circuits — the single-phase and three-phase power factor correctors, bidirectional active rectifiers, AC voltage choppers. Simultaneously with the examination of power circuits of the converters the most often applied control principles and circuits implementing it are discussed — in analogue as well as in digital versions.

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1st part

1. ELEMENTS OF SEMICONDUCTOR CONVERTERS AND THE BASIC CONCEPTIONS 1.1. Fundamental Electrical Parameters In the circuits of semiconductor converters often a non-sine form current passes through and therefore it has to be exactly coordinated with value and application of voltage and current parameters. If for instance AC sine-form voltage is supplied to a circuit with an ideal diode VD and resistor R (Fig.1.1) connected in series, then an instantaneous meanings of current i in one half-cycle corresponds to the shape of sine-form input voltage for one half-wave but for the second half-cycle they are at zero level. If instantaneous meanings for currents of the both elements are equal then for voltages at accepted positive directions they are different. i

Im

V uV u = Um sinωt

Iav R

0 uR

i

uV 0 uR 0

p

2p

wt wt

URm = Um

–Um

wt

Fig.1.1. Diagrams of instantaneous values of current (i) and voltages of elements (uV, uR) of the circuit under consideration

Instantaneous meanings will be denominated with small letters i and u, peak values of current and voltages with capital letters and index m: for current i its peak value is Im, for diode voltage uV its maximum voltage at reverse biasing is Um, resistor’s voltage uR maximum value is U Rm = U m . If the circuit is provided with magneto-electric measure devices (galvanometer type) with its pointer’s moving in direction in accordance with the polarity of current passing through then the average values, or DC component, within one cycle of the signal changing is measured as:

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1st part average current I av =

1 2π

π

∫I

m

sin ωtdωt =

0

Im I [− cos ωt ]π0 = m ; (1-1) 2π π

average voltage of the diode U Vav

1 = 2π

2π

∫U

m

sin ωtdωt =

π

Um U [− cos ωt ]2ππ = − m ; (1-2) π 2π

average voltage of the resistor U Rav = U R(d ) =

1 2π

π

∫U

m

sin ωtdωt =

0

Um . (1-3) π

Average values also will be denominated with the capital letters and indices av or (d) (d — direct). The average value of input AC per cycle is zero. For instantaneous and average values the Kirchhoff’s voltage law is valid in the algebraic form: u = uV + uR ; U av = U Vav + U Rav . (1-4) If electrical circuit is provided with electromagnetic system measurement devices the pointer system of which is attracted with magnetic field then the rms (Root-MeanSquare) values are measured (denominated with capital letters) as follows: for current 1 I rms = 2π

π

∫

π

I m2

2

sin ωtdωt = I m

0

I 1 ωt sin 2ωt − = m ; (1-5) 4 0 2 2π 2

for voltage across the diode U Vrms

1 = 2π

2π

∫U

2 m

π

sin2 ωtdωt =

Um ; (1-6) 2

for voltage of the resistor Um . (1-7) 2 Whereas sine-form supply voltage has rms value U rms = U m / 2 , where 2 is an amplitude factor of sine-form signal, then it can be stated that in the case U Rrms =

2 2 U rms = U Vrms + U R2 rms, (1-8)

i.e., for rms values Kirchhoff’s voltage law can not be applied in algebraic form.

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1st part Values of instantaneous power are obtained by multiplication of instantaneous meanings of voltage and current. As it can be seen for an ideal diode an instantaneous power for each time instant is at zero level (when current is passing through diode a voltage across diode is zero; when reverse voltage exists then the current is zero). If for the source and resistor instead of instantaneous meanings of power an averaged per cycle values are applied then averaged power is: for the source Pav =

π

1 2π

∫U

m Im

0

sin2 ωtdωt =

U m I m U rms I rms = ; (1-9) 4 2

for the resistor U m Im = U Rrms I rms = Pav . (1-10) 4 The both values are equal since the power losses for the ideal diode are zero. In (1-10) it is calculated that rms value of the source voltage is 2 times higher than that of the resistor. It can be seen that due to a non-sine-form of the current the average power can not be calculated by means of simple multiplication of rms values of supply voltage and circuit current. PRav =

Non-sine-form signals can be decomposed into a set of harmonic components with frequencies w, 2w, 3w, 4w, …, kw and it can be done separating non-sine function f(x) into sine and cosine components: f (x ) =

∞ A0 + ( Akm cos kωt + Bkm sin kωt ), 2 k =1

∑

(1-11)

where the coefficients 1 Akm = π Bkm =

1 π

2π

∫

f (x )cos kωtdωt , (1-12)

0

2π

∫

f (x )sin kωtdωt , (1-13)

0

but A0 presents a doubled value of the DC component. If we are examining the current through diode (Fig.1.1), then A0 = 2 I m / π, but fundamental (basic) harmonic (k = 1) has only sine component with coefficient

but

1 B1m = π A1m =

1 π

π

∫I

m

0

sin ωt sin ωtdωt =

Im ; (1-14) 2

π

∫I

m

sin ωt cos ωtdωt = 0.

0

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1st part For the second harmonic (k = 2) π

1 B2 m = π

∫ 0

1 A2 m = π =

I I m sin ωt sin 2ωtdωt = m π

sin(1− 2)ωt sin(1 + 2)ωt π = 0; − 2(1− 2) 2(11 + 2) 0

π

∫I

m

sin ωt cos 2ωtdωt =

0

cos(1− 2)ωt cos(1 + 2)ωt π 2I − = − m . (1-15) − 2(1− 2) 2(1 + 2) 0 3π

Im π

For the third harmonic (k = 3) both amplitudes are of zero values. For the fourth harmonic B4 m =

π

1 π

A4 m =

∫I

m

sin ωt sin 4ωtdωt = 0;

0

1 π

π

∫

I m sin ωt cos 4ωtdωt =

0

Im π

cos(−3)ωt cos 5ωt π 2I = − m . (1-16) − 15π 6 10 0

Therefore: i(ωt ) =

2I Im Im + sin ωt − m π π 2

cos 2ωt cos 4ωt cos 6ωt + + + ... . (1-17) 15 35 3

Example A circuit with a diode and resistor R = 5 Ω connected in series is supplied with the rectangular-form symmetrical AC voltage with amplitude 300 V. Calculate average and rms values of load current, active, apparent and reactive power supplied from the network as well as amplitude and rms values of the fundamental harmonic of current! 1. Whereas a pulsating unidirectional current of rectangular-form with the amplitude of 300 / 5 = 60 A is flowing through the circuit, then the average current is 1 I av = 2π

π

∫ 60dϑ = 0

60 = 30 A . 2

2. RMS value of such current is 1 I rms = 2π

π

∫ 60 dϑ = 2

0

60 2

= 42.4 A .

3. Active power supplied from the network is 1 P= 2π

π

∫ 300 ⋅ 60dϑ = 9000 W; 0

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1st part

Supplied from the network apparent power is (rms value for such form AC voltage is equal to the amplitude, i.e., it is 300 V) S = U rms I rms = 300 ⋅ 42.4 = 12 720 VA ;

Reactive power supplied from the network Q = S 2 − P 2 = 8990 var;

Power factor χ=

P = 0.707. S

4. Amplitudes of the fundamental harmonic components of the current are 1 π

A(1)m = B(1)m =

1 π

π

∫ 60 cos ϑdϑ = 0; 0

π

∫ 60 sin ϑdϑ = 0

120 ; π

The amplitude of fundamental harmonic is

120 = 38.2 A; π RMS value of fundamental harmonic of the current I(1)m = A(21)m + B(21)m =

I(1)rms =

I(1)m 2

= 26.9 A.

1.2. Properties of the Passive Reactive Elements In electrical circuits of converters reactive elements such as inductors and capacitors are often applied. An inductor with the changing of its magnetic field impedes to the changes of its current i. Instantaneous voltage across an ideal inductor with inductance L is di . (1-18) dt If resistance of the inductor winding is negligibly small then in the periodic processes an average value of the voltage across the inductor is zero. If within the time interval t1 constant voltage U1 is applied to an inductor (Fig.1.2) then its current rises in linear way for the value of uL = L

∆I =

U1t1 , L 15

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1st part but when the polarity of voltage is reversed to a negative –U2, which in accordance with zero averaged voltage rule is applied in time interval U1t1 , U2 the current of inductor linearly decreases for the same value (Fig.1.2). In such way at the periodically changing of polarity of the applied voltage the current of inductor is deviating around an average meaning of current but the range of deviations (current ripple range) depends in reverse proportion on the inductance of inductor — if the inductance is higher than at constant other parameters ripple range is lower, i.e., a current is better smoothed. t2 =

uL

iL uL

0

L

t2

U1 t1

U2

t

iL Iav

DI

t

Fig.1.2. Variations in periodic process instantaneous meanings of inductor voltage and current

If the ideal inductor (resistance-less) is introduced in AC circuit with the ideal diode (Fig. 1.3), then switched to the sinus shape voltage at its initial zero point the changes of current in cycle time can be calculated from the expression U m sin ωt = L

di . (1-19) dt

Accounting an initial value i = 0 A at ωt = 0, the expression for the current changes is i=

Um (1− cos ωt ) . (1-20) ωL

The peak value of this current will be at ωt = π, when I m = 2U m / ωL, but the current decreases to zero value only at the end of full cycle, i.e. the diode conducts current continuously. The average value of the current is I av =

1 2π

2π

∫ 0

Um U (1− cos ωt )dωt = m ; (1-21) ωL ωL

but its rms value

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1st part U I rms = m ωL

1 2π

2π

∫ (1− 2 cos ωt + cos

u = Um sinωt

L

ωt )dωt =

0

u V

2

i

0

Um ωL

3 . (1-22) 2

Um

p

2p

wt

i

0

wt

Fig.1.3. Waveforms of current in AC voltage circuit with inductor and diode

Whereas the instantaneous meanings of voltage across the inductor correspond to those of the supply network, the average value of voltage across the inductor is really zero. In some way it proves that the average voltage across the ideal inductor is always of zero value. This property of inductor allows provide a simplification of the investigation of the complex converter circuits. Regarding to instantaneous power p = ui curve for the estimated circuit, it is symmetrical in respect to the time axis and is bipolar: in the interval from 0 to p power is consumed from the network, in the interval from p to 2p — it is transmitted to the source. Active power averaged within the cycle is zero. As it can be seen from (1-20) the amplitude of the inductor current depends in reverse proportion on the value of inductance as well as the frequency of signal. With the increasing of inductance of the inductor a ripple range of its current is decreasing, i.e. inductor favors smooth current wave. This property of inductor is widely used in power electronics for smoothing of current. Real inductor has a low resistance RL . Then the average voltage drop across the inductor is U Lav = I Lav RL , (1-23) and current conductivity angle in cycle through the diode in Fig.1.3 will be lower than 2p.

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1st part

Example The circuit in Fig.1.3 with in series connected ideal diode and inductor with L = 20 mH is connected to AC voltage of sine-form of 50 Hz and rms value 200 V. Find an amplitude of the current, its average and rms value and reactive power consumed from the grid! 1. Peak value of current corresponds to angle ωt = π: 2U m 2 2 ⋅ 200 ⋅103 = = 90.4 A. ωL 314 ⋅ 20 2. Average value of the current Im =

I av =

Um 2 ⋅ 200 ⋅103 = = 45.2 A . ωL 314 ⋅ 20

3. RMS value of the current I rms =

Um ωL

3 = 55.4 A. 2

4. Consumed reactive power Q = U rms I rms = 200 ⋅ 55.4 = 11080 var. Capacitor creates impedance to change of its plates voltage uC level. Instantaneous value of capacitor current duC . (1-24) dt Averaged in cycle value (DC component) of the current through capacitor is zero. If through capacitor within time interval t1 rectangular shape current pulse with amplitude I1 flows, then the voltage of capacitor is rising to the value iC = C

I1t1 , C but changing polarity of the current to negative one –I2, which in accordance with zero averaged current condition has to flow in interval ∆U =

I1t1 , I2 voltage is decreasing for the same value (Fig.1.4). Therefore in this case the voltage of capacitor is deviating in linear way up and down around the averaged value of the capacitor voltage. Ripple range of the capacitor voltage is in reverse proportion to the applied capacitance of the capacitor and frequency too, i.e. if capacitance is larger then the voltage wave amplitude across the capacitor is smaller — the voltage is better smoothed. t2 =

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1st part iC

iC u C +

–

0

C

t2

I1 t1

I2

t

uC Uav

DU

t

Fig.1.4. Changes of capacitor current and voltage in periodical process

It is possible to assemble a circuit with series connected diode and capacitor where diode is by-passed with a transistor reverse to the diode (Fig.1.5). Transistor can be controlled so that it can be turned-on duty for all positive half-cycle of AC voltage applied to the circuit. Then the voltage across the capacitor in the positive half-cycle is uC = U m sin ωt , but its current iC = ωCU m cos ωt .

(1-25)

The averaged value of capacitor’s voltage in cycle U Cav

1 = 2π

π

∫U

m

sin ωt =

0

Um , (1-26) π

but the averaged current of capacitor ICav =

1 2π

π

∫ ωCU

m

cos ωtdωt = 0 . (1-27)

0

This also proves that an averaged in cycle value of the current through capacitor is zero. This property of capacitors also allows to simplify the analysis of complex circuits of power converters. The form of instantaneous power is also symmetrical to the time axis, and averaged in cycle value of active power is zero.

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1st part u

VT

V u = Um sinωt

iC uC

C

Um

0

p

2p

wt

uC iC uC 0 iC

wt

Fig.1.5. Diagrams of currents and voltages in circuit with diode, capacitor and transistor

Example Capacitor with C = 50 µF in circuit with series diode and transistor by-passing it (Fig.1.5) is connected to AC voltage with frequency 50 Hz and rms voltage 350 V. Find an rms value of the current in supply grid, averaged value of capacitor voltage as well as reactive power consumed from the network! 1. RMS value of network current 1 I rms = 2π

π

∫ 0

I m2

(ωCU m )2 π 314 ⋅ 50 ⋅ 2 ⋅ 350 = 3.9 A. cos ϑdϑ = π − = 2π 2 106 ⋅ 2 2

2. Averaged value of the voltage across capacitor Um 2 ⋅ 350 = = 158.3 V . π π 3. Consumed reactive power U Cav =

Q = U rms I rms = 350 ⋅ 3.9 = 1365 var.

1.3. Properties of Diode and Its Characteristics Diode is a structure of two silicon layers having different sorts of conductivities: the left-side layer (see Fig.1.6) contains majority of positive polarity charges (p), the rightside layer in its turn contains the majority of negative polarity charges (n). Such p-n junction is named as bipolar. 20

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1st part a

b

p n + – + +– A + –– + – K + – + – + – + –

V

p + A + + – +

c

n + –– K – –

A + + + R

V

– U +

– K – – uV

iV

R

+ U –

Fig.1.6. Structure of the semiconductor diode in a neutral (a), in reverse biased (b) and forward biased (c) modes

A surplus of one or another charge is obtained adding to silicon material an admixture of active doping materials: e.g. if the 4-valent silicon is doped with a 5-valent element (for instance — antimony) then majority of negative charges is obtained, but if admixture is a 3-valent element (for instance — aluminum) then majority of positive charges takes place. Nevertheless both layers contain minority charges — the charges opposite to those of major polarity. In the neutral mode a part of p majority charges being influenced by electrical field move to n layer, but a part of n majority charges move to p layer. But crossing the border between layers further motion of p charges is stopped under double-side influence of n charges of n layer and those which crossed the border. Similar situation arises also with n charges having crossed the border. In such way border electrical field — potential barrier — is created with potential difference for silicon material of about 0.7 V. If a negative pole of electrical source is applied to p layer but the positive one — to n-layer then the reverse biasing mode is obtained, when potential barrier is destroyed and a middle part of structure has very low conductivity level for the majority charges. Current through the device is close to zero and depends only on the level of concentration of minority charges in the structure. Voltage of such polarity is called a reverse biasing one, but the current is called a leakage one. If reverse biasing voltage is increased then the leakage current is rising too. At some rather high voltage (some hundreds or thousands V) leakage current should be so high that power losses in the structure as well as the temperature of p-n structure depen dent on them succeed the admissible value (for diodes usually 130 °C). Exceeding the admissible temperature a p-n structure can be destroyed in the full extent — so called break-down of the device takes place. Intensification of motion of minority charges at reverse biasing of the structure can be also enhanced with external heating of diode, with influence of light flow to structure as well as influence of ionization flow. If the positive terminal of external supply source is applied to p layer, but the negative terminal — to n layer (Fig.1.6,c), then p majority charges move to negative terminal, 21

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1st part but n charges — to the negative terminal, i.e. majority charges start to flow — forward electrical current arises U −U V , (1-28) R where UV is the voltage drop across the diode which has to be higher than the voltage of potential barrier, i.e. higher than 0.7 V. It means that for silicon diodes the current will flow at the supply voltage higher than 0.7 V only. IV =

Arrow in the schematic denomination of diode is directed from p layer to n and shows possible direction of forward current passing from anode A (p layer) to cathode K. Fig.1.7 presents the volt-ampere characteristic of diode (VAC), i.e., dependence of current on voltage difference between A and K. In the quadrant of forward biasing F for power electronic applications current usually is measured in amperes A, voltage — in volts V within the range from 0 to about 2.5 V to 3 V. In the reverse biasing quadrant R current is measured in mA, voltage — in hundreds volts scale. VAC in the forward quadrant can be linearized as U V = U 0 + I V Rd, (1-29) where U0 is the threshold voltage of the characteristic (about 0.9 V), Rd — dynamic resistance.

IV , A

F tga = Rd

·100 V URbd break-down

UN

0

U0

UV , V

R

Fig.1.7. Volt-ampere VA characteristic of diode

This linearized forward branch of the VAC is very essential for the calculations of conductivity losses. If diode is an element of the circuit presented in Fig.1.1 and its instantaneous value of forward current in the interval from 0 to p is iV = I m sin ωt , then the averaged in cycle interval power losses of diode can be calculated as

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1st part ∆PVav = =

1 2π

π

∫I

m

sin ωt (U 0 + Rd I m sin ωt ) dωt =

0

I mU 0 I m2 + Rd = I avU 0 + I r2ms Rd . π 4

(1-30)

This is typical expression for calculation of conductivity power losses through diode. The losses determine temperature of the diode’s structure in stationary operation case: Θpn = Θ0 + ∆PVav Rth, (1-31) where Q 0 is an ambient temperature, Rth — thermal resistance of structure and its heat sink system (measured in °C/W). For restriction of the break-down the Qpn can not exceed 130 °C. If parameters Rth, Rd, U0 are known then applying different meanings of Qpn and Q 0 it is possible to find the admissible current of a diode as continuously admissible ave raged meaning of current through the diode in presented in Fig.1.1. the basic investigation circuit: −2U 0 + 2 U 02 + I d(max) =

2

(Θpn − Θ0 )π2 Rd

π Rd

Rth

. (1-32)

If Rth will be smaller, i.e., cooling system is better, then Id(max) is higher. If the rated parameters Rth, Qpn, Q 0 are applied then from (1-32) it is possible to calculate the rated current of diode. About half of the break-down reverse biasing voltage URbd is applied to define the rated voltage of diode which usually is divided into 100 V classes. If, for instance, U Rbd = 1550 V , then rated voltage U N = 700 V (the diode is of the 7-th class for continuously applied periodical reverse voltage). It has to be mentioned that the reverse break-down depends on instantaneous meaning of the reverse biasing voltage, i.e. of the peak value of reverse voltage. For instance, if diode operates in a circuit supplied with AC rms voltage of 220 V then the peak value of sine-form reverse voltage exceeds 300 V. In the case at least diode of the 4-th class by voltage has to be applied. If the AC rms supply voltage is of 380 V then the peak voltage of sine-form reverse voltage exceeds 500 V and a diode of the 6-th class has to be applied by admissible reverse voltage.

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1st part

Example DC source of voltage U = 10 V supplies a circuit of current with series connected diode having parameters U 0 = 0.9 V, Rd = 30 mΩ, and resistor R = 0.5 Ω. Find current through the circuit, power losses of the diode, as also voltage across the resistor! Calculate a temperature of the diode if Θ0 = 35 °C and Rth = 0.7 °C/W! 1. Current through the circuit U −U 0 − IRd = I= R

10 − 0.9 − I ⋅ 0. 5

30 103 ;

hence I = 39.56 A. 2. Power losses in the diode ∆PV = U 0 I + I 2 Rd = 0.9 ⋅ 39.56 + 39.562 ⋅

30 = 82.56 W. 103

3. Voltage across the resistor U R = U −U 0 − IRd = 10 − 0.9 − 39.56 ⋅

30 = 7.91 V. 103

4. Temperature of the diode Θpn = Θ0 + ∆PV Rth = 35 + 82.56 ⋅ 0.7 = 92.8 °C .

Very essential are the dynamic operation processes of diode arising at the very beginning and end of its conductivity interval when the polarity of voltage across the structure changes. At the very beginning of conductivity interval a small increase of voltage across the diode can be observed but at the end (Fig.1.8) during a very short time interval trec0 diode in full extent conducts the current in reverse direction. The last situation can be explained with increased concentration of majority charges in opposite majority charge layers in duration of conductivity mode (in p layer n majority charges and otherwise). Therefore with the changing of polarity of the applied voltage the time interval — recovery time — arises during which the layers are «cleaned». In the interval trec0 the voltage across the diode is with polarity according to the direct conductivity and value of forward current flow but the value of voltage is gradually decreasing and at the end of recovery interval is at zero level. In the second part of recovery process the reverse current through the diode is very fast decreasing and reverse voltage across the diode is succeeding to supply network instantaneous meaning.

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1st part

Example Diode with series connected inductor L = 10 mH and resistor R = 1 Ω is connected to the AC network with frequency 50 Hz and rms value U = 200 V. Find power losses of the diode as well as its temperature if the diode has U 0 = 0.9 V, Rd = 30 mΩ, Θ0 = 35 °C and Rth = 0.7 °C/W! Calculating it assume that in the positive half-cycle of network voltage an one of the inductor is equal to the difference between averaged value of network voltage in the half-cycle and averaged value of voltage across resistance in the conductivity interval but in the conductivity interval of the diode in the negative supply voltage half-cycle — with the summative of the both voltages! It also has to be accepted that the diode does not influence values of current in interval of its conductivity! 1. The peak value of current in its rising part in the positive half-cycle can be found from expression I I U L m = − m R; 0.5T 1.11 2 then 200 Im = = 120.1 A . 10 1.11 3 ⋅ 2 ⋅ 50 + 0.5 10

From computer model I m = 120.5 A .

2. Current decreasing interval t2 in the negative half-cycle can be calculated as L

Im I U = + m R; t2 1.11 2

Whence t 2 = 5 ms, which accords with the computer model parameters. 3. Averaged value of the current I I Vav = m (0.5T + t 2 ) = 45 A. 2T In the computer model this current is 49 A. 4. Effective (rms) value of the current T / 2 t2 1 2 I rms = dt + I m 2 T T 0 5 ( . ) 0 5. Power losses in the diode

∫

t2

∫ 0

I m2 1−

2 t dt = I m ⋅ 0.5 = 60.05 A . t 2

2 ∆PV = U 0 I V + I rms Rd = 0.9 ⋅ 45 + 60.052 ⋅

6. Temperature of the diode

10 = 76.56 W. 103

Θpn = Θ0 + ∆PV Rth = 35 + 76.56 ⋅ 0.3 = 57.97 °C . 25

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1st part u

u trec0

V uV u = Um sinωt

Um

iV

uV

0

wt

R iV 0

trec

Qrec

wt

Fig.1.8. Operation of diode in recovery dynamic mode

The reverse current shape in time forms the very characteristic dynamic para meter — recovery charge Qrec. If the diode is of higher quality then both trec and Qrec are lower. For diode of the normal quality all interval trec is about 5 ms, for very good quality powerful diode this indicator is about 1 ms and even lower. Whereas in way of recovery interval the diode performs the function of unidirectional conductivity it can be stated that at certain high frequency of applied voltage diode stops to keep reverse biasing function. If for instance t rec = 5 µs then such situation will take place in AC voltage half-cycle of 5 ms or cycle time 10 ms. Admissible frequency of the applied voltage then is 106 = 100 kHz. 10 Whereas p-n barrier is a base of all types of semiconductor devices then it can be concluded that all range of the devices available have similar static and dynamic characteristics, i.e. semiconductor elements are not the ideal conductors and are not absolutely fast operating ones. f max =

Example Through diode a triangular shape current pulses are flowing (see Fig.) with amplitude of 500 A and declared in Fig. time parameters. Find an amplitude of reverse recovery current and duration of recovery time if the current in the first part of recovery interval continues to change with its decreasing speed at end edge of the forward conductivity and recovery charge for the triangular shape recovery current is Qrec = 100 µC !

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1st part 1. Amplitude of reverse direction recovery current can be retrieved from expression 500 ⋅ t rec ⋅103 ; 0. 2 2. Recovery time interval can be retrieved from expression I VRm =

Qrec =

2 500 ⋅ t rec ⋅103 I VRm ; t rec = 2 0. 4

whence t rec =

Qrec ⋅ 0.4

= 8.94 µs . 500 ⋅103 3. Then the amplitude of reverse recovery current 500 ⋅ 8.94 ⋅103 = 22.35 A. 106 ⋅ 0.2 In this task the form of the diode currtent pulse can be reflected like in this figure: I VRm =

iV

500 A

trec

0.2 ms 0

0.1 ms

Q

IVRm

t

Regarding to the maximum available parameters of powerful diodes then nowadays it is possible to produce diodes with rated current up to some kA and rated voltage up to 15 kV. To reduce power losses and created by its temperature of structure an effective natural and forced cooling systems with ventilation as well as liquid flows cooling are applied. If diode has to be used in circuits with a voltage higher than rated then the diodes have to be connected in series. Nevertheless at such connection it has to be observed that volt-ampere characteristics for reverse biasing quadrant of all diodes even of same type are not absolutely equal. Therefore at common value iVR of reverse leak current for all circuit with series connected diodes each diode can be under different value of reverse voltage. For equalization of reverse voltages across each diode the one has to be by-passed with a resistance of high value Rs (Fig.1.9).

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1st part URmax +

iVR

Vn

Rs

Rs

V1

Rs

Rs

– Is1

Is Fig.1.9. Realization of series connection of diodes

The rated value of by-pass resistor Rs can be chosen with the following assumption: one of the diodes (here in the Fig. diode V1) is accepted as an ideal one without a leakage current, but other (n – 1) diodes are under leakage current IVR, but reverse voltage across the ideal diode is not allowed to be over the rated voltage of one diode UVN. Then I VR + U Rmax −U VN Rs = U VN Rs (n −1) and Rs =

nU VN −U R max . (1-33) (n −1)iVR

Here n is the number of series connected diodes, UVN is rated voltage of diode, URmax is the maximum value of reverse voltage across all circuit of diodes. Power of the by-pass resistor can be calculated taking into account the form of reverse voltage across the circuit of diodes and their conductivity time interval. If reverse voltage is of sine half-wave form and with the amplitude UVN and the half-cycle long conducti vity interval between reverse waves exists then rms value of the voltage across diode is 0.5UVN and then power losses in resistor are ∆PRs =

2 U VN . 4 Rs

Example Maximally possible AC sine-form voltage across the circuit of series connected diodes has amplitude of 20 kV. Rated voltage of one diode is 3 kV, but typical reverse leakage current at rated voltage is 5 mA. Find the number of necessary in series connected diodes as well as the parameters of by-pass resistors! 1. Number of diodes connected in series n≥

U Rm ax 20 ⋅103 = = 6.66. U VN 3 ⋅103

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1st part

It’s accepted n = 7.

2. Nominal value of the by-pass resistor Rs =

7U VN −U Rmax (7 ⋅ 3 ⋅103 − 20 ⋅103 ) ⋅103 = = 33.2 ⋅103 Ω. 6 I VR 6⋅5

3. Power losses in one of the by-pass resistors ∆PRs =

(3 ⋅103 )2 = 67.77 W. 4 ⋅ 33.2 ⋅103

1.4. Bipolar Transistor (BT) as an Element of Powerful Converter Bipolar transistor comprises 3 semiconductor layers: two end sides p and one middle part n, creating a p-n-p version transistor (Fig.1.10), or two end sides n and one middle part p, creating a n-p-n version transistor. One of the end side layers is called emitter (E) and is depicted with arrow in the direction of current conductivity from E to the middle layer (current flows from positive pole of source to its negative one). Middle layer in p-n-p transistor together with emitter create bipolar p-n control barrier, i.e., arrow is directed from emitter to the middle layer which is called base (B). If through this diode a current is flowing then in thin middle layer the number and concentration of p charges increase and the second p-n potential barrier is weakened. Therefore through the second barrier being influenced by the external supply voltage a part of emitter’s p charges from the middle layer is overhauled to the operation voltage circuit. This circuit through a load resistor is connected to the second end layer terminal which is called collector (C) (Fig.1.10). Direction of an arrow of transistor determines necessary polarity of control and operation voltage source: for p-n-p transistor an emitter E directly or through resistor has to be connected to a positive pole, but B and C accordingly — to the negative poles of control and operation voltage (Fig.1.10,a); for n-p-n transistor E directly or through resistor has to be connected to negative poles but B and C — to positive poles of corresponding voltages (Fig.1.10,b). To provide the overhauling or the transistor-effect a voltage of operation circuit supply must be higher than that of control barrier UEB or UBE which accords to the voltage drop across conducting diode. Presented in Fig.1.10 circuits correspond to so called common emitter assemblage which is the most often used in power electronics. Currents of transistor structure in accordance with Kirchhoff current law are mutually connected as iC = iE − iB, (1-34) where all currents have to be with its positive signs. 29

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1st part Relation between collector and emitter current is called current transfer factor iC < 1, (1-35) iE

α=

which shows how efficiently the majority charges of the emitter are overhauled to a collector circuit. Relation α = f (iE ) is very essential for creating a complex structure of p-n barriers (Fig.1.11). For silicon this relation has a small initial zone of insensibility at low emitter currents but further with higher emitter current a is rising reaching to its maximum at about half of the rated current of transistor. a

VT

E iE

C

b

R

R

iC

B

iC

Uco

–

VT

E

B

iE

+

Uco

iB

iB +

C

Ud –

+ Ud

–

Fig.1.10. Application of voltages to the p-n-p (a) and n-p-n (b) transistors

Relation of collector current to a base one is accepted as a stationary current gain b of transistor: β=

iC α = . (1-36) iB 1 − α

a 1.0 Ge Si 0

iE

Fig.1.11. Dependence of stationary current transfer factor for silicon and germanium transistors on the emitter current

Whereas a is a function of iE, then factor b depends on iE. The maximum of current gain factor accords to a max. If, for instance, α max = 0.95, then βmax = 19. For qualitative bipolar transistors b succeeds a value of 100 and more. But if high collector currents have to be processed then relative large base current must be applied too and that complicates an application of the BT. 30

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1st part To characterize an operation of a BT the relations of collector current IC versus collector-emitter voltage UCE (n-p-n transistor!) at different base currents are examined (Fig.1.12). At constant base current and low UCE (saturated condition of transistor) collector current rises fast but further it is practically at constant level with the increasing of collector voltage UCE. If a base current is zero then the collector voltage at small forward leakage current IC0 is about 85 % of that for operation circuit Ud (point A). Nevertheless the voltage across transistor can be high enough as well as power losses. For instance, if forward leakage current I C0 = 10 mA , but U CE0 = 400 V, then at IB = 0 power losses ∆P0 = 4 W, that is relatively high for a transistor. For better closing of transistor circuit a polarity of control voltage can be changed, and then power losses are low.

IC

Ud R B

ICsat

C

IB > 0 A

IC0

IB = 0 UBE < 0

0

DUCE

UCE0

Ud

UCE

Fig.1.12. Dependence of collector current of n-p-n BT on collector voltage

With the rising of a base current at constant supply voltage Ud, collector voltage is decreasing and with sufficiently large IB the transistor is opened in full extent and the value of collector current ICsat is only little lower than the current of short-circuiting of operation circuit U d / R. It is accepted to denominate such operation situation on the characteristics point B as saturated one. Forward voltage drop across transistor DUCE in the case of deep saturation can be even smaller than the base-emitter control voltage and it shows that base-collector diode is biased in forward direction, i.e., in the collector layer there is a large concentration of emitter’s n charges but in the base layer — a large concentration of positive ones exists, and the B-C diode conducts from the base to collector (n-p-n transistor!). Power losses of transistor in the saturated operation case are relatively low. For instance if U d = 100 V, ∆U CE = 2 V , R = 10 Ω, then the collector current is 9.8 A, but 31

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1st part losses of the transistor ∆Psat = 19.6 W. If transistor will operate at point C of the characteristic (Fig.1.12) with U CE = 0.5U d, then the current is 5 A, but losses 250 W, i.e. very high.

Example n-p-n transistor is an element of the circuit of DC voltage of U d = 200 V. A load resistor of R = 10 Ω is introduced into the collector circuit. Current gain factor is β = 15. Calculate with what base current the power losses in the transistor will be of maximum value! Calculate at what base current the transistor will be saturated! 1. Voltage between collector and emitter of transistor is U CE = U d − I B βR. 2. In the saturated case U CE » 0. Then base current has to be Ud 200 = = 1.33 A. βR 15 ⋅10 3. Power losses in transistor can be calculated as I Bsat =

∆PV = U CE I C = (U d − βI B R)βI B . Extreme maximum of these losses will be at base current I B = I Bm which should be retrieved from d(∆PV ) = βU d − 2 I Bm β2 R = 0. dI B Here from the base current for an extreme maximum of power losses Ud 200 = = 0.666 A 2βR 2 ⋅15 ⋅10 and maximal power losses will be ∆PVm = 999 W! I Bm =

Taking into account very low power losses in the both extreme positions — saturated and turned-off, in power electronic systems the control effect is reached using only these extreme positions — transistors are operating in so called «switch» mode (Fig.1.13). Then the losses in switching cycle comprise losses for an extreme positions — turned-off DPoff and saturated (turned-on) DPsat — and losses DPsw in transient situations from one switch position to another (commutation losses). If transistor is operating with resistance load and the time intervals for the both transient situations are the same, then power losses are ∆P = ∆Poff + ∆Psat + 2∆Psw = I C 0U d

t off t I U t + I Csat ∆U CE on + 2 Csat d com . (1-37) T T 6T

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1st part Here IC0 and ICsat are collector currents for the turned-off and turned-on positions accordingly; tcom — transient intervals times — current rise time tr and fall tf ones (usually 1...2 ms long); ton, toff, T — time intervals for control signal applying for turned-on and turned-off parts of switching cycle T (Fig.1.13). It must be mentioned that the operation mode of transistor with resistive load is not a characteristic for operation of powerful converters. Usually then the load is activeinductive and with that it is clamped with reverse diode for fly-wheeling of electromagnetic energy of load at turned-off switch. Then the switching transient processes are essentially different from those discussed above but much closer the processes will be discussed in the part of switching converters. uBE

ton

toff

0 T

uCE iC

t tcom uCE

ICsat

Ud

iC

0

DUCE

IC0 t

Fig.1.13. Simplified diagrams of collector voltage and current of BT transistor switch at operation on resistive load

If switching frequency is increased then the impact of the transient intervals in shortened cycle T is rising and influence of the third component of losses — switching ones DPsw — is increasing too. At some tenth kilohertz frequency switching losses can reach up to 50 % to 70 % from the admissible for transistor losses DP TRadm. For calculation of the thermal process of transistor expression (1-31) should be applied.

Example n-p-n transistor is switched with frequency 5 kHz in the circuit of DC source U d = 200 V and load resistor R = 5 Ω. Duration of the turn-on condition is 0.4 part of the cycle and voltage across main circuit of turned-on transistor is 1.5 V. At turned-off condition the leakage current through the load is 10 mA. The rise time of transistor is 1 ms but the fall one is 2.5 ms. Calculate power losses in the transistor! 33

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1st part 1. Power losses of the transistor in its conductivity position ∆Psat =

U d − 1. 5 ⋅1.5 ⋅ 0.4 = 23.8 W. R

2. Losses of transistor at its turned-off position 10 ⋅ 0 . 6 = 1 . 2 W. 103 3. Losses in transistor in the rise time ∆Poff = U d ⋅

∆Pr =

(U d −1.5)U dt r f (200 −1.5)200 ⋅1⋅ 5 ⋅103 = = 6.62 W . 6R 6 ⋅ 5 ⋅106

4. Losses in transistor in the fall time t ∆Pf = ∆Pr f = 16.55 W . tr 5. Summary losses of power ∆P = ∆Psat + ∆Poff + ∆Pr + ∆Pf = 48.17 W. For each transistor an extreme admissible collector voltage in turned-off condition as also admissible collector current are defined. Likewise the static current gain factor as a function of current IC, voltage drop DUCE at ICmax, rise and fall times and some other parameters are defined too. Main disadvantages of the BT is a relatively small current gain factor, dependence of losses on the base current value as well as relatively long rise and fall times. For enlargement of current static gain factor the special transistor connection circuits are applied. For two transistors such circuit is called a Darlington circuit (Fig.1.14). IC1

IC C

IB VT1

B

IC2

b1 V

VT2 b2 E

Fig.1.14. The Darlington circuit

Here emitter current of the first transistor VT1 is the base current of the second transistor: 34

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1st part I B2 = I B + I C1 = I B (1 + β1 ). (1-38) Summary collector current is I C = I C1 + I C 2 = β1 I B + β2 I B2. (1-39) Taking into account (1-38), the total current gain factor is β=

IC = β1 + β2 + β1β2 . (1-40) IB

It can be seen that the total gain factor b is higher than the multiplication product of gain factors of the both transistors. The second drawback of the transistor is a limited overloading endurance. If at constant base current a collector current is rising over calculated value, i.e., a loading line in Fig.1.12 is turned clockwise around the point Ud, then the collector voltage and correspondent power losses are increasing. As a result the admissible temperature of the structure is exceeded and the latter can burn down. To exclude such possibility an operative control of collector voltage at an active control signal has to be introduced. With the increasing of collector voltage over the admissible, in such situation the control signal has to be disconnected. Nowadays BT of some hundred amperes and some hundred volts are achievable.

1.5. Thyristors Thyristor is a controllable switch with the properties of diodes. Thyristor has been developed on the base of the specific connection circuit of two silicon transistors (Fig.1.15). Transistor VT1 is of p-n-p type with current transfer factor a1 and its collector is applied as a base of the second n-p-n mode transistor VT2 and this point is also a common control terminal of all structure G (Gate). In its turn the base of the first transistor is connected to the collector of the second one. If forward positive voltage is applied to the structure (positive pole to the extreme left p layer — anode A, negative one to the extreme right n layer — cathode K), then in four-layers p-n-p-n structure of thyristor (Fig.1.15) two side-placed diodes are with forward biasing, but the middle diode — in a reverse bias mode and the last is sustaining all voltage applied. Through this later diode a leakage current of minority carriers IC0 flows.

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1st part A

a

b

iV a1 VT1

1

p

IC0 a2

VT2

n

p

n

A

K

2 iG G

G

iV K Fig.1.15. Substitution circuit of thyristor and its four-layers structure

It has to be discussed the case when control current iG = 0 A. Collector current of the second transistor VT2 is equal to a 2iV . Base current of transistor VT1 is iV (1−α1 ). Thus taking into account a leakage current IC0 a current equality expression for the point 1 is formed as iV (1− α1 ) = I C 0 + α 2iV . (1-41) Hence a current of a thyristor is iV =

IC0 . (1-42) 1 − α1 − α 2

As it can be seen in Fig. 1.11 for silicon transistor at small emitter currents factor a is zero. This property allows to keep non-conductive stability of thyristor structure at low leakage currents. When leakage current reaches a certain value (for powerful thyristors about 30 mA to 50 mA) a1 and a 2 start to increase, that initiates increa sing of current iV and next the increase of a1 and a 2, i.e., turning-on process keep an avalanche character at the end of which (α1 + α 2 ) > 1, both transistors of the structure are in saturated position and leakage current IC0 turns its sign. Leakage current can be raised if 1) voltage between anode and cathode is increasing, 2) temperature of the structure is increased, 3) strong light flow is influencing the structure, 4) radiation flow is affecting the structure. The first possibility is applied for the development of low power diodes switched with voltage influence — dinistors. The third possibility to increase IC0 to the necessary value using strong light flow upon structure is applied for the development of optically controlled thyristors. Optically controlled thyristor comprises Light Emitting Diode LED VS.1 at input which is electrically insulated of that a four-layer structure VS.2 (Fig.1.16,b). When through diode VS.1 a relatively low DC current is passing (0.1 A 36

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1st part to 0.2 A) then it radiates a strong light causing an increasing of leakage current IC0 of the structure VS.2 and as a result a transient of VS.2 to conducting stage, if a forward direction voltage is applied across A and K. Special heating of thyristor is not operation mean applied in practice but the existence of such process should be taken into account within the operation process of thyristor. It means that during operation the value of forward voltage at which leakage current can reach the critical value for uncontrolled turning-on really decreases. The boundary temperature at which structure of thyristor can sustain an influence of temperature usually is accepted 120 °C. The second case is turn-on process of thyristor using a control current iG > 0 A . Taking into account that the emitter current of VT2 is (iV + iG ) an expression for current’s equilibrium for point 2 is I C0 + α1iV + iG = (1− α 2 )(iV + iG ). (1-43) Hence iV =

I C 0 + α 2iG . (1-44) 1 − α1 − α 2

As it can be seen an increasing of current transfer factors and with that opening of the thyristor circuit can be obtained with increasing of the transistor VT2 base current iG above the critical value. Usually control triggering pulse can be applied shortly (for 30 ms to 100 ms). Datasheets of thyristors define the necessary values of turningon control current and voltage. Usually necessary turning-on control current is less than 1 A, but the control voltage — about 4 V. Most common is triggering of thyristor through an insulating pulse transformer (Fig.1.16,a). In such way a low voltage control circuit is electrically insulated from powerful operation circuit of thyristor. Shortly switching the transistor VT the primary winding of transformer w1 is connected to DC voltage of 15 V to 20 V but in the secondary winding a similar short triggering pulse of current and voltage is induced passing gate current between terminals G and K of thyristor.

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1st part a

iG

b +

15 V…20 V

–

A

VS.1

VS.2 K

VT

A

+

w1

VS

TR G

A

VS.1.1

VS VS.1.2

w2

K

VT –

K

Fig.1.16. Control of thyristor with pulse transformer (a) and optically controlled auxiliary thyristor (b)

In Fig.1.16,b triggering of the main thyristor VS is provided by means of opening of low power optical controlled thyristor VS1 an executing structure VS1.2 of which for a short time mutually connects terminals A and G of the main thyristor (usually through a resistor). When LED VS1.1 of the optical controlled thyristor is shortly activated with DC current pulse then VS1.2 structure turns-on and if anode of the main thyristor is positive in respect to the cathode then a current flows through its gate circuit such opening the VS. Gating currents flow is stopped simultaneously with the opening of the main thyristor.

Example The current transfer factors 1 and 2 of the both transistors of substitution circuit of a thyristor change identically dependently on their emitter currents: they start to rise linearly from zero value at emitter current 50 mA and reach value of 0.7 at emitter current 2 A. Determine a necessary control current of thyristor for its switching, i.e., to provide through its main circuit current iV = 50 mA ! Leakage current IC0 is negligibly small! 1. Factor for the anode side transistor changes as (iV − 0.05) ⋅ 0.7 (iV ³ 0.05; α1 ≥ 0); 1.95 2. Factor of cathode side transistor changes as α1 =

(iV + iG − 0.05) ⋅ 0.7 (α 2 ≥ 0 ). 1.95 3. At iV = 0.05 A expression for thyristor’s current calculation is α2 =

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1st part α 2iG 0.7iG2 0.05 = = ; 1− α1 − α 2 1.95 − 0.7iG Whence the necessary control current has to be iG = 0.35 A. The VA characteristic of thyristor in stage of its current conductivity is similar to that of the diode (Fig.1.17). But forward voltage drop is higher because the threshold voltage U0 is 1.2 V to 1.3 V. It can be accounted that VA characteristic of forward direction does not start from zero value but from relative very small latching current ILO. If impedance of circuit can not provide current higher than ILO at the end of a triggering pulse then thyristor stays switched-off even after the triggering process.

IV IN

iG > 0 ILO UN

0

iG = 0 UV

Fig.1.17. VA characteristic of thyristor

If iG = 0 A then forward characteristic of thyristor is symmetrical to that of the reverse biasing and thyristor turns-on at its break-down at admissible forward voltage, i.e., iV (U V ) = −iV (−U V ). When forward voltage across the thyristor is above the admissible value, thyristor turns-on and its VA characteristic step-wise transforms to VA characteristic for turned-on condition with relatively low voltage drop across the thyristor. When the ordinary thyristor is turned-on and current is passing through then it can be turned-off only decreasing its current to the zero level. In such way current periodically crosses the level of its zero value in AC voltage circuits and then some extra means for interruption of circuit are not necessary. The mode of interruption of thyristor’s circuit applying in its reverse direction voltage of previous charged with certain polarity capacitor in parallel to the thyristor is named as forced one. If initial energy stored in the capacitor is sufficiently large then after turning-off of the thyristor a necessary reverse biasing voltage is sustained for some interval such providing the guaranteed turn-off process. 39

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1st part In the 80-ies the thyristors that could be turned-on and off by means of control pulses named GTO (Gate Turned Off) have been elaborated. As in open condition (α1 + α 2 ) > 1, then for turn-off of anode current iV (accepting that leakage current IC0 is at zero level) a control gating current has to be applied iG =

(1− α1 − α 2 )iV < 0, α2

(1-45)

which can reach 30 % to 70 % of turned-off anode current iV. If for instance anode current is iV = 200 A, α1 = α 2 = 0.7 , then the gating current necessary for proper turn-off will be iG =

−0.4 ⋅ 200 = −114 A . 0. 7

Although time interval for application of the negative gating current is quite short (about 5 ms), a structure of thyristor’s gate has to be considerably strengthened as well as a special circuit for negative direction gating current must be installed. Usually for that a discharging circuit of capacitor (Fig.1.18) is applied. The gate of thyristor through transistor VT1 can be connected to the capacitor C1 and that creates a turning-on gating pulse (with G positive to the cathode). In its turn activating transistor VT2, capacitor C2 can be connected to the gate circuit and such a turn-off gating signal is generated when cathode point is positive according to the gate G. Both capacitors C1 and C2 are connected in series and their common point is connected with cathode K. For the both transistors one common control base exists and its polarity of control voltage to cathode K defines a control case: if the base is positive to the K, VT1 is on, but if negative then VT2 is on and the thyristor is turning-off. A special case of thyristor structure is of bipolar conductivity controlled thyristor — triac. Such an element is useful for control of symmetrical processes in AC voltage circuits and can substitute a two-anti-parallel connected ordinary thyristors. The development of the bipolar transistor and thyristor is connected with the name of American scientist W. Shockley; development of symmetrical thyristor is connected with the formulated by R. Aldrich in 1959 principle of clamped emitter barrier. Triac has a single control gate G as well as one terminal PG connected with control of the basic circuit and one terminal P is not connected with control circuit (Fig.1.19). Gating signal is applied between points G and PG. It is possible to elaborate triacs with different cases of control current direction. One of the cases is when turning-on of triac is provided with bidirectional control current. Other case is a turning-on with constant polarity of gating current — either positive or negative one.

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1st part A

TR C1 +

UZ ~

VS

VT1 G

K

+

C2

VT2 + (–)

– (+)

Fig.1.18. Forming of gating current for control of GTO thyristor

G PG VS

P

Fig.1.19. Denomination of the triac

For creation of symmetrical device a structure n-p-n-p-n is elaborated with clamped barriers (Fig.1.20).

iG n1

+

–

G

PG

n1

p1 n2

p2

n3 P

Fig.1.20. Structure of the triac with positive direction gating current

If terminal P is positive in respect to the PG, then the ordinary p2-n2-p1-n1 structure of thyristor operates with gating through p1-n1 barrier. If polarity of power circuit is opposite then p1-n2-p2-n3 structure operates with gating through clamped p1-n1 barrier. VA characteristics of triac in the both directions are similar to those of the ordinary forward biased thyristor except that latching current is higher for turning-on process as it is for turning-off process at decrease of current. 41

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1st part Very special operation modes for thyristors are the dynamic ones: turning-on process, turning-off process, endurance to the fast rise of forward voltage. The turn-on process is characterized with gradually «opening» of an area of structure at triggering in direction from the gate with rather small speed of about 0.5 mm/ms. If the current of the main circuit of thyristor flows in full extent through the restricted initial area then inadmissible current density arises in the conducting area that initiates increasing of the temperature of the area. If the temperature reaches the values above an admissible ones then the structure can be irretrievably damaged. For providing a normal thermal condition of thyristor’s structure at turn-on it is ne cessary to limit the speed of current rise in the process proceeding for 5 ms to 7 ms. Two methods for that exist: by means of introducing a series linear inductor into the circuit and with the help of introducing a saturated inductor (Fig.1.21). VS

a

L

b L

VS

uV, iV iG

iG

0 uV iV

0

t Ua0 uV tr

iV LI V Ua0

t

uV iV

Ua0

IV

uV

IV

iV

tr tns

0

0 t t Fig.1.21. Turning-on process of thyristor in circuit with a linear inductor (a) and with a saturated inductor (b)

In the first case with the linear inductor using a presented in data-sheets admissible speed of current rise [di/dt] and taking into account initial anode voltage of the thyristor Ua0 an inductance of the inductor inserted in series with thyristor (Fig.1.21,a) can be calculate as L³

Ua0 . (1-46) [di / dt ]

Then extreme maximum power losses in duration tr of turn-on process will be ∆Pmr ≈

4 2 tr U a0 , (1-47) L 54

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1st part but averaged over switching cycle T = 1/ f power losses ∆Prav ≈

1 U a20t r2 f . (1-48) L 24

Example A thyristor is triggered in a circuit with supply voltage U a0 = 600 V; in series with it a linear inductor for reducing a current rise speed with inductance L = 20 µH is inserted. Calculate extreme maximum instantaneous power losses within turning-on interval if this is t r = 10 µs long! 1. If it is assumed that voltage across the thyristor is decreasing in linear way, i.e., t uV = U a 0 1− , t r then voltage of the inductor at that changes as t t uL = U a 0 −U a 0 1− = U a 0 . tr t r 2. Changes of current through inductor and thyristor can be found from di V t = Ua0 dt tr and then L

iV =

Ua0 2 t . 2Lt r

3. Instantaneous meanings of power losses in thyristor within turn-on interval U2 t2 t ∆pV = uV iV = a 0 1− . 2Lt r t r 4. Extreme maximum of the loss will be at time instant t1, which can be calculated from expression d(∆pV ) 2t = 0; t1 = r . dt 3 5. The extreme maximum of power losses will be ∆Pmr =

U a20 4t r 6002 ⋅ 4 ⋅10 ⋅106 = = 13333 W . 54 L 106 ⋅ 54 ⋅ 20

As it can be seen from the expressions the power losses in the process are decreasing with the increasing of the inductance. But such a linear inductor with rather large inductance is large too and besides a voltage drop across inductor in operation over 43

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1st part its stationary case arises. Therefore often instead of linear inductor a saturated one is applied with iron core which has a rectangular shape of magnetization curve (Fig.1.21, b). While a core of the inductor is not saturated an inductance of it is practically endless and current through the thyristor circuit is close to zero. Parameters of the inductor have to be chosen in such way that inductor’s operation in non-saturated condition has to be longer than the turning-on interval for the thyristor tr: t ns =

(Bs − B0 )sw ≥ t r , (1-49) Ua0

where Bs, B0 are flux density in T for saturation and at residual condition respectively; s, w — respectively cross-section area of a core (m2) and number of turns of its win ding. When inductor becomes saturated a structure of a thyristor will be fully opened and at that structure of thyristor can sustain fast increase of current without any damage even if inductance of the inductor is zero too. The second dynamic process is turning-off of thyristor (Fig.1.22). Similarly like for diode (Fig.1.8), after decreasing to the zero level during some interval the current of thyristor continues to flow in the direction opposite to the forward conductivity one and after time interval trec a recovery process takes place. At the end of the recovery interval a reverse biasing voltage arises gradualy changing its polarity according to the forward biasing case (Fig.1.22); and if this transient is too fast then the thyristor can be turned-on without permission on it and an emergency situation can take place. Such spontaneous switching process can be connected with the fact that at end of recovery interval not all majority carriers are returned to the initial positions for sustaining a forward biasing voltage. For accounting of this feature a passport of thyristor contains a special indicator — turn-off time toff.

iV uV iV tr 0

toff du/dt

trec

t

uV Fig.1.22. Diagram for succesful turn-off process of thyristor

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1st part In the case presented in figure 1.22 the turn-off process will be successful because reverse biasing voltage exists longer than toff time, i.e. t rec > t off . In case of application in AC circuits toff value restricts the maximum available operation frequency of thyristor. Really the half-cycle of voltage has to be longer than toff and then f max <

1 . (1-50) 2t off

If t off = 100 µs (not a high quality of thyristor), then f max < 5 kHz. For the thyristors of high quality toff is 20 ms to 30 ms, and such thyristors can operate in circuits with frequency 15 kHz to 25 kHz. The special importance of turn-off interval value is for application of thyistors in switching DC converter circuits where for turn-off process of thyristors the forced commutation is applied using previously charged capacitor for creation the reversebiasing voltage across turned-off thyristor. The larger is toff of the thyristor the higher value of capacitor should be used. For decreasing of size of commutation junction usually the thyristors of higher dynamic quality with smaller toff are applied. The third dynamic process — fast increasing of forward voltage — is connected with a restoration process of forward voltage of thyristor after its turn-off process. At the rise of forward voltage a biased in reverse direction the middle barrier diode of thyristor (Fig.1.15) has the certain capacitance Cnp. At restoration of the forward voltage a leakage current through this barrier’s capacitance depends on voltage rise speed: iC 0 = iCnp = Cnp

du . (1-51) dt

If rise speed du / dt is too large then the current value iC0 can be higher as the critical one for non-triggered turn-off process and current transfer factors a1 and a 2 start to rise from initial zero value. If such process starts then the thyristor is turned-on spontaneously and it is possible to get a break-down situation. If for instance the critical current IC0 = 30 mA, but capacitance of barrier is Cnp = 100 ⋅10−12 F, then the critical value of du / dt is 300 V/ms. For excluding emergency situations the special circuits of voltage rise speed damping are applied (Fig.1.23) comprising connected in parallel to the thyristor through diode V0 capacitor C0 charged in turning-off process with the commutated current. For restricting a passing of large currents at turning-on of thyristor a discharging of capacitor through it is provided through resistor R0. Such protection circuits are called RDC snubber circuits. Especially efficiently such circuits operate with in series to thyristor connected saturated inductor. In way of turning-off of thyristor and its recovery process a reverse current over-magnetizes a core of the inductor in reverse direction or to the zero flux density position. Therefore at restoration of the forward voltage a charging current of 45

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1st part capacitor C0 can not exceed the forward magnetizing one of the saturation inductor which is low enough and therefore voltage rise speed of thyristor du / dt is also low.

