ELECTRONIC POWER CONTROL The textbook Electronic Power Control consists of two volumes: Power Electronics and Electronic Motor Control. The fundamentals and applications of power electronics are thoroughly discussed with mathematical derivations, a lot of numerical examples and clarifying images. This enables both students and professionals to study power electronics in depth. The author has more than 40 years of experience in teaching on the subject. Thousands of students (engineering, electronics) found a useful companion in this book. The seventh edition of the Dutch textbook, which is the result of continual updating, is now also available in English. The book is in full colour and includes a glossary containing more than 660 technical terms translated in Dutch, German and Spanish. An indispensable companion for engineers and other specialists in the field of power electronics
Volume 1  Power Electronics ISBN 9789038217918 504 pages ! 33.00 (VAT incl., shipping and handling excl.) ! Click here to order
Volume 2  Electronic Motor Control ISBN 9789038219110 404 pages ! 30.00 (VAT incl., shipping and handling excl.) ! Click here to order
TABLE OF CONTENTS Volume 1  Power Electronics PART 1  SEMICONDUCTOR SWITCHES 1. PHILOSOPHY OF POWER CONTROL 1. Controlling of electrical energy using switches 2. Switching matrix 3. Controllable semiconductors 4. Properties of switches 5. Commutation 6. Power converters 7. Power frequency domain 8. Evaluation 2. POWER DIODES 1. Semiconductors 2. IV characteristic of a junction diode 3. Power diodes 4. Data of a power diode 5. Excerpts from data books 6. Evaluation 3. TRANSISTOR POWER SWITCHES 1. Bipolar junction transistor (BJT) 2. Power MOSFET 3. IGBT 4. Evaluation 4. THYRISTORS 1. Shockley diode 2. Silicon controlled rectifier (SCR) 3. Diac 4. Triac 5. Gate turn off thyristor (GTO) 6. Mos controlled thyristor (MCT) 7. Integrated gate commutated thyristor (IGCT) 8. Evaluation 5. NOTES 1. 2. 3. 4. 5. 6. 7. 8.
Electrical and mathematical notation Optoelectronics Hall effect sensors Heat dissipation from semiconductors Colour code for resistors / Noninductive resistors Normalisation of resistor and capacitor values Colour code for capacitors Type indication of semiconductors and integrated circuits
9. 10. 11. 12.
The skin effect EMCEMI Sisurface area of power switches Size winding wire (AWG = American wire gauge)
6. COMPUTER SIMULATIONS 1. Introduction 2. History: analogous simulations 3. SIMULINK: simulation with block diagrams 4. SPICE: simulation of an electric network 5. CASPOC: the multilevel model PART 2  POWER CONVERTERS 7. UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD 1. Current flow of an inductive circuit with a sinusoidal supply 2. Current and voltage waveforms of an inductively loaded singlephase rectifier 3. Approximations in the study of rectifiers 4. Halfwave threephase rectifier. Resistive load 5. Threephase bridge circuit 6. Table 73: singlephase rectifiers 7. Table 74: threephase rectifiers 8. Evaluation 8. LINEFREQUENCY PHASECONTROLLED RECTIFIERS 1. Halfwave singlephase controlled rectifier (E 1 ). Resistive load 2. Resistive and inductive loaded E 1 controller 3. Table 81: Singlephase power control 4. Harmonics in the case of phasecontrolled sine wave 5. Distortion power 6. Full wave B2controlled rectifier . Resistive inductive load 7. Fully controlled B6rectifier. Resistive load 8. Fully controlled B6rectifier. Resistive and inductive load 9. Full controlled B6rectifier: dead time, harmonics 10. Twelvepulse controllers 11. Evaluation 9. ACCONTROLLERS 1. ACcontroller with phasecontrol 2. ACcontroller with integral cycle control 3. Turnoff snubber for thyristors 4. Solid state relays 5. Radio interference suppression (RFI) of thyristors 6. Evaluation 10. CYCLOCONVERTERS 1. Continuous cycloconverter 2. Trapezium cycloconverter 11. CONTROL OF THYRISTORS 1. Firing pulses 2. Pulse transformers 3. IC for phase control of thyristors and triacs 4. Triac control with a diac 5. Evaluation 12. CHOPPERS 1. Operating principle of a chopper 2. Control modes 3. Resistive and resistiveinductive loaded choppers 4. Chopped resistor 5. Chopper control IC’s 6. Evaluation 13. SWITCHMODE POWER SUPPLIES 1. Basic principles of switchmode power supplies 2. Basic converter configurations 3. Isolated switchmode power supplies 4. Flyback converter 5. Forward converter 6. Converter control strategies 7. Twotransistor SMPS of the forward type 8. Forward with multiple outputs 9. Fullbridge of the buck type 10. Synchronous SMPS 11. SMPS components 12. Overview of SMPS up to 2500 W 13. Non ideal waveform 14. Digital control of an SMPS 15. Evaluation 16. The design of switchmode power supplies 14. INVERTERS A. THREEPHASE INVERTERS 1. Voltage source and current source inverter 2. Switching matrix of a voltage source inverter 3. 180°type inverter 4. Pulse frequency inverter with a constant DCvoltage 5. Pulse width modulation (PWM) 6. PWMstrategies
7. Harmonics in a PWMwave 8. 120째type inverter 9. Common switches used in inverters 10. Control circuit for threephase inverter bridge of the 180째type B. SINGLEPHASE INVERTER 1. Basic circuit of full bridge inverter 2. Unipolar and bipolar PWM 3. Full bridge with unipolar PWM 4. Harmonics with unipolar PWM 5. Evaluation 15. APPLICATIONS OF POWER ELECTRONICS 1. Uninterruptible power supplies 2. Highfrequency inductive heating 3. Power factor correction (PFC) 4. Lighting 5. Renewable energy A. Wind turbines B. Photo voltaic solar panels 6. Drive technology 7. Motion control 8. High frequency induction cooking plate
Volume 2  Electronic Motor Control 16. ELECTRIC MACHINES 1. Static transformers 2. DC commutator machines 3. Threephase asynchronous motors 4. Synchronous machines 5. Small appliance motors 17. DRIVE SYSTEMS 1. History 2. Control theory 3. Types of drive systems 4. Electronic drive technology 5. Useful mechanical formulas 6. Moment (of torque) and power of a motor 7. Run out test to determine moment of inertia of a load 8. Numeric examples 18. CURRENT, ANGULAR POSITION, SPEED TRANSDUCERS 1. Current sensors 2. Angular position sensors 3. Speed sensors 19. SPEED and (or) TORQUE CONTROL of a DCMOTOR A. DCMOTOR supplied from an AC POWER GRID 1. Control of an independently excited motor 2. Mn curves 3. Regulated single quadrant drive 4. Two quadrant and four quadrant operation 5. Functional control diagram of a single quadrant drive 6. Optimising controllers 7. Numeric example B. DCMOTOR supplied from a DC SOURCE 1. Chopper controlled drive 2. Chopper control of a series motor Traction service Line filter 20. SPEED and (or) TORQUEcontrol of THREEPHASE ASYNCHRONOUS MOTOR 1. Threephase asynchronous motor 2. Electronic control of an induction motor 3. Scalar regulation of induction motor speed 4. Slip control 5. Scalar frequency converters 6. Indirect frequency converter of the VSItype 7. Vector control 8. Microelectronics with power electronics 9. Softstarters 10. Indirect frequency converters with a current DC link (CSI) 21. ELECTRONIC CONTROL of appliance motors, switched reluctance motor, synchronous threephase motor, induction servo motor 1. Appliance motors Universal motor Single phase induction motor 2. Switched reluctance motor 3. Synchronous AC motor 4. Threephase induction servo motor 22. ELECTRICAL POSITIONING SYSTEMS 1. Servomechanisms 2. Electrical positioning systems. Definitions 3. Position control with DC servomotor 4. Position control with brushless DC motor
5. 6. 7. 8. 9.
Position control with stepper motor Position control with AC servomotor Position control with linear motor Computer guided motion control An integrated system (SIMOTION from Siemens)
23. eMOBILITY 1. Renewable energy 2. Electrical traction 3. Electrical automobiles 4. Electrical boats 5. Electrical bicycles
THE PHILOSOPHY OF POWER CONTROL
1.7
7. POWER FREQUENCY DOMAIN Fig. 1.1 illustrates the state of the art in power switches. Currently we see that with the exception of an SCR of 8.5 kV that thyristors are limited to 4 kV  4 kA. This is a maximum of 16 MW per switch. This is considered the mid power domain. These days an SCR is also known as Phase Controlled Thyristor (PCT). There are for example single IGBT’s of 1200V  3600A, 1700 V  2400 A and 3300V  1500A. In a half bridge using IGBT’s a typical rating would be 1700V 1000A. State of the art components use 125 mm Si wafers.
Intensity kA SCR
IGCT
GTO
4 kA
IGBT module
3 kA
Voltage kV 100 Hz
2 kA
1 kHz 10 kHz
MOSFET
1 kA
100 kHz 1 MHz
Hz
10 MHz 2 kV
4 kV
6 kV
8 kV
Frequency Fig 111: Properties of power switches
Transport of electrical energy over long distances can be more economical using DC transmission. For example when the distance is greater than 250 km then transmission of 400kV1200MW is more economical using HVDC (high voltage direct current). With submarine cables the distance is even shorter. In China and Brazil where large hydro electric power stations are more than a 1000 km from the big cities HVDC is used. Example: the Ultra High Voltage DC (UHVDC) of 800kV between Xiangjiala and Shanghai. The firm ABB developed a 6inch 8.5kV thyristor for this application. Another application of HVDC is the connection between AC  grids of different frequencies. An example is the Garabi back to back station which connects the 60 Hz grid of Brazil and the 50 Hz grid of Argentina.
3.14
TRANSISTOR POWER SWITCHES 2.3 Operation If a positive voltage VGS is applied (fig. 315a) then as a result of electrostatic induction an Nzone is created in the P structure between the two existing N + zones. Since the Nzone consists of minority charge carriers (from the P substrate) we refer to this as an inversion layer. Source (S) and drain (D) are now connected via the Nchannel and with a positive voltage VDS a current ID will flow from D to S. The larger VGS , the wider the Nchannel and the larger the current ID will be. In fig. 315b a schematic representation of the configuration of fig. 315a is shown. I
D
I
D
induced Nchannel
D D
R
D
–+ –+ –+ –+ –+ –+ –+ –+
G
V
GS
G S
V GS +
S
R
D
N+
V
P
N
V DD
DD
(b) depletion layer
(a)
Fig. 315: Operation and principal configuration of a Nchannel enhancement type MOSFET
2.4 Mosfet characteristics The characteristic in fig. 316a is analogous to the IC  VCE  characteristic of a bipolar transistor. In the case of a BJT we control the transistor with the base current, while with a MOSFET we work with voltage control. The drain current consists of one type of charge carrier therefore we refer to a MOSFET as a unipolar transistor in contrast to the BJT. I
D
(A)
I (A) D
V
GS
9V 10
10
8V
8
V = 30V DS
8
7V
6
6 6V
4
4 5V
a
2
2 4V
0
10
20
30 (a)
40
DI D
DV GS
V
DS
50
(V)
V
GS
0
1 2 3 4 5 6 7 8 9 4,75 7,75 Vthreshold (b)
Fig. 316 (a and b): Characteristics of an Nchannel enhancement MOSFET
(V)
3.15
TRANSISTOR POWER SWITCHES
From the output characteristic we can derive the transfer characteristic (fig. 316b). This transfer provides us with a very important specification , namely the transconductance: ∆ ID te _____ gfs = ( ∆ V ) with VDS = C ; gfs = tg α
(36)
GS
In fig. 316 b we see that the gate voltage must exceed a certain minimum VGS(th) before the MOSFET conducts. The SiO2 layer between the gate and source is responsible for the very high input resistance of the MOSFET (10 10 to 10 15 Ω ). A single elementary unit as shown in fig. 314 is only capable of handling a drain current of 100µA. The parallel combination of for example 100,000 similar elementary units results in a MOSFET with a drain current of several amps as shown in fig. 316a. I
D
(A) [ v
GS
V
GS(th)
V GS
= vDS ]
9V
10
8V
8 Ohmic limit curves
ve Acit area
7V
6
6V
4
V
GS(th)
5V
2
4V
0
10
20
30
40
(c)
V 50
60
70 80 CUT  OFF
90
DS (V)
100
BV
DSS
Fig. 316c : Operating regions of an Nchannel MOSFET Fig. 316c shows the complete I  V characteristic with the three distinct regions: 1. Cutoff: With the gate voltage below the turnoff voltage VGS(th.) there is no current flow and the MOSFET and functions as an open switch. In the characteristic the breakdown voltage BV DSS of the MOSFET is shown. This means that the applied drain source voltage must be less than this BV DSS BV = Breakdown voltage; DSS = DrainSource in the common source configuration. 2. Active region: In this region we have an almost linear relationship between drain current and gate source voltage ( iD = g fs . u GS ) 3. Ohmic area: With a BJT the border between the active area and saturation is given by VCE = V BE or VCB = 0, see fig. 35. In similar fashion for a mosfet the border between the active area and saturation is given by: VDG = 0 or VGS − VGS(th.) = VDS . When VDS is low, every value of VGS ends up on the border line in the ID − V DS characteristic. These border lines (fig. 316c) all have the same ramp irrespective of the value VGS . On the border line the relationship between ID and VDS is practically constant, the MOSFET behaves as a resistor RDS . We refer to an ohmic operating area. In this area or region the MOSFET functions as a closed switch. We refer to the resistance of this “closed” (MOSFET) switch as RDS(ON) . The value of RDS(ON) follows from the ramp of the border line in fig. 316c, but this is also to be found in the datasheet of the MOSFET.
3.36
TRANSISTOR POWER SWITCHES
3 IGBT As we know a bipolar transistor has a low VCEsat even with a large current (also with high voltage transistors). A power MOSFET in contrast requires only low gate power and can operate at high frequency. In the construction of an IGBT (Insulated Gate Bipolar Transistor) the manufacturers sought to combine the benefits of the bipolar transistor and the MOSFET. The first design of an IGBT dates from 1980.
3.1 Construction
Fig 3.35a shows the construction of a single IGBT cell. Here in the N +− substrate from the MOSFET of fig. 320 has been replaced with a Psubstrate. An IGBT is made up of thousands of such cells which are all connected in parallel. For further study a diagram of a single cell is shown and this is considered as being representagate G emitter E tive of the complete IGBT. Analogous to fig. 320 Al the internal parasitic components (T1 , T2 , etc) are SiO2 represented in fig. 335b. N+ N+ P P In fig. 335 b and 336 note that an Nchannel MOSFET is present and a PNP transistor T1 . The parasitic NPN transistor from fig. 320 is also present. If T2 and its corresponding RBE are neglected then Nfig. 337a and b are a rudimentary equivalent diagram P of fig. 336. collector C
(a) Construction
emitter E CGE
R1
gate G
Al SiO2 N+
N+ RBE
T2
P
P
CCE
CGC
RD T1
N
P collector C (b) IGBT with internal parasitic elements Fig. 335: Construction of an IGBT
3.37
TRANSISTOR POWER SWITCHES 3.2 Operation collector C
The equivalent circuit as drawn in fig.337b consists of a PNP transistor controlled by an Nchannel MOSFET in a pseudoDarlington configuration. Since there is a MOSFET in the input and a transistor in the output the symbols used in fig.337c are easily understood. In the text the bottom symbol of the three will be used. At the input the IGBT behaves as a MOSFET and at the output approximately the behaviour of a BJT.
T1 CGC gate G
RD
N
R1
CCE T2
CGE
RBE
emitter E Fig. 336: Equivalent diagram for fig. 335b C
collector C
I
C G
ID CCG
R gate G
E C T1
CCE
R1
T1
G
E
G
C
CGE G
(a)
emitter E
(b)
E
(c)
E
Fig. 337: Reduced equivalent diagram of an IGBT and symbols
The physical operation of an IGBT more closely resembles that of a BJT than a power MOSFET. The reason is that the Psubstrate of the IGBT is responsible for injecting the minority charge carriers in the N − layer so that conduction is much better than with a MOSFET that works with majority charge carriers. Since no high voltage exists across the MOSFET part we can for construction purposes use a low voltage type. Low voltage types have a low RDS(ON) as important characteristic. The base of the PNP has no external connection which is expressed in the upper symbol of the IGBT in fig. 337c. In the equivalent diagram of fig. 337a it is clear that the voltage drop is composed of two parts: on one side the diode voltage drop across the emitterbase junction of T1 and on the other side the voltage drop across RD and the gating MOSFET. Since it is a low voltage MOSFET the voltage drop across this FET is dependent on the gate control voltage (this is not the case with high voltage MOSFETS !). VCEsat = VBE(T1) + ID . (RD + RDS(ON) ) We write:
(38)
IC ___
with ID = h FE
3.51
TRANSISTOR POWER SWITCHES 3.5.2 Trench NPTIGBT
The NPTtrench structure was designed to reduce V CEsat and the switchoff losses E off (fig. 348). In contrast with a planar NPTIGBT the trench type has a lower VCEsat , lower switching losses (Eoff ) and a lower thermal resistance R th . As a result of the trench structure theRDS(ON) is eliminated from the parasitic MOSFET of fig. 337a so that (38) becomes: CEsat V = V BE(T1) + ID . R D
(312)
The trench has therefore a lower V CEsat than the NPT shown in fig. 346.
3.5.3 Trenchfieldstop (FS) structure Fig. 349 shows the implementation. In comparison to the trench implementation of fig. 348 the thickness of the N −base is much reduced and there is a lightly doped Nbuffer added. From the graph of the field strength we can see where the name field stop comes from. Since the N −base of a trench FS is much thinner than a trench the drift resistance R D is lower and expression (312) shows that VCEsat is reduced in comparison to a trench IGBT.
E Emitter
0
Emitter
Emitter
Emitter
N
N P
P
N Gate
E
0
N Gate
P
P N
N  buffer FS P
NP
Collector
Collector
Fig. 348: Trench structure of an IGBT
x
x Fig. 349: Trench fieldstop IGBT
5.16
NOTES
2. OPTOELECTRONICS 2.1 LED (Light Emitting Diode) The light emitting diode (LED) has been used for years as a signal lamp, as number indicator and as a light emitting transmitter in an optocoupler. In recent years the LED is becoming ever more popular as a light source. This aspect is dealt with in chapter 15 in the applications of power electronics. The operating principle of the LED rests on the release of energy by direct recombination of electrons and holes in a PNsemiconductor. This energy is released in the form of electromagnetic radiation. In general this is: (529) W = h . f with: W = energy (eV) h = Planck constant = 4.133 x .10 −15 eVs f = frequency of electromagnetic waves in Hz. In the specific case of a semiconductor we find: W = Wg = energy gap or band gap between the valence and conduction band. This energy gap is also temperature dependent. Si has a W = 1.12eV at 0 K( = −273° C) and this becomes 1.1 eV at 300K (room temperature). The wavelength of the released radiation is given by: 6 1239 300 . 10 6 300.10 x 4.133 x 10 −15 300 . 10 6 c __________ _________________________ ______ λ = f __ = __________ = .h = = (nm) Wg Wg Wg f Herein c = 300,000km/s = speed of light. To have continuous radiation we require continuous recombination to occur between electrons from the conduction band and holes from the valence band. This continuous process can be obtained by a forward polarised PNjunction (e.g. with IF = 20mA). To obtain visible light (λ = 400 to 700nm, see fig. 536) Wg must lie between 1.77 and 3.1eV. The semiconductor material most often used for the construction of LED’s is 35 crystals with the following configuration: group 3: Ga, Al group 5: As, P, N. The different combinations and impurities lead to different wavelengths of the radiated light. The spectrum of a galliumphosphide  LED (GaP) can be yellow ( λ = 575nm) to green (λ = 560nm). Galliumarsenidephoshide (GaAsP) emits yellow ( λ = 590nm) to red light ( λ = 650nm). GaAs has an energy gap of about 1.43eV so that λ = 900nm and GaAs can operate as an infrared transmitter (IRED = infrared emitter). Fig 535 shows the curves that accompany a GaAsP  LED.
.
!
.
Remarks 1. With nominal current the forward voltage drop of a LED is between 1.3 to 3V depending on the the type of semiconductor and the impurities. 2. With the passing of time LED’s start to age: the radiated power decreases. High currents and high ambient temperatures have a disastrous effect on the aging process. A load operates
classically withIF = 20mA, but if a lower light intensity is required an IF = 5 or 10mA is used.
3. Fig. 535d shows a LED as indicator. With V = 12V ; IF = 10mA and VF = 1.6V the value of V − VF ________ 12 − 1.6 ______ the current limiting resistor becomes: RV = I = 10 ≈ 1 kΩ F
5.17
NOTES intensity (%)
mW
IF
100
100
V
IF (mA)
10
V VF (V) F
IF (mA) l (nm)
590
(a) Spectral curve
RV
1,6
10 (b) Radiated power
(c) forward
characteristic
(d) LEDindicator
circuit
Fig. 535: Curves of a GaAsPLED
2.2 Spectral sensitivity
human eye (receiver)
700 680 600
Ga As (transmitter) 100%
CIE CURVE
500
75
400 silicon
300
0
VIOLET
300
400
500
600
25
infrared 700
800
Ga As P4
Ga P:N
LEDmaterials
RED
YELLOW AMBER
100
GREEN
200
900 Ga As Zn
Ga AsP82:N Ga As P6
ultraviolet
400
500
420 460 blue violet
525 green
600
50
700
595 650 yellow
800 nm
infrared
725 red
orange
Fig. 536: Human eye curves and spectral sensitivity of LEDs and Sidetectors
% relative sensitivity
(Lm/W)
BLUE
Lumens / Watt transformation
In fig. 536 we draw: 1. The standard human eye curve prepared by the Commission Internationale de l'Eclair (the CIEcurve). This curve not only provides the sensitivity of the eye at different wavelengths but also the lumens/watt conversion of the light source at that particular frequency. 2. The spectral curve of a GaAs transmitter (LED). 3. The wavelength of maximum sensitivity of other LEDmaterials. 4. The sensitivity of a silicon detector (diode, transistor). It is clear why GaAS LED’s are used with an (Si) transistor or diode as optocouple elements.
λ (nm)
1000
1100
1200
5.19
NOTES (lux)
d
V
d
metal
V
N
P
metal
N
P
E
(a)
b
R
I
(b)
Fig. 538: Siphotovoltaic cell
From fig. 536 it follows that Siphotovoltaic cells are sensitive for light with a wavelength of between 400 and 1100 nm. This cells are especially sensitive for red and infrared light. A logarithmic scale links photoemf ( V ) and illuminance (lux). If we connect a resistor R b across a photovoltaic cell then we can determine the operating point of this circuit (fig. 539). From the product of the photoemf and the current produced we find the power that the cell gives. In the example of fig. 539 we find a maximum power of 27.8mW with a load resistor of 2.52 kΩ . We always attempt to load a PV  cell with an optimal resistor so that maximum power is obtained at the existing illumination. This operating point is indicated as the MPP (maximum power point). Note the PVcell can be short circuited and then has a current of ISC . In fig. 529 this short circuit current ISC = 0.15 mA and the open circuit photovoltaic emf is about 0.45V. To increase the voltage and required current we form batteries of photovoltaic cells. These are called solar batteries or sun panels. These days they are being more and more promoted as a clean energy source. This will be dealt with in chapter 15. Photovoltaic cells are used in control circuits, light measurement of visible and almost infrared light, to power calculators and photoelectric relays, etc. I F (mA) 0
0,1
0,2
0,3
0,4
b
R = 10 kΩ
0,5
E=5000lux 16mW
b
0,05
b
0,10
R =1kΩ
19mW 0,15
Fig. 539: Optimum operating point of a loaded cell
b
R
R = 4 kΩ
= 2,52 kΩ
26mW
27,8mW MPP
VF (V)
5.20
NOTES
Photo Ixys (KWx): Solar BIT's are monocrytalline PVcells with a high degree of efficiency (17%). Spectral sensitivity 300 to 1100nm. The solarbit exists in different voltagecurrent configurations (see www.ixys.com). Examples: XOB1712x1: 0.69V  39mA with an MPP of 0.51V  39mA. XOB1704x3: MPP of 1.51V  11.7mA.
3. HALLEFFECT SENSORS 3.1 Halleffect If we place a current carrying conductor or semiconductor in a perpendicular magnetic field B (fig. 540a) then an electric field arises perpendicular to the IB surface. This effect is known as the Halleffect. This effect was discovered in 1879 by the American physicist Edwin Herbert Hall.
y
B 1 b
I
I(A) d
I
I b
VH
X
2 Z
Fig 540: Halleffect
B(Wb/m2)
I

