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Eletronica_Potencia_-_Daniel_Hart_-_Solutions_Manual/HartChapter10solutions.doc 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 turn-off losses, restrict the data to 2.5 µs to 4.3 µs. The energy absorbed is 132 µJ. For turn-on 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 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 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, The steady-state base current in the on state determines R2. From Eq. 10-2, 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, 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, The steady-state base current in the on state determines R2. From Eq. 10-2, 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, 10-5) a) From Eq. 10-5 through 10-7 for t < tf, For tf < t < tx, Time tx is defined as when the capacitor voltage reaches Vs (50 V.): 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, Snubber loss is determined by the amount of stored energy in the capacitor that will be transferred to the snubber resistor: 10-6) Switch current is expressed as Capacitor voltage at t = tf = 0.5 µs would be 100 volts, which is greater than Vs. Therefore, the above equations are valid only until vC reaches Vs: For tx < t < tf, 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 Snubber loss is determined by the amount of stored energy in the capacitor that will be transferred to the snubber resistor: 10-7) 10-8) 10-9) 10-10) 10-11) Using the snubber circuit of Fig. 10-12(a), Eq. 10-12 yields 10-12) Using the snubber circuit of Fig. 10-12(a), Eq. 10-12 yields 10-13) 10-14) 10-15) 10-16) 10-17) 10-18) 10-19) _1332947809.unknown _1332947817.unknown _1332947821.unknown _1332947825.unknown _1332947827.unknown _1332947829.unknown _1332947831.unknown _1332947832.unknown _1332947830.unknown _1332947828.unknown _1332947826.unknown _1332947823.unknown _1332947824.unknown _1332947822.unknown _1332947819.unknown _1332947820.unknown _1332947818.unknown _1332947813.unknown _1332947815.unknown _1332947816.unknown _1332947814.unknown _1332947811.unknown _1332947812.unknown _1332947810.unknown _1332947805.unknown _1332947807.unknown _1332947808.unknown _1332947806.unknown _1332947803.unknown _1332947804.unknown _1332947802.unknown Eletronica_Potencia_-_Daniel_Hart_-_Solutions_Manual/HartChapter1solutions.doc CHAPTER 1 SOLUTIONS (1-1) (1-2) In part (b), the voltage across the current source is reduced from 24 V by the switch resistance and diode voltage drop. � (1-3) (1-4) Eletronica_Potencia_-_Daniel_Hart_-_Solutions_Manual/HartChapter2solutions.doc CHAPTER 2 SOLUTIONS 2/21/10 2-1) Square waves and triangular waves for voltage and current are two examples. _____________________________________________________________________________________ 2-2) a) 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) b) c) _____________________________________________________________________________________ 2-5) a) b) c) _____________________________________________________________________________________ 2-6) _____________________________________________________________________________________ 2-7) a) b) c) _____________________________________________________________________________________ � 2-8) Resistor: Inductor: dc source: _____________________________________________________________________________________ 2-9) a) With the heater on, b) P = 1500(5/12) = 625 W. c) W = PT = (625 W)(12 s) = 7500 J. (or 1500(5) = 7500 W.) _____________________________________________________________________________________ 2-10) a) b) All stored energy is absorbed by R: WR = 0.648 J. c) d) No change in power supplied by the source: 16.2 W. _____________________________________________________________________________________ � 2-11) a) b) Energy stored in L must be transferred to the resistor in (20 - 11.1) = 8.9 ms. Allowing five time constants, _____________________________________________________________________________________ 2-12) a) i(t) = 1800t for 0 < t < 4 ms i(4 ms) = 7.2 A.; WLpeak = 1.296 J. b) _____________________________________________________________________________________ � 2-13) a) The zener diode breaks down when the transistor turns off to maintain inductor current. b) Switch closed: 0 < t < 20 ms. Switch open, zener on: Therefore, inductor current returns to zero at 20 + 30 = 50 ms. iL = 0 for 50 ms < t < 70 ms. c) d) _____________________________________________________________________________________ 2-14) a) The zener diode breaks down when the transistor turns off to maintain inductor current. b) Switch closed: 0 < t < 15 