R0

R0

V0

V0

C0 VS

L

C0 VS

Fig.1.23. Snubber circuits RDC without series saturated inductor and with one

For choice of the parameters of RDC circuit the dynamic parameters of thyristors are denoted in its data-sheet — for instance admissible [du / dt ]. Taking into account a load current Ild, the capacitance of snubber can be found as C0 >

I ld . (1-52) [du / dt ]

Resistance R0 should be of such value that capacitor C0 in periodical process should be fully discharged during the turned-on position of the thyristor. If I ld = 100 A and [du / dt ] = 200 V/µs , then C0 > 0.5 µF. If thyristor operates with frequency 5 kHz but duration of turned-on position of thyristor is 100 ms (the half-cycle long) and C0 = 0.5 µF, then R0 = 100 ⋅106 / (3C0 ) = 67 Ω. Here it is accepted that full discharging time of the capacitor in RC circuit takes 3 time constants. Attention has to be applied to a possible large enough value of power losses in snubber circuit. The power losses depend on stored energy in the capacitor and operation frequency for RDC circuit and can be calculated as PR0 =

2 C0U max f , (1-53) 2

where Umax is the maximum voltage of capacitor C0 charging. If in the previous example the U max = 550 V then the necessary power of the resistor is PR0 = 377.5 W. That is rather high power loss and size of the resistor will be large too. For decreasing a size of the RDC circuits a better quality thyristors have to be applied. Thyristors basically are applied for the equipment of relative large power. Nowadays it is possible to use thyristors for the currents of some kA, voltages for more than 10 kV, with admissible [di / dt ] >1000 A/µs , [du / dt ] >1000 V/µs and turn-off time less than 20 ms.

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1st part

1.6. Power MOSFETs The second realization version of transistors is connected with a conductivity channel creating in a transistor structure by influence of the external electrical field. In such way MOSFET version (see the Introduction part) have been created. Channel can be with the n-type majority carriers as well as with the p-carriers majority and then MOSFET is of n-channel or the p-channel mode. The field effect transistors are created upon the n or p base, in upper edge corners of which a contact terminal of the opposite conductivity are introduced, the size of which determines a thickness of conductivity channel (Fig.1.24). An insulating metal oxide layer is placed over the base but over that a metal plate — gate G to which one pole of control voltage is applied but the second one is connected with one of the contact terminals named source S which in its turn is connected electrically with the base. The second contact terminal is named a drain D. If the base is of n-type then applying electrical field between G and S the channel can be fulfilled with the minority p-type carriers of the base creating in such way p-channel MOSFET. For that to the gate a negative pole of control voltage have to be used but to the base and S — a positive one. In denomination circuit arrow is directed from p zone to n zone, i.e., in the case of p-channel transistor the arrow has to direct to the base but in the case of n-channel — in opposite direction — to the channel. Since the insulation of gate plate from the base at stationary control case current through the gate circuit will be zero. If the base is of p-type but contact terminals are of n-type then control electrical field between G and S can fulfill the channel with n minority carriers of the p-base creating a n-channel. For that a positive pole of control voltage has to be applied to the gate G but to the base and S — a negative pole (Fig.1.24.b). Such n and p-channel transistors without control voltage are not conducting current, i.e., they are normally turned-off but fulfillment of channels can be provided with the rise of control voltage (usually not above 20 V ) and such MOSFETs are called Enhancement Mode MOSFET. In such transistor structure in natural way a fast operating a reverse direction diode (in the n-channel diode from the base to the drain D; in p-channel — from D to the base) is created that can be useful at application of MOSFETs in converter circuits. For turn-on initiation of MOSFET a control voltage between G and S can be applied above the threshold value 2 V to 4 V but for turn-on transistor as a switch the control voltage must be up to 15 V to 20 V. In fully saturated condition of the transistor a voltage across its channel depends on the relation between resistance of channel to the full one of load-transistor circuit and supply voltage, i.e. as for voltage divider. If supply voltage is rising then the voltage drop across the channel also is rising and with it also power losses in channel increase. That is a large disadvantage of the MOSFETs. 47

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1st part In the data-sheets the resistance of turned-on channel RDS(on) is denoted and for instance for the 10 A MOSFET with rated voltage 100 V it is about 0.1 W, with 200 V voltage — 0.2 W, with 400 V voltage — 0.6 W, with 1000 V voltage — 1.2 W. If the admissible power losses of transistor is 300 W (almost typical parameter for power MOSFET), at supply voltage 1000 V the rated current can not be above 15 A to 20 A. The operation temperature of the MOSFET structure can reach 150 °C. a

– G

D

p+ + + + + + + p + + S n

+ UB

b

I

n– – – – – – – n D S – – p + UB

UV

VT S UGS UV G

VT

UV

D

G

DUV –

+ G

S –2 V

0

UGS

–

0

2 V

UGS

Fig.1.24. Enhancement Mode MOSFETs with the p-channel (a) and n-channel conductivity

MOSFET transistors are very fast operating and their switching transient intervals are measured in nanoseconds. Typical rise time is within the range of 20 ns to 50 ns, fall times — 200 ns to 250 ns long. It can be noted that transients times do not depend on temperature of the structure. Because of small transient times MOSFETs can operate with switching frequencies up to some hundreds kilohertz. It can be accounted that MOSFETs can operate in parallel connection since their defined channel resistances allow to equalize currents through the transistors. This property allows in rather easy way to increase a power of converter. In respect to practical realization of control it can be accounted that between gate and base layer of transistor really very small capacitance exists that arises charging and discharging currents through the gate circuit at front and back edges of triggering pulse. For limitation of that currents in series with gate a resistance (usually recommended as 10 W) has to be introduced. In the last time a new version of the field transistors — Conductivity Modulated FET or COMFET are developed. Circuits of the n-channel COMFET are represented in Fig.1.25.

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1st part a

D

b

c p

G +

G S

UGS

–

D

RSl

n p

n

+

G

Ud

n

–

S

S

Fig.1.25. COMFET substitution circuit (a), circuit of its structure (b) and electrical denominations (c)

If the circuit is examined it can be seen that really COMFET is a thyristor controlled with a field effect transistor. Therefore at turned-on position voltage drop across the COMFET is close to 2.5 V, i.e., to the parameter which accords to thyristors. But switching transient time is shorter than for the thyristors — rise time about 200 ns, fall time — 700 ns to 900 ns. Taking that into account that the voltage drop across turned-on COMFET does not depend on operation voltage the rated voltages of COMFETs can be raised up to 1 kV and more but rated currents up to 50 A to 100 A and more.

1.7. Insulated Gate Bipolar Transistors IGBT Integrating MOSFET, bipolar transistor and in some extent — thyristor in one structure it is possible to create a switch with zero stationary control current, large values of rated current and voltage and with relatively small switching transient times. Such transistor is named as insulated gate bipolar transistor — IGBT. The substitution circuits of the IGBT are presented in Fig.1.26 and are very similar to pseudo-Darlington connection comprising a p-n-p bipolar transistor VT1 and nchannel MOSFET transistor VT2. Between the source of VT2 and the base of VT1 the ordinary channel field effect transistor VT3 is introduced taking part of the forward voltage between emitter of the p-n-p transistor (accepted as collector of IGBT) and collector of the VT1 allowing in such way to reduce voltage and since that a resistance R DS(ON) of the control input MOSFET VT2. Application of MOSFET in control circuit allows to obtain zero control currents in the stationary situations as well as provides fast operation speed at switching transients. If the potential of the gate G is positive in respect to the emitter of IGBT at existence of the forward voltage, transistors VT2 and VT1 are both turned-on. The forward voltage 49

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1st part drop in turned-on position across the IGBT cannot be lower than the voltage drop across the diode between the emitter and base of the transistor VT1. The typical volt-ampere characteristics of IGBT are presented in Fig.1.27. It can be seen that at turned-on position the forward voltage drop across the transistor is low (here about 1 V), that in complex with easy control and fast transient speeds provides a wide application of the transistors in power electronics. It can be noted that an admissible temperature of structure of IGBT is +150 °C. As it can be seen a current through the transistor depends on the value of a control voltage. To provide a full turn-on state with small voltage drop (under 2 V) rather large control voltage close to the maximum admissible 20 V has to be applied. For obtaining a turned-off position a reverse control voltage up to minus 20 V can be applied.

C VT3

VT1

IC

VT4

VT2 G

C

G

R E

E

Fig.1.26. The principal substitution circuit of the IGBT

When an internal p-n-p transistor of the IGBT is turned-on it is not in the deep saturated situation since it is restricted by a circuit of a base current generating method in the transistor VT1. Therefore IGBT turn-off process is sufficiently fast — from 100 ns to 3000 ns. If turn-off process is fast then switching power losses are relatively low and it allows operating with high switching frequency. IGBTs can be divided into 3 groups: • S — with standard switching speed when full turn-on interval lasts for about 25 ns but turn-off process till 1600 ns; the common switching energy losses are about 7.0 mJ (for transistor of the rated current 10 A) and transistor can operate with switching frequency till 1 kHz; • F — fast IGBT with full turn-on time about 13 ns, turn-off time about 600 ns, ener gy losses in switching processes about 1,8 mJ (for transistor of rated current 10 A) and such transistor can operate with switching frequency up to 10 kHz; • U — ultra-fast IGBT with turn-off time 190 ns and energy losses in switching processes about 1 mJ which can work with tenth kilohertz switching frequency.

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1st part Switching energy losses can be obtained in the standard circuit Fig.1.28 in which at the first simultaneously transistors VT1 and VT2 are activated providing current IC through inductor L, and after turning-off of the both transistors this current closes through clamp diode V1 and Zener diode V2 creating across the transistor under investigation the initial and end collector voltage. When an investigating transistor VT3 is turned-on initially turn-on stage takes place at voltage of the Zener diode UC and current of the transistor at it is increased to the value IC, but after that at full current IC a voltage of the VT3 decreases from the value UC to zero. At turn-off in the its first stage the collector current is equal to the value IC when collector voltage is rising but after activation of the Zener diode a voltage across the transistor is equal to UC but current is decreasing to zero. Time intervals presented in Fig. correspond to the F group of IGBT. If in the experiment U C = 450 V, I C = 10 A , then a common switching energy losses are ∆Ek = (450 ⋅ 5 + 225 ⋅10)(15 + 600) ⋅10−9 = 2.77 mJ.

UCE, V

5

4

7

10

UGE = 15 V

3 2 1 0 –1 10

100

101

102

IC , A

Fig.1.27. Characteristics of IGBT transistor

Really in the structure of IGBT p-n-p-n barriers exist, i.e., the structure is similar to that of thyristor. If in the substitution circuit a sum of current transfer factors of VT1 and VT4 becomes equal to 1, then a structure turns-on without its triggering. It can take place at fast rise of forward voltage across a transistor. A continuously admissible current of IGBT can reach values of some tenth hundreds amperes, but the voltages – up to 5 kV to 6 kV.

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1st part a

b VT3

L

+

VT1 V1 U

iC

i V2 VT2

UC

UC uC

t IC iC

VT3

–

15 ns 15 ns

600 ns

600 ns

t

Fig.1.28. The standard circuit for investigation of a switching of IGBT: circuit (a), diagrams of current and voltage in switching process of an investigated transistor VT3

1.8. Thermal Mode Calculations of Semiconductor Elements Temperature of semiconductor elements depends on its power losses and parameters of its cooling system. In the stationary case the temperature is Θs = Θ0 + ∆Ps Rth, (1-54) where Q 0 is an ambient temperature, DPs is power losses in a semiconductor device (W), Rth — a stationary thermal resistance of the cooling system (°C/W). Power losses for semiconductor diodes and thyristors can be obtained using expression (1-30), but for transistors operating in switching modes — using an expression ∆Ps = U V I V D + ∆Esw f , (1-55) where UV, IV — voltage and current of a turned-on element, DEsw — energy losses in switching process (J), f — switching frequency (Hz), D — on-duty ratio for switching cycle T = 1/ f . If all parameters are known as well as an admissible temperature of the structure [Qs], then it is possible to obtain admissible switching frequency fmax. If for instance for an IGBT transistor Θs = 130 °C, ambient temperature is 30 °C, Rth = 3 °C/W, D = 0.5, I V = 10 A, U V = 2 V , ∆Esw = 2.77 mJ, then the admissible switching frequency is f max = 8 kHz. Similar method for calculation of a thermal situation can be applied also for other types of semiconductors. It has to be taken into account that a thermal resistance depends on the construction of heat sink as well as on a cooling method. Best of all heat sink construction is with double-sided heat sink in the case of pellet form of semiconductor device. If heat sink of semiconductor device is cooled with air flow with a speed 52

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1st part of 12 m/s, then usually a stationary thermal resistance is twice lower as it should be at operation in natural cooling case without forced ventilation (Fig.1.29). In many cases the rated current of semiconductor devices is accepted for a forced air ventilation of heat sink with speed of flow 12 m/s. Improved cooling can be obtained using liquid flow through a heat sink. Usually it is accepted that rated speed of liquid flow is 3 l/min. In many cases it is necessary to obtain temperature of a structure in intermittent operation cases with sequence of short power loss pulses. If it is shorter than necessary for continuous operation case which usually is about some hundreds seconds, then it should be possible for calculations to apply a transient thermal resistance rT (Fig.1.29) which depends on the process time. Calculations can be done using superposition principle regarding to the changes of temperature within each power loss pulse with endless lossless situation beyond it: starting with each a next pulse a positive power loss sum affects a structure; starting with each next pause endless long affects a summative of the positive and negative value power loss and at that influence of each pulse and pause upon change of temperature are summed up (Fig.1.29). Then temperature at end edge of the first pulse is Θ1 = Θ0 + ∆PrT 10, (1-56) after the first pause — Θ2 = Θ0 + ∆PrT 20 − ∆PrT 21, (1-57) after the second pulse — Θ3 = Θ0 + ∆P (rT 30 + rT 32 − rT 21 ), (1-58) after the second pause — Θ4 = Θ0 + ∆P (rT 40 + rT 42 − rT 41 − rT 43 ) (1-59) and so on, where rTba is transient thermal resistance for time interval t b - t a ; DP — power losses in the pulse. If number of pulses reaches high values then the process tends to be stationary.

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1st part a r T, °C/W

0.8

natural cooling

0.6

6 m/s stationary Rth 12 m/s

0.4 0.2 0

tstat 0.1

1

b

10

100

DP Q1

Q0 0

Q5

Q3

Q2 1

Q7 Q6

Q4 2

3

t, s

1000

4

5

6

7

t

+DP –DP +DP –DP +DP –DP +DP Fig.1.29. Dependence of transient thermal resistance on time (a) and influence of power loss pulses on heating of structure

Example A thermal resistance of a cooling system at natural cooling of a semiconductor diode changes as it is shown in Fig.1.29,a. Calculate what will be the temperature of structure after 10 s long power loss pulse created by rectangular shape current pulse with an amplitude of 300 A, if the pulse takes place after continuously operation of the device at stationary temperature Θs = 50 °C ! VA characteristic of the diode has U 0 = 0.9 V and Rd = 1 mΩ. 1. Power loss in duration of one pulse ∆PVi = (300Rd + U 0 )300 = 360 W. 2. Temperature of structure at the end of the power loss pulse 54

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1st part Θib = Θs + ∆PVi rT10, where s — the stationary temperature, rT10 = 0.18 °C/W— transient thermal resistance for 10 s long pulse: Θib = 50 + 360 ⋅ 0.18 = 114.8 °C.

1.9. Control and Protection of Transistors The bipolar junction transistors have to be controlled in switching process with bidirectional current flow. At turn-off of BJT a base current must pass in an opposite direction in respect to its turn-on state. Such control can be provided using two in series connected DC supply sources (Fig.1.30). In Fig.1.30 both of the sources can be presented as capacitors C1 and C2. A base of controlled transistor is connected through an emitter follower VT1,VT2 and resistance RB to the inverting output of logical gate D. The emitter of controlled transistor is connected to the common point of the both DC sources. If output of D is of high logical level +U1 in respect to the common point then the current is conducted by transistor VT1 and in a base circuit of controlled transistor VT flows current in a positive direction which turns-on the transistor VT. +U1

TR UZ

C1 +

VT1 D

~ C2 +

RB VT

VT2

iB

–U2 iB

(U1 – 2)/RB

0

t

Fig.1.30. Driver circuit for a bipolar junction transistor

If output of the D is supplied with low logical level signal –U2 in respect to the common point then the current shortly passes through the VT2 and base current of a controlled transistor is with a negative sign and transistor VT turns-off. When turn55

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1st part off process is finished a reverse biasing voltage U2 is applied to the base-emitter barrier of the VT. For decreasing of size of such device which usually is called a driver, a transformer of the device can be supplied from low power high-frequency converter. In its turn a generator of logical level signals D can be controlled with optically controlled transistor providing a galvanic insulation of the device. Similarly the drivers for MOSFET and IGBT transistors can be also produced taking into account that their control does not require a bipolar control voltage (it is sufficient to decrease base — gate current to the zero level). Very often transistors are applied as switches for changing polarity of an electrical chain when one of the transistors is introduced from a side of positive pole of a DC source but other — from a side of the negative pole of the source (Fig.1.31). Then for the both transistors a common point for control signal of the intermittent switched transistors does not exist. For the case for the upper-side transistor an insulated DC source driver and optical control of a driver can be applied. But such realization requires an insulated supply converter. Therefore in practice so called floating driver is applied where for control of the upper-side transistor VT1 use special so called boot-strap capacitor C which is charged through a boot-strap diode V1 from low voltage (≈18 V) DC supply source UB in time interval when lower-side transistor VT2 is turned-on. Control circuit of the transistor VT1 in a driver is activated through an optical coupling element with a common for the both transistors control signal H/L. Us

+

VT1 +UB

V1

+

C i

H/L

VT2

–

Fig.1.31. Introduction of the floating driver in an electrical circuit of intermittent switched transistors leg

In the circuit it is necessary to obtain situation when both transistors VT1 and VT2 should not be triggered simultaneously to avoid a large through-pass currents which can damage the transistors. For avoiding the situation in circuits of drivers very short 56

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1st part time pauses (some tenth nanoseconds) are provided called dead time between active triggering signals of the both transistors. For transistors the operation under short-circuiting conditions with an active triggering signal, when collector voltage across transistor can reach up to the supply one, is very dangerous. Such operation situation results in high power losses and damage of the transistor. To fight against such case a collector voltage has to be checked at existence of triggering signal and if that is above the admissible, triggering has to be stopped. It is sufficiently to decrease gate voltage for instance from 15 V to 8 V to avoid a break-down process of a structure of transistor. Such control device can be implemented in the content of driver and should be elaborated by circuit of Fig.1.32. The comparator DA compares a voltage UCʹ proportional to the collector voltage of the transistor VT with a reference one U0. If U C ′ > U 0 and output of triggering signal generator D is of high logical level (active one) then at output of AND logical gate D1 is an active high level triggering signal for protection transistor VT1. The transistor decreases gate voltage of controlled transistor VT to the zero level. For undisturbed turn-on process of the controlled transistor it is necessary to apply a RC chain to introduce some time delay (some tenth nanoseconds long) for response of the D1 on activation signal for on-switching of the VT. If overloading exists then the driver can transmit information about the case to the main control circuit. Very special protection element is RDC snubber. Inserting a capacitor in parallel with turned-off transistor (through a diode, see Fig. 1.23) it is possible to obtain that in way of the process at rising of collector voltage a collector current will be lower than that of the load as well as it is possible to reduce overvoltage across the transistor due to small leakage inductances of controlled circuit. Really operation of snubber circuits will be described in the part of switching converters. +Us

D

+U1 V0

UC

RB

VT

UC´ U0

DA

&

D1

VT1

Fig.1.32. Protection circuit of transistor against operation under short-circuiting conditions

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Review Questions

1st part

Verification questions on the material of the first part 1. What are the main power semiconductor devices applied in power electronic? 2. What is the main semiconductor material? 3. How can be silicon semiconductor layers with a p and n majority carrier concentration obtained?

4. What majority carrier layers form a semiconductor diode? 5. Why does a potential barrier at contact zone of the both p and n majority carriers layers create and what are the parameters of the silicon barrier?

6. What are the names of electrical terminals of a diode applied to the p and n layers of structure?

7. What is the direction of current through a diode and how high is the voltage drop across a diode when a forward direction current is passing through?

8. What is the form of diodeâ€™s Volt-ampere characteristics forward biasing part and how is it practically applied for calculation of conductivity power losses?

9. What are the names of two main parameters of diodeâ€™s VA characteristics forward biasing part?

10. How is the rated voltage of a diode obtained? What are the extreme maximum values of it?

11. How are the power losses of diode connected in practical solutions with its structure temperature and what is the admissible temperature?

12. How can be diodeâ€™s periodical conductivity current averaged and RMS values calculated?

13. How to calculate power losses of a diode at periodical conductivity of current? How to calculate the temperature of diodes structure?

14. How is a rated current of a diode obtained? 15. What are the cooling measures applied for power semiconductor elements? 16. What does a value of diodes leakage current depend on? 17. What processes take place at fast change of a voltage polarity across a diode and why?

18. How can be a series connection of diodes installed? 19. How many and in what way are the diodes unified in the bipolar structure of the Bipolar Transistor BT?

20. What conductivity modes of transistors create if structure diodes are unified with common cathodes and anodes?

21. How are the layers and electrical terminals of a BT called? How are they denominated in electrical circuits?

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Review Questions

1st part 22. Why is it possible to obtain a current flow through a contrary connected of the internal opposite diodes circuit in the BT?

23. How large is the ratio between a collector current of the BT in respect to the emitter one and how does this ratio depend on value of an emitter current?

24. What is static amplification factor of the BT and how it is dependent on the transfer factor of emitter current?

25. How large base current must be applied for providing a full-scale turn-on of the BT and how then should be calculate conductivity loss?

26. What is relation between collector voltage of the deep-saturated BT and its base voltage regarding the emitter?

27. In what operation mode are applied transistors in circuits of power electronic converters?

28. How large approximately are turning-on and turning-off transient intervals for BT operating in the switch mode?

29. In what way can be obtained the switching and conductivity power losses of BT? 30. How large are possible maximum switching frequencies for the power BT operating in the switching modes?

31. How many semiconductor p and n conductivity layers are in content of thyristor structure and what sorts of transistors combine a structure in the substitution circuit?

32. How many terminals has thyristor and what are its names and current conductivity directions through these terminals?

33. In what way it’s possible to obtain uncontrolled turn-on of thyristor and how this process depends on current transfer factors of the structure’s transistors?

34. What are necessary electrical parameters of the gating circuit for guaranteed

turn-on of thyristor using control signal? How it can be constructed a control circuit of thyristor at its operation in high voltage circuits?

35. How it is possible turn-off thyristor in electrical circuits of different voltage sort?

36. What are differences between volt-ampere forward characteristic of diode and one of thyristor?

37. What are maximum admissible meanings of voltage and current of thyristors? 38. What are static and dynamic parameters of thyristor? 39. Why it’s necessary to restrict maximum speed of current rise for thyristor and in what way such restriction can be provided?

40. Why it’s necessary to restrict forward voltage rise speed of thyristor and in what way such restriction can be provided?

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Review Questions

1st part 41. How large are turning-off transient times of thyristors and what meaning has the parameter for safe operation of thyristor switch?

42. What is MOSFET transistor and what are meanings of its abbreviation letters? 43. In what way is created a conductivity channel of MOSFET transistor and how current should be transmitted through it?

44. What are denomination names of a MOSFET transistor terminals and what are necessary polarities of voltages for n-channel and p-channel enhancement type versions ?

45. What are maximum admissible rated electrical parameters of power MOSFET transistors?

46. How large gate voltage have to be applied for safe full-scale turn-on of the MOSFET?

47. What are time intervals for switching processes of the MOSFET transistors? What are admissible switching frequencies of MOSFET transistors in the power converter circuits?

48. What is main disadvantage of the MOSFET transistors which restrict wide application of the transistors in power electronic devices?

49. What are IGBT transistors and give meaning of its abbreviator letters? 50. What are two main basic elements of IGBT transistorâ€™s structures and why these elements define efficiency of the transistors?

51. What are admissible electrical parameters of the IGBT transistors? How large are switching transient intervals and what high switching frequency they can provide?

52. How large gating voltage must be applied for full-scale turning-on of IGBT transistor?

53. How large is forward voltage drop across turned-on in full extent IGBT transistor?

54. What parameter characterizes switching power losses of IGBT transistor? 55. On what parameters depend heating of semiconductor element at short pulse mode loading processes?

56. What is driver for control of transistor and how it can be constructed? 57. In what way can be organized control of transistor in circuit of high voltage? 58. Why must be applied protection of transistors from switching on shortened circuit?

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Tasks

1 part st

Calculation tasks for the 1st part 1. Current of semiconductor diode is as periodical rectangular shape pulses

with amplitude of current (200 + mn) A, duration of each pulse for t i = 1 ms and switching frequency (300 + mn) Hz. Calculated power losses defined by parameters of an idealized VAC if the threshold voltage of its is U 0 = 0.9 V and differential resistance Rd = 1.5 mΩ!

Answer: at mn = 00 power losses are 72 W. 2. For parameters of task 1 calculate a temperature of diodes p-n barrier if thermal resistance of cooling system is 0.6 °C/W and an ambient temperature is +35 °C!

Answer: at mn = 00 the temperature is 78.2 °C. 3. For conditions given for tasks 1 and 2 (given parameters of load pulse, parameters of VAC, cooling and ambient temperatures) calculate an admissible amplitude of loading pulse current which provides operation at admissible temperature of p-n barrier equal the +130 °C!

Answer: at mn = 00 the amplitude is 364.72 A. 4. At experiments with the diode discussed previous at passing through sinus

shape half-wave current the admissible temperature of the diode +130 °C at ambient temperature +20 °C have been reached for averaged meaning of diode’s current in the stationary operation mode (250 + mn) A. Calculate thermal resistance of diode’s heat sink!

Answer: at mn = 00 thermal resistance is 0.24 °C/W. 5. The recovery charge of diode is Qrec = 100 µC . Calculate an amplitude of

reverse recovery current if amplitude of sinus shape forward current of the diode is (300 + mn) A but angular frequency of the current is ω = 628 s−1!

Answer: at mn = 00 the amplitude of reverse recovery current is 6.14 A. 6. For sustaining of the maximal reverse voltage of (10 + m) kV there are connec ted in series 10 diodes with they rated voltage (1500 + m ⋅100) V for each. Each diode is by-passed with similar resistor. Calculate necessary meaning of resistor’s nominal providing calculations for one ideal diode (without reverse biasing leakage current) when all other diodes are under same for all of them a reverse leakage current 10 mA! How large must be admissible power of each resistor?

Answer: at m = 0 resistance is 55.55 kΩ, power of resistor — 40.5 W.

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Tasks

1 part st

7. A base current of the bipolar transistor is 1 A, the static gain factor for the case with a common emitter is 20, resistance of load is 5 Ω, but voltage of circuit’s supply is (200 + mn) V. Calculate voltage difference between collector and emitter as also power losses of transistor!

Answer: at mn = 00 voltage difference is 100 V, power losses — 2 kW. 8. How large must be a base current for the previous task conditions for the fullscale saturation of the transistor?

Answer: at mn = 00 base current is 2 A. 9. Two bipolar transistors with defined previous parameters are connected in the Darlington circuit. Calculate necessary value of the base current for full-scale saturation of the transistors switch!

Answer: at mn = 00 base current is 0.0909 A. 10. Calculate switching losses for bipolar transistor operating in the switch mode

on resistive load if switching frequency is 10 kHz, supply voltage is (200 + mn) V, resistance of load is 5 Ω, current rise time of transistor is 1 ms, fall time is 3 ms, collector voltage of transistor in the transient processes is changing linearly versus time!

Answer: at mn = 00 switching power losses are 53.33 W. 11. The current transfer factors a1 and a 2 of the both transistors of a thyristor

structure are changing from its emitter currents identically: starts to rise from zero value at emitter current 30 mA and is linearly rising up to value 0.6 at emitter current 1.m A. Define the necessary gate current for triggering of thyristor’s turning-on, i.e., providing in all circuit of transistor a current 0.03 A! Leakage current of the thyristor must be accepted as of zero value!

Answer: gate current have to be 0.206 A at m = 0. 12. Calculate the necessary value of reverse gate current for turning-off of thyris-

tor which conducted a current (200 + mn) A, if the current transfer factors of the both transistors in turned-on situation of thyristor are 0.8 for each!

Answer: –150 A at mn = 0. 13. The threshold voltage of an idealized VACh of thyristor is 1.3 V, differential

resistance Rd = 3 mΩ. Through the thyristor an active half-wave of sinusoidal current with thyristor turn-on delay angle 30° and an amplitude of current (300 + mn) A is passing periodically. Calculate voltage meaning across the thyristor at current’s amplitude point as well power loss of thyristor and temperature of p-n structure if thermal

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Tasks

1st part resistance of the thyristor’s cooling system is 0,3 °C/W and an ambient one is +35 °C!

Answer: 2.2 V, 181.36 W, 89.4 °C at mn = 00. 14. Before turn-on instant of thyristor its anode voltage is (600 + mn) V, turning-

on time interval is 10 μs, an admissible rise speed of current is 150·106 A/s. Calculate time instant at which current through the thyristor exceeds value 300 A, at it assuming that the rise speed is restricted using a linear inductor, a voltage across the thyristor in turning-on process decreases in linear way and summary voltage of the circuit thyristor-inductor is (600 + mn) V! Calculate also an inductance necessary for the inductor and amplitude of the instantaneous power loss of thyristor in the process!

Answer: 6.32 ms, 4 μH, 66.24 kW at mn = 00. 15. Through thyristor and its reverse by-passing diode and inductor L periodi-

cally in oscillating way a large scale capacitor is over-charged. At end of current pulse through the reverse diode a voltage of the capacitor is (600 + mn) V in direction of thyristor conductivity and a forward voltage rise across takes place when its admissible voltage rise speed 100·106 V/s. Calculate a necessary capacitance of snubber capacitor across the thyristor if an inductance of inductor L is 20 μH!

Answer: 1.8 mF at mn = 00. 16. Define necessary capacitance of a parallel snubber capacitor for a bipolar junction, MOSFET and IGBT transistors which should provide the idealized its turn-off process with zero current at clamped with diode active-inductive load if turning-off processes long respectively for 2 μs, 50 ns and 200 ns and commutated voltage is (200 + mn) V but load current is 20 A!

Answer: at mn = 00 respectively 0.2 μF, 5 nF, 20 nF. 17. Define necessary inductances of series inserted with a bipolar junction, MOSFET and IGBT transistors inductors which should provide the idealized voltage-less turning-on process of the transistors at clamped with diode active-inductive load if turning-on processes long for 1 μs, 10 ns, 50 ns respectively! Commutated voltage is (200 + mn) V but load current is 20 A.

Answer: at mn = 00 respectively 10 μH, 100 nH, 500 nH. 18. Define a possible voltage peaks across the transistors of the two previous tasks in their turning-off processes at commutated voltage and current parameters defined in the tasks!

Answer: at mn = 00 respectively 341.4 V, 289.4 V, 300 V.

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2nd part

2. LINE-FREQUENCY CONVERTERS (LFC) The electromagnetic processes taking place in the line-frequency converters are determined by the sine-form network voltage with all its parameters: the factor of amplitude (the ratio of maximum and rms values of the voltage equal to 2), the factor of form (the ratio of rms and average values of the voltage equal to 1.11), angular frequency (ω = 314 s−1 at network frequency f = 50 Hz). The main types of LFC are rectifiers, AC controllers, cycloconverters. Rectifiers can be built on the basis of diodes and they can be controlled using controlled semiconductor switching elements — more often thyristors.

2.1. Diode Rectifiers 2.1.1. Rectifier Circuits Rectifiers are converters of network AC current into DC (in fact it is a quasi DC as the current flows in one direction but with periodical ripples). Rectifiers are classified according to the network AC voltage conversion way: if only one polarity half-cycle is converted, then it is a half-wave rectifier, if both polarities are converted, then — fullwave. More qualitative rectifiers are full-wave rectifiers. The single-phase circuit of half-wave rectifier is shown in Fig. 2.1,a for the case of active load operation, and in Fig. 2.3 — for the operation with the ideal inductive load. Real loads of the rectifiers are active-inductive and often with a counter EMF E connected in series (Fig. 2.1,b). When EMF is missed (Fig. 2.1,a), the instantaneous value of the current in the circuit is u1 = U1m sin ωt = L

di + iR, where i > 0 A; (2-1) dt

if EMF exists (Fig. 2.1,b), then U1m sin ωt − E = L

di + iR, (here i > 0 A; U1m > E ). (2-2) dt

Both expressions are valid if the load current i > 0 A, i.e., diode conducts the current. It mainly relates to the positive half-cycle of the network voltage, but as inductance has an ability to sustain current then partly it relates to the negative half-cycle of network voltage as well. The higher the load inductance is, the lower the current amplitude is. However, the current flows through the diode longer if active-inductive load exceeds half the period (Fig. 2.1,a), i.e., the angle of diode conductivity l is larger than p. From (2-1) it is obvious that current in LR circuit increases if the instantaneous values of the network voltage exceed those of the load resistor R. The maximum of the 65

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2nd part c urrent will be at the time instant when the instantaneous values of the voltages are equal, i.e., when I m R = u1 and it is close to the end of the half-cycle. After this moment u1 is lower than iR, and the current starts to decrease until it is zero, and the diode is supplied with its reverse voltage. If the circuit contains inductance, the total time interval of the current conductivity l is longer than p; therefore, knowing the average value of the current Id and resistance R angle l can be determined from the balance of the average voltages, where the average value of the diode voltage is U V = −I d R : 1 UV = 2π

2π

∫U

1m

sin ωt =

λ

U1m (cos λ −1) = −I d R, (2-3) 2π

where 2πI R d λ = arccos 1− . (2-4) U1m

It is obvious, if 1 < (2πI d R / U1m ) < 2 , then π < λ <1.5π; if (2πI d R / U1m ) < 1, then 1.5π < λ < 2π. Knowing l and assuming that the form of the current is close to triangle and I d = 0.5λI m / 2π, the amplitude Im as well as the necessary inductance can be calculated. However, the calculation is approximate enough due to imprecise form of the current.

uV

a

b

V u1

~

V

i R, L

u1 uV

0

L2 > L1 L3 > L2

0

Im

i

ud L E

p

l

i

u1

u1

R

2p

uV

L1

wt

u

u1

0 Id

i wt

0

i l

E

wt

wt

Fig. 2.1. Single-phase half-cycle rectifier: a — with RL load, b — with RL and EMF load

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2nd part Having rectification also in the second half-cycle, the instantaneous values of the load current with active-inductive load are always higher than zero, i.e., the load current is uninterrupted (continuous). If the load contains EMF, the current in the half-cycle circuit can start to flow only when the instantaneous value of the input voltage exceeds E (Fig. 2.1,b). Then current conductivity angle of the diode depends on the relation of E and U1m as well as on the inductance L: the closer E to U1m is, the higher inductance L is required to provide the conductivity angle λ = π, whch in the case of rectification in the opposite direction voltage half-wave will also provide continuous current through the load. The circuit of the half-cycle single-phase rectifier is applied rarely; therefore, other circuits will be considered in detail.

2.1.2. Single-Phase Bridge Rectifier In the case of single-phase supply voltage, the circuit of bridge rectifier containing four diodes is more frequently used (Fig. 2.2). When the potential of point 1 is positive according to point 2, i.e., u > 0 V, then the current is conducted by diodes V1 and V2, and the network current in Fig. 2.2,a is equal to that of the load. When u < 0 V current is conducted by diodes V3 and V4, and load voltage ud = −u . The ripples of the load current have double frequency of the network and with constant Rd and Ed its range is in opposite proportion to the inductance of the load Ld: the higher the inductance is, the lower the range of the ripples is. If inductance Ld is assumed to be of a very high value, then the load current ild is smoothed (ild = Id) and assumed to be DC. Then the current consumed from the network is of rectangular form (Fig.2.2b) and its rms value is I1 =

1 2π

2π

π

∫

I d2 dϑ +

0

∫ π

I d2 dϑ = I d . (2-5)

The average value of the load voltage (ideal) is 1 U d0 = π

π

∫U 0

m

sin ϑdϑ =

2U 2 2 π U ≈ 0.9U . (2-6) − cos ϑ 0 = π π

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2nd part a 1

i2

V1

V3

R, L, E

~ u

V2

Ld

u1 w1

ild

u2 w2 TR –

V4

2

+

Id

u ud

0

i1

b

i

iV1

ud V1, V2

V3, V4 u

0

Ud wt

ild

i2

Id 2p

p

0

wt –Id

Id

0

wt

Id

wt

Fig. 2.2. Circuit and diagram of the bridge diode rectifier operating with a complex load (a), connecting of the transformer secondary winding and form of the current of one diode with infinitely high load inductance (b)

Average value of the load current is U − Ed Id = d0 , (2-7) R where Ed is a counter EMF of the load. The average value of voltage across an inductance is zero. If a voltage-matching transformer is applied (Fig. 2.2,b), then its factor of transformation is kTR =

w1 U1 0.9U1 = = , (2-8) w2 U 2 U d0

rms value of the primary winding current is I1 =

IU I2 = d d 0 , (2-9) kTR 0.9U1

but the estimated power of the transformer is P S1 = I1U1 = d , (2-10) 0. 9 where Pd = U d 0 I d — power of the load in the case of smoothed current. Investigation of (2-10) gives an opportunity to state that in this case power factor of the rectifier is χ = Pd / S1 = 0.9. 68

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2nd part The estimated power of the secondary winding of transformer is of the same value as S1: S2 = I 2U 2 =

Pd , (2-11) 0. 9

but the total estimated power of the transformer is STR =

P S1 + S2 = d . (2-12) 2 0. 9

Example Single-phase bridge diode rectifier is supplied through a transformer, the primary voltage of which is 380 V, and the secondary — 250 V. Calculate the necessary power of the transformer, if the load of the rectifier contains series connected resistor R = 5 Ω and an inductor with a very high inductance. 1. The average value of the rectifier load voltage U d 0 = 0.9U 2 = 0.9 ⋅ 250 = 225 V . 2. Average value of the load current U d 0 225 = = 45 A. R 5 3. For the AC of rectangular form the rms value of the transformer secondary winding current is Id =

I 2 = I d = 45 A. 4. The estimated power of the transformer secondary winding S2 = I 2U 2 = 45 ⋅ 250 = 11 250 VA. 5. rms value of the transformer primary winding current I1 =

I2 IU = 2 2 = 29.6 A . kTR U1

6. The estimated power of the transformer primary winding S1 = I1U1 = 26.6 ⋅ 380 = 11 250 VA, i.e., the estimated power of the transformer STR = 11 250 VA, but Pd = 225 ⋅ 45 = 10125 W , i.e., χ = Pd / S1 = 0.9. Important parameter is the average value of the diode current. Taking into account the form of the current (Fig. 2.2,b) it can be defined as 69

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2nd part I Vav

1 = 2π

π

∫ I dϑ = 0.5I . (2-13) d

d

0

RMS value of this current is I Vrms =

1 2π

π

∫ I dϑ = I 2 d

d

0.5 . (2-14)

0

The reverse voltages of the diodes V3, V4 are formed with the half-cycle of the input AC voltage while V1, V2 are conducting and then a reverse direction voltage is connected to diodes V3 and V4 and they should sustain the amplitude value of the bridge input voltage, i.e., 2U. If the load is resistive, the form of the load current is of sine half-wave with the amplitude U m / R and rms value U / R, where U is the rms value of the network voltage. The load power is determined as P = U 2 / R. The rms value of the network current is the same as the load current, i.e., apparent power of the network S = U 2 / R , P = S and the power factor of the rectifier is one.

Example The supply voltage of the single-phase bridge rectifier is 220 V, 50 Hz, and load resis tance is 10 Ω. Calculate the active power of the rectifier operating for resistive load as well as with 100 mH inductance connected in series. 1. In the case of resistive load, the rms values of the load and network current are equal and U1 = 22 A ; R 2. Active power with resistive load I1rms = I ldrms =

I12rms R = 222 ⋅10 = 4840 W; 3. With inductance of 100 mH in the load its current can be assumed to be a smoothed one and then 0.9U1 2 I d2 R = R = 19.82 ⋅10 = 3920 W. R To achieve a smoothed current, a high enough inductance is required in the load circuit. Its minimum value providing a continuous current (boundary case) in the load can be calculated assuming that in the single-phase bridge rectifier with R, L, E load an increase in instantaneous values of the load current is determined by the difference 70

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2nd part of the average value of network rectified voltage in time intervals when it exceeds the average value of the load voltage Ud0 (Fig. 2.3) and then: π−ϑ1 did 1 L = U m sin ϑdϑ −U d 0. (2-15) dt π − 2ϑ1 ϑ1

∫

u Um

u

Ud0 id 0

J1

p–J1

Id J

Fig. 2.3. Simplified graph of the load current in the boundary case of conductivity

As U d 0 = 0.9U = Ed + I d R , then ϑ1 = arcsinU d 0 / U m ≅ 43°. In the boundary case when instantaneous value of the current «touches» a zero level, increased value of did can be replaced with 2Id, but dt = (π − 2ϑ1 ) / ω. Then taking into account (2-7), an approximate relationship can be obtained ωL 0.8U ≥ 0.29 . (2-16) R 0.9U − Ed A correcting factor 0.8 is introduced here defining that the increasing current is not linear but of wave form. As it is obvious, the higher the counter EMF connected to the load circuit opposite to the current direction the higher the necessary inductance, i.e., it requires connecting an inductor of higher inductance and size. Thus, if Ed is 89 % of the rectified voltage, i.e., Ed = 0.8U , then obtaining of the continuous current requires applying relationship ωL / R = 2.33. If Ed = 0.85U , then this relationship is already 4.64. Applying this expression it should be taken into account that it is quite approximate: if load does not contain E, then in accordance with the expression a large inductance is required; however, under these conditions the discontinuous current mode is not possible at all.

Example The supply voltage of the single-phase diode bridge rectifier is U = 220 V, 50 Hz. Calculate necessary inductance of the load inductor connected in series with Ed = 80 V 71

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2nd part and resistor R = 10 Ω and providing a boundary case of continuous load current. Evaluate approximately the mass of the inductor. 1. In this circuit, the inductor should meet the requirements L≥

0.23UR 0.23 ⋅ 220 ⋅10 = = 0.0137 H. ω(0.9U − Ed ) 314(198 − 80)

2. The mass of the inductor is approximately GL » LI d2 ; In this case Id =

0.9U − Ed 118 = = 11.8 A . R 10

Approximate mass of the inductor is

Current, A

GL ≈ LI d2 = 0.0137 ⋅11.82 = 1.9 kg . 25.00 20.00 15.00 10.00 5.00 0.00 –5.00 360.00

365.00

370.00

375.00

380.00 Time, ms

Setting the calculated inductance with the pre-set parameters, the minimum instantaneous current value in the modelled system is close to zero (see the picture above).

2.1.3. Single-Phase Centre-Tap Rectifier with Transformer This circuit (Fig. 2.4) contains 2 diodes only but does not operate without a transformer. The secondary winding of the transformer is divided into two equal parts with the number of turns w2 ′ = w2 ″ . In accordance with the polarity of the transformer halfwinding voltage, the load is connected either to one or to the other winding. During the positive half-cycle of the network voltage, the current is conducted (assume it to be fully smoothed) by diode V1 and ud is equal to the voltage of half-win ding w2’ with the current Id. In this interval, diode V2 is supplied with a reverse voltage −uV = (u2 ′ + u2 ″ ). 72

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2nd part During the negative half-cycle of the network voltage diode V2 is conducting and ud = −u2 ″, but V1 is supplied with the reverse voltage −uV = −(u2 ′ + u2 ″ ). As the form of the rectified voltage is the same as in the case of the bridge circuit, the average value of the rectified voltage is U d0 = 0.9U 2 ′, (2-17) where U2’ is the rms value of the voltage of one secondary half-winding.

u1

i1

TR

i2’ w2’ u2’ ud

V1

–

w2’’

u2’’ Ld

V2

+

u2 ud

ud V1

V1

V2

0

2p

p

Ud wt

u2’ = u2’’

i2’ Id 0 i1 0

Id/k TR

wt wt

Fig. 2.4. Single-phase centre-tap circuit with transformer and operational diagrams

Secondary winding rms current for the case of the smoothed load current I2 ′ =

1 2π

π

∫I

d ′ dϑ

= I d 0.5 , (2-18)

0

but the total estimated power of the transformer secondary winding is 2 0.5 I dU d 0 = 1.57 Pd . (2-19) 0. 9 If the transfer factor of the transformer is assumed to be kTR = U1 / U 2 ′, then rms value of the rectangular form current of the primary winding is I I1 = d , (2-20) kTR S2 = 2 I 2 ′U 2 ′ =

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2nd part but the estimated power of the transformer primary winding is S1 = U1 I1 = 1.11Pd. (2-21) The total estimated power of the transformer is STR =

S1 + S2 = 1.34 Pd. (2-22) 2

The diodes of the circuit should be calculated for the average current I Vav = 0.5I d and maximum reverse voltage 2 2U 2 ¢.

2.1.4. Three-Phase Half-Cycle (Centre-Tap) Rectifier The output voltage ud curve of three-phase half-cycle rectifier (Fig. 2.5) is formed with the instantaneous values of the AC voltage network. The current is conducted by the diode, a phase voltage instantaneous value of which is with higher positive potential according to zero neutral wire. From ωt = 30° to ωt = 150° phase A voltage is more positive (according to zero) and the current is conducted by diode V1, further B phase voltage is more positive and during further 120° diode V2 conducts, etc. Therefore, the conductance angle of the diode is λ = 120°. The load voltage as well as load current (even in the case of fully resistive load) is continuous, and its AC component is pulsing with triple network voltage frequency. Average value of phase current is I I phav = I Vav = d , (2-23) 3 and rms value in the case of the smoothed load current is I phrms =

1 2π

150°

∫ I dϑ = I 2 d

30°

d

1 . (2-24) 3

It is obvious that the phase current is of one polarity, i.e., it is not AC that unfavorably influences the supply network. To overcome this negative effect, the rectifier is connected to the network through transformer (Fig. 2.6). The transformer does not transfer DC component of the signal; therefore, in fact AC is consumed from the network. The network current is significantly damaged by unequal amplitudes of positive and negative polarities as well as by unequal intervals of current of positive and negative polarities in time. This current rms value in the case of the smoothed load current is I1phrms =

4 π/3 2 π/3 4 I 2 I d2 I 1 2 d d + . (2-25) ϑ = d 2 2 k 2π 9kTR 3 k 9 T R TR 0 0

∫

∫

where kTR = U1ph / U 2 ph = w1 / w2 is a transformation ratio. The voltage of the 74

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2nd part s econdary winding of the transformer forms a rectified voltage. DC component of it can be calculated as follows: U d0 =

3 2π

150°

∫U

2 phm

sin ϑdϑ =

3 2U 2 ph 2π

30°

− cos ϑ 150° = 3 6 U 2 ph = 1.17U 2 ph . (2-26) 30° 2π

Taking into account the obtained value of the network current, the power factor of this transformer based scheme can also be calculated χ=

U d0 Id 1.17 = = 0.827. 3U1ph I1phrms 2 iA

uA

0

V1

ild –

Ld

0

0

UBm

C

V2

V3

ud A

ud U Am

B

+

uph U Cm

A

p/6

B

C

A Ud

p

l

wt

2p

ild V1

V2

V3

Id V1

0

wt

iA V1

Id

V1

0

Id Id/3 wt

Fig. 2.5. Three-phase centre-tap rectifier and diagrams

The estimated power of the transformer secondary winding is S2 = 3I d

1 U d0 = 1.48Pd, (2-27) 3 1.17 75

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2nd part but the estimated power of the transformer primary winding is S1 = 3

I d 2 kTRU d 0 = 1.21Pd . (2-28) kTR 3 1.17

The total estimated power of the transformer is STR =

a

S1 + S2 = 1.347 Pd . 2

A

B iA

U1ph

b

C w1

ia

Id Id/3

U2ph

0

ia a

w2 b

c

2p/3

iA

2p 2I d 3k TR

0

+

q

4p/3

-Id 3k TR

q

ud –

Fig. 2.6. Connection of the transformer and current diagrams

However, the core of the transformer in this circuit is magnetized in one direction, i.e., the magnetic flux has DC component with subcomponent of each bar MMF = w2 (I d / 3). It means that the range of the magnetic flux density change is 2 ( Bs − B0 ), where Bs is density of saturation flux, B0 — the residual density of the formed flux. The range of the flux density change for an ordinary transformer is 2Bs. Therefore, calculating the transformer of the circuit under consideration this fact should be taken into account that results in the necessity to increase the size of the core. The current of the load in this case will be continuous only if the EMF of load Ed > U phm / 2. It is related to the fact that the lowest level of the rectified voltage instantaneous values is at U phm sin(π / 6) = 0.5U phm . Then the inductance of the inductor necessary for the continuous current (boundary case) can be calculated applying the approach similar to that for a single-phase bridge circuit:

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2nd part 0.1U ph U ph ωL ≥ , Ed > , (2-29) R 1.17U ph − Ed 2 where Uph is phase rms voltage at the input of the diode side. Comparing with the single-phase bridge circuit, a lower inductance L is required. Thus, if Ed composes 89 % of the rectified voltage, it is necessary to provide ωL / R ≥ 0.78 that is 3 times less than it is required for the single-phase bridge circuit.

Example Phase voltage of secondary winding of three-phase centre-tap Y/Y transformer is U 2 = 220 V, 50 Hz. Calculate the inductance of the load inductor connected in series with Ed = 220 V and resistor R = 10 Ω and providing a boundary case of discontinuous load current. 1. In this circuit, the inductor should meet the requirements L≥

0.1U 2 R 0.1⋅ 220 ⋅10 = = 0.0187 H . ω(1.17U 2 − Ed ) 314(1.17 ⋅ 220 − 220)

2. For this case Id =

1.17 ⋅ 220 − 220 = 3.74 A. 10

Approximate mass of the inductor is

Current, A

GL ≈ 0.0187 ⋅ 3.742 = 0.26 kg . 8.00 6.00 4.00 2.00 0.00 –2.00 355.00

360.00

365.00

370.00 Time, ms

The figure above presents the current diagram with the given parameters obtained from the computer modelling. The situation is obviously close to the boundary case of the discontinuous current.

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2nd part

2.1.5. Three-Phase Full-Bridge Rectifier In fact, this circuit combines two three-phase centre-tap circuits (Fig. 2.7), but output voltage instantaneous wave ud is formed from positive and negative half-cycles of a three phase-to-phase voltages uAB, uBC, uCA. When uBC < 0 V, i.e., C pole of the source is positive in respect to B pole the current is conducted by the diodes V1, V2. This interval lasts for 60° until voltage uAB is mainly positive when A pole is more positive than B pole. The current is conducted by V2 and V3. Period consists of 6 intervals with consecutively positive voltages –uBC, uAB, –uCA, uBC, –uAB, uCA. Each of six diodes operates within two adjacent intervals, i.e., the conducting angle of each diode is l = 120°. The ripples of the output voltage ud are six times higher frequency than that of the network voltage. If the load current is assumed to be smoothed one, then the network current is with 60° zero breaks within a half cycle (see Fig. 2.7,b), and the rms value of this current is I phrms =

1 π

120°

∫ I dϑ = I 2 d

2 . (2-30) 3

d

0

The rectified DC voltage is calculated as follows: U d0 =

3 π

120°

∫U

lm

sin ϑdϑ =

60°

3 2U l − cos ϑ 120° = 3 2 U l = 1.35U l, (2-31) 60° π π

where Ul is rms value of the phase-to-phase voltage at the input of the bridge. If the transformer with Y/Y connection is applied, the estimated power of the secondary winding is S2 = 3U 2 l I 2 f =

3U d 0 2 Id = 1.05Pd . (2-32) 1.335 3

The estimated power of the transformer primary winding as well as the total estimated power is the same. The power factor of the circuit: χ=

U d0 I d 3U l I ph

=

1, 35U l I d 2U l I d

= 0.955.

Load current can be discontinuous only if the circuit contains EMF Ed exceeding a minimum instantaneous value of the rectified voltage U lm sin60° = 0.866U lm . Then the continuous character of the current requires the inductance calculated in accordance with this approximate expression obtained by means of the previously-mentioned approach using the correction factor 0.7: 0.0105U l ωL ≥ . (2-33) R 1.35U l − Ed It is seen that the relation should be lower than that for the circuits considered above. 78

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2nd part a

A

B

iA

V6

u1

V3

V2

V5

V4

V1 –

UBC B

+

Ld

u1 –BC AB –CA BC –AB CA –BC

b

C

C

ud

U CA U AB

A0

p

2p

wt

ild, ud lV3

ild 0 iA 0

V1 V2

V3 V2

lV6

V3 V5 V5 V1 V1 V4 V4 V6 V6 V2 p 2p Id

2p/3

p/3

V3 V2

Id wt

wt –Id

Fig. 2.7. Three-phase full-bridge rectifier (a) and its diagrams (b)

Example The phase-to-phase AC voltage of three-phase full-bridge rectifier is U l = 380 V, 50 Hz. Calculate inductance L of the load inductor necessary for the continuous character of the current boundary case, if the the load circuit contains R = 10 Ω and Ed = 480 V connected in series. Evaluate approximately the mass of the inductor. 1. The inductance of the inductor can be calculated as follows: L≥

0.0105U l R 0.0105 ⋅ 380 ⋅10 = = 0.0038 H. ω(1.35U l − Ed ) 314(1.35 ⋅ 380 − 480)

2. Average value of the load current 1.35U l − Ed = 3. 3 A . R 3. The mass of the inductor Id =

GL ≈ LI d2 = 3.8 ⋅10−3 ⋅ 3.32 = 0.042 kg. The computer simulation demonstrates that the mode of the current in the circuit with the calculated inductance is close to discontinuous.

