ISM Introduction to Power Elecronics - Hart PDF

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CHAPTER 1 SOLUTIONS(1-1)(1-2)(1-3)Time0s 5us 10us 15us V(V2:-)-20V0V20V40V(3,-1)(3,-1)96,23)(1-4)T i m e0 s 2 u s 4 u s 6 u s 8 u s 1 0 u s 1 2 u s 1 4 u s 1 6 u s V ( V 2 : - ) 5 V 0 V5 V1 0 V1 5 V2 0 V2 5 V( 3. 8 3 3 3 u , - 1. 0 5 1 7 )( 8 0 0. 0 0 0 n , 2 3. 9 2 4 )( ) ( )( ) ( )6 10 140 0 6 107...

Description

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

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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

=

[170 sin ( 377t ) ]2 10

= 2890 sin2 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 50ms < t < 70ms 70 ms < t < 100 ms

0  p ( t ) = v ( t ) i ( t ) = 40 0  b)

P=

1T 1 v ( t ) i ( t ) dt = ∫ T 0 100 ms

70ms

∫

40 dt = 8.0 W .

50 ms

c) T

W = ∫ p ( t ) dt = 0

70ms

∫ 40 dt = 800 mJ .;

or W = PT = ( 8W ) ( 100 ms ) = 800 mJ .

50ms

_____________________________________________________________________________________ 2-5)

a)

70 W . − W  50 . 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)

1 P= T c)

T

1 ∫0 p ( t ) dt = 20ms

10 ms 14 ms  6 ms   ∫ 70 dt + ∫ ( −50 ) dt + ∫ 40 dt = 19 W .  0  6ms 10ms

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10 ms 14 ms T  6 ms  W = ∫ p ( t ) dt =  ∫ 70 dt + ∫ ( −50 ) dt + ∫ 40 dt = 0.38 J .; 0 6ms 10ms  0 

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

PR =

1 p ( t ) dt = 12.5 W . T 0∫

b)

di ( t ) −3 = 10 ( 10 ) ( 377 ) ( 1.0 ) cos 377 t = 3.77 cos 377 t V . dt ( 3.77 ) ( 1.0 ) p L ( t ) = v ( t ) i ( t ) = ( 3.77 cos 377t ) (1.0sin 377t ) = sin 754 t =1.89sin 754 t W. 2 T 1 PL = ∫ p ( t ) dt = 0 T 0 vL ( t ) = L

c)

p ( t ) = v ( t ) i ( t ) = ( 12 ) ( 1.0sin 377t ) = 12 sin 377t W . T

Pdc =

1 p ( t ) dt = 0 T ∫0

_____________________________________________________________________________________

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2-8)

Resistor:

v ( t ) = i ( t ) R = 8 + 24sin 2π 60t V . p ( t ) = v ( t ) i ( t ) = ( 8 + 24sin 2π 60t ) ( 2 + 6sin 2π 60t ) = 16 + 96 sin 2π 60t + 144sin2 2π 60t W .

1/60 1/60  1T 1  1/60 2 P = ∫ p ( t ) dt =  ∫ 16 dt + ∫ 96sin 2π 60t dt + ∫ 144sin 2π 60t 1/ 60  0 T 0 0 0 

= 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 ) = = 1500 W . → I m = 12.5 2 2 120 2

(

)(

)

p ( t ) = Vm I m sin 2ωt = 120 2 12.5 2 sin 2ωt = 3000sin 2ωt 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 dλ = 900t ∫ L 0.1 ∫0 iL ( 4 ms ) = ( 900) ( 4) ( 10)

−3

0 < t < 4ms .

= 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. _____________________________________________________________________________________

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2-11)

a)

2 ( 1.2 ) 1 2 2W = =15.49 A. W = Li , or i = L 2 0.010 i(t) =

t

t

1 1 v ( λ) d λ = 14 d λ =1400t A. ∫ L0 0.010 ∫0 15.49 =1400t on 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,

τ≤

L 8.9 ms = = 1.7 ms.; 5 R

R≥

10 mH L = = 5.62 Ω 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

Ind uctor current

SEL>> 0A I(L1) 10A Sou rce cu rrent 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

60m s

80ms

100ms

-W(Vcc) Time

_____________________________________________________________________________________

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2-13)

a) The zener diode breaks down when the transistor turns off to maintain inductor current. b) Switch closed: 0 < t < 20 ms.

v L = 12 V . = L

diL ( t )

dt 12 diL v L = = = 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 . di L v L −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

10m s

20ms

30m s

40ms

W(D1) Time

50ms

60ms

70ms

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d)

PL = 0. T

1 1 1  PZ = ∫ p Z ( t ) dt = 0.03) ( 64) = 13.73 W . (  T0 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.

v L = 20 V . = L

di L( t )

dt 20 diL vL = = = 400 A/s dt L 0.050 at t = 15ms , iL = ( 400 ) ( 0.015 ) = 6.0 A.

