CHAPTER 1 SOLUTIONS
(1-1)
(1-2)
25V
20V
15V
10V
5V
0V
-5V
0s
2us
4us
6us
8us
10us
12us
14us
16us
10us
12us
14us
16us
V(D1:2)
Time
25V
(1.4333u,23.800)
20V
15V
10V
5V
(4.0833u,-851.690m)
0V
-5V
0s
2us
4us
6us
8us
V(S1:4)
Time
In part (b), the voltage across the current source is reduced from 24 V by the switch resistance and diode
voltage drop.
(1-3)
40V
96.46n,23.94)
20V
(3.150u,-1.052)
0V
(3.150u,-1.052)
-20V
0s
5us
10us
V(V2:-)
Time
15us
(1-4)
25V
20V
(800.000n,23.924)
15V
10V
5V
(3.8333u,-1.0517)
0V
-5V
0s
2us
4us
6us
8us
V(V2:-)
Time
10us
12us
14us
16us
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
CHAPTER 2 SOLUTIONS
2/21/10
2-1) Square waves and triangular waves for voltage and current are two examples.
_____________________________________________________________________________________
2-2)
a) p  t   v  t  i  t  
v2  t 

R
[170sin  377t  ]2
10
 2890sin 2 377t W .
b) peak power = 2890 W.
c) P = 2890/2 = 1445 W.
_____________________________________________________________________________________
2-3)
v(t) = 5sin2πt V.
a) 4sin2πt A.; p(t) = v(t)i(t) = 20 sin22πt W.; P = 10 W.
b) 3sin4πt A.; p(t) = 15sin(2πt)sin(4πt) W.; P = 0
_____________________________________________________________________________________
2-4)
a)
0  t  50 ms
50 ms  t  70 ms
70 ms  t  100 ms
 0

p  t   v  t  i  t    40
 0

b)
70 ms
T
1
1
P   v  t  i  t  dt 
40 dt  8.0 W .
T 0
100 ms 50ms
c)
T
70 ms
0
50 ms
W   p  t  dt 

40 dt  800 mJ .; or W  PT   8W   100 ms   800 mJ .
_____________________________________________________________________________________
2-5)
a)
 70 W .
 50 W .

p t  v t i t  
 40 W .
 0
0  t  6 ms
6 ms  t  10 ms
10 ms  t  14 ms
14 ms  t  20 ms
b)
P
c)
1
T
T
 p  t  dt 
0
1 

20 ms 
6 ms

0
10 ms
70 dt 

40 dt  19 W .

10 ms
14 ms
  50  dt  
6 ms
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart

T
6 ms
W   p  t  dt  
0

10 ms
70 dt 

 50  dt 

40
dt
  0.38 J .;

10 ms

14 ms
 0
6 ms
or W  PT   19   20 ms   380 mJ .
_____________________________________________________________________________________
2-6)
P  Vdc I avg
a ) I avg  2 A., P   12   2   24 W .
b) I avg  3.1 A., P   12   3.1  37.2 W .
_____________________________________________________________________________________
2-7)
a)
vR  t   i  t  R  25sin 377t V .
p  t   v  t  i  t    25sin 377t   1.0sin 377t   25sin 2 377t  12.5  1  cos 754t  W .
T
1
PR   p  t  dt  12.5 W .
T 0
b)
vL  t   L
di  t 
dt
 10  10 
3
 377   1.0  cos 377t  3.77 cos 377t V .
pL  t   v  t  i  t    3.77 cos 377t   1.0sin 377t  
 3.77   1.0  sin 754t  1.89 sin 754t W .
2
T
PL 
1
p  t  dt  0
T 0
c)
p  t   v  t  i  t    12   1.0sin 377t   12sin 377t W .
Pdc 
1
T
T
 p  t  dt  0
0
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-8)
Resistor:
v  t   i  t  R  8  24sin 2 60t V .
p  t   v  t  i  t    8  24sin 2 60t   2  6sin 2 60t 
 16  96sin 2 60t  144sin 2 2 60t W .
1
1 
p  t  dt 


T 0
1/ 60 
T
P
1/60

1/60
16 dt 
0

96sin 2 60t dt 
0
1/60
 144sin
0
2

2 60t

 16  72  88 W .
Inductor: PL  0.
dc source: Pdc  I avgVdc   2   6   12 W .
_____________________________________________________________________________________
2-9)
a) With the heater on,
P
Vm I m
 1500   2   12.5 2
 1500 W .  I m 
2
120 2



p  t   Vm I m sin 2t  120 2 12.5 2 sin 2t  3000sin 2t
max  p  t    3000 W .
b) P = 1500(5/12) = 625 W.
c) W = PT = (625 W)(12 s) = 7500 J. (or 1500(5) = 7500 W.)
_____________________________________________________________________________________
2-10)
iL  t  
t
1
1
vL  t  dt 
90 d   900t

L
0.1 0
iL  4 ms    900   4   10 
3
0  t  4 ms.
 3.6 A.
a)
W
1 2 1
2
Li   0.1  3.6   0.648 J .
2
2
b) All stored energy is absorbed by R: WR = 0.648 J.
c)
PR 
WR 0.648

 16.2 W .
T
40 ms
PS  PR  16.2 W .
d) No change in power supplied by the source: 16.2 W.
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-11)
a)
W
1 2
Li , or i 
2
2  1.2 
2W

 15.49 A.
L
0.010
t
t
1
1
i  t    v    d 
14 d   1400t A.
L0
0.010 0
15.49  1400ton
ton  11.1 ms
b) Energy stored in L must be transferred to the resistor in (20 - 11.1) = 8.9 ms. Allowing five
time constants,

L 8.9 ms

 1.7 ms.;
R
5
R
L
10 mH

 5.62 
1.7 ms 1.7 ms
_____________________________________________________________________________________
2-12)
a) i(t) = 1800t for 0 < t < 4 ms
i(4 ms) = 7.2 A.; WLpeak = 1.296 J.
b)
10A
5A
Inductor current
SEL>>
0A
I(L1)
10A
Source current
0A
-10A
-I(Vcc)
1.0KW
Ind. inst. power
0W
-1.0KW
W(L1)
1.0KW
Source inst. power (supplied)
0W
-1.0KW
0s
20ms
40ms
60ms
80ms
100ms
-W(Vcc)
Time
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-13)
a) The zener diode breaks down when the transistor turns off to maintain inductor current.
b) Switch closed: 0 < t < 20 ms.
vL  12 V .  L
diL  t 
dt
diL vL
12
 
 160 A/s
dt
L 0.075
at t  20 ms, iL   160   0.02   3.2 A.
Switch open, zener on:
vL  12  20  8 V .
diL vL
8
 
 106.7 A/s
dt
L 0.075
t to return to zero :
i
3.2
t 

 30 ms
106.7 106.7
Therefore, inductor current returns to zero at 20 + 30 = 50 ms.
iL = 0 for 50 ms < t < 70 ms.
c)
40mW
Inductor inst. power
0W
-40mW
W(L1)
80mW
Zener inst. power
40mW
SEL>>
0W
0s
10ms
20ms
30ms
40ms
W(D1)
Time
50ms
60ms
70ms
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
d)
PL  0.
T
1
1  1
PZ   pZ  t  dt 
 0.03  64  13.73 W .

T 0
0.07  2

_____________________________________________________________________________________
2-14)
a) The zener diode breaks down when the transistor turns off to maintain inductor current.
b) Switch closed: 0 < t < 15 ms.
vL  20 V .  L
diL t 
dt
diL vL
20
 
 400 A/s
dt
L 0.050
at t  15 ms, iL   400   0.015   6.0 A.
Switch open, zener on:
vL  20  30  10 V.
diL vL
10


 200 A/s
dt
L 0.050
t to return to zero :
i
6.0
t 

 30 ms
200 200
Therefore, inductor current returns to zero at 15 + 30 = 45 ms.
iL = 0 for 45 ms < t < 75 ms.
c)
200W
Inductor inst. power
0W
-200W
W(L1)
200W
Zener inst. power
100W
SEL>>
0W
0s
20ms
40ms
W(D1)
Time
60ms
80ms
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
d)
PL  0.
T
1
1  1
PZ   pZ  t  dt 
 0.03  180  36 W .

T 0
0.075  2

_____________________________________________________________________________________
2-15) Examples are square wave (Vrms = Vm) and a triangular wave (Vrms = Vm/√3).
_____________________________________________________________________________________
2-16)
2
2
Phase conductors: P  I R  12  0.5   72 W .

Neutral conductor: PN  I 2 R  12 3
Ptotal  3  72   216  432 W .
RN 
PN
72

2
IN
12 3


2

2
 0.5  216 W .
 0.167 
_____________________________________________________________________________________
2-17) Re: Prob. 2-4
Vrms  Vm D  10 0.7  8.37 V .
I rms  I m D  4 0.5  2.83 A.
_____________________________________________________________________________________
2-18) Re: Prob. 2-5
 14
  8.36 V .
 20
Vrms  Vm D  10 
I rms 
1
0.02
0.006

0
0.01
7 2 dt 
  5
0.006
2
0.02
dt 
4
2
dt  27.7  5.26 A.
0.01
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-19)
2
 5 
 
 2 
Vrms  22  
3

2
2
2
 4.58 V .
2
 2   1.1

I rms  1.5  
 
  2.2 A.
 2 
2

V I
P  V0 I 0   m m cos   n  n 
2
n 1
2
 5 
2

  3  1.1

 cos  20  

 cos  115   7.0 W .
2
 2 
2

 2
Notethat  cos(4 60t  45) is cos  4 60t  135 
  2.0   1.5   
_____________________________________________________________________________________
2-20)
dc : V0  3  100   300 V .
1  2 60 : Y1  1/R  jC  0.01  j 0.0189
V1 
I1
4 0

 187  62.1
Y1  0.01  j 0.0189 
2  4 60 : Y2  1/R  jC  0.01  j 0.0377
V2 
I2
60

 153  75.1
Y2  0.01  j 0.0377 

Vm I m
cos   n  n 
2
n 1
P  V0 I 0  
 300  5  
 187   4  cos
2
 62.1  
 153  6  cos
2
 75.1 
 1500  175  118  1793 W .
_____________________________________________________________________________________
2-21)
dc Source:
 50  12
 114 W .
4 

Pdc  Vdc I avg  12 
Resistor:
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2
P  I rms
R
I rms  I 02  I1,2rms  I 2,2 rms
I 0  9.5 A.
I1 
30
 3.51 A.
4  j  4 60   0.01
I2 
10
 0.641 A.
4  j  8 60   0.01
2
I rms
 3.51  0.641

 9.5  
 

2
 2 
2
2
 9.83 A.
2
PR  I rms
R  386 W .
_____________________________________________________________________________________
2-22)
2
P  I rms
R
V0 6

 0.375 A.
R 16
5
I1 
 0.269 A.
16  j  2 60   0.025 
I0 
I2 
3
 0.0923 A.
16  j  6 60   0.025 
2
I rms
 0.269  0.0923

 0.375  
 

2 
2

2
2
 0.426 A.
2
I rms  0.623 A.; P  I rms
R   0.426   16   2.9 W .
2
_____________________________________________________________________________________
2-23)

Vm I m
cos   n  n 
2
n 1
P  V0 I 0  
n
Vn
In
Pn
∑Pn
0
20
5
100
100
1
20
5
50
150
2
10
1.25
6.25
156.25
3
6.67
0.556
1.85
158.1
4
5
0.3125
0.781
158.9
Power including terms through n = 4 is 158.9 watts.
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-24)

Vm I m
cos   n  n 
2
n 1
P  V0 I 0  
n
Vn
In
θn - ϕn°
Pn
0
50.0000
10.0
0
500.0
1
50.0000
10.0
26.6
223.6
2
25.0000
2.5
45.0
22.1
3
16.6667
1.11
56.3
5.1
4
12.5000
0.625
63.4
1.7
Through n = 4, ∑Pn = 753 W.
_____________________________________________________________________________________
2-25)

Vm I m
cos   n  n 
2
n 1
V V
50  36
I 0  0 dc 
 0.7 A
R
20
2
P0, R  I 02 R   0.7  20  9.8 W (dc component only )
P  V0 I 0  
PVdc  I 0Vdc   0.7   36   25.2 W
PL  0
Resistor Average Power
n
Vn
Zn
In
angle
Pn
0
50.00
20.00
0.7
0.00
9.8
1
127.32
25.43
5.01
0.67
250.66
2
63.66
37.24
1.71
1.00
29.22
3
42.44
51.16
0.83
1.17
6.87
4
31.83
65.94
0.48
1.26
2.33
5
25.46
81.05
0.31
1.32
0.99
PR = ∑ Pn ≈ 300 W.
_____________________________________________________________________________________
2-26)
a)
b)
c)
d)
THD = 5% → I9 = (0.05)(10) = 0.5 A.
THD = 10% → I9 = (0.10)(10) = 1 A.
THD = 20% → I9 = (0.20)(10) = 2 A.
THD = 40% → I9 = (0.40)(10) = 4 A.
_____________________________________________________________________________________
2-27)
a)
 170  10


 cos  30   0  0  736 W .
 2 
2
P   Pn  
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
b)
2
I rms
 10 
 
