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Power Electronics, 1st Edition Solution Manual

Power Electronics, 1st Edition Solution Manual simplifies textbook problem-solving by offering clear and effective solutions to every problem.

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    CHAPTER 1 SOLUTIONS ( 1 - 1 )

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    (1 - 2 )

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    T i m e 0 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 ( D 1 : 2 ) - 5 V 0 V 5 V 1 0 V 1 5 V 2 0 V 2 5 V T i m e 0 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 ( S 1 : 4 ) - 5 V 0 V 5 V 1 0 V 1 5 V 2 0 V 2 5 V ( 4 . 0 8 3 3 u , - 8 5 1 . 6 9 0 m ) ( 1 . 4 3 3 3 u , 2 3 . 8 0 0 ) In part (b), the voltage across the current source is reduced from 24 V by the switch resistance and diode voltage drop.

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    (1 - 3) Time 0s 5us 10us 15us V(V2:-) -20V 0V 20V 40V (3.150u,-1.052) (3.150u,-1.052) 96.46n,23.94)

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    (1 - 4) T i m e 0 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 V 5 V 1 0 V 1 5 V 2 0 V 2 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 )

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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) ( ) ( ) ( ) ( ) ( ) 2 2 2 [170sin 377 ] 2890sin 377 . 10 v t t p t v t i t t W R = = = = 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 sin 2 2πt W.; P = 10 W. b) 3sin4πt A.; p(t) = 15sin(2πt)sin(4πt) W.; P = 0 _____________________________________________________________________________________ 2 - 4 ) a) ( ) ( ) ( ) 0 0 50 40 50 70 0 70 100 t ms p t v t i t ms t ms ms t ms     = =        b) ( ) ( ) 70 0 50 1 1 40 8.0 . 100 ms T ms P v t i t dt dt W T ms = = =   c) ( ) ( )( ) 70 0 50 40 800 .; 8 100 800 . ms T ms W p t dt dt mJ or W PT W ms mJ = = = = = =   _____________________________________________________________________________________ 2 - 5) a) ( ) ( ) ( ) 70 . 0 6 50 . 6 10 40 . 10 14 0 14 20 W t ms W ms t ms p t v t i t W ms t ms ms t ms     −    = =         b) ( ) ( ) 6 10 14 0 0 6 10 1 1 70 50 40 19 . 20 ms ms ms T ms ms P p t dt dt dt dt W T ms   = = + − + =           c)

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    ( ) ( ) ( )( ) 6 10 14 0 0 6 10 70 50 40 0.38 .; 19 20 380 . ms ms ms T ms ms W p t dt dt dt dt J or W PT ms mJ   = = + − + =       = = =     _____________________________________________________________________________________ 2 - 6) ( )( ) ( )( ) ) 2 ., 12 2 24 . ) 3.1 ., 12 3.1 37.2 . dc avg avg avg P V I a I A P W b I A P W = = = = = = = _____________________________________________________________________________________ 2 - 7) a) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 2 0 25sin 377 . 25sin 377 1.0sin 377 25sin 377 12.5 1 cos 754 . 1 12.5 . R T R v t i t R t V p t v t i t t t t t W P p t dt W T = = = = = = − = =  b) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( )( ) ( ) 3 0 10 10 377 1.0 cos 377 3.77 cos 377 . 3.77 1.0 3.77 cos 377 1.0sin 377 sin 754 1.89sin 754 . 2 1 0 L L T L di t v t L t t V dt p t v t i t t t t t W P p t dt T − = = = = = = = = =  c) ( ) ( ) ( ) ( )( ) ( ) 0 12 1.0sin 377 12sin 377 . 1 0 T dc p t v t i t t t W P p t dt T = = = = =  _____________________________________________________________________________________

