Power electronics 1st edition hart solutions manual

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CHAPTER 7 SOLUTIONS 4/03/10

7-1)

 D  N2  0.4  1  a ) Vo = Vs  = 36     = 12 V .  1 − D  N1  0.6  2  b) I Lm = iLm =

Vo2 122 = = 1.67 A. Vs DR 36(0.4)(6)

Vs D 36(0.4) = = 1.44 A. Lm f 100(10) −6 (100, 000)

iLm = 2.39 A. 2 i I Lm ,min = I Lm − Lm = 0.947 A. 2 VD 12(0.4) c) Vo = o = = 0.16 V .

RCf 6(50)(10) −6 (100, 000) Vo 0.16 = = 1.33% Vo 12 I Lm ,max = I Lm +

7-2)

 D  N2  0.6  1  a ) Vo = Vs  = 4.5     = 16.9 V .  1 − D  N1  0.4  0.4  Vo2 (16.9) 2 b) I Lm = = = 7.03 A. Vs DR 4.5(0.6)(6) iLm =

Vs D 4.5(0.6) = = 1.08 A. Lm f 10(10) −6 (250, 000)

iLm = 7.57 A. 2 i I Lm,min = I Lm − Lm = 6.49 A. 2 D (0.6) c) Vo = = = 1.6% RCf 15(10)(10) −6 (250, 000) I Lm,max = I Lm +

7-3)

N 2 Vo  1 − D  3  1 − .32   D  N2 a) Vo = Vs   =   =   = 0.145 N1 Vs  D  44  .32   1 − D  N1 N or 1 = 6.90 N2 b) I Lm =

Vo2 32 = = 0.640 A. Vs DR 44(0.32)(1)

iLm = 0.4 I Lm = 0.4(0.640) = 0.256 A. Lm =

Vs D 44(0.32) = = 184  H . iLm f (0.256)(300, 000)

7-4) Example design

Vs = 24 V . Po = 40 W . Vo = 40 V .

N 2 Vo  1 − D  =   N1 Vs  D  Let D = 0.4 (arbitrary ) then N 2 40  1 − 0.4  N1 =  = 0.4  = 2.5, or N1 24  0.4  N2 I Lm =

 N2  Vo 40 (2.5) = 4.17 A  = (1 − D) R  N1  (1 − 0.4)40

Vo2 402 = = 40  P 40 Let iLm = 40% of I Lm = 0.4(4.16) = 1.67 A

where R =

Let f = 100 kHz VD 24(0.4) Lm = s = = 57.6  H iLm f 1.67(100, 000) C=

D 0.4 = = 20  F 40(.005)(100, 000)  Vo  R f  Vo 

7-5) For continuous current

iLm 2 2 V VD I Lm,min = 0 = o − s Vs DR 2 Lm f I Lm,min = I Lm −

Vo2 2 Lm f 52 (2)(500)(10−6 )40, 000 R= = = 11.7  (Vs D) 2 [24(0.385)]2 R  11.7   continuous current R  11.7   discontinuous current

7-6) Switch is closed for DT, current returns to zero at t = tx:

I Lm,max =

Vs DT Lm

Switch open : iLm (t ) =

diLm −Vo  N1  =   dt Lm  N 2 

t −Vo  N1  −Vo  N1  Vs DT    d  + iLm ( DT ) =   (t − DT ) + Lm  N 2  DT Lm  N 2  Lm

I LM (t = t x ) = 0 = tx =

−Vo  N1  Vs DT   (t x − DT ) + Lm  N 2  Lm

Vs DT  N 2    + DT Vo  N1 

7-7)

N  a) Vo = Vs D  2  = 100(.35)(1) = 35 V .  N1  Vo 1− D 1 − .35 = = = 0.16% 2 −6 Vo 8LxCf 8(70)(10) 33(10)−6 (150, 000)2 Vo 35 = = 1.75 A. R 20  N   D .35 iLx = Vs  2  − Vo  = [100(1) − 35] = 2.17 A. −6 (70)(10) (150, 000)   N1   Lx f 2.17 I Lx ,max = 1.75 + = 2.83 A. 2 2.17 I Lx ,min = 1.75 − = 0.67 A. 2 V DT 100(.35) c) iLm = s = 0.233 A. Lm 1(10)−3150, 000

b) I Lx =

d ) isw = I1 + iLm = i pri I sw,max = I Lx ,max (1/1) + I Lm,max = 2.83 + 0.233 = 3.06 A.