 
 2 
2
6 
 3
 

2
 2
2
 8.51 A.
 170
 8.51  1024 VA.
 2
P 736
pf  
 0.719
S 1024
S  Vrms I rms  
c)
DF 
I1,rms
I rms

10/ 2
 0.831
8.51
d)
2
 6 
3

 

 2 
2
THDI 
10/ 2
2
 0.67  67%
_____________________________________________________________________________________
2-28)
a)
 170  12


 cos  40   0  0  781 W .
 2 
2
P   Pn  
b)
2
I rms
 12 
 
 
 2 
2
5 
 4
 

 2
 2
2
 9.62 A.
 170
 9.62  1156 VA.
 2
P 781
pf  
 0.68
S 1156
S  Vrms I rms  
c)
DF 
I1,rms
I rms

12/ 2
 0.88
9.62
d)
2
 5 
4

 

 2 
2
THDI 
12/ 2
2
 0.53  53%
_____________________________________________________________________________________
2-29)
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
8
 5.66 A.;
2
I1,rms 
I 2,rms 
4
 2.82 A.;
2
I rms  5.662  2.822  6.32 A.; I peak  10.38 ( graphically )
a) P  V1,rms I1,rms cos  1  1    240   5.66  cos  0   1358 W .
b) pf 
P
P
1358


 0.895  89.5%
S Vrms I rms  240   6.32 
c) THDI 
d) DF 
I 2,rms
I rms
I1, rms
I rms


2.82
 0.446  44.6%
6.32
5.66
 89.6%
6.32
e) crest factor 
I peak
I rms

10.38
 1.64
6.32
_____________________________________________________________________________________
2-30)
I1,rms 
12
 8.49 A.;
2
I 2,rms 
9
 6.36 A.;
2
I rms  8.492  6.362  10.6 A.; I peak  18.3 A. ( graphically )
a) P  V1, rms I1,rms cos  1  n    240   10.6  cos  0   2036 W .
b) pf 
P
P
2036


 0.80  80%
S Vrms I rms  240   10.6 
c) THDI 
d) DF 
I 2,rms
I rms
I1, rms
I rms


6.36
 0.60  60%
10.6
8.49
 80%
10.6
e) crest factor 
I peak
I rms

18.3
 1.72
10.6
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-31)
5V: I = 0 (capacitor is an open circuit)
25cos(1000t ): Z  R  j L  j
I
1
1
 2  j1000(.001)  j
 2  j0
C
1000  1000  106
25
cos(1000t )  12.5cos(1000t ) A
2
10cos(2000t ): Z  2  j1.5 
I10 
10
 4  37 A.
2  j1.5
2
 12.5 
 
 2 
I rms  
2
4
  9.28 A
2
2
PR  I rms
R  9.282  2   172.3 W
PL  0
PC  0
Psource  172.3 W
_____________________________________________________________________________________
2-32) PSpice shows that average power is 60 W and energy is 1.2 J. Use VPULSE and IPULSE
for the sources.
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
Energy
(20.000m,1.2000)
2.0
0
S(W(I1))
400W
Avg Power (20.000m,60.000)
0W
Inst Power
-400W
W(I1)
AVG(W(I1))
I(I1)
4ms
V(V1:+)
20
0
SEL>>
-20
0s
8ms
12ms
16ms
20ms
Time
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-33) Average power for the resistor is approximately 1000 W. For the inductor and dc source,
the average power is zero (slightly different because of numerical solution).
2.0KW
Average Power
Resistor
1.0KW
Inductor
(16.670m,0.9998K)
(16.670m,-30.131u)
0W
Vdc
(16.670m,189.361u)
-1.0KW
0s
AVG(W(R1))
5ms
AVG(W(L1))
10ms
AVG(W(V1))
Time
15ms
20ms
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2.0KW
Instantaneous Power
Resistor
1.0KW
Inductor
0W
Vdc
-1.0KW
0s
W(R1)
W(L1)
5ms
W(V1)
10ms
15ms
20ms
Time
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-34)
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
Rms voltage is 8.3666 V. Rms current is 5.2631 A.
10V
Voltage
(20.000m,8.3666)
5V
0V
V(V1:+)
RMS(V(V1:+))
10A
(20.000m,5.2631)
Current
0A
SEL>>
-10A
0s
I(I1:+)
4ms
RMS(I(I1))
8ms
12ms
16ms
20ms
Time
_____________________________________________________________________________________
2-35) See Problem 2-10.
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
0W
(40.022m,-16.200)
Source Power
-100W
SEL>>
-200W
AVG(W(V1))
4.0
Inductor
2.0
(4.0000m,648.007m)
Resistor
(40.021m,647.946
0
0s
I(L1)
10ms
S(W(L1))
20ms
30ms
40ms
S(W(R1))
Time
The inductor peak energy is 649 mJ, matching the resistor absorbed energy. The source power is
-16.2 W absorbed, meaning 16.2 W supplied.
b) If the diode and switch parameters are changed, the inductor peak energy is 635 mJ, and the
resistor absorbed energy is 620 mJ. The difference is absorbed by the switch and diode.
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-36)
The inductor current reaches a maximum value of 3.4 A with the resistances in the circuit: I = 75/
(20+1+1) = 3.4 A.
4.0A
Inductor Current
2.0A
SEL>>
0A
I(L1)
4.0A
Source Current
0A
-4.0A
0s
20ms
40ms
60ms
80ms
-I(V1)
Time
Quantity
Inductor resistor average
power
Switch average power
Diode average power
Source average power
Probe Expression
AVG(W(R1))
Result
77.1 W
AVG(W(S1))
AVG(W(D1))
AVG(W(Vcc))
3.86 W each
81 mW each
-85.0 W
100ms
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
_____________________________________________________________________________________
2-37)
a) Power absorbed by the inductor is zero. Power absorbed by the Zener diode is 13.8 W.
4.0A
2.0A
Inductor Current
0A
I(L1)
4.0A
2.0A
SEL>>
0A
0s
Zener Diode Current
10ms
-I(D1)
20ms
30ms
40ms
50ms
60ms
70ms
Time
b) Power in the inductor is zero, but power in the 1.5Ω resistor is 1.76 W. Power absorbed by the
Zener diode is 6.35 W. Power absorbed by the switch is 333 mW.
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
3-38)
See Problem 3-37 for the circuit diagram.
a) Power absorbed by the Zener diode is 36.1 W. Power absorbed by the inductor is zero.
10A
5A
Inductor Current
SEL>>
0A
I(L1)
10A
5A
Zener Diode Current
0A
0s
20ms
40ms
60ms
80ms
-I(D1)
Time
b) Power in the inductor is zero, but power in the 1.5Ω resistor is 4.4 W. Power absorbed by the
Zener diode is 14.2 W. Power absorbed by the switch is 784 mW.
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-39)
40A
Total Current
20A
0A
-20A
0s
I(I1)
Quantity
Power
rms current
Apparent power S
Power factor
4ms
I(I2)
I(I3)
8ms
I(I4)
12ms
-I(V1)
Time
Probe Expression
AVG(W(V1))
RMS(I(V1))
RMS(V(I1:+))* RMS(I(V1))
AVG(W(V1)) / (RMS(V(I1:+))* RMS(I(V1)))
16ms
20ms
Result
650 W
14 A
990 VA
0.66
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-40)
DESIRED QUANTITY
ORIGINAL RESULT
NEW VALUES
Inductor Current
Energy Stored in Inductor
Average Switch Power
Average Source Power (absorbed)
Average Diode Power
AVG(W(D1))
0.464
W.
Average Inductor Power
Average Inductor Voltage
Average Resistor Power
Energy Absorbed by Resistor
Energy Absorbed by Diode
Energy Absorbed by Inductor
rms Resistor Current
max = 4.5 A.
max = 2.025 J
0.010 W.
-20.3 W.
0.464 W.
4.39 A
1.93 L
0.66 W
-19.9 W
.449 W
0
0
19.9 W.
1.99 J.
.046 J.
0
0.998 A.
0
0
18.8 W
1.88 J
.045 J
0
0.970 A
_____________________________________________________________________________________
2-41) Use the part VPULSE or IPULSE (shown). Here, the period is 100 ms, and the rise times
chosen are 20 ms, 50 ms, and 80 ms. The fall times are the period minus the rise times. Each rms
value is 0.57735, which is identical to 1/√3.
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
1.0A
(100.000m,577.350m)
0A
-1.0A
0s
20ms
-I(R1)
40ms
RMS(I(R1))
60ms
80ms
100ms
Time
_____________________________________________________________________________________
CHAPTER 3 SOLUTIONS
2/20/10
3-1)
a) I0
=
b) I rms
V0
=
Vmp
=
170/ p
R
R
Vrms Vm
=
=
R
2R
=
15
170
2(15)
=
3.60 A.
=
5.66 A.
c) P = I 2 R = 5.66 2 (15) = 480 W .
d) S
= Vrms
rms I rms
rms =
e) pf
=
P
=
S
170 �
�
(5.66) = 679 VA.
� �
�2�
480 W
=
679 VA
0.707 = 70.7%
3-2)
a) I 0
Vo
=
Vrms
N1
N2
= 12
A.; I 0
Vm
; Vm
p
754
=
=
2
240
533
b) I o �
= Io
=
V0
R
= Vop =
� V0
240 p
=
=
I 0 R = (12)(20) = 240 V.
754 V .
= 533 V .
=
0.45
N2
N1
=
12
0.45
=
26.7 A.
3-3)
a ) pf
P
=
S
=
P
Vs , rms I rms
; I rms
=
Vs ,rms
R
; VR , rms
Vm �
�
�2 �/ R
V
/R
2
� �
pf =
=
=
Vm �
Vs , rms I rms �Vm �
2
�
�2 �/ R
� �
� �
�2�
2
R , rms
b) Displa
Displacem
cement
ent pf
I1
=
V1
R
=
1 Vm
R 2
=
�0; pf
=
Vm
2
; Vs , rms =
Vm
2
1
=
2
cos(q1 - f ) = cos(0) = 1
=
cos(q1 - f1 ) DF ;
\
DF
=
1
2
3-4) Using Eq. 3-15,
Vm
a ) i(wt ) =
Z
=
q
=
wt
Z
sin(wt - q ) +
R 2 + (w L) 2
=
Vm
Z
(sin q ) e -wt /wt
12 2 + (377(0.012)) 2
= 12.8 W
�w L �
�377(0.012) �
tan -1 � �= tan -1 �
�= 0.361 rad
R
12
� �
�
�
=
wL
R
=
377(0.012)
12
=
0.377
i (wt ) = 13.2 sin(wt - 0.361) + 4.67 e -wt /0.377 : b
= 3.50
rad = 201 �
b) I avg
=
4.36 A. ( numerical integration)
c ) I rms
=
6.70 A. ( numerical integration) P = I r2ms R = (6.70) 2 (12) = 538 W.
d ) pf
=
P
S
=
538
(120)(6.70)
=
0.67
3-5) Using Eq. 3-15,
Vm
a ) i(wt ) =
Z
=
q
=
wt
Z
sin(wt - q ) +
R 2 + (w L) 2
=
Vm
Z
(sin q ) e -wt /wt
10 2 + (377(0.015)) 2
= 11.5 W
wL �
�
�377(0.015) �
tan -1 � �= tan -1 �
�= 0.515 rad
�R �
� 10
�
=
wL
R
=
377(0.015)
10
=
0.565
i (wt ) = 14.8 sin(wt - 0.515) + 7.27 e -wt /0.565 : b
b) I avg
= 5.05
c ) I rms
=
d ) pf
=
= 3.657
rad =209.5 �
A. ( numerical integration)
7.65 A. ( numerical integration) P = Ir2ms R = (7.65) 2 (10)
P
S
=
584
(120)(7.65)
=
= 584
W.
0.637 = 63.7%
3-6) Using Eq. 3-15,
Vm
a ) i(wt ) =
Z
=
q
=
wt
Z
sin(wt - q ) +
R 2 + (w L) 2
=
Vm
Z
(sin q ) e -wt /wt
152 + (377(0.08)) 2
= 33.7 W
wL �
�
�377(0.08) �
tan -1 � �= tan -1 �
�= 1.11 rad
�R �
� 15
�
=
wL
R
=
377(0.08)
15
=
2.01
i (wt ) = 10.1sin(wt - 1.11) + 9.02 e -wt /2.01 : b
= 4.35
rad = 250 �
b) I avg
=
4.87 A. ( numerical integration)
c ) I rms
=
6.84 A. ( numerical integration) P = Ir2ms R = (6.84) 2 (15) = 701 W.
d ) pf
=
P
S
=
701
(240)(6.84)
=
0.427 = 42.7%
3-7) Using an i!a" io! #o!", $  48 & 'o an a!ag! c*!nt o' 2 +.
8.0A
Current
Iavg = 2 A fr R = 48 !ms
4.0A
(1".#00m$2.00%0)
Average Current
0A
0s
5ms
I(R1)
10ms
AVG(I(L1))
Time
15ms
20ms
3-8) Using Eqs. 3-22 an 3-23,
a ) i(wt ) =
Vm
Z
sin(wt - q ) -
Vdc
R
+
Ae -wt /wt
V �a /wt
� Vm
sin(a - q ) + dc �
e
R �
� Z
A=�
Z
=
q
=
wt
a
=
R 2 + (w L)2
=
10 2 + (377(.075) 2
= 30.0 W
�w L �
�377(.075) �
tan -1 � �= tan -1 �
�= 1.23 rad
R
10
� �
�
�
=
wL
R
sin -1
=
377(0.075)
Vdc
Vm
=
10
100
240 2
=
2.83
= 0.299
rad
= 17. 1�
i (wt ) = 11.3sin(wt - 1.23) -10 + 21.2 e -wt /2.83; b
I avg
= 3.13
b) I rms
c ) pf
=
=
A. (numerical integration) Pdc
= Vdc I avg =
4.81 A. ( numerical integration) PR
P
S
=
313 + 231
(240)(4.81)
=
0.472 = 47.2%
= 3.94
=
rad
= 226 �
(100)(3.13) = 313 W.
2
I rms
R = (4.81) 2 (10) = 231 W.
3-9) Using Eqs. 3-22 an 3-23,
a ) i(wt ) =
Vm
sin(wt - q ) -
Z
Vdc
R
+
Ae -wt /wt
V �a /wt
� Vm
sin(a - q ) + dc �
e
R �
� Z
A=�
Z
=
q
=
wt
a
R 2 + (w L) 2
=
12 2 + (377(0.12) 2
= 46.8 W
wL �
�
�377(0.12) �
tan -1 � �= tan -1 �
�= 1.31 rad
�R �
� 12
�
=
wL
R
= sin
-1
377(0.12)
=
Vdc
=
Vm
12
48
120 2
=
3.77
= 0.287
rad
= 16.4 �
i (wt ) = 3.63sin(wt - 1.31) - 4.0 + 7.66 e -wt /3.77; b
I avg
= 1.124
b) I rms
c ) pf
A. ( numerical integration) Pdc
= 1.70
=
P
S
=
(120)(1.70)
=
rad
= 233 �
= Vdc Iavg = (48)(1.124) = 54.0
A. ( numerical integration) PR
54.0 + 34.5
= 4.06
=
2
I rms
R = (1.70) 2 (12) = 34.5 W.
0.435 = 43.5%
3-10) Using Eq. 3-33,
i (wt ) =
a
=
Vm
wL
(cos a
- cos wt ) +
Vdc
wL
(a - wt)
�
V �
� 48 �
sin -1 � dc �= sin -1 �
�= 0.287 rad.
V
120
2
�
�
�m �
i (wt ) = 4.68 - 4.50 cos(wt) -1.23wt A.; b
Io
=
1
2p
b
�i(wt )d (wt ) = 2.00 A.; P
a
dc
=
W.
= 4.483
I oVdc
=
rad
= 257 �
2.00(48) = 96 W.
3-32) B  75C. +"a  75 !g!!s gi!s 35  in t! c o"tag! so*c!. +n $on  0.01 'o t!
sitc an n  0.001 'o t! io! (i!a" #o!").
3-33) Ao# Eq. 3-61,
a ) i(wt)
Io
=
-w t /14.1
A., 0.873 �w t �4.24 rad
b
1
2p
b) I rms
= 5.99sin(wt -1.50) - 24.0 + 29.3 e
i (wt )d (wt ) = 1.91 A.,
�
Pdc
=
I oVdc
=
(1.91)(48) = 91.6 W.
=
2
I rms
R = (2.93) 2 2 = 17.1 W.
a
b
1
i (wt ) d (wt ) = 2.93 A.;
2p �
=
2
PR
a
3-34) B  81C
3-35)
L
di(t )
dt
di (t )
= Vm
sin wt - Vdc
1
EVm sin wt - Vdc F or
dt
L
di (wt )
1
=
EVm sin wt - Vdc F
d (w t ) w L
=
i (wt ) =
Vm
1
wL
wt
(V
�
m
sin wt - Vdc ) d (wt )
a
Io
=
(cos a - cos wt ) +
Vdc
(a - w t )
wL
wL
i (wt) = 4.34 - 7.58cos w t - 1.82wt A., 1.309 �wt �4.249
=
1
2p
b
i (wt )d (wt ) = 1.91 A.
�
a
3-39) $*n t! si#*"ation "ong !no*g 'o st!a-stat! !s*"ts. Ao# t! o! o*t*t, t!
co##*tation ang"! 'o# D1 to D2 is ao*t 16.5 !g!!s, an 'o# D2 to D1 is ao*t
14.7 !g!!s. Got! tat t! ti#! ais is cang! to ang"! in !g!!s !!.
3-40) +t Ht  I, D2 t*ns on, D1 is on !ca*s! o' t! c*!nt in < (s!! Aig. 3-17).
VL; vLS
diD1
d ( wt )
=
at wt = p
= Vm sin w t =
Vm
w Ls
+ u,
Ls
diD1
dt
= w LS
did 1
d (w t )
wt
sin(wt ) d (wt) + i
�
D1
(p )
p
iD1
=
cos(p
+ u ) = - cos u
\u =
cos -1 �
1-
0=
Vm
w Ls
�0=
� IL ( s �
�
� Vm �
E- 1 - cos(p + u)F + I L
Vm
w Ls
(-1 + cos u) + I L = -
Vm
w Ls
cos u + I L
3-41) +t Ht  B,
is (wt ) =
1
w Ls
wt
�
Vm sin(wt ) d (wt ) + 0 =
a
iD 2 ( *t ) = I L - is
iD 2 (a
-
+ u) =
I Lw Ls
Vm
=
IL -
0 = IL -
Vm
w Ls
Vm
w Ls
Vm
w Ls
Ecos a - cos wtF
Ecos a - cos wt F
Ecos a - cos(a + u)F
= - cos a + cos(a + u)
�
u = cos -1 �
cos a
�
-
IL ( s �
�- a
Vm �
3-42) + goo so"*tion is to *s! a conto""! a"'-a! !cti'i! it an in*cto in s!i!s it t!
48- so*c! an !sistanc! (Aig. 3-15). ! sitc i"" cang! t! !"a ang"! o' t!
<>$ to o*c! t! to !q*i! o! "!!"s. ! a"*!s o' t! !"a ang"! !!n on
t! a"*! s!"!ct! 'o t! in*cto. is so"*tion aois aing !sistanc!, t!!
aoiing into*cing o! "oss!s.
3-43) <!!a" cic*it can acco#"is tis oJ!cti!, inc"*ing t! a"'-a! !cti'i! o' Aig. 3-2a
an a"'-a! !cti'i! it a '!!!!"ing io! o' Aig. 3-7, !ac it !sistanc! a!.
+not! so"*tion is to *s! t! conto""! a"'-a! !cti'i! o' Aig. 3-14a *t it no
!sistanc!. ! ana"sis o' tat cic*it is "i! tat o' Aig. 3-6 *t ito*t  c. !
!s*"ting a"*! o' B is 75C, otain 'o# a <ic! si#*"ation. at so"*tion is goo
!ca*s! no !sistanc! is n!!!, an "oss!s a! not into*c!.
3-44 an 3-45) ! conto""! a"'-a! !cti'i! o' Aig. 3-15 (ito*t t! !sistanc!) can !
*s! to satis' t! !sign s!ci'ication. ! a"*! o' t! !"a ang"! !!ns on t! a"*!
s!"!ct! 'o t! in*cto.
(MHUFLFLRVFDSLWXOR (OHFWUyQLFDGH3RWHQFLD
'DQLHO:+DUW
CHAPTER 5 SOLUTIONS
3/9/10
5-1)
a)
Vm
Vo, rms 
2
Vo2,rms
P
c)
pf 
 sin(2 )
 sin(2 )