ms. Switch open, zener on: Therefore, inductor current returns to zero at 15 + 30 = 45 ms. iL = 0 for 45 ms < t < 75 ms. c) d) _____________________________________________________________________________________ 2-15) Examples are square wave (Vrms = Vm) and a triangular wave (Vrms = Vm/√3). _____________________________________________________________________________________ 2-16) Phase conductors: Neutral conductor: _____________________________________________________________________________________ 2-17) Re: Prob. 2-4 _____________________________________________________________________________________ 2-18) Re: Prob. 2-5 _____________________________________________________________________________________ � 2-19) _____________________________________________________________________________________ 2-20) _____________________________________________________________________________________ 2-21) dc Source: Resistor: _____________________________________________________________________________________ 2-22) _____________________________________________________________________________________ 2-23) 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) 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) 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) b) c) d) _____________________________________________________________________________________ 2-28) a) b) c) d) _____________________________________________________________________________________ 2-29) a) b) c) d) e) _____________________________________________________________________________________ 2-30) a) b) c) d) e) _____________________________________________________________________________________ � 2-31) _____________________________________________________________________________________ 2-32) PSpice shows that average power is 60 W and energy is 1.2 J. Use VPULSE and IPULSE for the sources. _____________________________________________________________________________________ � 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). _____________________________________________________________________________________ � 2-34) � Rms voltage is 8.3666 V. Rms current is 5.2631 A. _____________________________________________________________________________________ 2-35) See Problem 2-10. 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. Quantity Probe Expression Result Inductor resistor average power AVG(W(R1)) 77.1 W Switch average power AVG(W(S1)) 3.86 W each Diode average power AVG(W(D1)) 81 mW each Source average power AVG(W(Vcc)) -85.0 W _____________________________________________________________________________________ 2-37) a) Power absorbed by the inductor is zero. Power absorbed by the Zener diode is 13.8 W. 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. 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) Quantity Probe Expression Result Power AVG(W(V1)) 650 W rms current RMS(I(V1)) 14 A Apparent power S RMS(V(I1:+))* RMS(I(V1)) 990 VA Power factor AVG(W(V1)) / (RMS(V(I1:+))* RMS(I(V1))) 0.66 _____________________________________________________________________________________ 2-40) DESIRED QUANTITY ORIGINAL RESULT NEW VALUES Inductor Current max = 4.5 A. 4.39 A Energy Stored in Inductor max = 2.025 J 1.93 L Average Switch Power 0.010 W. 0.66 W Average Source Power (absorbed) -20.3 W. -19.9 W Average Diode Power AVG(W(D1)) 0.464 W. 0.464 W. .449 W Average Inductor Power ( 0 0 Average Inductor Voltage ( 0 0 Average Resistor Power 19.9 W. 18.8 W Energy Absorbed by Resistor 1.99 J. 1.88 J Energy Absorbed by Diode .046 J. .045 J Energy Absorbed by Inductor ( 0 0 rms Resistor Current 0.998 A. 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. _____________________________________________________________________________________ _1334392376.unknown _1334392392.unknown _1334392400.unknown _1334392408.unknown _1334392412.unknown _1334392417.unknown _1334392419.unknown _1334392421.unknown _1334392422.unknown _1334392420.unknown _1334392418.unknown _1334392414.unknown _1334392415.unknown _1334392413.unknown _1334392410.unknown _1334392411.unknown _1334392409.unknown _1334392404.unknown _1334392406.unknown _1334392407.unknown _1334392405.unknown _1334392402.unknown _1334392403.unknown _1334392401.unknown _1334392396.unknown _1334392398.unknown _1334392399.unknown _1334392397.unknown _1334392394.unknown _1334392395.unknown _1334392393.unknown _1334392384.unknown _1334392388.unknown _1334392390.unknown _1334392391.unknown _1334392389.unknown _1334392386.unknown _1334392387.unknown _1334392385.unknown _1334392380.unknown _1334392382.unknown _1334392383.unknown _1334392381.unknown _1334392378.unknown _1334392379.unknown _1334392377.unknown _1334392368.unknown _1334392372.unknown _1334392374.unknown _1334392375.unknown _1334392373.unknown _1334392370.unknown _1334392371.unknown _1334392369.unknown _1334392364.unknown _1334392366.unknown _1334392367.unknown _1334392365.unknown _1334392362.unknown _1334392363.unknown _1334392361.unknown Eletronica_Potencia_-_Daniel_Hart_-_Solutions_Manual/HartChapter3solutions.doc CHAPTER 3 SOLUTIONS 2/20/10 3-1) 3-2) 3-3) 3-4) Using Eq. 3-15, 3-5) Using Eq. 3-15, 3-6) Using Eq. 3-15, 3-7) Using an ideal diode model, R = 48 Ω for an average current of 2 A. 3-8) Using Eqs. 3-22 and 3-23, � 3-9) Using Eqs. 3-22 and 3-23, 3-10) Using Eq. 3-33, � 3-11) 3-12) L ≈ 0.14 H for 50 W (51 W). 3-13) Using Eq. 3-34, a) b) n Vn Zn In 0 54.02 12.00 4.50 1 84.85 25.6 3.31 2 36.01 46.8 0.77 4 7.20 91.3 0.08 The terms beyond n = 1 are insignificant. 3-14) Run a transient response long enough to achieve steady-state results (e.g., 1000ms). The peak-to-peak load current is approximately 1.48 A, somewhat larger than the 1.35 A obtained using only the first harmonic. (The inductance should be slightly larger, about 0.7 H, to compensate for the approximation of the calculation.) 3-15) a) b) A PSpice simulation using an ideal diode model gives 0.443 A p-p in the steady state. This compares with 2(I1)=2(0.199)=0.398 A p-p. 3-16) 3-17) a) τ = RC = 10310-3=1 s; τ/T = 60. With τ >> T, the exponential decay is very small and the output voltage has little variation. b) Exact equations: c) Approximation of Eq. 3-51: � 3-18) a) R = 100 Ω: τ = RC (100)10-3 = 0.1 s; τ/T = 6. b) R = 10 Ω: τ = RC (10)10-3 = 0.01 s; τ/T = .6. In (a) with τ/T=6, the approximation is much more reasonable than (b) where τ/T=0.6. 3-19) a) With C = 4000 µF, RC = 4 s., and the approximation of Eq. 3-51 should be reasonable. b) With C = 20 µF, RC = 0.02, which is on the order of one source period. Therefore, the approximation will not be reasonable and exact equations must be used. � 3-20) a) With C = 4000 µF, RC = 2 s., and the approximation of Eq. 3-51 should be reasonable. b) With C = 20 µF, RC = 0.01, which is on the order of one source period. Therefore, the approximation will not be reasonable and exact equations must be used. 3-21) From Eq. 3-51 3-22) Assuming Vo is constant and equal to Vm, From Eq. 3-51 3-23) Using the definition of power factor and Vrms from Eq. 3-53, 3-24) 3-25) 3-26) 3-27) 3-28) α ≈ 46°. Do a parametric sweep for alpha. Use the default (Dbreak) diode, and use Ron = 0.01 for the switch. Alpha of 46 degrees results in approximately 2 A in the load. 3-29) α ≈ 60.5°. Do a parametric sweep for alpha. Use the default (Dbreak) diode, and use Ron = 0.01 for the switch. Alpha of 60.5 degrees results in approximately 1.8 A in the load. � 3-30) From Eq. 3-61, 3-31) From Eq. 3-61, � 3-32) α ≈ 75°. Alpha = 75 degrees gives 35 W in the dc voltage source. An Ron = 0.01 for the switch and n = 0.001 for the diode (ideal model). � 3-33) From Eq. 3-61, 3-34) α ≈ 81° 3-35) � 3-36) v0 = vs when S1 on, v0=0 when D2 on 3-37) PSpice: Use a current source for the constant load current: � D1 to D2 D2 to D1 3-38) u = 20°. Run the simulation long enough for steady-state results. From the Probe output, the commutation angle from D1 to D2 is about 20 degrees, and from D2 to D1 is about 18 degrees. Note that the time axis is changed to angle in degrees here. 3-39) Run the simulation long enough for steady-state results. From the Probe output, the commutation angle from D1 to D2 is about 16.5 degrees, and from D2 to D1 is about 14.7 degrees. Note that the time axis is changed to angle in degrees here. 3-40) At ωt = π, D2 turns on, D1 is on because of the current in LS (see Fig. 3-17). 3-41) At ωt = α, 3-42) A good solution is to use a controlled half-wave rectifier with an inductor in series with the 48-V source and resistance (Fig. 3-15). The switch will change the delay angle of the SCR to produce the two required power levels. The values of the delay angle depend on the value selected for the inductor. This solution avoids adding resistance, thereby avoiding introducing power losses. 3-43) Several circuit can accomplish this objective, including the half-wave rectifier of Fig. 3-2a and half-wave rectifier with a freewheeling diode of Fig. 3-7, each with resistance added. Another solution is to use the controlled half-wave rectifier of Fig. 3-14a but with no resistance. The analysis of that circuit is like that of Fig. 3-6 but without Vdc. The resulting value of α is 75°, obtain from a PSpice simulation. That solution is good because no resistance is needed, and losses are not introduced. 