+
X
VH
5.40
NOTES
Z
A second method involves a metal enclosure which includes a bridging capacitor (line bypass capacitor) Cy as shown in fig. 5–55. The disturbance flows via the signal lines and via Cy back to the casing and via the stray capacitance to the source of the disturbance.
L
signal source
noise N
C
y
C
y
(straycapacitance)
(line bypass capacitor)
metal package reference ground Fig. 555: Application of a metal enclosure
Fig. 5–56 shows an example of an EMIfilter commonly used in the supply cable on the ACinput side of equipment (for example a computer).
y y
2mH
C 2200pF
x
x
mains
C 0.015µF
2mH
C 0.015µF
SMPS
load
C 2200pF
Fig. 556: EMIfilter at the supplyside of equipment
!
Remarks 1. In the case of a single phase mains the differential mode current consists of the 50Hzcurrent together with the disturbance. The common mode disturbances normally have higher frequencies (10kHz to 10MHz). They flow via parasitic capacitances back to the source of noise. 2. A C y is placed between every line and the metal enclosure. This "line bypass capacitor" is usually only a few nF, and this is to limit the leakage current V.ω.C y . 3. Practical values for the common mode choke are for example 500µH up to 2mH. 4. Cx (acrosstheline capacitor) suppresses the differential mode disturbance. A practical value is for example 0.015µF. 5. In the case of a frequency converter, depending on whether it is a single phase or threephase supply an appropriate filter is included in the input.
6.5
COMPUTER SIMULATIONS ANALOGUE SIMULATORS INDIRECT
MECHANICAL planimeter nomograph ...
DIRECT
HYDRAULICALLY
ELECTRIC
unusual
MECHANICAL
HYDRAULICALLY
scale models
opamp (differential amplifier)
wind tunnel ...
(analogue computer) ...
hydraulical labs
ELECTRIC equivalente circuits
river bed dam model ...
Fig. 65: Classification of analogue simulators
network analyser ...
2.2 Analogies between physical systems 2.2.1 Table 61: Similarities between different physical quantities Parameter
Mechanical
Electric
Translation Rotation
Applied force
F
Speed
v
p
P
v
Q (mass flow)
v. dt
Q (mass)
dϴ ___
__ dx
c = dt
t
∫
s = v. dt
ϴ
__ dv
a = dt t
Pulse
i
ω = dt
0
Acceleration
Acoustic
M
t
Displacement
Hydraulically
t
∫
___ dω dt
∫
∫
0
t
__ dv dt
∫
t
x
t
∫
__ dc dt
∫
F. dt
M. dt
i. dt
p .dt
P. dt
0
0
0
0
0
Power
F . v
Inertia
m . dt
M . ω
__ dv
___ dω
J. dt
v . i
q . p
__ dv
P. c __ Ma .dt dc
C. dt
Elasticity
K
K
L
CH
Ca
Coefficient of friction
f
f
R
RH
ra
Momentum (of torque) Kinetic energy Potential energy Dissipated energy
___ dω
dϴ ___
J. dt ; f. dt ; K.ϴ 2 _ 1 2 . m. v 2 _ 1 2 . K. s 2 _ 1 2 . f .v
2 _ 1 2 . J. ω 2 _ 1 2 . K. ϴ 2 _ 1 2 . f. ω
2 _ 1 2 . C. v 2 _ 1 2 . L. i 2 _ 1 2 . R. i
2 _ 1 2 CH . p _ 2 1 2 .RH . p
2 _ 1 2 . Ma . c
Ma .x 2 _ 1 2 . ra . P
K = compliance; K = rotation compliance; Ma = acoustic inertia; Ca = acoustic capacitance; f = translation friction; f = rotational friction; R H = hydraulically resistance; ra = acoustic resistance; J = moment of inertia; M = momentum (of torque); v = speed; v = voltage
LINEFREQUENCY PHASECONTROLLED RECTIFIERS
8.3
.. .
4. Disadvantages of phase control: poor power factor; thyristor has to deal with a large di/dt ; results in harmonics causing radio and TV disturbance.
1.5 Voltage form across the thyristor As long as the thyristor is not fired (between 0° and α ) we can compare its behaviour to an open switch. The voltage across its terminals follows faithfully the supply voltage. See fig. 81e. The forward voltage drop across a conducting SCR is approximately 1.6 V. After the thyristor switches off (is extinguished) because the io has fallen below IH , the voltage again follows the supply voltage.
vT i0
SCR
(a)
v
vs
v0
Rb
v0 v^ (b) 2�
� 0
α
4�
ωt
β v ωt°
0
vs (c)
180
360
540
720
(v)
5
ωt 0
α
2 �+α
i0
v^ Rb (d)
IH
0 (e)
vT
0
�
α
IH
2�
ωt
≈ 1,6V
α
�
2�
v^
Fig. 81: Circuit and waveforms of a resistive loaded E 1  controller.
ωt
LINEFREQUENCY PHASECONTROLLED RECTIFIERS Numeric example 84:
Given:
Fig. 81 with R b = 10 Ω ; V line = 230 V  50 Hz ; α = 90° ; VT = 0. Required: Determine P, S, displacement factor, distortion factor, power factor, Q1 , D .
.
Solution: Fundamental harmonic of current p. 811: A1 = 81.32 V ; B 1 = − 51.77 V ; X 1 = 96.4 V ; V1 = 68.16 V B I1 = 6.816 A ; φ1 = bgtg A__ 1 = − 32°48’ ; I1P = I1 . cos φ1 = 5.75 A
. .
1
RMS value of current 115 Table 81: VRMS = 0.5 x 230 = 115 V ; IRMS = ___ = 11.5 A R b
Active power:
P = VRMS . IRMS = 115 x 11.5 = 1322.5 W
or: P = Vline . I1P = 230 x 5.75 = 1322.5 W
. . .
Total apparent power S = V . I = 230 x 11.5 = 2645 VA
Apparent power of fundamental harmonic S1 = V . I1 = 230 x 6.816 = 1567.68 VA
.
.
.
Displacement factor 1322.5 P __ cos φ1 = S = ________ = 0.8436 1567.68 1 Distortion factor S1 1567.68 _______ cos δ = ___ = 0.5926 S = 2645 Power factor 1322.5 P ______ λ = S__ = = 0.5 2645
Reactive__________ power of fundamental harmonic 2 2 Q1 = S 1 − P = 841.8 var
.
√
Distortion power __________ 2 2 D = S − S 1 = 2130.35 var
√
8.17
LINEFREQUENCY PHASECONTROLLED RECTIFIERS
8.57
11. EVALUATION 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15 8.16 8.17 8.18 8.19
In which specific situation does an SCR require a double pulse? Why? What are the disadvantages of using phase control with thyristors? For an SKT600 (Semikron) we assume that the maximum ambient temperature is 50°C and that R th ja = 0.11°C/W. What is the maximum permissible IT(AV) ? What is then the power dissipation in the thyristor? The SKT600 data can be found on p. 4.15 and p. 4.24/25. Why are some power electronic circuits called natural commutating circuits? What is the maximum RMS anode voltage that is allowable with an SKT10/08D configured as a controlled rectifier (E1 controller)? In the configuration shown in fig. 810 Vline = 400V.What is the average rectified output voltage when α = 135° and for α = 100°. What is meant by “missfiring” of an inverter? How can this be prevented? What does the expression “distortion power" means? What is the maximum power factor of a full controlled bridge? In fig. 820 the following information is provided: Rb = 10Ω , Lb = 150mH , transformer and thyristor are ideal, line voltage 230V  50Hz. Determine with a firing angle of α = 0° and 60°: the output DC voltage; output current, power factor, displacement factor, distortion factor and reactive power. Which specifications should the thyristors in fig. 820 have given the information in the previous question. Determine the maximum inverse voltage of the semiconductors in the situations shown in fig. 819 and 821. Determine the DC output voltage for the circuit shown in fig. 89 if the supply voltage is 230V  50Hz and the firing angle is respectively 60° and 105°. In fig. 83 the supply voltage is 230V  50Hz and the load is comprised of 7.255Ω and 40mH. Determine the average output voltage if the firing angle is 60°. In addition determine the specifications the thyristor needs to comply with. The full controlled B 2 controller of fig. 89 is connected to a 230V  50Hz supply. The load is comprised of 7.255Ω and 40mH. Calculate the average output voltage if the firing angle is respectively 30°, 60°, 90°. Determine the largest inverse voltage across the thyristor in fig. 81. What is the largest possible blocking voltage across the SCR? For which value of α does this occur? Set α = 120° in the configuration of fig. 820. Can the bridge be brought into conduction? Confirm this with a graphic such as fig. 821 or 823. Consider the speed control of an independently excited 100kW motor. The current protection is set at 1.5x de nominal armature current. How much reactive power (kvars) does this motor draw from the net at startup? We consider a half wave and full wave controlled rectifier, both with resistive load. Let α = 30° and 150° respectively. Determine for the rectifiers the output power with respect to the maximum possible output power (when α = 0°). Use table 81.
8.58
LINEFREQUENCY PHASECONTROLLED RECTIFIERS 8.20 8.21 8.22
Determine for the data used in question 8.10 with α = 60°: the RMS value of the fundamental harmonic of the input line current of the bridge, RMS values of the fifth and seventh harmonic current components of the line current, RMS values of the 300Hz and 600Hzvoltage components of the output. An E1 controller connected to a 230V  50Hz has a resistive load of Rb = 100Ω. Calculate the amplitude of the 150Hz current in the load when the firing angle is 60° and 90° respectively. A full controlled B 6 controller is connected to a three phase supply of 3x400V  50Hz with a highly inductive load of 5Ω  280mH. Calculate the RMS value of the 350Hz current in the supply line when the firing angle is 30°.
8.23
Sketch the output voltage waveform with a firing angle α for a highly inductive load in the following situations: a) full wave full controlled single phase bridge b) half wave full controlled three phase bridge (M3, see fig. 840 lower down). Sketch in a similar fashion to fig. 834, a periodic voltage vo = f(x) and derive an expression for ωt = f(x).
824 Determine the maximum output voltages for a B2  controller,M3 and B6 controller. The starting point is expression (825). 8.25 Given a full control highly inductive loaded B 6 controller. Calculate the ratio of the RMS values of the harmonics in the output when α = 90° in relation to α = 0°. Check your answer with the values from table 84.
v1
i1
v2
i2
v3
i3
Th1
L1
L2
L3
Th2
Th3
v0