 Vrms 1  

2

2
  60  Vo ,rms  431 V ; I o ,rms 
Vm  480 2
b)
1
R

431
 8.61 A
50
4312
 3708 W
50

R
Vo,rms
P
P
3708


 0.897
S Vrms I rms  480   8.61
d ) I avg , SCR 
I rms ,SCR 
Vm
480 2
 1  cos   
 1  cos 60  3.24 A
2 R
2  50 
I o ,rms
2

8.61
 6.1 A
2
 480
  8.0 A
 50 
e) I1, rms  0.84 
THDI 
2
I rms
 I1,2rms
I1, rms

8.612  8.02
 0.38  38%
8.0
_____________________________________________________________________________________
5-2)
Vm
Vo,rms 
a)
2
Vo2,rms
P
c)
pf 
 sin(2 )
 sin(2 )

 Vrms 1  

2

2
  45  Vo, rms  114.4 V ; I o, rms 
Vm  120 2
b)
1

R
Vo , rms
R

114.4
 5.72 A
20
114.42
 655 W
20
P
P
655


 0.953
S Vrms I rms  120   5.72 
d ) I avg , SCR 
Vm
120 2
 1  cos   
 1  cos 45   2.30 A
2 R
2  20 
I rms , SCR 
I o, rms
2

5.72
 4.05 A
2
 120
  5.53 A
 20
e) I1,rms  0.92 
THDI 
2
I rms
 I1,2rms
I1,rms

5.72 2  5.532
 0.26  26%
5.53
_____________________________________________________________________________________
5-3)
P
Vo2.rms
R

from Fig . 5.2,
I o, rms 
Vo, rms
I SCR ,rms 
pf 
R

I o, rms
2
Vo.rms  PR 
 800   35 
 167.3 V
167.3
 0.7    92
240
167.3
 4.78 A
35

4.78
2
 3.38 A
P
800

 0.70  70%
S 120  4.78 
_____________________________________________________________________________________
5-4)
120
 0.5 ; a  115 from Fig.5.2
240
 sin(2 )
or solving Eq.5-3, 120  240 1  
 0    1.99 rad  114

2
With the 240-Vsource, Vo ,rms  120V ;
240 V source : Vo, peak  2  240  sin  114   310 V
120 V source : Vo, peak  Vm  2  120   170V
_____________________________________________________________________________________
5-5)
For P  200 W , Vo ,rms  PR  200  40   89.4 V
Using Eq. 5-3,
89.4  120 1 
pf 
 sin(2 )

 0    1.48 rad  85

2
P
P
200


 0.75  75%.
S Vrms I rms  120   89.4 / 40 
For P  400 W , Vo ,rms  PR  400  40   126 V
Since 126 V > 120 V of the source, 400 W is not possible.
The maximum power available is
1202
 360 W. The pf is 1.0 for 360 W.
40
_____________________________________________________________________________________
5-6)
Using the circuit of Fig. 5-1a,
For P  750W , Vo,rms  PR  750  32   154.9 V
Using Eq. 5-3,
154.9  240 1 
 sin(2 )

 0    1.703 rad  97.6

2
For P  1500 W , Vo , rms  PR  1500  32   219 V
219  240 1 
 sin(2 )

 0  0.986 rad  56.5

2
Maximum SCR currents are for 1500 W: I SCR ,rms 
I SRC ,avg 
Io
2

219 / 32
 4.84 A
2
2  240 
Vm
 1  cos   
 1  cos  56.5    2.62 A
2 R
2  32 
Vmax  2(240)  340 V
_____________________________________________________________________________________
5-7)
Using the circuit of Fig. 5-1a,
For R  20, Vo,rms  PR  1200  20   154.9 V
Using Eq. 5-3,
154.9  240 1 
 sin(2 )

 0    1.703 rad  97.6

2
For R  40, Vo ,rms  PR  1200  40   219 V
219  240 1 
 sin(2 )

 0  0.986 rad  56.5

2
Maximum SCR currents are for R = 20 : I SCR ,rms 
I SRC , avg 
Io
2

154.9 / 20
 5.48 A
2
2  240 
Vm
 1  cos   
 1  cos  97.6   2.34 A
2 R
2  20 
Vmax  2(240)  340 V
_____________________________________________________________________________________
5-8)
R
V 2 1202

 144 
P
100
a ) P  75 W : Vrms 
 144   75 
 103.9V
From Fig.5-3,   1.16 rad  66.2
b) P  25 W : Vrms 
 144   25 
 60 V
From Fig.5-3,   1.99 rad  114
_____________________________________________________________________________________
5-9) S1 is on from α to π, and D2 is on from π to 2π.
vo  t   Vm sin t
Vo,rms 

2
1
2
Vm
2
for   t  2
  V
1
sin t  d  t   Vm
2
m
1  sin  2 


2 4
8
 sin  2 

2
4
0  

Vm
V
 Vo ,rms  m
2
2
_____________________________________________________________________________________
5-10)
vo (t )  Vm sin t
Vo.rms
1 

