HartChapter9solutions.doc

CHAPTER 9 SOLUTIONS 3/13/10

9-1)

0 

1  1.83(10)6 rad / s LC

L  1.83 C I L t1  0 r  0.5  s V3 Z0 

  I 0 Z 0     2.35  s  Vs    C V (1  cos 0 (t2  t1 )) t3  t 2  r s  0.845  s I0 t2  t1 



1 0

1  sin 

 t1   (t2  t1 )  (t3  t2 )  5.17 V .  2 

V0  Vs f s  9-2)

0 

1  1.69(10)6 rad / s LC

L  0.845 C I L t1  0 r  0.083  s Vs Z0 

t2  t1 

1 0



 I 0 Z 0     1.94  s  Vs  

1  sin 

 C V (1  cos 0 (t2  t1 )) t3  t 2  r s  8.36  s I0 f max  1/ t3  1/ (0.083  1.94  8.36)  s  96.3 kHz  t1    t2  t1    t3  t 2   17.9 V  2  for Vo  5V ,

Vo  Vs f s 

fs 

V0  t  Vs  1  (t2  t1 )  (t3  t2 )  2 

 26.9 kHz

9-3) a ) 0 

1  108 LC

L 1 C I L t1  0 r  1.39 ns Vs Z0 

  I 0 Z 0     32.8 ns  Vs    C V (1  cos 0 (t2  t1 )) t3  t 2  r s  143.3 ns I0 t2  t1 

1 0



1  sin 

 t1   (t2  t1 )  (t3  t 2 )  4.77 V .  2  V 36 b) I L , peak  I o  s  5   41 A. Z0 1

Vo  Vs f s 

VC , peak  2Vs  72 V .  12    1.89 MHz  4.77

c ) f s  750 kHz 

9-4) I L , peak  I o 

Vs 50  3  9 A.  Z 0  8.33  Z0 Z0

0 

1 1  C L02 LC

Z0 

L L   L0 C 1/ L02

L

Z 0 8.33   0.833  H 0 107

C

1  12 nF L02

t1 

I 0 Lr  50 ns Vs

 1  1  I 0 Z 0  sin      366 ns 0   Vs   C V (1  cos 0 (t2  t1 )) t3  t 2  r s  373 ns I0 t2  t1 

fs 

Vo  t  Vs  1  (t2  t1 )  (t3  t2 )  2 

 945 kHz

9-5) For I o  0.5 A. f 0  503 kHz t1  0.05  s t2  t1  1.04  s t3  t 2  3.97  s fs 

Vo  t  Vs  1  (t2  t1 )  (t3  t2 )  2 

 99.2 kHz

For I o  3 a. t1  0.30  s t2  t1  1.388  s t3  t 2  0.439  s fs 

Vo  t  Vs  1  (t2  t1 )  (t3  t2 )  2 

 253 kHz

 99.2 kHz  f s  253 kHz 9-6) V 15 RL 5   2; o   0.5 Z 0 2.5 Vs 30 From Fig . 9  1g ,

0 

fs  2 (200)(10)3  0.27  0  s   4.65(10) 6 rad / s f0 0.27 0.27

1 1  LC  2 0 LC

L  L  Z 02C C 1 1 1 LC  Z 02C 2  2  C    0.086  F 0 Z 0 0 (2.5)(4.65)(10)6

Z0 

L

1  0.538  H 02C

9-7) a) The circuit is shown with diode D2 added to make the switch unidirectional.

20 V(C) Average voltage (6.6839u,5.1605) I(L)

10

0

-10 0s

2.0us V(D1:2) I(L1)

4.0us V(S1:1)

6.0us 8.0us AVG(V(D1:2)) Time

10.0us

12.0us

(a) The average output (capacitor) voltage is 5.16 V, agreeing with the 5.17 V computed analytically. (b) Peak capacitor voltage= 20 V.; (c) Inductor currents: peak = 10.5 A.; average = 2.59 A.; rms = 4.54 A.

9-8)  t   Vo  Vs  1  f s  t3  1 2    1  Vo / Vs 1  15 / 20 fs    182 kHz t3  t1 / 2 1.46(10) 6  0.188(10) 6 9-9)

0 

1 1   (10)8 rad / s 8 (10) Lr Cr

Z0 

Lr  10  Cr

t1 

Vs Cr  2 ns Io

 Vs  1  1   sin      t1  35.4 ns 0   I o Z 0  LI t3  r o [1  cos 0 (t2  t1 )]  t2  134.4  s 0 t2 

 t   Vo  Vs  1  f s  t3  1  14.7 V . 2    I L, peak  I o  10 A. VC , peak  Vs  I o

L  114.7 V . C

9-10)

