Power Electronics, 1st Edition Solution Manual

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   v 2  t  [170sin  377t ]2   2890sin 2 377t W . R 10 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  p  t   v  t  i  t   40 0  0  t  50 ms 50 ms  t  70 ms 70 ms  t  100 ms b) T 70 ms 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) 6 ms 10 ms 14 ms T  1 1  p t dt  70 dt   50 dt  40 dt   19 W .         T 0 20 ms  0  6 ms 10 ms 10 ms 14 ms T  6 ms  W   p  t  dt    70 dt    50  dt   40 dt   0.38 J .; 0 6 ms 10 ms  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 1 PR   p  t  dt  12.5 W . T 0 b) di  t  3  10 10   377 1.0  cos 377t  3.77 cos 377t V . dt  3.77 1.0  sin 754t  1.89sin 754t W . pL  t   v  t  i  t    3.77 cos 377t 1.0sin 377t   2 T 1 PL   p  t  dt  0 T 0 vL  t   L c) p  t   v  t  i  t   12 1.0sin 377t   12sin 377t W . T Pdc  1 p  t  dt  0 T 0 _____________________________________________________________________________________ 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 . P T 1/60 1/60 1/60  1 1  p t dt  16 dt  96sin 2  60 t dt  144sin 2 2 60t       T 0 1/ 60  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 1500  2   12.5 2 Vm I m  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 0  t  4 ms. iL  4 ms    900  4 10   3.6 A. 3 a) 1 1 2 W  Li 2   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. _____________________________________________________________________________________ 2-11) a) W 2 1.2  1 2 2W Li , or i    15.49 A. 2 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 > 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 _____________________________________________________________________________________ 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 : t  i 3.2   30 ms 106.7 106.7 Therefore, inductor current returns to zero at 20 + 30 = 50 ms. iL = 0 for 50 ms < t > 0W 0s 10ms 20ms 30ms 40ms W(D1) Time 50ms 60ms 70ms d) PL  0. 1 1 1 PZ   pZ  t  dt   0.03 64   13.73 W .  T 0 0.07  2  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. 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 : t  i 6.0   30 ms 200 200 Therefore, inductor current returns to zero at 15 + 30 = 45 ms. iL = 0 for 45 ms < t > 0W 0s 20ms 40ms W(D1) Time 60ms 80ms d) PL  0. 1 1 1 PZ   pZ  t  dt   0.03180   36 W .  T 0 0.075  2  T _____________________________________________________________________________________ 2-15) Examples are square wave (Vrms = Vm) and a triangular wave (Vrms = Vm/√3). _____________________________________________________________________________________ 2-16) Phase conductors: P  I R  12  0.5  72 W . 2 2  Neutral conductor: PN  I R  12 3 2 Ptotal  3  72   216  432 W . RN    0.5  216 W . 2 PN 72   0.167  2 2 IN 12 3   _____________________________________________________________________________________ 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  Vrms  Vm D  10    8.36 V .  20  0.006 I rms  0.01 0.02 1 2 7 2 dt    5  dt   42 dt  27.7  5.26 A.  0.02 0 0.006 0.01 _____________________________________________________________________________________ 2-19) 2 2  5   3  Vrms  22       4.58 V .  2  2 2 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    2.0 1.5      cos  20      cos  115   7.0 W .  2  2   2  2  Note that  cos(4 60t  45) is cos  4 60t  135  _____________________________________________________________________________________ 2-20) dc : V0  3 100   300 V . 1  2 60 : Y1  1/R  jC  0.01  j 0.0189 I 40 V1  1   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   V I P  V0 I 0   m m cos  n  n  2 n 1  300  5   187  4  cos 62.1  153 6  cos 75.1     2 2  1500  175  118  1793 W . _____________________________________________________________________________________ 2-21) dc Source:  50  12  Pdc  Vdc I avg  12   114 W .  4  Resistor: 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 2  3.51   0.641  I rms  9.5       9.83 A.  2   2  2 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 2  0.269   0.0923  I rms  0.375       0.426 A. 2   2   2 2 I rms  0.623 A.; P  I rms R   0.426  16   2.9 W . 2 _____________________________________________________________________________________ 2-23)  V I P  V0 I 0   m m cos  n  n  2 n 1 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. _____________________________________________________________________________________ 2-24)  V I P  V0 I 0   m m cos  n  n  2 n 1 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)  V I P  V0 I 0   m m cos  n  n  2 n 1 V V 50  36 I 0  0 dc   0.7 A R 20 P0, R  I 02 R   0.7  20  9.8 W (dc component only ) 2 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) THD = 5% → I9 = (0.05)(10) = 0.5 A. b) THD = 10% → I9 = (0.10)(10) = 1 A. c) THD = 20% → I9 = (0.20)(10) = 2 A. d) THD = 40% → I9 = (0.40)(10) = 4 A. _____________________________________________________________________________________ 2-27) a)  170  10  P   Pn     cos  30   0  0  736 W .  2  2  b) 2 2 2  10   6   3  I rms         8.51 A.  2  2  2  170  S  Vrms I rms    8.51  1024 VA.  2 P 736 pf    0.719 S 1024 c) I 10/ 2 DF  1,rms   0.831 I rms 8.51 d) 2 2  6   3       2  2 THDI   0.67  67% 10/ 2 _____________________________________________________________________________________ 2-28) a)  170  12  P   Pn     cos  40   0  0  781 W .  2  2  b) 2 2 2  12   5   4  I rms         9.62 A.  2  2  2  170  S  Vrms I rms    9.62  1156 VA.  2 P 781 pf    0.68 S 1156 c) I 12/ 2 DF  1,rms   0.88 I rms 9.62 d) 2 2  5   4       2  2 THDI   0.53  53% 12/ 2 _____________________________________________________________________________________ 2-29) 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 . P P 1358    0.895  89.5% S Vrms I rms  240  6.32  b) pf  c) THDI  I 2,rms I rms I1,rms d) DF  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 _____________________________________________________________________________________ 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 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. 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 _____________________________________________________________________________________ 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 2.0KW Instantaneous Power Resistor 1.0KW Inductor 0W Vdc -1.0KW 0s W(R1) W(L1) 5ms W(V1) 10ms 15ms 20ms Time _____________________________________________________________________________________ 2-34) 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. 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. _____________________________________________________________________________________ 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 _____________________________________________________________________________________ 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 Zener Diode Current SEL>> 0A 0s 10ms 20ms 30ms 40ms 50ms 60ms 70ms -I(D1) 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. _____________________________________________________________________________________ 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. 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 _____________________________________________________________________________________ 2-40) DESIRED QUANTITY ORIGINAL RESULT NEW VALUES Inductor Current Energy Stored in Inductor Average Switch Power Average Source Power (absorbed) Average Diode Power 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. AVG(W(D1)) 0 0 19.9 W. 1.99 J. .046 J. 0 0.998 A. 4.39 A 1.93 L 0.66 W -19.9 W .449 W 0.464 W. 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. 1.0A (100.000m,577.350m) 0A -1.0A 0s 20ms -I(R1) 40ms RMS(I(R1)) 60ms 80ms 100ms Time _____________________________________________________________________________________