3-44 and 3-45) The controlled half-wave rectifier of Fig. 3-15 (without the resistance) can be used to satisfy the design specification. The value of the delay angle depends on the value selected for the inductor. _1326827428.unknown _1327078834.unknown _1328170120.unknown _1328177941.unknown _1328178073.unknown _1328178156.unknown _1328178005.unknown _1328177762.unknown _1327079924.unknown _1327081353.unknown _1327090710.unknown _1327091871.unknown _1327088878.unknown _1327081079.unknown _1327079431.unknown _1326829229.unknown _1327078632.unknown _1327078668.unknown _1327076667.unknown _1326827703.unknown _1326828757.unknown _1326827535.unknown _1326820338.unknown _1326825264.unknown _1326826894.unknown _1326827094.unknown _1326826749.unknown _1326823742.unknown _1326824632.unknown _1326820973.unknown _1326396145.unknown _1326651867.unknown _1326653677.unknown _1326396590.unknown _1326395319.unknown _1326395711.unknown _1325958918.unknown _1326394156.unknown Eletronica_Potencia_-_Daniel_Hart_-_Solutions_Manual/HartChapter4solutions.doc CHAPTER 4 SOLUTIONS 2/17/10 4-1) Load: Each diode: 4-2) 4-3) 4-4) 4-5) b) Power is determined from the Fourier series. Using Eq. 4-4 and 4-5. n Vn, V. Zn. Ω In, A. 2 72.0 27.1 2.65 4 14.4 47.7 0.302 4-6 b) Power is determined from the Fourier series. Using Eq. 4-4 and 4-5. n Vn, V. Zn. Ω In, A. 2 72.0 19.3 3.74 4 14.4 32.5 0.444 4-7) � 4-8) Load: 4-9) 4-10) � 4-11) b) Fourier Series n Vn, V. Zn. Ω In, A. 2 72.2 11.7 6.16 4 14.4 22.8 0.631 _____________________________________________________________________________________ 4-12 b) Fourier Series n Vn, V. Zn. Ω In, A. 2 144.3 30.6 4.72 4 28.9 60.5 0.477 4-13) 4-14) a) Continuous current; P=474 W. b) Discontinuous current; P=805 W. 4-15 b) Fourier Series n Vn, V. Zn. Ω In, A. 2 72.0 30.4 2.37 4 14.4 60.5 0.238 � 4-16 b) Fourier Series n Vn, V. Zn. Ω In, A. 2 72.0 45.5 1.58 4 14.4 90.6 0.159 _____________________________________________________________________________________ 4-17) The current with the 100 μH inductor is discontinuous. 4-18) 4-19) 4-20) C ≈ 3333/2 = 1667 µF. Peak diode currents are the same. Fullwave circuit has advantages of zero average source current, smaller capacitor, and average diode current ½ that for the halfwave. The halfwave circuit has fewer diodes, and has only one diode voltage drop rather than two. 4-21) > 1 ( continuous current < 1 ( Calculated Vo is slightly larger than initial estimate. Try Vo=120 V.: Therefore, 119 < Vo < 120 V. (Vo=119.6 with more iterations.) c) PSpice results: R = 7 results in continuous current with Vo = 108 V. R = 20 results in discontinuous current with Vo = 120 V. The simulation was done with C = 10,000 μF. 4-22) PSpice results with a 0.5 Ω resistance in series with the inductance: For Rload = 5 Ω, Vo=56.6 V. (compared to 63.7 volts with an ideal inductor); for Rload = 50 Ω, Vo=82.7 V. (compared to 84.1 volts with an ideal inductor). 4-23) � 4-24) 4-25) a) α = 15° : Check for continuous current. First period: b) α = 75° Check for continuous current. First period: � 4-26)a) α = 20°: Check for continuous current. First period: b) α = 80°: Check for continuous current. First period: 4-27) The source current is a square wave of ±Io. � 4-28) Note that the turns ratio could be lower (higher secondary voltage) and α adjusted accordingly. 4-29) � 4-30) � 4-31) Choose L somewhat larger, say 120 mH, to allow for approximations. � 4-32) In Fig. 4-14, Pac = Pbridge = -VoIo = 1000 W. Using Vdc = -96 V gives this solution: _____________________________________________________________________________________ � 4-33) 4-34) 4-35) � 4-36) 4-37) There are no differences between the calculations in Problem 4.36 and the PSpice results. The power absorbed by each diode ia approximately 1.9 W. 4-38)Equation (4-46) gives values of of I1 = 28.6 A, I5 = 5.71 A, I7 = 4.08 A, I11 = 2.60 A, and I13 = 2.20 A. All compare well with the PSpice results. The total harmonic distortion (THD) is 27.2% when including harmonics through n = 13. _____________________________________________________________________________________ 4-39) c) 4-40) c) _____________________________________________________________________________________ � 4-41) 4-42) � 4-43) 4-44) _____________________________________________________________________________________ � 4-45) 4-46) _____________________________________________________________________________________ 4-47) _____________________________________________________________________________________ 4-48) _____________________________________________________________________________________ 