(v1 = v.cosωt) Lb
+
i0 Rb
Fig. 840: M3 controller (half wave full controlled three phase bridge)
CONTROL OF THYRISTORS
11.7
3. CONTROL IC FOR SCR AND TRIAC A number of companies sell linear IC’s that can form part of a control module for thyristors. It is possible to build a temperature controller for an oven, a speed controller for a motor, etc...with a minimum number of components. We consider just such a control IC, the industry standard TCA785 from Infineon.
3.1 AC controller using a TCA785 L
R 4,7k 9W
RSYNC
(bulb als a load)
220k
1N4005
0,47µF 10k
470 µF 16V
0,22µF
15V
16
2
15
3
14
4
4.7k BAY 61
150
5
12
6
11
triac
BAY 61
13
TCA 785
BZY97C15
230V
BAY BAY 61 61
1
2.2µF (MKH)
10k 2.2k
250V
7
10
8
9
inhibit 22k R9
0,1µF 100k
C10 C12 47 nF
150 pF
Mp
Fig. 115: Phase control with a TCA785 for triacs with a gate current up to 50mA
The operation of the TCA785 is explained under heading 3.2. The IC has 16 pins. The IC is supplied with 15V DC via pin 16 and pin 1. In fig. 115 the DC supply is created with rectification (1N4005), filtering (470µF) and zener stabilisation (BZY97C15). To detect zero crossover of the supply, pin 5 is connected to the supply voltage via a resistor Rsync = 220kΩ . The value of R 9 (resistor connected between pin 9 and ground!) determines the charging current I10 that linearly charges capacitor C10 .At pin 10 we have a ramp voltage in which the slope is determined by R9 and C 10 . The voltage at pin 11 is regulated with a potmeter of 10kΩ and as a result the firing angle α can vary between 0° and 180° . Pin 15 delivers a firing pulse during the positive half period of the supply voltage and pin 14 does the same during the negative half period. Bypass capacitors are connected to pins 1, 11 and 13. In this circuit an AC current is phase controlled as shown in fig. 91.
12.12
CHOPPERS
5. CHOPPER CONTROL IC’S 5.1 Description of block diagram +24V
+ 100µF 
15 16
3 oscillator (out) 470k
470
1k
6
v
7
Oscillator
Rb
68 5W
+5V 4 FLIPFLOP
t
150
150
Vref.
voltage 1 regulator
Q Q
NOR
8
12
T
1
9,1V 3W
D
1
Rb
Lb
M
11 2 +5V
t + 3 7 +5V comparator +5V 1 + fault 7 2 amplifier + 6 0,1µ 1k 5 10 shutdown 10k
NOR
9
T2
13
4 5 Compensation
8
D2
S
14 270 470
9
BC140
IRFN530
2N2904 470
Fig. 1214: A didactic model of a chopper with a “1524” control IC. This IC works with a supply between 8 and 40V(max), has an oscillator frequency between 100Hz and 500kHz and has two output transistors (100mA) that can handle a maximum collector voltage of 60V
The “1524” and its derivatives is a much used ICcontroller. It first appeared in 1976 as the “SG1524” (SG = Silicon General) and quickly became the industry standard and to this day is offered by many chip producers (CA1524, LM1524, UC1524, XR1524,…). Fig. 1214 shows a didactic model which allows us to study a chopper circuit when PWM is used. With switch S we can choose between a resistive load , an inductive load or a motor as load. The interface between IC and switch S we recognise as a control circuit from the study of the powerMOSFET. The “1524”is a pulse width modulated controller for voltage controllers. The frequency is determined from the following expression: _____ 1.18 f = R . C (1210) 6
7
Here in is f in kHz with R6 in kΩ and C7 in µF. With R 6 and C 7 we mean the discrete components that are connected between pin 6 and 7 of the IC. Practical values are: 1.8 kΩ < R6 < 500 kΩ 0.001µF < C7 < 0.1µF The value of R 6 determines the constant current which linearly charges C 7 . The sawtooth of +1V to +3.5V is applied to the (+) terminal of comparator 3. A flip flop makes sure that only one of the NOR gates 8 or 9 receives a low input. The NOR gate which has all inputs low will saturate its respective output transistors T1 or T2 . For this to occur the following must be the case: V3 = 0V and the output of comparator 3 should also be 0V.
CHOPPERS v3 (V) 3
t (ms) 0
8
18
v7 (V) 3.5 1.5 1
t (ms) 8
0
10
18
20
1.5
t (ms) 0
vGS (V) t (ms) 8
0
10
18
20
v11/14 (V) 22
v11
v14
t (ms) 0
8
10
18
20
8
10
18
20
vDS (V)
24
t (ms)
1 0
vR (V) b 22
t (ms) 0
8
10
Fig. 1215: Waveform associated with fig. 1214
18
20
An internal voltage controller 1 regulates the 5V(± 1%) supply of the IC and which can be externally loaded to 20mA. The output pulse V3 of oscillator 2 is a “blanking” pulse who’s function is to prevent T1 and T2 conducting at the same time. This is necessary if the transistors are not in parallel as shown in fig. 1214, but where for example they operate separately to control a balance circuit. In this case the frequency of the output voltage is half of the sawtooth frequency.
5.2 Operation of comparator 3
v9 (V)
8
12.13
7 provides a sawtooth voltage of +1 to 3.5V to C the non inverting terminal (+) of comparator 3. On the inverting input (−) of the comparator a voltage is applied with four possibilities: A. The inverting input is smaller than +1V so that the output of the comparator is high (+5V) and both output transistors are blocking. This can be achieved by: 1) saturating transistor 5 via pin 10 so that VCE < 1V and the IC blocks within 200ns (shut down protection). 2) using opamp 7 as a current limiter. B. A voltage of between +1V and +3.5V is added to the inverting input of the compara tor. As a result the pulse width at the IC output is controlled. We can achieve this by: 3) linearly controlling the duty cycle via the fault amplifier (6). 4) adding a voltage V 9 as shown in fig. 1214 which also regulates the duty cycle. In this manner we see in fig. 1215 that with V 9 = +1.5V there is conduction between 8 and 10ms ( V11 ) and between 18 and 20 ms ( V14 )
13.11
SWITCHMODE POWER SUPPLIES 2.3 Buckboost converter 2.3.1 Operation of buckboost converter
Fig. 139 shows the basic configuration. Closing S causes a current iL to flow and magnetic energy 2 1 is stored in the coil ( __2 . L . iL ). When the switch opens, the magnetic energy can only be discharged in the load and this results in a reverse polarity of the output voltage Vo with respect to Vi . The capacitor once again functions as a filter capacitor. As with a boost converter we first store energy in the coil and pump this energy to the load when the switch is open. This explains why both converters are classified as flyback converters. +
D
S is
vi
iL
L
C
vL
Rb
iD +
D
S vs
vi
vD vL
L
iD