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2nd part

2.1.6. Six-Phase Rectifier If the secondary winding of the three-phase transformer is divided into two parts and a diode is connected to each part, then connecting cathodes of all 6 diodes and zero points of the windings results in a six-phase rectifier (Fig. 2.8). During a period the current is conducted in order by each of 6 diodes but only when the corresponding half-winding supplied voltage is the highest one in respect to the diode anode.

a iA

A

B

WA TR iA

WB iB

u2A

b u2 –u 2B

C WC iC

u2B

V4

V3

V6

iV1

V5

V2 +

–U 2B (V6) U 2C (V5)

U 2A (V1)

–u2C

u2B

–u2A

u2C

ud

u2C 0

V1

u2A

–

–U 2C (V2)

Id

60°

wt Id

0 iV4

Id

0 iA

2Id/3

wt

wt

Id/3 U 2B (V3)

0

wt

–U 2A (V4)

Fig. 2.8. Six-phase rectifier and its diagrams

If network phase voltage uA is higher, then voltage u2A of the phase A secondary halfwinding is also higher and current is conducted by diode V1. If network phase voltage uC is higher, then voltage u2A of the phase C secondary half-winding is also higher and current is conducted by diode V2, etc. Therefore, the period is divided into 6 equal typical intervals with duration of 60° each. The load current is assumed to be fully smoothed. Then the rms value of the current of transformer secondary half-winding and a correspondent diode is I 2 ′ = I Vrms = I d

1 , (2-34) 6

but the form of the rectified voltage is similar to that of the three-phase full-bridge rectifier; therefore, 80

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2nd part U d0 = 1.35U 2 ′, (2-35) where U2ʹ is the rms value of the voltage of the secondary half-winding. The estimated power of the transformer secondary winding is S2 = 6U 2 ′ I 2 ′ = 6

U d0 1 Id = 1.81Pd. (2-36) 1.35 6

At the primary winding side, the amplitude of the phase current with the same number of turns in both primary and secondary half-windings is 2 I d / 3 and its form is stepwise with lowest step of I d / 3. Within the conducting interval of the phase secondary half-winding, the parameters of the primary current can be determined using equations of MMF of both magnetic contours of the transformer core, e.g., when V1 is conducting and the transfer factor is 1, then the utmost left contour has a balance iA − i2 A − iB = 0, right-side contour iB − iC = 0, and iA + iB + iC = 0. Then iA = 2 I d / 3, but iB = iC = −I d / 3. The rms value of this stepwise primary current is I1 =

Id 2 , (2-37) 3kTR

where the transformation ratio U1ph

1.35U1ph

. U2 ′ U d0 The estimated power of the transformer primary winding is kTR =

=

S1 = 3U1 I1 =

3kTRU d 0 I d 2 = 1.05Pd , (2-38) 1.35 3kTR

but the total estimated power of the transformer is STR =

S1 + S2 = 1.43Pd . (2-39) 2

The relationship of the diode rms current of six-phase circuit to its average current reaches 2.4. As the rms value of diode is equal to that of transformer secondary halfwinding, then it is important to decrease this relationship. It can be achieved dividing the six diodes shown in Fig. 2.8 into two groups connecting a compensating inductor between them (Fig. 2.9). This forms the six-phase rectifier with an equalizing inductor.

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2nd part V1

V3

V5

V2

L

0.5Id

Id

V4

V6

0.5Id

Fig. 2.9. Currents of the six-phase rectifier with equalizing inductor

In fact, an equalizing inductor is a transformer with two series connected windings with equal number of turns w1 = w2. Then, if the magnetic core is not saturated the both windings have equal current (on average 0.5Id), but the load voltage ud is half part of the total summative voltage of the both groups. Although each diode now conducts within a time interval λ = 120° (Fig. 2.10), the amplitude of this current is 0.5Id only and the current rms value of the secondary halfwinding is much lower: I Vrms = I 2 ′ = 0.5I d

1 . (2-40) 3

The value of the rectified voltage in the six-phase circuit with an inductor can be calculated in the interval from 0° to 30° (Fig. 2.10) as follows: 150° 90° U c2m ′ 1⋅ 6 U b2 m ′ U d0 = sin ϑdϑ + sin ϑdϑ = π 2 2 120° 60° 6 2U 2 ′ 1 3 −1 + (2-41) = = 1.17U 2 ′ . π 4 4

∫

∫

The estimated power of the transformer secondary winding is S2 = 6U 2 ′ I 2 ′ = 6

U d0 1 0. 5 I d = 1.48Pd . (2-42) 1.17 3

The rms value of the transformer primary winding (Fig. 2.10 iA) is I1 =

0. 5 I d kTR

2 , (2-43) 3

where kTR = U1 / U 2 ′. The estimated power of the transformer primary winding is S1 = 3U1 I1 =

3U d 0 kTR 0.5I d 2 1.17kTR 3

= 1.05Pd. (2-44)

The estimated power of the transformer is STR = 1.27 Pd. Power factor related to network χ = 0.95. 82

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2nd part u

−ub2 ′ ua2 ¢ −uc2 ′ ub2 ¢ −ua2 ′ uc2 ¢ ud

0

30°60°

iV1 0 iV4 30° 0 iA

q

0.5Id 150°

q q

0.5Id/k TR

0

q

Fig. 2.10. Diagrams of the rectified voltage and currents of the six-phase rectifier with an inductor

The relation of rms and average values of the current of the diode in this circuit is the same like those for the 3ph bridge rectifier, i.e., 3. Diode is selected in accordance with the amplitude value of the breakdown voltage 2 2U 2 ¢ that can be supplied to it while another diode or diodes are conducting. As the circuit with an inductor is more effective in comparison with the one considered above without an inductor, the latter is not used very often.

2.1.7. Twelve-Pulse Rectifier If a transformer has two secondary windings, one of which has Y connection, but the second — delta connection (Fig. 2.11), then applying for each group its own threephase full-bridge rectifier a double rectified voltage 2Ud0 can be obtained at the output, the instantaneous value of which is pulsating with 12 times higher frequency than the network frequency. The phase-to-phase voltages of the groups with delta connection lag by 30° behind similar voltages of groups with Y connection: phase-to-phase voltage uABʹ in Y group leads to voltage uABʹʹ as the delta connected group of phase-to-phase voltages is formed from the transformer secondary phase winding Aʹʹ. 83

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2nd part Therefore, both line input currents of the rectifier groups iAʹ and iAʹʹ are similar in form, but mutually shifted by 30°, with the current iAʹʹ lagging behind in phase (Fig. 2.12). RMS value of these currents is I d 2 / 3 . Id w2ʹ

iAʹ

w1

A B C

Aʹ Bʹ Cʹ

w2ʹʹ

2Ud0

iaʹʹ Aʹʹ iAʹʹ iBʹʹ

ibʹʹ

Bʹʹ Cʹʹ

icʹʹ Fig. 2.11. Twelve-pulse rectifier

iAʹ Id 0 iAʹʹ

J 30°

0

J

iaʹʹ

iBʹʹ 0 Id +

iAʹ

2I d

k TR1 0

Id 3k TR1

3

Id +

Id

J

3

k TR1 J

Fig. 2.12. Time diagrams of the phase currents of both rectifier groups and current of the transformer primary winding

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2nd part The instantaneous values of phase currents of the winding of delta connection can be calculated using the following equations: ia ″ = ic ″ + iA ″ , ib ″ = ia ″ + iB ″ , (2-45) ia ″ + ib ″ + ic ″ = 0 , where iaʹʹ, ibʹʹ, icʹʹ are the instantaneous values of phase current of w2ʹʹ winding, iAʹʹ and iBʹʹ — instantaneous values of the phase current of the second rectifier input. Phase current iBʹʹ lags behind iAʹʹ by 120°. Solving the expressions for currents iA ″ − iB ″ . (2-46) 3 This current has stepwise form with steps Id/3 and 2Id/3; its rms value is ia ″ =

I arms ″ =

60° 120° 4 I d2 1 I d2 2 d + ϑ , (2-47) = Id π 9 9 3 0 0

∫

∫

i.e., it is 3 time lower than rms value of the rectifier input phase current. If the rectified voltages of the both groups are the same, the calculation of the estimated power of transformer secondary windings is the following: U d0

S2 ′ = S2 ″ = 3

Id

2 = 1.05Pd ′, (2-48) 3

1.35 3 where Ud0 is the voltage of one rectifier group, but Pdʹ is the rectified power of this group. Thus, the total estimated power of the secondary windings is S2 = 2.1Pdʹ.

The curve of the instantaneous values of the transformer primary winding current is formed in the mutual interaction of corresponding phase winding currents of the both groups: iA ′ =

iA ′ i ′ + A , (2-49) kTR1 kTR 2

where kTR1 = w1 / w2 ′ and kTR2 = w1 / w2 ″. If secondary phase-to-phase voltages of the both groups are equal, then kTR1 = kTR 2 3 , i.e., iA ′ =

iA ′ + 3iA ″ . (2-50) kTR1

The curve of this current in the half-cycle consists of steps, the most typical of which

(

)

are I d / 3kTR1 (two steps in the interval of 30° each), I d 1 + 1 / 3 / kTR1 (in the interval

(

)

of 30°), I d 1 + 2 / 3 / kTR1 (one step in the interval of 60°). Therefore, rms value of the 85

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2nd part primary winding current is 2 2 30° 30° 2 60° 2 2 I + I + 3 1 3 2 1.58 I d d I 1 d d I1rms = dϑ + 2 dϑ + dϑ = . (2-51) 2 2 2 2 k π k k k 3 3 3 TR 1 1 1 TR 1 TR TR 0 0 0 The estimated power of the transformer primary winding is

∫

S1 = 3U 2 f kTR1

∫

(

)

∫

(

)

1.58 I d U d0 =3 1.58 I d = 2.03Pd ′. (2-52) kTR1 31.35

The total estimated power of the transformer is STR = 2.065Pd ′. Power factor of the system related to network is χ = 0.985.

2.1.8. Ripple Factor of the Rectified Voltage The ripple factor of the rectified voltage is determined by the results of harmonics analysis. The ripple factor of the k-th harmonic is called the relation of the harmonic amplitude Akm to the rectified voltage Ud0: K pk =

Akm . (2-53) U d0

However, the characteristics of the rectifiers usually contain the factor of basic harmonic pulsation of AC components. The curves of the output voltage of the single-phase full-wave rectifier are not symmetrical in respect to the time axis, i.e., x(ωt ) ≠ −x(−ωt ). Therefore, the amplitude of sine-form harmonic Bkm = 0, but Akm is searched from the equation 2π π 1 Akm = U m sin ϑ cos kϑdϑ − U m sin ϑ cos kϑdϑ . (2-54) π π 0 Solution of this equation shows that there are only even harmonics

∫

U d (ϑ) =

∫

4U m 1 1 1 1 − cos 2ϑ − cos 4ϑ − cos 6ϑ.... (2-55) π 2 3 15 35

The fundamental harmonic of this curve is the second one as the number of ripples in the period m2 = 2, and the ripple factor of this harmonic K p2 =

4U m 4 2 = = 0.667 . 3πU d 0 3π0.9

The fundamental harmonic for the case of three-phase centre-tap circuit is the third one, as m2 = 3, and the ripple factor of this harmonic K p3 = 0.25.

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2nd part The fundamental harmonic for the case of three-phase full-wave circuits (including those with an inductor) is the sixth, as m2 = 6, and the ripple factor of this harmonic K p6 = 0.057 . Table 2.1 summarizes the data about the calculated parameters of all the circuits under consideration. Here m2 means the number of rectified voltage pulses per cycle, Kpk — ripple factor of the fundamental harmonic AC component of the rectified vol tage. The highest increase of the transformer estimated power is required for the sixphase circuit without an inductor, which is a reason for its limited application. The amplitude of the fundamental harmonic of the rectified voltage can be used for the approximate calculations of the load current pulses in the case of load R and L series connection. Then the amplitude of AC component is K pkU d 0 I dpm = . (2-56) R2 + (kf 2πL)2

Example The supply voltage of the single-phase bridge is 220 V, 50 Hz. Load circuit contains EMF in series connection with resistor R = 10 Ω and inductor with inductance L = 13.7 mH. Calculate a full range of the load current pulsations using the ripple factor of fundamental harmonic of the rectified voltage K p2 = 0.667 . 1. The average value of the rectified voltage U d 0 = 0.9 ⋅ 220 = 198 V ; 2. Amplitude of the fundamental harmonic of the voltage U d2m = 0.667 ⋅198 = 132 V; 3. Amplitude of the load current pulsation with 2 f = 100 Hz I d2m =

132 2

10 + (628 ⋅13.7 ⋅10−3 )

= 10 A;

4. Full range of the pulsation ∆I ld = 2 I d 2 m = 20 A. In the current diagram in Sub-chapter 2.1.2, the range is 20.5 A for this circuit with the given parameters.

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2nd part

2.1.9. Form of the Network Current The network current is of sine-form only in case of single-phase full-wave rectifier operation upon the resistive load. All the other circuits irrespectively of the type of load give non-sine form of the network current, i.e., it is distorted. In the case of singlephase full-wave rectifier with load when Ld = ∞, the form of the network current is rectangular (Fig. 2.13). The rms value of this current is I1 = I d , but that of the 1st harmonic component is I1(1) = 0.9 I d . a

i1

b

Id

0

i1

Id

0

wt

2p/3

–Id

wt

p/3

–Id

Fig. 2.13. The form of the rectifier input current: for the full-wave single phase (a), three-phase bridge (b) schemes

The rectifier input current of the full-wave single phase rectifier also contains other odd harmonic components: Number of the harmonic n rms value of the current I(n)/Id

1

3

5

7

9

11

13

15

0.897 0.299 0.179 0.128 0.0997 0.0815 0.069 0.0598

The quality of the current form is characterized by the so-called the factor of Total Harmonic Distortion (THD) ∞

THD =

∑I 2

I(1)

2 (n )

=

I 2 − I(21) I(1)

I 2 = −1 , (2-57) I(1)

which in this case is 0.493. The ratio of rms value of the fundamental harmonic to the total rms value, in its turn, is the factor of current distortion ν=

I(1) I

≤ 1. (2-58)

The multiplication of this factor with the network current fundamental harmonic shift angle cosj p (displacement factor) forms the power factor expression χ = ν cos ϕ p . (2-59) Thus, the diode bridge type single-phase rectifier has approximately χ = 0.9, current THD is 0.897 and displacement factor is 1. 88

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2nd part If the three-phase centre-tap circuit has no transformer, then for the smoothed load current I1 = 0.577 I d , the rms value of the 1st harmonic component of this network current is 0.385Id, containing even as well as odd harmonics, and THD = 1.12. Connecting input transformer with transformation factor 1, the network current I1 = 0.47 I d, rms of the 1st harmonic I1(1) = 0.388 I d, but THD = 0.684. In the case with a transformer, the distortion factor is ν = 0.825, power factor is approximately of the same value and the displacement factor is 1. The harmonic composition of the network current is of great importance. Thus, for example, the network current of the three-phase full-bridge (Fig. 2.13,b) consists of rectangular pulses of double polarity with amplitude Id, the duration of each pulse is 2p / 3, and interval between the pulses p / 3 time. The amplitude of the first harmonic is about 1.1Id, this current does not contain even harmonics; the next important harmonic is the 5th harmonic with the amplitude of about 0.22Id, then the 7th harmonic goes with the amplitude of about 0.158Id. The network current has other odd harmonics: 1

Number of the harmonic n

5

0.777 0.155

rms value of the current I(n)/Id

7

11

13

0.11

0.07

0.06

17

19

15

0.046 0.041 0.0598

THD = 0.32 , which is better than for the single-phase and three-phase centre-tap circuits. To improve the network current form, at the input three-phase bridge should contain the filters of the 5th and 7th current harmonics, i.e. for voltage distortion of (m2 −1) 5th harmonics and (m2 + 1) 7th harmonics. Applying such filters, the network current can be achieved very close to sine-form. Parameters of Diode Rectifiers Table 2.1

Ud0 m2 Diode average current IVav/Id Diode rms current (Ld=∞) IVrms/Id Breakdown voltage of the diode UVm Kpk S1/Pd0 S2/Pd0 STR/Pd0

Singlephase bridge

Singlephase centre-tap

Threephase centre-tap

Threephase bridge

Six-phase Y/YY

Six-phase with inductor

0.9U 2

0.9U2ʹ 2

1.17Uph 3

1.35Ul 6

1.35U2ʹ 6

1.17U2ʹ 6

0.5

0.5

0.33

0.33

0.166

0.166

0.707

0.707

0.577

0.577

0.4

0.288

2U

2 2U 2 ¢

6U ph

6U l

6U 2 ¢

6U 2 ¢

0.66 1.11 1.11 1.11

0.66 1.11 1.57 1.34

0.25 1.21 1.48 1.345

0.057 1.05 1.05 1.05

0.057 1.05 1.81 1.43

0.057 1.05 1.48 1.265

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2nd part

2.1.10. The Process of Current Commutation Table 2.1 provides the data for an ideal operation mode of rectifier. The supply network with a rectifier (e.g., with an input transformer) has significant active and inductive resistance. The output voltage is influenced by the equivalent input inductive resistance causing the so-called effect of current commutation significantly decreasing the value of the rectified voltage. Let us consider the commutation influence in the single-phase circuit with a transformer (Fig. 2.14).

u2

ud

w1 0

TR u2’ La

u2’’

i1

Rd id

V1

i2

La V2

Du g

p

2p

J

u2 i1

Id

0 i2 0

J J

Fig. 2.14. The diagrams of the commutation process for the single-phase centre-tap rectifier with transformer

The commutation effect exists at the moment when input voltage u2 changes its polarity. If the load current is smoothed (Ld = ∞) and the diodes circuit contains the inductances La, both currents i1 and i2 cannot change stepwise, but their sum according to Kirchhoff’s law is i1 + i2 = I d. (2-60) When u2’ changes to positive polarity, the current i1 is gradually increasing, but i2 — gradually decreasing, i.e., during some time interval g both switches conduct. The following voltage equations are valid within this commutation time interval: di1 + ud , dt di u2 ″ = −La 2 − ud . (2-61) dt u2 ′ = La

Taking into account (2-60) and assuming that t = ϑ / ω, from (2-61) the following equation can be obtained Xa

di1 = u2 = U 2 m sin ϑ. dϑ

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2nd part Applying this for u2 in the first equation of (2-61) makes it obvious that during the commutation time ud = 0 V . Therefore, the curves of the rectified voltage are decreased (cut out) for the shaded area, that within the half-cycle time decreases the average rectified voltage for DU. Current i1 during this process is changed (at ϑ = 0 current i1 = 0 A) as i1 =

U 2m (1− cos ϑ), (2-62) Xa

but the commutation interval is determinable at i1 = I d : I X γ = arccos 1− d a . (2-63) U 2m

The higher Id and Xa are, the higher g is. The cut out area can be calculated as follows: ∆U =

1 π

γ

∫U

2m

sin ϑdϑ =

0

Xa I d m2 Xa I d = . (2-64) π 2π

Therefore, the real rectified voltage is U d = U d0 −

m2 Xa I d . (2-65) 2π

In the single-phase bridge circuit, all 4 diodes are conducting during the commutation interval (ud = 0 V ) and the change of the source current i from one polarity i = I d to the opposite polarity of Id takes place. Going to the voltage of positive direction the changes of the source current i are described as follows: u = U m sin ϑ = Xa

di . (2-66) dϑ

Taking into account that at ϑ = 0 current i = −I d, solution of the equation is i=

Um (1− cos ϑ) − I d. (2-67) Xa

At the end of the commutation when i = I d 2I X 2I X 1− cos γ = d a and γ = arccos 1− d a . (2-68) U U m

m

The loss of the voltage during the commutation time 1 ∆U = π and

γ

∫U 0

U d = U d0 −

m

sin ϑdϑ =

2I X Um (1− cos γ) = d a (2-69) π π

2m2 I d Xa . (2-70) 2π

As it is seen for the application of (2-64) in the case of single-phase bridge, a double value of the real source inductance should be assumed. The commutation in a 91

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2nd part s ingle-phase circuit takes place only with an active-inductive load as operating with an active load only the current at the end of half-cycle is zero. If the input of a rectifier is a transformer with two windings, then Xa can be approximately calculated with the help of basic rated parameters of the secondary winding U2N, I2N and and impedance voltage of the transformer U k*, that is the ratio of the voltage of the primary winding to the rated voltage for the case if short-circuit secondary winding is loaded with its rated current. Then Xa =

U k*U 2 N . (2-71) I2N

The impedance voltage ratio is usually from 0.05 to 0.07. If the parameters of the supply transformer of the single-phase rectifier are U 2 N = 100 V , I 2 N = 50 A , U k* = 0.07, then 0.07 ⋅100 = 0.14 Ω. 50

Xa =

Example A voltage of the primary winding of the bridge rectifier transformer is 380 V, but rated voltage of the secondary winding is 100 V. The rated current of the secondary winding is 100 A, impedance voltage is 0.07. Calculate a commutation angle of the rectifier and the losses of voltage of the process while operating for the load with series connection of resistance of R = 1 Ω and large scale inductor! 1. Inductive reactance of the transformer’s windings in respect to the rectifier Xa =

U k*U 2 N 0.07 ⋅100 = = 0.07 Ω. I2N 100

2. Load current of the rectifier Id =

U d 0 − ∆U γ

Here from

R

=

0.9U 2 N − 2 I d R

Xa π ;

0.9U 2 N = 86.16 A. Xa R +2 π 3. Commutation angle 2I X 2 ⋅ 86.16 ⋅ 0.07 = 23.78°. γ = arccos 1− d a = arccos 1− U 2 ⋅100 Id =

2m

4. Loss of the voltage due to commutation ∆U γ = 2

I d Xa = 3.84 V . π

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2nd part In the three-phase centre-tap circuit, the commutation between phases A and B starts from the instant when the voltage of phase B uB is higher than uA and the current is conducted by V1 and V2. Besides, in the three-phase circuit commutation also takes place with only active load as at the instant of commutation the instantaneous value of the load current is not zero. With the smoothed load current (Fig. 2.15) during the commutation −uA + Xa

di A + ud = 0 , dϑ

−uB + Xa

diB + ud = 0 , (2-72) dϑ

i A + iB = I d . As dI d / dϑ = 0, then ud =

u A + uB , (2-73) 2

i.e., during the commutation time the rectified voltage is a half of the phase commutating voltages.

a

Id

b

TR

u

A

B

uB

w2

uA

iA

Xa iB

Xa

V1

V2

ud

C ud

0

p

iV

0

V1 V3

g Id

V2

J

J

Fig. 2.15. Illustration of three-phase centre-tap rectifier between phases A and B (a) and the diagrams (b)

Introducing (2-71) in system (2-70) and accepting that at the very beginning of commutation ϑ = ωt = 0, it can be written that 93

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2nd part 2 Xa

diB 5π π = uB − uA = U m sin ϑ + −U m sin ϑ + (2-74) 6 6 dϑ

2 Xa

diB = U m 3 sin ϑ. (2-75) dϑ

or

Here Um is an amplitude value of the phase voltage. From the previous expression, taking into account that at ϑ = 0, iB = 0 A, iB =

Um 3 (1− cos ϑ), (2-76) 2 Xa

but commutation (overlapping) angle can be expressed from the equation 2 I d Xa

. (2-77) Um 3 Loss of the rectified voltage in the commutation interval is (1− cos γ) =

3 ∆U γ = 2π

γ

∫ 0

u − uA + uB dϑ = 3 B 2 2π

γ

∫ 0

U BAm sin ϑ dϑ = 2

mI X 3I X 3U m 3 (2-78) (1− cos γ) = d a = 2 d a . 4π 2π 2π Voltage drop DU can also be found similarly for the other possible patterns of commutation. It can be accepted that a rectifier should be substituted with an ideal generator Ud0 of the DC voltage with an internal resistance defined by the commutation process (Fig. 2.16). In this circuit =

U d = U d0 − Id

m2 Xa . (2-79) 2π

+

Ud0

Id

Ld

Ud m2 X a 2p

–

Fig. 2.16. Substitution circuit of a rectifier

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2nd part

Example A phase-to-phase voltage of the primary winding of 3ph Y/Y connection transformer is 380 V, but a rated phase voltage of the secondary winding is 100 V. The rated current of the secondary winding is 100 A, impedance voltage is 0.08. Calculate a commutation angle for the 3ph bridge rectifier as well as loss of voltage by the process while operating on load with series connection of resistance of R = 1 Ω and large scale inductor! 1. Inductive reactance of the transformer windings in respect to the rectifier U k*U 2 phN 0.08 ⋅100 Xa = = = 0.08 Ω I 2 phN 100 2. Load current of the rectifier Id =

U d 0 − ∆U γ R

=

1.35U 2 lN − R

6 I d Xa 2π ;

Here from Id =

1.35 ⋅ 3U 2 phN 233.8 = = 217.2 A. 6 Xa 1.076 R+ 2π

3. Commutation angle 2 ⋅ 217.2 ⋅ 0.08 2 I d Xa = 30.89°. γ = arccos 1− = arccos 1− U 6 ⋅ 100 2 3 2 phN 4. Loss of the voltage due to commutation ∆U γ =

m2 Xa I d = 16.6 V. 2π

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Review Questions

2nd part

Verification questions 1. What are the main circuits of uncontrolled rectifiers? 2. What is polarity of rectifier load voltage in respect to the common points of anodes and cathodes of the bridge mode rectifier?

3. How large is the average value of voltage across the inductor in the periodical processes?

4. DC voltage of a load of rectifier is 200 V; the load contains a resistor of 5 W and an inductor connected in series. How large is the average current through the load and does it depend on the inductance of inductor?

5. How does the value of inductance of inductor influence the change in load current instantaneous values in a periodic process?

6. Load of the single-phase bridge rectifier should be accepted either as resistive

or with resistor connected in series with a large scale inductor. How large is an instantaneous value of load current at supply voltage zero crossing point in each case of the load circuit?

7. In the previous case, the amplitude of supply voltage is 312 + mn V. How large is

the average and rms value of load voltage, as well as rms value of supply current in the both mentioned cases of load if resistance of load is 5 W? How large is active power in the both cases? Answer: if mn = 00, then the average value of load voltage Ud0 = 198 V, Urms = 220 V; I1rms is 44 A (R load) and 39.6 A, respectively for RL load; active power is 9680 W (R load) and 7841 W. 8. How large is the DC component in the voltage of single-phase bridge type diode rectifier in an ideal case, if rms value of the supply sine-form AC voltage of 50 Hz frequency is (200 + mn) V? Answer: for mn = 00, answer is 180 V.

9. How large is the DC component of the load current for the above-mentioned

rectifier in the case of active-inductive load with 10 W + 50 mH? Could the load current be continuous or discontinuous in such conditions? Answer: for mn = 00, Id = 18 A; it can be continuous as the load contains high enough inductivity.

10. How large is the DC component of the load current for the above-mentioned

rectifier if a back-EMF of (150 + mn) V is connected in series with the load? Is the load current continuous in this case? Answer: for mn = 00, Id = 3 A; the current is discontinuous. 11. Is it possible in a general case to obtain the mode of discontinuous current operating with active load in the circuit of a single-phase bridge rectifier?

12. What is an AC component frequency of the load current of a single-phase bridge 96

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Review Questions

2nd part rectifier and in what way does the range of full oscillations of it depend on the load inductance? What approaches can be applied for an approximate calculation of this range?

13. Is the active power of a single-phase bridge rectifier higher in the case of activeinductive or fully active load?

14. How large is the estimated power of transformer for the above-mentioned

cases of the load of a single-phase bridge type rectifier, if the supply voltage is Um = 312 + mn V, R = 5 Ω, the inductance of the load is infinitively high and the secondary winding of the transformer contains twice less turns than the primary one. Answer: for mn = 00, STR = 2178 VA for active-inductive load and STR = 2420 VA for active load.

15. What is the power factor of a single-phase bridge type rectifier for the case of ideally smoothed load current? What is the distortion factor and THD indicator for this operation mode?

16. At what parameters of a single-phase bridge rectifier is the process of network

current commutation formed, and how many diodes are conducting this time? Is the current commutation possible with active load?

17. What parameter of a transformer defines an inductance of transformer coordinating rectifier to the supply network?

18. What is required for the formation of a single-phase centre-tap rectifier circuit? 19. What winding of the transformer of a single-phase center-tap rectifier has higher estimated power and why?

20. Secondary half-winding of the transformer of a single-phase centre-tap rectifier

contains two times less turns than that of the primary winding. Calculate the rms values of the half-winding current of the supply source and secondary winding for both cases defined in question 14, if the amplitude of the network voltage is 312 + mn V! Answer: for mn = 00, with LR load I2ʹ = 14 A, I1 = 9.9 A; with R load I2ʹ = 15.6 A, I1 = 11 A.

21. How large is the estimated power of a single-phase transformer for the cases mentioned above for a single-phase centre-tap rectifier, if secondary winding of the transformer contains twice less turns than the primary one. Answer: for mn = 00, with LR load STR = 2629 VA; with R load STR = 2926 VA.

22. Can three-phase centre-tap rectifier operate without a transformer? What are the particularities of this case operation? Why does this circuit require the transformer?

23. In what way does the input voltage form the load voltage of three-phase centre-tap rectifier, and what is the frequency of AC component of this voltage form?

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Review Questions

2nd part 24. How large is the DC component in the load voltage of three-phase centre-tap rectifier in an ideal case, if rms value of the supply AC voltage is (200 + mn) V? How large is the DC component of the current of active-inductive load, if R = 10 Ω and L = 50 m? Answer: for mn = 00, Ud0= 234 V, Id = 23.4 A.

25. Could the load current of three-phase centre-tap rectifier operating for an ac-

tive load be discontinuous? Is the network current commutation possible in this circuit operating for an active load?

26. How large should the value of back-EMF of the load of three-phase centre-tap rectifier be to get a discontinuous current of the load?

27. Can the network current commutation exist in the circuit of three-phase centretap rectifier if it operates for a fully active load?

28. Does the active power of three-phase centre-tap rectifier depend significantly on the active-inductive load current ripples?

29. Is the operation of the three-phase bridge type rectifier possible without a sup-

ply transformer, and what kind of supply voltage forms the AC component of the rectified voltage? What is the frequency of this AC component?

30. How large is the DC component of the load voltage of three-phase ideal bridge

type rectifier, if supply AC phase-to-phase voltage is (350 + mn) V? How large is the DC component of active-inductive load, if R = 10 Ω and L = 50 mH? Answer: for mn = 00, Ud0 = 472.5 V, Id = 47.25 A.

31. Could the load current of three-phase bridge rectifier be discontinuous? Is the

network current commutation possible in this circuit operating for an active load?

32. How large should the value of back-EMF of three-phase bridge rectifier be to get a discontinuous load current?

33. Can the network current commutation exist in the circuit of three-phase bridge type rectifier if it operates for a fully active load?

34. Does the active power of three-phase bridge-type rectifier depend significantly on the active-inductive load current ripples?

35. Which of the diode rectifier circuits provides better network current THD indicator and power factor for the condition of smoothed load current?

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Tasks

2 part nd

Calculation Tasks 1. The output current of a single-phase bridge-type diode rectifier with infini-

tively high load inductivity and resistance R = 5 Ω is Id = (20 + m + n) A. Calculate the estimated power of the supplying transformer if the network voltage is 220 V.

Answer: for m = n = 0 the estimated power is 2220 VA. 2. The output current of a three-phase centre-tap diode rectifier with infinitively high load inductivity and resistance R = 5 Ω is Id = (20 + m + n) A. Calculate the power of the supplying transformer if the network voltage of the primary winding of the transformer is 380 V.

Answer: for m = n = 0 the estimated power is 2685.8 VA. 3. The output current of a three-phase bridge type diode rectifier with infinitively high load inductivity and resistance R = 5 Ω is Id = (20 + m + n) A. Calculate the power of the supplying transformer if the network line-to-line voltage is 380 V.

Answer: for m = n = 0 the estimated power is 2108 VA. 4. The output current of a single-phase centre-tap diode rectifier (with transformer) with infinitively high load inductivity and resistance R = 5 Ω is Id = (20 + m + n) A. Calculate the power of the supplying transformer if the network voltage is 380 V.

Answer: for m = n = 0 the estimated power is 2674 VA. 5. The output current of the six-phase diode rectifier without an equalizing inductor with high load inductance and resistance R = 5 Ω is Id = (20 + m + n) A. Calculate the estimated power of the supply transformer if the supply voltage is 380 V (line-to-line)!

Answer: for m = n = 0 the estimated power is 2860 VA. 6. The output current of the six-phase diode rectifier with a smoothing inductor

with high load inductance and resistance R = 5 Ω is Id = (20 + m + n) A. Calculate the estimated power of the supply transformer if the supply voltage is 380 V (line-to-line)!

Answer: for m = n = 0 the estimated power is 2533 VA. 7. Short-circuit (impedance voltage) ratio of the transformer of a single-phase

diode bridge rectifier (in accordance with the data of the 1st problem) is 5 %. Calculate the value of the commutation angle and average value of the load voltage, if the load current is (20 + m + n) A. How large should the load resistance be to provide the defined current?

Answer: 0.376 rad, 96.46 V, 4.82 Ω for m = n = 0. 99

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Tasks

2 part nd

8. Short-circuit ratio of the transformer of a three-phase diode centre-tap recti-

fier (in accordance with the data of the 2nd problem) is 5 %. Calculate the value of the commutation angle and average value of the load voltage, if the load current is (20 + m + n) A! How large should the load resistance be to provide the defined current?

Answer: 0.2862 rad, 97.95 V, 4.89 Ω for m = n = 0. 9. Short-circuit ratio of the transformer of a three-phase diode bridge type

rectifier (in accordance with the data of the 3rd problem) is 5 %. Calculate the value of the commutation angle and average value of the load voltage, if the load current is (20 + m + n) A! How large should the load resistance be?

Answer: 0.316 rad, 97.5 V, 4.87 Ω for m = n = 0. 10. The load of a three-phase diode rectifier includes high inductance and resistance R = 5 Ω. Load current is (20 + m + n) A. Without taking into account the commutation process calculate the amplitudes of the first harmonic component of the network current and others (to the 7th).

Answer: for m = n = 0 (1st) = 22.07 A; (2nd, 3rd, 4th, 6th) = 0; (5th) = 4.4 A; (7th) = 3.12 A. 11. The input voltage of a single-phase bridge type rectifier is (100 + mn) V, 50 Hz, load resistance is 1,m Ω, back-EMF of the load is (75 + mn) V. Calculate the load inductance necessary for the continuous mode of the load current.

Answer: for mn = 00 inductance is 6.1 mH. 12. The input phase-to-zero voltage of a three-phase centre-tap rectifier is

(100 + mn) V, back-EMF of the load is (100 + mn) V, load resistance is 1,m Ω. Calculate the load inductance necessary for the continuous mode of the load current.

Answer: for mn = 00 inductance is 1.87 mH. 13. The input phase-to-phase voltage of a three-phase bridge type rectifier is

(100 + m + n) V, back-EMF of the load is (120 + mn) V, load resistance is 1,m Ω. Calculate the load inductance necessary for the continuous mode of the load current.

Answer: for mn = 00 inductance is 0.34 mH.

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2nd part

2.2. Line-Frequency Phase Controlled Converters As a basic line-frequency converter one can consider a controlled rectifier (CR) on the basis of which a network inverter, reversible rectifier as well as cycloconverter can be implemented. The direct AC controllers are formed in a way different from the principles of controlled rectifier implementation. As a rule, these systems apply thyristors as a controlling and switching element. A controlled rectifier (CR) is a system where the diodes are partly or fully replaced with controlled semiconductor elements (e.g., thyristors) and the regulating effect is achieved changing the conductivity time interval of controlled element within the possible control interval. CR is implemented with all rectifier circuits examined above, replacing diodes with thyristors. As the thyristor in AC circuit is switched off when instantaneous value of alternative current is decreasing to zero, the control effect is reached changing the instant of thyristor switching on.

2.2.1. Control Characteristics Let us study the operation of single-phase bridge-type controlled rectifier with the load of two types: fully active resistance and with Ld = ∞ (Fig. 2.17). The shift of the thyristor switching-on instant in the analogue control circuits is implemented by means of comparison of a reference saw-tooth voltage ust with a slowly changing control voltage Uco (Fig. 2.17). Then if the saw-tooth voltage is linearly increasing to the value Ustm, the delay angle a of the thyristor firing from the possible control starting point is calculated as follows: α=

U co π. (2-80) U stm

Ustm is the amplitude of the saw-tooth voltage. When U co = 0 V, then α = 0; when U co = U stm , then α = π . As the control effect is obtained shifting up-down Uco, the control approach is called vertical, but angle a is a firing delay (or control) angle. Thyristors V1, V2 of the bridge-type rectifier require one saw-tooth voltage, thyristors V3, V4 — the other (dotted line in Fig. 2.17). If the load is active, the load voltage ud is of single polarity, as the current through the load is decreasing till zero at the angles ϑ = π, 2π, 3π, . The average value of the rectified voltage in this case is 1 Ud = π

π

∫U α

m

sin ϑdϑ =

Um 1 + cos α . (2-81) (1 + cos α ) = U d 0 π 2

U d0 = 0, 9U is ideal idle-run voltage of the diode bridge rectifier. Obvious: if α = 0, 101

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2nd part then U d = U d0 ; if α = π , then U d = 0 V. The conductivity angle of each thyristor λ = π − α . In the time intervals at ud = 0 V (id = 0 A) voltage across each thyristor of the bridge is half of supply one (see Fig. 2.17). The other circuits of controlled rectifiers with active load have other expressions for the calculation of output voltage. Thus, if angle α is counted from the crossing point of the phase voltage curves, then the output voltage of the three-phase centre-tap topo logy is equal to zero at α = 5π / 6. For the three-phase bridge-type circuit with active load U d = 0 V if α = 2π / 3. The control characteristics U d = f (α ) for the case of fully active load is given in Fig. 2.19.

Example Calculate the average value of the load current for the controlled single-phase bridgetype rectifier if the rms value of the supply voltage is 380 V, the load contains resistor R = 10 Ω but the firing delay angle of the rectifier is α = 60°. 1. Average value of the load voltage Ud =

U d0 0.9 ⋅ 380 (1 + cos α ) = (1 + cos 60°) = 256.5 V. 2 2

2. Average value of the load current Id =

Ud = 25.65 A . R

Example Calculate the average value of the load current for the thyristor three-phase centretap rectifier if the rms value of the AC supply phase voltage is 220 V, the load contains resistor R = 5 Ω, and the firing delay angle of the rectifier is α = 60°. 1. Average value of the load voltage π π 3 2U ph 1⋅ 3 Ud = U phm sin ϑdϑ = (− cos ϑ) π = +α 2π π 2π 6

∫

6

+α

3 2U ph 1 + cos π + α = 149.23 V. 6 2π 2. Average value of the load current =

Id =

U d 149.23 = = 29.85 A. R 5

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2nd part

Example Calculate the average value of the load current for the thyristor three-phase bridgetype rectifier if the rms value of the AC line-to-line supply voltage is 380 V, the load contains resistor R = 5 Ω, and the firing delay angle of the rectifier is α = 60°. 1. Average value of the load voltage Ud =

6 2π π

π

∫

3

π

U lm sin ϑdϑ =

+α

6 2U l (− cos ϑ) π = +α 2π 3

6 2U l 1 + cos π + α = 257.7 V. 3 2π 2. Average value of the load current =

Id =

U d 257.73 = = 51.55 A . R 5

The processes are different if the load contains large enough inductance. In this case, the load current is continuous (uninterrupted) and at the end of the half-cycle of voltage the conducting thyristors do not switch off, but continue conducting current until the second thyristor pair is on. It results in the negative sign of the load voltage ud instantaneous value after the end of the half-cycle. The commutation of the thyristors is forced by means of reverse voltage (Fig. 2.17). Each thyritor conducts during all possible conductance interval, i.e., λ = π.

active load Rd u1 uV1

u V4 u1

V1 uV1

0

p

V2 V3 id

ust Ld

p

J

a

Ustm

Uco

0

J

a

ud 0

active-inductive load Ld = ∞

V1 a V2

V3 a V4

l Rd

l V1 V2

V3 V4

J Ld = ∞ Rd

Fig. 2.17. Control of single-phase bridge-type rectifier operating with active and active-inductive load 103

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2nd part For the single-phase bridge circuit, the rectified voltage with this type of load is Ud =

1 π

π+α

∫U

m

sin ϑdϑ =

α

2U m cos α = U d 0 cos α. (2-82) π

In the cases of other topologies with the load with large inductance, load voltage is U d = U d0 cosα. In the case of three-phase centre-tap circuit (Fig. 2.18) thyristor V1 conducts from ϑ = π / 6 = 30°, but control system provides delay angle a, and thyristor V1 is on when ϑ = π / 6 + α . Thyristor V1 is switched off with the forced commutation when V2 is on, i.e., when ϑ=

π 2π 5π +α + = + α. 6 3 6

The average value of the rectified voltage ud is Ud =

0

3 2π

5π/6+α

∫

π/6+α

B

uph A V1

Ld

U m sin ϑdϑ =

C V2

3U m 3 cos α = U d 0 cos α. (2-83) 2π

uph

A

B

a

V3

C

a

a

iV id = Id

0

Ud0cosa

p 6

5p 6

J

Id

0 ≤α ≤ 3X a 2p

ud

π 2

ust

Ustm a p Uco

0 iV 0

J

Id V3

V1

V2

V3 J

Fig. 2.18. Three-phase centre-tap circuit, its principles of control and equivalent substitution circuit

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2nd part Therefore, if the load contains large inductance the rectified voltage can be calculated with the help of unified expression. If α = 0, then U d = U d0 , but U d = 0 V, if α = π / 2. If a is set higher than p/2, the generator of the rectified voltage in the equivalent circuit (Fig. 2.18) changes its sign, but the voltage of this polarity cannot produce current in the passive load because of the equivalent diode in the direction of the rectifier conductivity. The dependence of the average value of the idle-run rectified voltage on the firing angle a is called a control characteristic (Fig. 2.19).

Ud Ud0 1 Ld = 0 Rd > 0 m2 = 6 m2 = 3 m2 = 2

Ld = ∞

0

p 2

2p 3

5p 6

p

a

Fig. 2.19. Control characteristics of the controlled rectifier

2.2.2. Inverting Operation Mode of the Controlled Rectifiers The operation mode with α > π / 2 is possible if the load circuit contains its own source of voltage, only the direction of EMF of this source should be set in accordance with the possible direction of current in the CR circuit (Fig. 2.20). In the case given in Fig. 2.20, a the rectifier is set with α < π / 2 and the outside source generates energy as the direction of current which corresponds to that of both EMF. The current is limited by the resistance of the circuit: Id =

U d 0 cosα + Ed . (2-84) Rrec + Rd

This operation mode is not of high importance. However, if the conditions α > π / 2 (Fig. 2.20,b) and (U d 0 cosα ) < Ed are applied, then the outside voltage source is a generator but CR is becoming an energy consumer, because its EMF and current are in

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2nd part the opposite direction. The controlled rectifier recuperates the obtained energy to the AC network. As, in fact, the rectifier transmits the DC voltage of the outside source into network AC voltage, this mode is called a rectifier inverter mode.

a

b

Id Ud0cosa

Rd

π 2

Ld

α<

–

Rrec

Ed

+

Id α>

π 2

–

Rd Ld –

Rrec +

Ed

+

Fig. 2.20. The substitution circuits of the rectification and invertation modes

Let us analyse the electromagnetic processes in this mode assuming that current id is fully smoothed (Ld = ∞) and CR is a single-phase bridge-type topology (Fig. 2.21). At α < π / 2, the average value of output voltage Ud is with a positive sign; at α = π / 2 it is zero (if Rrec = 0 Ω), but at α > π / 2 it has a negative sign (Fig. 2.20). It is convinient that the sign and value of the rectifier control voltage Uco is selected in accordance with the sign and value of voltage Ud, i.e., at α < π / 2 U co > 0 V , at α = π / 2 U co = 0 V, and at α > π / 2 U co < 0 V . For this purpose, the reference saw-tooth voltage should be of gradually decreasing front (Fig. 2.21) crossing zero axis at α = π / 2. Then α = π / 2 (1−U co / U stm ), where Ustm is the amplitude of the saw-tooth voltage of positive polarity. Even if during the first half-cycle the network voltage is with a positive sign (Fig. 2.21), the current is conducted by V3, V4 until V1, V2 are on, the sign of voltage ud is opposite to that of the network instantaneous voltage u, but the direction of the network current is opposite to the voltage u polarity, i.e., AC source consumes power (p < 0 W). When at angle α > π / 2 V1 and V2 are on the reverse voltage produced from the network voltage (shaded areas in ud curve) is in the interval p – a of switch-off thyristors V3, V4. The polarity of the network current i1 and voltage u within this interval of reverse voltage is the same, i.e., the network generates positive power. The polarity of the network voltage in the second half-cycle is negative, but while the on signals are supplied to the thyristors V3, V4 the current is conducted by V1, V2 providing the instantaneous network voltage with the opposite current and p < 0 W (source consumes energy). As soon as V3, V4 are switched on, thyristors V1, V2 are supplied with reverse voltage and they are turned-off, but the network current changes 106

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2nd part a sign into the negative one and as u < 0 V power p becomes positive, i.e., within the reverse voltage interval the network generates power. Thus, intervals of negative and positive network power are periodically changed during the process, but if the first interval is longer, then on the average the power is supplied to the network from outside source Ed. The value of the inverted current can be calculated by means of (2-84) taking into account that the internal voltage of the rectifier is negative. For the safe off condition of the thyristors, interval p – a cannot be infinitely small. It should be longer than the rated turn-off time of the thyristor toff, i.e., π −α > t off . ω It means that the maximum value of α is α max < π − ωt off . (2-85) Thus, if ω = 314 s−1, t off = 100 µs , then α max < 178.4°. For the description of the inversion mode, a reverse voltage existence angle b is often used, which is called an inverting angle (within this interval inversion takes no place, since power is positive). In the case of the circuit under consideration β = (π − α ) < 90°.

u

u1 ~

ud

u

i1 V1

V4 V2

V3 ud

0

p/2

id

2p

–Ud

b

ust Ld = ∞

p

Ustm

0

Ed

a i1 0

J –Ed

p > 0 –

–Uco +

–

p < 0

Id +

+ –

J

–

J

Fig. 2.21. The diagrams of the invertation mode of the single-phase bridge-type controlled rectifier

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2nd part

Example Calculate the value of outside voltage source Ed that is necessary to provide an inverted to the supply current I d = 100 A, if the controlled three-phase bridge-type rectifier is supplied with network voltage 380 V, its internal resistance is Rrec = 0.3 Ω, CR firing delay angle is 130°, but the circuit of the outside source contains only inductor and Rd = 0 Ω. 1. An equivalent EMF of the controlled rectifier is Erec = U d 0 cos α = 1.35 ⋅ 380 ⋅ cos 130° = 329.7 V. 2. Outside source required EMF is Ed = I d Rrec + Erec = 100 ⋅ 0.3 + 329.7 = 359.7 V .

2.2.3. Commutation Processes in the Controlled Rectifiers Controlled rectifiers have the same commutation processes like the diode rectifiers with the only difference that they start conducting not at diode natural conductivity beginning instant but with the angle delay a. The commutation processes are determined by the supply phase inductive reactance Xa = ωLa . Let us consider the commutation influence in the single-phase circuit with transformer (Fig. 2.22). To examine commutation, the load current should be continuous, i.e., Ld should have a large enough value. Inductance is assumed Ld = ∞. Until thyristor V1 is off, current Id is conducted by V2. With V1 on, the current of thyristor V2 cannot instantly decrease, but the current of V1 is increasing as it is influenced by inductance La.

u

ud

inet TR u Xa

i1 V1

p a

u Ld = ∞ id

g

Xa i2

iv

V2

0

V2

V1

2p

J = wt

u V2

V1

inet j (1) 0

J J

Fig. 2.22. Commutation processes in the single-phase centre-tap controlled rectifier with transformer 108

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2nd part An interval of commutation is formed with the both thyristors switched on and then voltage ud = 0 V (similarly to the diode circuits). The process is described as follows: i1 + i2 = I d , di di di 2u + Xa 2 − Xa 1 = 2u − 2 Xa 1 = 0. (2-86) dϑ dϑ dt Assuming that at the beginning of the commutation process ϑ = 0 the voltage of the secondary half-winding u is U m sin(α + ϑ). Then the current i1 of thyristor V1 is increasing from zero to Id as follows: i1 =

Um cos α − cos(α + ϑ) , (2-87) Xa

but angle g should be find from Id =

Um cos α − cos(α + γ) , Xa

From here

I X γ = arccos cos α − d a − α . (2-88) U m For the case of three-phase controlled rectifier the similar expression is applied: cos α − cos(α + γ) =

2I d Xa , (2-89) U lm

where Ulm is the amplitude of phase-to-phase voltage (the analogue of this voltage in Fig. 2.22 is 2Um, then equation 2-89 is universal). The losses of voltage within half-period (shaded area in Fig. 2.22) are calculated as follows: 1 ∆U γ = π

α +γ

∫U α

m

sin ϑ =

I X mX Um cos α − cos(α + γ) = d a = 2 a I d. (2-90) π π 2π

It should be noted that the obtained expressions are related to the case when the load current is fully smoothed. If it is not the case, then current Id should be replaced with the value of the load current during the process of commutation. In the three-phase circuits, the commutation process can also take place with fully active load. It will be in 3ph centre-tap circuit if α < 30°, but in 3ph bridge-type circuit — if α < 60°. Calculating g from (2-89), the commutation angle is obviously increasing along with the increase in Id and Xa. If current Id achieves value 2U m cosa / Xa , then g is equal to p and a rectifier is permanently short-circuited as the conductivity angle of each thyristor in the circuit could be 2p. In three-phase centre-tap circuit (Fig. 2.18), the same phenomenon takes place if g is

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2nd part increased to 2p / 3, when the current is permanently conducted by two thyristors — V1 and V2, followed by V2 and V3, and then by V3 and V1, the angle of conductivity of each thyristor is 240°, but the diagram of the output voltage is formed with the sum of two phase voltages. With the rated load currents g usually does not exceed 30° to 40°. In the three-phase bridge-type circuit with rated load current 2 thyristors and du ring the commutation intervals 3 thyristors conduct current (Fig. 2.23 with γ = 0° and γ = 30°), for example, V1, V2, V3 then during the next commutation period V2, V3, V4, etc. With the increase in current, g achieves p / 3 when simultaneously three thyristors conduct, i.e., conductance interval of each thyristor is p. In its turn, if a exceeds 30°, the intervals of conductivity of 4 simultaneous thyristors can take place when the instantaneous value of the load voltage is zero causing a decrease in the rectified voltage. With γ = 120° the current is permanently conducted by 4 thyristors, i.e., a short circuit takes place when each thyristor conducts during the interval of 240° (Fig. 2.23). If 0° ≤ α < 30° maximum g value is 60° = π / 3, as the fourth thyristor of the bridge cannot switch on because of reverse biasing voltage. At α = 30° and γ = 60° border point 1 of the load characteristic is formed (Fig. 2.23). When the current is increased above the border point real angle a is increased and operation with γ > 60° takes place. If the current Id is not so high to cause the border mode of the commutation process, then the load voltage of the three-phase controlled rectifier is calculated as follows U d = U d 0 cosα − I d

6 Xa . (2-91) 2π

Commutation process is of high importance for the implementation of safe invertation. Figure 2.21 demonstrates that in fact the opposite voltage across the thyristor is formed in the interval p – a – g and the maximum angle a should be set taking into account the commutation angle α max < (π − ωt off − γ). (2-92)

Example Calculate commutation angle of single-phase centre-tap CR with transformer if its secondary half-winding has U 2 ′ = 150 V, 50 Hz, Xa = 0.2 Ω, the resistance of the RL load is R = 5 Ω, but the firing angle of the CR is 60°. 1. Load voltage of the rectifier U d = U d 0 cos α − I d

m2 Xa = U d 0 cos α − 2π

U d 0 cos α − I d R

m2 Xa 2π m2 Xa , 2π

where U d = I d R . 110

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2nd part Therefore, the load current with m2 = 2 and U d 0 = 0.9U 2 ′ = 135 V is Id =

mX U d 0 cosα ⋅1− 2 a 2πR 2

m X 1 R − 2 a 2π R

=

2 ⋅ 0.2 135 ⋅ 0.5 ⋅1− 2π ⋅ 5 2

2 ⋅ 0.2 1 5 − 2π 5

=

66.64 = 13.35 A. 4. 99

2. The commutation angle can be calculated from Id = or

Um [cos α − cos(α + γ)] Xa

cos(α + γ) = 0.487; α + γ = 60.82°; γ = 0.82°.