2 


 Vm
Vo.rms 
Vm
2
for 1  t   and for    2  t  2
  Vm sin t 
1
2
d  t  
2


 Vm sin t  2 d  t 
  2


1 1   2 sin  21   sin  2 2 


2
4
8
1 
sin  21   sin  2 2 
1   2

2
4
_____________________________________________________________________________________
5-11) a) Using Eq. 5-9,
Z  21.3 ;   0.561 rad ;
  0.628
i  t   7.98sin  t  0.561  19.25e t /0.628 A
  60  1.047 rad ,   3.696 rad  212
b) I rms  4.87 A
c) I rms , SCR 
4.87
2
 3.44 A
2
d ) P  I rms
R   4.87   18   427 W
2
_____________________________________________________________________________________
5-12) Using Eq. 5-9,
Z  26.7 ;
  0.601 rad ;
  0.685
i  t   6.36sin  t  0.601  6.10et /0.685 A
  50  0.873 rad ,   3.738 rad  214
b) I rms  4.18 A
c ) I rms , SCR 
4.18
2
 2.95 A
2
d ) P  I rms
R   4.18 
2
 22   384 W
_____________________________________________________________________________________
5-13) Using Eq. 5-9,
  0.646 rad ;
Z  15.0 ;
  0.754
i  t   11.3sin  t  0.646   158e t /0.754 A
  115  2.01 rad ,   3.681 rad  211
I rms  2.95 A
_____________________________________________________________________________________
5-14) Using Eq. 5-9,
Z  14.2 ;
  0.561 rad ;
  0.6.28
i  t   11.98sin   t  0.561  54.1e t /0.628 A
  70  1.222 rad ,   3.691 rad  212
I rms  6.69 A
2
P  I rms
R   6.69   12   537 W
2
PSpice: P = AVG(W(R)) in Probe gives 523 W (read at the end of the trace). The difference
between PSpice and the theoretical output is because of the nonideal SCR model in PSpice. The
PSpice result will be more realistic. The THD is 22.4% from the PSpice output file using Fourier
terms through n = 9.
_____________________________________________________________________________________
5-15) Use the PSpice circuit of Example 5-3. The .STEP PARAM command is quite useful for
determining α. (a) α ≈ 81° for 400 W. (b) α ≈ 46° for 700 W.
SINGLE-PHASE VOLTAGE CONTROLLER (voltcont.cir)
*** OUTPUT VOLTAGE IS V(3), OUTPUT CURRENT IS I(R) ***
**************** INPUT PARAMETERS *********************
.PARAM VS = 120
; source rms voltage
.PARAM ALPHA = 81
; delay angle in degrees
.STEP PARAM ALPHA 10 90 20
; try several values of alpha. Modify the range for more precision
.PARAM R = 15
; load resistance
.PARAM L = 15mH
; load inductance
.PARAM F = 60
; frequency
.PARAM TALPHA = {ALPHA/(360*F)} ; converts angle to time delay
.PARAM PW = {0.5/F}
; pulse width for switch control
***************** CIRCUIT DESCRIPTION *********************
VS 1 0 SIN(0 {VS*SQRT(2)} {F})
S1 1 2 11 0 SMOD
D1 2 3 DMOD
; forward SCR
S2 3 5 0 11 SMOD
D2 5 1 DMOD
; reverse SCR
R 3 4 {R}
L 4 0 {L}
**************** MODELS AND COMMANDS ********************
.MODEL DMOD D(n=0.01)
.MODEL SMOD VSWITCH (RON=.01)
VCONTROL 11 0 PULSE(-10 10 {TALPHA} 0 0 {PW} {1/F}) ;control for both switches
.TRAN .1MS 50MS 0MS 1u UIC ; one period of output
.FOUR 60 I(R)
; Fourier Analysis to get THD
.PROBE
.END
_____________________________________________________________________________________
5-16) Modify the PSpice circuit file of Example 5-3. Use the .STEP PARAM command (see Prob. 5-15)
for determining α. (a) α ≈ 80° for 600 W. (b) α ≈ 57° for 1000 W.
_____________________________________________________________________________________
5-17) The single-phase voltage controller of Fig. 5-4a is suitable for this application. Equation (5-9)
applies for each half-period of the input sine wave. For 250 W delivered to the load, each half period
must deliver 125 W. Therefore, the rms value of the current in Eq. (5-9) must be 2.28 A, found by using
I2R = 125. A closed-form solution is not possible, but trial-and-error numerical techniques give α ≈ 74°. A
similar but perhaps easier method is to use PSpice simulations using the PSpice A/D circuit file in
Example 5-3. Modifying the diode model to .MODEL DMOD D(n=.01) to represent an ideal diode, and
with trial-and-error values of α, gives α ≈ 74°.
The average and rms currents are determined from a numerical integration of the current expression from
Eq. (5-9) or from a PSpice simulation. ISCR,avg = 1.3 A, ISCR,rms = 2.3 A. The maximum voltage across the
switches is 120√2sin(74°) = 163 V.
5-18) The PSpice circuit file is shown below. The total average load power is three times the power in
one of the phase resistors. Enter 3*AVG(W(RA)) in Probe. The results are (a) 6.45 kW for 20°, (b) 2.79
kW for 80°, and (c) 433 W for 115°. Note that the .STEP PARAM command can be used to run the three
simulations at once.
THREE-PHASE VOLTAGE CONTROLLER -- R-L LOAD (3phvc.cir)
*SOURCE AND LOAD ARE Y-CONNECTED (UNGROUNDED)
********************** INPUT PARAMETERS ****************************
.PARAM Vs=480
; rms line-to-line voltage
.PARAM ALPHA=20
; delay angle in degrees
.STEP PARAM ALPHA LIST 20 80 115
.PARAM R=35
; load resistance (y-connected)
.PARAM L = 1p
; load inductance
.PARAM F=60
; source frequency
********************** COMPUTED PARAMETERS **************************
.PARAM Vm={Vs*SQRT(2)/SQRT(3)} ; convert to peak line-neutral volts
.PARAM DLAY={1/(6*F)}
; switching interval is 1/6 period
.PARAM PW={.5/F} TALPHA={ALPHA/(F*360)}
.PARAM TRF=10US
; rise and fall time for pulse switch control
*********************** THREE-PHASE SOURCE **************************
VAN 1 0 SIN(0 {VM} 60)
VBN 2 0 SIN(0 {VM} 60 0 0 -120)
VCN 3 0 SIN(0 {VM} 60 0 0 -240)
***************************** SWITCHES ********************************
S1 1 8 18 0 SMOD
; A-phase
D1 8 4 DMOD
S4 4 9 19 0 SMOD
D4 9 1 DMOD
S3 2 10 20 0 SMOD
D3 10 5 DMOD
S6 5 11 21 0 SMOD
D6 11 2 DMOD
; B-phase
S5 3 12 22 0 SMOD
; C-phase
D5 12 6 DMOD
S2 6 13 23 0 SMOD
D2 13 3 DMOD
***************************** LOAD **********************************
RA 4 4A {R}
; van = v(4,7)
LA 4A 7 {L}
RB 5 5A {R}
LB 5A 7 {L}
; vbn = v(5,7)
RC 6 6A {R}
; vcn = v(6,7)
LC 6A 7 {L}
************************* SWITCH CONTROL *****************************
V1 18 0 PULSE(-10 10 {TALPHA}
{TRF} {TRF} {PW} {1/F})
V4 19 0 PULSE(-10 10 {TALPHA+3*DLAY} {TRF} {TRF} {PW} {1/F})
V3 20 0 PULSE(-10 10 {TALPHA+2*DLAY} {TRF} {TRF} {PW} {1/F})
V6 21 0 PULSE(-10 10 {TALPHA+5*DLAY} {TRF} {TRF} {PW} {1/F})
V5 22 0 PULSE(-10 10 {TALPHA+4*DLAY} {TRF} {TRF} {PW} {1/F})
V2 23 0 PULSE(-10 10 {TALPHA+DLAY} {TRF} {TRF} {PW} {1/F})
************************ MODELS AND COMMANDS *************************
.MODEL SMOD VSWITCH(RON=0.01)
.MODEL DMOD D
.TRAN .1MS 50MS 16.67ms 10US UIC
.FOUR 60 I(RA)
; Fourier analysis of line current
.PROBE
.OPTIONS NOPAGE ITL5=0
.END
_____________________________________________________________________________________
5-19) The PSpice input file from Example 5-4 is used for this simulation. In Probe, enter the expression
3*AVG(W(RA)) to get the total three-phase average power in the load, resulting in 368 W. Switch S 1
conducts when the current in phase A is positive, and S4 conducts when the current is negative.
_____________________________________________________________________________________
5-20) The smallest value of α is 120°. The conduction angel must be less than for equal to 60°. The
extinction angle is 180°, so α is 120° or greater.
_____________________________________________________________________________________
5-21)
THREE-PHASE VOLTAGE CONTROLLER -- R-L LOAD
*MODIFIED FOR A DELTA-CONNECTED LOAD
*SOURCE IS Y-CONNECTED (UNGROUNDED)
********************** INPUT PARAMETERS ****************************
.PARAM Vs=480
; rms line-to-line voltage
.PARAM ALPHA=45
; delay angle in degrees
.PARAM R=25
; load resistance (y-connected)
.PARAM L = 1p
; load inductance
.PARAM F=60
; source frequency
********************** COMPUTED PARAMETERS **************************
.PARAM Vm={Vs*SQRT(2)/SQRT(3)} ; convert to peak line-neutral volts
.PARAM DLAY={1/(6*F)}
; switching interval is 1/6 period
.PARAM PW={.5/F} TALPHA={ALPHA/(F*360)}
.PARAM TRF=10US
; rise and fall time for pulse switch control
*********************** THREE-PHASE SOURCE **************************
VAN 1 0 SIN(0 {VM} 60)
VBN 2 0 SIN(0 {VM} 60 0 0 -120)
VCN 3 0 SIN(0 {VM} 60 0 0 -240)
***************************** SWITCHES ********************************
S1 1 8 18 0 SMOD
; A-phase
D1 8 4 DMOD
S4 4 9 19 0 SMOD
D4 9 1 DMOD
S3 2 10 20 0 SMOD
D3 10 5 DMOD
S6 5 11 21 0 SMOD
D6 11 2 DMOD
; B-phase
S5 3 12 22 0 SMOD
; C-phase
D5 12 6 DMOD
S2 6 13 23 0 SMOD
D2 13 3 DMOD
***************************** LOAD **********************************
RA 4 4A {R}
;
LA 4A 2 {L}
RB 5 5A {R}
LB 5A 3 {L}
;
RC 6 6A {R}
LC 6A 1 {L}
;
************************* SWITCH CONTROL *****************************
V1 18 0 PULSE(-10 10 {TALPHA}
{TRF} {TRF} {PW} {1/F})
V4 19 0 PULSE(-10 10 {TALPHA+3*DLAY} {TRF} {TRF} {PW} {1/F})
V3 20 0 PULSE(-10 10 {TALPHA+2*DLAY} {TRF} {TRF} {PW} {1/F})
V6 21 0 PULSE(-10 10 {TALPHA+5*DLAY} {TRF} {TRF} {PW} {1/F})
V5 22 0 PULSE(-10 10 {TALPHA+4*DLAY} {TRF} {TRF} {PW} {1/F})
V2 23 0 PULSE(-10 10 {TALPHA+DLAY} {TRF} {TRF} {PW} {1/F})
************************ MODELS AND COMMANDS *************************
.MODEL SMOD VSWITCH(RON=0.01)
.MODEL DMOD D
.TRAN .1MS 50MS 16.67ms 10US UIC
.FOUR 60 I(RA)
; Fourier analysis of line current
.PROBE
.OPTIONS NOPAGE ITL5=0
.END
40A
Ia
0A
SEL>>
-40A
I(RA)
50A
Source
A
current
0A
-50A
15ms
20ms
-
25ms
30ms
35ms
40ms
45ms
50ms
I(VAN)
Time
_____________________________________________________________________________________
5-22) The PSpice circuit file modification must include a very large resistor (e.g., one megaohm)
connected between the neutral of the load to ground to prevent a “floating node” error because of the
series capacitor. The steady-state phase A current has two pulses for each of the switches, assuming that
the gate signal to the SCRs is continuously applied during the conduction interval. The rms current is
approximately 5.52 A. The total average power for all three phases is approximately 1.28 kW. The THD
for the load current is computed as 140% for harmonics through n = 9 in the .FOUR command. However,
the current waveform is rich in higher-order harmonics and the THD is approximately 300% for n = 100.
It should be noted that this load is not conducive for use with the voltage controller because the load
voltage will get extremely large (over 5 kV) because of stored charge on the capacitor.
40A
S1
S1
(1.0000,5.5229)
0A
Phase A current
SEL>>
-40A
S4
I(RA)
S4
RMS(I(RA))
2.0KW
(1.0000,1.2811K)
1.0KW
Total average power
0W
0.980s
0.984s
AVG(W(RA))*3
0.988s
0.992s
0.996s
1.000s
Time
_____________________________________________________________________________________
5-23) With the S1-S4 switch path open, the equivalent circuit is as shown. The current in phase A is zero,
so the voltage across the phase-A resistor is zero. The voltage at the negative of V 14 is then Vn, and the
voltage at the positive of V14 is Va. The voltage across the phase B resistor is half of the voltage from
phase B to phase C, resulting in
Vn  Vb 
Vb  Vc Vb  Vc

2
2
Therefore,
V14  Va  Vn  Va 
Vb  Vc
2
CHAPTER 6 SOLUTIONS
5/17/10
6-1)
Po  Vo I o  Vo I s ; Ps  Vs I s

Po Vo I o Vo


Ps Vs I s Vs
6-2)
Io 
P 100

 3.33 A.
Vo 30
a ) Ps  Vs I o  (100)(3.33)  333 W .;  
Po 100

 30%
Ps 333
b) PQ  VCE I o  (70)(3.33)  233 W .
1 yr.=8760 Hr.; W = (233)(8760)=2044 kW-Hr,
c) e.g., @10 cents/kW-Hr, cost = $204.40/yr.
6-3)
a ) Vo  Vs D  (100)(0.6)  60 V .
b) Vo,rms  Vm D  (100) 0.6  77.5 V (see Example 2-4)
Vo2,rms
77.52
c) P 