iL
+
C
Rb
Vo
Io
Fig. 139: Basic principle of buckboost converter
Once again we distinguish between two operating modes, namely continuous and discontinuous current mode in the coil.
2.3.2 Continuous current in the coil Fig. 1310a shows the associated waveforms. When the switch closes a voltage results across the (I1 − I2 ) ∆i __ _______ coil: v L = V i = L . ∆t = L . δ . T or: L . (I1 − I2 ) = δ . T . V i (1314). ( I − I ) ∆i 2 1 ___ _________ When the switch opens after neglecting vD : v L = V o = L ∆t = L .(1 − δ ) . T or:
L . (I2 − I1 ) = Vo . (1 − δ ) . T From (1314) and (1315) it follows:
(1315)
δ
____ Vo = − Vi . 1 − δ
(1316)
13.12
SWITCHMODE POWER SUPPLIES 2.3.3 Discontinuous current in the coil In this case the energy in the coil is completely discharged before the transistor switch recloses. Once again there is a dead time td (discontinuous current). Fig. 1310b shows the associated waveforms. Closing the switch S results in a voltage across the coil, whereby I1 ∆i __ ___ Vi = vL = L . ∆t = L . δ.T or: L . I1 = δ . T . Vi
(1317)
After the switch opens during the time interval (1 − δ ) . T − td almost all the energy in the coil is gone. The voltage across the coil is now V i and the current changes fromI1 to zero, so that: (0 − I1 ) _______________ Vi = vL = L . (1 − δ ) . T − →→ L . I1 = − Vo . [ (1 − δ ) . T − t ] d td ] [
(1318)
From (1317) and (1318) it follows that: δ . T . Vi = − Vo . [ (1 − δ ) . T − td ] from which:
δ _________ Vo = − Vi . 1 − δ − t /T
(1319)
d
!
2.3.4 Remarks 1. 2. 3. 4.
From (1316) and (1319) it follows that the buckboost converter results in voltage inversion. Depending on the value of δ we have  V o  ≤ or ≥  V i  . This is therefore a stepup /stepdown converter. The maximum voltage across the (transistor) switch is V i + V o as can be seen in fig. 1310 at the bottom . The maximum reverse voltage across the diode is V i + V o .
2.3.5 Numeric example 133: For a buck boost converter with continuous current mode we are given: input voltage: Vi = 3 to 15V max. output current: Io = 3A output voltage: Vo = 9V ± 0.1% chopper frequency: f = 100kHz
..
..
Required: determine the values of L and C Solution: δmin Vo 9 _________ _________ From (1316) →→ V o = Vi max . (1 − →→ δmin = V + = ______ = 0.375 . V δmin ) 15 + 9 o i max
Assumption ∆iL max = 0.2 x Io = 0.6A − 5 (1 − δ min ) . T __________________ 9 . (1 − 0.375) . 10 ____________ From (1315) →→ L max = V o . = = 93.75µH ∆iL 0.6 ∆iL 0.6 ___________ ______________ From (137) →→ C min = 8 . f . ∆ = = 83.4µF vC max 8 . 10 5 . 9 . 10 − 3 ∆vo max _______ 0.1% . 9 _______ and: E SR max = ∆i = 0.6 = 0.015 Ω L
SWITCHMODE POWER SUPPLIES
iS
iS
I1
I1
I2 t ∆ iL
iL I1
t iL I1
I2
t vL
t vL Vi
Vi t
t Vo
Vo iD I1
iD I1
I2
t
t vS
vS
Vi + Vo
Vi + Vo
Vi
Vi t (1  δ).T
δ.T T
t td (1  δ).T
δ.T
T
(a) Continuous current Fig. 1310: Waveforms of fig. 139
13.13
(b) Discontinuous current
13.32
SWITCHMODE POWER SUPPLIES
13. NON IDEAL WAVEFORM In fig. 1326 we redraw the flyback converter, but now a number of parasitic elements are included. These have been drawn in green. Where: cb L : 0 : L lk : L prim C : sec C : tr : C D : C o : C
parasitic self induction of the printed circuit board; magnetizing inductance of the transformer; leakage inductance of transformer; capacitance of the primary windings; capacitance of the secondary windings; feedback capacitance between secondary and primary transformer; diode capacitance; output capacitance of transistor.
Fig. 1326 is an approximated drawing since these parasitic elements do not have a constant value and are often voltage and frequency dependent. Fig. 1327a shows for example the voltage across the MOSFET. It is clear that the oscillations cause the voltage across the MOSFET to increase. A snubber across the MOSFET can prevent it being destroyed. If we take a look at the first two periods of oscillation (fig. 1327b), then we see in this example a period of 100ns. This corresponds to 10MHz. A rule of thumb indicates that the oscillation frequency is approximately 100 times the switching frequency. Fig. 1327b is the result of resonance between the leakage inductance of the transformer on the one hand and the self inductance of the current loop on the printed circuit and on the other hand the output capacitance of the FET and the transformer capacitances. The distortion of the sinusoids is the result of the voltage dependent output capacitance of the MOSFET.
C L cb
C
tr.
C
L 0
C sec.
prim. L
lk C 0
Fig. 1326: Flyback converter with parasitic elements
D
SWITCHMODE POWER SUPPLIES
V DS
v
(V)
13.33
(V)
80
140
60
120
40
100
20
80 60
0 20
40
40
20
t (ns) 40
80
120
140
60
t (Âľs)
0
2
4
6
8
10
12 (b)
(a)
Fig.1327a: Waveform on the FET in fig. 13.26
Fig. 1327b: First period of the oscillations in fig. 1327a
14. DIGITAL CONTROL OF AN SMPS With a new type of digital signal controller (DSC) there is now enough calculation power on board to execute the software quickly enough to handle closed loop control. This DSC has in addition builtin intelligent modules for PWM, and an analogue to digital converter (ADC). L
T 1
IN
T 2
PWM1
C
PWM2
R
OUT
We look at the example of the synchronous buck converter from fig. 1324 and apply a voltage between the points A and B. This brings us to fig. 1328. The analogue control circuit from fig. 1324 has now been replaced with a digital control system with a DSC (fig. 1329).
Fig. 1328: Basic configuration of a synchronous buck converter
IN
S&H
+
PID
PBM
T 1 T 2
ADC
S&H
DSC Fig. 1329: Digitally controlled buck converter
OUT
13.34
SWITCHMODE POWER SUPPLIES The processing in the digital control loop results in delays. A PID controller that runs in the DSC has a typical calculation time of 1µs. The PID control itself takes about 2µs. The analoguedigital converter (ADC) takes samples and converts in about 500ns. Consider in addition a transistor switching time of 100ns, then we arrive at a total delay of 3.6µs. From this we can determine an optimum sampling rate of 277kHz. In order not to loose any information, the Shannon theorem tells us the sampling frequency has to be at least twice the highest frequency that Fourier analysis reveals in the analogue signal. With this “ twice” sampling rate the phase displacement is 180° . Due to the associated delay in the system we exceed the maximum allowed 180° required for a stable system. For this reason in digital control systems the analogue signal is over sampled with a factor ten.
15. EVALUATION 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 13.10 13.11 13.12
With which isolated converter is the supply voltage across the transistor and for which converter is the double supply voltage across the transistor. What does discontinuous current mean for a converter? Which types of flyback converter are you familiar with? What is the difference between a flyback and a forward type? What does “feed forward” control of an SMPS involve? What type of output diode is used for an SMPS with a high output voltage ? What is the big advantage of ferrite core transformers compared to standard iron core transformers? What does offtheline mean? Why should flyback converters never have an open output, or be without a control loop? Up to which power level are linear power regulators used? What is the efficiency of such a regulator? a) What does EMI stands for? b) What for the most part determines the weight and volume of a linear stabilised power supply? Prove that expression (1321) is equivalent to expression (1319).
16. THE DESIGN OF SWITCHMODE POWER SUPPLIES This can be found on the website of CASPOC. www.caspoc.com/education under the heading books “Pollefliet”.
14.16
INVERTERS L L
V
t
2 1
v
0,78 0,612 linear area
overmodulation
square wave m
0
1
3,2
Fig. 1414: Output line voltage as a function of modulation depth m (for N = 15)
6. PWM STRATEGIES Up until 1980 the pulse patterns for PWM were produced using analog techniques. Between 1980 and 1990 digital technology started to be used to produce the PWM waveforms in threephase inverters for motor control. In the early years 8 bit microprocessors were used. After that DSP’s (digital signal processors) appeared on the scene.
6.1 Analog technology ( system with modulated output voltage) This system with a modulated output voltage was common in early analog technology. A reference voltage was compared with a carrier wave of higher frequency. This is comparable to the didactic example of fig. 1411.
6.2. Digital technology (system with calculated voltage pattern) The desired condition of the inverter switches is calculated using a DSP. For the three voltages there is a varying single step shaped reference, this can for example be derived from a “lookup” table. This (hypothetical) sinusoidal variable reference is corrected during every pulse with a “zero sequence” value. This zero sequence value is determined per pulse by dividing by two the sum of the most positive value and the most negative value and adding this zero sequence value to the three basic references. In this way the zero pauses at the start and end of every pulse are synchronised.
6.3 Value of the output voltage of a PWM inverter Fig. 145 shows the output voltage of an inverter with a quasi square wave in the output. __ 2 . √ 3 . Vt _________ ^ From (828) the amplitude of such__ a waveform can be found: v 1 = , π __ ____ √ 6 so that the RMS value is: V 1 = π . V t . Here in : V t ≈ ^ v = √ 2 . V net . eff.
APPLICATIONS OF POWER ELECTRONICS
15.7
. software control: via RS232 and a local network. Via the built in web server the UPS can be
remotely monitored and controlled.
1.3 Static UPS without auxiliary network 1.3.1 Passive standby topology When there is no auxiliary network available when faults occur the batteries have to supply the energy. These batteries are referred to as the backup or redundant current source. For computer power supplies extra requirements may be made. In the supply network a number of disturbances may be present due to atmospheric influences (lightning,…) faults in the supply network (short circuits,…) industrial disturbance (welding machines, motors, lifts, fluorescent lamps,…). Over voltage and under voltage can occur as well as micro interruptions or long interruptions. These variety of disturbances are not compatible with good operation of (micro) computers. The supply network of a computer should be free from the following: frequency and voltage variation voltage drop short power interruptions electromagnetic faults. In addition there should be suitable autonomy to enable saving of data and the completion of calculations in the event of a power failure. An autonomy of between 10 to 15 minutes is usually the maximum. It takes several ms for the autonomy to kick in. Fig. 155 shows the basic schematic. A filter neutralizes the high frequency peaks. Sometimes an ultra isolator is included, This is a transformer with one or more screens so that short duration high frequency faults may be filtered.
. . . .
S.B.
INVERTER
RECTIFIER
230V
CONSUMER
GRID
O.L.
+ 24V Fig. 155: UPS of the passive standby type