Example The rms value of the phase voltage of three-phase centre-tap CR is 220 V, 50 Hz, inductive reactance of the phase is Xa = 0.5 Ω. Calculate the commutation angle γ and average value of the load voltage with R = 5 Ω, if α = 60°. 1. The load current can be calculated from U d = I d R = U d 0 cos α −

U d 0 cos α − I d R

m2 Xa 2π m2 Xa ; 2π

taking into account m2 = 3, 3 ⋅ 0.5 mX U d 0 cos α R − 2 a 220 ⋅1.17 ⋅ 0.55 − 2π 2π 612.76 Id = = = 24.56 A . = 2 2 9 4 24 . . m X ⋅ 3 0 5 25 − R2 − 2 a 2π 2π 2. The commutation angle can be calculated from 2 I d Xa or

3U 2 2

= cos α − cos(α + γ)

2 ⋅ 24.56 ⋅ 0.5

= 0.5 − cos(α + γ); 3 ⋅ 220 ⋅ 2 cos(α + γ) = 0.4545; γ = 2.96°. 111

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2nd part 3. The load voltage U d = I d R = 24.57 ⋅ 5 = 122.85 V or ∆U γ =

3 Xa 3 ⋅ 0. 5 Id = 24.57 = 5.86 V . 2π 2π

Example Three-phase bridge-type controlled rectifier input AC phase-to-phase voltage is 380 V, 50 Hz, phase inductive reactance Xa = 0.5 Ω. Calculate maximum admissible inverted current if α = 150°, but turn-off time of the thyristor is 150 μs. 1. The reserve angle of the thyristor switching off is ϑ TR = ωt off = 314 ⋅150 ⋅10−6 = 2.7°. 2. Maximum admissible commutation angle is γ max = 180°− α − ϑ TR = 27.3°. 3. Maximum inverted current is I d max =

U lm [cos α − cos(α + γ max )] 380 2[−0.866 + 0.999] = = 71.7 A. 2 Xa 2 ⋅ 0. 5

Taking into account everything mentioned above, the load characteristics of real controlled rectifiers U d = f ( I d ) are the falling characteristics (Fig. 2.23) limited by the currents in the rectification mode and expression (2-92) in the invertation mode. Angle g is increasing along with the increase in current. The given loading characteristics correspond to the case of continuous (uninterrupted) current. However, with low load currents, especially if the load contains back EMF, it can be discontinuous, i.e., periodically can have shorter or longer intervals of zero value. This discontinuous current mode takes place if about a half of the load current full ripple is equal to the average value of the load current, i.e., 0.5idmax = I d , where idmax is the amplitude of the load current instantaneous value. If the mode of the current is discontinuous, the average value of the rectified voltage Ud is increased above U d0 cosa. With I d = 0 A the maximum rectified voltage is equal to Ud in the case of active load (Fig. 2.17). Therefore, the loading characteristic at very low Id differs a lot from a straight line (Fig. 2.23 shaded area). Approximately the average value of rectified current Idb in the boundary case between 112

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2nd part the continuous and discontinuous current mode can be defined assuming that the current instantaneous value linearly increases from zero till 2Idb within the time interval when the instantaneous values of the rectified voltage exceed the average rectified voltage U d0 cosa. 120° g = 0°

g = 30° g = 60°

g = 90° g = 120°

3

1 6

2

6 1

1 6

5 4

2 1

4

2

2

6

3

5 4

0

3

a = 90°

Id

a = 120°

2 1

6

a = 30° a = 60°

3

1

a = 0° g = 60° 1

3

1

5

Ud0

2

6

3

J

1 6

3 2

2

5 4

Ud

3

6

3 2

6

1

4

1

6

5

3 2

–Ud0

a = 150°

Fig. 2.23. Conductivity diagrams of the three-phase bridge-type rectifier with different commutation angles and load characteristic

Then the process is described with the approximate differential equation π−arcsin

2LωI db = U dm

U d0 cos α U dm

∫

α +ϑ s

U cosα − ϑ s , (2-93) sin ϑdϑ −(I db R + E )π − α − arcsin d0 U dm

where L is the load inductance, R — resistance, E — EMF, Udm is the amplitude of the rectified voltage, J — angular frequency of the network voltage, J s — delay angle from the beginning of the half-wave of the positive polarity voltage to the moment of angle a counting (ϑ s = 0° for single-phase bridge; 30 degrees for the circuit of three-phase centre-tap and 60 degrees for the three-phase bridge). To evaluate the validity of expression (2-93) for the calculations, a single-phase bridgetype rectifier has been modelled by means of software PSIM 6.0 and the results of the modelling are presented in Fig. 2.24 as the diagrams of the load current id = f (t ) for the boundary case, taken with L = 5 mH, R = 3 Ω, E = 0 V, the supply voltage of the rectifier 220 V, 50 Hz and α = 30°, with which according to expression (2-93) I db = 57.45 A (see example), but modelling — 57.6 A (see Fig. 2.24).

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2nd part id, A

R = 3 W L = 5 mH

100.00 80.00 60.00 40.00 20.00 0.00 –20.00

ud, V

300.00 200.00 100.00 0.00 –100.00 –200.00 20.00

a = 30°

25.00

30.00 Time (ms)

35.00

40.00

Fig. 2.24. The boundary case obtained in PSIM software for the load current of the single-phase bridge-type rectifier with Idb = 57.6 A (approximately calculated Idb = 58.3 A)

2.2.4. Power Factor of the Controlled Rectifier Controlled rectifier operates with the shift angle between current and voltage. Even with fully active load the basic harmonic component of non-sine form network current is shifted for a particular angle depending on control angle a. When CR operates with active-inductive load, the reactive energy is periodically recuperated through the network. The shift angle of the current first harmonic is close to α + γ / 2 (see Fig. 2.22). A notion of power factor is used for the description of the CR operation mode χ = νcosϕ(1), (2-94) where ν = I1(1) / I1 is the factor of distortion of the network current, I1 and I1(1) — rms values of the current and its basic (fundamental) harmonic component, respectively, cosφ(1) — cosine of the shift angle between current first harmonic and network voltage (Displacement Factor). Improvement of the power factor is one of the most important tasks. The simplest method is to by-pass an active-inductive load in rectifier with a diode (Fig. 2.25) that eliminates the discharge of the load reactive energy through the supply source.

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2nd part

Example Calculate the average value of the load current in the boundary case Idb of single-phase bridge-type CR with supply voltage U = 220 V, 50 Hz, if α = 30°, load inductance L = 5 mH, R = 3 Ω and E = 0 V. 1. For the calculations (2-93) can be applied, if we replace there U dm = 2U = 312 V , U d 0 = 0.9U = 198 V and ϑ s = 0°: π−arcsin

0 ,9 cos α 2

∫

2LI db ω = U dm

α +ϑ s

0.9 cos α . sin ϑdϑ − I db R π − α − arcsin 2

2. Taking the numbers π−0.58 −3

2 ⋅ 5 ⋅10 ⋅ 314 ⋅ I db = 312

∫ π 6

5π sin ϑdϑ − 3I db − 0.58 = 532 − 3I db ⋅ 2.04; 6

Therefore, I db = 57.45 A .

Example Calculate the average value of the load current in the boundary case Idb of three-phase centre-tap CR with supply voltage U ph = 220 V , 50 Hz, if α = 30°, load inductance L = 5 mH, R = 1.15 Ω and E = 182 V. 1. For the calculations (2-93) can be applied, if we replace there U dm = 2U ph = 312 V , U d 0 = 1.17U ph and ϑ s = 30°: π−arcsin

2LI db ω = U dm

1.17 cos α

∫ α+

2

π 6

1.17 cos α π sin ϑdϑ − (RI db + E ) π − α − arcsin − . 6 2

2. Taking the numbers π−0.794

2 ⋅ 5 ⋅10−3 ⋅ 314 ⋅ I db = 312

∫ π 3

2π sin ϑdϑ − (1.15I db + 182) − 0.794 = 3

= 374 −1.495I db − 236.6 ; Therefore, I db = 29.64 A. With these parameters the computer calculated I db = 34.69 A. 115

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2nd part In this case, the load voltage ud cannot be of negative polarity. As soon as the voltage crosses zero point to the direction of negative polarity diode V0 is on, the load current flows through the diode, and the reactive energy is not recuperated to the network. The rms value of the network current 1 I1 = π

π

∫ I dϑ = I 2 d

α

d

π −α , (2-95) π

amplitude of the fundamental harmonic component of this current I1(1)m =

2Id 2(1 + cos α ) , (2-96) π

obtained taking into account A1m and B1m values of this harmonic. Then the distortion factor of the current i1 is: ν=

I1(1) I1

=

2

1 + cos α

π

π −α

, (2-97)

but the shift angle of the 1st harmonic is ϕ(1) = 0.5α . The distortion factor depends on a, but it is close to the case without diode, when ν = 0.897 (see Section 2.1.9), but the shift angle is two times less than that of a simple controlled rectifier with active load. At α = 60° and γ = 0° operating with active inductive load the power factor of a typical controlled rectifier is 0.45 but for the circuits with by-passing diode χ = 0.93 ⋅ 3 / 2 = 0.804. The average current of the by-pass diode is I V 0av = I d

π −α . (2-98) π

If in a typical reversible controlled rectifier the rectification mode takes place from α = 0 to α = π / 2, then for this circuit a can change to p, i.e., the invertation mode is not possible. The output voltage of the circuit with diode U d = U d 0 (1 + cosα ) / 2 , and the active power of the rectifier at smoothed load current P = U d I d = U1 I d ⋅ 0.45(1 + cosα ), (2-99) reactive power Q = U1 I d

π −α − 0.45(1 + cos α ) , (2-100) π

but the total apparent power is

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2nd part S = U1 I d

π −α . (2-101) π

If we assume U1 I d = P = const , then this power can be described with relative values as in Fig. 2.26. It is obvious that reactive power of the rectifier with diode is much lower than that of a typical controlled rectifier, the power S of which is constant within all the control range at I d = const . ud

u1 ud i1 V1

V4 u1

0 a

V2

iV0

V3 iV0

V0 Ld

id ud

a

J u1

Id

0 i1

Id

J

0

J

j (1) Fig. 2.25. Voltages and currents of the controlled rectifier with by-pass diode

S*

1.0

with the by-pass diode P*

Ud*

Q*

0.5

Q* 0

20

cosj (1) ≈ I1* = S*

Ud*

40

P*

cosj (1) 60

80

100

120

140

160

180°

a

Fig. 2.26. Dependence of relative power and load voltage on α of a typical CR with by-pass diode

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2nd part Application of fully commutated switches in the rectifiers gives fully new opportunities. The first step in this direction was the application of CR with thyristors commutated with capacitors (Fig. 2.27). In this circuit, the thyristors of the cathode group V1, V3, V5 are switched off with thyristor V7 connecting to the group through the secondary winding of transformer capacity C charged with positive voltage polarity, but the anode group — with thyristor V8 connecting to the capacitor charged with negative voltage polarity.

Id

++

V7 uC

IA

V3

0

V5 A

C

w2 TR

V8

A iA V3

B

V6

C V3 p

V6

V1 B

C

uph iA

ud uI BC AB CA BC AB CA BC AB CA

V2

a <0

V4 0

–

J

p

J

A1

C1 B1

w1

V1 V2 V3 V4 V5 V6 V7

uC

V7

V7

V7

0

J V8

V8

V8

V8

V8

Fig. 2.27. Controlled rectifier with forced thyristor commutation, diagrams of operation with a < 0

For example, if current Id is conducted by V1, V2 and capacitor C is charged in positive direction (left plate +), then the transition to the next conducting pair V2, V3 is 118

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2nd part implemented with turn-on of thyristor V7. When thyristor V1 is switched off, the current Id is conducted by V2, V7 for some time (Fig. 2.27) and capacitor C is recharged with the opposite polarity. When the recharging process is completed, V3 is on and V2, V3 conduct. After 60° with thyristor V8 on and capacitor C charged in the negative direction V2 is off and the current is conducted by V8, V3. When the capacitor is recharged in the positive direction, V4 is on and the current is conducted by V3, V4. During the period, 6 commutations take place and the input AC current contains the third harmonic component. In order not to transmit it to the network, the primary winding of the transformer has the delta connection. Applying the forced commutation, the operation can be implemented with α < 0° in the rectification mode (Fig. 2.27) and with α > 180° in the invertation mode. Therefore, the basic harmonic of the network current can either lag behind the voltage or lead it (in the second case CR is a generator of reactive power, i.e., operates like a capacitor). This CR is also a load current source type rectifier if the load is with large inductance. More opportunities are obtained replacing thyristors with GTO fully controlled thyristors (Fig. 2.28). Comparing to the circuit with forced commutation, this circuit is simplified, but the input should be connected through the capacitor batteries eliminating over-voltages at the moments of thyristors switching off and compensating the inductance of the network. Application of the GTO thyristors gives an opportunity to implement multiple thyristor switching on within the interval of conductivity. Then the bridge-type circuit with the load of current source (smoothed current) can have the order of commutation as in Fig. 2.29. Modulating in such a way, the primary current harmonic composition is improved and the form of the current in the supplying network is close to sine-wave.

V3

A B C

iA

V5

V1

V2

V4

Id

iB iC V6

Fig. 2.28. CR with GTO thyristors and smoothed load current

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2nd part 5

1 1 6

4 6

1

5 1 2 2

3 3 2

3

6 2

1 3 4 4

5 5 4

2 4

5

3 5 6 6

1 1 6

4 6

1 2

iA 0

J

iB 0

J

Fig. 2.29. The order of the switch commutation in CR with GTO thyristors

2.2.5. Reversible Rectifiers Reversible rectifiers consist of two groups of rectifiers, one of which provides conditionally positive, but the second — negative direction of the rectified voltage (Fig. 2.30). The left group of thyristors in the circuit (V1, V2, V3) provides the direction of the load current Id signed downwards, but the right group of thyristors — in the direction signed upwards. The transition from the operation of one group to that of the second group can be implemented when the current is decreased to zero only. The on-state of the turned-on thyristors of the second group when the current flows through those of the first group can result in short-circuit between phases with all possible consequences.

0

A

V1

V2

B

C

V6

V3

V5

V4

L + –

Id

–

+

load

Fig. 2.30. Reversible three-phase centre-tap rectifier circuit 120

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2nd part To avoid this short-circuit, it is necessary to apply either separate control with the sensor controlled current or coordinated control when the equivalent back-EMF of the both groups are equal and opposite directed in the way that the average value of the circulating contour current is close to zero. In the case of separate control, diodes VD1 and VD2 can be applied as sensors of contour current (Fig. 2.31). One or the second group can be switched on only if the direct voltage drop of the opposite group diode is zero signalling that no current flows through the diode. Operation of group II can be initiated only if voltage across VD1 is zero, but of group I — if voltage across VD2 is zero. In the case of coordinated control, both groups are controlled simultaneously but with different control laws. For its description, let us use the equivalent substitution circuit of the reversible rectifier (Fig. 2.32).

i1

I V1 I1

t

II V2

Id

I2

to control board

V2 0 V1 0

i2 t t

Fig. 2.31. Separate control of reversible rectifier

The first group of the rectifier is shown as a series connection of the internal resistance R1 of the voltage generator U d 0 cosa1 and equivalent diode of the conductance direction V1. If the first group is operating as a source of energy, then the positive direction of its generated EMF should coincide with the conductance direction of diode V1, and it can be achieved with the positive values of cosa1, i.e. if α1 ≤ π / 2 . The second group is shown as a series connection of the inside resistance R2 of the voltage generator U d 0 cosa 2 and equivalent diode of the conductance direction V2. If the second group is the source of energy, then the positive direction of the EMF gene rator at α 2 ≤ π / 2 should coincide with the conducting mode of diode V2 (downward direction in Fig. 2.32).

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2nd part i12

1 Ls

V1 R1

2

+

Ls Id

Ld

I1 Ud0cosa1

a1 uco a1

α1 <

π 2

V2

I2

R2 α2 >

Ud0cosa 2 –

3

ust

0 a2

a2

π 2

Uco wt –Uco

Fig. 2.32. Substitution circuit of the coordinated control of the reversible rectifier

If both groups operated in rectification mode, i.e., were the sources of energy, then the total generated voltage U d 0 (cosα1 + cosα 2 ) would result in inadmissible high currents in the circuit of sources, limited by R1 and R2 resistances only. To avoid this situation, both EMF should operate contrary. If the source of energy is in the first group and its α1 < π / 2 , then EMF of the second group should be in its negative direction, i.e., opposite to diode V2, and it can be provided with α 2 > π / 2. Additionally, in ideal case cosα1 + cosα 2 = 0 that provides almost zero average value of the current between groups, i.e., I1 = I d and I 2 = 0 A, and the average value of the load voltage is with the polarity indicated in Fig.2.32. In the opposite case when the second group is an energy source with α 2 < π / 2 , the first group should be controlled with α1 > π / 2 . Then the average value of the load voltage is with the polarity opposite to that indicated in Fig. 2.32. The following equation provides ideal conductivity control law: α1 + α 2 = 180° = π. (2-102) To provide this control law, both groups should be controlled with the saw-tooth voltage of double polarity and equal control voltages Uco with opposite signs (see Fig. 2.32). π U α1 = 1− co , 2 U stm U π α 2 = 1 + co (2-103) 2 U stm and α1 + α 2 = π. 122

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2nd part Thus, the control voltage Uco is supplied to one group directly, but to the other — through a polarity inverting element. If direct Uco is supplied to the first group, then at U co > 0 V the first group is operating in the rectification mode and current I1 = I d , but at U co < 0 V the second group is operating as a rectifier and current is I d = −I 2. In reality, the instantaneous values of the circulating current between the groups are not zero because the instantaneous values of the generated voltages do not coincide in time. Figure 2.33 shows the voltage curves of the both groups of three-phase centretap reversible rectifier at α1 = 30° and α 2 = 150°. The voltage difference u12 between the voltage curves of the voltage generators changes along with triple network frequency and produces current i12 circulating between the groups. To limit the circulating current, special inductors Ls (Fig. 2.32) between the groups are connected.

a1

u

0

wt a2

I group (rectifier) II group (waiting inverter)

u12 0

i12 wt

Fig. 2.33. Determining voltage between the groups

If the load contains a consumer with the source of EMF (e.g., DC motor), it can also operate with rectifier as a source of energy relating to the AC network. If in Fig. 2.34 EMF Em of electric motor with rotating shaft is lower than EMF of both rectifier groups, then group I operates as a source of energy (I1 = I d ), but group II — as a wai ting inverter. If Uco is decreasing, then a1 is increasing, but a 2 is decreasing, EMF of both groups is becoming lower than Em and group I breaks its operation and the electric motor transfers from the motor mode to that of generator inverting energy to the network through group II (I 2 = −I d).

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2nd part 1

L1

L2

I1

I2

2

0 a

Id a1 Em Ud0cosa1

α2 >

3

Uco

π 2

0 Ud Em 0 I

a2 a1

p p/2

t

Em I1 = Id

0

t

t I2

I1 t

Fig. 2.34. Coordinated operation of reversible rectifier with motor load

When the motor operates in the generator mode it is braking and Em = kn is decreasing. As soon as it is lower than new EMF of both groups, the operation is started by I group again. In that way, this circuit allows for operative and fast control of the motor speed.

Example Both groups of the reversible three-phase centre-tap controlled rectifier with phase voltage U = 220 V, 50 Hz, (see Fig. 2.30) are coordinated with α1 = 30° and α 2 = 150°. Calculate the amplitudes of the voltage curves of the inductors L1 and L2 (Fig. 2.34) between the groups and, having this curve linearised, the amplitude value of the circulating current if L1 = L2 = 5 mH. 1. The voltage across the inductors u12 = u13 − u23 appears, if off-state situations of each two-phase voltage curves cover intervals, i.e., between p and a 2 for inverting group and from 0 to a1 for the generating group. 2. When ω = 0 s−1, then u13 = U m sin (π − α1 − 30°) = 2 ⋅ 220 ⋅ 0.866 = 270.2 V, but u23 = 0 V. Therefore the amplitude of the voltage between the groups U12 m = 270.2 V. 3. Assuming that approximately tω 2α u12 = U12 m 1− , 0 ≤ t ≤ 1 , α ω 1

the inductor circulating current between the groups is calculated from the expression di (L1 + L2 ) 12 = u12 ; dt 124

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2nd part

solution of it gives t 2 ω U , i12 = 12 m t − L1 + L2 2α1 where at time t = 0 s, current i12 = 0 A.

4. The amplitude of the current i12 is at time moment t1 calculated from t ω di12 U = 0 = 12 m 1− 1 ; dt L1 + L2 α1

therefore, t1 =

and I12 m =

α1 ω

U13m α1 270.2 ⋅103 0.523 = ⋅ = 22.5 A. L1 + L2 2ω 10 2 ⋅ 314

2.3. Cycloconverters If both groups of the reversible rectifier are controlled intermittently and with equal time in the rectification mode, then the active inductive load can be supplied with a periodic AC voltage, the frequency f 2 of which is lower than that of the supply network f1. Let us consider single-phase cycloconverter consisting of two bridge rectifiers (Fig. 2.35) with coordinate control by means of a bipolar control voltage Uco changed periodically linearly and with small frequency in time. If the saw-teeth voltage is of double polarity, then group I is supplied with control voltage Uco but group II — with opposite polarity, i.e., minus Uco. Figure 2.35 demonstrates that the amplitude of the output voltage basic harmonic and its frequency depends on the amplitude and frequency of the control voltage. This converter can operate with load of any power factor. When the control voltage of the correspondent group changes its polarity from positive to negative, this group, accor ding to the time constant of the load, continues to conduct current during some time and operates with α > π / 2, i.e., in the inversion mode. The second group begins the work when the load current is zero. This converter is called a cycloconverter with natural commutation. The circuit in Fig. 2.30 is a three-phase single-phase centre-tap cycloconverter. Two three-phase bridge rectifiers are also used to build a three-phase to single-phase voltage cycloconverters. The determination of harmonic components of the output voltage is a complicated task, but to obtain basic expressions condition f 2 << f1 is assumed. Then the voltage of the fundamental harmonic component with frequency f 2 is described as follows: 125

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2nd part u2 = U d 0 cosα (t ), (2-104) where Ud0 is the maximum output voltage at α = 0. If u2 is assumed to be equal to U 2 m sinw2t , then a is changed as follows: α (t ) = arccos(±sinω2t ), 0 ≤ α ≤ π. (2-105) With U 2 m / U d 0 = 1 and 0 £ t £ T2 / 2 α(t ) =

π 2πt − , (2-106) 2 T2

where T2 is the period of the output voltage fundamental harmonic component. It is obvious that within the positive load voltage half-period the control angle a should linearly decrease from 0.5p to zero in the middle of the half-period, i.e., control voltage should be increased from zero to Ustm. During the second part of this half-period a should again increase to 0.5p (Fig. 2.36). u

u

0 V1.1

I

V1.4

V1.3 V1.2

ust

V2.2

II

V2.1

Rd

V2.4 Ld

Ucom

1/2 3/4

–Uco

0

I V2.3

wt

1/2 3/4

wt Uco

II

wt

ud id

wt

0

id ud

wt id

Fig. 2.35. Single-phase cycloconverter and diagrams of its operation

In the negative half-period in its turn a(t) should be linearly increased from p/2 to p in the middle of the half-period, but in the end decreased to p/2 (Fig. 2.36). This a change can be achieved if the double-polarity saw-teeth voltage has low-angle decrea sing front, then π U α = 1− co . 2 U stm 126

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2nd part It means that control voltage should be changed as it is in Fig. 2.36. Then the fundamental harmonic component of the output voltage with frequency f 2 can be changed as follows:

πU com , (2-107) 2U stm where U stm = const is the amplitude of saw-tooth voltage, Ucom — amplitude of the control voltage. u2(1) = U d 0 sin

u2(1)

U2m

0

T2

T2/2

a

t p p/2

0 uco

t

Ucom = Ustm

0

t

Fig. 2.36. Form of control voltage for cycloconverter

id, A

100.00

f2 = 5 Hz

50.00

0.00

–50.00

–100.00 0.10

U1 = 220 V, 50 Hz Ld = 30 mH Rd = 2 W 0.20

0.30 Time (s)

0.40

0.50

Fig. 2.37. The form of the load current of single-phase cycloconverter operating with active inductive load with Rd = 2 Ω, Ld = 30 mH and f2 = 5 Hz, if Ucom = 6 V un Ustm = 12 V 127

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2nd part To test the latter expression, a computer modelling of a single-phase bridge circuit of cycloconverter has been implemented; the circuit operates from the supply of U1 = 220 V, 50 Hz for the active-inductive load with Rd = 2 Ω and Ld = 30 mH. The frequency of the control voltage is f 2 = 5 Hz, with the amplitude U com = 6 V , but the amplitude of the saw-tooth voltage is U stm = 12 V . Figure 2.37 presents the form of the load current. The amplitude of the basic harmonic is close to 60 A, but its calculated value is 63.36 A (see calculation example). The rms value of the load voltage significantly depends on the inductor inductance — the higher the inductance is, the lower the voltage is. If the inductor inductance is low, then the voltage is high causing the increase in the network current.

Example Cycloconverter with two single-phase bridge rectifiers operates from the network AC voltage U1 = 220 V, 50 Hz, the load contains series connection of resistor R = 2 Ω and inductor with L = 30 mH. The amplitude of the thyristor control double polarity sawtooth voltage is U stm = 12 V , the frequency of the converter control double-polarity saw-tooth voltage f = 5 Hz, amplitude – U com = 6 V . Calculate the amplitude and rms value of the load current fundamental harmonic component. 1. The amplitude of the fundamental harmonic component of the load voltage is at minimum a: π U α min = 1− com = 45°. 2 U stm

2. The amplitude of the fundamental harmonic component of the load voltage U 2 pm = U d 0 cos α min = 0.9U cos 45° = 140.07 V . 3. The amplitude of the basic harmonic component of the load current U 2 pm 140.07 I 2 pm = = = 63.36 A. 2 2 R + (2πf 2 L) 4 + (2π ⋅ 5 ⋅ 30 ⋅10−3 )2 4. The rms value of the basic harmonic component of the load current I 2 pm I2p = = 44.62 A. 2 If we consider the case with active load, it is seen that output voltage frequency is lower than that of the network and (see Fig. 2.38) T2 T1 T = + n 1 , (2-108) 2 2 m2 128

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2nd part where T1 and m2 are the period and number of the pulsations of the network voltage at the output of the rectifier, respectively, but n = 0, 1, 2, 3, are integer numbers. As f1 / f 2 = T2 / T1 = (2n + m2 ) / m2, then f2 =

f1m2 . (2-109) 2n + m2

If m2 = 2, then f 2 according to n can be equal to f1; 0.5f1; 0.33f1; 0.25f1, etc. If, in its turn, m2 = 3 (reversible rectifier contains two three-phase rectifiers with centre-tap groups), then f 2 can be equal to f1; 0.6f1; 0.33f1; 3f1/11; 3f1/13, etc. In a general case, the implementation of direct frequency converter or cycloconverter requires replacing the thyristor with a bi-directional conducting switch (Fig. 2.39). Then a single-phase to single-phase circuit with m2 = 2 (a), three-phase single-phase centre-tap circuit and m2 = 3 (b), three-phase single-phase with m2 = 6 (c) can be obtained, as well as it is possible to implement this circuit on the basis of three-phase to three-phase cycloconverter with m2 = 2 or m2 = 3, or m2 = 6 (in Fig. 2.39 d m2 = 3). For the latter (m2 = 6) implementation, the control voltages should be shifted to 120 degrees. u

m2 = 3 T1/m2

T1/m2

T1/2

0 T1/m2

T1/m2

t

T1/2

n = 2

f2 = 3f1/7

Fig. 2.38. Form of the output voltage of the cycloconverter with active load and m2 = 3 (three-phase to single-phase centretap circuit) a

CJ

c

d

V1 V2 e

b CJ CJ

CJ

Fig. 2.39. Partly (a, b, c) and fully controlled (d, e) semiconductor switches; CJ – commutation junction 129

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2nd part a

c S1

S3

u2(1)

~f1

A

S4

B

C

S1

S1

S3

S3

S5

S5

S2 u2(1)

b

A S1

B

C

S2

N

N

d

S3 S1

S2

S3

S4 S5

S6

S7

S8

S9

u2(1)

Fig. 2.40. Simplified circuits of the direct frequency converter

The switches can be implemented on the basis of thyristors with special commutation circuits for their switching off while switching-on is implemented by firing pulses as well as on the basis of modern transistors, on- and off-states of which are both controllable (Fig. 2.39). Simple thyristor switches give an opportunity to implement cycloconverters with natural current commutation only. Applying GTO thyristors or devices of special commutation with capacitors allows getting modulated cycloconverters with improved form of the output voltage, obtaining almost sine-form of the load current and improving the power factor when the fundamental harmonic of the network current is not shifted to the angle a according to its voltage. However, the application of transistors gives more effective solutions. Let us consider implementation of modulation in bridge circuit (Fig. 2.40,c) The switches of each group operate without breaks, i.e., S1, S3, S5, S1…, as well as S2, S4, S6, S2…. Such turn of switching provides an operation in the mode of load current source. Modulation period is used TM = 1/ f M, where f M ³ f1. Each switch is on during the third part of TM, but operating switches S1, S2, S3… work with time shift TM / 6. Using this modulation principle, the frequency of the load voltage fundamental harmonic f 2 = f M − f1, i.e., selecting corresponding f M, the load AC voltage can be at frequency lower as well as higher than that of the network.

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2nd part Figure 2.41 represents the case with f M = 150 Hz . As it is obvious, the frequency of the load voltage is f 2 = 100 Hz. However, the form of the load voltage is considerably distorted.

TM

S1 S4

t

S3

t

S6 S5 S2 u

–BC

AB

–CA

BC

–AB

CA

–BC

AB

u2

t t t t fM = 150 Hz

T2 T1

0 T1 2

t f2 = fM – f1

Fig. 2.41. The diagrams of the switch operation and curves of the output voltage for the three-phase to single-phase bridgetype cycloconverter at fM = 150 Hz

The lower the modulation frequency is, the closer load voltage form is to a sine-wave. Figure 2.42 represents the case with f M = 66.66 Hz. As it is seen, the form of load voltage is much closer to the basic harmonic component, but the voltage frequency is 16.66 Hz. The main disadvantage of the circuit under consideration is a considerably distorted form of the network current. It is determined connecting the conducting intervals of the two switches of each phase with the curve of the load current instantaneous values. For example, in phase A current flows when S1 and S4 conduct. When S1 conducts, then current iA is equal to load current ild, but when S4 conducts iA = −ild. It results in the discontinuous network current with a considerable impact of the load current fundamental harmonic component (see Fig. 2.42).

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2nd part TM

S1 S4 S3 S6 S5 S2 u

fM = 66,66 Hz

90°

–BC

AB

–CA

BC

–AB

CA

0

–BC

T1

iA f2 =

t t t t t t AB

t

0,25T2

f1 3

0

t

Fig. 2.42. Formation of the output voltage curve with the frequency of switch modulation fM = 66,66 Hz

If the primary windings of three single-phase transformers are connected to a threephase network by means of modulated switches, but the secondary windings are connected in series (Fig. 2.43), then the total voltage of the secondary windings u2 can be obtained with variable frequency similarly to the above-mentioned circuit. u2 TR2

TR1

TR3

S3

S2

S7

S6

S11

S10

S1

S4

S5

S8

S9

S12

A

B

N

C

Fig. 2.43. Formation of output voltage curve with summation of three modulated voltages 132

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2nd part Increasing the number of the switches and varying the input voltage allow for the formation of the so-called matrix converters with almost ideal sine-form voltage at the output. In fact, on the basis of the circuit (Fig. 2.39,d) an operating mode of matrix converter can be obtained as the switches can be controlled in three modes: 1) phase of each network voltage is connected to its own load, 2) each phase is connected to two load groups, 3) each phase is connected to all three loads. Modulating the time intervals of the possible modes, the frequency of output voltage as well as its rms value can be changed.

2.4. AC Voltage Controllers AC voltage (current) control can be implemented by means of semiconductor elements without transformers providing small size of the devices. The simplest way is to switch into the AC circuit two opposite parallel connected thyristors (Fig. 2.44). With the change in the control angle a, the control time of each thyristor within half-period is changed, so does the rms value of the output voltage. The rms value of the output voltage in the case of single-phase active load U2 =

1 π

π

∫U

2 m

sin2 ϑdϑ = U1 1−

α

α sin 2α + . (2-110) π 2π

If the load is active-inductive, then with the change in the instantaneous value of the input voltage the current is further conducted by the thyristor corresponding to the previous polarity for some time and the voltage and current across the load correspond to the initial part of the network voltage half-period (see Fig. 2.44,b). The rms value of the output voltage is increasing and U2 = Um

1 π

δ

∫ 0

π 2

sin ϑdϑ +

∫ α

δ − α sin 2α − sin 2δ sin2 ϑdϑ = U1 1 + , (2-111) + π 2π

where d is the angle of the current «overlapping» according to the half-wave of the voltage of corresponding polarity. Angle d is changed from d = 0 with active load to ϕ = arctg(ωL / R) for the case of continuous sine-form current.

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2nd part a

b u 1

V1 i u1 = Um sinϑ

V2

0

u2 R

a V2

0 a V1

p

a V2

a V1

J

i

u1 0

a V1

J

J

u1

u2 0 0

d

J

J

Fig. 2.44. Single-phase AC control with a thyristor controller

As it is obvious from Fig. 2.44,b the control angle a can be changed from the maximum value p to the minimum d when the current is of sine form and, therefore, α min = ϕ. Taking into account these features, the control system should be provided with sufficiently long control signals (the best way — «longer» than j) that at a, lower than j, can provide continuous sine-form output current and U 2 = U1. The control characteristics corresponding to this case are given in Fig. 2.45. This thyristor controller is series, and, therefore, the rms values of the current at the input and output are equal. As active power in the case of Fig. 2.44,a is U 2 I , but the apparent power of all the system is U1 I , then the power factor is χ=

P U2 α sin 2α . (2-112) = = 1− + S U1 2π π

It is obvious that if a is increasing, c is decreasing. In the case of active load, the average value of the thyristor current in this single-phase circuit, according to which the semiconductor element is selected, is as follows: I Vav =

Im (1 + cos α ). (2-113) 2π

The maximum value is at α = 0 when the current is of sine form and I Vav =

2I = 0.45I , (2-114) π

where I is the rms value of the circuit sine-form current.

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2nd part U2 U1

1

ϕ = 90° ϕ = 60° ϕ = 30°

j= 0

0

30°

60°

90°

a

Fig. 2.45. Characteristics of the single-phase AC controller

To verify everything mentioned above, the modelling of single-phase AC voltage controller is performed in PSIM software with load R = 5 Ω, L = 30 mH with control angle α = 70°. The voltage and current of the load are shown in Fig. 2.46. 400.00

u2, V

R = 5 Ω

200.00

L = 30 mH

a = 70°

0.00 –200.00

a = 70°

–400.00 30.00 i A 20.00 ld, 10.00 0.00 –10.00 –20.00 –30.00 15.00

20.00

25.00

30.00

35.00 Time (ms)

40.00

45.00

Fig. 2.46. The simulated diagrams of the load voltage and current of the single-phase AC controller model

Example Single-phase AC network with U1 = 220 V, 50 Hz supplies through AC thyristor controller a load with R = 5 Ω and L = 30 mH. Calculate the rms value of the load voltage at α = 70°, linearising characteristic U 2 = f (α ) between α = ϕ and α = 180°. Calculate the rms value of the load current, active power and total power consumed from the network, as well as power factor.

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2nd part 1. Angle ϕ = arctg

ωL 314 ⋅ 30 = 62°. = arctg 3 R 10 ⋅ 5

2. The rms voltage of the load is approximately calculated as follows: U 2 ≈ U1

180°− α 110 = 220 = 205.08 V. 180°− ϕ 118

3. Approximate value of the load current I ld =

U2 2

2

R + (ωL)

=

205.08 25 + 88.73

= 19.23 A.

4. Active power consumed from the network P = I ld2 R = 19.232 ⋅ 5 = 1848.9 W. 5. Total power consumed from the network S = U1 I ld = 4230.6 VA. 6. Power factor P = 0.437. S For the load cosϕ = 0.469. χ=

Different types of three-phase AC voltage regulator implementation are possible (Fig. 2.47). The most traditional and widely applied type is switching of the opposite parallel thyristors into each phase. If the load has Y connection and its zero point is connected to the network centre-tap, the three-phase circuit is obtained connected to three phase voltages. However, to avoid zero component in the network zero point is not applied. Then the processes differ from those considered for the single-phase circuit because the instantaneous values of the load phase voltage can be 1) equal to the line-to-zero voltage instantaneous value, if at the correspondent time moment all three thyristors conduct or 2) equal to the half of line-to-line voltage instantaneous value if at the correspondent moment the current is conducted by two phase thyristors. The order of one or the other case is determined with the value of control angle a.

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2nd part a

N

A

B V3

V1

V2

C

b

V4 V5

uA

V6

V1

V2 V3

Fig. 2.47. Circuits of three-phase thyristor AC controllers

In the process of formation of the load voltage in the circuit (Fig. 2.47,a) without zero wire in the case of active load the control angle a is 45° according to the network phase voltage initial point and the control signal lasts until the phase voltage half period is over (Fig. 2.48). As it is seen, after thyristor V1 is switched from 45° to 60° the current is conducted by three thyristors, each in its own phase and the voltage of A phase in this interval corresponds to the supply A phase voltage. In phase C from angle 60° until thyristor V6 is on, the current is conducted by two thyristors and A phase voltage is a half of AB lineto-line voltage. Then again the current is conducted by three thyistors for 15° interval and A phase voltage is formed from the network phase voltage of A phase, etc., i.e., 3 and 2 conductance intervals of the phase thyristors are periodically changed. Figure 2.48 demonstrates: if 60° < α < 120°, the load voltage is formed from half of line-to-line voltage only. In its turn, if α > 120° the thyristor conductance intervals are not overlapped, and the load voltage is zero. Thus, with the control approach when the control signals are stopped with the end of the correspondent half-cycle, the wide control of the output voltage is not possible. With a lower than 120° the rms value of the output voltage with active load is about 44 % of the network phase voltage. Figure 2.49 demonstrates the curve of A phase voltage for the same three-phase circuit without zero wire at α = 45° (similar to Fig. 2.48) obtained in the modelling by software PSIM with active load. It is obvious that the curve obtained in modelling corresponds to the theoretical considerations above. Control can be improved if the thyristors are controlled after the end of the half-period for 30° interval. Then a maximum value can be 150° and rms value of the output voltage can be controlled till zero.

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2nd part u1

A

B

0

V1 V2 V4 V3 V5 V6

C

p

2p

V1

wt

V2

V4

V3

V5

V4

V6

V5

wt wt wt

uA 3 0

a

2

3

2

3

uA uAB uA uAC uA a 2 2

wt

Fig. 2.48. Form of load voltage of the three-phase AC controller (shaded areas are the intervals of the control pulse formation)

400.00

uldA, V

R load

200.00

a = 45°

a = 45° 0.00

–200.00

–400.00 35.00

40.00

45.00

50.00 55.00 Time (ms)

60.00

65.00

70.00

Fig. 2.49. Voltage curve for the three-phase circuit obtained from the model phase

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2nd part If there is no necessity to control output voltage in a wide range, then a circuit with additional voltage source can be applied (Fig. 2.50) providing much better power factor. Here the load voltage basic curve is formed from the network voltage but switching V3 and V4 an additional voltage and network voltage sum are formed. In this case, the rms value of the output voltage is higher than the network voltage.

V1

UR

V2

UR

u1

TR

~ V3

R

0

V1 a V3

u1

V2

V4

V3

wt

V4

Fig. 2.50. Voltage controller with an additional voltage transformer

Applying the controllable switches makes it possible to modulate the curve of the output voltage. For this control implementation, a transformer is required to decrease the efficiency of this device. The secondary winding of the transformer is made open (Fig. 2.51) and its both outputs are connected to a three-phase diode bridge (one output to UZ1, the other — to UZ2).

TR u

U1

a

S1

b

0 S2 0 u

c

UZ1

lS1 wt wt

UZ2 0

S1

wTM

wt

S2

Fig. 2.51. Circuit for the modulation of the output AC voltage

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2nd part If switch S1 is on, but S2 is off then the secondary winding is connected to Y and the output voltages are ua, ub, uc. If S1 is off, but S2 is on then the circuit of the transfor mer secondary winding is open, but the output voltage is zero. Changing the relation of on- and off-states of S1 and S2, the rms value of the controlled voltage is changed. U ldrms = U1

λ S1 , (2-115) ωTM

where TM is modulation period, but lS1 is relation of the conductivity angle of switch S1 to the network voltage curve with angular frequency w.

2.5. Control Principles of the Line-Frequency Converters Line-frequency converters are mostly based on the naturally commutated thyristors with control of only angular switching-on delay a in accordance with the closest possible switching instant in the corresponding circuit. For each thyristor this instant is defined individually and time interval is started to be counted corresponding to angle a. As it has already been mentioned, a angle calculation can be implemented in the analogue form comparing reference voltage ust with slowly changed control voltage Uco. In a general case, the channel of the control system consists of synchronization element SJ determining the moment of calculation a start, when the saw-tooth voltage generator STV is on, the element of comparison CC comparing the saw-tooth ust and control Uco voltages, and the element of control signal forming SF (Fig. 2.52).

Un

1 / 0 SJ

STV

1 / 0

ust

CC

SF

thyristor

Uco Fig. 2.52. Block-diagram of the control system channel

Usually the number of the control system channels corresponds to the number of the controlled thyristors in the converter. However, in some cases simultaneous control of two thyristors from one channel can take place as in the case of single-phase bridgetype rectifier (Fig. 2.53).

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2nd part V1

V4 u1

u1

V2

~

u1 > 0 u1 < 0

V3

SJ1 SJ2

1/0

STV1

1/0

STV2

ust1

CC1

ust2

CC2

Uco VT1

R

V1 V2

SF2

V3 V4

+UB VTO1.2

V1

VTO2.1

DA2 –Us

1/0

SF1

VTO1.1

C DA1

1/0

DA3 +Us

ust

DA4 Uco

&

DD1

VTO2.2

V2

Fig. 2.53. Block-diagram of the single-phase bridge rectifier control

In this case, the saw-tooth voltage of the upper control channel is formed, if u1 > 0 V, but for lower channel if u1 < 0 V. Upper channel controls V1, V2 thyristors and lower — V3, V4 thyristors. With network the control systems are usually connected through low power voltage decreasing transformer or by means of resistances with very high resistivity. A possible implementation of the control channel for single-phase rectifier is shown in Fig. 2.53. It demonstrates the implementation of saw-tooth voltage, with which α = 0 at Ucom, α = 0.5π at U co = 0 V and α = π at –Ucom. When the network voltage u1 > 0 V (the upper terminal with positive polarity), then the output of analogue comparator DA1 is with a high logic level signal that maintains transistor VT1 in the off-state and allows for the charging of capacitor C through resistor R. At the output of operational amplifier DA2 a voltage linearly increasing in time is formed for transformation at the next amplifier cascade DA3 into saw-tooth voltage ust of the necessary type. In the comparator DA4 the saw-tooth voltage is compared with control voltage Uco, and, when U co > ust , the thyristors are supplied with the control signal. When the supply voltage is u1 < 0 V , then the output of DA1 comparator is of negative polarity and transistor VT1 is on, maintaining the voltage of capacitor C at a zero level but the signal inverse to that of the comparator DA1 output initiates the saw-tooth voltage formation of the second channel (not demonstrated in the circuit). In the three-phase centre-tap CR (Fig. 2.54), the input of the thyristor V1 is defined when uA is higher than uC, and then the formation of the saw-tooth voltage is provided for this channel. Then the formation of saw-tooth voltage for thyristor V2 channel is provided when uB is higher than uA. When uC is detected higher than uB, the 141

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2nd part formation of saw-tooth voltage is allowed for thyristor V3. The relation of the voltages uA, uB, uC is controlled by comparators DA1, DA2, DA3, but the output signals (as in Fig. 2.53) are controlled with the formation of the corresponding channel saw-tooth voltage. The formation of the output signals of all three channels is controlled with one and the same slowly changing control voltage Uco. The control system of the three-phase bridge-type rectifier should contain 6 similar channels, the operational order of which is defined by means of 6 comparators comparing the CR supply voltage mutual relation (Fig. 2.55).

DA1

N

V1

A

V2

B

V3

C

+ –

u 0

V1

V2

uA uC

V1

+ – DA2 + –

DA3

uC

V3 Uco

V3

uB

V2

uA q

Fig. 2.54. Control system of three-phase centre-tap rectifier connection to the supply network

N AB C

V4 V6 V3

+ –

V3 V5

uB > uC

+ DA3 –

uB > uA

+ –

uC > uA

+ –

u –BCAB –CABC –ABCA –BC

V6 V1 V2 V3 V4 V5 V6

uA > uC

+ DA2 –

V1

0

DA1

DA4

DA5

uC > uB

+ DA6 –

q

uA > uB

V1 V2 V3 V4 V5 V6

Uco

Fig. 2.55. Control system of three-phase bridge-type rectifier connection to the supply network

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2nd part The control channel of thyristor V1 starts operation when line-to-line voltage uAB is the highest one. It happens when the sign of line-to-line voltage uCA is changed to negative, i.e., when uA is higher than uC. Condition uA > uC defines an operation range of V1 channel and it is implemented by comparator DA1. Comparator DA2 controls condition uB > uC , when the operation of thyristor V2 channel is started as in the case when line-to-line voltage uAC is the highest one. Comparator DA3, in its turn, controls condition uB > uA , when the operation of thyristor V3 channel is started as in the case when line-to-line voltage uBC is the highest one, etc. Within the voltage period each channel can form «long» or «short» control signal for the thyristor. If we implement the second and more economic case, then each channel should provide pulses for two thyristors: DA1 control channel — to thyristors V1, V6; DA2 — thyristors V2, V1; DA3 — thyristors V3, V2, etc. The control approach under consideration with saw-tooth voltage is the so-called analogue control. For the same angles a for all thyristors all saw-tooth voltages should be the same. As in practice it is not possible, control asymmetry takes place. More accurate control can be implemented using a digital approach, when the delay angle is defined by the number of pulses of the clock generator constant frequency. For example, if the clock frequency is 100 kHz, then the period of the pulses is 10 ms. Angle α = 30° (in time form with network frequency of 50 Hz it takes 30 ⋅ 20 ⋅10−3 / 360 = 1.66 ms) requires counting from the beginning of synchronization 1.66 ⋅106 = 116 clock pulses. 103 ⋅10 It is more convenient to implement the digital control with the help of microprocessor, the operation algorithm of which is defined by the program. There is no necessity to control each synchronization voltage but controlling only one provides the others with the known period. For example, for three-phase centre-tap rectifier, first of all, it is necessary to define when uA is positive, then it is possible to count 30°, then — angle a and supply signal to V1, after that it is possible to count 120° angle and supply signal to V2. After that again it is possible to count 120° and supply signal to V3. Further the system returns to the uA > 0 V control operation and repeats the process mentioned above. N=

This control approach with the synchronization of one voltage in multi-phase systems is called equidistant.

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Review Questions

2nd part

Review Questions 1. What is the basic semiconductor element of the controlled line-frequency converter and why?

2. How is the effect of load voltage control achieved in controlled rectifiers? 3. What is the saw-tooth voltage of the control system and how is it connected with the rectifier supply AC voltage?

4. In what way does the character of the load influence the control parameters of the load voltage?

5. How large is the maximum control angle in the single-phase circuit with active load?

6. How large is the maximum control angle in the three-phase centre-tap circuit with active load and why?

7. How large is the maximum control angle in the three-phase bridge-type circuit with active load and why?

8. How is the connection of load voltage of the controlled rectifier with control angle described for the cases of active and active-inductive loads?

9. How large is the maximum control angle of the controlled rectifier operating as an energy source and why?

10. What is the inverter mode of the controlled rectifier and what should the load be for the implementation of this mode?

11. How large should the control angle be in the inverter mode? 12. What should the relation be between the controlled rectifier equivalent EMF and that of the load?

13. What type of saw-tooth voltage is preferable for the controlled rectifier for the operation in the source as well as inverter mode?

14. Does the value of the control angle influence the commutation process of the CR? 15. What limits the maximum acceptable control angle of CR in the inverter mode? 16. What is power factor of CR and how is it connected with a control angle? 17. How can the power factor of CR be improved? 18. What is a reversible rectifier? 19. What is separate and coordinated control of the reversible rectifier? 20. What is the basic drawback of the separate control? 21. What is the basic condition regarding the control angle of the reversible rectifier? 22. What is a cycloconverter and on the basis of what type of rectifier is it implemented?

23. How are the control voltages of the cycloconverter groups changed to achieve the 144

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Review Questions

2nd part load voltage basic harmonic close to sine form?

24. What switches are required for the direct AC voltage transformation into the regulated voltage of other frequency and phase number?

25. How are the double-direction controlled switches implemented? 26. What is the block-diagram of the direct frequency converters? 27. What is AC controller and how is it implemented with thyristors? 28. In what range is the control angle of thyristors changed for the single-phase AC controller operating for active and active-inductive load?

29. Why is the minimum control angle limited in the operation for active-inductive load?

30. How is three-phase AC controller implemented? How many thyristors are necessary for this implementation?

31. How large is a maximum control angle for three-phase AC voltage controller with Y connected load?

145

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Tasks

2 part nd

Calculation Tasks 1. Calculate the an ideal rectifier load voltage controlled in a single-phase bridge mode at its active and active-inductive loading at large volume inductance if AC supply voltage is (220 + mn) V, but delay angle is α = 60°!

Answer: at mn = 0 load voltage is respectively 148.5 V and 99 V (RL load). 2. For the previous task calculate load currents if resistance of load is R = 5 Ω and in series with load a counter-EMF with E = 40 V is introduced. Calculate for the both loading cases RMS value of network current!

Answer: at mn = 0 load currents are 25.84 A and 11.8 A, respectively; network currents are 35.17 A and 1.8 A, respectively.

3. Calculate the load voltage of the three-phase neutral point controlled rectifier

at loading with smoothed current I d = (20 + m + n) A , if transformer secondary phase voltage is 220 V, 50 Hz, inductance of the phase winding is L = 1 mH, but delay angle is α = 1.2 rad!

Answer: at m = n = 0 voltage is 90.12 V. 4. Calculate the maximum possible inverter current in the inverter mode if

three-phase centre-tap rectifier has α = 150°, thyristor turnoff reserve angle is 2°, but the input inductive resistance of the rectifier is Xa = 1.5 Ω! Phase-to-zero supply vol tage of the inverter is (200 + mn) V. Answer: at mn = 00 maximum current is 21.85 A.

5. Bridge-type rectifier with Xa = 1 Ω and output voltage 300 V from the three-

phase network of 380/220 V, 50 Hz supplies current I d = (50 + m + n) A to other threephase bridge inverter with Xa = 1.5 Ω that is connected to a network of 660/380 V 50 Hz. Calculate the control angles of both rectifiers a1 and a 2, correspondingly to provide this operation.

Answer: at m = n = 0 correspondingly 47.34° and 104.9°. 6. Single-phase controlled rectifier operates with delay angle α = 50° with a load of

resistance R = (5.m) Ω and large inductance. Calculate approximate apparent, active and reactive power components and power factor if the network voltage is 380 V and Xa = 0!

Answer: at m = 0 power S = 16.71 kVA; P = 9.67 kW; Q = 13.62 kvar; cos j = 0.58. 7. Single-phase controlled rectifier operates under the condition of the previous task, but the load circuit is by-passed with a diode. Calculate the parameters given in the the previous task.

Answer: S = 18.146 kVA; P = 15.865 kW; Q = 8.8 kvar; cosj = 0.874 with m = 0. 146

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Tasks

2 part nd

8. One group of the reversible single-phase rectifier operates with delay angle

α1 = (30 + m + n)°, but the second group — in coordination with angles 180°−α1. Supply voltage is 220 V. What is the amplitude of the full voltage across the inductor between the groups if the load contains large inductance? Do not take into account a commutation angle.

Answer: at m = n = 0 the amplitude of the inductor voltage is 312 V. 9. Using the conditions of the previous task calculate the amplitude of the

smoothing current in the inductor if its inductance is (20 + m + n) mH, resistance is zero and the changes of the inductor voltage instant value in time are linear — between positivze and negative amplitude values. Frequency of the network voltage is 50 Hz.

Answer: at m = n = 0 the amplitude of the current is 12.99 A. 10. The control of a single-phase bridge type cycloconverter is implemented

comparing a triangle double polarity control voltage, the amplitude of which is (10 + m + n) V and the period of repetition is (100 + mn) ms, with double polarity sawtooth voltage of a control system of thyristor rectifier. The amplitude of the saw-tooth voltage is (15 + m + n) V. What is an approximate rms value of the load current if the load inductance is 0.1 H, resistance is 5 Ω, but the supply network voltage is 380 V? Do not take into account the commutation process.

Answer: at m = n = 0 the rms value of the current is 25.95 A. 11. Single-phase AC controller operates with active load and delay angle α = (50 + m + n)°. RMS value of the network voltage is 220 V, load resistance is (5 + m + n) Ω. Calculate the rms value of the current as well as ratio of the consumed active and apparent power. Answer: at m = n = 0 the current is 41.37 A and power ratio is 0.94. 12. Single-phase AC thyristor controller operates with active-inductive load, the

resistance of which is R = (5 + m + n) Ω and inductance L = (10 + m + n) mH. Delay angle α = 100°, rms value of the network voltage is 220 V, frequency 50 Hz. Replace the control characteristic with a straight line between the coordinates α = 180°; 0 V and α = ϕ; 220 V, where j is load current and voltage shift angle is in the sine mode. Calculate rms value of the load voltage from the curve and rms value of the load current.

Answer: 120 V and 20.3 A at m = n = 0. 13. Three-phase AC controller operates with load zero wire and supply voltage of 380/220 V. Calculate apparent power, active and reactive power consumed from the network using the phase load and control parameters from the previous task.

Answer: 13.4 kVA, 6.18 kW, 11.89 kvar at m = n = 0. 147

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Tasks

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3rd part

3. SWITCH-MODE DC-DC CONVERTERS Basically switch-mode converters operate from DC primary or secondary sources. Therefore, first of all DC switches and systems of control and then the autonomous inverters are to be considered.