 600 W .
R
10
d ) Results are not dependent on frequency.
6-4)
a ) Vo  Vs D  (24)(0.65)  15.6 V .
Vo 15.6

 1.56 A.
R
10
V
15.6
1
iL  o (1  D )T 
(1  0.65)
 2.18 A.
6
L
25(10)
100, 000
i
2.18
I L ,max  I L  L  1.56 
 2.65 A.
2
2
i
2.18
I L ,min  I L  L  1.56 
 0.47 A.
2
2
b) I L  I R 
Vo (1  D)
15.6(1  0.65)

 0.182
2
8LCf
8(25)(10) 6 (15)(10) 6 (100, 000) 2
Vo
or
 1.17%
Vo
c) Vo 
6-5)
a ) Vo  Vs D  9 V .
b) I L  1.8 A.; iL  2.4 A.
iL
 3.0 A.
2
i
 I L  L  0.6 A.
2
I L,max  I L 
I L,min
c)
Vo
 0.44%
Vo
6-6)
a) D 
Vo
 0.5
Vs
b) I L  I R 
Po 125

 5 A.
Vo 25
V
iL
 1.25; iL  2.5 A.  o (1  D)T
2
L
V
25
1
L  o (1  D)T 
(1  .5)
 50  H .
iL
2.5
100, 000
I L,max  6.25 A. 
c) 
Vo
1 D
 5%  .005 
Vo
8 LCf 2
C

8

1 D
1  .5

 25  F .
6
2
8(.005)(50)(10)
(100,
000)
Vo
2
 Lf
Vo 
6-7)
a) D 
Vo 1.5

 0.25
Vs
6
V 1.5
b) average : I L  I R  o 
 0.5 A.
R
3
2
rms : I L ,rms
 0.5625 / 2
 0.5  
  0.526 A.
3 

2
iL  0.5625
 1 1  D
 1

1  .25

 1.5


  0.781 A.
6
 R 2 Lf 
 3 2(5)(10) (400,000)

peak : I L,max  Vo 
 1 1  D

  0.219 A.
 R 2 Lf 
I L,min  Vo 
c ) Ps  Po  Vs I s  Vo I R  I s 
Vo I R 1.5(0.5)

 0.125 A.
Vs
6
d ) I D ,max  I L ,max  0.781 A.
I D  I o  I s  0.5  0.125  0.375 A.
6-8)
Io  I L 
D
Po 25

 1.25 A.
Vo 30
Vo 20

 0.667
Vs 30
iL
2
iL  ( I L  I L ,min )2  (1.25  0.31)2  1.88 A.
I L,min  (0.25)(1.25)  0.31 A.  I L 
Vo
(1  D)T
L
V
1
20
1
L  o (1  D ) 
 89  H
 1  .667 
iL
f 1.88
40000
iL 
6-9)
Lmin 
D
(1  D) R
2f
Vo
20
20
; Dmax 
 0.4; Dmin 
 0.33
Vs
50
60
IL  IR 
Po
75
125
; I R ,min 
 3.75 A.; I R ,max 
6.25 A.
Vo
20
20
Vo
202
202
R  ; Rmax 
 5.33 ; Rmin 
 3.20 
P
75
125
(1  Dmin ) Rmax (1  .33)(5.33)
Lmin 

 17.76  H
2f
2(100, 000)
6-10)
(1  D)( R)
2f
Lmin 
f  200 kHz
Vo=5 V
Vs, V
D
I, A.
R, Ω
Lmin, µH
10
0.5
0.5
10
12.5
10
0.5
1.0
5
0.5
10
6.25
16.7 (worst case, D = 1/3, R =
10)
1.0
5
8.33
15
15
1/3
1/3
L
(1  Dmin ) Rmax
2f

Lmin
1
 1   (10)
3

 16.67  H
2(200 k )
6-11)
Example design:
D
Vo 15

 0.3125
Vs 48
Let f  100 kHz ( for example)
 Vo
 15
 0.40    0.75 A

 8
 R
Let iL  40% of I L  0.40 
L
C
 Vs  Vo  D  48  15 0.3125

 137.5  H
 iL  f
 0.75 100,000
1 D
1  0.3125

 12.5  F
 Vo
8  150  106 (0.005)100,000
8L 
 f
 Vo 
Other values of L and C are valid if the inductor current is continuous with margin.
6-12) (Based on the example design in 6-11)
Vmax, switch = Vs = 48 V
Vmax, diode = Vs = 48 V
Imax, switch = ILmax = 1.5 + 0.75/2 = 1.875 A
Iavg, switch =
Irms, switch 
Vo Io 15  1.875 

 0.586 A
Vs
48
1
T
DT
i
2
L (t) d t
 1.06 A (numerically)
0
Imax,diode = ILmax = 1.875 A
Iavg,diode =IL- Iavg,switch = 1.875 – 0.586 = 1.289 A
T
Irms,diode

1
i 2L (t) d t  1.56 A (numerically)
T DT

6-13) Example design:
D
Vo 15

 0.625
Vs 24
Let f  400 kHz ( for example)
Let iL  40% of I L  0.40  2   0.8 A
L
C
 Vs  Vo  D  24  15 0.625

 17.6  H
 iL  f
 0.8 400,000
1 D
1  0.625

 1.67  F
 Vo
8  17.6  106 (0.01)400,000
8L 
 f
 Vo 
6-14) Example design:
D
Vo 12

 0.667
Vs 18
Let f  200 kHz ( for example)
I L  Io 
Po 10W

 0.833 A
Vo 12V
Let iL  40% of I L  0.40  0.833  0.333 A
L
C
 Vs  Vo  D   18  12  0.667  60  H
 iL  f
 0.333 200,000
1 D
1  0.667

 3.5  F
 Vo

 6 0.1
8L 
 f 8  60  10  12 200,000


 Vo 
Other values of L and C are valid if the inductor current is continuous with margin.
6-15)
n  1  V1  30.27
Using ac circuit analysis, Vo1  0.048 V  2(0.048)  0.096 V p  p
0.096 0.096

 0.48%
Vo
20
Vo
 0.469%
Vo
Using Eq. 6 -16,
The output voltage is mainly the dc term and the first ac term.
6-16)
a ) rC  0.5 , iL  2.88 A  iC
Vo ,ESR  iC rC  2.88(0.5)  1.44 V .
Vo 1.44

 8%
Vo
18
b)
Vo
 0.5%
Vo
Vo  Vo ,ESR  iC rC  rC 
rC 
50(10) 6
C
 C
Vo 0.005(18)

 0.031 
iC
2.88
50(10) 6 50(10) 6

 1600  F .
rC
0.031
6-17)
Vs
20

 50 V .
1  D 1  .6
Vs
20
b) I L 

 10 A.
2
(1  D ) R (1  .6) 2 (12.5)
Vs
V DT
20
20(.6) / (200,000)
I max 
 s


 13 A.
2
2
(1  D ) R
2L
(1  .6) (12.5)
2(10)(10) 6
Vs
V DT
I min 
 s
 7.0 A.
2
(1  D ) R
2L
Vo
D
0.6
c)


 0.6%
Vo
RCf 12.5(40)(10) 6 (200,000)
a ) Vo 
d ) I D  Io 
Vo
50

 4.0 A.
R 12.5
6-18)
Inductor current: (see Example 2-8)
2
4.61/ 2
 I L / 2
2
  10 

3 
3


I L,rms  I L2  
Capacitor current: (define t=0 at peak current)
2
 10.09 A.
I C ,rms


1

6 
 25(10) 
10  s

0
 
( 4.61(10)5 t  8.3) 2 dt   (4) 2 dt

10  s
 
25  s
1/2
 4.97 A.
6-19)
Vo 
Vs
V
5
 D  1  s  1   0.667
1 D
Vo
15
Vo2 152

9 
25 25
Vs
5
IL 

 5 A.
2
(1  D) R (1  .667) 2 (9)
I L,min  0.5(5)  2.5 A.  I L  5 A.
R
I L 
Vs DT 5(.667) / 300

 2.22  H
I L
5
From Eq. 6  27, C 

R

D
0.667

 24.7  F .
9(.01)(300, 000)
Vo
  f
Vo
6-20) Example design:
D 1
R
Vs
12
 1   0.333
Vo
18
Vo2 182

 16.2 
P
20
Vs
IL 
1 D
2
R

12
 1  .333 2 16.2
 1.67 A
Let f  200 kHz
Let iL  40% of I L  0.4  1.67   0.667 A
L
12  0.333
Vs D

 30  H
 iL  f (0.667)200,000
 Lmin
for continuous current  6  H 
D
C
 Vo
 f
 Vo 
R

0.333
 20.6  F
16.2  0.005  200,000
6-21)
Using C  48  F , R  50 , ton  0.6T 
0.6
 24  s
25000
vo (t )  Vo ,max e  t / RC
vo (24  s )  Vo ,max e 24/[(50)(48)]  Vo,max  0.99005
Vo ,max  vo (24  s )  Vo  Vo ,max  0.99005Vo ,max  0.01Vo ,max
Vo
 0.01  1%
Vo
6-22)
6-23)

D
0.6

 12
 18 V .



 1  D
 1  0.6
Vs D
12(.6)
b) Eq. 6  31: I L 

 4.5 A.
2
R (1  D )
(10)(1  .6) 2
Vs D
V DT
12(.6) / 200, 000
I L,max 
 s
 4.5 
 6.3 A.
2
R (1  D)
2L
2(10)(10) 6
Vs D
V DT
I L,min 
 s
 2.7 A.
2
R (1  D)
2L
Vo
D
0.6
c)


 0.015  1.5%
Vo
RCf 10(20)(10)6 (200, 000)
a ) Vo  Vs 
6-24)
Inductor current: (see Example 2-8)
2
I L ,rms
3.6 / 2
 I L / 2
2
 I 
  4.5 

3 
3


2
 4.62 A.
2
L
Capacitor current: For convenience, redefine t = 0 at the peak current. The current is then
expressed as
 
iC  t   4.5  1.8 106 t A
for 0  t  2 s
for 2  s  t  5 s
 1.8 A
T
1 2
i (t )dt
T 0
I rms 
I C ,rms

1 

6 
 5(10) 
T
2s
1
1

 5  106   5 s
f 200,000
  4.5  1.8  10 
0
6
2
 
dt   (1.8) dt

2s
 
5 s
2
1/2
 2.30 A.
6-25)
a ) From Eq. 6-48, D 
Vo
36

 0.6
Vs  Vo 24  36
Vs D
24(.6)

 9 A.  I L ,min  0.4(9)  3.6 A.
2
R (1  D) 10(1  .6) 2
I L  2(9  3.6)  10.8 A.
IL 
From Eq. 6-28, L 
Vs DT
24(.6)

 13.3  H
I L
10.8(100, 000)
b) From Eq. 6-36, C 

R

D
0.6

 120  F
10(0.005)(100, 000)
Vo
 f
Vo 
6-26) Example design:
Vo
Using Eq. (6-48),
D=
Using Eq. (6-49),
IL 
R
Vs  Vo

50
 0.556
40  50
P
75

 3.375 A.
Vs D 40  0.556 
Vo2 502

 33.3 
P
75
Letting f = 100 kHz (designer's choice),
Lmin 
 1  D  2 R  1  0.556  2 33.3

 32.9  H
2f
2  100, 000 
Choose L at least 25% larger than L min (41  H). A common practice is to select L such that
i L = 40% of I L  0.40  3.375   1.35 A. Using Eq. (6-45),
L
40  0.556 
Vs D

 165  H
iL f 1.35  100, 000 
Using Eq. (6-54),
D
C
 Vo
 f
 Vo 
R

0.556
 16.7  F
33.3  0.01  100, 000 
6-27) Example design:
Using Eq. (6-48),
and D 
D=
Vo
Vs  Vo
, D
15
 0.556 for the 12-V source,
12  15
15
 0.455 for the 18-V source.
18  15
Using Lmin 
 1  D 2 R
2f
, the worst case is for D = 0.455 for the 18-V source.
 1  0.455 2 15
Letting f = 100 kHz (designer's choice), Lmin 
 22.3  H
2  100, 000 
Choose L at least 25% larger than L min (28  H).
Alternatively, a common practice is to select L such that i L = 40% of I L .
Il 
Vo2
152

 1.83 A
Vs RD 18  15   0.455 
iL  0.40  1.83  0.73 A. Using Eq. (6-45),
L
18  0.455 
Vs D

 112  H (100  H will be fine)
iL f 0.73  100, 000 
Using Eq. (6-54),
C
C
D
 Vo
R
 f
 Vo 
0.556
 37  F
15  0.01  100, 000 
, so base C on D = 0.556, (12-V source):
6-28) Using the equations
Vo
D
R
Vs  Vo
Vo2
P
 1 D
Lmin 
IL 
2
R
2f
P
Vs D
D
C
 Vo
 f
 Vo 
R
and using f = 100 kHz (designer’s choice), results are shown in the table.
Vs, (V)
10
10
14
14
P (W)
10
15
10
15
D
0.545
0.545
0.462
0.462
R (Ω)
14.4
9.6
14.4
9.6
Lmin (µH)
14.9
9.9
20.9
13.9
IL (A)
1.83
2.75
1.55
2.32
C (µF)
37.9
56.8
32.1
48.1
The value of L should be based on Vs = 14 V and P = 10 W, where Lmin = 20.9 µH. Select the
value of L at least 25% larger than Lmin (26.1 µF). Using another common criterion of ΔiL = 40%
of IL, again for 14 V and 10 W, L = 104 µH.
The value of C is 56.8 µF for the worst case of Vs = 10 V and P = 10 W.
6-29)

D
0.6

 12
 18 V .