FILTER ULTRAISOLATOR SWITCH OVER
In normal service the consumer is supplied via a filter and ultraisolator. When a voltage disturbance becomes too extreme the UPS switches to an inverter output. This inverter output is in continuous service together with the rectifier and battery. The positions of the switch are indicated with S.B. (standby) and O.L. (on line). Under 2kW UPS are used almost exclusively to supply computers.
APPLICATIONS OF POWER ELECTRONICS
15.21
As a light source, LEDs have a number of interesting properties: 1. 2. 3. 4. 5. 6. 7. 8. 9.
!
In contrast to other light sources, LEDs do not suddenly fail or burnout. The luminous intensity of LEDs reduces gradually. They still emit 70% of their initial luminous intensity after 50,000 hours (on for 12 hours per day for 11 years!). The lifetime of LEDs is between 50,000 and 100,000 hours. By contrast, an incandescent lamp has a lifetime of about 3000 hours and a CFL (compact fluorescent lamp) lasts for up to 10,000 hours. There is therefore no necessity to replace a LED lamp, which is important for critical applications such as beacons, buoys, lighthouses, emergency exits, etc. High efficiency. With current technology we have UHBLEDs (ultrahigh brightness LED) which have 10 times the lumen/watt ratio of a halogen lamp. A LED headlight of 50 W in a car is therefore equivalent to a 500 W halogen lamp. A UHBLED currently provides up to 100 lm/W (white light), and prototypes exist which provide 200lm/W. The original LEDindicator in the early 70s delivered 0.1 lm/W. LEDs emit little heat. They are flat elements with small dimensions. Can withstand shocks and vibrations, which is important in cars. They are semiconductors, which means there is no gas, mercury or filament present. The LED therefore has an environmental advantage. Often multiple LEDs are connected in series, with a constant current. If one LED gets shortcircuited the rest continue to operate since they are supplied with a constant current source. By using a control IC, it is possible to control the luminous intensity using PWM. In practice, it is possible to control from 0 to the full luminous intensity. Using PWM it is possible to control the luminous intensity, while maintaining full colour integrity. Operation is at low voltage levels and high voltage connections are therefore avoided. A LED requires about 3V (between 2.5 and 4.47V depending on the type) so that a series connection of tens of LEDs can easily be controlled from an IC that is supplied with an input voltage of 40 to 50V. The LT3595 (Linear Technology) is a 16 channel LEDdriver capable of handling 160 LEDs from a DC source of 45V. This IC can have a duty cycle (dimming) of 5000:1.
Remarks 1. Another interesting application of LEDs is that an RGBcoloured light source can be built. To achieve this often two green LEDs with one blue and one red LED are used. This com bination takes the sensitivity of the human eye to colour into account. A LED has a nomi nal wavelength and colour with a specific current. That’s why we use a current source with the LEDs. Using colour mixing LEDs can create a continuous colour spectrum. To achieve this, the luminous intensity of each LED is controlled. This is not achieved by linearly controlling the current of each LED since this would change the colour. The solution is PWM of the current. Four channel LEDdrivers exist in IC form to realize this RGBluminous flux (for example the STP04CM596 from the firm ST, see www.st.com under “lighting”).
15.22
APPLICATIONS OF POWER ELECTRONICS 2. 3. 4. 5. 6. 7.
Another example of lighting control using LEDs is the application of the control IC IRS2540 from International Rectifier. This IC can operate with a supply of 200V. A PWMcontrol signal uses burst mode control at 175kHz. The duty cycle of the PWM control is 100%. An important niche for LEDs is the car industry. European cars were the first to use blue/ green/white/amber coloured LEDs in the instrument panel. American and Japanese luxury car brands followed this trend. A brand such as Lexus uses 50W UHBLEDS for the headlights. LEDs have started to replace halogen and Xenon lamps. Advantages include the streamlined designs and the longer lifetime of the lamp. LED technology is no longer just used in the luxury car market but is also used in the middle class segment. Applications vary from head lights to interior lighting. The LT3755 from Linear Technology is a controlIC intended for 50W headlights. This IC can boost the 12V battery voltage to 60 V to control a series circuit of 14 LEDS drawing 1A. Another important terrain for LEDS is “backlighting” of HDTV LCD screens. For backlighting a 46” LCD TV requires about ten LT3595’s per HDTV. ICmanufacturers produce families of LEDdrivers that comply with specific requirements for specific applications. Infineon for example has a low cost LEDdriver (BCR450) for control of high power LEDs. In combination with an external transistor this BCR450 is capable of precise control of the current. In addition there is over current and over voltage protection, etc. A LED cannot be dimmed using a triac dimmer. National has developed an IC (LM3445) that makes it possible to dim a LED in combination with a triac dimmer. (www.national. com/powerwise) More and more public authorities are considering installing LED street lighting. A streetlight is routinely required to deliver 10,000 lumens. Current LED streetlights use between 50 and 200 LEDs with a current of 350mA. A solution for street lighting is to connect a number of series LED circuits in parallel. According to IEC specifications a transformer needs to be placed between the LEDs and the supply net and the secondary voltage may not be greater than 120V. Often 48V is used. In this manner 12 InGaNLEDs can be connected in series (12 x 3.3V = 39.6V). A buck converter is recommended for the supply. It is the cheapest, simplest and most efficient of the classic converters. If a converter is used for every series circuit of LEDs then they can be dimmed independently. A disadvantage of this setup though is the cost price. When the buck converters are connected in parallel then EMC problems result. To combat this, filters containing at least a coil and a capacitor are required in the input of each converter. This increases the cost price even further. To preserve the quality of the supply, network power factor correction is also implemented. Street lighting is usually classified as HPWA (high power wide area lighting). In addition to energy efficient solutions for interior lighting, streets and tourist attractions at the World Expo 2010 in Shanghai, OSRAM together with its mother company Siemens, installed 150,000 LEDs in the pavilions and streets around the Expo.
15.23
APPLICATIONS OF POWER ELECTRONICS
5. RENEWABLE ENERGY
A. WIND TURBINES 5.1 History Charles Brush (18491929) was one of the pioneers of the American electrical industry. In the winter of 18871888 Brush built the first automatic wind turbine to produce electricity. A rotor of 17m diameter with 144 rotor blades made of cedar wood drove a DC generator of 12kW. The turbine worked for 20 years. Poul La Cour (18461908) was the pioneer of modern aerodynamics and built his own wind tunnel to use for experiments. In 1897 La Cour had experimental wind turbines in the Askov Folk High school (Denmark). In 1918 there were about 120 local wind turbines of between 20 and 35kW in Denmark. All the generators were DC machines. In 1951 a DC generator was replaced with a 35kW asynchronous machine. In 198081 a 55kW generator was developed that ultimately was responsible for the break through of the modern industrial wind turbine. In special cases wind turbines up to 6MW are possible. On the other extreme of the power range there is also a market for small wind turbines (100kW) for private use on farms etc. The majority of the wind turbines have a power capability between 500kW and 1MW. The 6MW turbine from REpower installed in 2009 on the German Danish border has a rotordiameter of 126m, rotor blades with a length of 61.5m and a hub height of 100m. Currently 1% of all the worlds electricity is produced from wind energy, 16% from nuclear energy and 16% from hydroelectricity.
5.2 Construction of a wind turbine 1
2
7
9
3 6 5
4
8
Fig. 1522: Cross section of a 2MW wind turbine. Courtesy of Vestas Wind Technology A/S
The parts are :
1: rotor 2: main shaft 3: gearbox
4: alternator 7: measurement of wind speed and direction 5: radiator 8: tower 6: mechanical brake 9: gondola
16.6
ELECTRIC MACHINES 1.4 Impedance transformation 1.4.1 Transformation formula Fig. 167a shows an ideal transformer, loaded with a series RLC circuit. An ideal transformer is a transformer without losses. Ip
Ip
Is
R’ = k2 . RS
RS Vp
NP
NS
Vp
Vs
L’ = k2 . LS
LS
C’ = CS / k2
CS
(a) Fig. 167: Ideal transformer, loaded with a series RLC circuit: impedance transformation t dis __ 1 __ Secondary: v s = Rs . is + L s. dt + C ip . dt s
∫ 0
di
(b)
t
∫
p 1 Application of (168) gives: vp = k .R s.i p + k .L s. ___ + k 2 . ___ i p . dt
2
2
dt
Cs
0
For an ideal transformer, the secondary load can be represented as an equivalent circuit seen from the primary side (fig. 167b), as long as: R’ = k 2 . Rs ; L’ = k 2 . Ls ; C’ = CS / k 2 . More generally: an ideal transformer with secondary impedance Zsec . may be seen as a
( ) Np
primary impedance:
2
. Z ( Ω ) Z’ prim. = ______ sec. Ns
(1611)
From (1611) it follows that we can transform a primary impedance to an equivalent secondary impedance:
(
N
)
2
s . Z Z’ sec. = _______ prim. ( Ω ) N p
(1612)
1.4.2 Numeric example 161: 1. An electrical oven is supplied with 46 volt and has a power of 4 kW. The supply network is 230V50 Hz. If we had an ideal transformer available, what is then: a) the transformation ratio b) the primary and secondary current c) impedance seen from the 230 V50Hz net?
16.7
ELECTRIC MACHINES Solution: NP
230
___ a) k = ___ N = 46 = 5 S
PS
4000
b) secondary current: I S= ___ V = ____ 46 86.95A = S
IS 86.95 primary current: IP = __ = _____ 5 17.39A =
Transformed impedance seen from the source: Zprim. = k 2. Zsec. = 5 x 0.529 = 13.23Ω
Proof: Z prim. x I P= 13.23 x 17.39 = 230V !!
2.
If maximum power transfer is required from the generator to the consumer, then the consumers impedance should be the complex conjugate value of the generator impedance. We have a power amplifier with an output resistance of 48Ω and wish to connect a loudspeaker with the following characteristics: 30 W  4Ω . Maximum power transfer is possible by placing an impedance transformer between amplifier _____
k VS 46 c) load impedance: ZS = __ I = _____ 86.95 = 0.529 + j.0 Ω S
2
√
Z
___
NP prim. 48 ____ and loudspeaker. The turns ratio should be k = ___ = ___ 4 = 3.46. N = Z S
√
sec.
1.5 Magnetizing inductance
NP .I µ
With a magnetising currentI µthe magnetic field strength in the core is H = _____ l and the k magnetic induction is B =µ 0.µ r. H so that the flux in the core of the transformer is: µ0 . µr .A k
Φ0 = B . Ak = ________ l . Np . Iµ
. Φ (Wb): noload flux ≈ resulting flux (Φ − Φ ) with load . B (Wb/m²): magnetic induction in the core . µ : relative permeability of core material . µ : = 4 . π . 10 H/m . l (m): average length of field line in the core . A (m²): crosssectional area of core . I (A): magnetising current of the transformer. k
Whereby:
1
0
r
2
−7
0
k
k
µ
If we call L 0 the self inductance of the primary with respect to the flux Φ0 in the core, then we may µr . µ0 . Ak
write: Np . Φ0 = L0 . Iµ so that: Np . Φ 0= ________ l . Np .I µ= L0 . Iµ 2
k
from which follows:
µr . µ0 . Ak
L0 = N p . ________ l 2
k
(H)
(1613)
16.8
ELECTRIC MACHINES Numeric example 162: 1. A ring core transformer (fig. 168) consists of:
average radius 60 mm; crosssection of torus 45 mm; µr =1600.
Core:
Insulation layer: 1 mm thick
Primary:
3 layers: respectively 201, 189 and 140 windings AWG 18.
Each layer is separated by 1 mm thick insulation.
Insulation layer: 4 mm thick
Secondary:
two layers: 50 and 22 windings AWG 10, separated by 1 mm of insulation
2. Extract from winding wire table (AWG = American wire gauge) diameter (with insulation) in mm min. max.
AWG
resistance (per 100 m) Ω
admitted current (on base of 2A/mm²) A
10
2.64
2.69
0.3276
10.38
18
1.08
1.11
2.095
1.624
insulation
60 mm
primary
secondary
core ø 45 mm
1mm 1,11
ø 45 mm
1
(a)
(b)
Fig. 168: Ring core transformer (a) crosssection (b) core and windings
Question: 1. Resistance of primary and secondary coil 2. Magnetising inductance Solution: 1.1 Primary resistance LAYER 1: length of one winding: π x 0.04811 = 0.15114 m
total length:
1,11 1 1,11
201 x 0.15114 = 30.38 m
2,69 1mm 2,69 4mm
ELECTRIC MACHINES
LAYER 2: length of one winding: total length: LAYER 3: length of one winding: total length: Total length primary winding: 86.32 m
Resistance: R p = _____ 100 x 2.095 = 1.8 Ω
1.2
Secondary resistance LAYER 1: length of one winding: π x 0.06835 = 0.2147 m total length: 50 x 0.2147 = 10.73 m LAYER 2: length of one winding: π x 0.07537 = 0.238 m total length: 22 x 0.238 = 5.234 m Total length of secondary winding: 15.96 m
Resistance: Rs = _____ 100 0.3276 = 0.0523 Ω x
16.9
π x 0.05233 = 0.1644 m 189 x 0.1644 = 31.07 m π x 0.05655 = 0.1776 m 140 x 0.1776 = 24.87 m
86.32
15.96
2. Magnetizing inductance µ0 . µr . Ak
−7
2
4 x π x 10 x 1600 x π x 0.0225
L 0= Np . ________ l = 5 30 2. __________________________ 2 x π x 0.06 = 2.38 H 2
k
1.6 Leakage inductance From the viewpoint of voltage loss leakage inductance is undesirable. Transformers are therefore constructed to minimise the leakage fluxes. Fig. 169 shows, for example, how a coaxial implementation of primary and secondary coils minimises the leakage reactance by minimising the distance between consecutive coils. On the other hand, possible short circuit currents are limited by the leakage reactance, which can form a protection for the transformer. In practice distribution transformers are constructed with sufficient leakage reactance, so that shortcircuit current is limited to 8 or 10 times the full load current. In electronic power supplies ring core transformers are frequently used. Due to the construction method they have a minimum leakage reactance. Electronic technicians talk about “hard” transformers since large variations in the load coupled with low leakage inductance can produce large current spikes. These varying load conditions occur for example during commutation of one rectifier element to another on the secondary side of threephase transformers.
16.10
ELECTRIC MACHINES
Ø0
llp Øls
Ølp
Ølp
Øls lls
Ø 0
secondary
Als
primary
Alp
Ø0
core
Fig. 169: Leakage fluxes by a coaxially wound transformer
To determine the leakage inductance, we consider the primary leakage flux (the same reasoning is valid for the secondary side). We can not make an accurate calculation since the crosssectional area through which the flux flows can not be accurately determined. It is possible to make an approximate calculation. If the crosssectional area where in the leakage flux flows is Alp and the average length of the field line is l lpthen similar to expression (1613), it may be written as: Alp 2 Llp = sP = NP . µ0 .___ l
(1614)
lp
The field lines of the leakage flux complete their circuit through the air (µr = 1) instead of through the ferromagnetic core (µr ), which explains the difference with expression (1613). Numeric example 163: We reuse the data of numeric example 162 ensure the possible parts of the leakage fluxes in fig. 1610a and fig. 1610b.
23,5 22,5
Øls Alp
50 x 2,69
Ø lp
ø 4 mm
Als
23,5 22,5
CORE ø 45
primary
28,83 32,83
Ølp Fig. 1610a: Primary leakage flux of transformer in fig. 168a
Øls Fig. 1610b: Secondary leakage flux fig. 168b
16.11
ELECTRIC MACHINES Primary leakage inductance µ0 . Alp
4 . π . 10 . π . ( 23.5 − 22.5 ) .10
lp
201 x 1.11 x 10
−7
2
2
−6
2
−6
_____________________________ sp = Np . ______ l 530 2. = = 228 µH −3 2
Secondary leakage inductance −7
2
. µ0 . Als
4 . π . 10 . π . ( 32.83 − 28.83 ) . 10
ls
50 x 2.69 x 10
ss = Ns _______ l = 72 2 . _______________________________ = 37.42 µH −3 2
If we realise that the magnetising inductance for this transformer is 2.38 H then we see that the leakage inductance is indeed minimal. It is clear that the path of the leakage fluxes depends upon the practical implementation of the transformer windings. The present numeric example gives us a rough idea of the relative magnitude of the leakage inductance.
1.7 Energy losses 1.7.1. Copper losses
In the primary and secondary windings energy losses occur. IfR Pand RS are the respective resis2 2 tances of the windings then the losses may be written as RP . IP andR S. IS . The sum of both is the total energy loss. This is referred to as the copper losses of the transformer.
1.7.2 Iron losses In ferromagnetic materials, subjected to a varying magnetic field, hysteresis losses occur: n ^
Ph = kh . f .B
W/kg
(1615)
whereby: kh = material constant of the ferromagnetic material used in relation to hysteresis losses. f = frequency (Hz) ^ B = amplitude of the magnetic induction (T = Wb/m²) n = empirical constant for the magnetic material (1 < n <3). Since the magnetic circuit of a transformer is constructed from metal plates, hysteresis losses occur. To limit these losses, it is desirable that the material constant be as small as possible. A possibility in this case is an iron alloy using silicon (e.g. 3% silicon). If the core was made from solid iron, then considerable eddy currents would occur. These can be dramatically limited by making the magnetic circuit from plates which are insulated from each other and the surface of which is in the direction of the flux. As result of this the path of the eddy currents is limited. The eddy current losses Pw can be determined with a formula in the following form:
^2
Pw = kw . δ 2.f 2. B
W/kg
(1616)
Here in : kw = material constant with respect to the eddy current losses δ = plate thickness in mm. By adding silicon the electrical resistance is also increased as a result of which the eddy current losses are reduced. According to the last formula, it is advantageous to have the plates as thin as possible. Typical plate thickness lies between 0.3 and 1mm for 50 Hz operation. The plates can be 0.02 mm for high frequencies. For band wound cores thicknesses of 0.003 to 0.3mm are possible.
16.42
ELECTRIC MACHINES Numeric example 168: We reconsider a stator with 12 slots (fig.1649). The coils N1to N6are in series and form one phase. Assume that one such complete phase winding has 1200 windings. We wish to distribute these windings as sinusoidal as possible. Determine the number of windings for these six coils, as well as the mmf in the airgap if the current is 1A. N1 N2
α
N3
N6
N5 N4 Fig.1649: Stator iron wound with one phase coil S1 ( =U1U2)
Solution: In a stator with m slots we are able to determine the number of windings in slot n by taking the 2.π ___ ideal sinusoidal distribution over an angle m around slot n: 2πn/m
∫
Nk =
2πn/m N ___S n S1 . dα= 2 sin α . dα
∫
2π( n − 1)/m
with n = 1 , 2 , ..., 6
(1643)
2π( n − 1 )/m
With m = 12 and NS = 1200 we find: π/6
∫
π/3
∫
1200 1200 _____ _____ N3 = N6 = 2 sin α . dα= 80 ; N2 = N5 = 2 sin α . dα= 220 ; 0
π/6
π/2
∫
1200 _____ N1 = N4 = 2 sin α . dα = 300 π/3
With iS1 =1A we find as a result of N1 and N4 in each air gap an mmf of 300Aw across a width of 5.π/6 . The coils N2and N5 produce a field of 220 Aw across π/2 radians. N6 and N3 are responsible for an extra 800 Aw across π/6 radians. This has been drawn in fig.1650.
16.43
ELECTRIC MACHINES
F 600 500 400 200 100
�
�
0
2
α
3� 2
2
Fig.1650: Magnetomotive force in the airgap of the configuration of fig. 1649
Application of the Fourier series shows that only odd cosine terms can exist. π/2
We calculate these Fourier components:
F
S1 , α
∫
4 __ = π F . cos k α . dα with k = 1, 3, 5, ... 0
The fundamental harmonic (k = 1) has an amplitude: F
1 (S1)
[


π/4
5.π/12
 ] π/12
4 __ = π . 300 . sin α + 220 . sin α + 80 . sin α = 593.39 Aw 0
0
0
The amplitude of the third harmonic (k = 3) is given by:
[
4


15.π/12
F 3 (S1)= ____ 3 . π . 300 . sin α
3.π/4
o
o
[
π/4 o
Fifth, seventh and ninth harmonics prove to be zero. 4
 ]
+ 220 . sin α + 80 . sin α = 0 Aw