3.1. DC/DC Converters 3.1.1. Step-Down (Buck) Converter Pulse control with a semiconductor switch is the only method to transform DC value without significant power losses. If it is required to decrease voltage in comparison with source, the switch is connected in series with the load (Buck converter), if an increasing voltage is required, then the switch is connected in parallel to the source (Boost converter). Let us consider step-down (or Buck) converter replacing the semiconductor switch with an ideal contact switch (Fig. 3.1). The switch operates at high enough frequency f (higher than some hundreds Hz) and duty ratio according to the switching period D=

t on , (3-1) T

where ton — the on-time of the switch during the switching period T = 1/ f and f — switching frequency. The circuit of the load is bypassed by diode V0, which in the case of active-inductive load gives to the current the way to flow during the off-state of the switch S. If the time constant of load is at least 3...4 times higher than the process period T, then the load current is smoothed well enough and continuous, i.e., during the off-pause its instantaneous meanings do not achieve zero. Then the instantaneous values of the voltage across the load are either U1 (when S is on) or zero (when S is off). The average voltage value during the period is 1 U ld = T

t on

∫ U dt = U D, (3-2) 1

1

0

but the average value of the active-inductive load current is I ld =

U ld U = D 1 , (3-3) Rld Rld

as the average voltage across the inductance within the period is zero. 149

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3rd part The ideal switch has no power losses; therefore, input average power P1 = I1U1 is equal to the load average power Pld = U ld I ld , where I1 is the average value of the current of the supply source. Taking into account (3-2) from the power balance the average value of the source current is I1 = DI ld, (3-4) but in accordance with the Kirhhoff’s current law at point 1 (see Fig. 3.1) the current of the diode V0 I V 0 = I ld − I1 = (1− D ) I ld . (3-5) The control characteristics U ld = f (D ), I ld = f (D ) and I1 = f (D ) in the case of activeinductive load are demonstrated in Fig. 3.2. The control effect is achieved changing D within the range from 0 to 1. Some cases are possible how to implement the changing of D. If the frequency of the switch operation is constant, the D is changed increasing or decreasing the time of its on-state. This approach is called Pulse Width Modulation (PWM). It is a common method of control.

+

U1

uld

f, D

S i1 uld

1

iV0 V0

Rld

ton

U1

ild Lld

Uld 0 ild

0

t

T

–

DIld

ton

Ild

T

t

Fig. 3.1. Step-down (buck) converter

Pulse frequency modulation is less popular, when ton is unchangeable but the effect of control is achieved either decreasing period till T = t on when D = 1 or increasing T till a very high value when D is close to zero. Figure 3.1 demonstrates the voltage pulse influence on the current’s ripples. If one assumes that the voltage drop across the load active resistance is unchangeable in time and equal to I ld Rld = DU1, then the ripples of the current are described with the 150

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3rd part differential equation dild = U1 − DU1 = (1− D)U1, (3-6) dt which refers to on-state of the switch. Lld

Full change of the load current during the on-state ton of switch S is ∆I ld =

U1 D(1− D)T . (3-7) Lld

Uld, Ild Uld

U1 Ild U1 Rld

I1 DIld 0

0.5

1

D

Fig. 3.2. Control characteristics of the buck pulse converter with active-inductive load

In the case of PWM the maximum load current’s ripples are at D = 0.5 (Fig. 3.2) when ∆I ldm ax =

U1 . (3-8) 4 fLld

At D = 0 and D = 1 PWM gives zero current ripples. If pulse frequency modulation is applied then ∆I ld = U1 (1− D )t on / Lld, i.e., with unchangeable ton maximum ripples take place if D is close to 0, but if D = 1 the load current is not pulsating at all. If the average value of the load current is higher than the half of the current ripples range DIld, then the load current is continuous (uninterruptable). In the opposite case, it is discontinuous. Usually continuous current is preferable to obtain and it can be achieved selecting a high enough PWM frequency as well as high enough Lld. Converter can also be controlled by means of load current instantaneous values checking and maintaining it within the range of Ildmax and Ildmin, i.e., operating with ∆I ld = const. This is called a double position control: S is turned on when the instantaneous value is decreased to that minimum Ildmin, but it is turned off when ild is increased to Ildmax. From (3-7) it is obvious that in this case the frequency is variable: f=

U1D(1− D) , (3-9) Lld ∆I ld

achieving maximum value at D = 0.5. 151

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3rd part

Example The load circuit of the DC buck converter contains series connection of resistor R = 3 Ω and inductance L = 10 mH. Commutation frequency is f = 1000 Hz, duty ratio D = 0.5, supply voltage U1 = 100 V. Define whether the load is operating in the mode of continuous current and calculate the average values of the load current and voltage. 1. The full ripple range of the load current within the control is ∆I ld = therefore,

D(1− D)U1 ; Lf

0.5 ⋅ 0.5 ⋅100 ⋅103 = 2.5 A. 10 ⋅103 2. The average value of the load voltage for the continuous load current ∆I ld =

U ld = DU1 = 0.5 ⋅100 = 50 V ;

average value of the load current I ld =

U ld 50 = = 16.6 A. R 3

3. As 0.5∆I ld < I ld, then the load current is continuous. Decreasing the value of the load average current till the half of the current ripples, the boundary case can be calculated (1− Db ) = If (1− D ) >

2Lld f . (3-10) Rld

2Lld f , then the discontinuous current mode takes place. Rld

3.1.2. Step-Up (Boost) Converter The parallel connection of the switch to the source provides boost DC conversion. Inductance L is connected in series to the source as an accumulator of energy (Fig. 3.3). The average voltage across the switch is equal to the source voltage U1 (as the average voltage across the inductance within the period is zero). This average voltage is formed from two instantaneous values (Fig. 3.3) — zero, when switch S in the interval DT is on, and load voltage Uld, when in the interval (1 – D)T the switch is off. Therefore, U1 = (1− D )U ld (3-11) 152

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3rd part and U1 . (3-12) 1− D Taking into account the equality of the input and output powers, an averaged supply current U ld =

I1 =

I ldU ld I = ld . (3-13) U1 1− D

If D increases, then output voltage Uld, according to U1, and the source current also increase. As each source has its own inside resistance R1 voltage U1 decreasses with an increasing in I1 and, in practice, the load voltage cannot infinitely increase. If one applies a value of idle-run voltage U10, then U1 = U10 − I ld R1 / (1− D ). Applying this U1 value to (3-12), U ld ∗ =

U ld 1− D = , (3-14) U10 (1− D)2 + R *

where R * = R1 / Rld . D derivative and equalization this to 0 (searching for maximum)

gives the solution of maximum voltage at Dm = 1− R1* , but the load voltage is zero at D = 1 (see Fig. 3.3). Maximum possible load voltage value is U ldmax =

+

U1

2 R1*

. (3-15)

V0

L

a

U10

i1 uS

b u1

S

+

Rld

f, D – C

R1* = 0.028

R1 = 0 *

0 i1

t

T I1

2 0.055

0

t

0.063

1 0

U1

DT

Lld

–

c * Uld 3

Uld

Uld

0.5

1

D

Fig. 3.3. DC boost converter: circuit (a), diagrams (b) and characteristics (c)

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3rd part Equating half of the supply current i1 ripple range ∆I1 = DU1 / (Lf ) to the source current average value I1 gives connection for the discontinuous source current boundary case D(1− D)2 = 2

If D (1− D ) >

2Lf . (3-16) Rld

2Lf , then the discontinuous source current mode takes place. Rld

3.1.3. Improvement of the Converter Characteristics with Autotransformer Autotransformer application in the circuit of a converter (Fig. 3.4) gives an opportunity to correct its control characteristics. In the buck circuit (a) with switch S in off-state and current flowing through the diode V0 the sum of the voltages of both autotransformer windings w1 and w2 is equal to the load voltage that is smoothed connecting high capacity C: u10 + u20 = u10 (1 + k ) = U ld, (3-17) where k = w2 / w1 — turns ratio of transformer. a

w1

S +

U1

u1

u2

b C

w2 V0

–

V0 +

Ld U ld

U1

w1 u1

S

w2 u2

Uld

+ C

Ld

–

Fig. 3.4. DC converter with correcting autotransformer: a — buck, b — boost

Within the interval DT when switch S is on voltage of winding w1 u11 = U ld −U1, but the total average voltage of the winding within the period is zero: u10 (1− D)T + u11DT = 0, and it results in U ld =

D(1 + k )U1 . (3-18) 1 + Dk

In accordance with the obtained expression Fig.3.5,a demonstrates the control characteristics of this circuit with different factor k. 154

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3rd part a * Uld 1

b

k = 2

Uld*

k = 1

k = 0

3 k = 2

0.5

k = 0 2

k = 1

0

0.5

1 D

1

0

0.5

1 D

Fig. 3.5. Control characteristics of the converter with autotransformer: a — buck, b — boost

The sum of both winding voltages in circuit (b) with switch off is equal to U ld -U1, i.e., u10 (1+ k ) = U ld −U1. With switch on u11 = −U1. As the average voltage of the winding within the period is zero, i.e., (U ld −U1 )(1− D) −U1D = 0, 1+ k then load voltage U (1 + Dk ) U ld = 1 . (3-19) 1− D These equations are valid for the ideal case, when U1 = const.

3.1.4. Buck-Boost Converter 3.1.4.1. Conventional Topology Changing in some way the place of switch S in the circuit provides a converter that can step down as well as step up the supply voltage (Fig. 3.6). This circuit is called conventional topology buck-boost. The average voltage across the inductor L is zero and it is formed of the positive voltage U1 at switch on and negative Uld when switch S is off within the interval (1− D )T . Therefore, U1D = U ld (1− D ), (3-20) and the voltage of the load and the stand-by capacitor is D U ld = U1 = U C. (3-21) 1− D The average current of the source D I1 = I ld . (3-22) 1− D 155

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3rd part That of the inductor L I L = I ld + I1 =

I ld I = 1 , (3-23) 1− D D

and it is equal to the amplitude of switch current. Inductance L is selected to get a half of the full ripple of the current less than IL . Thus, assuming that current iL has a linear increment of DIL in the interval DT, inductance must be L>

U1 D 2 . (3-24) 2 I1 f

The stand-by capacitor, in its turn, should be provided with the acceptable range of ripple of the load voltage ∆U C < 2U ld , which is determined by the capacitor discharging time interval DT when its current is Ild. Therefore, I ld D . (3-25) f ∆U C

C>

a

S +

i1

U1 –

cU

ld

f, D L

b uL

V0

iL uL

–

Uld

+ C

DT

Rld 0

U1 T(1 – D)

t –Uld

*

2

iL

S(on) DIL

1

IL I1

0

0.5

D0

1 D

0

t

Fig. 3.6. DC buck-boost converter: circuit (a), diagrams (b) and characteristics (c)

If U1 is constant, then at 0 £ D £ 0.5 the load voltage is changed from zero to U1. With higher D load voltage exceeds U1. Taking into account the active resistance of the source and amplitude of the switch current, the load voltage in terms of the induc tance zero average values D(1− D) , (3-26) (1− D)2 + R1* D where the relative values are in respect of the source idle-run voltage and its load resistance. This circuit has also a limited maximum load voltage and this will take U ld ∗ =

156

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3rd part place at D0 =

1− R1* 1− R1*

.

3.1.4.2. Cuk Converter The conventional buck-boost circuit has some drawbacks — the source current is of pulse mode and influencing the operation of the source; polarity of the load voltage has a reversing character in accordance with that of the source; the load should be bypassed by large enough capacitance that can take its volume dependence on the inductance L current pulses independently of the type of load. Taking it into account another circuit has been developed with an opportunity to get a discontinuous source current and not to bypass the load by high volume capacitance. This circuit has been elaborated by S. Cuk, and it is called after his name (Fig. 3.7). uC

i1 +

L

–

C Lld

U1

ild

V VT

–

Rld +

Fig. 3.7. Cuk converter circuit

In this circuit, the pulses of the load voltage with amplitude UC are generated with capacitor through on-turned switch — transistor VT. The average value of the load voltage is U ld = DU C , but U C = U1 + U ld . It results in the opportunity to calculate the load average value by the same expression like that for the conventional circuit (3-21) and the polarity of this voltage is also reversible to that of the source. The average value of the capacitor voltage is U U C = 1 , (3-27) 1− D but the amplitude of the switch current is I I VT = I ld + I1 = 1 . (3-28) D The inductance L should be selected to get a half of the full ripple of its current less 157

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3rd part than the average current of the inductance I1. This way the required inductance can be calculated L>

DU1 . (3-29) 2 fI1

It is obvious that in the conventional circuit the storage inductance should be smaller but its average current — higher. The capacitor, in its turn, is also discharged with current Ild within the time interval DT, i.e., it is necessary that its capacity can be described with the same expression (3-25) like for the conventional circuit but ∆U C < 2U C.

3.1.4.3. Buck-Boost Converter of the Load Voltage of the Same Polarity This circuit (Fig. 3.8) called SEPIC (Single-Ended Primary-Inductor Converter) provides a load voltage of the same polarity as the source has. uC

i1 +

L1

C1 uL2

U1 VT

V

Lld

L2

ild

+

C2 Rld

–

Fig. 3.8. Circuit of buck-boost converter with the same load voltage polarity regard to the source

Taking into account that the average voltage of the inductance is zero, the average value of capacitor C1 voltage can be stated to be equal to U1. When transistor VT is turned on, the value of this voltage is supplied to inductance L2 with its negative polarity but within the DT time interval the pulse of this voltage does not influence the load circuit due to the diode V. When the switch is off, the pulse of the inductance L2 positive voltage U1D / (1− D ) is supplied to the load circuit with parallel capacitor C2 through diode V and this voltage pulse determines the voltage of the load in the same way as in the two above-mentioned circuits. The source current is smoothed, and its pulsations are determined with the volume of inductance L1 that should be selected in accordance with (3-29). Capacity C2 of the load junction should be selected from (3-25). As the average value of the capacitor current is zero then that of L2 current is — Ild but its inductance should be selected according to (3-29) replacing I1 with Ild. Capacity C1 should be selected from (3-25) but ∆U C < 2U1. 158

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3rd part The amplitude of the switch current is formed with the sum of current I1 and load current Ild, i.e., it is determined with (3-28). Therefore, the number of reactive elements in SEPIC circuit is two time more than in the circuit with the load voltage of reverse polarity.

3.1.4.4. Buck-Boost Converter of Commutated Configuration The load and source voltages of coordinated polarity can also be obtained with the help of buck-boost converter of commutated configuration (Fig. 3.9) with two transistor switches VT1 and VT2. Switch VT1 is operating in buck case, while VT2 is permanently off. The load voltage is D1U1, where D1 is duty ratio of the switch VT1 in the switching cycle. In boost operation mode, switch VT1 is permanently on, but VT2 operates with duty ratio D2, which results in the load voltage equal to U1 / (1− D2 ) that is higher than U1. In buck and boost modes, the current of inductance L is continuous that is obtained with correctly selected value of inductance, which should provide a ripple range of the current lower than double average value of the inductor current. In the same way, the value of capacity should provide the ripple range of voltage lower than a double load voltage value.

+

U1

VT1

L

V2 Rld

V1 –

VT2

+

C

Fig. 3.9. Circuit of buck-boost converter of commutated configuration

3.1.5. Input and Output Filters Input and output filters are applied in order to obtain the source and load currents as close to DC in form as possible. Buck converter (Fig. 3.1) is usually provided with filter in DC supply circuit. Boost converter (Fig. 3.3) usually has the filter in load circuit. Filter in the circuit of the supply source is also applied for the traditional as well as commutated buck-boost converter (Figs. 3.6 and 3.9). An input filter is formed connecting capacitor CF in parallel with supply source and 159

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3rd part inductor LF in series with that (Fig. 3.10,a). Thus, the capacitor processes the ripples of the current but the inductance provides the impedance for the AC component of supply current iL . The filter with the capacitor processing the current pulses can be calculated under the following assumptions: • the periodic filter load current i passing through the switch is of rectangular form with amplitude I and «length» of the pulse ton; • the current of LF in the first approximation is smoothed and equal to the average value of the pulse form load current. I LF = t on I / T = DI , where D is relative length of the switch pulse current within the period (duty ratio). These assumptions can be proved with the fact that the average value of the capacitor current within the period is zero. Then current of a negative direction ( I − I LF ) = I (1− D) is flowing through the capacitor within the pulse time DT of the load current. Taking into account the filter load current duration DT the voltage of capacitor will change down for I (1− D)DT , (3-30) CF this changing will take place around the capacitor voltage value U Cav = U (as the ave rage voltage value of LF per period is zero). ∆U CF =

As it is seen (Fig. 3.10,a), the capacitor voltage during the time of the current pulse (0 £ t £ DT) is decreasing linearly: ∆U CF t − ∆U CF + U . (3-31) 2 DT The instantaneous value of LF current during this interval is uCF = +

iLF1 =

1 LF

∫

(U − uCF )dt = −

∆U CF ∆U CFt 2 t+ + I LF10, (3-32) 2LF 2LF DT

but during time interval (1- D)T, when uCF is increasing (Fig. 3.10,a), i LF 2 =

1 LF

∫

∆U CF ∆U CF ∆U CFt 2 t + dt = + t− + I LF20 . (3-33) − ∆U CF (1− D)T 2 2LF 2LF (11− D)T

ILF10 and ILF20 are the instantaneous values of the inductor current at the initial moments of the first and second intervals. Replacing t = DT in (3-32) gives a value ILF10 at the end of the interval. As inductor current does not change stepwise, I LF20 = I LF10 (Fig. 3.10,a). This initial value can be determined taking into account that the average 160

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3rd part current DI of the inductance is the sum of the average values of the first and second interval currents. Usually the current ripples that can be detected as a difference of the both interval extremes are of interest. Besides, the extremes are at uC = U , i.e., in the middle of the intervals. Therefore, ∆I LF = I LF max − I LF min =

iLF

a +

CF

–

ton

S

+

iCF

U

S

f, D

i

I

0

t

ILF (1 – D)T

uCF

t –(I – ILF)

t on T

DUCF

t

CF

iCF

IldR = Uld –

I DT

ILF

0

iCF

t

(I – ILF)

0 uCF

U

+

LF

–

ILF = iCF 0

iLF

i

b

f, D

LF

U

i

i

∆U CF I (1− D)D . (3-34) = 8LF f 8CF LF f 2

Uld

–ILF

t

t

iLF

iLF ILF ILF10

t

ILF20

ILF

t

Fig. 3.10. Diagrams of input for buck (a) and output for boost converter (b) filters

The maximum value of product (1− D ) D is when D = 0.5. Then (3-30) and (3-34) give information that maximum values of ripples are at D = 0.5: ∆U CF max = ∆I LF max =

I ; (3-35) 4CF f

I . (3-36) 32LFCF f 2

In the case of buck-boost converter, LC filter can also be connected into the circuit supply voltage U1 (Fig. 3.6,a). But in the case of boost converter, the filter can be connected into the circuit of load (Fig. 3.10,b). The stresses of the pulse form current i 161

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3rd part through the diode V in the latter are consumed by the parallel capacitor CF that with inductance LF forms a filter in the load circuit of the boost converter (Fig. 3.10,b). Like before with the switch in on condition (0 £ t £ DT) the voltage of the capacitor is decreased linearly around the average value Uld and increased when the switch is off. The pulsations of the voltage are determined by the load current equal to I LF = I (1− D ) and can be calculated in accordance with (3-30). The pulsations of the current in the filter inductance are formed with the difference of the instantaneous values of the voltages between capacitor and load, where that of the latter is almost constant; therefore, the pulsations can be calculated using (3-34). If the filter is formed in the supply circuit of boost (Fig. 3.11,a) and load one of buck (Fig. 3.11,b) converter, then the current of inductor LF is changed linearly in time — it increases when the switch is on during interval DT, and the current is changed around the average value ILF iLF = I LF −

∆I LF t + ∆I LF . (3-37) 2 DT

As the changes of the inductor current are defined by expression uLF = LF then

diLF , (3-38) dt

U D DT . (3-39) LF Here UD is the amplitude of the inductor voltage when the switch is on. In the boost circuit a it is equal to the source voltage U1, but in the buck circuit — to (U1 −U CF ) = (U1 −U 2 ). ∆I LF =

The AC component of the inductor current causes changes in the CF capacitor voltage: • the switch is on in circuit b (0 £ t £ DT) uCF1 =

1 CF

∫

iCF dt = −

∆I LF ∆I t 2 t + LF + U CF1; (3-40) 2CF 2CF DT

• in the same circuit when the switch is off uCF2 = +

∆I LF ∆I LFt 2 t− + U CF2. (3-41) 2CF 2CF (1− D)T

Both initial values UCF1 and UCF2 are also equal, but the ripple of the capacitor voltage is ∆U CF =

∆I LF U D D . (3-42) = 8CF f 8LFCF f 2

This expression is valid for the both circuits. The application of input capacitor CF in 162

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3rd part the boost converter makes sense only if the inductance of the source is taken into account. Besides U D = U1, and ∆U CFa =

U1D . (3-43) 8LFCF f 2 iLF

a +

U1

b +

iCF LF

ILF

–

uLF

+

S f, D

uCF

CF

S

U2

LF

f, D

U1

–

iLF +

CF

–

uLF

U1 DT

–

ILF U2 = uCF

(U1 – U2) DT

0

T (U1 – U2)

iLF

DILF

ILF

t

0 iLF

t

uCF

ILF

T –U2 DILF

t

t

uCF DUCF

U1

t

U2

DUCF

t

Fig. 3.11. Influence of the pulse-form voltage on the input filter in the cases of boost (a) and load one for buck (b)

In the buck converter U D = U1 −U 2, where U2 is the voltage of the load (U 2 = DU1). Therefore, ∆U CFb =

U1D(1− D) . (3-44) 8LFCF f 2

The maximum value of the ripple is when D = 0.5: ∆U CFb max =

U1

32LFCF f 2

. (3-45)

All the expressions describing the filters are valid if the current of the filter inductor is continuous, i.e., the instantaneous values are higher than zero. It will take place if a half of the inductor current’s ripple range is lower than the average value of the current ILF. 163

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3rd part

Example The average value of the load current of the buck converter is I ld = 100 A, the value of the supply DC voltage U = 100 V. The commutation frequency of the converter f = 500 Hz. Calculate the necessary input parameters of the input filter capacitor and inductor so that the maximum values of the voltage and current’s ripple range correspondingly would not exceed 10 % of those average. 1. The maximum values of the voltage of filter capacitor as well as the current ripples of the inductor are when D = 0.5: ∆U CF max =

I ld 0.5 ⋅ 0.5 ; I 0. 5 ⋅ 0. 5 . ∆I LF max = ld CF f 8CF LF f 2

2. Taking into account that ∆U CFmax = 0.1U = 10 V, CF =

100 ⋅ 0.25 = 5000 µF. 10 ⋅ 500

3. Taking into account that ∆I LFmax = 0.1I LF = 0.1⋅ 0.5I ld = 5 A, the inductance of the inductor LF =

100 ⋅ 0.25 ⋅106 = 0.5 mH. 5 ⋅ 8 ⋅ 5000 ⋅ 5002

Example The value of the supply voltage of the buck converter is U = 100 V, the commutation frequency f = 500 Hz, the resistance of the load — R = 3 Ω. Calculate the necessary parameters of LC filter in the load junction so that the maximum values of the voltage and current’s ripple range at D = 0.5 would not exceed 10 % of those average. 1. Average value of the load voltage U 2 = DU = 0.5 ⋅100 = 50 V. 2. The average value of the load and filter inductor current I ld = I LF =

U2 = 16.66 A. R

3. The necessary inductance of the filter inductor is when ∆I LF = 0.1I LF = 1.66 A LF =

(U −U 2 )D 50 ⋅ 0.5 = = 30.1 mH. f ∆I LF 500 ⋅1.66

4. The necessary capacitance of the filter is when ∆U CF = 0.1⋅ 50 = 5 V CF =

UD(1− D) 100 ⋅ 0.25 ⋅103 = 100 µF. = 8LF f 2∆U CF 8 ⋅ 30.1⋅ 5002 ⋅ 5

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3rd part

3.1.6. Multi-Phase Buck Converters Multi-phase buck converters apply some parallel load legs correspondingly synchronised and isolated in respect to the load using the inductors (Fig. 3.12). In this way without an increase in the frequency of a particular switch, the total operation frequency of the circuit switches can be increased in respect to the supply element as well as to the load. Besides, the form of the input current is improved and the power losses in the source resistance are decreased without the input filter. This two-phase buck converter is illustrated in Fig. 3.12. The active operation of VT1 and VT2 is shifted for half-period resulting in the double increase of the input and output current ripple change frequency in comparison with that of the switch operation. Two operation modes are considered in the circuit: 1. when D < 0.5 for each switch; 2. when D > 0.5.

iL1

1

i1

L1 iL2

VT1

+

u10 iL1

2

L2

Lld

VT2

U1

V2

V1

Rld

0

ild u 20 iL2

–

0

0 ild

U1

iL1

0.5Ild

VT1 0.5T

U1

T

t iL2

0.5Ild

VT2 Ild

0

t

t

Fig. 3.12. Two-phase buck converter and its diagrams when D < 0.5

In the first case, supply current i1 is formed as pulses with amplitude 0.5Ild and frequency 2f. Average value of each of the inductor currents is 0.5Ild and the frequency of the current change is f. As average values of the voltages of inductors L1 and L2 are 0, the average value of the load voltage is U ld = DU1 = I ld Rld . At the turn-on interval of the switch, the increase of the inductor’s L1 current can be defined as (VT1 is on) L

∆I ld f ∆I L1 f = U1 (1− D) − Lld . (3-46) D D

However, when, for example, VT1 is on and current ild is increasing, then VT2 is off 165

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3rd part and iL2, in its turn, is decreasing. The decrease of current iL2 at this time is determined by the voltage uld with ild that is almost linearly increasing: L

∆I ld f ∆I L 2 f . (3-47) = −DU1 − Lld D D

The sum of both currents iL1 and iL2 define the change of current ild U1D(1− 2D) . (3-48) f (L + 2Lld )

∆I ld =

It is obvious that DIld maximum value is when D = 0.25, but when D = 0.5 load current is fully smoothed. Change of current iL1 is of more complex form: in the interval D1T the current is increased for the value ∆I L1 =

U1D(L + Lld − DL) , (3-49) Lf (L + 2Lld )

then in both pauses it decreases with the influence of DU1 − Lld

∆I ld f 0, 5 − D

voltage, but when VT2 is on — it decreases faster with the influence of DU1 + Lld

∆I ld f D

voltage. However, the ripples of the inductor current in the first case can be calculated according to DIL1 expression. In the second case, current i1 is formed as DC current 0.5Ild basis with Ild pulses of (D − 0.5)T time long. The average value of the source current in both cases is I1 = DI ld, but the factor of the source current form is better than in the case of traditional buck circuit without input filter. The parameters necessary for the input filter can also be significantly decreased. In the second case for the calculation of the load current change, a similar expression for DIld can be applied: ∆I ld =

U1 (1− D)(2D −1) . (3-50) f (L + 2Lld )

The full change of the inductor current per period T is calculated in the interval T (1− D ), when only one switch is on: L

∆I ld ∆I L = −DU1 + Lld , (1− D)T (1− D)T

placing into it expression for DIld the following can be obtained: ∆I L =

U1 (Lld + DL)(1− D) . (3-51) Lf (L + 2Lld )

The inductance and size necessary for the inductor can be significantly decreased if an 166

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3rd part inductor is replaced with a current dividing transformer. In this case if the number of half-winding turns w1 = w2 and there is a magnetic link between them, then the currents in the both half-windings are similarly changing in time, i.e., all the time iL1 = iL 2 and load current also changes in accordance with the winding current change. If D < 0.5, then with the transistor on each half-winding is supplied with voltage 0.5U1, but when both transistors are off — voltages of the windings are zero.

Example The voltage of DC source is 100 V, load resistance Rld = 5 Ω, load inductance Lld = 10 mH, the switch frequency of two-phase system with coupling inductor with inductance L = 5 mH is 5 kHz. Calculate average values of load and source current as well as their rms values when D = 0.25 and D = 0.75. With these D values calculate also ripples of load current and select the inductance of the coupling inductor. 1. Average value of the load current I ld =

0.25U1 0.75U1 = 5 A with D = 0.25; I ld = = 15 A with D = 0.75; Rld Rld

2. Average value of the source current I1 = DI ld = 1.25 A with D = 0.25; I1 = 0.75I ld = 0.75 ⋅15 = 11.25 A with D = 0.75; 3. RMS value of the source current 1 I1rms = 0.5T

DT

∫ 0.25I

2 ld dt

= I ld 0.5D = 1.77 A;

0

0.5T ( D−0.5)T 1 2 2 I1rms = I ld dt + 0.25I ld dt = I ld 1.5D − 0.5 = 11.85 A . 0.5T ( D−0.5)T 0 4. Load current’s ripple range with D = 0.25 as well as with D = 0.75 are the same:

∫

∆I ld =

∫

U1D(1− 2D) U1 (1− D )(2D −1) 100 ⋅ 0.25 ⋅ 0.55 ⋅103 = 0.1 A. = = f (L + 2Lld ) f (L + 2Lld ) 5 ⋅103 ⋅ 25

5. The ripple range of the coupling inductor current with D = 0.25 as well as with D = 0.75 are the same: ∆I L =

U1D(L + Lld − DL) U1 (1− D)(Lld + DL) = = 2.2 A. fL(L + 2Lld ) fL(L + 2Lld )

As the half of the inductor current ripple range is less than the average value 0.5Ild, we can conclude that the current is continuous and the selection of inductance of the inductor is correct.

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3rd part

3.1.7. Discontinuous Load Current Operation Mode of Buck Converter The mode of discontinuous load current appears in the case when the load contains back-EMF. The most convenient way is to investigate this operational mode for the armature circuit of DC motor with separate excitation (Fig. 3.13). If the armature is rotating, then the voltage across the armature in the mode of discontinuous current is equal to its E = cenΦ, where ce is a constructive parameter, n is the rotation speed 1/min, F is magnetic flux produced from the separate excitation winding.

a

iM

S +

f, D

U1

iM

b +

uM

U1

M

–

iM

–

DT

IMm t1

0 uM

–iM tp

f, D S

uM

DT tp

t1 T

U1

t

0 uM

E 0

M

U1

T

t

E t

0

t

Fig. 3.13. Discontinuous current operational mode in motor (a) and generator (b) operation

From Fig. 3.13,a it is obvious that the average value of armature voltage uM in the mode of discontinuous current is higher than that in the continuous current mode when U Mav = DU1: tp U Mav = U1D + E , (3-52) T where tp is the current pause time per period. Assuming that the instantaneous values of the current change linearly in time, the full time of the current decrease in the circuit t1 =

LI Mm , (3-53) E + 0.5I Mm R

where L, R are the inductance and active resistance of the armature circuit, IMm — amplitude of the current in the discontinuous mode. 168

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3rd part The current amplitude in a circuit is (U1 − E )DT . (3-54) L + 0.5RDT Taking into account IMm, the time of the current decrease (τ = L / R ) I Mm =

t1 =

τ(U1 − E )DT . (3-55) τE + 0.5DTU1

In the boundary case t1 = (1− D )T , 0.5I Mm = I MR, but E = DU1 − 0.5I Mm R . Taking into account these equations, the average value of the load current initiating the beginning of the discontinuous mode is calculated as follows: I MR =

(1− D)DU1 . (3-56) 2Lf

It is obvious that the higher the inductance L is, the lower load current initiates the discontinuous mode. A similar situation takes place with frequency: the higher the frequency is, the lower opportunity of the discontinuous current is obtained. In the case of the generator operational mode, the average voltage in the discontinuous current mode is (1− D)T − t p t p U Mav = U1 + E; T T the time of the current decrease t1 =

LI Mm ; U1 − E + 0.5I Mm R

amplitude of the current I Mm =

EDT . (3-57) L +0.5RDT

Taking into account IMm, the time of the current decrease t1 =

τEDT . (3-58) (U − E )(τ + 0.5DT ) + 0.5EDT

In the boundary case, when t1 = (1− D )T , then 0.5I Mm = I MR, but E = (1− D )U1 + I MR R, the average value of the motor current is D(1− D)U1 . (3-59) 2Lf As it is seen in the motor as well as generator operational mode, the boundary case for the discontinuous current is calculated in the same way. The characteristics U M = f ( I M ) of the converter with different values of D according to these expression are illustrated in Fig. 3.14. Shaded zones correspond to the discontinuous mode of load current. I MR =

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3rd part D = 0 0.25

UM generator

D = 1

motor

0.75

0.50

0.50

0.75

0.25 IM

–IM 0

Fig.3.14. Zones of discontinuous and continuous current of converter

For the validation of the obtained expressions, computer modelling has been implemented with U1 = 200 V, motor R = 0.5 Ω, L = 15 mH, E = 100 V, commutation frequency f = 200 Hz and D = 0.4 , i.e., in correspondence with the data of the example below. Figure 3.15 fully aproves the calculation results of the processes and motor current and voltage parameters.

200.00

uM , V

150.00 100.00

E = 100 V

50.00 0.00 –50.00 i ,A 15.00 M 12.50 10.00 7.50 5.00 2.50 0.00 –2.50 82.00

IM = 5 A tp = 1.1 ms 84.00

86.00

Time (ms)

88.00

90.00

92.00

Fig. 3.15. Diagrams of the motor voltage and current with the parameters of the calculation example

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3rd part

Example The armature of DC motor is connected into the circuit of buck converter. The value of the supply voltage is U1 = 200 V, parameter of the motor magnetic flux ceΦ = 0.1 V ⋅ min, resistance of the armature is R = 0.5 Ω, inductance — L = 15 mH. The average value of the motor load current is I M = 5 A, commutation frequency of the converter is f = 200 Hz, D = 0.4 . Calculate the rotation speed of the motor shaft. 1. The rotation speed of the shaft is calculated as follows: n=

U M − IM R . ceΦ

2. If the current mode is continuous then the average value of the motor voltage is U M = DU1; if the current is discontinuous then UM depends on a E = ceΦn of the armature. 3. To determine the current mode, the ripples of the current for the continuous mode should be defined as follows: ∆I =

U1D(1− D) 200 ⋅ 0.4 ⋅ 0.6 ⋅103 = = 16 A, 15 ⋅ 200 Lf

as it exceeds the average value of the armature current for more than two times, then the load current is discontinuous.

4. For this case U M = DU1 + E

tp

= 80 + 0.1⋅ 200nt p, T where tp is the time of the current break t p = (1− D)T − t1, t1 is the time of the current iM decrease τ(U1 − E )DT 30 ⋅10−3 (200 − 0.1n)0.4 ⋅ 5 ⋅10−3 12 − 6 ⋅10−3 n = ; = τE + 0.5DTU1 30 ⋅10−3 ⋅ 0.1n + 0.2 ⋅ 5 ⋅10−3 ⋅ 200 3n + 200 therefore, t1 =

tp = 5. As n=

15n ⋅10−3 −11.4 . 3n + 200

DU1 − I M R = ceΦ(1− t p f )

then n = 1000 min−1.

775 , 3n − 2280 1− 3n − 200

As a result, E = 100 V, t p = 1.1 ms , t1 = 1.9 ms. 171

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3rd part

3.1.8. Full-Bridge DC-DC Converter Full-bridge dc-dc converter provides an opportunity to obtain across the load dc voltage components of the both polarities. For the implementation, the circuits of bridgetype and half-bridge can be applied (Fig.3.16).

a

b

+

V11

VT1

U1

VT3

V13 U1

Ld C V14

â€“

VT4

uld,ild

+

VT2

V12 â€“

C1

C2

VT3

V11

Ld VT2

V12

Fig. 3.16. The circuits of reversible DC-DC converters: bridge-type (a), half-bridge (b)

The control of circuit a is possible as separated and coordinated modulation. In circuit b, for providing DC voltage component over load, the both capacitors must be supported by two similar voltage sources, as the capacitors cannot be the dividers of dc-voltage. The separated modulation can be implemented with one pair of transistors, and it does not usually differ from typical step-down circuit operation. If one of the transistors of the pair (for example, VT1) is on all the time but the second one (VT2 in this case) is periodically changing its state from on to off then during the pause time the load current flows through VT1 and diode V13 and the load voltage has two instantaneous values: U1, when both transistors are on, and zero, when VT2 is off. Thus, the positive load voltage (according to the direction defined in Fig. 31.6,a) is obtained with average value DU1, where D is the relative on-time (duty ratio) of a pair of transistors per period. In the case of separated control, the opposite polarity of the load voltage can be obtained with the second pair of transistors VT3, VT4, one of which is permanently on. This type of control is called single-polarity modulation, which is illustrated in Fig.Â 3.17. In the separated control, double-polarity modulation is also possible when instantaneous values of both voltage polarities are possible across the load. This mode of operation is available if both transistors of a pair are simultaneously off. When turned off the current is conducted by diodes (for example, after the transistors VT1 and VT2 simultaneously turn off diodes V14 and V13 continue to conduct), the current flows

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3rd part through the supply source or the input capacitor C, and the polarity of voltage during this time period is opposite to that when the transistors are turned on (Fig. 3.17,b). The case of double-polarity modulation of separated control can cause a mode of discontinuous load current. Therefore, this control is not widely used.

a VT1

b DT

VT2 uld

T

t

VT2 uld

t

U1

VT1

DU1 0

0

Ustm

0 VT1 VT2 VT3 VT4 uld

Uco t

ust

t t

U1

Im

ild t

c ust

T

DT

t

V13 V14 –U1

t t

U1 ild

0

T

t–

t+

t

–U1 Fig. 3.17. Time diagrams of the separated (a, b) and coordinated (c) control

Example In the case of double-polarity modulation of the separated control of the full-bridge DC-DC converter, the relative time of the on-state of transistors pair is D = 0.45. Calculate the average value of the load current if the supply voltage U1 = 200 V, the load parameters — R = 5 Ω, L = 10 mH, frequency of transistor commutation f = 1 kHz. 1. The load current is discontinuous (Fig. 3.17,b); therefore, its amplitude can be calculated as follows: Im + 0.5I m R = U1; DT therefore L

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3rd part Im =

U1DT 200 ⋅ 0, 45 = = 8.09 A. L + 0.5RDT 10 + 0.5 ⋅ 0.45 ⋅ 5

2. The time of the current decreasing to zero in the loop “diodes-source” is calculated using Im = U1 + 0.5I m R; t1 therefore L

t1 =

LI m = 0.4 ms. U 1 + 0. 5 I m R

3. Assuming the change of the current in time as linear the average value of the load current is I ldav =

I m DT + t1 = 3.438 A. T 2

In the case of coordinated control, both pairs of transistors are controlled all the time intermittently and the load is supplied with instantaneous voltages of the both polarities. If transistors VT1, VT2 are longer in the on-state, then the load is more supplied with voltage U1 of positive polarity and its average value is with a positive sign. When VT1 and VT2 are off and the control is supplied to VT3 and VT4 they sometimes should not conduct the current. It depends on the polarity of the load current instantaneous value: if ild > 0 A , then VT3 and VT4 do not conduct the current, but diodes V13, V14 do. The control is implemented with the help of saw-tooth voltage of double-polarity ust with the linear changed front (Fig. 3.17,c). When slowly changing control voltage Uco is higher than the instantaneous value of the saw-tooth voltage, the transistors VT1, VT2 conduct the current (or diodes V11, V12). If ust is higher than Uco, then transistors VT3,VT4 (or diodes V13,V14) conduct. The average value of the load voltage U ldav =

U U1 (t + − t− ) = U1 co , (3-60) U st T

where Ustm is the amplitude of the saw-tooth voltage, U t− = 0.51− co , t + = T − t− U stm

are respectively the time intervals when the saw-tooth voltage exceeds the control voltage Uco and vice versa — when it is lower. It is obvious that if U co = U stm , then U ldav = U1. If U co = 0 V, then U ldav = 0 V , but, if U co = −U stm, then U ldav = −U1. Thus, the value of the load voltage is changing in a

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3rd part linear way in accordance with Uco, changing also the polarity along with the that of the control voltage. To calculate the load current ripples, the active voltage drop across the load is assumed to be ild R = I ld R . If the load is active-inductive, the average value of the load current is I ldav = U ldav / R. Then the equation for the case with transistors VT1 and VT2 in the on-state is U1 = L

dild di U + U ldav = L ld + U1 co . (3-61) dt dt U stm

Taking into account that the time of on-state of these transistors is U t VT1,2 = 0.5T 1 + co , (3-62) U stm

load current full-range ripple is 2 U co U1 1− U stm . (3-63) ∆I ld = 2Lf

From (3-63) it is obvious that the largest ripple range is at U co = 0 V, when ∆I ldmax = U1 / (2Lf ). The dependence of the current ripple range on U co / U stm is illustrated in Fig. 3.18.

DIld D 1 Ildmax

–1

–0.5

0

0.5

1 Uco Ustm

Fig. 3.18. The dependence of load current ripple range on the control voltage

The average value of the network current is calculated from the balance of active power: I1av =

2 I ldef R , (3-64) U1

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3rd part where the rms value of the smoothed load current is equal to the average value of the load current, but when the ripples are higher the influence of DIld on the rms value of the load current should be taken into account: I ldef = I ld2 +

∆I ld2 . (3-65) 12

In the half-bridge circuit (Fig. 3.16,b) at the before mentioned voltage support, the positive and negative amplitude of the load voltage is 0.5U1. It means that the average value of the load voltage is 0.5U1U co / U stm, and the ripple of the load current DIld is half less than for the full bridge circuit.

Example The amplitude of double polarity symmetric saw-tooth voltage is U stm = 15 V, frequency — 1 kHz. Control voltage U co = 5 V is compared with saw-tooth voltage. Calculate how long within the period is the load supplied with the voltage of positive polarity in the dc-dc converter with coordinated control and how long — with negative; what are the average values of the load current and voltage and how large are the current ripples in the load, if U1 = 200 V, load R = 5 Ω, L = 10 mH. Do the diodes conduct current? What is the average value of network current? 1. The time intervals in the period when the instantaneous value of the load voltage is with positive polarity U 0.5 5 t + = t VT1,2 = 0.5T 1 + co = 3 1 + = 0.666 ms; U 10 15 stm

with negative polarity — t− = T − t + =

1 0.666 − 3 = 0.333 ms. 3 10 10

2. Average value of the load voltage t U t U ldav = U1 + − − = U1 co = 66.6 V. T T U stm 3. Average value of the load current I ldav =

U ldav 66.6 = = 13.33 A . R 5

4. Load current’s ripples 2 5 2 U co U1 1− 200 1− 103 U stm 15 = ∆I ld = = 8.88 A. 2Lf 2 ⋅10 ⋅103

5. As I ldav > 0.5∆I ld , the current is not conducted by the diodes. 176

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3rd part 6. RMS value of the load current is 8.882 = 13.57 A. 12 7. Average value of the source current I ldrms = 13.332 +

I ldav =

13.572 ⋅ 5 = 4.605 A. 200

The dc-dc converter with coordinated control has been modelled by means of compu ter software at U1 = 200 V, f = 1 kHz, R = 5 Ω, L = 10 mH, U co = 5 V and U stm = 15 V. As Fig. 3.19 illustrates, the parameters of the load current correspond to those obtained in the calculation pattern.

200.00

uld, V

100.00 0.00 –100.00 –200.00 20.00

ild, A

15.00

Ildav

DIld

10.00 5.00 0.00 –5.00 88.50

89.00

89.50 Time (ms)

90.00

90.50

Fig. 3.19. Diagrams of the computer model of dc-dc converter with the coordinated control using the parameters of calculation pattern

3.1.9. Implementation of qZ Source Network One of the problems of dc-dc converter application is related to the unified type of the input circuit filter — the output of this circuit in the buck circuit should possess the properties of voltage source, but in the boost circuit — the properties of current source. This contradictions can be solved applying the so-called qZ (quasi Z) source network. In the boost circuit, source inductor is replaced with two inductors L1 and L2, and in the buck circuit the filter capacitor is replaced with two capacitors C1 and C2 (Fig. 3.20), f orming the so-called qZ source network.

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3rd part iC1 C1

i1 +

L1

iL2 L2

V1

U1

iC2

–

1

C2

V S

f, D

Ild C

R

0

Fig. 3.20. Application of qZ source network in boost converter

In this circuit the load is affected by two currents formed from the two LC circuits L1-C1 and L2-C2. As the both circuits contain capacitor and inductor, the circuit in general has both properties of voltage as well as current source and it can operate as buck and as boost converter and their combination. However, the circuit influences the parameters of the dc-dc converter. Let us analyse the operation of circuit in boost mode. Assuming that the current of input inductor L1 and inductor of the circuit L2 is continuous, the load voltage in this system can be calculated from the balance of currents for node 1. The average current of the boost circuit diode V is I V = I ld , because the average current of all the capacitors is zero. By this principle the average currents of inductors L1 and L2 as well as of the diode V1 of qZ circuit are equal to the source current I1. During the time when switch S of the boost circuit is off, the balance of currents for node 1 is I1 = IC1(off) + I V(off), where in the expression the right side currents are average values of the capacitor C1 and boost circuit diode V when S is off. As during the on-interval of the switch S the voltage drop across inductor L1 decreases the potential of the diode V1 anode, the diode within this interval is off and current I1 flows through capacitor C1, i.e., iC1(on) = −I1. It means that the average value of the capacitor C1 current during the off-time of the switch is IC1(off) =

DI1 . (3-66) 1− D

The current of the diode V at the same time is I V(off) =

I ld (3-67) 1− D

Then the relation between the average value of the source current I1 and the load current is I1 =

I ld . (3-68) 1− 2D

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3rd part As the power at the input and output of the system is equal, i.e., U1 I1 = U ld I ld then U ld =

U1 . (3-69) 1− 2D

If one compares it with typical boost circuit where 0 < D < 1, then in this case 0 < D < 0.5 and the equal voltage increase in boost qZ input system is achieved at D being two times less. The average voltage of the capacitor C1 is the difference between the average voltage of the switch U10av = U ld (1− D ) and the source voltage U1: U1 D . (3-70) 1− 2D

U C1av =

The average voltage of the capacitor C2 is equal to U10av: U1 (1− D) . (3-71) 1− 2D

U C2av =

The full ripple range of the capacitor voltage can be calculated on the basis of the current diagrams as follows: DI ld . (3-72) C1 f (1− 2D) DI ld ∆U C 2 = C2 f (1− 2D) ∆U C1 =

The ripple range of the current in inductor is calculated on the basis of its voltage diagrams: when switch S is on U L 2 = U C 2av and ∆I L2 =

D(1− D)U1 , (3-73) (1− 2D) fL2

at the same time U L1 = U1 + U C1(on) = U1 (1− D ) / (1− 2D ) and ∆I L1 =

D(1− D)U1 . (3-74) (1− 2D) fL1

In terms of current and ripple balance in the both inductors, we can assume that two similar inductors should be applied, i.e., L1 = L2. If 0.5∆I1 < I1, then the current of L1 is continuous. Applying equations for DI1 and I1 the boundary case D b can be calculated from Db (1− 3Db ) + 2Db3 =

2L1 f . (3-75) R

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3rd part For example, if 2L1 f / R = 0.05, then the discontinuous current mode is within the range from D = 0.06 to D = 0.39, but the continuous mode at D > 0.39 and D < 0.06.

3.1.10. DC-DC Converters with Thyristors A large amount of DC-DC conversion is still implemented by means of thyristor converters, one of the basic elements of which is commutation circuit with capacitor (Fig. 3.21). To switch off the basic thyristor VS1, capacitor C charged with the polarity marked in Fig. 3.21 is connected to it in parallel through thyristor VS2 producing a reverse voltage. The reverse voltage across VS1 should be supplied for a longer time than it is necessary for full recovery of the thyristor non-conductivity properties, i.e., longer than its rated turn-off time toff. As after switching on VS2, the voltage of capacitor is added to the supply voltage, then during the entire period of the capacitor charging from source the reverse voltage is supplied across the load by-passing diode V0 and it does not conduct the current. Therefore, the load current flows through capacitor and its charging takes place at constant current: I t uC = −U C 0 + ld . (3-76) C a

VS1

+

U1 –

b

V2

VS2 – L1

L2 +

C uld iC , uC

Ild

0 uC U1

Ild V0

iC

Ld

0

V1

t

trev

t

–UC0 uld U1 0

t

Fig. 3.21. Circuit and diagrams of the thyristor dc-dc converter

The charging will be completed when uC is equal to U1. Then the voltage across diode V0 is zero and the diode starts current conduction. That happens after the capacitor charging interval 180

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3rd part t Cp =

(U1 + U C 0 )C , (3-77) I ld

but the reverse voltage across thyristor VS1 is applied for interval trev, at the end of which uC = 0 V: U C 0C . (3-78) I ld

t rev =

This time should be higher than toff. Therefore, in the simplest case the necessary capacity is calculated as follows: C³

t off I ld . (3-79) UC0

Initial reverse voltage UC0 depends on the parameters of the coordinating circuit of the capacitor polarity. In Figure 3.21 the polarity coordination causes the oscillations in the contour C-V1-L1-VS1-C, the angular frequency of self-oscillations of which is 1 L1C

ω1 =

and wave impedance is L1 . C In the ideal case, the capacitor current in this circuit during the charging time ρA =

iC = −

U1 sin ω1t , (3-80) ρA

where w1t interval influenced by the diode V1 is within the range from zero to p. Therefore, the full time interval of the coordination process t p1 =

π = π L1C . (3-81) ω1

During this time period, the main thyristor VS1 must be turned on. This restricts the minimum possible switch on-duty state during the switching period. Calculating duty ratio Dmin in accordance with the relation of the load average voltage and source voltage, the influence of the post-commutation capacitor charging process on the average load voltage should be taken into account. Therefore, C = π L1C + (U C1 + U C 0 ) f . (3-82) I ld T It is obvious that Dmin either directly depends on the capacitance or reversely depends on the load current. To ensure the operation of the converter at high enough frequency (e.g., 500 Hz), the capacitance should be low but this is possible only if toff of the thyristor has a small value. Dmin =

t p1 + tCp

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3rd part Disadvantage is the dependence of Dmin on the load current, when at low load currents Dmin is highly increasing. To remove this dependence, the second inductor L2 with d iode V2 should be connected in parallel with thyristor VS1 (as it is in the circuit). Then during the time of commutation charging the second oscillation circuit C-L2-V2-VS2-C is formed with the oscillation frequency ω2 =

1 L2C

and wave impedance is L2 . C

ρ2 =

Then the equation of capacitor current is iC =

UC0 sin ω2t + I ld cos ω2t , (3-83) ρ2

but its voltage expression is uC = −U C 0 cos ω2t + I ld ρ2 sin ω2t . (3-84) The time interval of reverse voltage is determined when uC = 0 V: U 1 t rev = arctg C 0 . (3-85) ω2 I ld ρ2 This time period lesser depends on the value of Ild. Even at I ld = 0 A the time of reverse voltage is: π t rev = . 2ω 2 In the circuit under consideration (Fig. 3.21) voltage Uco in the ideal case is equal to U1, but in fact it is 10 % to 15 % less because the circuit of coordination over-charging has active resistance determining the loss of voltage.

Example The thyristor DC-DC converter with the resonance charging circuit is operating in the system at supply voltages U1 = 200 V, duty ratio D = 0.5 and commutation frequency f = 500 Hz. Calculate the necessary commutating capacity and the inductance of charging inductor, if the load resistance is R = 5 Ω and inductance is very high, Dmin = 0.1, the turn-off time of the thyristor is t off = 50 µs . 1. The average value of the load current DU1 0.5 ⋅ 200 I ld = = = 20 A. R 5 182

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3rd part 2. Capacity t I 50 ⋅ 20 C ≥ off ld = = 5 µF; U1 200 ⋅106 Assuming C = 7 µF. 3. The inductor inductance is calculated using DminT = π L1C + whence L1 =

2U1C ; I ld

2 D T − 2U1C min I ld

π2 C

=

3600 ⋅106 = 52 µH. 1012 ⋅ π2 ⋅ 7

To make UC0 equal to U1 the bridge-type junction of capacitor connection can be applied (Fig. 3.22). Thus, in each period the capacitor is connected with the help of two auxiliary thyristors: when uC < 0 V, VS2, VS3 are used; when uC > 0 V VS4, VS5 are used. After each commutation the capacitor changes its polarity and it is adjusted to the further commutation. In this situation, t p1 = 0 s , which stipulates less possible Dmin, and the operation frequency of the capacitor is two times lower than in the circuit in Fig. 3.21. V2 L2 VS1 +

VS2

U1 –

C VS5

VS4 uC VS3

Ld V1

Fig. 3.22. DC-DC converter thyristor circuit with a bridge-type commutator

3.1.11. Operation of Transistor Switch in Real Circuit 3.1.11.1. Process of Transistor Switching The real operation conditions of a transistor switch are demonstrated in Fig. 3.23. The source U1 supplies series RL load with a clamp diode V and IGBT transistor VT. The 183

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3rd part circuit is operating in the switch periodic mode and the current of RL load Ild is sufficiently smoothed. When the control signal UGE is not supplied, the load current flows through the clamp diode V. When the control signal is applied and after the turning-on delay time td(on) the transistor current iC starts increasing while the diode current is decreasing, then in the rise time tr iCr = I ld

t , 0 £ t £ t r . (3-86) tr

During this rise time the diode conducts current and its voltage is almost zero, i.e., uCEr = U1. When the diode current achieves zero, the voltage of the transistor uCE decreases to zero during almost the same time interval, which is needed for iC to increase. If we assume the dynamic properties of the diode to be ideal, the approximate loss of the switching energy is tr

Etr = U1

∫ 0

t I ld dt + I ld tr

tr

∫ 0

t U1 1− dt = I ldU1t r . (3-87) t r

As soon as the control signal is zero, the transistor voltage starts increasing within the fall time tf. Within this interval the transistor current is equal to Ild. After switching off delay interval td(off) the transistor voltage achieves value U1, diode V starts conducting and simultaneous transistor current is decreasing and diode current is increasing during diode fall time tf. Total losses of energy within the off-process are Etf = I ldU1t f . (3-88) The total energy losses in the commutation processes within a period are Ets = Etr + Etf = I ldU1 (t r + t f ). (3-89) The losses of power are calculated as follows: ∆Psw = Ets f , (3-90) where f is the switching frequency.