0 

1  (10)7 rad / s Lr Cr

Z0 

Lr  10  Cr

t1 

Vs Cr  16.7 ns Io

t2 

1 0

t3 

Lr I o [1  cos 0 (t2  t1 )]  t 2  1.54  s 0



  Vs      t1  348 ns  I o Z 0 



1  sin 

 t   Vo  Vs  1  f s  t3  1  1.17 V . 2    1  Vo / Vs For Vo  2.5, f s   326 kHz. t3  t1 / 2

9-11)

0 

1  1.414(10) 7 rad / s Lr Cr

Z0 

Lr  7.07  Cr

t1 

Vs Cr  12 ns Io

t2 

1 0

t3 

Lr I o [1  cos 0 (t2  t1 )]  t2  1.07  s 0



  Vs      t1  246 ns  I o Z 0 



1  sin 

 t   Vo  Vs  1  f s  t3  1  5.6 V . 2    For I o  8 A., t1  15 ns, t2  252 ns, t3  911 ns fs 

1  Vo / Vs  394.1 kHz. t3  t1 / 2

For I o  15 A., t1  8 ns, t2  238 ns, t3  1.48  s f s  645.4 kHz 394.1 kHz  f s  645.4 kHz

9-12) VC , peak  Vs  I o Z 0  Z 0  Z0 

Lr  Lr  Z 02C Cr

0 

1 1  Cr  Lr02 Lr Cr

VC , peak  Vs Io



40  15  6.25  4



Z 1  6.25  Lr  0   3.91  H 2 0 1.6(10)6  Lr0

Lr  Z 02Cr  Z 02  C

1  0.1  F . Lr02

t1 

Vs Cr  0.375  s Io

t2 

1 0

t3 

Lr I o [1  cos 0 (t2  t1 )]  t2  4.62  s Vs

fs 

1  Vo / Vs 1  5 /15   150 kHz. t3  t1 / 2 (4.62  0.375 / 2)(10) 6

Io 

Vo 15   3 A. RL 5



  Vs      t1  2.74  s  I o Z 0 

1  sin 



9-13)

V 15 RL 5   0.2; o   0.5 Z 0 25 Vs 30 From Fig . 9  2 g ,

fs f 100 kHz  0.37  f 0  s   270 kHz. f0 0.37 0.37

0  2 f 0  1.70(10)6 rad / s  Z0  Cr 

1 Lr Cr

Z Lr 25  Lr  0   14.7  H Cr 0 1.70(10)6 Lr  23.5 nF . Z 02

9-14) A suitable circuit is shown. The values of the output filter components L1 and C2 are not critical. The load resistor is chosen to give 10 A. The switch must be open for an interval between t2 and t3; 50 ns is chosen.

400u Energy from source per period (149.500u,-72.676u) 0

SEL>> -400u S(W(V1)) (149.470u,120.125) 100V Capacitor 50V

Output 149.088u,14.578)

0V 149.0us 149.2us 149.4us V(INPUT,D3:2) V(R1:1)

149.6us Time

149.8us

150.0us

Results from Probe for steady-state output: a) Vo ≈ 14.6 V., b) VC,peak= 120 V., c) Integrate instantaneous power, giving 72.7 μJ per period (supplied). 9-15) V1  80 2  113 V .  THD 

4Vdc  Vdc  88.9 V . 

V3  V3  (0.05)(113)  5.66 V . V1

V1 113   37.7 V (input to filter ) 3 3 Vo,3 5.66 1    Q  2.47 2 Vi ,3 37.7 0  2  30 1 Q     0 30 1 C  13.4  F . Q0 R For a square wave, V3 

L

1  11.8 mH . 02C

VC , peak 

V1 V 113  280 V .; I L, peak  1   9.43 A. 0 RC R 12

9-16) V1  100 2  141 V .  THD 

4Vdc  Vdc  111 V . 

V3  V3  (0.1)(141)  14.1 V . V1

For a square wave, V3  Vo,3 Vi ,3



14.1  47

V1 141   47 V (input to filter ) 3 3

1  30 0  1  Q2     0 30

C

1  13.9  F . Q0 R

L

1  1.27 mH . 02C

2

 Q  1.19

200

100

(10.416m,3.6212) 0

-100

-200 10.0ms V(V1:+)

10.4ms V(OUT)

10.8ms I(R1)

11.2ms

11.6ms

12.0ms

Time

The output file shows that the THD is 10.7%. Increase Q by increasing L, and adjust C accordingly. L=1.4 mH and C=12.6 µF gives THD=9.8%. Switching takes place when load (and switch) current is approximately 3.6 A.

9-17) P

V1,2rma R

 V1,rms  PR  500(15)  86.6 V .

V1  86.6 2  122.5 V .  THD 

4Vdc  Vdc  96.2 V . 