55.π/12
F 11 (S1) = _____ 11 . π . 300 . sin α 0

11.π/4
 ] 11.π/12
+ 220 . sin α + 80 . sin α 0
0
= 53.94 Aw
F 13 (S1) = 35.84 Aw 1 __ The eleventh harmonic is 11 ≈ 9% of the fundamental.
According to fig. 1650 the maximum mmf is 600Aw. The amplitude of the fundamental according to our calculation is 593.39 Aw. Usually we define an equivalent sinusoidal winding as : 593.39 NSe = 1200 . ______ 600 = 1184.78 Aw
16.45
ELECTRIC MACHINES
F µ,α =
^ NSe . i µ 3 _______ __ . .
[ 2
2
]
cos ( ωS . t + φµ − α )
^
3 . NSe . i µ _________ F = . cos ( ω S . t + φ µ − α ) µ,α 4
(1644)
Expression 1644 indicates that the magneto motive force is sinusoidally distributed in space and ^ rotates with an angular velocity of ω S. The amplitude is determined by NSe. i µ . The maximum of the magnetic field at each instant occurs at an angle α , given by: α = ωS . t + φµ (fig.1652).
Iµ
ωs
Fµ,α φµ
α
stator reference axis
(t = 0)
Fig. 1652: Space vector of the mmf in the air gap of an asynchronous motor
^ . cos ω .t and The angleφ µis the displacement angle between the applied phase voltage vS1 = v S S the magnetizing current i µ. With the choice of phase order UVW we obtain a counter clockwise rotating stator field. If the connection to any two windings are reversed (e.g. UWV) then the stator field rotates clockwise.
3.1.4 Magnetising inductance
→ µ0 . F µ,α → _________ The induction in the air gap is calculated with: B ag = l ag
From (1644) we can calculate the effective value of F µ,α so that we can write for B ag : →
3 . N . µ →
0 Se B = ________ 4 . l . I µ ag
(Tesla)
(1645)
ag
We consider fig. 1651. ^ We are looking for the instant at which i S1= i µ . cos (ωS . t + φ µ), this occurs at ω S . t = − φµ . At that instant the magnetic induction (vector) is directed according to the reference shaft of the stator (this is coil U 1U2). The value of the mmf follows from (1644): F µ,α =
^ 3 . NSe . i µ µ0 . F µ,α 3 .NSe . µ0 ^ __________ ________ __________ . cos α so that: B = = . i µ . ag,α 4 lag 4 .lag
^
cos α = B ag . cos α
The magnetic induction is sinusoidally distributed in space just like the winding.
16.46
ELECTRIC MACHINES We will now determine the flux in the coil S1 and that leads us to the magnetizing inductance. Assume in the first case a winding of coil U1 U2 at an angle α (and α + π) to the stator. For a stator with internal radius r and axial length l the flux of one winding is: ^
Φwinding = 2 . B ag . l . r . sin α To find the maximum of all the linked flux of a winding U1 U2we have to integrate Φ winding across the winding distribution: +π/2 ^ S1 Φ = nS1 . Φwinding . dα
∫
−π/2
+π/2 N ^ ^ Se ___ Applying (1639): Φ S1 = 2 . sin α . ( 2 . B ag . l . r . sin α ) . dα
∫
−π/2
^
^
π __
S1 = 2 . NSe . B ag . l . r Φ
(Wb)
(1646)
If we replace (1645) in expression (1646), then we find the effective value of the stator flux, generated in one coil (S1): µ . l . r →
→
3.π 0 2 ____ Φ S1 = 8 . NSe . ______ l . I µ
(1647)
ag
→
The linked flux Φ S1 is proportional to the magnetizing current. Flux and current are in phase. We →
Φ S1 refer to ____ =L 0 as the magnetizing inductance: → I µ
3.π
µ0 . l . r
______ L0 = ____ .N Se . 8 l 2
(1648)
ag
Fig. 1653 shows the magnetizing inductance. The induced emf in one phase winding is given by: →
→
→
E S1 = j . ω S. Φ S1 = j . ωS . L0 . I µ
(1649)
Øs1
β
tg β = L0
Iµ 0 Fig. 1653: Magnetising curve of an asynchronous machine
ELECTRIC MACHINES
16.47
Numeric example 169: A two pole threephase machine has a rotor with a length of 100mm. The radius of the rotor is 50 mm while the air gap has an effective length of 0.6 mm. The effective number of sinusoidally distributed windings is 160. The motor is connected in star to a 400V50Hz power grid. Question: 1. The nominal induction in the air gap 2. The magnetising inductance and the magnetising current. Solution: If we neglect the resistance R S of the stator coil, then: VS1 _______ 230 ___ (1649): Φ S1= ω = 2 . π . = 0.732Wb 50 S __
400 √ 3
E S1 = VS1 = ___ __ = 230V
__
2 . √ 2 . ΦS1 _________________ 2 . √ 2 . 0.732 ^ _________ (1646): Bl = π .N . l . r = π x 160 x 0.1 = 0.824Tesla x 0.05 Se
A value of 0.8 to 1 T is normal. The maximum inductance in the stator teeth is usually 1.6 to 1.8T. If the stator teeth are as wide as the slots, then the average inductance is 0.8 to 0.9T. 3.π 4 x π x 10 −7x 0.1 x 0.05 ____ ____________________ L0 = 8 . 160 2. = 315mH 0.0006 VS1 230 _____ Magnetising current (1649): Iµ = ω . L = _________________ = 2.324A 2 . π . 50 . 315 . 10 −3 0 S
Magnetising inductance (1648):
1LE1 Standard induction motor 0.55 to 200 kW 9.9 to 1546 Nm 50/60 Hz 246 poles USA ( also 8 poles) General purpose in aluminium and "Severe Duty" in cast iron for more demanding industry. Rotor in cast copper for higher efficiency (IE1 to IE3) Modular design: a number of options are easly to add (see p.20.15). Photo Siemens: Cross section of a 1LE1 asynchronous motor
18.8
CURRENT  , ANGULAR POSITION  , SPEED TRANSDUCERS
1.6.2 Closed loop sensor In fact the construction of a closed loop sensor with fluxgate is comparable with the magnetic circuit in fig. 184 in which a Hall sensor is now replaced by a fluxgate sensor. This is shown in fig. 189. The current to be measured I1 produces a flux Φ1 and a current I2 is sent through the secondary coil (flux Φ2 ) in such a way that Φ 2 = − Φ1 . Here N 1 .I1 = N2 .I2 . N2 __ It follows that:I 1= N . I2 . 1 We are continually trying to bring the fluxgate out of saturation to the point of symmetry of the hysteresis loop so that the resulting flux (Φ1 + Φ 2 ) is zero. As was the case with the zero flux transducer (fig.184), the voltage drop of I2 over a resistor produces the output voltage of the current sensor. Since this is a floating output it is sometimes fed into a differential amplifier with opamps. The signal generator which supplies the winding Nf is comprised of a comparator circuit with hysteresis (Schmitttrigger). The current change in If produces noise in the primary of the current sensor as a result of the transformer operation. A filter is required to remove the noise.
I1
Φ 1 YES
I =
N2 . I 2 N 1
Φ + Φ2 = 0
1
Φ 2
1
fluxgate N
f
N1
NO N2
I2
2
adjust I
Fig. 189: Adjusting the closed loop
The firm LEM supplies the market with the following fluxgate sensors CAS/CASR, CFSR and CTSR (see photo’s on p. 20.13 and p.20.82).The photo at the top of p. 189 shows a didactic model of a fluxgate current sensor.
CURRENT  , ANGULAR POSITION  , SPEED TRANSDUCERS
18.11
1.6.3 Current measurement in a photovoltaic installation As discussed in chapter 5 and 15 we want a PVinstallation to operate at the MPP (maximum power point). This means for every PV topology we need to measure the DCoutput (current and voltage) of the solar panel. In addition a current measurement in the input of the control loop is necessary for protection against short circuit and over current. In installations without transformer the maximum DCcurrent that the PV installation can send into the power grid is a maximum of 10mA to 1A. The value depends on the standard used in the different countries. (IEC 61727,…VDE 01261). From all these requirements current sensors used need to have: an accuracy better than 1%, low offset, low drift in amplification, operate well with DC and low frequencies. A sensor with closed loop fluxgate technology meets the specified requirements. The firm LEM has developed their CTSRseries for this type of application. In addition to the mentioned requirements (accuracy, low drift,…) the CTSRseries has a number of additional functions such as: reference pin (for self testing), demagnetizing function (also via the reference pin). The CTSR can be used for single and threephase voltages. The CTSR also has a version with four individual primary conductors. Three conductors are used for threephase systems and the fourth conductor is used for testing the operation or as neutral conductor of the threephase net. In addition the magnetic core of the CTSR has two magnetic screens to protect the fluxgate from external magnetic fields. The reference pin provides access to the 2.5 V reference voltage. This V ref. can be used as a reference for an ADconverter.
Fig. 1810 (photo LEM): Example of a leakage current
Fig. 1810 shows an example of how a leakage current can flow in a PVinstallation without transformer. The capacitance between solar panel and the roof can be the cause of the leakage current which can result in the solar panel rising to the net potential. It is therefore necessary that any leakage currents are measured contact free and in a galvanically isolated manner.
CURRENT  , ANGULAR POSITION  , SPEED TRANSDUCERS
18.17
The photo below from the firm Heidenhain shows once again the transparent principle of an optical encoder.
Photo Heidenhain: Principle configuration optical encoder
The light beam emitted by L (e.g. a LED) is concentrated by a lens K and thereafter split by a detection plate into five fields. If a glass ruler M with an incremental scale C (e.g. 20µm) comprised of chrome strips (C/2 wide) moves with respect to the detection plate A the light which shines through is sinusoidally modulated (due to the special construction of the detection plate). Four of the five fields produce via phototransistors sinusoidal signals displaced by 90° with respect to each other. Since the four sinusoids are not symmetrical with respect to the zero line two resulting symmetrical sinusoids S1 and S2 are produced via a push pull circuit which are 90° out of phase with each other. This is the SIN/COS output. A typical amplitude is 1 V peak to peak. The sinusoids can also be converted internally to block waves (as in fig. 1815) which then appear on the output at TTL or HTL level. The fifth field serves as a reference R (one per revolution of the shaft). Encoders may be single turn and multi turn types. Single turn systems provide the actual position within one revolution. Multiturn pulse sensors operate with multiple revolutions.
19.12
SPEED AND (OR) TORQUECONTROL OF A DCMOTOR
3. CONTROLLED SINGLE QUADRANT DRIVE 3.1 Operation A good speed controller must meet two requirements: 1. n should be independent of the load 2. I ashould be limited to an adjustable maximum. The block diagram of the control circuit for the speed control with armature control is shown in fig. 1911. GRID 3
speed reference (set value) speed regulator
current limiting circuit
ε
+
current regulator
control unit
+

PI (D)
Ia
PI (D) reference
controlled rectifier
Ia actual
nactual
tacho
M
n
+
Im
Φ
Fig. 1911: Block diagram of a controlled single quadrant drive
With a potmeter the desired speed is set. A tachometer produces a voltage proportional with the actual speed of the motor. Setpoint and process value in the form of voltages are compared with each other in a comparator. At the output of this comparator the difference ε appears. If ε ≠ 0 the speed of the motor is not equal to the setpoint, in other words there is an error. The fault ε is referred to as the error signal. The error ε is transferred to the first controller. If this controller is a proportional amplifier, then the output signal will be proportional to the difference between setpoint and the actual speed of the motor. In other words the output of the speed controller indicates if the motor should accelerate or decelerate. For a positive ε the motor should accelerate and for a negative ε the motor should decelerate. An acceleration or deceleration corresponds with an increased or decreased motor torque. This is achieved by an increased or decreased current in the motor.
SPEED AND (OR) TORQUECONTROL OF A DCMOTOR
19.13
The output of the speed regulator is therefore a measure of the desired armature current of the motor. We compare this desired armature current with the actual armature current in a current comparator. The output of this comparator is now the input for a second controller to achieve an as exact as possible current control. The output of this second controller sends a pulse (control signal), which for its part controls the thyristor bridge. The output voltage of the bridge is controlled as a function of the input voltage of the pulse generator. As we shall see later the controllers are not just proportional controllers but normally PIcontrollers. In exceptional cases PIDcontrollers may be encountered.
3.2 Advantages of internal current control The current control circuit is especially active with changes in the motor load torque. The short control time (7 or 10 ms) in the current circuit do not allow the load variations to affect the speed control since the armature current is quickly adjusted to the required value by the new load torque. With this internal current control the maximum motor current is easily controlled. If we limit the input current controller to for example 120% of the nominal value of the armature current, then thyristor bridge and motor are protected against overload and against a high start current surge of the motor.
Current limitation is especially important when quick speed changes are required from the machine. If the motor load has a large moment of inertia then large (required) speed changes cannot be followed. If the speed setpoint is suddenly raised, then the controller output will want a much higher armature current than the nominal armature current. The current limitation sees to it that the nominal value is not exceeded. With constant flux and nominal armature current during acceleration the motor will maintain an almost constant torque M ≈ Mem = k 2. Ia . Φ. If the counter torque of the mechanical load isM t, then the difference M v = M − Mt will cause the machine to accelerate according to M v = Jm . (dω/dt). Here dω/dt = acceleration; J m = moment of inertia. If M v is constant then the machine will accelerate linearly to the desired speed. If the desired speed is less than the actual speed then with single quadrant operation the motor will run till the desired speed is reached and this with zero armature current.
!
Remark Why is the armature current limit set to 120% of the nominal armature current? Very simply since a fully loaded motor (Ia =I a nominal) could no longer accelerate if we should limit the current to 100%.
19.14
SPEED AND (OR) TORQUECONTROL OF A DCMOTOR Numeric example 192:
. . . .
Given: DCmotor: 220V / 10A / 1.76kW / 1000 rpm / Ri = 1Ω tacho generator: constant is 10V / 1000 rpm B 6  controlled rectifier with impulse module: Vcontrol = 0 to 10V gives V diα = V a = 0 to 220V control characteristic: fig. 1912 To enable maximum use of this numeric example a number of values are indicated in this characteristic. Vdi α = Va 220 162,5 110 57,5
α 0
30
42
60 74°50' 90
120
150
180
Vcontrol (V) 10
5,3
3,33 1,683
0
Fig. 1912: Control characteristic of a B 6 Bridge
Required: 1. With n = 250 rpm and with 50% motor load determine the input and output values of the different “blocks “in fig. 1911. The internal current control and current limitation are inactive. 2. The same question but now with n = 500 rpm, 50% motor load and active current control as in fig. 1911.
SPEED AND (OR) TORQUECONTROL OF A DCMOTOR
19.15
Solution: To make the solution as short as possible we will ignore the transient behaviour, and only concentrate on the final steady state value. The transient behaviour of a single quadrant drive is studied under optimising the controllers (p.19.34). From the name plate of the motor we obtain the following: at full load, V a = 220V and n = 1000 rpm: Va − E _______ 220 − E Ia = 10 = ______ R = 1 or: E = 210V at 1000 rpm i The normalised emf e N is: eN = 0.21V/rpm 9550 . P _______
The nominal torque is: M =
n
9550 x 1.76 _________ = 16.8 Nm 1000
=
1. n = 250 rpm  50% motor load M t = 50% . 16.8 = 8.4Nm I a = 50% . 10 = 5A n = 250 rpm →→ E = 250 . eN = 250 . 0.21 = 52.5V V a= E + Ia . Ri = 52.5 + 5 x 1 = 57.5V Fig. 1912: Vdiα = 57.5 = 220 . cosα → → α = 74°50’ →→ Vcontrol = 1.683V
. . . . . . The single quadrant controller without internal current loop is drawn in fig. 1913. GRID 3 + 10 V
(n*)
ε
2,5V +

0V
speed regulator PI
pulse unit Vcontrol
α
1,683V
74°50'
B6  controlled rectifier
2,5V
V = 57,5V a
Ia
(n)
1W
5A 5V
V
5A E = 52,5V
I
{ 00  10A 10V
M Φ
8,4 Nm M 250 tr/min
Mt 8,4 Nm
2,5V
Fig. 1913: Speed controlled DCmotor at 250 rpm
Ri
T
n
Im
F1
F
2
19.16
SPEED AND (OR) TORQUECONTROL OF A DCMOTOR 2. n = 500 rpm  50% motor load * To set the desired speed n = 500 rpm we adjust the wiper of the potmeter from 2.5 V to 5V. In the first instant the motor remains rotating at a speed of 250 rpm and the tachometer produces 2.5 V. The error voltage ε becomes 5 − 2.5 = 2.5 V. The ncontroller immediately makes a jump at its output (proportional part of the PI controller) and at the same time the controller begins to integrate (I  part) so that the output voltage of the ncontroller reaches 10V in a number of ms. With Vcontrol =10 V then α = 0° and Vdiα =V a= 220 V. Due to the inertia of the motor with its drive in the first instant E remains practically unchanged at E = 52.5V so that: a − E ________ V 220 − 52.5 ______ Ia = R = = 167.5A ! 1 i
Too much for our 10 A motor and also for the associated SCR bridge . We are compelled to limit the armature current and maintain it within safe operating values. The only way to do this is to measure the current Ia with a current sensor and to control this current to below an adjustable (acceptable) maximum. For this reason the internal current loop of fig. 1911 is now active. This is shown in fig. 1914 with the current limited to 9 A, and speed setpoint n*= 500 rpm with the motor 50 % loaded. The current sensor is followed by an I/V converter that for example converts 010 A to 010 V. GRID 3 limited at 9V (I = 9A!)
+ 10 V
(n*)
5V
ε1
+
a
speed regulator 5V
ε2
5V +


5V
5V
current regulator
control unit Vcontrol
0V
α
60°
3,33V
B6  controlled rectifier Va = 110V
(n)
Ia 5A 5V
V
Ri 1W
5A E = 105V
I
M Φ Mt
Im
8,4 Nm
8,4 Nm
500 tr/min 5V
T
M n
Fig. 1914: Single quadrant DCmotor at n = 500 rpm, Mt = 0.5 x M = 8.4 Nm andI a max= 9A
E = 500 x eN = 500 x 0.21 = 105V; Va = E +Ia . Ri = 105 + 5 x 1 = 110V Vdiα = 110 = 220 . cosα →→ α = 60° →→ Vcontrol = 3.33V
F1
F2
20.18
SPEED AND TORQUECONTROL: 3PHASE ASYNCHRONOUS MOTOR 6.3 Control range smallest speed by nominal flux _____________________________
We define the control range =
largest speed (in field weakening area )
(2011)
. Electromechanical torque
In the assumption that the nominal motor current I nis flowing, it follows from (206) that the torque M em proportionally changes with Φ . constant _______ In the field weakening area Φ = k 1. VS /fS = k1 . f , so that the torque: M em ~ 1/fS ~ 1/n. S This is indicated by the hyperbole in fig. 2017.
. Maximum torque
In the field weakening area VS is constant. From (208) it follows that in this area the maximum torque quadratically decreases with increasing frequencyf S. Above F7 the motor cannot run with constant power, see fig. 2017. The range wherein the motor power remains constant (up to f 7 ) depends upon Mn /Mpo in the area with nominal flux Φn . Often for a motor M n /M po= 0.5, so that : f7 ≈ 2 . f4 . From f7 the motor can only deliver a certain percentage of its maximum torque because of the losses. In this area of “higher speed” M emis inversely proportional to fS 2 .
M (In = constant) nominal flux
Mpo
field weakening
1 2 fs
Mmax.
Mn
Mem
1 fs
M
em
0
f1
f2
f3
f4
constant torque
Fig. 2017: Mn curves of a threephase asynchronous motor
f5 constant power
f6
fs (Hz)
f7 higher speed
n (rpm)
1 fs2
20.46
SPEED AND TORQUECONTROL: 3PHASE ASYNCHRONOUS MOTOR 7.5 Coordinate transformation 7.5.1 Transformation formulas The flux creating and torque creating components of the fictive coils D and Q are in reality obtained from two variable (e.g. sinusoidal) currents iα and iβ in the real coils of our twophase motor. These coils are fixed in the stator. We draw the αβ coordinate system that is related to the stator in such a way that the coil α produces a flux according to the αaxis and coil β according to the βaxis. (fig. 2042).
β
α
ωS
α
λ(t)
β
β
ΦR
α
Fig. 2042: Equivalent twophase stator coordinate system
We know from fig. 2031 that the rotating field in the stator has an angular velocity ωS . As with the threephase motor the rotor is pulled along with a speed n < nS . The rotor has a rotor flux ΦR which also rotates at ω S rad/s, similar to the threephase motor. If we now allow the fictive coils D and Q (and therefore the dq coordinate system) to rotate with the rotor flux ΦR as in fig. 2041, then we can draw fig. 2043.
SPEED AND TORQUECONTROL: 3PHASE ASYNCHRONOUS MOTOR
20.47
β q
d
iβ
iq
β
id
α
β
0
ωs
id α
ΦR
λ
iα
α
iq
α
ia
β iβ Fig. 2043: dq coordinate system, linked to the rotor flux
The current iα produces the components in:
. ddirection: i = i . cos λ . qdirection: i = − i . sin λ . ddirection: i = i . sin λ . qdirection: i = i . cos λ dα
α
qα
The current iβ produces the components in:
α
dβ
β
qβ
β
. .
The currents iα and iβ from the stator coordinate system (α, β) provide currents id and iq in a coordinate system that rotates with the rotor flux ΦR : id = idα + idβ iq = iqα + iqβ or:
i d= iα . cos λ + i β . sin λ
(2020)
i q= iβ . cos λ − iα . sin λ
(2021)
Conclusion: The expressions (2013) and (2014) give for a threephase induction motor the currents iα and iβ as transformed components of the threephase rotor current (iS1 , iS2 , iS3 ). This equivalent system αβ is fixed to the stator. We call it the αβ stator coordinate system. The coordinate system dq with its components id and iq are called the field coordinate system since they are fixed to the (rotor) flux. The expressions (2020) and (2021) describe the transformation from stator coordinate system to (rotor) field coordinate system. This is known as the Park transformation.
20.56
SPEED AND TORQUECONTROL: 3PHASE ASYNCHRONOUS MOTOR
iµ
calculation module sinλ iq
cosλ
i*s1