Example IGBT transistor has t r = 50 ns , t f = 400 ns . Bypassed by a diode the load current is 50 A, the source voltage is 600V. Calculate the commutation power losses if the switching frequency is 4 kHz. 1. The energy losses in the commutation processes within a period are Ets = 50 ⋅ 600 ⋅(50 + 400)⋅10−9 = 13.5 mJ; 184

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3rd part 2. Commutation power losses are ∆Psw = Ets f = 13.5 ⋅10−3 ⋅ 4 ⋅103 = 54 W. a

b uGE 20 V 0 iV

+

iC

C G –

t1 0 iC

iV

U1

t U1

uCE

VT E

td(off)

Ild

Ild

V

t

td(on)

uCE

Ild iC

trec 0

tr

tf t(Etr)

t

t(Etf )

Fig. 3.23. Circuit (a) and diagrams (b) of the transistor switch real operation condition

Everything mentioned above is valid when the clamp diode is ideal. In fact, when the voltage polarity across it is changed from that corresponding to the diode conducti vity direction to the reverse biasing one, both zones of the diode structure collect the charges of the opposite polarity and at the instant of the voltage polarity change they are returning to their basic zones, i.e., during some time interval the diode conducts the reverse current. This recovery charge has a particular value depending on the quality of the diode. The higher the quality is, the higher the operation speed is and the lower recovery charge Qrec of the diode is. Taking into account the dynamic pro perties of the diode, the transistor current increases above the load current value during the recovery time trec (Fig. 3.23,b) and it leads to a significant increase in the losses of the transistor. The amplitude of the transistor current depends on the speed of current increase during the turn-on process, i.e., on I ld / t r ratio. Assuming the triangular form of the diode reverse current, the time of its increase is t1 =

2Qrect r (3-91) I ld t rec

and the amplitude of the transistor current is 185

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3rd part t 2Q I VTmax = I ld 1 + 1 = I ld + rec . (3-92) t r t rec

Example The clamp diode has good enough parameters: Qrec = 1 µC, t rec = 0.1 µs . The load current is 50 A, the transistor has t r = 50 ns . Calculate the amplitude of the switch-on current. 1. Time intervals of the reverse current increase 2 ⋅1⋅ 50 ⋅106 = 20 ⋅10−9 s. 106 ⋅109 ⋅ 50 ⋅ 0.1 2. Expected amplitude of the transistor current t1 =

2 ⋅1⋅106 = 70 A. 106 ⋅ 0.1 Total increase of the switching energy losses for I VTm ax = 50 +

∆Et(rec) = 0.5(I ld + I VTm ax )U1t1 and if U1 = 600 V, then ∆Et(rec) = 0.72 mJ . If we compare with the previous calculation data, then DEt(rec) is approximately a half of the on-switch process energy losses. However, applying a worse diode, IVTmax and DEt(rec) increase. The total power losses in the transistor also depend on the conductive losses that are calculated according to the collector current IC in the on-state and voltage across collector and emitter, which, in its turn, is a function of IC, i.e., UCE(IC). The duty ratio D should also be taken into account. Then ∆PtD = I CU CE (I C )D + Ets (I C ) f . (3-93)

Example Calculate total power losses, if the collector current of 70 A transistor is 50 A, U CE (50) = 1.8 V, frequency f = 4 kHz, D = 0.5, supply voltage is 800 V, t r = 50 ns , t f = 500 ns, clamp diode is assumed to be ideal. Calculate the surface temperature of the structure, if the thermal resistance between the structure and radiator is Rth(js) = 0.9 °C/W. 1. Conductive losses

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3rd part ∆PtDco = I CU CE ( I C )D = 50 ⋅1.8 ⋅ 0, 5 = 45 W; 2. Energy losses during the commutation processes Ets (I C ) = U1 (t r + t f ) = 800 ⋅ (50 + 550) ⋅10−9 = 22 mJ; 3. Total power losses ∆PtD = 45 + 22 ⋅10−3 ⋅ 4 ⋅103 = 133 W. 4. Surface temperature of the structure Θpn = ∆PtD Rth(js) = 119.7 °C. This is a high temperature, and the most effective way to decrease it is the decrease of the commutation frequency.

3.1.11.2. Safe Operation Area (SOA) The process of the transistor switching is graphically represented in Fig. 3.24. At the first stage of the turn-on process from point 1 to point 2 the process takes place at constant voltage and rising collector current. At the second stage from point 2 to 3 this current remains constant, but the voltage is decreasing almost to zero.

uCE B turn-off process U1

1

2

A turn-on process C 0

3 IC = Ild

iC

Fig. 3.24. Trajectory of the transistor switching process

At the beginning of the turn-off process from point 3, the transistor voltage is increased but when it achieves the value of the supply voltage U1 at point 2, the collector current starts decreasing. The instantaneous power losses uCEiC are changed in accordance with the trajectory — at points 1 and 3 they are zero, but the highest value is achieved at point 2, then the instantaneous power losses are U1IC. For example, if 187

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3rd part U1 = 800 V and I C = 50 A, the power losses at point 2 are 40 kW. However, these losses exist for a very short period of time not providing any significant heating of the structure, but it is preferable to decline from this trajectory to the direction of lower power losses. In fact, the processes and trajectory significantly differ from the rectangular form represented in Fig. 3.24. The transistor circuits have leakage inductance Lσ and with the increase of the current during the turn-off process the transistor voltage is slightly lower than U1, as positive Ls diC / dt voltage is subtracted from U1. During the recovery time of the load clamp diode, the transistor current can significantly exceed Ild (trajectory point A in Fig. 3.24). Within the turn-off process when current iC decreases ( diC / dt < 0 ) the voltage across leakage inductance Lσ is with a negative sign and the transistor voltage that is the difference of U1 and Lσ voltage can grow over U1 (trajectory point B in Fig. 3.24). As a result, the power losses in the transistor at over-current as well as over-voltage can exceed the rated value determined for point 2. It can result in irreversible damage of the transistor.

Example The total leakage inductance of the transistor circuit is 0.5 μH, supply voltage is 600 V. Current of the transistor is decreasing from 50 A to zero within the time of 200 ns. Calculate the expected maximum voltage and maximum instantaneous power losses. 1. The speed of current decrease diC −50 ⋅109 = = −0.25 ⋅109 A/s; dt 200 2. Maximum voltage of the transistor U tm = U1 − Lσ

diC 0.5 ⋅ 0.25 ⋅109 = 600 + = 725 V ; dt 106

3. Maximum instantaneous power losses ∆Ptm = I CU tm = 50 ⋅ 725 = 36.25 kW. The trajectory can be improved if part of the commutation power losses is directed to an additional damper circuit — snubbers. During the turn-off process, the first part of the process with almost linear increase of the collector-emitter voltage can be eliminated applying a capacitor in parallel with the transistor that significantly decreases the transistor current during the turn-off process. During the first part of the turn188

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3rd part on process, the transistor voltage can be decreased if the circuit has negligible low inductance. The so-called safe operating area (SOA) of the transistor should meet the switching process trajectories. An admissible zone of collector voltage and current is determined for the operation of BT, MOSFET and IGBT transistors without damage and usually it is given in the specification of the parameters. The voltage and current are represented in the logarithmic scale (Fig. 3.25). SOA limits the admissible values of the current and voltage; thus, all the parameter combinations outside this zone are not allowable.

iC , A 100 ICmax

10

1 10

100

1000 uCE, V UCEmax

Fig. 3.25. Example of the transistor safe operating area

The decreasing restricting zone determines the maximum admissible power losses that are approximately defined as 0.3I C maxU CE max , where ICm and UCEm are admissible maximum collector current and voltage, respectively. The decreasing zone of transistor with I Cmax = 75 A and U CEmax = 800 V corresponds to SOA (see Fig. 3.25). Therefore, the decreasing zone limiting the power losses corresponds to 18 kW. The trajectory of SOA marked with dotted lines very often corresponds to the accepted law — applying only half of the transistor maximum current and voltage, the maximum point of the power loss trajectory will always be within the range of SOA.

3.1.11.3. Improvement of the Turn-Off Process The snubber circuit for turn-off process has two tasks — to decrease the losses in the process and to limit the over-voltage. Let us consider its operation. A circuit of such Cs-Rs-Vs snubber parallel to the transistor is illustrated in Fig. 3.26.

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3rd part a

b

+

iVT iCs Ild

iVT

Ild

V

0

U1

uVT

Vs

iCs tf

t U1

Rs iVT

Cs

VT

–

iCs

0

t

Fig. 3.26. Snubber circuit of the turn-off process (a) and diagrams (b)

The diagrams present an ideal case when the capacitor is selected with such accuracy that the voltage at the end of the turn-off interval tf is exactly U1. Then the process of the capacitor voltage increase can be described as follows: 2 1 I − I 1− t dt + A = I ld t + A, uVT = (3-94) ld ld t f Cs 2Cst f A = 0, uVT = 0 V when t = 0 s.

∫

As in the end the voltage is U1, the selected capacity should be Cs =

I ld t f . (3-95) 2U1

In such an ideal case, the power losses of instantaneous turn-off process within the interval tf are changed as follows t t2 ∆ptf = iC uVT = I ld2 1− , 0 £ t £ t f (3-96) t 2C t f

s f

reaching the maximum at 2 t fm = t f , 3 and the maximum power losses are Ptfm = I ld2

4I U 2t f = ld 1 . (3-97) 27Cs 27

The energy losses during the process are tf

∆Etf =

∫ 0

ptf dt =

I ld2 t f2 I Ut = ld 1 f , (3-98) 24Cs 12

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3rd part i.e., they are 12 times lower than those for the case without damper circuit. In Fig. 3.27 the trajectory of the ideal process is curve 1. However, at lower load current the capacity Cs can be too high, or at higher current — too low. In the first case, at the end of the tf interval uVT < U1, the charging of Cs continues with current Ild, until uCs = U1. The power losses of the transistor in this additional interval are zero, but the total energy losses are lower than the ones mentioned above. In Fig. 3.27 this trajectory of the process is curve 2.

uVT U1

3 1 2

0.25U1 0

0.5Ild

Ild

iVT

Fig. 3.27. Trajectories of the turn-off process with different capacities of the snubber circuits

In the case when capacity is too low, the voltage of capacitor Cs reaches the level of U1 before the end of the interval tf. The current at this moment is conducted by the clamp diode V, but during the time of the further decrease of the transistor current to zero the transistor voltage is U1. The process is described by curve 3 in Fig. 3.27. In the last case, the energy losses of the turn-off process are higher than in the ideal case. The charging of the capacitor to voltage U1 is completed after the interval tf1 =

2U1Cst f , (3-99) I ld

and the transistor current at this moment is t I VT1 = I ld 1− f 1 > 0. t f After the interval tf1 the capacitor current is zero and the transistor current flows to the clamp diode, during the same time transistor voltage is U1. When the transistor is on again, the capacitor is discharging through the resistor Rs, the average power losses are ∆PRs = 0.5CsU12 f , (3-100) 191

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3rd part i.e., they do not depend on the resistance of the circuit. The minimum duration of the on-state of transistor should be long enough for the full discharging of the capacitor Cs, i.e., t VT(on ) > 3Cs Rs. The rated resistance Rs, in its turn, should be higher than U1 / (0.1I ld ), which limits the initial value of the discharging current.

Example The supply voltage is 600 V, load rated current is 50 A, for the transistor t f = 400 ns . Calculate the parameters of the turn-off process RCD snubber circuit and the power losses during the operation at rated current and its increase to 75 A, if the frequency is 4 kHz. 1. Capacity Cs Cs =

I ld t f 50 ⋅ 400 = = 16.66 nF ; 2U1 109 ⋅1200

it is assumed that Cs = 20 nF.

2. Rated value of resistor Rs Rs ≥

U1 600 = = 120 Ω. 0.1I ld 5

The power losses of the resistor are

10 ⋅ 6002 ⋅ 4 ⋅103 = 14.4 W. 109 Resistor at the rated power of 20 W should be selected. ∆PRs = 0.5CsU12 f =

3. The energy losses of the turn-off process at the rated load current I ldU1t f 50 ⋅ 60 ⋅ 400 = = 1 mJ . 12 109 ⋅12 Power losses of the turn-off process ∆EtfN =

∆PtfN = ∆EtfN f = 4 W. 4. The first interval of the turn-off process at the increased current 2U1Cst f 2 ⋅ 20 ⋅ 600 ⋅ 400 = 375 ns. = I ldmax 109 ⋅109 ⋅ 75 5. The energy losses of the first stage of the turn-off process at the increased load current t1 =

∆Et1f = I ld2

t f31 = 1.33 mJ. 24Cst f

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3rd part 6. The energy losses of the second stage of the turn-off process at the increased load current t ∆Et 2 f = 0.5I ld 1− f 1 U1 (t f − t f 1 ) = 0.1 mJ . t f 7. The total energy losses of the turn-off process at the increased current ∆Et f = 1.33 + 0.1 = 1.43 mJ

and power losses ∆Ptf = 1.43 ⋅10−3 f = 5.72 W.

The second task of the snubber circuit is to compensate the influence of the commutation circuit leakage inductance Lσ on the maximum value of the transistor voltage. When the transistor current is zero at the end of the process, the current in the lea kage inductance Lσ and series capacitor Cs is equal to Ild. As at the initial instant the capacitor Cs voltage is U1, the current in Ls-Cs circuit starts decreasing but the current in the clamp diode is increasing. It results in the voltage U1 supply to the entire circuit and the process is described using the expression Lσ

di 1 + d t Cs

∫ idt +U

s

= U s, (3-101)

the initial value of which at t = 0 s is i (0) = I ld and di / dt(0) = 0. The solution of the second order equation is i = A sin ωt + B cos ωt or i = I ld cos ωt , (3-102) the voltage of capacitor is changing as follows: uCs = U1 +

1 Cs

∫ idt = U

1

+

I ld sin ωt . (3-103) ωCs

When ωt = 0.5π, current i = 0 A, uCs reaches a higher value U Cs max = U1 + I ld Here ω =

Lσ . (3-104) Cs

1 is taken into account. LσCs

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3rd part

Example The current of the load is 50 A, supply voltage is 600 V. Capacity of the snubber circuit is 20 nF, leakage inductance is 0.5 μH. Calculate the maximum voltage of the transistor after the turn-off. U Cs max = U1 + I ld

Lσ 0.5 ⋅109 = 850 V. = 600 + 50 Cs 106 ⋅ 20

It is obvious that the increase is high enough, i.e., after the turn-off process the previously selected capacity of the snubber circuit of 20 nF is too low. With Cs increased to 50 nF, U Csmax = 758.1 V. Some work should also be done with the decreased leakage inductance (it is necessary to shorten the length of the connecting wires as much as possible, optimise the position of the elements in the circuit, etc.).

3.1.11.4. Improvement of the Turn-On Process The turn-on process is relatively short, and the power losses of it are much lower than those of turn-off process; however, in the operation at high frequencies these power losses can significantly increase. Therefore, sometimes snubber circuits for the turn-on process are used consisting of additional series inductance in transistor or the clamp diode circuit (Fig. 3.28). In the first case, during the time of the transistor current increase a voltage drop across the inductance Ls is formed that results in the transistor voltage uVT = U1 − Ls

I VT max , (3-105) t r + t rec

where I VTmax = I ld +∆I rec, DIrec is the current increase range during the diode V recovery time trec. In the second case, the transistor current can slowly increase (if Ls is high enough) as well as DIrec can decrease. Additionally, the voltage of transistor over the process can decrease to a very low value. In the first case (Fig. 3.28, a), to decrease the voltage to zero the inductance Ls should be selected as follows: Lsmax =

U1t r . (3-106) I ld

In fact, Ls < Lsmax. In the second case (Fig. 3.28,b), the transistor current increases as follows: iVT =

U1t , (3-107) Ls

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3rd part where 0 £ t £ Ls I ld / U1. The interval of current increase t1r = Ls I ld / U1 can be extended so that it is much times higher than t r + t rec . In order to weaken the energy collected in Ls, discharge resistance R Ls is connected in parallel to Ls through the reverse diode. Then during the off-state of the transistor that should be longer than 3Ls/R Ls, the current of inductance decreases almost to zero. At the end of the turn-on process, the overvoltage produced by this snubbing circuit can achieve Ild R Ls. The power losses of the resistor R Ls are calculated as follows: ∆PRLs = 0.5Ls I ld2 f , where f is frequency. a

b

+

+

RLs Ild V

Vs

U1

U1

Vs Ls

V

RLs

iVT –

Ild

Ls

iVT –

VT

VT

IVTmax Ild

U1

U1

Ild iVT

uVT uVT 0

tr

trec

t

0

t1r

t

Fig. 3.28. The damping circuits of the turn-on process of the transistor (a) and clamp diode (b)

Example Calculate the inductance of the damping circuit at the turn-on process, if the supply voltage is U1 = 600 V, load current is 50 A, transistor t r = 30 ns , frequency is 50 kHz. Maximum duty ratio of the transistor is Dmax = 0.9 . 1. Inductance of the damping circuit is Ls =

U1t r 600 ⋅ 30 = 9 = 0.36 µH. I ld 10 ⋅ 50 195

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3rd part 2. Minimum time of the transistor off-state 1− Dmax tt(off)min = = 2 µs . f 3. Discharging resistor 3L 3 ⋅ 0.36 ⋅106 = 0.54 Ω. Rs = s = tt (off) 106 ⋅ 2 4. Power losses of the resistor 0.36 ⋅ 0.5 ⋅ 502 ⋅ 50 ⋅103 ∆PRLs = 0.5Ls I ld2 f = = 22.5 W. 106 It is obvious that the power losses are high enough; therefore, the parameters of this snubber circuit should be carefully selected. The inductive snubber circuit of the turnon process is highly necessary for bridge-types circuits when without this inductance in the case of turn-on process of one transistor a capacitor of the R-C-D circuit of the opposite transistor snubber is charged. If the circuit does not contain an inductor, the charging current can be high enough and can influence the transistor heating during the turn-on process. Inductor in the circuit limits the charging current.

3.1.12. Isolated DC-DC Power Supply DC-DC converter can be connected in series with a transformer, thus obtaining voltages and currents of pulse form in primary and secondary windings, a that the latter has galvanic isolation from a supply source. These supply sources are widely used as they are simple enough. Different circuits are available, but in practice the ones illustrated in Figs. 3.29 and 3.30 are mostly applied. The first circuits are the forward DC-DC converters (3.29,a and b) when the energy is supplied to the load when the primary winding of the transformer with demagnetizing winding is connected to the supply voltage. The second type is the magnetizing energy fly-back type when to the load energy is transmitted from the supply source after the disconnection of the transformer primary winding. The transformer winding of such a circuit has a single-direction current and the operation point of the magnetizing curve is moving only within the first quadrant changing the flux density between residual value B0 and saturation value Bs, which should ask for air-gap in magnetic core of transformer. The main problem of the first type circuit is related to the discharging of the transformer magnetizing current after the disconnection of the primary winding from the DC supply source. For this purpose, the circuit in Fig.3.29,a contains the third winding w3 that is connected to the source through diode V3 after the turn-off process of 196

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3rd part transistor VT when the voltage of the primary winding changes its sign. Then current i3 produced from the magnetizing current flows through diode V3 to the source and the reverse voltage of the primary winding is U1w1 / w3 . As the average value of the transformer winding should be zero but the active condition takes place during the interval DT, where D is the duty ratio of VT, the reverse voltage can be supplied during interval (1− D )T only, i.e., in the boundary case D=

w1 (1− D). w3

Therefore, a maximum admissible D can be determined by the given ratio of the turns or vice versa.

i1

a +

V3

U1

TR w1 uW1

i3 –

c

b

V1 w2 i2

L V2

+ +

C

R

VT uVT

U1 –

w3

VT1 V1

TR w1

w2 i2

V2

i3

V3 L V4

+

C

R

VT2

uw1 imag 0

i3

t

DT

T

ILm I3

0

t

Fig. 3.29. Forward DC-DC insulated converters: with the winding of magnetizing current discharge (a), double-switch circuit with discharge to network (b), diagrams of the processes in b circuit (c)

Circuit (Fig. 3.29,b) should be considered typical one. After both VT1 and VT2 are simultaneously switched off, the magnetizing current is discharged through the diodes V1 and V2 to the supply network. As the average value of the transformer winding is zero, at equal value of the direct and discharging voltage the time interval of the discharge should be equal to the time interval DT of the direct connection to the supply source, where D is the duty ratio of the both switches. It means that value D cannot exceed 0.5 and current of the load inductor L is discontinuous. 197

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3rd part When both switches are on, during the time interval DT a voltage signal at amplitude U1 / n is formed in the secondary winding, where n is a factor of transforming — the ratio of number of turns of the primary winding to the secondary one w1 / w2. This voltage pulse results in the increase of the current in the inductor L: L

I Lm U1 = −U ld. (3-108) DT n

Taking into account that the average value of the inductor current is equal to load current Ild, the following can be stated: t I I ld = Lm D + d , (3-109) T 2 where td is the time of the inductor current decrease to zero after the turn-off process of the switch that depends on the load voltage: td =

LI Lm . (3-110) U ld

It results in U ld =

(

U1D −DR + D 2 R2 + 8RLf 4 Lnf

). (3-111)

Example The supply voltage is 50 V, load resistance is 200 Ω. Calculate the load voltage and amplitude of the current in load inductance for a double-switch forward circuit if the frequency of switching is 5 kHz, D = 0.3, inductance of the load is 5 mH, factor of transformation is n = 2. 1. The load voltage U ld =

(

0.3 ⋅ 50 −0.3 ⋅ 200 + 0.32 2002 + 8 ⋅ 200 ⋅ 5 ⋅10−3 ⋅ 5000 4 ⋅ 5 ⋅10−3 ⋅ 2 ⋅ 5000

) = 11.16 V.

2. Amplitude of the current in inductor U1 −U D ld n (0.5 ⋅ 50 −11.16) ⋅ 0.3 = I Lm = = 0.1166 A. Lf 5 ⋅10−3 ⋅ 5000 The computer modelling gives the following results: U ld = 11.11 V and I Lm = 0.16 A.

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3rd part The example of the circuit of the second type — with energy discharge to the load — is represented in Fig. 3.30,a. When transistor VT is on, the current iT1 in the energy transmission part of the transformer primary winding and i2 in the secondary win ding are equal to zero, but the current in the magnetizing inductance Lm of the transformer iL is increasing to a value (Fig. 3.30,c) ∆I L =

U1t1 U1DT , (3-112) = Lm Lm

where t1 = DT is the time of the on-state of transistor in the period T, D — the duty ratio. V1

TR

a +

U1

w1 uw1

w2 +

V1

b C

Uld

+

w3 w1 iL Lm

U1

R

i2

+

C

Uld R

iVT

VT VT

–

–

c

w2

T

VT iL uw1

U1

DT

ILm

t

t2

0

iVT 0 –i2

t Uld

ILm

ILm

w1 w2

w1 w2

0

t

t

Fig. 3.30. Circuit with the energy discharge to the load — fly-back (a), circuit (b), diagrams (c)

When the transistor is turned off, the current of the magnetizing inductance is closed through the element transforming the energy of the primary win ding, i.e. iT 1 = −iL = −i2w2 / w1, but voltage across the magnetizing inductance is uL = −U ld w1 / w2.

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3rd part As the average voltage across the inductance within the period is zero, U ld =

U1t1w2 U1DTw2 , (3-113) = t 2w1 t 2w1

where t2 is possible time of the inductance current decrease. Figure 3.30,c illustrates the diagrams of the discontinuous current of magnetizing inductance Lm for the case when the time of the current decrease t2 is smaller than the off-state time of the transistor (T − t1 ) = T (1− D ) and ∆I L = I Lm . The load power depends on the energy collected in inductance Lm and the commutation frequency: Pld = WL f , (3-114) where: WL =

L1 I L2m L1U12 D 2T 2 U12 D 2T 2 = = . (3-115) 2 2Lm 2L2m

Pld =

U ld2 , R

As

then

and

U12 D 2T U ld2 = , 2Lm R U ld = DU1

RT . (3-116) 2Lm

In the case of continuous current of the magnetization induction iL , time t2 is equal to T (1− D ), and then from (3-113) the voltage of the load is U ld =

DU1w2 . (3-117) (1− D)w1

The mode of the discontinuous current will start when t2 is equal to (1− Db )T , but ILm is calculated in accordance with (3-112). Then 1 − Db =

w2 w1

2Lm f , 0 < Db < 1. (3-118) R

It is obvious that the mode of the continuous current of the magnetizing inductor can take place at high enough commutation frequency and high magnetizing inductance. If, for example, w2 = w1, R = 10 Ω, Lm = 500 mH , then at Db = 0.5 the frequency should be 2.5 kHz. For this mode, at the supply voltage U1 = 10 V the load power is 10 W.

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3rd part

Example The transistor commutation frequency and duty ratio of the insulated fly-back type DC-DC converter supply source are f = 5 kHz and D = 0.2, respectively. The supply voltage is U1 = 50 V, the number of turns of the primary and secondary windings of the transformer w1 = 100, w2 = 50 , the magnetizing inductance of the primary win ding Lm = 500 µH. Load resistance is R = 10 Ω. Calculate the voltage and power of the load. 1. At first, the mode of the current in the circuit of magnetizing inductance should be examined whether it is discontinuous or not. If w2 2Lm f > 1 − D, R w1 then the current is continuous.

In the task

50 2 ⋅ 500 ⋅ 5 ⋅103 = 0.35 < (1− D) = 0.8. 100 106 ⋅10 Therefore, the current is discontinuous.

2. For this case U ld = DU1

R 10 ⋅106 = 14.2 V . = 10 2Lm f 2 ⋅ 500 ⋅ 5 ⋅103

3. Load power U2 Pld = ld = 20.2 W. R The average value of the current consumed from the supply source at continuous current U2 I1 = ld , (3-119) RU1 but the average value of the load current is DU1w2 I ld = . (3-120) (1− D)w1R

3.1.13. Calculation of the DC Smoothing Inductor Inductor is an important element of DC-DC converters and its available data and parameters are limited. That is why sometimes special inductors produced for particular case are required; in these cases a methodology for the calculation of the smoothing inductor parameters should be applied. A simplified calculation approach may be used and it is further considered. 201

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3rd part The parameters of the smoothing inductor are calculated in accordance with the following expression: L=

Φw Bs Sw = , (3-121) Im Im

where Bs is a magnetic flux density of the ferromagnetic core material saturation (measure unit for density of the magnetic flux T = V ⋅ s/m2), S — the cross-section area of the core within the zone associated with the winding, w — a number of turns of the winding, Im — maximum instantaneous current of the inductor. As I m w is usually high enough that the core is saturated, in the cores of these inductors an air gap is formed with total length d calculated as follows: I m w I m wµ 0 ≈ , (3-122) Hδ Bs where µ 0 = 4π ⋅10−7 Ω⋅ s/m. δ=

If the inductor is operating at a comparatively low frequency (no higher than some kHz) in the current’s ripple zone, then the core can be made of iron with Bs = 1.2 T. The core can be selected from the catalogue or self-produced. If the frequency is higher, then the materials for the core are either iron powder with Bs = 1.35 T or ferrite core from a catalogue with Bs = 0.25 T . As the calculation methodology a iron core can be considered, the geometry of which can be self-formed and a base of the construction is an armoured type of core. The goemetry of the core is assumed to meet the data in Fig. 3.31. Thus, the average crosssection of the core is S = 2a2 ; area of the magnetic conductor aperture is 4a2. This area is filled with the current conductors by 35 % to 40 % only; therefore, the possible number of turns 4a2 ⋅ 0.35 4a2 ⋅ 0.35 ⋅ j . (3-123) = Sv Ir Ir is the rated rms value of the inductor current (heating) (I r < I m ), j — calculated current density, which approximately can be assumed as 2 ×106 A/m2. Then w=

Bs ⋅ 2a2 ⋅ 4a2 ⋅ 2 ⋅106 ⋅ 0.35 , (3-124) Im Ir geometric parameter a is L=

a= 4

LI r I m

8 Bs ⋅ 0.35 ⋅ 2 ⋅106

. (3-125)

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3rd part

a/2 4a

a

a

2a

Fig. 3.31. Parameters of the core of the DC smoothing inductor

The further calculations are those of number of turns, resistance of the inductor, power losses in the inductor, its cooling area and specific power losses of the cooling area. If these specific power losses are lower than 1200 W/m2, then the density of the current can be increased over 2 ×106 A/m2; if they are higher — the current density should be decreased. The calculations of the parameters should be made in both cases.

Example Calculate parameters of a smoothing inductor of low frequency with inductance L = 5 mH and I r = 16 A , I m = 30 A. 1. Width of the active part of the core 5 ⋅10−3 ⋅16 ⋅ 30 = 2.44 ⋅10−2 m. 6 8 ⋅1.2 ⋅ 0.35 ⋅ 2 ⋅10 2. Number of turns of the inductor a= 4

4 ⋅ 2.442 ⋅10−4 ⋅ 0.35 ⋅ 2 ⋅106 = 104 turns. 16 3. Resistance of the inductor (copper wire with specific resistance 0.02 ⋅10−6 Ω ⋅ m) w=

Rinduc = w ⋅ 0.02 ⋅10−6

l ⋅ 2 ⋅106 = 0.063 Ω. Ir

The average length of a turn of the inductor l = 10a.

4. Losses in the inductor ∆P = 162 ⋅ 0.063 = 16.24 W. 5. Cooling area of the inductor Scool = 4 ⋅14a2 = 0.032 m2. 203

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3rd part 6. Specific power of losses of the cooling area q = 487 W/m2. 7. The conclusion is that the thermal load of the inductor is low (the loaded inductor has q = 1200 W/m2) and j can be increased by 10 % to 15 %. 8. Mass of the inductor, mass of iron (density 7600 kg/m3) M Fe = [2a 2 ⋅ 4a + a2 (9a + 7a)]7600 = 2.65 kg;

mass of copper (8900 kg/m3) MCu = wl

Ir 8900 = 1.8 kg , j

total mass is M = 4.46 kg.

9. The necessary air gap is δ=

30 ⋅104 ⋅ 4π ⋅10−7 = 3.26 mm. 1. 2

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Review Questions

3rd part 1. How can the controlled electronic switch reduce the value of load DC voltage in accordance with DC supply source?

2. What is the switching frequency of an electronic switch and what is duty ratio D? 3. What is «Pulse Width Modulation»? 4. Why is an inductor connected into a load circuit of buck converter? 5. In what way do the ripples of load current of a buck converter depend on switching frequency, duty ratio D and inductance of the load?

6. What is the ratio of supply source active power and load active power for buck type converters?

7. Why are LC filters used in the circuits of supply source of buck converters and what is their purpose?

8. How large is the average voltage across a reactor of DC-DC converter? Haw large is average current flowing through a capacitor of DC-DC converter?

9. What is the dependence of buck converter input filter supply voltage ripples on

the parameters of control system and how does the level of ripples influence the ripples of the network current?

10. What does a notion «discontinuous load current» mean and can the current of active-inductive load be discontinuous?

11. What element should the load circuit contain to get a discontinuous current? 12. What are the instant values of load voltage during the period of armature current zero pause in case if DC motor with separate excitation is used as a load?

13. What gives an opportunity to exclude of load current discontinuous mode? 14. What effect does DC-DC converter with several phases give? 15. What is the ratio of multi-phase DC-DC converter source and load smoothing filter operation frequency to the frequency of electronic switch?

16. At what position to the source should the switch be inserted to get an increase of load voltage according to DC source? What element should be connected to the source circuit?

17. Do the boost DC converters require source current smoothing filters? 18. How large are the average voltage and average current of the boost converter switch?

19. Is it possible to increase infinitely the load voltage of boost converter? Reason? 20. Is it possible to increase infinitely the load voltage of the buck-boost converter? 21. What value of duty ratio D of buck-boost converter gives a load voltage equal to that of the source?

22. What is the polarity of the load voltage according to that of the source in the conventional buck-boost converter?

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Review Questions

3rd part 23. Does the conventional buck-boost converter require the connection of smoothing capacitor in parallel with the load?

24. What is the difference of the source and load circuits of the conventional and Cuk buck-boost converters?

25. What buck-boost circuit has harder operation mode of the same power transistor — conventional or Cuk — and why?

26. What is connected in parallel with the load in Cuk buck-boost circuit? 27. Is Cuk buck-boost converter characterized with the polarity of load voltage reverse to that of the source?

28. Is it possible to implement a buck-boost circuit with the polarity of the load voltage coordinated with the source? What is this circuit?

29. Does SEPIC buck-boost circuit require a smoothing filter of the source current and smoothing capacitor for the load voltage?

30. What is the ratio of the average value of load by-pass inductor current to the load current in SEPIC buck-boost converter?

31. What is two-quadrant 2Q DC-DC converter and how is it applied in the control of DC motor speed?

32. What connection of the semiconductor switches should be used for reversing of load voltage polarity?

33. At what part of supply voltage do the bridge-type and half-bridge circuits operate according to the load?

34. What is the reason to by-pass each s/c switch of unidirectional conductivity with a reverse diode?

35. How to control the diagonal transistors of a full bridge DC-DC converter to

obtain single polarity load voltage? Does a half-bridge circuit allow for singlepolarity modulation?

36. Is the energy accumulated in the load inductance in the case of single-polarity modulation «discharged» through the bridge diodes to the source?

37. What is the double-polarity coordinated modulation of a reversible DC-DC converter and how is it implemented?

38. What does the average value of the load voltage of coordinated double-polarity modulation in reversible pulse converter depend on?

39. How large is the average value of load voltage in the case of double-polarity

modulation of reversible pulse converter if within the modulation interval the transistor switches of both diagonals are switched alternately and for same time intervals?

40. Within what range can the control DC voltage signal be changed and what is

206

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Review Questions

3rd part the reaction of the load voltage average value in each case if the reversible pulse converter is controlled with the help of double-polarity symmetric high frequency signal of saw-tooth form?

41. What are typical systems with the load isolated from the source by transformer? 42. In what switch case the energy from the source is transmitted to the load in the isolated supply source of forward type?

43. What is the maximum duty factor of turn-on process of the switch if the energy of transformer primary winding in a forward type converter is reversed to the source after the turn-off process of the switch?

44. In what switch case is the energy from the source transmitted to the load in the isolated supply source of fly-back type?

45. What element of fly-back transformer accumulates electromagnetic energy with a switch turned on?

46. What is the current of the fly-back transformer magnetizing inductance — dis-

continuous or continuous — if the inductance is large and switching frequency is high?

47. What parameters of the system does the load voltage depend on if fly-back magnetizing current is continuous?

48. What is the description of the transistor switching process in the switch mode

with the help of trajectory? What is the load connection to investigate the transistor switching process?

49. Why do the transistor switching processes — either switching on or off — have two stages?

50. How large are the time intervals of current rise and fall for the transistors of different types?

51. What parameter characterises the transistor switching process from the power perspective?

52. What technical solution is used to improve a transistor switching trajectory while switching it off?

53. What two tasks does an RCD damping circuit solve in the turn-off process of transistor?

54. What are the opportunities to improve the trajectory of turn-on process of transistor?

55. How can the thyristors be applied as switches in DC circuits?

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Tasks

3 part rd

1. The supply voltage of DC buck converter system is (200 + mn) V, active resis tance of the load is (5 + m + n) Ω, inductance is 5 mH, commutation frequency f = 200 Hz, duty ratio D = 0.5. Is the load current continuous?

Answer: at m = n = 0 the load current is discontinuous. 2. The conditions of the 1st task for inductance are increased to 20 mH. Deter-

mine whether the load current curve is continuous and calculate the average value of the load current as well as full ripple of the current.

Answer: at m = n = 0 the current is continuous, the average value is 20 A and ripples are 12.5 A.

3. How large should the load inductance be to obtain the load current ripple 10 % of this current average value under the conditions mentioned before?

Answer: at m = n = 0 inductance is 125 mH. 4. Under the conditions of the 2nd task calculate the average and rms values of

the supply source current if the load current linearly rises in time with the switch on.

Answer: at m = n = 0 10 A and 14.37 A, respectively. 5. The supply voltage of boost converter in idle-run is (100 + mn) V, inside resistance of the supply source is 0.05 Ω. Calculate the load current if the load resistance is 10 Ω and duty ratio of the switch is D = 0.6. The network current is smoothed.

Answer: at mn = 0 the current of the load is 24.2 A. 6. To smooth the DC supply current inductor L of the filter is connected in series with the source, but in parallel with the buck converter input — capacitor of the filter C. Calculate values of L and C if the load current is (200 + mn) A, the voltage of the supply is (600 + mn) V, commutation frequency is 300 Hz, duty ratio is 0.5 and maximum possible source current and input voltage ripple is 10 % of the average values.

Answer: at mn = 0 L = 2.5 mH, C = 2777 mF. 7. To smooth buck converter load current, it is connected in parallel to the filter

capacitor C and in series — to the filter inductor L. Calculate the necessary L and C if the load resistance is 5 Ω , supply voltage — (300 + mn) V, commutation frequency is 300 Hz and duty ratio is 0.5. The current of the inductor should be with 20 % ripples, the full ripple of the capacitor voltage — 10 % of the load average voltage.

Answer: at mn = 0 L = 41.66 mH, C = 166.6 mF. 8. DC electric motor with separate excitation at magnetic constant

ce φ = 0.1 V ⋅ min is supplied from the buck converter from network supply voltage 208

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Tasks

3 part rd

(200 + mn) V. Commutation frequency of the converter is 200 Hz, duty ratio of the switch is 0.5. The inductance of the motor armature is 10 mH, resistance is 1 Ω. Calculate the speed of the shaft rotation and armature current for the bordercase of its continuous mode. Linearise the change of the armature current at both stages of the switching period.

Answer: at mn = 0 speed is 875 1/min, current – 12.5 A. 10. The supply voltage of a conventional buck-boost converter is (100 + mn) V, the

inside resistance of the source can be neglected. The frequency of the switch operation is 2 kHz, resistance of the load R = 10 Ω. Calculate the average value of the load current, average value of the network current, average value of the current of storage inductor L and its inductance necessary to obtain full ripple of the current up to 10 % of its average value for the cases of duty ratio D = 0.25 and D = 0.75. Calculate the amplitude of the transistor current and capacity of the load for the above-mentioned condition of the voltage ripple level.

Answer: at mn = 0 and D = 0.25 Ild = 3.33 A (at D = 0.75 30 A); I1 = 1.109 A (90 A);

IL = 4.439 A (120 A); L0.25 = 28.2 mH (3.125 mH); IVTm = 4.31 A (126 A); C0.25 = 375 mF (125 mF).

11. The supply voltage of a buck-boost Cuk converter is (100 + mn) V, the inside

resistance of the source can be neglected. The frequency of the switch operation is 2 kHz, resistance of the load R = 10 Ω, duty ratio D = 0.25 and D = 0.75. For both these cases calculate the necessary inductance of the source inductor so that full ripple of the current is up to 10 % of its average value, average value of the capacitor voltage and its capacity at similar value of ripples of its voltage, amplitude of the transistor current and inductance of the load inductor for the above-mentioned condition of the current ripple level.

Answer: at mn = 0 Ls0.25 = 112.7 mH (4.17 mH); UC0.25 = 133.2 V (400 V); C0.25 = 31.22 mF (281.25 mF); Lld0.25 = 37.5 mH (12.5 mH); IVT0.25 = 4.31 A (126 A).

12. The supply voltage of a buck-boost SEPIC converter is (100 + mn) V, the in-

side resistance of the source can be neglected. The frequency of the switch operation is 2 kHz, resistance of the load R = 10 Ω, duty ratio D = 0.25 and D = 0.75. For both cases calculate the voltage of the source inductor and load, inductance of the source inductor necessary to maintain the current ripple at the level of 10 % of its average value, average value of the capacitor, capacity of the load capacitor under the abovementioned condition of the voltage ripple level, amplitude of the transistor current.

Answer: at mn = 0 Ls0.25 = 112.7 mH (4.17 mH); Lf0.25 = 37.5 mH (12.5 mH); UC0.25 = UC0.75 = 100 V; C0.25 = 41.6 mF (1125 mF); C ld0.25 = 125 mF (375 mF); IVT0.25 = 4.31 A (126 A).

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Tasks

3 part rd

13. Isolated DC-DC converter with energy transmission to the load with trans-

former primary winding disconnection from the supply source (fly-back) at 30 V voltage operates with load resistance R = 10 Ω, bypassed by high capacity. The number of turns is the same in both primary and secondary windings of the transformer with the inductivity (300 + mn) mH. Commutation frequency of the transistor switch is 5 kHz, on-state time is 60 ms per period. How large is the load voltage?

Answer: at mn = 0 magnetizing inductivity current is discontinuous and load voltage is 16.4 V.

14. Isolated DC-DC converter with energy transmission to the load with trans-

former primary winding connection to the supply source and discharge to the source (forward two-switch circuit) with voltage (30 + m + n) V operates with load resistance R = 10 Ω, connected through LC filter with L = 5 mH. The number of turns is the same in both primary and secondary windings of the transformer. The commutation frequency of the transistor switch is 5 kHz , on-state time is 60 ms per period. How large is the load voltage?

Answer: at m = n = 0, the load voltage is 12.55 V. 15. Full-bridge transistor DC-DC converter with the load of an electric motor of

separate excitation is connected to the supply source with voltage (200 + mn) V. The converter is controlled by means of comparison of double-polarity symmetric sawtooth voltage signal of 15 V amplitude and 2 kHz frequency with DC control voltage signal of positive polarity slowly changed in time of (7 + 0.5m) V. Motor factor ce φ = 0.1 V ⋅ min, resistance of the armature is 1.n Ω, inductivity of the armature is 10 mH, current of the load is 15 A. How large is the speed of motor shaft rotation and full ripple of the motor current? Calculate current amplitude of the operating transistors, the average value of the diode current and average value of the supply source current.

Answer: at m = n = 0 speed is 783 1/min, ripple of the current is 3.91 A, amplitude of the transistor current — 16.95 A, average current of the diodes — 4 A, average current of the source is 7 A.

16. Under the conditions of the previous task, the left leg of the converter is controlled comparing saw-tooth voltage with control DC voltage of (7 + 0.5m) V, but the right leg — comparing the same saw-tooth voltage with DC voltage of negative pola rity, i.e., (–7 – 0.5m) V. If the saw-tooth voltage is higher than the control, lower transistor of the leg is on. Calculate those parameters required in the previous task.

Answer: at m = n = 0 speed is 783 1/min, ripple of the current is 1.24 A, amplitude of the transistor current is 15.62 A, average current of the diodes — 8 A, average current of the source is 7 A. 210

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Tasks

3 part rd

17. Calculate the necessary capacity of the DC-DC thyristor converter commutation capacitor if the supply voltage is (200 + mn) V, load current is 100 A and turn-off time of thyristor is 70 ms.

Answer: at mn = 0 capacity is 35 mF. 18. How large is the minimum load voltage of the thyristor buck-type converter

under the conditions of the previous task if the capacitor is discharged through main thyristor within the circuit with series connected inductor with L = (20 + m + n) µH? Commutation frequency of the converter is 300 Hz, the remaning data are given in task No. 17.

Answer: at m = n = 0 the load voltage is 13.4 V. 19. Calculate switching power and energy loss of the three mode transistors — BT,

MOSFET, IGBT, if commutated at (2 + m + n) kHz current is (50 + m) A, voltage of DC source is (600 + 3.mn) V, load is clamped with an ideal diode, switchings are doublestage — first stages over rise and fall times take places respectively at full source voltage and full load current. The rise times are respectively 1 μs; 20 ns; 50 ns long, but fall times — 3 μs; 100 ns; 400 ns long.

Answer: BT 240 W, 12 mJ; MOSFET — 7.2 W, 360 μJ; IGBT — 27.0 W, 1.35 mJ, if m = n = 0.

20. Calculate admissible meanings of switching frequencies at the parameters given in the task 19, if an admissible power loss is (20 + m + n) W.

Answer: 1.66 kHz, 83.3 kHz, 20 kHz, if m = n = 0.

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3rd part

3.2. Voltage and Current Source Inverters (VSI and CSI) Inverters convert DC voltage or current into AC voltage. Applying the controlled rectifier, it is possible to develop inverter, output of which should be connected to AC conventional grid, and this features the order of electromagnetic processes in inverter itself. Using completely controlled semiconductor switches, it is possible to develop inverters, which convert DC to the independent AC voltage applied to an independent load. The frequency, shape, number of AC voltage phases and other parameters characterizing quality of AC voltage should be accepted arbitrary. Therefore, such inverters are denominated as autonomous. The DC supply voltage should be obtained from conventional AC grid through a rectifier and then such system rectifier-autonomous inverter is called a frequency converter.

3.2.1. Single-Phase Voltage Source Inverters (VSI) Periodically operated switches should be provided for voltage source inverters. For example, in case of two series-connected DC supplies at voltage of 0.5Ud each, the inverter can be built with the zero-point where 2 alternatively operating switches S1 and S2 (Fig. 3.32) are available (i.e. as half-bridge inverter). Within a half-cycle S1 is turned «on» and S2 is turned «off», the active inductive load voltage u takes a conditionally positive polarity; however, within the second half-cycle S2 is turned on and S1 is turned off, it takes a conditionally negative polarity.

a +

i1

Ud/2

i, u

– +

Ud/2

c S1 S2 S1

b

–Ud/2 0

Ud/2 +

Ud/2

VT1

i, u

–

L

R

–

VT2

V1 V2

t

V2 VT2

i

i1

+

t t

0

S2

i2

–

T

Ud/2

u

L

R

0.5T

t1

0.5T

t

V1 VT1 i1 0

t

i2

Fig. 3.32. Simplified single-phase voltage source inverter circuit (a), transistor circuit (b) and diagrams 212

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3rd part As the changes of the active inductive load current lag behind the time of voltage changes, in the beginning, once S1 is turned on, both the load current i and the switch S1 current are negatively polar and only in the second half of the half-cycle they have a positive polarity of load voltage. Similarly, once S2 is turned on and the load voltage carries negative polarity, in the beginning, the load current i tends towards positive direction and the switch S2 current tends towards its negative direction. Again after about half-time of the half-cycle the current changes direction respective to the load voltage polarity. It follows from the foregoing that each inverter voltage switch should feature a bidirectional conductivity. As the controlling semiconductor switch (usually a transistor) features only a unidirectional conductivity (this direction is shown by the arrow element in reference symbols), a diode is connected to the switch in parallel and in opposite direction (see Fig. 3.32). The actual circuit b illustrates that while the load current is still negatively polar after the activation of transistor VT1, it is conducted by the diode V1 till the time instant t1. Transistor VT1 controls the current from the time instant t1 till the end of a positive half-cycle at t = 0.5T . Once the transistor VT2 has been activated, the load current positive direction closes through V2, and only from the point when the load current polarity becomes negative transistor VT2 takes the control (see Fig. 3.32). Subject to the foregoing, it is evident that in order to set up a diode current circuit the voltage inverter supply sources should have bidirectional conductivity, for example, as in batteries. However, actual DC supplies are often conducting one-way, where, for example, the circuit is supplied from a rectifier. To ensure proper inverter operation, a sufficient high-value capacitor is to be connected to the supply source in parallel, which takes control over the source current pulses of opposite polarity. a

b

id

w2 TR

+

VT1 V1 V4

V3

i, u

U1

R VT4

–

w1’

VT3 L VT2

w1’’

+

V2

V1 V2

U1

VT1

VT2

–

Fig. 3.33. Single-phase voltage source inverter bridge circuit (a) and with neutral point transformer circuit (b)

Thanks to the semiconductor switches, the single-phase voltage source inverter can be configured either with a bridge circuit or with neutral point transformer inverter 213

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3rd part circuit (Fig. 3.33). In the first case, 4 semiconductor switches are required (circuit a), in the second case (circuit b) — only two, which are supposed to be rated for a doublevoltage supply source plus transformers. In the bridge-type circuit, the current for one half-cycle is simultaneously conducted by VT1, VT2 (or V1, V2) and for the second half-cycle by VT3, VT4 (or V3, V4). The current of the second circuit is alternatively conducted by VT1 (V1) and VT2 (V2). Both circuits are widely popular. Therefore, their characteristics deserve detailed consideration.

3.2.2. Features of Bridge-Type Single-Phase Voltage Source Inverter Currents and voltages of elements of the circuit (Fig. 3.33,a) vary as shown in the curves (Fig. 3.34) of full-bridge voltage source inverter. To obtain the approximate circuit characterizing equations, it is assumed that the load current creates a constant voltage drop at its positive values across the load resistance, which equals 0.5ImR, but for the negative values of current it equals –0.5ImR, where Im is the load current amplitude value.

a

b u i

id +

iV1

VT1 V1

V3

i, u

U1 V4

iVT1

R VT4

–

0 VT3

L

VT2 V2

Im t1

U1

T/2

–Im Im

iVT1 iVT2 0 iV1 Im iV2 0 id

T

t

–U1

t

t1

V2 0 V1 t1

Im

t

VT1 VT2 t

Fig. 3.34. Diagrams of voltage source inverter load voltage and current of a single-phase bridge-type circuit

Under positive load voltage assumed at the beginning of a half-cycle, the load current i changes according to the differential equation 214

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3rd part U1 = L

di − 0.5I m R, (3-126) dt

but after time instant t1, when the current polarity really becomes positive, according to the equation U1 = L

di + 0.5I m R . (3-127) dt

These equations in each interval have constant derivatives di/dt. In (3-126), the current i changes from the start value –Im to zero, and it takes place in the interval t1 =

LI m . (3-128) U 1 + 0. 5 I m R

Whereas within the time interval (0.5T – t1) the current i increases from zero to Im according to (3-127). The current amplitude value is to be found from the equation Im =

(U1 − 0.5I m R)(0.5T − t1 ) . (3-129) L

By solving the equation, Im =

−8LU1 + 2U1 16L2 + R2T 2

, R2T but taking into account the short-circuit current U1/R as well as introducing the concept «relative time constant τ* = (L / R ) / T», the current amplitude is I m∗ = −8τ∗ + 64τ∗2 + 4 . (3-130) Substituting the value Im* in (3-128), the diode relative control time in the period t1* =

t1 τ* I m* = , (3-131) T 1 + 0.5I m*

where the current and voltage angular displacement is ϕ = 2πt1*. The load voltage rms value in the form of a rectangular curve is equal to its amplitude, i.e., U = U1 or U * = 1. The load current rms value is t1 1 I= 0.5T 0

∫

2 −I + tI m dt + m t 1

0 ,5T −t1

∫ 0

I m2

t2 1 dt = I m 2 3 (0.5T − t1 )

or I* =

IR 1 = I m* . (3-132) U1 3

In turn, based on the source and load power equality, the average value of the power source current is 215

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3rd part Id =

I m2 R . (3-133) 3U1

The derived expressions do not work out in each case because the active inductive load is assumed. Referring to (3-130), it is fair to say that the expressions are workable at τ* = (τ / T ) > 0.25, otherwise Im* will be greater than 1, which is impossible. It can be concluded that the load time constant and the switching period ratio should be sufficiently large. As can be seen from Fig. 3.34, the load current lags in phase behind the load voltage by angle ϕ = ωt1 = 2πt1*. Therefore, the load active power is also approximately determined Pld = I 2 R = UI cos ϕ, (3-134) where U = U1 is the rms value of load voltage, ϕ = 2πft1. This power, disregarding the power loss in switches, is equal to the source input power Pd = U1 I d . Equating the both powers shows that I d = I cos ϕ, (3-135) i.e., whenever the load becomes more inductive the source and load current ratio decreases, but at the pure inductive load I d = 0 A. At a strong approach, (3-134) and (3-135) will be in force only if the real power of load is calculated using fundamental harmonics of current and voltage. However, the expression works out for qualitative relationship evaluation. At the same time, the expression I 2 R = U1 I d is accurate. From here I d* =

I m*2 . (3-136) 3

In order to select elements, it is essential to find out the diode average current: * I Vav

1 = T

t1

∫I 0

m

t dt = 0.5I m* t1*. (3-137) t1

As the output voltage change curve takes a rectangular form with the amplitude U1, as well as is symmetrical against the time axis, to determine the first harmonic (1) only B1m should be calculated: 2π π 4U 1 U1(1)m = B1m = U1 sin ωtdωt − U1 sin ωtdωt = 1 . (3-138) π π π 0

∫

∫

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3rd part This is the amplitude of output voltage fundamental harmonic. Consequently, the amplitude of the load current fundamental harmonic is I1(m ) =

4U1 π R2 + (2πfL)2

,

or applying relative units it is I(*1)m =

4 π 1 + 4 π2 τ * 2

. (3-139)

Example The supply voltage of a single-phase bridge-type voltage source inverter is U1 = 200 V, switching frequency f = 2 kHz, loads R = 5 Ω, L = 5 mH. Determine the rms value of the load current and the average value of the source currrent! 1. The amplitude of the load current wave I m* U1 (−8τ* + 64τ*2 + 4 )U1 = , R R

Im = where τ* = Thereby,

L 5 ⋅ 2 ⋅103 f= = 2. R 103 ⋅ 5

(−16 + 260 )200 = 4.98 A . 5 2. Load rms currrent Im =

I ld = I m

1 = 2.88 A . 3

3. DC source average currrent Id =

I ld 2 R 2.882 ⋅ 5 = = 0.21 A. U1 200

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3rd part 10.00 7.50 5.00 2.50 0.00 –2.50 –5.00 –7.50 200.00 10.00 0.00 –100.00 –200.00

ild, A

uld, V

id, A

10.00 5.00 0.00 –5.00 –10.00 91.00

91.25

91.50

Time (ms)

91.75

92.00

92.25

Fig. 3.35. Diagrams of a single-phase voltage source inverter in reference to the example data calculations

In order to verify the accuracy of the expression-based calculations, a computer-generated model referring to the example data simulation has been developed. Figure 3.35 shows the resulting curves; their parameters comply with the calculated ones.