7-8)

N  1 a) Vo = Vs D  2  = 170(0.3)   = 5.1 V .  10   N1  Vo 1− D 1 − 0.3 = = = 0.175% 2 −6 Vo 8LxCf 8(20)(10) 10(10)−6 (500, 000)2 b) The currents in the converter are shown below. The winding currents are for the windings in the ideal transformer model, not the physical windings. The physical primary winding current is the sum of winding #1 and Lm currents.

c) iLm =

Vs DT 170(0.3) = = 0.3 A. Lm 340(10) −6 500, 000

Peak energy in Lm : Wmax = P=

1 1 2 Lm ( iLm ) = (340)(10) −6 (0.3) 2 = 15.3  J 2 2

W = Wf = 15.3(10) −6 (500, 000) = 7.65 W .

T

7-9)

N   N  V 50 a ) Vo = Vs D  2   D  2  = o = = 0.625  N1   N1  Vs 80 N If 3 = 1, then D  0.5 N1 Let D = 0.3, then

N 2 0.625 N = = 2.08 or 1 = 0.48 (not unique) N1 0.3 N2

Vo2 502 V 50 = = 10 ; I Lx = o = = 5 A. P 250 R 10 V (1 − D) 50(1 − .3) iLx = o = = 3.5 A. Lf 100(10) −6100, 000 3.5 I Lx ,min = 5 − = 3.25 A.  0  continuous current 2 Vo 1− D 1 − 0.3 b) = = = 0.058% 2 −6 Vo 8LxCf 8(100)(10) 150(10) −6 (100, 000) 2 R=

7-10)

N  1 a) Vo = Vs D  2  = 100(0.25)   = 5 V . 5  N1  using Lx = 20  H , Vo 1− D 1 − 0.25 = = = 0.33% 2 −6 Vo 8LxCf 8(20)(10) 10(10)−6 (375, 000)2 b) The currents in the converter are shown below. The winding currents are for the windings in the ideal transformer model, not the physical windings. The physical primary winding current is the sum of winding #1 and Lm currents.

c) iLm =

Vs DT 100(0.25) = 0.20 A. Lm 333(10) −6 375, 000

Peak energy in Lm : Wmax P=

1 1 2 Lm ( iLm ) = (333)(10) −6 (0.2) 2 = 6.66  J 2 2

W = Wf = 6.66(10) −6 (375, 000) = 2.5 W .

T

7-11)

N   N  V D 125(0.3) a ) Vo = Vs D  2    1  = s = = 0.75 50  N1   N 2  Vo V 50 b) I Lx = o = = 2 A.; I Lx ,min = (0.4)(2) = 0.8 A.; iLx = 2(2 − 0.8) = 2.4 A. R 25 V (1 − D)T V (1 − D)T 50(1 − 0.3) iLx = o  Lx = o = = 58.3  H .

Lx iLx 2.4(250, 000) c)

Vo 1− D =  C= Vo 8 Lx Cf 2

1− D 1 − 0.3 = = 4.8  F . −6  Vo  2 8(58.3)(10) (0.005)(250, 000) 2 8 Lx  f  Vo 

7-12)

Let

N1 = 1, then D  0.5 N3

Let D = 0.35, then N1 Vs D (170)(.35) = = = 1.2396 N2 Vo 48 Rounding , let Then D =

N1 = 1.25 N2

Vo  N1   48   =  (1.25) = 0.353 Vs  N 2   170 

Let f = 200 kHz , and design for iLx = 40% of I Lx Lx =

Vo (1 − D) 48(1 − 0.353) = = 124  H 0.4 I Lx f 0.4(3.125)200, 000

where I Lx = I o =

Po 150 = = 3.125 A. Vo 48

Alternatively, solving for the minimum Lx for continuous current , I Lx ,min = 0 = I Lx − Lx ,min =

iLx Vo Vo (1 − D) = − 2 R 2 Lx f

(1 − D) R (1 − 0.353)(15.36) = = 24.9  H 2f 2(200, 000)

where R =

Vo2 482 = = 15.36  Po 150

Lx must be greater than 24.9  H with margin, (e.g ., 25% greater ) making Lx  31  H Using Lx = 124  H , C

1− D 1 − 0.353 = = 1.63  F  Vo  2 8(124)(10) −6 (0.01)(200, 000) 2 8 Lx  f  Vo 

7-13)

150  Vs  175 V .

Vo = 30 V .