Switch open, zener on:

vL = 20 − 30 = − 10 V . − 10 di L v L = = = − 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

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d)

PL = 0. T

1 1 1  PZ = ∫ p Z( t ) dt = 0.03 ) ( 180) = 36 W . (  T0 0.075  2  _____________________________________________________________________________________ 2-15) Examples are square wave (Vrms = Vm) and a triangular wave (Vrms = Vm/√3). _____________________________________________________________________________________ 2-16)

2

2

Phase conductors: Pφ = 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 Ω

_____________________________________________________________________________________ 2-17) Re: Prob. 2-4

Vrms = Vm D = 10 0.7 = 8.37 V . Irms = Im D = 4 0.5 = 2.83 A. _____________________________________________________________________________________ 2-18) Re: Prob. 2-5

 14 Vrms = Vm D = 10  ÷ = 8.36 V .  20 Irms =

1 0.02

0.006

∫ 0

7 2 dt +

0.01

0.02

0.006

0.01

2 2 ∫ ( −5 ) dt + ∫ 4 dt

= 27.7 = 5.26 A.

_____________________________________________________________________________________

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2-19) 2

Vrms =

 5   3 2 +  ÷ + ÷  2   2 2

2

2

= 4.58 V . 2

 2   1.1  I rms = 1.52 +  ÷ +  ÷ = 2.2 A .  2  2 ∞ V I P = V0I 0 + ∑ m m cos ( θ n − φn ) 2 n =1  5  2   3  1.1 = ( 2.0 ) ( 1.5 ) +  ÷  ÷ cos ( −20° ) + ÷ ÷ cos ( −115° ) = 7.0 W. 2  2 2   2  Note that − cos(4π 60t + 45° )is cos ( 4π 60t − 135° ) _____________________________________________________________________________________ 2-20)

dc : V0 = 3 ( 100 ) = 300 V .

ω1 = 2π 60 : Y1 = 1/ R + jωC = 0.01+ j0.0189 V1 =

4∠ 0 I1 = = 187∠ − 62.1° Y1 ( 0.01+ j 0.0189 )

ω 2 = 4π 60 : Y2 = 1/ R + jωC = 0.01+ j0.0377 V2 =

6∠ 0 I2 = = 153∠ − 75.1° Y2 ( 0.01 + j 0.0377 ) ∞

V mI m cos ( θ n − φn ) 2 n =1

P = V 0I 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 Pdc = Vdc I avg = 12  = 114 W .  4  Resistor:

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2 P = I rms R

Irms = I 20 + I1,2rms + I 22,rms I 0 = 9.5 A . 30 I1 = = 3.51 A. 4 + j ( 4π 60) ( 0.01) I2 =

10 = 0.641 A. j 4 + ( 8π 60) ( 0.01) 2

 3.51  0.641  Irms = 9.52 +  ÷ + ÷ 2  2  2 PR = I rms R = 386 W .

2

= 9.83 A.

_____________________________________________________________________________________ 2-22) 2 P = I rms R

V0 6 = = 0.375 A. R 16 5 I1 = = 0.269 A. 16 + j ( 2π 60) ( 0.025) I0=

I2 =

3 = 0.0923 A. 16 + j ( 6π 60 ) ( 0.025 ) 2

 0.269  0.0923  I rms = 0.375 2 +  ÷ + ÷ 2  2   2 Irms = 0.623 A.; P = I rms R = ( 0.426)

2

2

= 0.426 A.

( 16) = 2.9 W .