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    2 - 8) Resistor: ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 2 1/60 1/60 1/60 2 0 0 0 0 8 24sin 2 60 . 8 24sin 2 60 2 6sin 2 60 16 96sin 2 60 144sin 2 60 . 1 1 16 96sin 2 60 144sin 2 60 1/ 60 16 72 88 . T v t i t R t V p t v t i t t t t t W P p t dt dt t dt t T W        = = + = = + + = + +   = = + +     = + =     Inductor : 0. L P = dc source: ( )( ) 2 6 12 . dc avg dc P I V W = = = _____________________________________________________________________________________ 2 - 9) a) With the heater on, ( )( ) ( ) ( )( ) ( ) ( ) 2 2 2 1500 2 1500 . 12.5 2 2 120 2 sin 120 2 12.5 2 sin 3000sin max 3000 . m m m m m V I P W I p t V I t t t p t W    = = → = = = = = = b) P = 1500(5/12) = 625 W. c) W = PT = (625 W)(12 s) = 7500 J. (or 1500(5) = 7500 W.) _____________________________________________________________________________________ 2 - 10) ( ) ( ) ( ) ( )( )( ) 0 3 1 1 90 900 0 4 . 0.1 4 900 4 10 3.6 . t L L L i t v t dt d t t ms L i ms A  − = = =   = =   a) ( )( ) 2 2 1 1 0.1 3.6 0.648 . 2 2 W Li J = = = b) All stored energy is absorbed by R: W R = 0.648 J . c) 0.648 16.2 . 40 16.2 . R R S R W P W T ms P P W = = = = = d) No change in power supplied by the source: 16.2 W. _____________________________________________________________________________________

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    2 - 11) a) ( ) ( ) ( ) 2 0 0 2 1.2 1 2 , 15.49 . 2 0.010 1 1 14 1400 . 0.010 15.49 1400 11.1 t t on on W W Li or i A L i t v d d t A L t t ms    = = = = = = = = =   b) Energy stored in L must be transferred to the resistor in (20 - 11.1) = 8.9 ms. Allowing five time constants, 8.9 10 1.7 .; 5.62 5 1.7 1.7 L ms L mH ms R R ms ms   = =  = =  _____________________________________________________________________________________ 2 - 12) a) i(t) = 1800 t for 0 < t < 4 ms i(4 ms) = 7.2 A.; W Lpeak = 1.296 J. b) Time 0s 20ms 40ms 60ms 80ms 100ms -W(Vcc) -1.0KW 0W 1.0KW Source inst. power (supplied) W(L1) -1.0KW 0W 1.0KW Ind. inst. power -I(Vcc) -10A 0A 10A Source current I(L1) 0A 5A 10A SEL>> Inductor current _____________________________________________________________________________________

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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. ( ) ( )( ) 12 . 12 160 / 0.075 20 , 160 0.02 3.2 . L L L L L di t v V L dt di v A s dt L at t ms i A = = = = = = = = Switch open, zener on: 12 20 8 . 8 106.7 / 0.075 : 3.2 30 106.7 106.7 L L L v V di v A s dt L t to return to zero i t ms = − = − − = = = −   −  = = = − − Therefore, inductor current returns to zero at 20 + 30 = 50 ms. i L = 0 for 50 ms < t < 70 ms. c) Time 0s 10ms 20ms 30ms 40ms 50ms 60ms 70ms W(D1) 0W 40mW 80mW SEL>> Zener inst. power W(L1) -40mW 0W 40mW Inductor inst. power

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    d) ( ) ( )( ) 0 0. 1 1 1 0.03 64 13.73 . 0.07 2 L T Z Z P P p t dt W T =   = = =      _____________________________________________________________________________________ 2 - 14) a) The zener diode breaks down when the transistor turns off to maintain inductor current. b) Switch closed: 0 < t < 15 ms. ( ) ( )( ) 20 . 20 400 / 0.050 15 , 400 0.015 6.0 . L L L L L di t v V L dt di v A s dt L at t ms i A = = = = = = = = Switch open, zener on: 20 30 10 . 10 200 / 0.050 : 6.0 30 200 200 L L L v V di v A s dt L t to return to zero i t ms = − = − − = = = −   −  = = = − − Therefore, inductor current returns to zero at 15 + 30 = 45 ms. i L = 0 for 45 ms < t < 75 ms. c) Time 0s 20ms 40ms 60ms 80ms W(D1) 0W 100W 200W SEL>> Zener inst. power W(L1) -200W 0W 200W Inductor inst. power