20  Po  50 W  0.667  I o  1.667 A. Example design : N Let 1 = 1, then D  0.5 N3 Let D = 0.3 for Vs = 150 V .

Then

N1 Vs D (150)(0.3) = = = 1.5 N2 Vo 30

For Vs = 175 V ., D =

Vo  N1   30   =  (1.5) = 0.35 Vs  N 2   175 

 0.3  D  0.35, which is an acceptable range of D. Other choices are possible.

Using the design criterion of iLx = 40% of I Lx , Lx =

Vo (1 − D) 0.4 I Lx f

The worst case is for the smallest D and the smallest I Lx .

Letting f = 250 kHz (arbitrary ), 30(1 − 0.3) Lx = = 315  H 0.4(0.667)(250, 000) 1− D 1 − 0.3 C = = 2.22  F −6  Vo  2 8(315)(10) (0.002)(250, 000) 2 8 Lx  f  Vo 

7-14) The current in the physical primary winding is the sum of iL1 and iLm in the model. The physical currents in windings 2 and 3 are the same as in the model.

7-15)

N  a) Vo = 2Vs  s  D = 2(50)(0.5)(0.35) = 17.5 V . N   p V 17.5 b) I Lx = o = = 2.19 V .

R 8 V 17.5 iLx = o (0.5 − D)T = (0.5 − 0.35)150, 000 = 0.29 A. Lx 60(10) −6 iLx 0.29 = 2.19 + = 2.33 A. 2 2 i 0.29 I Lx ,min = I Lx − Lx = 2.19 − = 2.04 A. 2 2 Vo 1 − 2D 1 − 2(0.35) c) = = = 0.018% 2 Vo 32 Lx Cf 32(60)(10) −6 39(10) −6 (150, 000) 2 I Lx ,max = I Lx +

7-16)

7-17)

Sw1 closed : vP1 = Vs = 50 = vLm = Lm

diLm dt

diLm VLm 50 = = = 25(10)3 A / s −3 dt Lm 2(10) 0.35 = 2.33  s  iLm = 25(10)3 2.33(10) −6 = 58.3 mA. 150, 000 Sw2 closed : vP 2 = −Vs , VP1 = −Vs For DT =

diLm = −25(10)3 A / s dt Both switches open : VP1 = 0 

diLm =0 dt

7-18) The input voltage vx to the filter is Vs(Ns / Np) when either switch is on, and vx is zero when both switches are off. (See Fig. 7-8.) The voltage across Lx is therefore

N  vLx = Vs  s  0  t  DT N   p = −Vo DT  t  T / 2  N  T   1  VLx = Vs  s  DT − Vo  − DT    =0 2    T / 2    N p  N  Vo = 2Vs  s  D N   p 7-19)

 Vs   N s    1  24 Vo =   =       = 17.1 V .  2(1 − D)   N p   2(1 − 0.65)   2  I Lx =

Vo2 17.12 = = 1.22 A. Vs R 24(10)

N  VSw,max = 2Vs  p  = 2(24)(2) = 96 V .  Ns  7-20)

N p / N s 50  Vs   N s  V Vo =  =   o =   Vs 2(1 − D) 30  2(1 − D)   N p  Let D = 0.7 ( D  0.5) N s  50  =   (2)(1 − 0.7) = 1.0 N p  30  Vo2 502 Vo2 502 R= = 62.5; I Lx = = = 1.33 A. Po 40 Vs R 30(62.5)

7-21)

Ps = Po Vs I Lx = Vo I o  N   2(1 − D)T  I x = I o = I Lx  p    T   Ns    N   Vs I Lx = Vo  I Lx  p  2(1 − D )    Nd    N   1 Vo = Vs  S    N p   2(1 − D)   

7-22) The simulation is run using a Transient Analysis with a restricted time of 3 to 3.02 ms, representing two periods of steady-state operation. The steady-state output voltage has an average value of approximately 30 V and peak-to-peak ripple of approximately 600 mV, ignoring the negative spike. The average transformer primary and secondary currents are 912 mA and 83.5 mA, respectively. The output voltage is lower than the predicted value of 36 V because of the nonideal switch and diode, mostly from the switch. The output voltage ripple is 2%, matching the predicted value. The converter would operate much better with a switch that has a lower on resistance.