_____________________________________________________________________________________ 2-23) ∞

Vm Im cos ( θ n − φn ) 2 n =1

P = V0 I0 + ∑

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. _____________________________________________________________________________________

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2-24) ∞

Vm Im cos ( θ n − φ n ) 2 n =1

P = V0 I0 + ∑

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 Im cos ( θ n − φn ) 2 n =1 50 − 36 V −V I0 = 0 dc = = 0.7 A 20 R 2 P0, R = I02 R = ( 0.7 ) 20 = 9.8 W (dc component only )

P = V0 I0 + ∑

PVdc = I0Vdc = ( 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  P = ∑ Pn =  ÷  ÷ cos ( 30 ° ) + 0 + 0 = 736 W .  2 2

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b) 2

2

 10   6   3 Irms =  ÷ + ÷ + ÷  2  2   2

2

= 8.51 A.

 170 S = Vrms I rms =  ÷ 8.51= 1024 VA.  2 P 736 pf = = = 0.719 S 1024 c)

DF =

I 1,rms 10/ 2 = = 0.831 8.51 Irms

d) 2

THDI =

 6   3  ÷ + ÷  2  2 10/ 2

2

= 0.67 = 67%

_____________________________________________________________________________________ 2-28)

a)

 170 12  P = ∑ Pn =  ÷ ÷ cos ( 40 °) + 0 + 0 = 781 W .  2 2 b) 2

Irms

2

 12   5   4 =  ÷ + ÷ + ÷  2  2   2

2

= 9.62 A.

 170 S = Vrms I rms =  ÷ 9.62 = 1156 VA.  2 P 781 pf = = = 0.68 S 1156 c)

DF =

I 1,rms I rms

=

12/ 2 = 0.88 9.62

d) 2

THDI =

 5   4  ÷ + ÷  2  2 12/ 2

2

= 0.53 = 53%

_____________________________________________________________________________________ 2-29)

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8 = 5.66 A.; 2

I1,rms =

I 2,rms =

4 = 2.82 A.; 2

Irms = 5.662 + 2.822 = 6.32 A.; I peak ≈ 10.38 (graphically ) a) P = V1,rms I 1,rms cos( θ1 − φ1) = ( 240) ( 5.66) cos( 0) = 1358 W . b) pf =

1358 P P = = = 0.895 = 89.5% S Vrms I rms ( 240 ) ( 6.32 )

c) THDI =

d) DF =

I2, 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 Irms

=

10.38 = 1.64 6.32

_____________________________________________________________________________________ 2-30)

I1,rms =

12 = 8.49 A.; 2

I 2,rms =

9 = 6.36 A.; 2

Irms = 8.492 + 6.362 = 10.6 A.; I peak ≈ 18.3 A. (graphically ) a) P = V1,rms I 1,rms cos( θ1 − φn ) = ( 240) ( 10.6) cos( 0) = 2036 W . b) pf =

2036 P P = = = 0.80 = 80% S V rmsI rms ( 240 ) ( 10.6 )

c) THDI =

d) DF =

I2, rms I rms

I1, rms I rms

=

=

6.36 = 0.60 = 60% 10.6

8.49 = 80% 10.6

e) crest factor =

I peak Irms

=

18.3 = 1.72 10.6

_____________________________________________________________________________________

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2-31)

5V: I = 0 (capacitor is an open circuit) 25cos(1000 t): Z = R + jω L − j I=

1 1 = 2 + j1000(.001) − j = 2 + j0 1000 ( 1000 ) 10−6 ωC

25 cos(1000 t) =12.5cos(1000 t) A 2

10cos(2000t): Z = 2 + j1.5 Ω I 10 =

10 = 4∠ − 37° A. 2 + j1.5 2

2

 12.5  4 I rms =  ÷ + ÷ = 9.28 A 2  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.

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Energy (20.000m,1.2000)

2.0 0 S(W(I1)) 400W

Avg Power (20.000m,60.000)

0W Ins t Powe r -400W W(I 1)

AVG(W(I1) )

I(I 1)

4 ms V(V1:+)

20 0 SEL >> -20 0s

8ms

12m s

16ms

20ms

Time

_____________________________________________________________________________________

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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

(16.670m,0.9998K) Resistor

1.0KW

Inductor

(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

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2.0KW

Instantaneous Power

Resistor 1.0KW

Inductor

0W Vdc

-1.0KW 0s W(R 1)

W(L1)

5ms W(V1)

10ms

15ms

20ms

Time

_____________________________________________________________________________________

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2-34)

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Rms voltage is 8.3666 V. Rms current is 5.2631 A.<...