 1  D
 1  0.6
P
V 2 R 27
 o  o 
 1.5 A.
Vo Vo 18
Vo  Vs 
I L2
I L1 
Po 27

 2.25 A.
Vs 12
iL1 
Vs D
12(.6)

 0.14 A.
6
Lf
200  10  (250,000)
i L 2 
Vs D
 0.29 A.
L2 f
6-30)
D
1
1

 0.333
V
20
1 s 1
Vo
10
I L 2  I o  1 A.
Vo
 10
I L 2    (1)  0.5 A.
Vs
 20
VD
VD
20(0.333)
iL1  s
 L1  s 
 1.33 mH
L1 f
iL1 f 0.10(.5)(100,000)
I L1 
i L 2 
Vs D
VD
20(0.333)
 L2  s

 0.667 mH
L2 f
iL 2 f 0.10(1)(100,000)
6-31) Example design:
Vo
D
30


 1.2  D  0.5455
Vs
1  D 25
I L2 
25  0.5445 
Po
VD
60

 2.0 A;  iL 2  0.4(2.0)  0.4 A  L2  s

 341  F
Vo 30
iL 2 f 0.4  100,000 
I L1 
Ps 60

 2.4 A;
Vs 25
 iL1  0.4(2.4)  0.48 A  L1 
25  0.5445 
Vs D

 284  F
iL1 f 0.48  100,000 
Let f = 100 kHz (designer's choice).
C2 
1 D
1  0.5455

 1.67  F
2
6
 Vo
0.01
8
341
10
100,000
2





 8L2 f
 Vo 
 
VC1  Vs  Vo  25   30   55 V

 vC1  0.05  55   2.75 V
Using R = Vo2 / P   30  / 60  15 ,
2
C1 
30  0.5455 
Vo D

 3.97  F
Rf vC1 15  100,000  2.75
6-32)
D
Vo
12

 0.706
Vo  Vs 12  5
I L1 
Vo 2 122

 7.2A
Vs R 5(4)
i L1 
Vs D
(5)(0.706)

 3.53A
L1f 10(10)-6 (100,000)
I L1,max  7.2 
3.53
 8.96A
2
I L1,min  7.2 
3.53
 5.44A
2
I L2 
Vo 12

 3A
R
4
i L1 
Vs D
(5)(0.706)

 1.765A
L2f 20(10)-6 (100,000)
I L2,max  3 
1.765
 3.88A
2
I L2,min  3 
1.765
 2.12A
2
6-33)
Vo 
Vs D 3.3(.7)

 7.7V
1  D 1  .7
I L1 
Vo 2
7.7 2

 3.6A
Vs R 3.3(5)
i L1 
Vs D
(3.3)(0.7)

 1.925A
L1f 4(10)-6 (300,000)
I L1,max  3.6 
1.925
 4.56A
2
I L1,min  3.6 
1.925
 2.64A
2
I L2 
Vo 7.7

 1.54A
R
5
i L1 
Vs D
(3.3)(0.7)

 0.77A
L 2 f 10(10)-6 (300,000)
I L2,max  1.54 
0.77
 1.925A
2
I L2,min  1.54 
0.77
 1.155A
2
VC1  VC2 
Vo D
(7.7)(.7)

 0.0719V
RCf 5(50)(10) 6 (300,000)
6-34) Equation (6-69) for the average voltage across the capacitor C1 applies:
VC1  Vs .
When the switch is closed, the voltage across L2 for the interval DT is
v L2  vC1
Assuming that the voltage across C1 remains constant at its average value of Vs
vL2  VC1  Vs
(switch closed)
When the switch is open in the interval (1 - D)T,
vL2  Vo
(switch open)
Since the average voltage across an inductor is zero for periodic operation,
v
L2, sw closed
  DT    v
L2, sw open
  1  D T  0
Vs  DT   Vo  1  D  T  0
resulting in

D 

 1  D
Vo  Vs 
6-35)
I L1  I s 
Vo 2
62

 1.2A
Vs R (15)(2)
D
Vo
6

 0.286
Vo  Vs 6  15
L1 
Vs D
(15)(0.286)

 35.7H
( i L1 )f 0.4(1.2)250,000
I L2  I o 
L2 
Vo 6
  3A
R 2
Vs D
(15)(0.286)

 14.3H
( i L2 )f 0.4(3)250,000
VC2  Vo  6
VC2  Vo 
Vo D
RC2f
C1  C2  28.6F
or C2 
D
0.286

 28.6F
2(.02)250,000
 Vo
R
f
 Vo 
6-36)
D
Vo
2.7

 0.231
Vo  Vs 2.7  9
R
2.7
 2.7
1
I L1 
Vo 2
2.72

 0.30A
Vs R 9(2.7)
L1 
Vs D
9(0.231)

 57.7H
 i L1  f 0.4(0.30)300,000
I L2  I o  1A
L2 
6-37)
Vs D
9(0.231)

 14.2H
 i L2  f 0.4(1)300,000
iC  I L ,max  2.7 A.
VO , ESR  iC rC  (2.7)(0.6)  1.62 V .
VO , ESR
VO

1.62
 0.054  5.4%
30
Worst case : VO  VO ,C  VO , ESR  0.3  1.63  1.92 V .  6.4%
6-38)
Switch closed : vL  Vs  VQ
Switch open :
vL  VO  VD
avg (vL )  0 : (Vs  VQ ) DT  (VO  VD )(1  D)T  0

D 

 1  D
 VO  VD  (Vs  VQ ) 
6-39)
(1  D) R
L 2 f 97.5(10) 6 (2)(40000)
 Rmax 

 12.5 
2f
1 D
1  .375
b) For R=20Ω, current is discontinuous:
a ) Lmin 





2D
2(0.375)
  21.4 V .
  48 
Vo  Vs 
6 


8L 
8(97.5)(10)
2
2
 0.375  (0.375) 

 D D 

RT 

20 / 40000 

 18  Vo  21.4 V .



(1  D) R (1  0.375)(20)

 6.25
2
2
6.25
6.25
Increase Lf min : e.g., Lmin 
 157  H or f min 
 64.1 kHz
40000
97.5 mH
c) Lf min 
6-40)
2 Lf
2(120)(10)6 (25000)
a ) Rmax 

 62.5  for continuous current
D(1  D) 2
0.6(1).6) 2
b) For R=100Ω, the current is discontinuous:

 1 1
Vo  Vs 




2 D 2 RT 

L 

2


2(.6) 2 (100) / 25000

120(10) 6
  36 V .

2



 36 for 25  R  100
 1 1
Vo  12 




 30  Vo
c ) Lf min  4.8  L 
4.8
4.8
 192  H or f 
 40 kHz.
25000
120(10) 6
6-41) Discontinuous current for the buck-boost converter: Let DT be the time that the switch is
closed and D1T be the time that the switch is open and the current in the inductor is positive. For
a lossless converter, the output power is the same as the input power.
 I max D

 2 
Ps  Vs I s  Vs 
Vs DT
L
2 2
V DT
Ps  s
2L
2
V
Po  o
R
2 2
Vs D T Vo2

2L
R
Vo
RT
D
Vs
2L
I max 
6-42) When switches “1” are closed, C1 and C2 are connected in series, each having Vs/2 volts.
When the “1” switches are opened and the “2” switches are closed, Vo = Vs of the source plus
Vs/2 of C1, making Vo = 1.5Vs.
6-43)
20W
p(t) for the MOSFET
10W
0W
(1.000m,405.19m)
Average P = 405 mW
-10W
0.990ms
W(M1)
0.992ms
AVG(W(M1))
0.994ms
0.996ms
Time
0.998ms
1.000ms
6-44) Simulate the buck converter of Example 6-1 using PSpice. (a) Use an ideal switch and ideal
diode. Determine the output ripple voltage. Compare your PSpice results with the analytic results
in Example 6-1. (b) Determine the steady-state output voltage and voltage ripple using a switch
with an on resistance of 2 Ω and the default diode model
Using Ron =0.01 for the switch and n=0.01 for the diode, the p-p ripple voltage is 93.83 mV.
93.83/20 = 0.469%, agreeing precisely with the analytical results.
With Ron = 2 ohms, the p-p ripple is 90 mV, with a reduced average value.
6-45)
Note that for each converter topology, the average voltage across each inductor is zero,
and the average current in each capacitor is zero.
Buck Converter:
Show from Eqs. (6-9) and (6-17)
Vo  Vs D
Io 
and
Is
D
From the averaged circuit of Fig. 6.33b,
I L  Io  i c
Vap  Vs
Is  Di c
and

Vo  DVap
and
Io 

Is
D
Vo  DVs
Boost Converter:
Show from Eqs. (6-27) and (6-28) that
Vo 
Vs
1 D
I o  Is  1  D 
and
From the averaged circuit of Fig. 6.33c,
DVap  Vs  Vo
and
Io  i c  Di c  i c  D  1
Vap  Vo
and

Is  i c
Vo 

Vs
1 D
I o  Is  1  D 
Buck-Boost Converter:
Show from Eqs. (6-47) and (6-49) and preceding equations that

D 

 1  D
Vo   Vs 
Is  I L D
and
From the averaged circuit of Fig. 6.33d,
IL  ic
Is  Di c
and
Vap  Vs  Vo
and

Is  I L D
Vo  DVap


D 

 1  D
Vo   Vs 
Ćuk Converter:
Show from Eqs. (6-59) and (6-61) that
I L1
D

I L2 1  D

D 

 1  D
Vo  Vs 
and
From the averaged circuit,
Vap  Vs  Vo
and
DVap  Vo
i c  Di c  I L2
and
I L1  Di c



D 

 1  D
Vo  Vs 
I L1
D

I L2 1  D
CHAPTER 7 SOLUTIONS
4/03/10
7-1)

D  N 2  0.4
  1
 36

   12 V .
 1  D N1  0.6
  2
a ) Vo  Vs 
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
 4.5


  16.9 V .
 1  D N1
 0.4

 0.4
a ) Vo  Vs 
b) I Lm
Vo2
(16.9) 2


 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)
D  N2
N 2 V o 1  D  3 1  .32

 

  
  0.145
N1 V s
D
 1  D N1
 44 .32


a ) Vo  Vs 
or
N1
 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
 N 2
Vo
40
(2.5)  4.17 A

 
(1  D) R  N1
(1  0.4)40
V 2 402
where R  o 
 40 
P
40
Let iLm  40% of I Lm  0.4(4.16)  1.67 A
I Lm 
Let f  100 kHz
VD
24(0.4)
Lm  s

 57.6  H
iLm f 1.67(100, 000)
C

R

D
0.4

 20  F
40(.005)(100, 000)
Vo
 f
Vo 
7-5) For continuous current
iLm
2
2
V
VD
0 o  s
Vs DR 2 Lm f
I Lm ,min  I Lm 
I Lm ,min
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 = t x:
I Lm ,max 
Vs DT
Lm
diLm Vo  N1



dt
Lm  N 2
 1
Vo  N1 t
V o N
Vs DT
iLm (t ) 

  d   iLm ( DT ) 

 (t  DT ) 
Lm  N 2 DT
Lm N
Lm
 2
V  N 
V DT
I LM (t  t x )  0  o  1 (t x  DT )  s
Lm  N 2
Lm
Switch open :
tx 
Vs DT  N 2

  DT
Vo  N1
7-7)
 N 2
  100(.35)(1)  35 V .
 N1
a ) Vo  Vs D 
Vo
1 D
1  .35


 0.16%
2
6
Vo
8Lx Cf
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 2
 1
  170(0.3)    5.1 V .
 10
 N1
Vo
1 D
1  0.3


 0.175%
2
6
Vo
8Lx Cf
8(20)(10) 10(10) 6 (500, 000) 2
a ) Vo  Vs D 
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 2
 N
 2
Vo 50

 0.625
  D
 
Vs 80
 N1
 N
 1
a ) Vo  Vs D 
If
N3
 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
8 Lx Cf
8(100)(10) 150(10) 6 (100, 000) 2
R
7-10)
 N 2

  100(0.25) 

 N1
a ) Vo  Vs D 
1
  5 V.
5
using Lx  20  H ,
Vo
1 D
1  0.25


 0.33%
2
6
Vo
8Lx Cf
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 2
 N
 1
Vs D 125(0.3)

 0.75
 
 
Vo
50
 N1
 N
 2
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)
a ) Vo  Vs D 
c)
Vo
1 D