30.5V Output voltage (3.0041m,30.295)

(3.0200m,30.057)

30.0V

(3.0141m,29.697)

29.5V 3.000ms 3.004ms 3.008ms V(Output) AVG(V(Output))

3.012ms

3.016ms

3.020ms

Time

4.0A Primary current

(3.0200m,912.072m)

2.0A

0A I(TX1:1)

AVG(I(TX1:1))

200mA

0A

(3.0200m,83.489m)

Secondary current SEL>> -200mA 3.000ms I(TX1:3)

3.005ms AVG(I(TX1:3))

3.010ms Time

3.015ms

3.020ms

7-23) Using a nonideal switch and diode produces lower values for the currents. For iLx, the maximum, minimum, and average values in PSpice are 1.446 A, 0.900 A, and 1.17 A, compared to 1.56 A, 1.01 A, and 1.28 A, respectively. However, the peak-to-peak variation in iLx in PSpice matches that of the ideal circuit (0.55 A).

2.0A (3.6286m,900.720m) (3.6114m,1.4463)

SEL>> 0A I(Lx) 2.0A

(3.6114m,1.5068)

0A

(3.6287m,603.330m)

-2.0A I(L1) 2.0A (3.6114m,1.4463)

0A

(3.6287m,903.647m)

-2.0A -I(L2) 1.0A 0A

(3.6115m,539.288m)

-1.0A 3.60ms I(L3)

3.61ms

3.62ms

3.63ms Time

3.64ms

3.65ms

3.66ms

7-24) Design for θco= -210° and a gain of 20 dB for a cross over frequency of 12 kHz.

From Eq. 7 − 85, K = 3.73 : co = 2 12000 = 75400 radis

co

75400 = 20200 rad / s K 3.73 R For gain = 20 dB, 2 = 10 R1

z =

=

Let R1 = 1 k , R2 = 10 k  C1 =

1 R2z

= 4.95 nF ; C2 =

1 = 355 pF  p R2

7-25)

 comp = 45 − (−105) = 150    150  K = tan  comp  = tan   = 3.73  2   2  Gain = 15 dB + 9.5 dB = 24.5 dB G = 1024.5/20 = 16.8 R2 = 16.8 R1 Let R1 = 1 k  and R2 = 16.8 k  C1 =

K 3.73 = = 7.07 nF 2 f co R2 2 (5000)(16,800)

C2 =

1 1 = = 508 pF K 2 f co R2 3.73(2 5000)(16,800)

7-26) Using Vs = 6 V as in Example 7-8, the frequency response of the open-loop system shows that the crossover frequency is approximately 16.8 kHz. The phase angle at the crossover frequency is 17°, which is much less than the desired value of at least 45°. Therefore, the system does not have the desired degree of stability.

120

Phase 80

40

Magnitude (dB) (16.814K,16.866) (16.814K,48.439m)

0

-40 10Hz 100Hz DB(V(error)) P(V(error))

1.0KHz

10KHz

Frequency

7-27) a) A frequency response of the circuit yields Vo ≈ -2.5 dB and θ ≈ 103° at 10 kHz.

100KHz

40 Magnitude

-0

(10.000K,-2.5181)

Phase

-40

-80

(10.000K,-102.646)

-120 1.0Hz 10Hz DB(V(Output))

100Hz 1.0KHz P(V(Output)) Frequency

10KHz

100KHz

b) With Vp = 3, the gain of the PWM function is 20log10(1/3) = -9.54 dB. The required gain of the compensated error amplifier is then 2.5 + 9.54 = 12.06 dB, corresponding to a gain magnitude of 4.0. The phase angle of the compensated error amplifier at the crossover frequency to give a phase margin of 45° is

comp = 45 − (−103) = 148

From 7-75, 7-85, 7-86, and 7-87,

Let R1 =1kΩ, then R 2 = 4 kΩ.   comp   148  K = tan   = tan   = 3.49 2  2    C1 =

K 3.49 = = 13.8 nF 2 f co R2 2 (10,000)(4000)

C2 =

1 1 = = 1.14 nF K 2 f co R2 (3.49)2 (10,000)(4000)

c) Referring to Example 7-9, the PSpice simulation results are shown indicating a stable control system. The switching frequency was not specified, and 50 kHz was used here.