3.2.3. Features of Neutral Point Transformer Based Single-Phase Inverter In a single-phase neutral point transformer circuit, the currents and voltages change over time as shown in Fig. 3.36. Here, it is assumed that the transformer winding lacks leakage inductance. Transistor VT1 connects to the primary half-winding w1ʹ, the supply DC source, providing the negative polarity output voltage corresponding to the specified initial points of windings, and, on average, the same polarity load current. Transistor VT2 is connecting DC source to the other primary half-winding w1ʹʹ, which provides the positive polarity output voltage and, on average, the load current is tending to the same polarity as well. Although at the starting point the transistor VT1 is activated, it does not conduct the current at once, because at the active inductive load its current lags behind the voltage in phase, its initial value is positive, and according to this polarity the current is conducted through w1ʹ and supply source by the diode V1. In addition, in the diode conducting mode, the initial point of a half-winding w1ʹ is connected to the supply source negative pole, but the winding end point — to the positive pole of supply, which provides negative polarity voltage across the load. Just as the diode V1 current becomes equal to zero, the current turns to be conducted 218

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3rd part by the transistor VT1, the polarity of the supply source current starts to be positive, but polarity of the load current — negative. In the middle of the period T, just as the transistor VT1 is switched off, the transistor VT2 is activated. In the start instant, the transformer secondary winding current carries its negative polarity and respectively can be conducted only by the diode V2 through the half-winding w1ʹʹ. Since in the start instant this half-winding initial point is connected to the positive terminal of the supply source, but its end point — via diode V2 — to the negative pole, the polarity of the output voltage is positive, which makes the load current change in positive direction. Once the current becomes positive, it turns to be conducted by VT2. Thus, alternately conducting (VT1-V2-VT2-V1-VT1) the rectangular-type voltage of the active inductive load circuit is created, as well as the respective load current. The source and the load current and voltage ratio corresponds to the transformer ratio between the number of turns of the primary half-winding and the number of turns of the secondary winding w2. The number of turns of both primary half-windings is equal and, in general, to reduce leakage inductance both windings should be wound with parallel wires. In this circuit, each transistor is exposed to, at least, double supply voltage effect. If controlled by VT1, the voltage of the half-winding w1ʹ transforms to the half-winding w1ʹʹ and the latter voltage is added to the source voltage, creating a double voltage to the turned-off VT2.

a

b VT1 VT2

L

R w2

i, u

i Im u

TR w1’ i1’

id +

0 V2

V1 VT1

U1 –

w1’’ i1’’

iV1

iVT1

VT2 iVT2

iVT1

t1

U1 T/2

t t

w2 w1¢¢

T

t

–Im w Im 2 w1¢

0 iV1

t

0

t

Fig. 3.36. Single-phase neutral point transformer inverter circuit and signal diagrams

To calculate parameters of the transformer, it is essential to know rms half-winding current. Let us take a look at the curve i1ʹ of the current of the half-winding w1ʹ 219

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3rd part (Fig. 3.36): from the starting point till the point t1 the current carries a negative polarity, which varies from −I m w2 / w1 ′ at t = 0 s to zero at t = t1; from the moment t1 the current increases from zero to I m w2 / w1 ¢, where Im is the amplitude value of the load current. Thus, the half-winding rms current is t1 1 I1 ′ = T 0

∫

0.5T 2 2 w2t w 1 dt = I m 2 dt + I m . (3-140) w1 ′(0.5T − t1 ) w1 ′ 6 1 t1

−I w2 + I w2t m w ′ m w ′t 1

1

∫

The circuit values Im, t1, Id and other parameters are derived from the same expressions applied to the bridge circuit, only here the base short-circuit is U1 (w2 / w1 ′) / R .

3.2.4. Single-Phase Voltage Source Inverter Control Very often it is necessary to regulate the load voltage at the inverter output. For this purpose, the input voltage control can be arranged. However, the method requires an additional control device. Therefore, the method are used at which U1 = const. On such terms, there are two feasible approaches: • pulse width half-wave adjustment; • pulse width modulation. The first approach may be implemented in two ways: • changing the actual pulse width with the help of the control circuit (Fig. 3.37a); • shifting in phase the conducting periods of the opposite bridge circuit legs (Fig. 3.37b). According to the first approach, by setting the delay time between the switch-off of one transistor pair and switch-on of the other pair and changing the delay time tv in the half-wave period, the interval (t v − t 2 ) is achieved, at which voltage on the load u and current i is zero. Once t v = t 2 , the maximum rms value of the output voltage is gained. By increasing tv, the rms value of the output voltage is decreased. According to the second approach, by changing the shift time tv of the switching moment of the bridge-type circuit legs S1, S4 and S3, S2, the interval is achieved, at which upper or lower transistors of legs are simultaneously switched on and then the instantaneous value of load voltage is zero because the load circuit is short-circuited. Unlike the first approach, the load current within this interval keeps almost constant amplitude value Im (Fig. 3.37,b). Whenever t v = 0.5T , the maximum rms value of the output voltage is gained. By decreasing tv, the rms value of voltage is also decreased: 2 U= T

tv

∫ U dt = U 0

2 1

1

2t v . (3-141) T

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3rd part a

u i 0

tv

Im 0.5T

tv

U1

S1 0

T

tv

b u10

t2 i 0

–U1

0.5T

0.5T

t u20

Im

S3

0 S2 t

t2

u i

0.5T – tv

t

U1/2

tv

0.5T

T

S4

Im

S2

t

U1 I1

+

S1 1

U1

S4

–

i, u L

R

S3

0

S2

t

t1

2 –I1

–Im

Fig. 3.37. Voltage control by means of dead-time (a) and switch leg operation shift (b)

Analysing the load voltage harmonious composition, the amplitude of the first harmonic is determined 2U U (1)m = 1 2(1− cos ωt v ) , (3-142) π but the amplitude of the current components generated by the voltage of the first harmonic, which is very close to the actual load current amplitude, is I(1)m ≈ I m =

2U1 π

2(1− cos ωt v ) R2 + X L2

. (3-143)

Here ω = 2πf , X L = 2πfL. Applying a similar method as in Section 3.2.2 as well as assuming that over the shor tening load current is i = I m , one can find the approximate expression to determine the time interval t1 and the load current amplitude Im: t1* =

−(2τ * + t v* ) + 4τ*2 + 8τ *t v* + t v*2

I m* =

2 −(2τ* + t v* ) + 4τ*2 + 8τ*t v* + t v*2 4τ* + t v* − 4τ*2 + 8τ*t v* + t v*2

, . (3-144) 221

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3rd part Figure 3.38 shows the curves presenting the rms value of load voltage and the first harmonic amplitude, as well as dependence of the amplitudes of load current and its first harmonic on the shift relative time t v * = t v / T at the load relative time constant t * = Lf / R = 1. As can be seen from Fig. 3.38, by decreasing the shift time from a half-wave (t v * = 0.5) to zero, one can smoothly decrease the load voltage and current.

a

U*

U(1)m*

U(1)m 1.0 *

U*

0.5

0

b

Im*

Im*

I(1)m* 0.2

I(1)m*

0.1

0.1

0.2

0.3

0.4

0.5

tv*

0

0.1

0.2

0.3

0.4

0.5

tv*

Fig. 3.38. Estimated load voltage rms U* and its first harmonic U(1)m* amplitude as well as dependence of the load current amplitude Im* and the amplitude of its first harmonic on tv* in the circuit with a switch leg shift

A similar expression can also be derived for a single-phase bridge-type inverter through the application of the first control approach. Only here by increasing tv* to 0.5, rms values of voltage and current are decreased. The sinusoidal high-frequency pulse width modulation has become widely popular (PWM — Pulse Width Modulation), which allows obtaining practically sinusoidal current on load. The PWM method can be well explained by studying the half-bridge inverter circuit (Fig. 3.39). Control of switches S1 and S2 is arranged mutually comparing the high-frequency bipolar reference saw-tooth voltage ust with the sine control voltage uR = U Rm sinωt required for output voltage fundamental wave frequency and amplitude, where ω = 2πf is the inverter AC output current angular frequency. As the instantaneous value of the saw-tooth voltage exceeds the value of control voltage curve uR, the switch S2 is turned on and the negative polarity voltage –0.5U1 is supplied to the load. As the instantaneous value of the saw-tooth voltage is less than the value of curve uR, the switch S1 is turned on and the positive polarity voltage 0.5U1 is supplied to the load. In the control voltage half-wave period 0.5TR the N switch-over cases of switches takes place: 0.5TR N= Tst 222

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3rd part where time length of switch S2 in the on-state is U sin ωt , (3-145) t− = 0.5Tst 1− Rm U stm

but the time length of switch S1 in the on-state is U sin ωt . (3-146) t + = 0.5Tst 1 + Rm U stm

a

b

+

S1

U1 0

i, u

–

S2

Tst

T/2 URm

Ustm

ust

0 uR

c +

VT1

V1

u

V2

i VT2

–

U1/2 Uavst

0

U1 0

t

t –U1/2

Fig. 3.39. Pulse width sine-waveform modulation in half-bridge circuits and diagrams

Here, Tst << TR is the saw-tooth voltage repeating period, but Ustm is the positive value of the saw-tooth voltage amplitude. The average value of the load voltage within the time of one saw-tooth voltage period is U avst = 0.5U1

t+ U sin ωt t − 0.5U1 − = 0.5U1 Rm , (3-147) U stm Tst Tst

where t values are obtained discretely for every Tst. As shown, the average value of the load voltage within the time of the teeth-form voltage period changes in accordance with the change in time uR, i.e. in a quasi-sinusoidal manner and with the frequency f = 1/ TR . In addition, the resulted average voltage values configure the sinusoidal curve of the output voltage fundamental harmonic, i.e., uld =

0.5U1U Rm sin ωt = U m sin ωt . (3-148) U stm

This load voltage configures the load current ild = I m sin(ωt − ϕ), (3-149)

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3rd part the amplitude value of which is Im =

0.5U1U Rm U stm R2 + X L2

, (3-150)

but the phase shift with regard to the control voltage wave depends on R and XL: ϕ = arctg

ωL . R

The real load current curve is only approximately sinusoidal, but actually includes ripples around the sinusoidal principal curve. The higher modulation frequency is, the smaller background ripples are. In modern systems equipped with the IGB transistor switches, the modulation frequency for the rated 50 Hz output voltage frequency is approximately 10 kHz, which provides about 200 over-switches for output voltage switching within a half-wave length. As already mentioned, the switches S1, S2 shown in Fig. 3.39 configure the opposite parallel circuit with the help of transistor and diode (Fig. 3.39,c). Once the transistor VT1 is closed, which conducted at i > 0 A, the load current of this direction goes to the diode V2 and is closed through the diode and a supply source. Thus, under positive load current polarity the current change is controlled by the transistor VT1 and diode V2. Under negative load current polarity, the current change is controlled by VT2 and diode V1.

Example The single-phase bridge-type inverter voltage is regulated with the switching shift period t v = 0.3T of the switch legs. The inverter supply DC voltage U1 = 200 V, switching frequency f = 1 kHz, load resistance R = 5 Ω, L = 1 mH. Determine the load voltage and the amplitude and rms value of the current fundamental harmonic, as well as the current amplitude! 1. The amplitude of the load voltage fundamental harmonic is 2U1 400 2(1− cos 2πft v ) = 2(1− cos 2π ⋅ 0.3) = 206.12 V ; π π the rms value of this fundamental harmonic is U (1)m =

U (1)ef =

U (1)m

= 145.5 V . 2 2. The amplitude of the load current fundamental harmonic is I(1)m =

U (1)m R2 + (2πfL)2

=

206.12 25 + 6.282

= 25.67 A;

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3rd part

the rms value of this fundamental harmonic is I(1)rms =

I(1)m

= 18.07 A. 2 3. Load current amplitude at t v* = 0.3 and t * = 0.2 −(2τ* + t v* ) + 4τ*2 + 8τ*t v* + t v*2 U1 Im = ⋅ = 25.14 A. R 4 τ * + t * − 4 τ * 2 + 8 τ * t * + t *2 v

v

v

To verify how the inverter leg switching shift influences the control parameters, a computer-generated model referring to the calculation of the circuit parameters given in the example has been developed. Figure 3.40 illustrates the load current curve resulted in a computer generated simulation. As can be seen, the current curve is similar to that shown in Fig. 3.40, but the current amplitude is only a few percent greater than the calculated one. It proves the equation quality of the calculation method for control approach.

30.00

ild, A

20.00 10.00 0.00 –10.00 –20.00 –30.00 5.0

5.50

6.00 Time (ms)

6.50

7.00

Fig. 3.40. Load current curve of a single-phase inverter with the leg shift control referring to the calculated example data

3.2.5. PWM-Regulated Single-Phase Full-Bridge Voltage Source Inverter The bridge-type voltage inverter may be regulated by the sinusoidal PWM method in two ways: by simultaneous control of the diagonal transistor pairs VT1, VT2 and then VT3, VT4 (Fig. 3.41,a) and by the opposite phase control of bridge both legs (Fig.3.41b). In the first case, there is only one sinusoidal control voltage uR and the 225

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3rd part bipolar modulation is implemented: when the instantaneous value of the saw-tooth voltage ust exceeds the value of the sinusoidal control voltage uR, the switches S3 (VT3, V3) and S4 (VT4, V4) are turned on; when the instantaneous value of the saw-tooth voltage becomes less than the value uR, the switches S1 (VT1, V1) and S2 (VT2, V2) are turned on. Once the instantaneous value of the saw-tooth voltage is less than value uR, the switches S1 (VT1, V1) and S2 (VT2, V2) are activated. In case of the active inductive load, only transistors VT1, VT2 are actually switched on at the positive polarity uR and the same polarity load current i, but once both of them are simultaneously switched off, the current control is taken by the diodes V3 and V4. The inverter output to the loads receives bipolar modulated voltage at the amplitudes +U1 (through the transistors) and –U1 (through the diodes), where the average value of voltage over modulation intervals roughly corresponds to the ratio of the control voltage uR and the saw-tooth voltage amplitude Ustm: U avst = U1

U Rm sin ωt . (3-140) U stm

As uR carries its negative polarity and the active inductive load current i displays a negative polarity as well, in fact, only transistors VT3, VT4 are switched on, but once they are simultaneously switched off, the load current is controlled by the diodes V1 and V2. Controlled by the diodes, the on-load voltage of positive polarity +U1 then is created, but when the current is controlled by the transistors, there will be the voltage of negative polarity –U1. For opposite phase control of the both bridge legs (Fig. 3.41,b), two counter-phase control voltages uR and –uR are required. At intersection of the control voltage uR with the saw-tooth voltage, control of the transistor leg VT1, VT4 is provided, but at intersection of the opposite phase control voltage –uR — control of the transistor leg VT3, VT2. Once the both transistors connected to the supply source positive (VT1, VT3) or negative pole (VT4, VT2) are simultaneously switched on, the instantaneous value of the load voltage is equal to zero. Once the diagonal transistors VT1, VT2 (when direct control voltage uR carries its positive polarity) or VT3, VT4 (when at positive polarity there is the reverse control voltage –uR) are simultaneously switched on, the instantaneous values of the load voltage are +U1 or –U1, respectively. The average value of the modulation period voltage can be derived from the same expressions (3-140). By this control the unipolar modulation is created.

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3rd part a +

U1

ust

ust uR V1 1 V4

VT3

VT1

i, u VT4 VT2

–

V3 2 V2

0 uR

VT1 VT2 VT3 VT4 u

t U1 t

0

b

ust uR

Tst

t –U1

T/2

Ustm ust –u

R

0 uR VT1 VT4 VT3 VT2 u

t

t

t

U1 ild

0

t

Fig. 3.41. Bipolar (a) and unipolar (b) pulse width sine-waveform modulation

The difference between the two modulation types can be seen in the change of the instantaneous values of the network current id, as well as the elements are loaded. In the case of the bipolar modulation, either positive or negative sign of the input current id is carried. If the load current is described as i = I m sinωt , then the instantaneous value of the switched transistor currents id coincides with the load current absolute value, but, when controlled by the diodes, the current id = − i . In the case of the unipolar modulation, the current id is equal to the load current direct (where uR > 0 V ) or reversed (uR < 0 V ) absolute value only when conducting the two diagonal transistors, but in the absence of their control current id = 0 A. This means that compared to the bipolar modulation there is no load reactive energy exchange with the supply source, but the load reactive energy is suppressed in the switching circuit.

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3rd part

Example A single-phase bridge-type inverter is controlled by the sinusoidal PWM, compared to the bipolar saw-tooth voltage with amplitude U stm = 15 V and frequency f = 1 kHz, with one sinusoidal control voltage uR = 10sin (314t ) or two such voltages formed at the opposite phase. The inverter supply voltage U1 = 300 V, loads R = 5 Ω, L = 10 mH . Determine the rms value of the load current fundamental harmonic and the average value of the supply source current! 1. The amplitude of the load voltage 50 Hz (ω = 314 s−1) fundamental harmonic is U (1)m = U1

U Rm 10 = 300 ⋅ = 200 V . U stm 15

2. The value of the load current fundamental harmonic amplitude is I(1)m =

U (1)m 2

2

R + (ωL)

=

200 25 + 3.142

= 33.87 A;

this current rms value is I I(1)rms = (1)m = 23.85 A . 2 3. The average value of the supply source current is Id = 40.00

2 I(1)rms R

U1

= 9.48 A .

ild, A

20.00 0.00 –20.00 –40.00 40.00

id, A

20.00 0.00 –20.00 –40.00 280.00

290.00

300.00 Time (ms)

310.00

320.00

Fig. 3.42. Curves of load current and supply source current in the computer-generated model referring to the calculated example (in simulation model value Id is 9.8 A)

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3rd part

3.2.6. Three-Phase Voltage Source Inverter Operation Principle DC voltage conversion to three-phase AC voltage can be affected by six bridge-type connected switches S1…S6, which, as follows from the option where voltage is not switch-regulated (rectangular or square version), are usually sequentially turned on for a half-cycle interval, where in each leg switches (for example, S1 and S4) are turned on alternatively (Fig. 3.43). As can be seen from Fig. 3.43, each instant 3 switches are enabled, which connect two phase terminals of the load to one supply voltage U1 pole, terminal of the third phase to the second pole. Thus, in situation 1, when the switches S1, S5, S6 are turned on, phases A and C are connected to the supply source plus terminal, but phase B — to the source minus pole. In situation 2, only phase A is connected to the source plus terminal, but phases B and C — to the source minus terminal. This means that the load interphase (or line) voltage is U AB = U1 in both situations. At the same time in situation 1, when phases A and C are connected to the source positive pole, the line voltage is U CA = 0 V. In the second interval, the phase C terminal is connected to the source negative terminal and U CA = −U1. In the course of the period, there are 6 situations formed, each lasting for T/6. In three of them, each phase terminal is connected to the source positive pole, while in the other three — to the source negative pole. This means that each switch conducts the current for one half-cycle, i.e., the switch conductivity angle is 180°. Line voltage rms value is 2 Ul = T

T /3

∫ U dt = U 2 1

0

1

2 ≈ 0.816U1, (3-152) 3

but the amplitude of this voltage first harmonic is calculated in the following order: 5π/ 3 2 π/3 3U 1 B(1)m = U1 sin ωtdωt − U1 sin ωtdωt = 1 ; π π π 0 5π/ 3 2 π/3 3U1 1 A(1)m = U1 cos ωtdωt − U1 cos ωtdωt = ; π π π 0

∫

∫

∫

∫

U l(1)m = B(21)m + A(21)m =

2 3 U1 = 1.1U1. (3-153) π

If the load has Y-type connection, then the curve of the load phase impedance voltage (or phase voltage) is formed as 1/3 of the source voltage U1 in the situation, where the phase impedance in question is connected in parallel to other phase impedance, and as 2/3 of U1, where phase impedance is not connected in parallel to other phase impedance (Fig. 3.43). As can be seen in Fig. 3.43, there are 4 intervals in the load phase A voltage curve, when voltage is 1/3 of U1 (2 positive, 2 negative polarity), as well as 2 intervals when the voltage is 2/3 of U1. 229

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3rd part a

S1 S2 S3 S4 S5 S6 uAB

T/6

T

id

b +

U1 1

t

2

S1

+

A

C iA

2/3U1

2 t

B A

+

t 2/3U1 1/3U1

iC

U1

–

0

0

C S6 S2

t

uCA

uA

S4

uA 1

0

B

C

t

uBC

uA

A

–

0

S3 S5

uA

U1

–

B

C

Fig. 3.43. Three-phase square-type load voltage source inverter implementation with semiconductor switches

The phase voltage rms value is U ph = U1

2 = 0.471U1, (3-154) 9

and it is 3 times less than the the line voltage rms value. The amplitude of the phase voltage first harmonic is also less than Ul(1)m: 2U 2 Bph(1)m = 1 ; Aph(1)m = 0; U ph(1)m = U1. (3-155) π π Under active inductive load, each phase current first harmonic lags behind the phase voltage first harmonic by angle j: iA(1) = I ph(1)m sin(ωt − ϕ); 2π iB(1) = I ph(1)m sin ωt − − ϕ ; (3-156) 3 4π iC (1) = I ph(1)m sin ωt − − ϕ. 3 In view of the supply source current id, it is clear that it changes the same way in each of 6 intervals and, for example, in interval 1 it is formed as follows:

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3rd part 4π id = iA + iC ≈ I ph(1)m sin(ωt − ϕ) + sin ωt − − ϕ . (3-157) 3 The dependence of the current id instantaneous value on the shift angle ϕ = arctg(ωL / R) can be either positive all the time or bipolar: part of the interval can be positive, but part — negative. If id instantaneous value is only positive, it means that the load switched phase electromagnetic energy is not recovered to the source like it happens in a single-phase circuit, but it is transferred to other load phases. Figure 3.44 illustrates the load phase current and current supply source circuit. a uA 0 iA 0 id 0

b uA

Ud/3 t 2Ud/3

0 iA

t Id t

0

c uA

2Ud/3 Ud/3

0

t

t

iA 0

t

t

id

id Id 0

t

0

t

Id

Fig. 3.44. Diagrams of phase voltage, phase current and network current of 3-phase inverter with the load of Y-type connection: at j less than 60° (a), equal to 60° (b) and greater than 60° (c)

As can be seen, if ϕ = 60°, there is a boundary case, where, for example, at the beginning of interval 1 the instantaneous value of the current iA = I ph(1)m sin(−ϕ) and current −iC = −I ph(1)m sin(−4π / 3 − ϕ) is with the opposite sign and equal, and then id = 0 A. Addressing the expression sin(−ϕ) + sin − 4π − ϕ > 0, 3 one can determine: if ϕ ≤ 60°, then the instantaneous value of the current id will always be greater than zero (as clearly seen in Fig. 3.44,a). In turn, if ϕ > 60°, i.e., the ratio of the load phase active and apparent power is less than 0.5, then the switched phase reactive current returns partially to the network (Fig. 3.44,c). One part of the switched phase reactive current deviates to other phase, but the other deviates to the supply source. In case of a unidirectional conducting rectifier type supply source, the input capacitor shall be provided in the circuit to capture negative-orientated current id . To verify nature and form of the circuit current, a computer-generated model of the circuit parameters has been developed, which shows the agreement with the calculated 231

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3rd part example. The curve of phase A current and the curve of the supply source current are shown in Fig. 3.45. As can be seen, the phase current form is truly quite close to a sine wave, and because the shift angle between the current and voltage at the parameters applied is about 70°, the current id is bipolar.

Example The supply voltage of the three-phase voltage source inverter with square-wave operation without modulation is U1 = 513 V, Y-type load phases R = 5 Ω, L = 5 mH. Determine the rms values of the phase voltage and current, the average value of the supply source current and this current instantaneous value at the beginning and end of onesixth of the specific period, if the switching frequency is 500 Hz! 1. RMS value of phase voltage U ph = 0.471U1 = 241.62 V . 2. RMS value of phase voltage fundamental harmonic 2U1 U ph(1) = = 230.1 V. π 2 3. Current rms value U ph I ph = = 14.66 A ; 2 R + (2πfL)2

fundamental harmonic rms value I ph(1) = 13.97 A ;

this current amplitude I ph(1)m = 19.84 A.

4. Average value of supply source current 2 3I ph R 3 ⋅14.662 ⋅ 5 Id = = = 6.28 A . 513 U1 5. Load current shift angle 2πfL ϕ = arctg = 72.33° . R 6. Instantaneous value of supply source current at the beginning of one-sixth of the period 4π idi = I ph(1)m sin(−ϕ) + sin − − ϕ = 19.84 ⋅ (−0.2135) = −4.24 A . 3 7. Instantaneous value of supply source current at the end of one-sixth of the period π π 4π ide = I ph(1)m sin − ϕ + sin − − ϕ = 19.84 ⋅ 0.9528 = 14.67 A. 3 3 3 232

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3rd part From calculation it is evident that in the square-wave operation system at the supply DC voltage 513 V, typical of three-phase 380/220 V bridge-type rectifier output, one can gain the rms value of the load phase voltage U ph = 241.62 V, which provides three-phase load in the rated voltage mode. This is a very good feature of the inverter, which allows developing more advantageous control systems of inverters.

30.00 20.00 10.00 0.00 –10.00 –20.00 –30.00 30.00

iA, A

id, A

20.00 10.00 0.00 –10.00 –20.00 426.00

427.00

428.00

Time (ms)

429.00

430.00

431.00

Fig. 3.45. Load phase A current and source current id waveform of three-phase square wave operation inverter circuit in a computer-generated model 110

010

100

101

0

011

001

Fig. 3.46. Characteristic 6-state space vector pattern of phase A voltage fundamental harmonic vector of square-wave operation 3 phase inverter with the state characteristic numbers of the inverter bridge upper swithes S1, S3, S5

The state of the circuit switches and their influence over the instantaneous values of the phase A voltage fundamental harmonic in the period can be specified by a six-state space vector chart (Fig. 3.46), wherein each star shifts by 60°, and each shift is met by the corresponding number inferring the specific state of the bridge upper switches, where the on-state of the switch is specified as 1, off state — as zero. The cycle starts 233

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3rd part with a horizontal and right-orientated space vector characterized by the switch state number 101 (S1 and S5 — on, S3 — off). After 60° counter-clockwise, the space vector configuration is specified by the number 100 (S3 and S5 — off), after that — by number 110, followed by 010, 011, 001 and after that the cycle is ended.

3.2.7. Three-Phase Voltage Source Inverter Control Control is required to ensure the conformity of the rms value of output voltage to the output voltage frequency of the inverter. In addition, only the control technique able to satisfy an application of the constant DC supply source voltage is to be considered. The pulse width control (PWC) technique is employed for the oldest designs; the pulse width modulation (PWM) technique is employed for recent designs, and space vector control technique is employed for the most recent designs. Referring to the circuit with 6 switches (Fig. 3.43), PWC can be implemented if, for example, in the period of one half-wave each switch control is constantly turned on in the first and third 60° intervals, but in the second interval it is turned on only in the interval section (60°−λ), where l is the angular duration of the switch-off pause. This switching sequence is illustrated in Fig. 3.47. As shown, the rms value of line voltage depends on angle 0 ≤ λ ≤ π / 3: π 2 − λ 60−λ 3 λ 2 2 Ul = U12 dϑ = U1 = U1 − 2 . (3-158) π π π 3

∫ 0

When λ = 60°, the output voltage is zero; when λ = 0°, the maximum output voltage is obtained. a

1 2 3 4 5 6

uAE

60°

l 1

b

360°

l

l

+

1 U1

2

0

U1

3

4

5

6

5

3

A B

C

t 4

6

2

–

t

Fig. 3.47. Pulse width control by means of adjustable middle part of switch cconductivity interval

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3rd part Another PWC case is also feasible: to create a pause, turn off both switches simultaneously connected to the pole of one supply source, for example, in interval 1 switches 1 and 5 are simultaneously turned off, while switch 6 is kept in the on-state. This control technique can be called the PWC at fundamental frequency. 1 2 3 4 5 6

uAB

0

U1

t

t

Fig.Â 3.48. PWC with double switching of switches in the middle part of conductivity interval 1 2 3 4 5 6 uAB 0

t

t

uA 0

t

Fig. 3.49. Partially modulated three-phase inverter switching and voltage curves

Harmonic composition of the output voltage is rather poor and is getting worse for law frequency area. Harmonic composition can be improved, provided that the switching-over of switches is performed k times with a period of 60Â° / k in the middle 235

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3rd part c onductivity interval (as shown in Fig. 3.48, where k = 2). However, the option above is still insufficient to improve voltage harmonic composition. Harmonic composition can be considerably improved implementing the pulse width modulation technique. Without the use of high-frequency modulation, associated with the increased power loss at switches, PWM can be partially applied over switches in their on-state in the first and third part (60° long) of the conductivity half-cycle (see Fig. 3.49). If these thirds are divided, for example, into 4 stages, and each stage is subject to a reduced duration of active switch operation, then the curve of the phase voltage average value during the stage is approaching the curve of voltage first harmonic. It is difficult to provide a wide control range over rms values of output voltage through the use of this option; therefore, the option has failed to gain a widespread application. RMS voltage and frequency control can be achieved through three-phase highfrequency pulse width modulation technique. The option employs three sinusoidal control voltages mutually shifted in phase by 120 el. degrees compared to the reference saw-tooth voltage (Fig. 3.50). Switches S1 and S4 (transistors VT1 and VT4, respectively) control under the control voltage related to phase A, switches S3 and S6 control under the control voltage related to phase B and switches S5 and S2 — related to phase C. As soon as the instantaneous value of the respective control voltage exceeds the instantaneous value of the saw-tooth voltage, a switch, corresponding to a certain phase and connected to the supply source plus pole, is turned on; when the situation is reverse, the low-side switch, connected to the minus pole, is turned on. u

ust uB

0

uA uC

S4 S1 S2 S5 S6 S3 uA

Ud/2

0

t

t

t –Ud/2

Fig. 3.50. Three-phase PWM curves with the high-frequency saw-tooth voltage and A phase voltage

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3rd part As shown in Fig. 3.50, the average value of the load phase voltage in the period of modulation depends on a half of the supply voltage and interrelation between the sawtooth and sinusoidal control voltage: Ud uphavst = U Rm sin ωt , (3-159) 2U stm where Ustm — value of the saw-tooth voltage amplitude, URm — control voltage amplitude. Line voltage is 3 times greater. RMS value of the phase voltage fundamental harmonic UU U ph(1)ef = d Rm . (3-160) 2 2U stm In fact, both relations directly refer to the case when neutral point of the load is connected to that of the supply capacitor divider. Without this connection the form of the phase-to-zero voltage is more complicated, but the relations are valid. As can be found, the maximum phase voltage with the sinusoidal load current is achieved at U Rm = U stm, when the voltage rms value at U d = 513 V is only 180.6 V, i.e., much less than needed to ensure the load nominal rating. This is a significant drawback of the sinusiodal PWM method; the drawback can be partially eliminated by increasing URm over Ustm values, i.e., providing the constant on-state of a respective switch in a certain half-cycle area; therefore, moving partially to a quazi-square mode. It is evident that the load current of this interval then does not follow a sine-waveform and the load current THD is getting worse; however, the rms value of load phase voltage exceeds the maximum value as determined by (3-160). Such type of modulation is called the overmodulation PWM technique.

Example The supply voltage of a three-phase voltage source inverter with sinusoidal PWM is U1 = 513 V, the Y-type load phase R = 5 Ω, L = 5 mH. The amplitude of bipolar sawtooth voltage of control system is U stm = 15 V, frequency 5 kHz, the control voltage to one phase changes uR = 10sin (3140t ). Determine the rms values of line voltage and phase voltage as well as the rms value of phase current! 1. RMS value of phase voltage is UU U phrms = 1 Rm = 120.42 V. 2 2U stm 2. RMS value of line voltage is U l = U ph 3 = 208.33 V . 3. RMS value of phase current is U ph I ph = = 7. 3 A . 2 R + (2πfL)2 237

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3rd part To verify the above-mentioned calculation, a computer-generated model of a threephase circuit with the parameters given in the example has been developed, wherein the resulted phase current curve is presented in Fig. 3.51. As can be seen, the load current is practically a sine-waveform, its rms value is 7.25 A, which confirms the accuracy of derived expressions.

15.00 10.00 5.00 0.00 –5.00 –10.00 –15.00 15.00

iA, A

id, A

10.00 5.00 0.00 –5.00 –10.00 434.50

435.00

435.50

436.00 436.50 Time (ms)

437.00

437.50

438.00

Fig. 3.51. Curves of phase A current and supply current id in a computer-generated model of a three-phase sine-wave form modulated inverter

Despite the possibility of obtaining high-quality current in load circuit, the PWM method has a major drawback — at every instant of high-frequency switching — all circuit switches are exposed to modulation that causes increased switching loss, makes operation of the switches more complicated and disrupts their level of safety. In addition, as mentioned above, three-phase circuits do not allow obtaining nominal voltage at the standardised supply DC voltage without the distortion of the load current sinewaveform character. Besides, it is not entirely appropriate that the instantaneous value of the load voltage continuously changes the sign from positive to negative and vice versa, which can severely affect, for example, the operation of an electrical machine. The situation can be easily improved by the implementation of the control technique based on modulation of the square-wave controlled inverter state vector — space vector control technique. Using this technique, new vectors can be defined for each of six sectors (Fig. 3.46) generated by two adjacent one-sector boundary vectors. Thus, for example, a new vector can be obtained in the first sector with any angle of inclination in the range of 60 degrees against the initial vector with state 101. To this end, according to the angle α either vector with state 101 or vector with state 100 (Fig. 3.52) is alternately used in one high-frequency modulation interval. The greater the angle 0° ≤ α ≤ 60°, the more vector with state 100 is used and less — with state 101. In order 238

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3rd part to reduce the resultant vector length (phase voltage amplitude), zero-state 000 is introduced into a modulation interval, when only all low-side switches of bridge are turned on or 111 state is introduced (all high-side switches are turned on). In addition, as seen from the numbers of states, periodically only one pair of switches is switched-over in each sector — the switches S5 and S2 in this sector, i.e., only one bridge leg low-side switch (turned on to reach state 101) and high-side switch (to reach state 100). Zero state is better to be turned on at the end of one modulation interval and at the beginning of the next interval (Fig. 3.52) V2

100

V ref T V2 2 Tm

a

2U1 V1 V1 = 3 101

T V1 1 Tm Tm A B C

T0 2

Tm

1

1

1

1

0

1

1

0

0

0

0

0

T1

T2

T2

T1

T0 2

T0 2

T0 2

t

Fig. 3.52. Diagram of a new vector pattern and changes of switch states in the period of two adjacent modulation intervals (T1 — state 100 is implemented, T2 — state 101)

The intervals T1 and T2 in this area (Fig. 3.52) in the time of modulation of high-frequency interval Tm can be expressed as follows: π sin − α 3 T1 = DTm ; π sin 3 T2 = DTm

sin α ; (3-161) π sin 3 239

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3rd part T0 = Tm −T1 −T2 ; V ref T + T2 D= . = 1 2 Tm U1 3 Here D is the ratioof the summative active interval duration against modulation interval time, but V ref is synthesised vector length — phase voltage amplitude at angle a; U1 = Ud. Moving to another area, the area number is to be taken into account 1 £ n £ 6: 3Tm V ref nπ sin − α ; T1 = 3 U1 3Tm V ref sin α − n −1 π . (3-162) T2 = U1 3

3.2.8. Multi-Level Voltage Source Inverters One of the ways to produce a relatively good quality voltage inverter is to generate a voltage curve from a number of equal or different level voltage sources. Figure 3.53 presents a 3-level single-phase inverter configuration. Here the output DC voltage is divided in half with the help of the capacitor divider C1, C2, and it is possible to connect to the load a conditionally positive polarity voltage +U / 2 from the capacitor C1 through two closed series transistors VT1, VT2 and a conditionally negative polarity voltage -U / 2 generated by capacitor C2 through closed series transistors VT3, VT4. If at the first case the transistor VT1 is turned off, then the load current is closed through diode V1 and transistor VT2, thus, forming a zero level voltage. Similarly, if at the second case the transistor VT4 is turned off, then negatively oriented load current is closed through transistor VT3 and diode V2, forming a zero level voltage in the same way. As a result, if only a square-wave operation is used, the rms value of the load voltage in such a single-phase system is U ldrms = U

ft1 , (3-163) 2

where t1 is the interval where the voltage is controlled by the both series transistor switches for a half-wave period.

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3rd part a

b VT1 VT2

VT1

+ +

C1

V1

VT3 VT4 uld

VT2

U +

V2

VT3

R Ld

C2 VT4

–

ild uld

L

T/2

t2

t1 t

U/2

0

t

ild 0

–U/2 V1

V1 t

V2

Fig. 3.53. Three-level single-phase inverter configuration and operation characteristic curves

PMW control method can also be applied to the given circuit, where each arm switch is turned over in the period of high-frequency modulation interval Tm << T , in addition, interval t1 = DTm ≤ Tm and in a half-wave period can be changed so that the interval average voltage value changes after the sine law. Then the curve of the load voltage fundamental harmonic is formed uld(1) = Dmax

U sin 2πft . (3-164) 2

The load voltage curve is generated with unipolar modulation in a half-wave period and the voltage rms value is U ldrms =

DmaxU

. (3-165) 2 2 Following a similar principle, a 3-phase multi-level inverter can be configured (see Fig. 3.54). Here, the full circuit consists of 3 equal 3-level single-phase inverter arms, each for its phase of star connected load; besides, neutral point of load can be connected to the supply capacitor zero point or can also be isolated. Analogue PWM control is implemented comparing two different polarities of the same frequency and the same phase saw-tooth voltages — the positive polarity for control of the transistor VT1 according to the respective supply phase, but the negative polarity voltage — control of the transistor VT4 according to the respective supply phase — with a sine-wave control voltage (Fig. 3.54). When positive polarity saw-tooth voltage is less than the positive polarity control voltage, the high-side switch VT1 is turned on, but, when the negative polarity saw-tooth voltage is more than the negative polarity control voltage, the lowside switch VT4 is turned on. Since the three-phase inverter system includes three phases for 120° shifted c ontrol,

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3rd part three control voltages are applied, and it is preferable to have an odd number of pulses for each load voltage in a half-wave period. Then it can be concluded that the saw-tooth voltage frequency in relation to the control voltage frequency should be an odd number multiplied by 3 — i.e., 3f, 9f, 15f, 21f, 27f …, where f is the control voltage frequency. Research reveals that the best option is under saw-tooth voltage frequency 15f, where each load voltage in a half-wave has 7 pulses (Fig. 3.54.).

0

t uVa

uan 0

U/2 t

VT1a VT4a

–U/2

Fig. 3.54. Explanatory pattern of 3-phase 3-level inverter control system operation at its one phase

3.3. Current Source Inverters 3.3.1. Operation Principle of Current Source Inverter The second type of inverters is a current source inverter that can in due time switch over the polarity of the low-variable DC supply current Id in the load circuit. The DC supply source of this type of an inverter should feature special current source properties well displayed whenever high inductance choke Ld is brought into the circuit (see Fig. 3.55). As the switches S1 and S2 are turned on, the inverted current takes a positive sign (iin = I d ); as the switches S3, S4 are turned on, it takes a negative sign (iin = −I d ). Since the load is not always efficient to work with a stepwise alternating current, it is imperative to shunt the load with the capacitor C, which is capable to accommodate the stepwise alternating current pulses. The capacitor voltage uC lags behind (in phase) the switching point instants by the time interval t1 (see Fig. 3.55,b). In its turn, the inductive load current ild lags in phase to the capacitor voltage uC. The capacitor current here is

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3rd part iC = iin − ild (3-166) and it changes stepwise in the moments of switching over. a Ld +

iC

Id

C

S1

Ud iin S4 –

S3

b S1, S2 S3, S4 iin

T T/2

0

ild Ld

S2

t

Id

UCm

uC ild

t

–Id uC ild

0

t

t1 iC

0

t

Fig. 3.55. Simplified circuit of a single-phase current source inverter (a) and diagrams (b)

Referring to the diagrams (Fig. 3.55,b), the currents and voltages are non-sinusoidal. However, in the first approximation, signal changes can be referred to as sinusoidal. Below the current source inverter processes are represented by a vector diagram shown in Fig. 3.56. UC I in j1 j ld IC

I ld

0

Fig. 3.56. Current source inverter vector diagram

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3rd part Using this assumption, the current source inverter power correlation can be found from the expressions: U d I d = I ld2 R =

U C2 R z ld2

, (3-167)

U d I d = U C I ld cos ϕ ld = U C I in cos ϕ1, (3-168) where Ild, UC — rms values of the load current and voltage; Iin — rms value of the inverted current; R, z ld, cosj ld — load characteristic parameters; cosj1 — a parameter defining the shift angle between the inverted current and capacitor voltage. Here j ld can be calculated as follows: ϕ ld = arctg

ωL , R

where ω = 2πf = 2π / T , T — switching period of switches, L — load inductance. As Id and Iin are equal in value, referring to (3-157), one can obtain that UC =

Ud . (3-169) cos ϕ1

The last expression fairly well describes the load voltage dependence. In order to ensure stabile UC (load voltage) adaptive to any load changes, constant angle j1 should be taken into account at constant Ud. If the angle increases, the output voltage increases too. In this case, it is relevant to bring an additional inductive load in order to reduce angle j1. Referring to the vector diagram (Fig. 3.56.), taking into account Iin and Id dependence, it is possible to write 2

2

I d2 = ( I ld cos ϕ ld ) + ( IC − I ld sin ϕ ld ) . Calculating Id value, a rough correlation between Id and Ud depending on the load parameters can be achieved: Id =

(

U d 1 + ω2C 2 z ld2 − 2ωC z ld2 − R2 R

) , (3-170)

and calculating UC value — an approximate relationship between Uld and Ud can be obtained: U C = U ld =

U d z ld 1 + ω2C 2 z ld2 − 2ωC z ld2 − R2 . (3-171) R

In turn, cosj1 can be calculated as follows: cos ϕ1 =

R z ld 1 + ω

2

C 2 z ld2

− 2ωC

z ld2

−R

2

. (3-172)

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3rd part Using the obtained expressions, it is possible to calculate the relative input choke current I d * = I d R / U d = f (z ld / R ), relative capacitor voltage U C * = U C / U d and cosj1 versus the ratio of the load impedance and resistance z * = z ld / R . Figure 3.57 illustrates the relationships above at R = 5 Ω and various values wC. As seen from the curves, the dependence of the parameters Id* un UC * on z* is sufficently complicated, though if z* goes up, then Id* un UC * are starting to rise strongly. The smaller wC value is, at higher z* values the rise starts. As to cosj1, while z* increases, cosj1 decreases reaching a very low value at z * = 10 and more. The curves above confirm that in the context of work stabilization the current source inverter greatly needs the constant control of load parameters, the essence of which has already been defined above: the need to stabilize angle j1. Then UC * is constant, but Id* varies in the inverse proportion to z*2. Thus, if R = const, along with the increase of the load inductance, Id value decreases. As the load is shunted with the capacitor, the reverse voltage is generated across the switches, which are turned-off. Thus, by turning on S1, S2 and simultaneously turning off S3, S4, the reverse voltage is generated across the latter ones, positive polarity of which is across the switches lower poles. Such polarity reverse voltage remains until the voltage of the capacitor C (Fig. 3.55) under this on/off action does not change polarity (initially prior to turning on S1, S2, the right plate is positively charged relative to the left). Likewise S3, S4 is turned on and S1, S2 are simultaneously turned off, the reverse voltage is generated across the latter two. The existence time of the reverse voltage depends on angle j1: tR =

ϕ1 . ω

Considering that the thyristor, turned-off in the DC circuit, needs to be connected in parallel to the bias voltage direction of fully-charged capacitor, the current inventor switches can be configured as thyristor-based (see Fig. 3.58). Every time after a halfwave when a new pair of thyristor is brought, the reverse bias voltage corresponding to the capacitor voltage is formed, action time of which needs to exceed the thyristor turn-off time, i.e., tR =

ϕ1 > t off . (3-173) ω

As it is obvious, if w goes up and j1 is constant, the bias voltage action time remains smaller and may become less than toff. In this case the inverter is short-coming, which should be cut off for safety. For example, if the angle is maintained ϕ1 = 45° and t off = 50 µs , the maximum inverter operating frequency is f max = 2, 5 kHz . It is a relatively low maximum operating frequency. 245

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3rd part a Id*

b cosφ 1 R = 5 Ω ωC = 0.05 1

ωC = 0.2 ωC = 0.1

R = 5 Ω ωC = 0.02

2

a U C* 3

.2 = 0

0 1

ωC

0.5

1 ωC = 0.02 2

3

4

5

6

7

8

9 10 z*

ωC = 0.2 ωC = 0.1 ωC = 0.05

0 1

2

3

ωC = 0.1 4

5

6

ωC = 0.05

7

8

9 10 z*

R = 5 Ω

2 ωC = 0.02

1 0 1

2

3

4

5

6

7

8

9 10 z*

Fig. 3.57. Input relative current of a single-phase current source inverter, dependence of the load relative voltage and inverter circuit cosj1 on the load impedance and resistance ratio at various wC Ld +

Id VS1

C

VS3

Ud VS4

Ld

VS2

–

Fig. 3.58. Single-phase thyristor current source inverter circuit

Example The supply voltage of a single-phase current source inverter is U1 = 200 V, load resistanceR = 5 Ω, inductance L = 10 mH, load is shunted with a capacitor C = 15 µF. Determine the rms values of the load current and voltage, as well as the average value of the supply source current Id, if switching frequency is f = 500 Hz! 246

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3rd part 1. Load current shift angle against a capacitor voltage 2πfL = arctg6.28; R

ϕ ld = arctg

from here

ϕ ld = 80.95°. 2. RMS value of the capacitor current IC = U C ωC = 0.0471U C. 3. RMS value of the load current UC

I ld =

2

R + (2πfL)2

= 0.0315U C .

4. RMS value of the inverted current 16.76 U C. 103 5. Shift angle between the inverted current and capacitor voltage I in = (I ld cos ϕ ld )2 + (IC − I ld sin ϕ ld )2 =

ϕ1 = arcsin

IC − I ld sin ϕ ld = 72.58°. I in

6. RMS value of the load voltage U C = U ld =

Ud = 668.4 V . cos ϕ1

7. RMS value of the load current I ld = 0.0315 ⋅ 668.4 = 21.05 A . 8. Average value of the supply source current Id =

I ld2 R = 11.08 A. U1

To verify the accuracy of the expression, a computer-generated model referring to the calculated example parameters has been developed. The resulted curves are displayed in Fig. 3.59. As shown, the capacitor voltage amplitude is 1 kV, and rms value is 704 V, which is almost in line with the estimates.

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3rd part 15.00 10.00 5.00

iinv, A Id

0.00 –5.00 –10.00 –15.00 1.50 1.00 0.50

uc, kV t1 =

ϕ1 ω

0.00 –0.50 –1.00 –1.50 44.00

45.00

46.00 Time (ms)

47.00

48.00

Fig. 3.59. Curves of the inverted current of a single-phase current source inverter and capacitor voltage in a computergenerated model

3.3.2. Three-Phase Current Source Inverters Three-phase current source inverters are configured with 6 switches (usually thyristors are used as switches), out of which 3 are connected to a negative supply source terminal, forming a cathode group, and other 3 are connected to the blank terminal of the supply in-circuit choke Ld, forming an anode group (Fig. 3.60). Among 3 line voltages uAB, uBC, uCA the respective capacitors C1, C2, C3 are engaged, which capture step-like inverted current of the respective load phase circuits. The load circuit may be, for example, Y-type. Ld +

Id

S1 A

S3

C1

iinvA

Ud

iA, uA

S5

B

C2

iC1 iC2 C3

iC3

C S4

S6

S2

–

Fig. 3.60. Three-phase current source inverter circuit

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3rd part As the supply source current is constant, each of the switches can be turned on only with the conductivity angle 120° over the period. In addition, there are only two switches always simultaneously turned on: one out of the anode group and one out of the cathode group. The switching sequence is as follows: S1 + S2; S2 + S3; S3 + S4; S4 + S5; S5 + S6; S6 + S1… Line (inter-phase) voltages are equal to the voltage of capacitors, but variable instantaneous values of the latter are determined by the currents, formed in the anode group as Id, and respective load phase difference plus one more capacitor current. For example, when S1 is turned on, the current of capacitor C1 is iC1 = I d − iA + iC 3, when S3 is turned on, then iC 2 = I d − iB + iC1, when S5 is turned on, then iC 3 = I d − iC + iC 2 . Figure 3.61 shows the inverter line voltage uAB, as well the curves of capacitor C1 and the inverted current of phase A at U d = 300 V, f = 500 Hz, R = 5 Ω, L = 5 mH and C = 10 µF. The instantaneous value of inverted current of phase A depends on the operation sequence of the switches S1 and S4 and the source current Id. Referring to the curves, the rms value of the inverted current may be defined as follows: I invA = I d

2 . (3-174) 3

Proceeding from the equlity of the inverter input and output active power, it is safe to write that Pd = U d I d = 3U A I A cos ϕ ld =

3U A2

R, (3-175) z ld2 where UA, IA — the rms values of phase voltage and current, j ld — shift angle between phase voltage and current fundamental harmonic, R — load phase resistance, z ld — load phase impedance.

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3rd part S1 S2 S3 S4 S5 S6

iinvA, A

20.00 10.00 0.00 –10.00 –20.00 i ,A 30.00 C1 20.00 10.00 0.00 –10.00 –20.00 –30.00

uAB, kV

1.00 0.50 0.00 –0.50 –1.00 75.50

76.00

76.50

77.00 77.50 Time (ms)

78.00

78.50

79.00

Fig.3.61. Curves of phase A inverted current, line voltage uAB and capacitor C1 current of a three-phase current source inverter in the example at Ud = 300 V, f = 500 Hz, load R = 5 Ω, L = 5 mH, C = 10 μF

As Fig. 3.61 shows, the line voltage curve of this inverter is close to a sine-wave. Therefore, the capacitor current is practically sine-shaped. This facilitates a rough calculation of the circuit by means of a pilot sine-wave circuit calculation method. Thus, phase A connection point given in the vector diagram (Fig. 3.62) can be used in the calculation. A UA

0

IA

j ld IA

C I C3

I invA I C1 - I C3

B I C1

Fig. 3.62. Vector diagram of 3-phase current source inverter calculation

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3rd part From the vector diagram for A point the current balance is as follows: I invA − I A = I C1 − I C 3, where I A = U A / z ld is load phase A current, but summary current of capacitors is 3U A wC . Applying trigonometric calculation, the inverted current of phase A can be expressed as follows: I invA = I d

1 2ωC sin ϕ ld 2 . (3-176) = 3U A2 2 + 3ω2C 2 − 3z ld z ld 3

From here one can determine the source current Id and, using relationship (3-175), define the rms voltage value of the load phase: U ph =

z ldU1 R2 1 1.5ω2 z ld2 C 2 + − Cz ld ω 1− 2 . (3-177) R 6 z ld

To verify the validity of the obtained expression, the example gives the estimates based on the computer simulation parameters shown in Fig. 3.61. As it is obvious, the source current oscilloscope trace is slightly below 15 A, but as per calculation — 13.42 A; line voltage rms value is 650 / 2 = 457.8 V, but as per calculation 270 3 = 467 V, i.e., voltage coincidence by itself is very good, but the source current — by 10 % incorrect, which is quite satisfactory in such approximate calculation. Three-phase current source inverters can also be configured substituting switches with thyristors. The line voltages define reverse voltage of thyristors during their off-period, starting from the turn-off instant of the thyristor. This provides especially satisfactory switching conditions.

Example Calculate the operating parameters of a 3-phase current source inverter circuit, if U d = 300 V, f = 500 Hz, load R = 5 Ω, L = 5 mH, C = 10 µF. 1. Load phase impedance z ld = 52 + (2π ⋅ 500 ⋅ 5 ⋅10−3 )2 = 16.48 Ω. 2. Load power factor 2πfL ϕ ld = arctg = 72.33°; R sin ϕ ld = 0.953 3. Supply source current relationship with phase voltage Id =

3U f 2

2ωC sin ϕ ld 1 + 3ω2C 2 − = 0.0497U ph. 2 z ld 3z ld 251

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3rd part 4. RMS value of the load phase voltage U ⋅ 0.0497 ⋅16.482 U ph = d = 270 V 3⋅ 5 5. Source current value I d = 0.0497 ⋅ 270 = 13.42 A .

3.3.3. Control of the Voltage of Current Source Inverter In case of a single-phase self-commutated current source inverter, the load voltage depends on both the DC source voltage Ud and the load character disclosed as the shift angle between the inverted current first harmonic and capacitor voltage: Ud U ld = . cos ϕ1 As can be seen, when U d = const , the rms value of the output voltage depends only on angle j1: if it is constant, then the output voltage is also constant. This mode can be secured with a control device composed of the capacitor and controlled choke connected in parallel. The latter may be set up by connecting the thyristor AC controller into the choke circuit (see Fig. 3.63). Id +

Ld

VS1

QC

VS3

C L QL

Ud

Qld VS4

Ld

Q VS2

–

Fig. 3.63. Reactive power flow in a single-phase current source inverter

As the active power is Pd = U d I d, and the total power is S = U C I1in, 252

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3rd part where I1in is the rms value of the inverted current, then. cos ϕ1 =

Pd . S

The reactive power required can be calculated as follows: Q = S 2 − Pd2 = I d U C2 −U d2 . (3-178) In turn, Q consists of (Fig. 3.48) the three components: Q = −QC + QL + Qld, (3-179) where QC is the capacitor reactive power, QL — control choke reactive power, Qld — load reactive power. Taking into account the dependence (3-167), the choke reactive power can be calculated as follows: QL = I d U C2 −U d2 + QC − I ldU C sin ϕ ld . (3-180) As it is necessary to stabilize UC, Ud can be assumed constant, and QC at constant UC is constant too, but QL directly depends only on Id and I ld sinj ld. To make control automatic, the following algorithm is defined: whenever the load voltage goes up, the control choke reactive power should be increased, i.e., the regulating angle a of AC controller should be decreased. It is definitely essential to control the minimum acceptable value of the angle j1, which ensures normal thyristor switching.

3.4. Resonant Inverter In resonant inverter (RI), a choke, a capacitor and the load or inverter elements are connected in series to the load or inverter circuit. The processes of the circuit, periodically switched over by the semiconductor switches, are defined by the interrelation of parameters and control actions specified by these switches. As an example, let us consider the circuit shown in Fig. 3.64, where the energy from DC supply source is supplied only for one half-wave operating period of the inverter, once the transistor VT1 is switched on, i.e., the circuit can be qualified as a half-wave RI. As soon as VT1 is switched on, the capacitor C is getting charged through the choke L and the load resistor R with the conditionally positive polarity terminal voltage UC1 (the left plate is positively charged in relation to the right). In the second halfwave period, the transistor VT2 is switched on and the discharge of the capacitor C is implemented through the choke L, transistor VT2 and series connected load. Once the current i takes a minus sign, it results in the recharge of the capacitor, at the end of the operation the polarity of the plates is opposite to the initially. The capacitor recharged final voltage UC0 takes a conditionally negative polarity.