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    d) ( ) ( )( ) 0 0. 1 1 1 0.03 180 36 . 0.075 2 L T Z Z P P p t dt W T =   = = =      _____________________________________________________________________________________ 2 - 15) Examples are square wave (Vrms = Vm) and a triangular wave (Vrms = Vm/√3). _____________________________________________________________________________________ 2 - 16) Phase conductors: ( ) 2 2 12 0.5 72 . P I R W  = = = Neutral conductor: ( ) ( ) 2 2 12 3 0.5 216 . N P I R W = = = ( ) ( ) 2 2 3 72 216 432 . 72 0.167 12 3 total N N N P W P R I = + = = = =  ________ _____________________________________________________________________________ 2 - 17) Re: Prob. 2 - 4 10 0.7 8.37 . 4 0.5 2.83 . rms m rms m V V D V I I D A = = = = = = _____________________________________________________________________________________ 2 - 18) Re: Prob. 2 - 5 ( ) 0.006 0.01 0.02 2 2 2 0 0.006 0.01 14 10 8.36 . 20 1 7 5 4 27.7 5.26 . 0.02 rms m rms V V D V I dt dt dt A   = = =     = + − + = =    _____________________________________________________________________________________

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    2 - 19) ( ) ( )( ) ( ) ( ) ( ) 2 2 2 2 2 2 0 0 1 5 3 2 4.58 . 2 2 2 1.1 1.5 2.2 . 2 2 cos 2 5 2 3 1.1 2.0 1.5 cos 20 cos 115 7.0 . 2 2 2 2 cos(4 60 45 ) cos 4 60 135 rms rms m m n n n V V I A V I P V I W Note that t is t      =     = + + =             = + + =         = + −       = + −  + −  =             − +  −   _____________________________________________________________________________________ 2 - 20) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) 0 1 1 1 1 1 2 2 2 2 2 0 0 1 : 3 100 300 . 2 60 : 1/ 0.01 0.0189 4 0 187 62.1 0.01 0.0189 4 60 : 1/ 0.01 0.0377 6 0 153 75.1 0.01 0.0377 cos 2 187 4 153 6 300 5 cos 62.1 cos 7 2 2 m m n n n dc V V Y R j C j I V Y j Y R j C j I V Y j V I P V I          = = = = = + = +  = = =  −  + = = + = +  = = =  −  + = + − = +  +  ( ) 5.1 1500 175 118 1793 . W  = + + = _____________________________________________________________________________________ 2 - 21) dc Source: 50 12 12 114 . 4 dc dc avg P V I W −   = = =     Resistor:

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    ( )( ) ( )( ) 2 2 2 2 0 1, 2, 0 1 2 2 2 2 2 9.5 . 30 3.51 . 4 4 60 0.01 10 0.641 . 4 8 60 0.01 3.51 0.641 9.5 9.83 . 2 2 386 . rms rms rms rms rms R rms P I R I I I I I A I A j I A j I A P I R W   = = + + = = = + = = +     = + + =         = = _____________________________________________________________________________________ 2 - 22) ( )( ) ( )( ) ( ) ( ) 2 0 0 1 2 2 2 2 2 2 6 0.375 . 16 5 0.269 . 16 2 60 0.025 3 0.0923 . 16 6 60 0.025 0.269 0.0923 0.375 0.426 . 2 2 0.623 .; 0.426 16 2.9 . rms rms rms rms P I R V I A R I A j I A j I A I A P I R W   = = = = = = + = = +     = + + =         = = = = _____________________________________________________________________________________ 2 - 23) ( ) 0 0 1 cos 2 m m n n n V I P V I    = = + −  n V n I n P n ∑P n 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) ( ) 0 0 1 cos 2 m m n n n V I P V I    = = + −  n V n I n θ n - ϕ n ° P n 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, ∑P n = 753 W. _____________________________________________________ ________________________________ 2 - 25) ( ) 0 0 1 cos 2 m m n n n V I P V I    = = + −  ( ) ( )( ) 0 0 2 2 0, 0 0 50 36 0.7 20 0.7 20 9.8 ( ) 0.7 36 25.2 0 dc R Vdc dc L V V I A R P I R W dc component only P I V W P − − = = = = = = = = = = Resistor Average Power n V n Z n I n angle P n 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 P R = ∑ P n ≈ 300 W. _____________________________________________________________________________________ 2 - 26) a) THD = 5% → I 9 = (0.05)(10) = 0.5 A. b) THD = 10% → I 9 = (0.10)(10) = 1 A. c) THD = 20% → I 9 = (0.20)(10) = 2 A. d) THD = 40% → I 9 = (0.40)(10) = 4 A. _____________________________________________________________________________________ 2 - 27) a) ( ) 170 10 cos 30 0 0 736 . 2 2 n P P W    = =  + + =       

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