 C
Vo
8Lx Cf 2
1 D
1  0.3

 4.8  F .
6
8(58.3)(10) (0.005)(250, 000) 2
 Vo
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
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
Then D 
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
6
2
8(124)(10)
(0.01)(200,
000)
 Vo
2
8 Lx 
 f
V
 o
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
8(315)(10) (0.002)(250, 000) 2
 Vo
2
8 Lx 
 f
 Vo 
7-14)
The current in the physical primary winding is the sum of i L1 and iLm in the model. The physical
currents in windings 2 and 3 are the same as in the model.
7-15)
 N s
D  2(50)(0.5)(0.35)  17.5 V .
 N

p


a ) Vo  2Vs 
Vo 17.5

 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
b) I Lx 
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 L x is therefore
 N s
0  t  DT
 N

 p
vLx  Vs 
 Vo
DT  t  T / 2
  N
1
 T
 
VLx   Vs  s DT  Vo   DT 
 0
2
T
/
2




  N p



 N
Vo  2Vs  s D
 N p


7-19)

Vs 

 2(1  D)
Vo  
 N s



 N p



24 
 1
    17.1 V .
2(1  0.65)
 2

2
o
V
17.12

 1.22 A.
Vs R 24(10)
I Lx 
 N p
  2(24)(2)  96 V .
 N s
VSw,max  2Vs 
7-20)

Vs 

 2(1  D)
Vo  
 N s
N / N s 50
V
 o  p




Vs 2(1  D) 30
 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 p  2(1  D)T


T

 N s 
I x  I o  I Lx 


Vs I Lx  Vo  I Lx 


 N
Vo  Vs  S
 N
 p

N p
 2(1  D)
N d


1 

 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 i Lx, 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 i Lx 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
Let R1  1 k , R2  10 k 
1
1
C1 
 4.95 nF ; C2 
 355 pF
R2 z
 p R2
z 
7-25)
 comp  45  (105)  150
  comp
 150
  tan 
  3.73
 2 
 2 
Gain  15 dB  9.5 dB  24.5 dB
K  tan 
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
  tan 
  3.49
2
 2 


K  tan 
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
  tan 
  3.17
2
 2 


K  tan 
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
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 
co
1
 115 
KC3
7-30)
 1

 20 log10 

 V p

 
Ramp function gain = 20 log10 
Total gain = - 8 dB - 9.54 dB = -17.54 dB
17.54
20
G  10
 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 
R2 
K
 2.93 k 
C1 
K
 9.31 nF
co R2
C2 
1
 1.41 nF
co R2 K
C3 
K
 27.3 nF
co R1
R3 
7-31)
G R1
co
1
 151 
KC3
1
  9.54 dB
3
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
-100
-200
1.0Hz
10Hz
DB(V(Output))
(10.000K,-133.095)
100Hz
1.0KHz
P(V(Output))
Frequency
10KHz
100KHz
CHAPTER 8 SOLUTIONS
4/24/10
8-1)
Load: I avg  0, I rms  10 A.
Switches: I avg  5 A., I rms  I m D  10 0.5  7.07 A.
Source : I avg  I rms  10 A.
(See Example 2-4)
8-2)
I max
V
 dc
R
 1  e T /2
L 0.1
1 V
96
;  
 0.02s.; T  ; dc 
 19.2 A.

T /2
R 5
60 R
5
 1 e

 0.341
  3.94 A.
 1.66)
I max  19.2 
b) From Eq. 8  1: io (t ) 
io (0)  0  A  
Vdc
 Ae  t /
R
Vdc
R
Vdc
1  e  t / 

R
i (T / 2)  19.2  1  e 1/2.4   6.54 A.
io (t ) 
c) PSpice results are consistent with parts (a) and (b). The current waveform reaches
steady state after approximately 100 ms, corresponding to 5 time constants.
8.0A
(8.3333m,6.5486)
(158.333m,3.9485)
4.0A
0A
-4.0A
0s
40ms
80ms
120ms
160ms
I(L)
Time
8-3)
Vdc 150
L 40 mH

 7.5 A.;   
 2 ms;
R
20
R
20 
a)
T 1 / 60

 4.167
2 4 ms
Using Eq (8  8),
 1  e 4.167
 7.271 A.
4.167
 1 e

I max  7.5 
I min   I max  7.271 A.
Using Eq. (8-5),
for 0  t  8.33 ms
 7.5  14.8e  (t 1/120)/.002
for 8.33 ms  t  16.7 ms
io  
 7.5  14.8e t /.002

200ms
b)
c ) I peak  7.271 A.
d ) Vmax  Vdc  150 V .
8-4)
Vdc 125
L 25 mH

 6.25 A.;   
 1.25 ms;
R
20
R
20 
a)
T
1 / 60

 13.33
2 1.25 ms
Using Eq (8  8),
 1  e 13.33
 6.25 A.
13.33
 1 e

I max  6.25 
I min   I max  6.25 A.
Using Eq. (8-5),
 6.25  12.5e t /.00125

for 0  t  8.33 ms
 6.25  12.5e  (t 1/120)/.00125
for 8.33 ms  t  16.7 ms
io  
b) Using the first half-period,
1

120
I rms
1/120
  6.25  12.5e
 t /.00125
0

2
dt  5.45 A.
2
c) P  I rms
R   5.25  20  594 W .
2
Is 
P 594

 4.75 A.
Vdc 125
8-5)
Z1  152   2  400   0.01
a)

2
 29.3 

V1  I1Z1  8 2  29.3   331 V .
V1 
4Vdc

b)
 Vdc 
Vn 
4Vdc
;
n
n
1
3
5
THDI 
8-6)
 V1
 260 V .
4
Z n  R 2   2 400 L 
Vn
331
110
66
1.022  0.37 2
 0.136  13.6%
8.0
2
;
Zn
29.3
77
127
In 
Vn
Zn
; I n ,rms 
In,rms
8.0
1.02
0.37
In
2
a ) Z1  2.52   2  120   0.025

2
 31.3 

V1  I1Z1  2 2  31.3  88.6 V .
V1 
4Vdc

b) Vn 
 Vdc 
4Vdc
;
n
 V1
 69.6 V .
4
Z n  R 2   2 120 L 
n
1
3
5
Vn
88.6
29.5
17.7
THDI 
Using PSpice,
2
;
In 
Vn
Zn
Zn
31.3
61.8
97.5
0.342  0.132
 0.185  18.5%
2.0
; I n ,rms 
In,rms
2.0
0.34
0.13
In
2
FOURIER COMPONENTS OF TRANSIENT RESPONSE I(L_L)
DC COMPONENT = -3.668708E-06
HARMONIC FREQUENCY FOURIER NORMALIZED PHASE
NORMALIZED
NO
(HZ)
COMPONENT COMPONENT (DEG)
PHASE (DEG)
1
2
3
4
5
6
7
8
9
1.200E+02
2.400E+02
3.600E+02
4.800E+02
6.000E+02
7.200E+02
8.400E+02
9.600E+02
1.080E+03
2.830E+00
5.377E-06
4.778E-01
3.589E-06
1.818E-01
2.858E-06
9.427E-02
2.523E-06
5.743E-02
1.000E+00 -3.716E+01 0.000E+00
1.900E-06 -1.203E+02 -4.594E+01
1.688E-01 -6.658E+01 4.490E+01
1.268E-06 -1.223E+02 2.629E+01
6.422E-02 -7.587E+01 1.099E+02
1.010E-06 -1.162E+02 1.068E+02
3.331E-02 -8.028E+01 1.798E+02
8.913E-07 -1.095E+02 1.878E+02
2.029E-02 -8.292E+01 2.515E+02
TOTAL HARMONIC DISTORTION = 1.847695E+01 PERCENT
8-7)
Using a restricted time interval of 33.33 ms to 50 ms to analyze steady-state current, the peak
value is 8.26 A and the rms value is 4.77 A. The THD from the output file is 32%.
20A
10A
Peak
(35.134m,8.2603)
rms
(50.000m,4.7738)
i(t)
0A
-10A
30ms
I(R)
35ms
RMS(I(R))
40ms
45ms
50ms
Time
voltage (100 V)
10
current
S1, S2
D3, D4
0
D1, D2
S3, S4
-10
30ms
I(R)
35ms
V(A)/10
0
40ms
Time
45ms
50ms
8-8)
 4Vdc
 cos   
  
a ) V1  
 90 
 V1 
1
 55.6

  cos 

 4Vdc
 4  125 
  cos 1 
 4Vdc
Vn  
 cos  n 
 n 
n
1
3
5
;
Z n  R  jn0 L ; I n 
|Vn|
90
51.6
4.43
THDI 
8-9)
V1 
a)
Vn
Zn
; I n ,rms 
Zn
12.5
24.7
39
In,rms
5.08
1.5
0.08
1.52  0.082
 0.29  29%
5.08
4Vdc 4  200 

 255 V .


Z1  R  j0 L  10  j 2 60  0.035   16.6 
V1 255

 15.3 A.
Z1 16.6
I1 
I1,rms 
b)
15.3
2
 10.9 A.
At 30 Hz ,
Z1  10  j 2 30  0.035   12.0 
V1  I1Z1   15.3  12.0   184 V .
 4Vdc
 cos   
  
V1  
8-10)
 184 
 V1 
1
 43.7

  cos 

4
200


 4Vdc


   cos 1 
In
2
α = 30°
Using the FFT function in Probe shows that voltages at frequencies at multiples of n = 3 are
absent.
b) α = 15°
Using the FFT function in Probe shows that voltages at frequencies at multiples of n = 5 are
absent.
8-11)
From Eq. (8-22),

90 90

 12.86
n
7
Using the FFT function in Probe, the n = 7 harmonic is absent.
8-12)
Letting T = 360 seconds and taking advantage of half-wave symmetry,
Vrms 
2 

360 
Vrms  Vm
8-13)
54

30
114
Vm2 dt 

66

150
Vm2 dt 
V
2
m dt
126


1
  54  30    114  66    150  126   0.730Vm
180 
The VPWL_FILE source is convenient for this simulation. A period of 360 seconds is used,
making each second equal to one degree. A transient simulation with a run time of 360 second
and a maximum step size of 1m gives good results. The FFT of the Probe output confirms that the
3rd and 5th harmonics and their multiples are eliminated.
0
30
30.01
54
54.01
66
66.01
114
114.01
126
126.01
150
150.01
210
210.01
234
234.01
246
246.01
294
294.01
306
306.01
330
330
360
0
0
1
1
0
0
1
1
0
0
1
1
0
0
-1
-1
0
0
-1
-1
0
0
-1
-1
0
0
8-14)
a)
b)
Vm 
4Vdc
 cos  n1   cos  n 2   cos  n 3 
n 
Vdc  48 V ; 1  15 ;  2  25 ;  3  55
n
Vn
c) M i 
1
149.5
3
0
cos  1   cos   2   cos   3 
3
5
-2.79
7
-3.04
9
-14.4
 0.815
8-15)
To eliminate the third harmonic,
cos(31 )  cos(3 2 )  cos(3 3 )  0
Select two of the angles and solve for the third.
Examples:
α1
15
20
10
10
α2
25
30
30
30
α3
55
40
50
70
Mi
0.815
0.857
0.831
0.731
8-16)
This inverter is designed to eliminate harmonics n = 5, 7, 11, and 13. The normalized coefficients
through n = 17 are
n
Vn/Vdc
1
4.4593
3
-0.8137
5
0.0057 ≈ 0
7
-0.0077 ≈ 0
9
-0.3810
11
0.0043 ≈ 0
13
-0.0078 ≈ 0
15
-0.0370
17
0.1725
The coefficients are not exactly zero for those harmonics because of rounding of the angle values.
8-17)
8-18)
V1  V1,rms 2  54 2  76.8 V .
ma 
V1 76.8

 0.8
Vdc
96
Z n  R  jn0 L  32  jn 2 60  .024   32  jn9.05
From Table 8-3,
mf
mf - 2
mf + 2
n
1
17
15
19
Vn/Vdc
0.8
0.82
0.22
0.22
Vn
76.8
78.7
21.1
21.1
Zn
33.3
157
139
175
In=Vn/Zn
2.30
0.50
0.151
0.121
2
2
0.121
 0.50  0.151
 


 
 

2 
2 

 2
THD 
2.30
2
2
 0.23  23%
8-19)
V1  V1, rms 2  160 2  226.3 V .
ma 
V1 226.3

 0.9
Vdc
250
Z n  R  jn0 L  20  jn 2 60  .050   20  jn18.85
From Table 8-3,
mf
mf - 2
mf + 2
2
n
1
31
29
33
Vn/Vdc
0.9
0.71
0.27
0.27
2
0.108
 0.305  0.122
 


 
 

2 
2 

 2
THD 
8.18
2
Vn
225
178
67
67
Zn
27.5
585
547
622
2
 0.044  4.4%
In=Vn/Zn
8.18
0.305
0.122
0.108
8-20)
The circuit “Inverter Bipolar PWM Function” is suitable to verify the design results. The
parameters are modified to match the problem values.
Transient Analysis and Fourier Analysis are establish in the Simulation Setup menu:
The output file contains the THD of the load current, verifying that the THD is less than 10%.
TOTAL HARMONIC DISTORTION = 9.387011E+00 PERCENT
8-21)
Example solution:
Let ma  0.9,
Vdc 
Vm 120 2