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3rd part The operation of a resonant inverter can be efficiently performed only on condition that the circuit parameters provide the oscillation process 2 1 R > = δ 2, (3-181) LC 2L

where d is assumed to be called a damping factor. The circuit self-oscillations frequency is ω=

1 − δ 2 . (3-182) LC

If the control angular frequency ωco = 2π /T is less than w, the load current obtains a discontinuous operating mode, where switches of the transistor turn off and on at zero instantaneous current values, thus ensuring the operation of the switches without switching losses. If ωco = ω , the boundary option is achieved, where the current is continuous and practically sinusoidal. This is also a preferred operating mode, known as «normal»; therefore, this option is directly analysed below.

a

+

b VT1 VT2

VT1 i

Ud VT2

L

uC

t

i

C

0

t

R uC

–

UC1 T

0

t

UC0 Fig. 3.64. Resonant inverter circuit and operation characteristic diagrams at ωco < ω

As the transistor VT1 is switched on, then the current value is i = 0 A, but voltage value is uC = U C 0 . Thus, the changes of the current instantaneous value are described in the equation i = e−δt ( A sin ωt + B cos ωt ), (3-183) which subject to the initial conditions can be transformed as follows: i = e−δt

U d −U C 0 sin ωt . (3-184) ωL

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3rd part In turn, the choke voltage is uL = L

di δ = e−δt (U d −U C 0 )cos ωt − sin ωt . (3-185) ω dt

At the end of first half-wave period, the voltage of the capacitor C is U C1 = U d − (uL )

t=

π. ω

Taking into account the value uL , the voltage is U C1 = U d + (U d −U C 0 )e

−

δπ ω . (3-186)

The current of second half-wave period (it is accepted for this half-period that i = 0 A as well) U i = e−δt C1 sin ωt . (3-187) ωL Since the average value of the capacitor current, i.e., current i, is zero, integrating both expressions for current U d −U C 0 = U C1, as well as taking into account the expression UC1, it can be stated, that U C0 = −U d

e

−δ

π ω

−δ

π ω

. (3-188)

1− e The amplitude value of the current i in first half-wave period is Im = e

−δ

π 2ω

Ud

π 1− e−δ ω ωL

, (3-189)

The rms value of the current is I rms =

Im

2 and the rms value of the load voltage is U R = I rms R =

Im 2

R. (3-190)

The expressions above allow calculating the required parameters. For example, if U d = 300 V, L = 10 mH, C = 100 µF and R = 5 Ω, then at ωco = ω = 937.5 s−1, at the expressed capacitor extreme voltage U C1 = 528.8 V and U C0 = −229 V the rms value of the load current is 26.1 A, the rms value of the load voltage is 130.7 V. Figure 3.65 shows the curve of the load current and capacitor voltage of the example in q uestion with given parameters. As it is evident, the coincidence is satisfactory. 255

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3rd part Provided that the operating control mode is «normal» and the current curve is continuous, the calculations can be approximated assuming that the instantaneous vol tage across the section L-C is zero; therefore, the whole resonant circuit input voltage fundamental harmonic is equal to the load resistor voltage harmonic: 2U d . (3-191) π It is assumed that the resonance circuit input voltage may be approximately expressed as follows: Im R =

U d 2U d + sin ωt + . 2 π Then I m = 2U d / (πR ), the capacitor C average voltage is 0.5Ud, U C1 = 0.5U d + 2U d / (πRωC ), U C0 = 0.5U d − 2U d / (πRωC ). The calculation based on the given parameters gives I m = 38.2 A, U C1 = 557.86 V , U C0 = −257.86 V. As can be seen, the results are close to those previously calculated and simulated. uVT2 =

40.00

i, A

Im

20.00 0.00 –20.00 –40.00

u ,V 600.00 C

UC1

400.00 200.00 0.00 –200.00 –400.00 70.00

UC0 75.00

Time (ms)

80.00

85.00

Fig. 3.65 Curves of the load current and capacitor voltage in a computer-generated model referring to the calculated example

If the control frequency is lower that the circuit resonance frequency, the rms value of load voltage decreases with respect to the calculated one, but the load current is interrupted in this circuit. The circuit in Fig.3.64 is not operable at ωco > ω, because as soon as the transistor is switched off, the instantaneous value of load current is interrupted. To ensure the circuit operation under the above-mentioned control mode, each transistor is shunted with a reverse diode; moreover, the supply source circuit needs to ensure a bidirectional conducting capability (it is possible to shunt the circuit input with a high-value capacitor). 256

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3rd part Once the circuit transistors are shunted with a diode, ωco < ω and UC1 is exceeding Ud, the capacitor oscillation discharge takes place through the transistor VT1 shunted diode and the supply voltage circuit. Such discharge decreases the absolute value of the capacitor voltage UC0, whereas the capacitor polarity after the oscillation process through VT2 can be even positive, which significantly alters the course of the process. Resonant inverter can also be configured according to other circuits (for example, bridge). It is important to point out that under «normal» control operation or reduced control frequency the turn-on/off process takes place at zero current values, which significantly facilitates the operation of the switches. It is one of the positive aspects of RI application.

3.5. Commutated Converter Control Systems The key component of commutated converter systems is a clock generator, wherein the pulse frequencies generated are easily controlled in a sufficiently wide range. In modern power energy and electronic technology, a timer chip can be regarded as a quality clock generator, in the input of which charge and discharge of a capacitor take place through the resistor circuits, but the output terminal periodically switches up and down subject to definite capacitor voltage levels. The second important component is the saw-tooth voltage unit, wherein, depending on the converter functions, unipolar or bipolar saw-tooth voltage is generated. In general, the saw-tooth voltage is generated through charging-discharging (in case of unipolar voltage) or recharging of the capacitor (in case of bipolar voltage). Provided that a conventional transistor switch controller is controlled, the control circuit can be configured as displayed in Fig. 3.66. Here, the clock generator (timer DD1) output 3 generates periodic and short-duration low-level voltage signals. The timer output 3 via the resistor R1 is connected to the saw-tooth voltage generator by the operational amplifier DA1 of the generator capacitor C discharge transistor VT1 base. If there is a high level signal in the timer output, VT1 is closed and the charging process of the capacitor C is triggered via the resistor R2 and the current defined by the supply voltage –UB. The voltage across the capacitor plates increases linearly till the moment when a zero level signal appears in the timer output, which provokes the VT opening and virtually instant discharge of the capacitor C. Thus, the periodic saw-tooth voltage is generated with the amplitude U stm =

U BT , (3-192) R2C

where the signal period T of the clock generator is accepted less than R2C.

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3rd part +UB

a

DD1 4 7

8

VT1 R1

C

3 6 2 1

R2

+

DA1

R4

– +

u3

0 ust

VT

uVT

Uco

–UB b

DR

DA2

R3

ust

– +

+UB

T

t

Ustm Uco

0 uVT 0

t

t

Fig. 3.66. Transistor switch controller control system circuit (a) and signal diagram (b)

The comparator DA2 compares the saw-tooth voltage with the DC voltage control signal Uco received from the potentiometer R4. As soon as the instantaneous value of the saw-tooth voltage is less than Uco, the DA2 output high logic level signal activates DR «drivers», feeding a signal to the control transistor VT base-emitter that turns it on. Therefore, the relative duration (duty ratio) of the VT turn-on period is D=

U co , (3-193) U stm

where 0 £ U co £ U stm. The bipolar symmetric saw-tooth voltage should be generated to control the reversible switch controller. Also, in this case, a timer can operate as the clock generator; however, to produce a symmetrical saw-tooth voltage the complete symmetrical bipolar rectangular-form voltage must be supplied via the resistor to the operating amplifier input in the frame of the capacitor C feed-back (integrating) circuit. Such voltage is difficult to obtain. Thus, it is better to accept the operating amplifier as the bipolar saw-tooth voltage generator, which operates in the periodic multi-vibrator mode (see Fig. 3.67). Here, 258

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3rd part the symmetrical bipolar practically linear alternating voltage is generated across the capacitor C, amplitude value of which is defined by the amplifier supply voltage multiplied by the resistor nominal ratio R3 / (R3 + R2 ) of the voltage divider R3, R2. R1

a ust

uC

UB

– +

C

b uC = ust uout

R2

R3 R2 + R3

0

R3

R3 −UB R2 + R3

t

Fig. 3.67. The bipolar saw-teeth voltage generator curcuit obtained across the capacitor C terminals (a) and the saw-tooth voltage diagram (b)

In case of the pulse width sinusoidal modulation, the bipolar saw-tooth voltage is to be compared with the controlled frequency and amplitude sine-wave voltage. To generate such a voltage, it is quite a challenging task, especially, if 3 voltages shifted in phase by 120° are required for three-phase systems. Under analogue control the voltage similar to a sinusioidal can be generated by providing the capacitor-included (integrating) operational amplifier input with the bipolar saw-tooth voltage of the required amplitude Ustm and frequency (see Fig. 3.68).

a

b ust

C ust

R – +

Ustm

0 us

us

t Usm

0

t

Fig. 3.68. Quasi-sinusoidal voltage generation circuit (a) and diagram (b)

Here U sm =

U stmT , 8RC

(3-194)

but within the limits 0 £ t £ T / 2 the voltage us changes as follows: U 2t 2 . (3-195) us = stm t − RC T 259

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3rd part The resulted curve THD is close to zero, i.e., the curve is sufficiently close to a sinewave. The commutated converters of the inverter type are designated to solve the problem of the periodic commutation of switches. The task can be solved through the use of counters together with a decoder in three-phase systems or — a frequence divider in single-phase systems, which serves as a clock trigger. The single-phase inverter control circuit can be configured as presented in Fig. 3.69. Here the clock generator is configured on the timer DD1 base with an operating frequency 2f; the timer output signal activates the clock trigger DD2, wherein the direct output via AND logical element DD3 controls one diagonal transistors (or thyristors) VT1, VT2 of the inverter, but the inverse output via AND logical element DD4 – other diagonal transistors (or thyristors) VT3, VT4. The logical AND element engagement into the circuit allows implementing the dead-time between activation instants of the lower and upper transistors of one arm, which excludes the possibility of the shortcircuit in the moments when the lower transistor has been already turned on while the upper transistor stands by in its turned-on position. The dead time is a mandatory requirement for practical application of commutated converters in real world. a

+UB DD1 4 7

8 3

C

6 2 1

DD3 & VT1 DD2

VT2 Q1

T

Q2 DD4 & VT3 VT4

b

u3 0 Q1

T

t

0 VT1 VT2

t

0 VT3 VT4

t

0

t

Fig. 3.69. Single-phase inverter control circuit (a) and voltage diagrams (b)

Provided that the three-phase inverter bridge transistors require the sequence commutation with the control angle 180°, the clock generator frequency is assumed to be 6f, the binary counter CT2 with the output decimal equivalent from 0 to 5 is introduced, as well as the decoder DC with the active outputs from 0 to 5, which according to 6 transistor control OR logic circuits connected in respective sequence provide cascade transistor connection (Fig. 3.70).

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3rd part a

b 1

TG

0 D0 1 D1 CT2/6 DC 23 D2 4 5

1 1 1 1 1

VT1

D0

VT2

D1

VT3 VT4 VT5 VT6

1

2

3

4

5

0

1 t t

D2 VT1 VT2 VT3 VT4 VT5 VT6

t

Fig. 3.70. Three-phase inverter transistor control system (a) and signal diagram (b)

It is required to provide each switch control angle 120° in a three-phase current source inverter. That sort of control can also be delivered in a circuit employing a counter and decoder. Only in this case the thyristor V1 OR element input is to be connected to the decoder output 0, 1; thyristor V2 OR element input — to outputs 1, 2; V3 OR element input — to outputs 2, 3; V4 OR element input — to outputs 3, 4; V5 OR element input — to outputs 4, 5; V6 OR element input — to outputs 5, 0.

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Review Questions

3rd part 1. What is the rms value of the load voltage of a single-phase square wave operation bridge transformerless inverter and what is the rms value of this voltage funda mental harmonic?

2. Which out of the two rms voltages mentioned in the question above determine the rms value of RL load current?

3. Which of both RL load elements release active power and what is the approximate

relationship between the DC source supplied active power and the load distributed?

4. How much would the average value of the DC source supplied current (DC com-

ponent) be, if the load is ideally inductive with R = 0 Ω? Can the rms value of this load current be greater than zero?

5. What is the rms value of the load voltage of the single-phase square wave opera-

tion half-bridge transformerless inverter and what is the rms value of this voltage fundamental harmonic?

6. Why is it necessary to shunt each transistor (thyristor) of the voltage source

inverter in a reverse direction to a diode? What load parameter does the time of the current flow through a diode depend on? In what way does the diode current influence the DC supply source current? In what manner shall the path for the inverter diode-generated current be provided if the supply source features unidirectional conductivity with a diode in circuit?

7. What is the value of the voltage of the diode shunted transistor at the moment of its switch-on, if the load is of RL type?

8. How many transistor switches are simultaneously engaged in a single-phase

transformative zero-point inverter operation? How much is the maximum voltage of these switches in relation to the supply source?

9. Why does the primary half-winding transistor-related current «jumps» imme-

diately after the transistor switch-off to the second half-winding in the previous circuit operated with RL load? In what way does it flow across it?

10. What is the relationship between the previous circuit primary half-winding rms current and the load current? What does this relationship depend on?

11. How could you regulate the rms value of a single-phase voltage source inverter

load voltage, if the switching frequency of the inverter switches corresponds to the inverted AC voltage fundamental harmonic frequency?

12. What does a single-phase inverter voltage PWM sine modulation mean? What is the ratio between the switching frequency of the inverter switches and the inverted voltage fundamental frequency? What is the variation range of the instantaneous value of the transformerless inverter load voltage?

13. What is the relationship between the single-phase transformerless inverter load

rms voltage controlled by a sine PWM and the supply DC voltage? Does it depend on PWM parameters?

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Review Questions

3rd part 14. If the above-mentioned load rms voltage is independent of PWM parameters, then why does the rms value of load current depend on them?

15. In which way is the fundamental harmonic of the single-phase inverter load

voltage controlled by PWM formed? How does it depend on PWM parameters?

16. Is the PWM controlled inverter load current purely sinusoidal? How can the current form distortion be affected?

17. How many semiconductor switches are needed to configure a three-phase voltage source inverter? What is the ON state duration of each switch within a period, if such an inverter is in the square-wave operation mode with 3 inverter switches being simultaneously in operation?

18. What time and instantaneous voltage steps in a half-period time do the output line voltage and phase voltage curves of the previously defined 3-phase voltage source inverter consist of?

19. How much do the rms values of the inverter output line voltage and phase volt-

age depend on the supply source, if the above-mentioned 3-phase inverter load is Y-type?

20. At what phase shift angle between the 3-phase inverter Y-type RL load voltage and current is the instantaneous value of DC source positive polarity equal to zero at the moment of the switch position change?

21. How do any changes in state of the 3-phase square-wave operation voltage source inverter switches affect the one-phase voltage state characteristic vector pattern within a period? How can these states be displayed in a diagram?

22. How could you control the rms value of the 3-phase voltage source inverter load voltage, if the switch-over frequency of the switches coincides with the load voltage fundamental harmonic frequency?

23. How could you implement the 3-phase voltage source inverter load voltage sinewave pulse width modulation?

24. How large is the variation range of the instantaneous values of the line and phase voltages of the 3-phase sine PWM controlled voltage source inverter Y-type load within the time of one load voltage fundamental harmonic period? What is the relationship between the amplitude and rms value of the greatest possible load line voltage fundamental harmonic and the source?

25. How could you use the fixed states of the switches in the square-wave operation of 3-phase voltage source inverter phase voltage vector to obtain the derivative state vectors at any angular position?

26. How many out of 6 switches of a 3-phase inverter shall periodically change their state moving from one fixed vector 60 degree area to the next?

27. What are the voltage levels of 3-level single-phase multi-level inverter in relation to the source? How is it implemented in practice?

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Review Questions

3rd part 28. How could you reduce the derivative voltage vector length? What do the states

000 and 111 of the switches mean? How could you best introduce «zero» states into the control algorithm?

29. How many diodes are required to implement a «zero» state of the 3-level threephase multi-level inverter?

30. What element is mandatory for the current invertor DC source circuit? What functions does it perform?

31. Is it essential to provide the semiconductor switches of a current source inverter with the bidirectional conductivity capacity? Which of the semiconductor elements is better tailored for the current source inverter switch functions?

32. What is the purpose to shunt the current source inverter load to a capacitor? In what way does the capacitor affect the inverter thyristor operation?

33. What is the phase shift angle of the inverted current fundamental harmonic in

relation to the load voltage fundamental harmonic voltage? In what way does the angle degree value affect the load voltage?

34. What calculation method can be applied to calculate the current source inverter parameters? What variables can be approximately regarded as a sine curve?

35. How could you control the current source inverter load voltage? 36. How many thyristor switches and capacitors are required to implement a threephase current source inverter?

37. How could you determine the current inverter DC circuit current provided that the rms value of definite load current is given?

38. What kind of electrical circuit elements are required to implement a resonance process? What is the value of the ideal LC circuit resonance frequency?

39. What is the voltage value between the series resonant LC circuit endpoints? In

what way can these properties be used in the calculation of the commutated resonance circuits?

40. What form is the current in the LC circuit commutated with the resonance

frequency? How much is the instantaneous value of the commutated circuit current in switching moments?

41. What kind of relationship between the resonance frequency and LC circuit commutated switching frequency provides the highest current amplitude and the most efficient resonant inverter operation?

42. Is it feasible to develop a three-phase resonant inverter? 43. What element is regarded as the key element in a commutated converter control system and why?

44. What functions does a saw-tooth voltage generator perform under the commutated converter control procedure and how to create one?

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Tasks

3 part rd

1. The operating frequency of bridge-type single-phase transistor voltage source

inverter is 1 kHz, the supply DC voltage value is (300 + m) V, the load resistance is 5 Ω; the inductance is 1 mH. Through the linearization method of current changes (see Section 3.2.2) determine the time shift between the load voltage and current curves, load current amplitude, rms value, transistor by-pass diode average current and source current average value!

Answer: 0.129 ms and 57.67 A; 33.33 A; 3.74 A; 13.95 A, respectively at m = n = 0. 2. Between the both vertical arms of a single-phase transistor bridge-type voltage

source inverter the time shift 0.1 ms is introduced for each supply source pole connected to the transistor switch-over moment. At this time the both arm opposite transistors are turned on. Determine the rms value of load voltage, if the supply voltage is (200 + mn) V and the inverter operating frequency is 3 kHz! Through the linearization method of current changes determine the load current amplitude value, if the load is L = 5 mH, R = 20 Ω, assuming that in the load circuit at the shortened intervals the current instantaneous value remains the same!

Answer: 154.9 V, 2.14 A at m = n = 0. 3. The single-phase bridge-type voltage source inverter diagonal transistors

are modulated with the PWM method, comparing the bipolar triangular saw-tooth voltage with amplitude 20 V, frequency 5 kHz, with the sinusoidal control voltage (10 + m)sin314t . What is the approximate value of the load voltage fundamental harmonic amplitude, as well as the load current amplitude, if R = 10 Ω, L = 10 mH and the supply source voltage is (200 + mn) V? What is the load voltage frequency?

Answer: 100 V, 9.55 A, 50 Hz at m = n = 0. 4. The three-phase transistor voltage source inverter, wherein the transistor on-

position control time lasts for a half-cycle, is operated with the frequency 2 kHz. How much is the load (star connected) phase and the line rms voltages, if the supply source voltage is (500 + mn) V? Define the rms value of voltage fundamental harmonic!

Answer: at mn = 0 the rms values of line voltage and its fundamental harmonic are

408.2 V and 387.9 V, respectively; phase voltage — 235.7 V and 224.3 V, respectively.

5. The three-phase transistor voltage source inverter with the previous data load phase inductance 1.m mH, resistance is (10 + m + n) Ω. Determine (according to the voltage fundamental harmonic) the rms value of phase current, amplitude and phase shift against the voltage fundamental harmonic! Estimate the DC source current potential polarity (unipolar or bipolar)!

Answer: at m = n = 0 13.96 A, 19.83 A, 51.47°; unipolar current.

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Tasks

3 part rd

6. Determine the average value of supply source current on the basis of the two previous assignments according to the fundamental harmonic!

Answer: at m = n = 0 average value is 11.7 A. 7. Each vertical leg transistor of a three-phase transistor voltage source inverter

is modulated by the sine law, comparing the triangular bipolar saw-tooth voltage, frequency of which is 5 kHz and amplitude is 20 V, with sinusoidal control voltages (10 + m)sin314t with each phase shifted by 120°. How much is the Y-type load line and phase fundamental harmonic rms values, if the supply source voltage is (500 + mn) V?

Answer: at m = n = 0 the line voltage is 152.5 V; phase voltage is 88.0 V. 8. The single-phase current source inverter input voltage is (200 + mn) V, thyris-

tor control frequency is 1 kHz. The active inductive load resistance is 10.m Ω, the inductance is 1.n mH. Determine the source current value and the load parallel capacitor C = 50 µF voltage amplitude, assuming that the inverter output current (either load or capacitor) is sinusoidal!

Answer: 215.7 A and 1101.6 V at m = n = 0. 9. What volume capacitor is required under the conditions of previous assignment, if the load voltage amplitude does not exceed 426 V (rms value 300 V)? How large is the source current at this voltage?

Answer: at m = n = 0 19.92 mF and 32.32 A. 10. What is the inductance value of the choke connected in parallel to the load, if the voltage amplitude of 426 V is achieved at the initial capacitor capacity 50 mF?

Answer: at m = n = 0 0.84 mH. 11. The half-wave period resonance invertor supply voltage is (300 + mn) V, the

choke inductance is 10 mH, the capacitor capacity is 100 mF, the load resistance is 10 Ω. Taking into account the sinusoidal operation mode, determine the control frequency, capacitor negative polarity voltage amplitude and load current rms value!

Answer: at mn = 0 137.9 Hz, 58.4 V, 11.66 A.

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4th part

4. NETWORK POWER CONVERTERS WITH FREE COMMUTATION CAPACITIES The development taking place in high-power transistor engineering gives an excellent opportunity to adjust certain processes that occur in the field of network power converters. The main adjustments tend to maximally approximate power waveform consumed by a network to a sine-wave, as well as to ensure that a network power converter consumes (generates) pure active power or reactive power required. At present, the application of switched network converters is rather limited due to the lack of complete theory formulation in the field of converters. However, certain types of converters are involved in a broad spectrum of application, for example, a practically non-controllable three-phase or single-phase rectifier with a sine-wave current at the input (known as a power conditioner or active rectifier) or a single-phase rectifier with a sine current wave and active current at the input (known as a power factor corrector). Some of those converters are discussed below.

4.1. Rectifier — Power Conditioner The circuit of this single-phase rectifier as shown in Fig. 4.1 consists of a bridgetype diode rectifier with an inductor L at the input and a capacitor at the output. Each diode is shunted to a reverse direction by a transistor (traditionally IGBT type). Similarly, by using certain diode-transistor elements a three-phase bridge-type circuit can also be configured, wherein an inductor L is connected into each network phase and on the DC side — a high-value capacitor is found. Nevertheless, to explain operation principles a single-phase circuit as an example is discussed below. By means of commutating transistors it is feasible to enable the current consumed by a network to be sinusoidal with a displacement angle against voltage required. If such a rectifier is not loaded with an active load, the current consumed by a network tends to become either leading or lagging by 90 degrees. An off-load rectifier operates only as a reactive power compensator and is able to consume only 90 degree leading or lagging current from the grid. If a rectifier is loaded, the device generally targets to achieve an active character sinusoidal current at the input or, if required, to ensure the phase displacement of network current less than 90 degrees (either leading or lagging the network voltage). To examine the performance of the circuit, let us take a look at a single-phase bridge circuit in the mode of a reactive power compensator with a leading network current 267

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4th part provided that there are no any other elements connected to a capacitor in parallel. Regardless of the direction of compensation, the following expression is valid U 12 = U n − I L X L , (4-1) where U n , I L represent respective vectors of the rms values of network voltage and inductor current fundamental harmonics, and U 12 represents the vector of the rms value of bridge input voltage fundamental harmonics. The equation-based vector diagrams are shown in Fig. 4.1 b and c. As can be seen, the vector of network voltage in a reactive power compensation mode and the vector of the bridge input voltage fundamental harmonics match in phase. In addition, in the case of the leading network current of an inductor by 90°, the bridge input voltage amplitude is greater than the network voltage amplitude, but in the case of a lagging current — less. Accordingly, the output capacitor voltage under the same rms current via an inductor in the first case is to be higher, in the second — less. In such a circuit, the capacitor voltage takes only the polarity indicated in Fig. 4.1 because the opposite polarity voltage is to be «nulled» by bridge diodes. Typically, high-volume capacitors are assumed and, therefore, to the first approximation it is possible to treat the capacitor capacitance as infinite large and its voltage as completely smooth — DC voltage Ud is to be found on the capacitor plates. The sine-wave curve of the bridge input voltage fundamental harmonics is formed by connecting each input terminal 1 and 2 alternatively to the upper or lower capacitor plates, thus m odulating DC voltage Ud in a sine-wave manner in relation to the bridge input terminals. If, for example, the current iL flows in the direction of an arrow indicated in Fig. 4.1 a and the transistors are switched off (namely, current flows via diodes V1, V2), then u12 = U d (mode A). When under the same current direction the transistors VT4, VT3 are switched on, u12 = −U d (mode B). a

b V1

un

L

V3 VT1

iL 1 V4

ULm

VT3

iC uC

+

u12

2

C

V2 VT4

VT2

U12m

U nm

c

U nm -ULm U12m

I Lm

ULm 0 ULm

0

I Lm

Fig. 4.1. Single-phase power conditioner circuit (a) and vector amplitude diagrams of leading current (b) and lagging current (c) modes

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4th part In the first mode, the instantaneous value of current iL decreases as the capacitor voltage is switched on in the direction opposite to that of the current iL , but in the second mode — it increases, as the capacitor voltage is switched on in series in alignment with the inductor current direction. Mode A, if direction of the input current iL is opposite to the depicted one, is applied when VT1, VT2 are on, whereas mode B is applied when the current is controlled by the diodes V4, V3. The curve of the bridge input voltage fundamental harmonics in modulation period results from the voltage average values of modes A and B within each modulation interval t t U12Tp = U d A − B , (4-2) T T where tA and t B stand for the mode duration in the modulation period T. The average voltage U12Tp of each modulation period T = t A + t B is to be equal to the sinusoidal instantaneous voltage value of fundamental harmonics, which in case of a leading input current is as follows: t t u2p = U nm sin ωt + ωLI Lm sin ωt = U d A − B . (4-3) T T It means that

(U + ωLI Lm )sin ωt 2t A −1 = nm . (4-4) T Ud In case of a high-volume capacitor and 90° leading current DC voltage of the capacitor C is determined as follows: di U d = ut − L L = U tm + ωLI Lm (4-5) dt max

therefore, the relative switching time t A sin ωt + 1 t 1− sin ωt = and B = . (4-6) T T 2 2 Figure 4.2 displays both relative switching time variations into 50 Hz frequency (ω = 314 s−1) wave taking into account the network AC voltage un of period Tn = 1/ f n.

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4th part 1.0 tB T

tA T

0.5

0

5

10

15

20 t, ms

Fig. 4.2. Variations of relative switching time within a network voltage period (fn = 50 Hz)

The capacitor current fundamental harmonic iCp is also a result of input current iL modulation: at mode A current iC = iL, but at mode B iC = −iL . Taking into account the relative switching time changes during the period (Fig. 4.2), the variations of current iC fundamental harmonic in the period take place as shown in Fig. 4.3. The frequency of fundamental harmonic iCp variations are twice higher than that of the network frequency: t 2t t iCp = I Lm cos ωt A − B = I Lm cos ωt A −1 = T T T = I Lm cos ωt sin ωt = 0.5I Lm sin 2ωt , (4-7) where ILm is an input current amplitude. Taking into account the changes of the capacitor current fundamental harmonics average value Ud, the curve of the capacitor instantaneous voltage uC also varies around Ud (Fig. 4.3). The capacitor voltage variations can be found by integrating curve of uC: uC =

1 C

∫i

Cp d t

+A=−

I Lm cos 2ωt + U d, (4-8) 4ωC

where A is determined at ωt = π / 4, when uC = U d. As long as the capacitor voltage matches the maximum instantaneous value UCm (4-5) at angle ωt = 0.5π, U Cm = U d +

I Lm , (4-9) 4ωC

i.e., the amplitude of the capacitor AC component is U Cmrip =

I Lm . (4-10) 4ωC

The expression can be used to calculate the capacitor capacitance provided for the values of allowable capacitor voltage AC component amount and pre-set ILm. For

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4th part example, if a capacitor at 2 f = 100 Hz can operate with the voltage AC component amplitude 50 V, but ILm can reach 100 A, then the capacitor is required of the volume C≥

100 = 1592 µF. 4 ⋅ 314 ⋅ 50

The inductor L inductance determines the range of variation of the current iL instantaneous value in the period of modulation, as well as under limited ripple range — switching frequency as well the capacitor voltage average value. To reduce the capacitor voltage in the leading current compensation case, it is relevant to assume the least possible inductance. Then in the case of a limited variation range of iL either the switching frequency gets higher or the variation range of current iL gets higher in the constant modulation period. The two mentioned parameters can be defined from the descriptive differential equation of the current iL variation in mode A: U nm sin ωt −U d = L

di L . (4-11) dt

Substituting diL for a complete change DI L and dt for the duration of mode A tA, it is feasible to determine the duration of mode A: tA =

L ∆I L U nm sin ωt −U d

, (4-12)

where DI L is the absolute value of the complete change of current iL in mode A. Taking into account (4-6), the transistor switching frequency at the defined constant DI L is calculated as follows: f=

(sin ωt + 1) U nm (sin ωt −1) − ωLI Lm

un iL

2L ∆I L

Unm

ILm

un iL

0

uC iCp

. (4-13)

t UCm uC

Ud

ICm 0

iCp

t

Fig. 4.3. Diagrams of instantaneous values of current and voltage fundamental harmonics 271

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4th part The maximum switching frequency of switches is at ωt = arcsin

ωLI Lm , (4-14) 2U tm

when it is f max =

(2U tm + ωLI Lm )2 . (4-15) 8U1m L ∆I L

The last expression can be applied provided that DI L and fmax of the defined inductance L are specified. Thus, if ∆I L = 5 A, U nm = 312 V (U n = 220 V), f max = 10 kHz, I Lm = 100 A , the applicable inductance L = 0.0055 H. f, kHz

tA , ms 0.30 0.25

8 0.20 f = f(t)

6 0.15 4 0.10

tA = f(t)

2 0.05 0

0

5

10

15

20

t, ms

Fig. 4.4. Mode A interval aplication time and modulation frequence variations in network voltage period at U1m = 312 V, w = 314 s–1, L = 0.005 H, ILm = 50 A, DIL = 5 A

The variations of mode A duration and the switching frequence within one period are displayed in Fig. 4.4.

Example A single-phase active rectifier operates with the input current leading by 90°. The rms value of a single-phase network voltage is U n = 220 V, f n = 50 Hz , input inductor inductance L = 5 mH, output capacitor capacitance C = 5000 µF, the control is implemented by the input current modulation in the period at constant total ripple (amplitude) ∆I L = 5 A and at the defined input current sine-wave amplitude I Lm = 50 A. Determine the capacitor voltage average value, capacitor voltage ripple total range and maximum frequency of transistor switching! 272

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4th part 1. Average value of capacitor voltage is U d = U n 2 + 2πfLI Lm = 220 2 + 314 ⋅ 5 ⋅10−3 ⋅ 50 = 390.5 V . 2. Total range of capacitor voltage ripple is ∆U C = 2U Cmk =

I Lm 50 ⋅106 = = 15.9 V. 2ωC 2 ⋅ 314 ⋅ 5000

3. Maximum frequency of transistor switching is f max =

(2U nm + ωLI Lm )2 (2 ⋅ 312 + 314 ⋅ 5 ⋅10−3 ⋅ 50) ⋅103 = = 7908 Hz. 8U nm L ∆I L 8 ⋅ 312 ⋅ 5 ⋅ 5

According to the example data, a computer-generated circuit simulation has been implemented, the results of which are shown in Fig. 4.5. As can be seen, the capacitor voltage average value is close to the calculated one. Similarly, the total amplitude of voltage ripple is about 16 V. The maximum commutating frequency is also similar to the calculated one.

60.00 30.00 0.00 –30.00 –60.00 400.00 300.00 200.00 100.00

iL , A fmax = 7 kHz

uC , V

un , V

400.00 200.00 0.00 –200.00 –400.00 730.00

740.00

750.00

760.00 Time (ms)

770.00

780.00

790.00

Fig. 4.5. Curves of inductor L current and capacitor C voltage in computer-generated simulation of a single-phase power conditioner in the leading input current mode subject to the calculated example parameters

Provided that the compensation current lags behind the voltage, the capacitor average voltage is as follows: U d = U nm −ωLI Lm, (4-16) i.e. the voltage is lower than the network voltage amplitude, but the AC component 273

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4th part amplitude of the capacitor voltage instantenous value is the same as in the previously considered compensation type. However, the transistor operating mode is easier because along with the same inductor inductance the maximum switching frequency is to be lower. Provided that a rectifier is loaded, i.e., a load consuming direct current Id (Fig. 4.6 a) is connected in parallel to the capacitor or DC source supplying the capacitor with the direct current Id is engaged, then the vector diagram in the case of conditioner active power operation mode shows that the fundamental harmonic vector of bridge input voltage U 12p is not in phase with the network voltage vector U n (Fig. 4.6 b) but lags behind (if Id is consumed) or leads the vector ahead (if Id is generated to the network from DC source).

a

V1 VT1

iL un

VT3

1

L

2 V4

cu n iL

b

V3

VT4

V2

i1

Ud

+

iC uC

– C

iL 0

Ld

I Lm

d un iL

2p

U12pm

ULm = X L I Lm

ILm p

Id

VT2

U1m

un

U nm

ULm

wt

un

0 iL

p

2p

wt

Fig. 4.6. Conditioner circuit in operation with a load (a), vector diagram of active current consumption (b), current and voltage time diagram in the case of consumption (b) and generation (d)

In the vector diagram, the amplitude value of voltage u12 fundamental harmonics 2 2 U12 pm = U nm + U L2m = U nm + ω2 L2 I L2m . (4-17)

On the other hand, if current iL is in phase with voltage un, then u12p = un − L

t di L t = U nm sin ωt − LωI Lm cos ωt = U d A − B . (4-18) T T dt

From here, the modulation interval time difference of modulation period T is 274

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4th part t A t B U nm LωI Lm − = cos ωt . (4-19) sin ωt − T T U UC C

If the capacitor voltage is U d = U Cm = U12 pm and relative relation U L* = (ωLI Lm ) / U nm, then

* *2 t A sin ωt −U L cos ωt + 1 + U L = . (4-20) T 2 1 + U *2 L

The fundamental harmonics of the bridge output current i1 is also the result of modulation of input current iL:

t t i1p = I Lm A − B sin ωt . (4-21) T T Taking into account the time difference of modulation interval, i1p =

I LmU nm 2 LωI L2m sin ωt − sin 2ωt , (4-22) UC 2U C

but the fundamental harmonics of a capacitor current is iCp = i1p − I d =

I LmU tm 2 ωLI L2m sin ωt − sin 2ωt − I d . (4-23) UC 2U C

Taking into account the assumed relative units iCp = −I d (cos 2ωt + U L* sin 2ωt ). (4-24) tA/T 1.0 0.9 0.8 UL* = 0.1

0.7

UL* = 0.2

0.6 0.5

UL* = 0.3

0.4

UL* = 0.4

0.3 0.2 0.1 0

0

2

4

6

8

10

12

14

16

18

20

t, ms

Fig. 4.7. Variations tA /T = f(t) of connection relative time tA within the network voltage period at different UL* 275

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4th part Figure 4.7 shows relative values t A / T of the connection time tA during the network voltage period at different values U L*. As can be seen, the relationship varies by the sine law with the network voltage frequency around an average value 0.5 and reaches either value 1 (point 1 of the circuit is completely connected to the capacitor upper plate) or value 0, when point 1 is completely disconnected from the upper plates of the capacitor. The variations of capacitor current relative values iCp / I d during the period at different U L* are shown in Fig. 4.8. As can be seen, the current changes with the two-fold network voltage frequency and amplitude, which is substantially equal to the load current Id.

iCp/Id 1.1

UL* = 0.1

0.9

UL* = 0.2 UL* = 0.3 UL* = 0.4

0.7 0.5 0.3 0.1 0 –0.1

5

10

15

20

t, ms

–0.3 –0.5 –0.7 –0.9 –1.1 Fig. 4.8. Variations of capacitor current fundamental harmonic relative values iСp/Id = f(t) within the period at different UL*

Applying the expression of the capacitor current fundamental harmonics, it is feasible to determine the curve of the capacitor voltage instantaneous variation: uCp =

1 C

∫

iCp dt + A =

I dU L* I cos 2ωt − d sin 2ωt + A. (4-25) 2ωC 2ωC

Assuming that at t = 0 s voltage uCp = U d =

I dU L* + A, 2ωC

it is possible to obtain the initial value A = Ud −

I dU L* . 2ωC

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4th part Taking into account A, the voltage instantaneous variation curve is described as follows: uCp = U d −

I U* Id sin 2ωt − d L (1− cos 2ωt ). (4-26) 2ωC 2ωC

The variations of the capacitor voltage instantaneous value within the period are demonstrated in Fig. 4.9. a iC

b uC

Id

Ud 0

2p wt

p

0

p Fig. 4.9. Capacitor current and voltage waveform in the active consumed current mode during the period

2p wt

When ωt = 135°, then 2 uCp = U Cm = U12 pm = U nm + ω2 L2 I L2m =

Id I U* − d L + U d. 2ωC 2ωC

From here the final capacitance of the average value of capacitor voltage is calculated as follows: U d = U nm 1 + U L*2 − As a result, C≥

(

Id 1−U L* ). (4-27) ( 2ωC

I d (1−U L* )

2ω U nm 1 + U L*2 −U d

)

. (4-28)

The calculation proves that if, for example, the 50 Hz frequency network voltage amplitude U nm = 600 V , U d = 500 V, U L* = 0.2 and I d = 200 A , then the required capacitance C = 0.00227 F. Since the input and output active power is equal, then I LmU nm = U d I d (4-29) 2 and the inductor inductance is equal to L = 1.14 mH at the assumed ratio U L* = 0.2 277

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4th part taking into account (4-29) at ω = 314 s−1. Taking into account (4-29), it is feasible to formulate the control operation calculating principle of the required amplitude of an inductor current. As can be seen, at the constant Ud the inductor current amplitude should be two times greater than the load current. A three-phase option provides that each AC voltage phase is supplied by an inductor and the bridge is configured with six dual conductivity (diode-transistor) switches (Fig. 4.10). Each inductor is attached to its switching leg, which comprises the upper and the lower switch. Control can be implemented in each inductor circuit by switching on a current sensor and comparing its signal with the phase current instantaneous value reference wave, at ϕ = 0° respectively iAr = I m sin ωt ; iBr = I m sin(ωt − π / 3); iCr = I m sin(ωt + π / 3), wheret ω = 314 s−1 at the network voltage frequency 50 Hz. If the inductor instantaneous current is a little higher than its reference current instantaneous value, then the upper switch of the leg is conducting and depending on the inductor current direction the control is exercised either by a diode or a parallel transistor. If the inductor instantaneous current is a little lower than the reference instantaneous value, then the lower switch of the leg is conducting. In both cases, the switch currents are coupled through the capacitor C0 circuit.

A

L

B

L

C

L

iA iAr iB

iL

C0

Ud

1 –

1

+ –

iBr iC

1

+ –

iCr

+

Fig. 4.10. Three-phase active rectifier — reactive power compensator circuit 278

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4th part Similar to a single-phase option, if an output capacitor is not connected to a load or a power source, then the system functions only as a reactive power compensator, i.e., with each inductor current displacement angle against the phase voltage by 90°. If a load is connected, then an angle of phase displacement can be either zero or equal to other values except 90°, when the load is to be disconnected. In turn, if a DC voltage source is connected to the capacitor C0, then via this active rectifier the power can be regenerated to the ac current network with the displacement angle of 180° or any other required angle. If the modulation frequency of a switch is high enough, then an inductor current is practically sinusoildal and the rms value of the switch bridge interphase voltage fundamental harmonics is determined as follows: 2 U12rms = 3(U ph − 2U phU L sin ϕ + U L2 ), (4-30)

where Uph — the rms value of network phase voltage, UL — the rms value of inductor voltage wIL , IL — the rms value of inductor currrent, angle φ is assumed to be positive at lagging current IL . The dc voltage of the capacitor C0 tends to this interphase voltage fundamental harmonic amplitude, that is, U d = 2 U12rms . (4-31) It is useful to remember that the active power of an active rectifier is equal at the input and output, i.e., the load resistance rating should secure the inductor active component current of an active rectifier in accordance with the load current.

4.2. Rectifier — Network Current Waveform Corrector The corrector should be configured, for example, by the way of a single-phase bridge rectifier, where an output load is connected through the DC voltage boost mode switch controller (Fig. 4.11). A switch controller consists of the inductor L connected in series to a rectifier output and transistor VT connected in parallel to a load junction, separated from the load by the diode V0. The transistor switches at a high frequency; besides, the inductor current is kept in a sinusoidal unipolar half-wave form that matches in phase with a rectifier output voltage ud half-wave (see Fig. 4.12).

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4th part id V1

V4 u1

V2 V3

V0

L

ud

uC

VT

ild

Fig. 4.11 Single-phase bridge-type rectifier with corrected network current

u i

UC

U1m ud Idm id

0

T1/2

t

Fig. 4.12 Corrected rectifier output inductor current waveform

In this circuit, the capacitor voltage is higher than the current instantaneous m eaning of rectifier voltage ud, because without the transistor switch the capacitor voltage will be directly related to the amplitude of ud, i.e., to be U C > U1m in this circuit. Let us assume that the capacitor has an infinitely high volume, then the capacitor voltage is constant in time and equal to uC = U C but higher than the amplitude value U1m of rectified voltage ud. Since ideally circuit elements suffer no power loss, the load power and input average power are equal: U1m I dm

1 π

π

∫ sin ωtdωt = 2

0

U1m I dm = U C I ld , (4-32) 2

where I ld = U C / R is load current with resistance R, which under smooth uC is also smooth, U1m is the supply network voltage amplitude, and Idm is the amplitude of an inductor current. From here the capacitor voltage is UC =

U1m I dm R > U1m . (4-33) 2

The average value of the diode V0 current fundamental curve iVp during the network 280

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4th part voltage half-cycle period T1 / 2 is equal to the load current Ild, because the average value of capacitor current is zero. The fundamental curve itself is the result of modulation of current id by a transistor: tB I dm sin ωt , (4-34) T

iVp =

where T is the switching period of transistor switch, but t B is the time interval during the period, when the transitor is in the off-state and the current id flows via a diode. The ratio t B / T in a half-cycle period may be defined from the relationship U1m sin ωt =

tB U C , (4-35) T

which characterises the voltage ratio in the case of dc voltage boost switch controller. As can be seen, t B U1m 2U1m = sin ωt = sin ωt < 1. (4-36) UC I dm R T This time interval ratio should be less than 1 because U1m < U C . It means that 2U1m < I dm R , i.e., in the process of control the current amplitude of the inductor L should be maintained as follows: I dm >

2U1m . (4-37) R

Using value t B / T , the diode current fundamental curve can be described as follows: iVp =

U1m I dm 2 sin ωt . (4-38) UC

The ratio of the switching time of transitor switch VT is defined as follows: t tA 2U1m = 1− B = 1− sin ωt < 1. (4-39) T T I dm R The expressions obtained support the calculation of all main parameters of stationary electromagnetic process. Thus, for example, if U1m = 312 V , ω = 314 s−1 and R = 15 Ω, then Idm should be greater than 42 A. The variations of capacitor voltage, load currents and switching relative times extreme values in relation to I dm > 42 A are shown in Fig. 4.13.

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4th part UC

Ild, A UC , V 450

30

(tB/T)max (tA/T)min Ild

400

1.0

(tB/T)max 0.5 (tA/T)min

25

350

20

300 45

55

65

75

85

0 95 Idm, A

Fig. 4.13. Dependence of capacitor voltage, load currents and switching relative time extreme values on the input inductor current amplitude taking into account the calculated data

Taking into account (4-36), the expression of current fundamental curve of the capacitor C in one network voltage half-cycle period is as follows: iCp = iVp − I ld =

U1m I dm 2 sin ωt − I ld . (4-40) UC

According to this expression, the diode and capacitor currents as well as the capacitor voltage demonstrated in Fig. 4.14 vary in the network voltage half-cycle period. The capacitor voltage starts to gain from the moment of time when iCp is zero, and the voltage continues to gain till the moment of time when iCp is zero again. These time instants in a half-cycle period can be defined from (4-40): t e1 =

I ldU C T 1 1 1 0.785 and t e2 = 1 − t e1. arcsin = arcsin = ω U1m I dm ω ω 2 2

i

0

iVp

te1

te2

uC 0

Ild T1/2

T1 iCp

t

UCm DUC

t

Fig. 4.14. Fundamental curves of diode and capacitor current and the curve of capacitor AC component variation

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4th part Taking into account increment time interval and roughly assuming that the voltage gain is defined by the side of the capacitor current fundamental curve positive amplitude, the capacitor voltage increment is calculated as follows: T1 − 2t U1m I dm − I 0.5 e1 ld U 2 C ∆U C = . (4-41) C This expression can be applied to calculate the required capacitance of a capacitor after a given maximum permissible voltage ripple range [DUC]: T1 1.57 U1m I dm − ω 2R 2 C≥ . (4-42) 2[∆U C ] For instance, if in the example considered above I dm = 65 A and [∆U C ] = 30 V, then capacitance of at least 2166 mF of a capacitor is required. The input inductor L is the second element that determines the features of the circuit. The increment of the inductor current instantaneous value within transistor switching-on interval A is ∆I L d = U1m sin ωt . (4-43) tA Applying (4-37), the total increment of the inductor current during interval tA is 2U1m sin ωt U1m sin ωt ⋅T 1− I dm R ∆I d = . (4-44) L Finding the maximum increment value in a network voltage half-cycle period from the dependence d(∆I d ) = 0, dt it can be concluded that the maximum corresponds to the moment of time t max =

1 1 I dm R arcsin , (4-45) 2 2U1m ω

but the value of current increment maximum is ∆I d max =

U1maxT 4L

I dm R . (4-46) 2U1m

From here the required inductance of an inductor at the given current maximum increment range is L≥

U1m I dm R 4 2 ∆I d max f sw

, (4-47) 283

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4th part where f sw = 1/ T — the switching frequency of the switch. If in the example above f sw = 5 kHz, [∆I dmax ] = 5 A, then at I dm = 65 A the inductor of inductance L = 3.9 mH is required. To verify the expression-based calculations, the computer-generated simulation of the circuit has been configured with I dm = 65 A, U1m = 312 V , 50 Hz, inductor L = 3.9 mH, capacitor C = 2166 µF and ∆I d = 5 A , load R = 15 Ω. As can be seen (Fig. 4.15), the average value of capacitor voltage is 390 V, which approximately corresponds to the calculated value. The maximum frequency is 4.8 kHz, the value is also in line with the calculated one, but ∆U C = 31 V.

80.00

id, A DId = 5 A

60.00 40.00 20.00

fmax = 4.8 kHz

0.00 –20.00

uC , V

420.00 400.00 380.00 360.00 340.00 320.00 300.00 280.00 75.00

DUC = 30 V

80.00

85.00 Time (ms)

90.00

95.00

Fig. 4.15. Curves of inductor current and capacitor voltage in computer-generated simulation taking into account the calculated data

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Review Questions

4th part 1. To what purpose a non-controllable diode rectifier is shunted by the transistor with the reverse conduction direction to a diode?

2. To what purpose a high-value capacitor is mounted to the dc voltage side of an active rectifier with diode-transistor elements?

3. What element should be connected to the ac voltage side of an active rectifier with diode-transistor elements and to what purpose?

4. What do we call rectifiers with diode-transistor elements in practice? 5. In which mode of operation does a rectifier work if there is no any other element as one parallel capacitor on the dc side of an active rectifier?

6. Is voltage of a capacitor higher or lower than the output dc voltage of a

uncontrolled diode rectifier of similar configuration, if an active rectifier operates as a lagging reactive power compensator?

7. Is voltage of a capacitor higher or lower than the output dc voltage of a

uncontrolled diode rectifier of similar configuration, if an active rectifier operates as a leading reactive power compensator?

8. What are the instantaneous values of the input AC voltage of an active bridge rectifier?

9. Which way shall we determine the reactive power of a compensating active rectifier?

10. Is the waveform of an inductor current on the AC voltage side of an active rectifier sinusoidal?

11. What transistor of an active rectifier leg is to conduct the current after the current ceases to be run by the upper diode of the same leg?

12. How high is the fundamental harmonic frequency of a capacitor current in a single-phase active rectifier circuit?

13. Is an active rectifier capable of consuming (generating) reactive power, if this rectifier is loaded with a resistor on the DC voltage side?

14. Is a resistor-loaded active rectifier capable of operating with respect to AC current network with a pure active power?

15. What shall have to be connected to the DC voltage side of an active rectifier to enable it to generate active power to the grid?

16. How much source voltage connected to the DC voltage side is required to enable an active rectifier to generate active power to the AC current network?

17. How big is the angle of displacement between the network phase voltage and its current to provide maximum active power to the grid?

18. What should be taken into account to decide on the required inductance of the network phase inductor?

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Review Questions

4th part 19. What should be taken into account to decide on the required capacitance of a DC voltage side capacitor?

20. In what way will you configure a three-phase active rectifier circuit? 21. What kind of voltage fundamental harmonics will define a three-phase active rectifier capacitor voltage?

22. What is the purpose of use of a power factor corrector electronic system? 23. At what DC voltage in DC voltage conversion mode does the corrector DC voltage side system operate?

24. What is the relationship of the capacitor voltage connected in parallel to the load against the network voltage amplitude in a corrector circuit?

25. Is a corrector capable of providing a sine-wave character to the network current if the load capacitor voltage is less than the network voltage amplitude?

26. What is the dependence of the load active power in a corrector circuit on the capacitor voltage?

27. To what extent does the switching ratio of a corrector regulating transistor change due to the rectifier output voltage instantaneous value in a half-cycle period?

28. What should be followed to decide on an inductor and capacitor volume choice in a corrector circuit?

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Tasks

4th part 1. A single-phase bridge-type reactive power compensator-active rectifier net-

work voltage is 220 V, 50 Hz, it operates with a leading input current. The maximum instantaneous value of the output capacitor voltage is (350 + mn) V, the input inductor sinusoidal current amplitude is (50 + m + n) A. Determine the applicable input inductor inductance, capacitor voltage average value as well as the required capacitance of a capacitor if the permissible ac component amplitude of the output dc voltage is 10 % of this voltage average value!

Answer: L = 2.4 mH, Ud = 318.2 V, C = 1138 mF at m = n = 0. 2. A single-phase bridge-type reactive power compensator-active rectifier net-

work voltage is 220 V, 50 Hz, it operates with a lagging input current. The average value of the output capacitor voltage is (150 + mn) V, the input sinusoidal current amplitude is (50 + m + n) A. Determine the applicable input inductor inductance, capacitor amplitude voltage, as well as the required capacitance of a capacitor if the permissible ac component amplitude of the output dc voltage is 10 % of this voltage average value!

Answer: at m = n = 0 L = 9.4 mH, UCm = 165 V, C = 2654 mF. 3. The network phase voltage rms value of a three-phase bridge-type active recti-

fier is 220 V, 50 Hz, it operates with a current leading its phase voltage run by each input inductor with inductance L = (1, 5 + 0, m) mH . Determine the output capacitor voltage value if the inductor current rms value is (50 + mn) A and the inductor current phase displacement angle is 90° and 30°! What is the rated load of a resistor to be connected in parallel to the capacitor providing the second phase displacement angle?

Answer: at m = n = 0 Ud = 599 V (at angle 90°) and 572, 25 V at 30°. Load R = 11,46 Ω. 4. The network current correction input voltage of a single-phase bridge-type

rectifier is 220 V, 50 Hz. Determine the current of the load R = 10 Ω if the input sinusoidal current amplitude is (75 + m + n) A! Determine the applicable capacitance of a capacitor if the permissible full load voltage ripple is 10 % of the load average voltage!

Answer: at m = n = 0 34.08 A and 3325 mF.

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REFERENCES 1. Raņķis, I. Power Electronics. Riga: RTU, 2002, 142 p. (Raņķis, I. Energoelektronika. Rīga: RTU, 2002, 142 lpp. – in Latvian) 2. Greivulis, J., Raņķis, I. Electronic Elements and Devices of Control. Riga: Avots, 1997, 287 p. (Greivulis, J., Raņķis, I. Iekārtu vadības elektroniskie elementi un mezgli. Rīga: Avots, 1997, 287 lpp. – in Latvian) 3. Blumbergs, E., Greivulis, J., et al. High Voltage Industrial Electronics. Riga: Liesma, 1974, 246 p. (Stiprās strāvas rūpniecības elektronika / Blumbergs E., Greivulis u.c. – Rīga: Liesma, 1974, 246 lpp. – in Latvian) 4. Chizhenko, I., Rudenko, I., Senko, V. Basics of Converter Devices. Moscow: Vischaja shkola, 1974, 430 p. (Чиженко, И., Руденко, В., Сенько, В. Основы преобразовательной техники. Москва: Высшая школа, 1974, 430 стр. – in Russian) 5. Raņķis, I.J. Basics of Converter Devices. Semiconductor Elements and Rectifiers. Riga: RPI, 1979, 110 p. (Ранькис, И.Я. Основы преобразовательной техники. Полупроводниковые элементы и выпрямители. Рига: РПИ, 1979, 110 стр. – in Russian) 6. Raņķis, I.J. Basics of Converter Devices. Thyristor Converters and Regulators. Riga: RPI, 1980, 110 p. (Ранькис, И.Я. Основы преобразовательной техники. Тиристорные преобразователи и регуляторы. Рига: РПИ, 1980, 110 стр. – in Russian) 7. Thorborg, K. Power Electronics – In Theory and Practice. Lund: Studentliteratur, 1997, 522 p. 8. Mohan, N., Undeland, T., Robbins, W. Power Electronics: Converters, Application, Design. NY: John Wiley and Sons, 1989, 667 p. 9. Rashid, M., H. Power Electronics. Handbook. London: Academic Press, 2001, 895 p.

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Power electronics

The textbook is developed for the students, who study power electronics and other close segments of science and technologies. It can be usef...

Power electronics

Published on May 16, 2018

The textbook is developed for the students, who study power electronics and other close segments of science and technologies. It can be usef...

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