Use initial conditions for the capacitor voltage at 8 V and the inductor current at 2 A.

10

Output voltage

5

Inductor current step change

0 0s

0.5ms V(OUTPUT) I(L1)

1.0ms

1.5ms Time

2.0ms

2.5ms

3.0ms

7-28) a) The gain at 8 kHz is approximately -2.44 dB, and the phase angle is -100°.

40 Magnitude

-0

(8.0358K,-2.4358)

Phase

-40

(8.0358K,-100.156)

-80

-120 1.0Hz 10Hz DB(V(Output))

100Hz 1.0KHz P(V(Output)) Frequency

10KHz

100KHz

b) This design is for fco = 8 kHz. With Vp = 3, the gain of the PWM function is 20log10(1/3) = 9.54 dB. The required gain of the compensated error amplifier is then 2.44 + 9.54 = 11.98 dB, corresponding to a gain magnitude of 3.97. The phase angle of the compensated error amplifier at the crossover frequency to give a phase margin of 45° is

comp = 45 − (−100) = 145

From 7-75, 7-85, 7-86, and 7-87,

Let R1 =1kΩ, then R 2 = 3.97 kΩ.   comp   145  K = tan   = tan   = 3.17 2  2    C1 =

K 3.17 = = 15.9 nF 2 f co R2 2 (8000)(3970)

C2 =

1 1 = = 1.58 nF K 2 f co R2 (3.17)2 (8000)(3970)

c) Referring to Example 7-9, the PSpice simulation results are shown indicating a stable control system. The switching frequency was not specified, and 50 kHz was used here.

Use initial conditions for the capacitor voltage at 8 V and the inductor current at 1.6 A.

10

Output voltage

5

Inductor current

step change 0 0s I(L1)

0.5ms V(Output)

1.0ms

1.5ms

2.0ms

2.5ms

3.0ms

Time

If designing for fco = 10 kHz, the gain of the converter is -4.38 dB, and θco = -98°. R1 = 1k, R2 = 4.97k, C1 = 9.58 nF, and C2 = 1.07 nF.

7-29) 2

  + 90     195 + 90   K =  tan  comp   =  tan    = 8.68 4 4       f co = 15 kHz  co = 2 f co = 94, 248 rad / s.

20 log10 (G ) = 15 dB G = 1015/20 = 5.62 Using Eq. (7 - 112) and letting R1 = 1 k  R2 =

GR1 = 1.91 k  K

C1 =

K = 16.4 nF co R2

C2 =

1 = 1.89 nF co R2 K

C3 =

K = 31.3 nF co R1

R3 =

1 = 115  co KC3

2

7-30)

 1  1 Ramp function gain = 20 log10   = 20 log10   = −9.54 dB V  3  p Total gain = - 8 dB - 9.54 dB = -17.54 dB G = 10

17.54 20

= 7.54

Using 45 for the phase margin,  comp = 45 − (−140) = 185 2

  185 + 90   K =  tan    = 6.61 4    co = 2 f co = 2 (15, 000) = 94, 248 rad / s Let R1 = 1 k 

7-31)

R2 =

G R1

C1 =

K = 9.31 nF co R2

C2 =

1 = 1.41 nF co R2 K

C3 =

K = 27.3 nF co R1

R3 =

1 = 151  co KC3

K

= 2.93 k 

Using Vs = 20 V , gain at 10 kHz = −9.16 dB  co = −133 1 gain of pwm = 20 log10   = −9.54 dB 3 Total gain = − 9.16 − 9.54 = −18.7 dB 18.7

G = 10 20 = 8.61

 comp = 45 − (−133) = 178 Using equations 7 - 104 and 7 - 112, K = 5.55 Let R1 = 1 k , R2 = 5.55 k  C1 = 10.3 nF C2 = 1.85 nF C3 = 37.5 nF R3 = 180  100

Magnitude 0

(10.000K,-9.1569)

Phase

(10.000K,-133.095) -100

-200 1.0Hz 10Hz DB(V(Output))

100Hz 1.0KHz P(V(Output)) Frequency

10KHz

100KHz