k2
M

L
R Φ
Φ* R
+ field weakening n

n*
+
n  controller
0
iµ
limiter
M*
+
M
Φ  conroller
i*d
i  controller i*q
ejλ
* iβ
i*α
2/3
i*s2
i*s3
+
GRID
CRPWM inverter
is1
is
ω
is2
3
T
M
Fig. 2054 shows finally a block diagram with associated PIcontrollers for field orientated speed control of an induction motor with imposed stator currents.
Fig. 2054: Speed control of induction motor in field coordinates (indirect method)
7.6.5 Applications with closed loop control Frequency controllers with vector control are used in: cranes, elevators, lifting equipment, extruders, paper machines, rolling mills in the production of steel, CNCmachines, cable layers (ship based), drums etc.
20.66
SPEED AND TORQUECONTROL: 3PHASE ASYNCHRONOUS MOTOR 7.8.3 Moment of work torque As shown in fig. 2065 the work torque in fig. 2069 is proportional to the area of the triangle LR → __ → → jδ formed by iS , L . iR . e and iµ . This area can also be described in fig. 2069 as: 0 1/2 .i q.i µ so that: M em = k2 .i µ.i q
(2039)
With Φ r =L 0.i µ is M em = k3 . iq . ΦR
(2040)
Another method is to assume that M em is proportional with the product of the rotor flux with the → component of iS that is perpendicular to this rotor flux. In fig. 2069 this leads to : Mem = k3 .i S . sin ε . ΦR , so that:
M em = k3 . iq . ΦR
(2040)
L0 3 . p ___ ____ Calculation leads to : k 3 = 2 . L with p = number of pole pairs per machine. R
7.8.4 Mathematical model in field coordinates → → d jλ jλ jλ __ With iµ = iµ . e (2037) becomes: τR . dt ( iµ . e ) + ( 1 − jω .τR ) . i µ. e = iS →
d iµ → dλ jλ jλ jλ __ τR . e . ____ dt + τR . iµ . j . . e + ( 1 − j ω . τR ) . iµ . e = iS dt →
d iµ → − jλ τR . ____ + τ . i . j . ω + ( 1 − j ω . τ ) . i = iS . e µ µ S R R dt →
→
jθ
− jλ
From (2024): iS (t) = iS . e it follows: iS . e
j (θ − λ)
= iS . e
→
jε = iS . e − jλ
This may be expressed in the complex dq coordinates system as: iS . e →
d i
(2041)
jε = iS . e = id + j.iq
µ ____ so that: τ R . + τ R . iµ . j . ωS + ( 1 − jω . τR ) . iµ = id + j . iq
dt Taking the real and imaginary parts separately: →
. . ( ω − ω ) . i . τ =i d i τR . dt + iµ = id µ ____
S
µ
R
q
(2042)
→
d ΦR ____ Multiplying (2042) with L0 : τ R . dt ΦR = L0 . id +
(2043) (2044)
SPEED AND TORQUECONTROL: 3PHASE ASYNCHRONOUS MOTOR
20.67
Rewriting of (2044) using Laplace notations: τR . s . ΦR = L0 .I d ΦR
1
________ ___ I =L 0. 1 + s . τ d
R
LR
This is the transfer function of a first order system with time constant τR = __ R (fig. 2070) R The flux ΦR is determined by id ! Id
ΦR
τR
L0
Fig. 2070: Transfer function of rotor flux
iq ωS − ω ________ ______ From (2043) the slip follows: g = ω = i . τ . ω R Φ ___ With iµ = L then 0
iq 0 L g = _____ . ___ ω ΦR . τ R S
From (20.43) we can derive:
S
µ
R
S
L0 . iq
ω S− ω = ______ Φ . τ R
(2045)
R
The mathematical model of a threephase squirrel cage induction motor in field coordinates (reference coordinates as field coordinates rotates with rotor flux) is given by:
M = k3 .i q . ΦR
(2040)
τR . ____ dt + ΦR = L0 . id
(2044)
ω S− ω = ______ Φ . τ
(2045)
dΦR
L0 . iq R
R
If we now determine the stator current components at every instant in a coordinates system that rotates with the rotor flux then the torque of the induction motor may be controlled exactly as with a DCmotor. In fact the expression Mem = k3 . iq . ΦR is comparable with the formula for the moment of work torque of a DCmachine: M em = k2 . Ia . Φ (192). With an induction motor ΦR can be kept constant or controlled via i das (2044) suggests. The speed n of the induction motor can be controlled with (2045) by influencing ωS andi q. In fact with field orientation via mathematical transformation the way is open to control the torque of an induction motor in the same way as with a DCmotor.
21.12
CONTROL APPLIANCE, SR, 3PH SYNCHR., INDUCTION SERVOMOTOR
Fig. 2117 shows a typical threephase SRconverter. Since the stator coils of an SRM are totally insulated from each other a fault in one stator coil will have no effect on other “legs” of the converter. The motor can remain operating even with a faulty coil, although with copper losses. Note also that the current through a phase winding always flows in the same direction. Opposing poles are provided with windings which are in parallel or in series and each forms a stator coil. The polarity of the motor torque is independent of the current direction through the stator coil. It is the position of the rotor teeth with respect to the stator that determines the direction of rotation. In fact the diodes operate as freewheel diodes. The transistors are always in series with a coil in which the current always flows in the same direction. The problem of protection is a lot simpler compared to the threephase inverter of an induction motor. These SRM converters are more robust than classic threephase inverters. +V cc
D2 T3
T1
D1
S1
T2
D3
D4 T5
S2
T4
D5
D6
S3
T6
Fig. 2117: Typical threephase inverter used to supply an SRM
2.5 Speed, position and torque control The topology of a closed control loop is comparable to that of a DCmotor. With an SRM in addition feedback of the rotor position is required for commutation of the inverter. Fig. 2118 shows the closed loop for a position controller. rotor position
θ*
position controller
n*
velocity controller θ/ n
i* +

current controller
inverter
i
SRM
i PG
θ
θ
Fig. 2118: Position control using an SRmotor
θ
ELECTRICAL POSITIONING SYSTEMS
!
22.3
Remarks The name “servomechanism” is often reserved in the literature for a positioning system.
For highdynamic positioning and torque control electrical servosystems used only DCservomotors or permanent magnet synchronous motors. Due to the technological developments in the previous years the flux vector control of an asynchronous motor has dramatically improved. The result is that this vector control is now capable in many applications of competing with traditional servosystems. The border between drive system and servosystem is becoming more vague. Certain company’s such as Yaskawa use a “radar” card on which the differences in performance between servosystems and flux vector control (VC) are indicated. This is shown in fig. 221.
speed regulation range (X : 1) 10,000 VO SER
acceleration time 0.01 (s)
0.1
1,000
TOR VEC ROL T CON
100 0.01
1 10
10
10
100
0.1 1
1
1000 cycles/min
0.1 0.01
position accuracy (°) Fig. 221: Points of difference servosystem with VCsystem
Source : Yaskawa and Control Engineering
0.1
1 10
1 10
speed regulation (%)
10
100 1,000
frequency respons (Hz)
22.16
ELECTRICAL POSITIONING SYSTEMS Fig. 2212 shows the principle configuration of a DCservosystem. µprocessorboard Ia*
ω
wave shaper
servo amplifier
+