V3  V3  (0.1)(122.5)  12.25 V . V1

V1 122.5   40.8 V (input to filter ) 3 3 Vo,3 12.25 1    Q  1.19 2 Vi ,3 40.8 2 30 0  1 Q     0 30 1 C  17.8  F . Q0 R For a square wave, V3 

L

1  5.68 mH . 02C

VC , peak 

V1 122.5   146 V . 0 RC (2 500)(15)(17.8)(10) 6

I L, peak 

V1 122.5   8.17 A. R 15

200

v(out) V(cap)

100 V(in)

(10.434m,8.1206) I(L)

0

-100

-200 10.0ms V(OUT)

10.4ms V(IN)

10.8ms V(L1:2,C1:2)

11.2ms I(L1) Time

11.6ms

From the output file, THD = 10.8%. From Probe: VC,peak=149 V.; IL,peak=8.12 A. 9-18) 2  839 kHz  f3  f 0 Lr Cr

f 0  20 

8 RL  8.11  2 s  2 f s  5.65(10)6 Re 

X L  s Lr  33.9  XC 

1  29.5  s Cr 





Vo 

Vs  2 

 1    

     1 1  10    4.38 V .  2 2   2  33.9  29.5  X L  X C  1       8.11    Re    

12.0ms

9-19) 2  1.33 MHz  f s  f 0 Lr Cr

f 0  20 

8 RL  4.05  2 s  2 f s  9.42(10)6 rad / s. Rs 

X L  s Lr  11.3  XC 

1  8.84  s Cr 





V  Vo  s  2

   10.25 V . 2  X L  XC   Rs    1

 1    



9-20) Vo 6   0.3 Vs 18

s  1.2  Q  3 from Fig . 9  5d 0   2 (800, 000) 0  s  s   4.19(10)6 rad / s 1.2 1.2 1.2 QRL 3(5) Lr    3.58  H 0 4.19(10)6 1 Cr  2  1.59(10) 8  15.9 nF . 0 Lr Let

A PSpice simulation using the circuit of Fig. 9-6(a) gives an output voltage of approximately 5.1 V.

9-21) Vo 18   0.36 Vs 50 Let

s  1.2  Q  2.1 from Fig . 9  5d 0

s s 2 (10)6    5.23(10)6 rad / s 1.2 1.2 1.2 QRL 2.1(9) Lr    3.61  H 0 5.23(10)6 1 Cr  2  10.1 nF . 0 Lr

0 

9-22)

Vo 15   0.375 Vs 40 Let

s  1.2  Q  1.9 from Fig . 9  5d 0

s 2 (800, 000)   4.19(10)6 rad / s 1.2 1.2 QRL 1.9(5) Lr    2.27  H 0 4.19(10)6 1 Cr  2  25.1 nF . 0 Lr

0 

A PSpice simulation using the circuit of Fig. 9-6a shows a steady-state output voltage of approximately 14.4 V, slightly less than the target value of 15 V. Note that the current in Lr and Cr is not quite sinusoidal.

9-23) Vo 55   0.367 Vs 150 Let

s  1.2  Q  2 from Fig. 9  5d 0

If f s  1 MHz , 0 

s 2  1, 000, 000    5.23(10)6 rad / s 1.2 1.2

Lr 

QRL 2(20)   7.64  H 0 5.23(10)6

Cr 

1  4.77 nF .  Lr 2 0

A PSpice simulation using the circuit of Fig. 9-6a shows a steady-state output voltage of approximately 53 V, slightly less than the target value of 55 V. Note that the current in Lr and Cr is not quite sinusoidal.

9-24) RL 2 10 2   12.3  8 8 1 0   2.53(10)6 rad / s Lr Cr Re 

0  403 kHz.  f s  f 0 2 X L  s Lr  4.08  1 XC   2.65  s Cr 4Vs Vo   9.60 V . 2 2     X XL  2  1  L   X C  R e  f0 

9-25) RL 2 15 2   18.5  8 8 1 0   5.66(10)6 rad / s Lr Cr Re 

0  901 kHz.  2 X L  s Lr  7.54  1 XC   6.12  s Cr 4Vs Vo  2 X L  2    1    X C   f0 

f s  f0

X L  R e

2

 25.9 

9-26) Vo 20   1.67 Vs 12 Let

f s s  2 500000   1.05  0  s   2.99(10)6 rad / s f 0 0 1.05 1.05

From Fig . 9  10c, Q  3.8 R 15 Lr  L   1.32  H 0Q 2.99(10)6 (3.8) Cr 

1  84.7 nF  Lr 2 0

9-27)