 189 V .
ma
0.9
Using Table 8-3, at n  m f , Vmf  0.71 189   134 V .
for THD  8%, I mf  0.08I1
I1 
V1
120 2
120 2


 13.6 A.
Z1 10  j 2 60  0.020 
12.5
I mf  0.08  13.6   1.09 A.
Z mf 
mf 
Vmf
I mf

134
 123   m f 0 L
1.09
123
123

 16.4
0 L 377  0.020 
Choose odd integer 19 or greater for m f .
8-22)
Example solution:
V1  V1,rms 2  100 2  141 V .
Let ma  0.9  Vdc 
I1 
V1 141

 157 V .
ma 0.9
V1
V1
141


 4.48 A.
Z1 R  j0 L 30  j 377  0.025 
THDI 
Z mf 
mf 
I mf
I1
Vmf
I mf

 0.10

0.71 157 
0.448
I mf  0.1 4.48   0.448 A.
 249   m f 0 L
249
 26.4
377  0.025 
Choose odd integer 29 or greater for m f .
8-23)
Use the bipolar PWM function circuit of Fig. 8-23a, and use the unipolar PWM function circuit
of Fig.8-26 with mf = 10. Use ma = 0.8 for V1 = 120 V from the 150-V dc source.
The THD for bipolar, mf = 21, is 10.2 %, for bipolar mf = 41 is 5.2%, and for unipolar mf = 10 is
5.9%.
Bipolar mf = 21:
Bipolar mf = 41:
Unipolar, mf = 10:
8-24)
a ) V1, L  N 
I1 
V1
159
159


 6.09 A.
Z1 25  j 377  0.020  26.1
I1,rms 
8-25)
2Vdc 
 
 2   
 500
 2  cos    cos
 

 3  159 V .
3 
 3

3  
 3
I1
2
 4.31 A.
Use Eq. (8-42) for Vn,L-N , Z n  R  jn2 fL , I n  Vn, L- N / Z n , and I n, rms  I n / 2.
For f = 25 Hz:
n
1
5
7
11
13
THDI 
VnL-N
255
50.9
36.4
23.1
19.6
Zn
11.1
25.6
34.5
52.8
62.0
In
23.0
2.0
1.06
0.44
0.32
In,rms
16.3
1.41
0.75
0.31
0.22
1.412  0.752  0.312  0.222
 0.10  10%
16.3
THDV 
For f = 100 Hz,
n
1
5
7
11
13
THDI 
50.92  36.42  0.222  19.62
 0.273  27.3%
255
VnL-N
255
50.9
36.4
23.1
19.6
Zn
21.3
94.8
132
208
245
In
11.9
0.54
0.27
0.12
0.08
In,rms
8.43
0.38
0.19
0.08
0.06
0.382  0.192  0.082  0.062
 0.0519  5.19%
8.43
The THD for current is reduced from 10% to 5.19% as f is increased from 25 Hz to 100 Hz. The
THD of the line-to-neutral voltage remains at 27.3%.
These results can also be determined from a PSpice simulation for the six-step inverter.
8-26)


At f  30 Hz , Z1  10.7 , V1  I1Z1  10 2  10.7   151 V .
V1, L  N 
Vdc 
2Vdc 
 
 2  
 2  cos    cos
   Vdc  0.637 
3 
 3

3 
V1, L  N
0.637

151
 237 V .
0.637


At f  60 Hz , Z1  19.5 , V1  I1Z1  10 2  19.5   276 V
Vdc 
276
 433 V .
0.637
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
CHAPTER 10 SOLUTIONS
3/20/10
10-1)
a) For the elementary MOSFET drive circuit, losses can be determined from the energy
absorbed by the transistor. In Probe, the integral of instantaneous power is obtained by
entering the expression S(W(M1)) to get the energy absorbed by the transistor. For turnoff losses, restrict the data to 2.5 µs to 4.3 µs. The energy absorbed is 132 µJ. For turnon losses, restrict the data to 5 µs to 5.6 µs. The energy absorbed by the MOSFET is 53.3
µJ. Power is determined as
T
1
1

 5 s
f s 20000
W 132  J

 26.4 W .
T
5 s
W 53.3  J
 
 10.7 W .
T
5 s
Pturn off 
Pturn on
For the emitter-follower drive circuit restrict the data to 2.5 µs to 2.9 µs, giving 21.3 µJ
for turn-off. Restrict the data to 5 µs to 5.3 µs, giving 12.8 µJ for turn-on. Power is then
W 21.3  J

 4.26 W .
T
5 s
W 12.8  J
 
 2.56 W .
T
5 s
Pturn off 
Pturn on
b) For the first circuit, peak gate current is 127 mA, average gate current is zero, and rms
gate current is 48.5 mA. For the second circuit, peak gate current is 402 mA (and -837
mA), average gate current is zero, and rms gate current is 109 mA.
10-2)
For R1 = 75 Ω, toff ≈ 1.2 μs, ton ≈ 0.6 μs, and PMOS ≈ 30 W.
For R1 = 50 Ω, toff ≈ 0.88 μs, ton ≈ 0.42 μs, and PMOS ≈ 22 W.
For R1 = 25 Ω, toff ≈ 0.54 μs, ton ≈ 0.24 μs, and PMOS ≈ 14 W.
Reducing drive circuit resistance significantly reduces the switching time and power loss
for the MOSFET.
10-3)
The values of Vi, R1, R2, and C must be selected for the BJT base drive circuit. First,
select Vi: let Vi=20 V. then, the value of R1 is determined from the initial current spike
requirement. Solving for R1 in Eq. 10-1,
R1 
Vi  vBE 20  1

 3.8 
I B1
5
The steady-state base current in the on state determines R2. From Eq. 10-2,
R2 
Vi  vBE
20  1
 R1 
 3.8  34.2 
I B2
0.5
The value of C is determined from the required time constant. For a 50% duty ratio at
100 kHz, the transistor is on for 5 µs. Letting the on time for the transistor be five time
constants, τ = 1µs. From Eq. 10-3,

R1 R2 
 C  (3.42)C  1  s
 R1  R2
C  0.292  F
  RE C  
10-4)
The values of Vi, R1, R2, and C must be selected for the BJT base drive circuit. First,
select Vi; let Vi = 20 V. then, the value of R1 is determined from the initial current spike
requirement. Solving for R1 in Eq. 10-1,
R1 
Vi  vBE 20  1

 6.33 
I B1
3
The steady-state base current in the on state determines R2. From Eq. 10-2,
R2 
Vi  vBE
20  1
 R1 
 6.33  25.3 
I B2
0.6
The value of C is determined from the required time constant. For a 50% duty ratio at
120 kHz, the transistor is on for 4.17 µs. Letting the on time for the transistor be five
time constants,
τ = 1 µs. From Eq. 10-3,

R1 R2 
 C  5.06 C  0.833  s
 R1  R2
C  0.165  F
  RE C  
10-5)
a) From Eq. 10-5 through 10-7 for t < tf,


t
t 
iQ  I L  1    4  1 
 4  8(10) 6 t
6


t f
0.5(10) 


I t
4t
iC  L 
 8(10)6 t
tf
0.5(10) 6
I Lt 2
4t 2
vC (t ) 

 8(10)13 t 2
6
6
2Ct f 2(0.05)(10) (0.5)(10)
For tf < t < tx,
iQ  0
iC  I L  4
vC 
I t
IL
(t  t f )  L f  8(10)7 (t  0.5(10) 6 )  20
C
2C
Time tx is defined as when the capacitor voltage reaches Vs (50 V.):
vC (t x )  Vs  50  8(10)7 (t x  0.5(10)6 )  20  t x  0.875 s
b) With tx > tf, the waveforms are like those in Fig. 10.12(d).
c) Turn-off loss is the switch is determined from Eq. 10-12,
PQ 
I L2t 2f f
24C

42 [0.5(10)6 ]2 (120000)
 0.4 W .
24(0.05)(10) 6
Snubber loss is determined by the amount of stored energy in the capacitor that will be
transferred to the snubber resistor:
1
0.05(10) 6 (50) 2 (120000)
PR  CVs2 f 
 7.5 W .
2
2
10-6)
Switch current is expressed as


t
t 
iQ  I L  1    4  1 
 4  8(10)6 t
6


t f
0.5(10) 


I t
4t
iC  L 
 8(10)6 t
6
tf
0.5(10)
vC (t ) 
I Lt 2
4t 2

 4(10)14 t 2
6
6
2Ct f 2(0.01)(10) (0.5)(10)
Capacitor voltage at t = tf = 0.5 µs would be 100 volts, which is greater than V s.
Therefore, the above equations are valid only until vC reaches Vs:
vC (t x )  Vs  50  4(10)14 t x2  t x  0.354  s
For tx < t < tf,
iQ  4  8(10)6 t
iC  0
vC  Vs  50
b) With tx < tf, the waveforms are like those of Fig. 10.12(b).
Equation 10-12 is not valid here because tx < tf. Switch power is determined from
T
T
T
1
1
PQ   p (t )dt   iQ vQ dt  f  iQ vC dt 
T 0
T 0
0


6






4

8
10
t
4(10)
t
dt

4

8
10
t
(50)
dt
  1.84 W .






 


 0

tx
120000 
tf
tx
6
14 2
Snubber loss is determined by the amount of stored energy in the capacitor that will be
transferred to the snubber resistor:
1
0.01(10 6 (50) 2 (120000)
PR  CVs2 f 
 1.5 W .
2
2
10-7)
C
I Lt f
2Vs

10(0.1)(10) 6
 3.33 nF .
2(150)
ton D / f
0.4 /100000


 240 
5C
5C
(5)3.33(10) 9
1
PR  (3.33(10) 9 (150) 2100000  3.75 W .
2
I 2t 2 f 102 [0.1(10) 6 ]2105
PQ  L f 
 1.25 W .
24C
24(3.33)(10) 9
R
10-8)
C
I Lt f
2V f

10(0.1)(10) 6
 6.67 nF .
2(75)
ton D / f
0.4 /100000


 120 
5C
5C
5(6.67)(10) 9
1
1
PR  CVs2 f  (6.67)(10)9 (150) 2100000  7.5 W .
2
2
2
I t f 102 [.1(10) 6 ]2105
PQ  L f 
 0.625 W .
24C
24(6.67)(10) 7
R
10-9)
C
I Lt f
2Vs

7(0.5)(10) 6
 10.3 nF .
2(170)
ton D / f
0.4 /125000


 62.2 
5C
5C
5(10.3)(10) 9
1
1
PR  CVs2 f  (10.3)(10) 9 (170) 2125000  18.6 W .
2
2
2 2
I t f 102 [0.5(10) 6 ]2125000
PQ  L f 
 6.2 W .
24C
24(10.3)(10) 9
R
10-10)
C
I Lt f
2V f

7(0.5)(10)6
 14.0 nF .
2(125)
ton D / f 0.4 /125000


 45.7 
5C
5C
5(14)(10) 9
1
1
PR  CVs2 f  (14)(10)9 (170) 2125000  25.3 W .
2
2
2 2
I t f 102 [0.5(10) 6 ]2125000
PQ  L f 
 4.56 W .
24C
24(14)(10) 9
R
10-11)
Using the snubber circuit of Fig. 10-12(a), Eq. 10-12 yields
I L2t 2f f
52 [0.5(10) 6 ]2 200000
C

 52.1 nF .
24 PQ
24(1)
ton D / f 0.35 / 200000


 6.72 
5C
5C
5(52.1)(10) 9
1
1
PR  CVs2 f  (52.1)(10) 9 (80) 2 200000  33.3 W .
2
2
R
10-12)
Using the snubber circuit of Fig. 10-12(a), Eq. 10-12 yields
C
I L2t 2f f
24 PQ

62 [1(10) 6 ]2100000
 75 nF .
24(2)
ton D / f 0.3 /100000


 8.0 
5C
5C
5(75)(10) 9
1
1
PR  CVs2 f  (75)(10) 9 (120) 2100000  54 W .
2
2
R
10-13)
a ) From Eq. (10-16), TJ  TA  P  R , JA   30  2  40   110 C
b) P 
TJ  TA 150  30

3W
R , JA
40
10-14)
a ) From Eq. (10-16), TJ  TA  P  R , JA   25  1.5  55   107.5 C
b) P 
TJ  TA 175  25

 2.73 W
R , JA
55
10-15)
TJ  P  R , JC  R ,CS  R , SA   TA  10  1.1  0.9  2.5   40  85 C
10-16)
TJ  P  R , JC  R ,CS  R , SA   TA  5  1.5  1.2  3.0   25  53.5 C
10-17)
TJ  P  R , JC  R ,CS  R , SA   TA
R , SA 
TJ  TA
110  40
 R , JC  R ,CS 
 0.7  1.0  2.19 C / W
P
18
10-18)
From Fig. 10.24 using the bottom curve for a single pulse,
Z , JC  0.013 C / W for a pulse of 10 5 sec.
TJ  Pdm Z , JC   500 W   0.013 C / W   6.5 C
10-19)
a) For 50 kHz and D = 0.1, the pulse width is 2s. From Fig. 10.24, Z , JC  0.11 C / W .
TJ  Pdm Z , JC   100   0.11  11 C
b) Using R  , JC  1.05 C / W ,
TJ  Pavg R , JC   Pdm D  R , JC  100  0.1 1.05  10.5 C.
Note that the value of Z from the graph is very rough,
and more precise evaluation in (a) is closer to the 10.5 of part (b).