Ia
ENCODER
µprocessor
* ωsc +
TACHO
D/A
MOTOR
Fig. 2212: DCservosystem with the three control loops
3.4.2 Controlled speed Once the value ωSC is reached the speed controller comes into action (SC = Speed Control). * * Depending on the setting for ωSC , the ω of the motor will be controlled so that ω = ωSC .
3.4.3 Delay In order to stop at the desired position a time t3 has to be determined at which the motor should start to decelerate. In reality we determine the number (k) of pulses that the encoder still needs to give until the motor shaft is at the setpoint location. Expression (2211) allows for calculation of the negative value Ia required to decelerate with a dω ___ value γ = − dt Fig. 2213a shows that ωSC = (t4 − t3) . γ. From fig. 2213b it follows that the shaft of the motor is ωSC ___ rotated by an angle θ = (t4 − t3 ) . 2 between t3 and t4 . Elimination of (t4 − t3 ) in both of the previous formulas gives 2
ωSC ___ ωSC ___ ωSC ____ ωSC θ = ( t 4− t3 ) . ___ = . = γ 2 2 2 . γ
2.π.n Assume that the speed is controlled at 3000 rpm, then: ωSC = ______ 60 314.1593 rad/s = If for example (according to the microcontroller countdown) we are still k = 3800 encoder pulses removed from the setpoint position and the encoder generates 2000 pulses/rev, then: θ = 11.938 rad.
ELECTRICAL POSITIONING SYSTEMS
22.17
2
ωSC ____ 2 The required delay is then: γ = 2 . θ = 4133.6925 rad/s
With JR + Jeq = 3.9 x 10 − 5kgm 2and K M = 5.4 x 10 − 2 Nm/A becomes: γ . ( JR + Jeq ) __________ I a= = − 2.985A K M
ω ω = ωSC
ωSC
ωSC γ= 
t = t3
dω dt real STOP
θ t
0
t3
t4
ts
(motor shaft)
STOP
encoder pulses
t = t4
k pulses (a)
(b)
Fig. 2213: Deceleration of the servomotor to a stop
3.4.4 Stopping It is clear that the calculation above is dependent on the linear operation during deceleration. This is difficult to know for amongst other reasons because of the digital control of the speed which is always in“steps”. Normally a positioning system is controlled in such a way that the rotor with only a few oscillations can be brought to a halt. The time t Sof these oscillations is known as the settling time. Mostly the speed controller is switched off in the region close to the angular position setpoint and only the position controller is operating so that when the stop position is achieved a good holding torque is maintained.
22.32
ELECTRICAL POSITIONING SYSTEMS B. Mn curves 1. Static torque The torque that is required for a stationary SM (zero steps and nominal current through one phase) to move one step is called the holding torque.
M Mh
0
1
2
steps ∆α
 Mh
Fig.2223: Static torque of a steppermotor
2. Dynamic torque M
(mNm)
holding torque
pullout torque
start without error 30 25 running
20 pullin torque 15 slew range 10 5
STARTSTOP RANGE pulse rate (Hz) 100 pull  out rate
200
300 pull  in rate
400
500
steps/s
maximum slew  rate
Fig. 2224: Path of the dynamic torque of a SM
With a higher step frequency the average stator current (and flux) will decrease. The torque decreases as a result.
22.40
ELECTRICAL POSITIONING SYSTEMS F. Microstepping A typical small step angle of a hybrid stepper is 1.8°(sometimes 0.9°). This corresponds to 200 (or 400) steps per revolution. To obtain a higher precision a reduction mechanism or gearbox can be used. An electronic solution for obtaining higher precision consists of “micro stepping”. A full step of the SM is divided into smaller steps. Assume an SM with p pole pairs on the rotor. For the full step drive we vary the phase current between + If and zero or between zero and − If . If we now change the phase current in k small steps then it is possible to orientate the flux in m = k.p different directions resulting in the rotor assuming m possible positions. This can be applied to a threephase SM. Fig. 2238a shows the basic cross section of such a stepper.
1
ΦS
N Z
1 3
1 3
2
2
(a)
2
3
(b)
(c)
Fig. 2238: Threephase SM with the stator flux vector
In fig. 2238b the three stator flux components are shown. These components are called positive if for example they induce a magnetic south pole on the rotor side of a stator tooth. With maximum positive and If2 = If3 = − 0.5 x I f1we obtain a resulting stator flux as shown in fig. 2238c. A practical application is to change the stator current in small steps and therefore the stator flux, and superimpose this on a sinusoidal signal as shown in fig. 2239a. With a (p =) 50pole machine and k = 40 we find that m = k.p = 2000 steps per revolution. This is shown in fig. 2239. Fig. 2239b shows the first 11 states of the stator flux. The marking 1, 2,...of the stator flux correspond to the indexes 1, 2... of the stator current changes in fig. 2239a. Position 11 in fig. 2239b corresponds with fig. 2238c. This mechanism is known as microstepping. In this way practically 10,000 to 25,000 steps per revolution are possible.
ELECTRICAL POSITIONING SYSTEMS If
1
θ
If
2
θ 0
If
3
θ 0
1 2 3 4 5 6 7 8 9 10
Fig. 2239a: Stator currents during microstepping
11
10
9
8 7 6
ΦS
5 4 3 2 1
Fig. 2239b: Stator flux during microstepping
22.41
22.54
ELECTRICAL POSITIONING SYSTEMS
9. INTEGRATED SYSTEM (SIMOTION FROM SIEMENS) As an example of an integrated positioning system we consider Simotion (Siemens motion control system). Integrated systems have multiple tools. The use of standard modules each with its own intelligence results in not only time saving during development and implementation of a project but in addition there is a multitude of application possibilities. The communication between the different modules occurs via Profibus and Profinet. In the upper part of fig. 2246 the Siemens philosophy behind Simotion is shown and in the lower figure the possibilities are shown (hardware platforms). There are three platforms (Simotion C, D and P).
Simotion D is a very compact solution. Here closed loop control, PLCfunctions and motion control are integrated into the module. This is comparable to a certain extent to the “standalone” solution of fig. 2245. Connection of the Simotion D (fig. 2248) occurs via the ET200. The ET200 (ET = external terminal) is a decentralized (periphery) module for communication (Profibus or Profinet) via a local network cable to avoid long signal cables. Connection of the Simotion D with Profibus, Profinet or Ethernet results in visualisation and operator control. HMI (fig. 2248) means “human machine interface”. PLC functionality The system philosophy of
Motion control (positioning, ...)
SIMOTION Motion control, PLC and technological functions are merged
Technological functions
(cams, temperature control,..)
Engineering system . programming . parameterizing . graphic or textbased programming . testing and diagnostics
Runtime software . PLC . Positioning . Synchronous operation . Camming . Interpolation
Hardware platforms . Controllerbased . PCbased . Drivebased
Fig. 2246: Siemens Simotion philosophy and hardware platforms
Fig. 2247: Photo of the Simotion D module, clicked
on to a Sinamics S120 power module
ELECTRICAL POSITIONING SYSTEMS
22.55
SIMATIC HMI PROFINET Industrial Ethernet
ET200 SIMOTION D
ET200 SIMOTION D
Fig. 2248: Simotion D topology (drive based)
Machines with multiple shafts are a challenge for every position control system. The reason is that every shaft increases the load on the system and on the bus. With Simotion the machine parts are split into different modules, each controlled by it’s own Simotion system. The individual systems are then connected via PROFIBUS or PROFINET. The Simotion D version is highly suited for this type of applications, on the one hand because of it’s compact design and on the other hand because of it’s fast communication within a multiple shaft system.
Simotion C is a modular solution and has the largest field of application. There are four interfaces on board for analogue and stepper drivers and also multiple digital input and outputs. Due to the segregated architecture, drives and IO ports can be connected via PROFIBUS or PROFINET.
Simotion P is an (open) system for applications where an open PCenvironment is required. To benefit fully from the options of Simotion there is an engineering system (SCOUT) that allows motion control, PLC, technical functions and drives to dealt with in one tool. Configuring, programming, testing and commissioning can all be graphically dealt with from one “work” space. After the drive has been commissioned all data can be requested via the central administration. In addition via LAD (ladder diagram), FBD (function block diagram) and text based ST (structured text) the positioning sequences can be programmed via MCC (motion control chart). More extensive control tasks can be implemented by using DCC (drive control chart).
23
23.1
E  MOBILITY
e  MOBILITY CONTENTS 1. Renewable energy 2. Electric traction 3. Electric automobile 4. Electric boats 5. Electric bicycles
At the start of the twentieth century there were already electric cars in existence. Due to the cheap availability of fossil fuels the electric car was unable to compete with the benzine engine. In addition the action radius of electric cars was small due to the problem of limited energy that could be stored in batteries. A problem that we still have today. Good battery technology is probably the most important condition for the mass breakthrough of electric automobiles. In addition to road transport there has always been electric traction (train/tram/metro). In addition electric bicycles, scooters and sailing vessels are candidates for electric drives. The complete packet of electrical transport is referred to as electromobility or in short emobility. Emobility deals with speeds from 25 km/h (bicycle) to more than 200 km/h (train) and this with powers of 250 W (bicycle) to more than 9MW (train). Electric cars are somewhere in between with powers of between 40 and 200 kW with maximum speeds between 80 and 200km/h. In this chapter we look at a number of examples of traction, cars and boats as applications of electronic motor control.
23.2
E  MOBILITY
1. RENEWABLE ENERGY Since fossil fuels (oil, gas, coal) are quickly becoming scarce, in the future emobility will be supplied by renewable energy. The energy on earth comes for a large part from the sun. The remainder comes from the heat of the earth, tide power and nuclear energy. The sun produces 3.89. 10 26 watt. The earth is at a distance of 150 million km from the sun and about 174 petawatt reaches the earth. Right now humanity consumes about 15 terawatt, this is less than one ten thousandth of 174 petawatt! The following table is a memory aid in connection with multiples and subdivisions of units.
Table 23.1 SUBDIVISIONS yocto
=
−24
kilo
−21
mega 10
−18
giga 10
−15
tera 10
−12
peta
−9
exa 10
−6
zetta
10
−3
yotta
10
10
zepto
10
atto
10
femto 10 pico
10
nano
10
micro 10 milli
. .
MULTIPLES
10
3
= 10 6 9
12 15
10 18 21 24
Can we satisfy our energy needs with solar panels and windmills? A small calculation shows: 174 petawatt of solar energy reaches the earth 25 petawatt hits the ground as radiation energy. If we could convert one thousand of one percent to electrical energy then we would have 25 TW, more than the current consumption of the entire population of the planet 0.5% of the suns energy is converted to air currents. If we could convert 2% of this with windmills into electrical energy then we would have 174 x 10 15 x 0.5% x 2% = 17.4 TW. This is still more than the required 15 TW of the entire world. We will generate renewable energy in the most favourable locations (solar panels in the Sahara and windmills in the sea and off the coast of Mexico!). Then the problem presents itself of an electricalnetwork capable of transporting the energy over vast distances. Such a network is called a supergrid. As transmission system HVDC (high voltage direct current) presents itself as a candidate. DCcables are cheaper than ACcables and the losses of HVDC are less than 3% per 1000 m. The disadvantage is the high investment costs of the converters at the beginning and end of the line. This is only viable for large distances and heavy loads. A supergrid requires international cooperation and enormous investment.
.
23.3
E  MOBILITY
2. ELECTRICAL TRACTION Electrical traction (tram, train, metro) occurs over a rail network (train) or over a local city network (tram, metro). In addition a number of local tracks exist for example in factories. The rolling stock is mostly comprised of a motorised part (locomotive), followed by passenger or freight carriages. Electric locomotives have speeds of up to 230 km/h and powers of more than 9MW. There are international differences in the supply via the overhead lines and rail. We find 1.5 kV 2 DC, 3kV DC , 15 kV16 __3 Hz, 25 kV50 Hz, mixed 3kV DC/25kV50 Hz. In addition there is a system with a DC third rail. This is usually in metro systems. 2 A number of examples of standard voltages: Germany(15 kV16. __3 Hz), France (25kV50 Hz), Belgium (3kV DC and 25 kV50 Hz) , Netherlands (1.5 kV DC). Turkey, India and the Congo also work with 25 kV50 Hz. The new electrification networks in Western Europe 90 % of make use of 25 kV50Hz. For ease of transport on international lines multiple voltage locomotives are used. Urban transport operates with DC: 400 V/600V/650 V/750V/900V. Most common are 600 V and 750 V. In addition to the “classic” train/tram/metro another term has become popular in recent years namely “light rail”. There is no clear definition for this term since “light rail” is more a collective term for fast trams (inter local), rail vehicles that operate on tram and train tracks and light weight trains (light and cheap trains). There is for example the Docklands Light Railway in London in which you can connect from a metro and ride to the Docklands. Special metro lines are for example the Thais “Special Skytrain Bangkok” with a maximum speed of 80 km/h and operating on 750 V DC  third rail, or the automatic metro in NurembergGermany which operates without driver. In addition to speed there is also the factor acceleration /deceleration. It is accepted that the comfort value of acceleration and deceleration is 1m/s 2. With regenerative braking the motor operates as generator. The energy is either returned to the power grid or stored in batteries or supercaps. Supercaps are capacitors which can store a large amount of energy. Table 232 provides an idea of the energy density and lifespan of batteries and supercaps.
Table 232 COMPONENT
ENERGY DENSITY
LIFE SPAN
Lead acid battery
30  50 Wh/kg
3  5 years
Liion
90  150 Wh/kg
5  10 years
Supercap
3  5 Wh/kg
unlimited
23.4
E  MOBILITY 2.1 The Flexity Outlook Tram (Brussels) by Bombardier Transportation Fig. 231 shows the block diagram of the primary electronics of the Brussels tram. The Flexity Outlook tram by Bombardier Transportation has two models. The “long” tram has a length of 43.2m and the “short” model is 31.8m long. The tram is constructed with four (six) asynchronous motors depending on the model. Every motor (photo lower right on p. 23.5) is designed for 3 x 430V60Hz and has a nominal power of 112 kW. Via pantograph PT, power switch DJ and input filter L10 C , the 650 V supply is applied to the inverter. This inverter is constructed with IGBT’s, operates according to the PWM method, has a switching frequency of 2.5 kHz and is driven by a computer board (with microprocessor MC68360, DSP 56203 and FPGA XC4013). The output frequency of the inverter is 0 to 300 Hz. At 150 Hz the tram speed is 70 km/h. During braking energy is regenerated as long as the catenary wire voltage is less than 920V. Above 920 V it is switched over to dissipation braking via the brake chopper to resistors of 2.6 Ω. On the lower left of p. 23.5 there is a photo of a traction converter.
Fig. 231: Block diagram of the primary electronics of a Flexity Outlook tram (Bombardier Transportation)
23.5
E  MOBILITY
Photo Flexity Outlook: Tram in Brussels (Bombardier transportation)
Photo Bombardier transportation:
Photo Bombardier transportation:
Traction converter of the tram shown above
Motor of the tram shown above. The grey part has
(Dimensions: b = 1600mm; h = 450 mm;
dimensions 850 x 385mm
d = 1255mm; Weight 424 kg)
EV.16
EVALUATION: SOLUTIONS
8.12
^ ; fig. 821: v ^ Fig. 819: v
8.13
Φ = 72°20’;
8.14
α = 60° →→ (813): V diα = 103.5V α = 105° →→ (811): V diα = 26.52V
Φ = 60°; fig. 84 →→ δ = 240°; (85) →→ Vdiα = 51.76V
^ v ___ (85) →→ V di = 2 π . (1 + 0.438) = 74.46V; IT(AV)max = 10.26A and V RRM ≥ 325V
Φ = 60° →→ α = 30° →→ (813) →→ Vdiα = 179.29V α = 60° →→ (813) →→ Vdiα = 103.5V α = 90° →→ δ = 233° →→ (811) →→ Vdiα = 62.30V ^ 8.16 a) v b) v^ with 90° ≤ α < 180° 8.17 No. Consider for example Th1 and Th6 with there driving voltage v 12 . At ωt = 120°, v12 8.15
is negative so that Th6 cannot fire. The same conclusion can be made for the other semi conductor pairs
Startup: α = φ1 = 90° →→ Q1α = Vdi . Id ( x 1.5) = 150kvar 2 V RMS 8.19 P = _____ R b HALF WAVE: α = 0° : Pmax = 0.5/R b ; α = 30° : P 1 = 0.4855/Rb
8.18
α = 150° : P2 = 0.0144/Rb for the ratios: P 1 /Pmax = 97% and P2 /Pmax = 3%
FULL WAVE: α = 0° : Pmax = 1/Rb ; α = 30° : P 1 = 0.97/Rb α = 150° : P2 = 0.03/Rb
for___the ratios: P 1 /Pmax = 97% and P2 /Pmax = 3% √ 6 . Idα I 1 _______ __ 8.20 I1 = π = 12.1A; Ik = k →→ I5 = 2.42A; I7 = 1.73A
21
Table 84 →→ 300Hz: Vkα = ___ = 65.22V; 100 . Vdi 10.29
600Hz: Vkα = _____ . Vdi = 31.96V; 100
8.21
α = 60° →→ X3 = 0.115 x 230 x √ 2 →→ i 3 = 0.374A
Fig. 86:
__
^
__
^
α = 90° →→ X3 = 0.153 x 230 x √ 2 →→ i 3 = 0.497A ___
8.22
Vdi √ 6 . Id . cosα ___ (829) and (830) →→ I7 = ____________ with I = π . 7 R d
(825) →→ V di = 540V →→ I7 = 10.42A
EV.17
EVALUATION: SOLUTIONS
8.23 Fig. 841
v0
v0 v1
v1
v2
ωt
2.�/p
α
ωt
2.�/p
α
v0
v0 X
0
v1
�/p
�/p
α  �/p
2.�
X
0
2.�
(a)
Fig. 841
v3
0
0
α  �/p
v2
(b)
The v o = f(x) axes is the same as previously found in fig. 834, so that the expression for ωt = f(x) in fig. 834 is also valid here
^ p π ____ π 2 . v 2 __ __ ^ . __ ^ __ 8.24 B2controller: p = 2 →→ v π sin p = v . π . sin 2 = π __ ^ p 3 √ 3 . v π _______ π 3 __ __ ^ . __ ^ . __ M3controller: p = 3 →→ v sin = v . sin = p π π 2.π 3 ^ p π ____ π 6 3 . v __ __ ^ . __ ^ __ B6–controller: p = 6 →→ v π sin p = v . π . sin 6 = π
8.25
V
__
Vdi
(k − 1)
√ 2 k0° _______ α = 0° →→ cosα = 1 →→ ____ = 2 .
α = 90° →→ cosα = 0 →→
__ ______________ √ 2 _______ 2 2 k − (k − 1) = 2
√
(k − 1)
__ __ ____ Vk90° k . √ 2 √ 2 _______ _______ ____ 2 V = 2 . k = 2 (k − 1) (k − 1) di
√
Vkα Vk90° ____ ____ = k. Table 84: example k = 2 →→ α = 0° →→ V = 47.14% Vk0° di Vkα ____ α = 90° →→ V = 94.2% di
VOCABULARY
ENGLISH abscissa AC controller acceleration torque ACcontroller active component active power aerodynamic air gap algorithm alternating current (AC) ambient temperature amplifier analogons analogue computer analogue
DUTCH
GERMAN
abscis Abszisse wisselstroominsteller Wechselstromsteller versnellingskoppel Beschleunigungsmoment AC controller ACKontroller actieve componente aktive Komponente aktief vermogen Wirkleistung aërodynamisch aerodynamisch luchtspleet Luftspalte algoritme Algorithmus wisselstroom Wechselstrom omgevingstemperatuur Umgebungstemperatur versterker Verstärker analogons analoges Modell analoge computer Analogrechner analoog analog angular displacement sensor hoekstandopnemer Winkelencoder angular position hoekpositie Winkelposition anode anode Anode anodizing geanodiseerd eloxiert antistatic packaging antistatische verpakking antistatische Verpackung apparent power schijnbaar vermogen Scheinleistung appliance motor huishoudelijke motor Haushaltmotor applications toepassingen Anwendungen arc furnace boogoven Lichtbogenofen armature reaction ankerreactie Ankerrückwirkung armature anker Anker asynchronous motor asynchrone motor Asynchronmaschine auxiliary pole hulppool Hilfspol auxiliary thyristor doofthyristor Löschthyristor auxiliary winding hulpwikkeling Hilfswicklung avalanche diode zenerdiode Zenerdiode average resistivity soortelijke weerstand spezifischer Widerstand average value gemiddelde waarde (arithmetischer) Mittelwert backtoback (SCRs) antiparallel (SCRs) antiparallele Ventilen ballast ballast Vorschaltgerät band gap (energy gap) bandafstand Energielücke (Bandabstand) bandwidth bandbreedte Bandbreite bank of accumulators accubatterij Akkumulatorenbatterie barrier layer sperlaag Sperrschicht base voltage basisspanning Basisspannung base basis Basis battery charger batterijlader Batterieladegerät biasing voormagnetisatie Vormagnetisierung bidirectional bidirectioneel bidirektional bipolar transistor bipolaire junctietransistor bipolarer Transistor bistable bistabiel bistabil block diagram blokschema Blockschaltbild block orientated blokgeoriënteerd blockorientiert
VO1
SPANISH abscisa Controlador AC torque de aceleración Controlador AC componente activo potencia activa aerodinámico entrehierro algoritmo corriente alterna temperatura ambiente amplificador modelo analógico ordenador analógico análogo sensor de posición angular posición angular ánodo anodizado Empaque antiestatico potencia aparente motor doméstico aplicaciones horno de arco reacción de armadura armadura motor asincrono polo auxiliar tiristor auxiliar arrollamiento auxiliar diodo zener resistividad media valor medio SCRs en antiparalelos balast banda prohibida anchura de banda bateria de acumuladores capa de barrera tensión de base base cargador de batería polarización magnética bidireccional transistor bipolar biestable diagrama duncional bloque orientación
The fundamentals and applications of power electronics are thoroughly discussed with mathematical derivations, a lot of numerical exampl...