Vo 36   0.8 Vs 45 Let

f s s  2 900000   1.1  0  s   5.14(10)6 rad / s f 0 0 1.1 1.1

From Fig . 9  10c, Q  1.9 R 20 Lr  L   2.05  H 0Q 5.14(10)6 (1.9) Cr 

1  18.5 nF  Lr 2 0

9-28) Vo 60   1.2 Let f s  500 kHz Vs 50 Let

f s s  2 500000 6   1.1  0  s   2.86  10  rad / s f 0 0 1.1 1.1

From Fig . 9  10c, Q  3.4 R 25 Lr  L   2.57  H 0Q 2.86(10)6 (3.4) Cr 

1  47.6 nF  Lr 2 0

9-29) Vs  100 V .; f s  500 kHz; RL  10 ; L  12  H ; Cs  C p  12  F

2 2 Re  RL  (10)  12.3  8 8 s  2 f s  2 500000  3.14(10)6 4Vs Vo  2 Cp   s L 1 2  2   1  s LC p    Cs Re s ReC     s

2

 37.7 V .

9-30) Sample solution Vo 5   0.417 Vs 12 Let

f s s   1.2  Q  2.7 from Fig . 9  11c f 0 0

s 2 800000 6   4.19  10  rad / s 1.2 1.2 QRL 2.7(2) L   1.29  H . 0 4.19(10)6 1 Cs  C p  2  44.2 nF . 0 L Check with Eq. 9  74 : Vo  4.97 V .

0 

9-31) Sample solution Vo V 5 5   0.25; RL  o   5  Vs 20 Io 1 Let

f s s   1.2  Q  5 from Fig . 9  11c f 0 0

s 2 750000   3.93(10)6 rad / s 1.2 1.2 QRL 5(5) L   6.37  H . 0 3.93(10)6 1 Cs  C p  2  10.2 nF . 0 L Check with Eq. 9  74 : Vo  5.08 V .

0 

9-32) Sample solution Vo 10 V 10   0.4; RL  o   10  ; Let f s  100 kHz Vs 25 Io 1 Let

f s s   1.15  Q  3.3 from Fig. 9  11c f 0 0

s 2 100000   546 krad / s 1.15 1.15 QRL 3.3(10) L   60.4  H . 0 5.46(10)5 1 Cs  C p  2  55.5 nF . 0 L Check with Eq. 9  74 : Vo  10.02 V .

0 

Using a circuit based on Fig. 9-11a but with a square-wave source implemented with Vpulse (see Fig. 9-6a), the result is approximately 9.4 V.

9-33) (a) A PSpice simulation using the circuit shown reveals that the capacitor voltage returns to zero at 15.32 μs and the switch must remain closed for 5.58 μs for the inductor current to return to 12 A. Initial conditions for the inductor (12 A) and for the capacitor (0 V) must be applied. Ideal models for the switch and diode are used. b) Using the expression S(W(V1)) for energy (S is integration), 15.7 mJ are supplied by the voltage source in one period. c) Average power is 754 W, obtained by entering AVG(W(V1)). d) Average resistor power is 104 W. e) With R = 0, the capacitor voltage returns to zero at 15.44 μs and the switch must remain closed for 5.45 μs. The source power and energy are not changed significantly.

20

(20.901u,12.000) I(L)

15.318u,16.390m) 0

-20 14.71us 16.00us V(switch) V(cap)

18.00us

20.00us

21.53us

I(L1) Time

200 V(capacitor) 0

SEL>> -200 0s

4us V(SWITCH) V(CAP)

8us I(L1)

12us Time

16us

20us

24us

9-34)

o  

1 1   2(10)5 rad / s 6 6 Lr Cr 250(10) 0.1(10)

R 1   2000 2 Lr 2(250)(10) 6

  o2   2  0 vC (t )  Vs  e  t [ Vs cos ot  o Lr ( I1  I o ) sin ot ]  75  e 2000 t [ 75cos(2(10)5 t )  2(10)5 250(10) 6 (7  5)sin(2(10)5 t )]  75  e 2000 t [ 75cos(2(10)5 t )  100sin(2(10)5 t )] vC (t x )  0  t x  22.3  s   V iL  t   I o  e  t  ( I1  I o ) cos ot  s sin ot o Lr     75  5  e 2000t  (7  5(cos(2(10)5 t )  sin(2(10)5 t ) 5 6 2(10) 250(10)   2000 t 5 5  2 cos(2(10) t )  1.5sin(2(10) t )  5e iL (t x )  iL (22.3  s )  3.14 A. t 

iL Lr (7  3.14)250(10) 6   12.9  s Vs 75

o

1 1   3.65(10)5 rad / s 6 6 Lr Cr 150(10) 0.05(10)



R 0.5   1667 2 Lr 2(150)(10) 6

9-35)

  o2   2  0