4-49) _____________________________________________________________________________________ � 4-50) _____________________________________________________________________________________ 4-51) Uncontrolled rectifiers with additional resistances added can also satisfy the specifications. However, adding resistance would increase power loss and decrease efficiency. _____________________________________________________________________________________ _1329737868.unknown _1329737885.unknown _1329737893.unknown _1329737901.unknown _1329737905.unknown _1329737909.unknown _1329737911.unknown _1329737913.unknown _1329737914.unknown _1329737912.unknown _1329737910.unknown _1329737907.unknown _1329737908.unknown _1329737906.unknown _1329737903.unknown _1329737904.unknown _1329737902.unknown _1329737897.unknown _1329737899.unknown _1329737900.unknown _1329737898.unknown _1329737895.unknown _1329737896.unknown _1329737894.unknown _1329737889.unknown _1329737891.unknown _1329737892.unknown _1329737890.unknown _1329737887.unknown _1329737888.unknown _1329737886.unknown _1329737876.unknown _1329737880.unknown _1329737883.unknown _1329737884.unknown _1329737881.unknown _1329737878.unknown _1329737879.unknown _1329737877.unknown _1329737872.unknown _1329737874.unknown _1329737875.unknown _1329737873.unknown _1329737870.unknown _1329737871.unknown _1329737869.unknown _1329737860.unknown _1329737864.unknown _1329737866.unknown _1329737867.unknown _1329737865.unknown _1329737862.unknown _1329737863.unknown _1329737861.unknown _1329737856.unknown _1329737858.unknown _1329737859.unknown _1329737857.unknown _1329737854.unknown _1329737855.unknown _1329737853.unknown Eletronica_Potencia_-_Daniel_Hart_-_Solutions_Manual/HartChapter5solutions.doc CHAPTER 5 SOLUTIONS 3/9/10 5-1) _____________________________________________________________________________________ � 5-2) _____________________________________________________________________________________ 5-3) _____________________________________________________________________________________ 5-4) _____________________________________________________________________________________ 5-5) _____________________________________________________________________________________ � 5-6) _____________________________________________________________________________________ 5-7) _____________________________________________________________________________________ 5-8) _____________________________________________________________________________________ 5-9) S1 is on from α to π, and D2 is on from π to 2π. _____________________________________________________________________________________ 5-10) _____________________________________________________________________________________ 5-11) a) Using Eq. 5-9, _____________________________________________________________________________________ 5-12) Using Eq. 5-9, _____________________________________________________________________________________ 5-13) Using Eq. 5-9, _____________________________________________________________________________________ 5-14) Using Eq. 5-9, 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 ; B-phase D3 10 5 DMOD S6 5 11 21 0 SMOD D6 11 2 DMOD 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} ; vbn = v(5,7) LB 5A 7 {L} 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 S1 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 ; B-phase D3 10 5 DMOD S6 5 11 21 0 SMOD D6 11 2 DMOD 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 _____________________________________________________________________________________ 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. _____________________________________________________________________________________ 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 V14 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 _1329917845.unknown _1329917849.unknown _1329917853.unknown _1329917855.unknown _1329917856.unknown _1329917857.unknown _1329917854.unknown _1329917851.unknown _1329917852.unknown _1329917850.unknown _1329917847.unknown _1329917848.unknown _1329917846.unknown _1329917843.unknown _1329917844.unknown _1329917842.unknown Eletronica_Potencia_-_Daniel_Hart_-_Solutions_Manual/HartChapter6solutions.doc CHAPTER 6 SOLUTIONS 5/17/10 6-1) 6-2) 6-3) 6-4) 6-5) 6-6) 6-7) 6-8) � 6-9) 6-10) 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 6.25 15 1/3 0.5 10 16.7 (worst case, D = 1/3, R = 10) 15 1/3 1.0 5 8.33 6-11) Example design: 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 Imax,diode = ILmax = 1.875 A Iavg,diode =IL- Iavg,switch = 1.875 – 0.586 = 1.289 A Irms,diode 6-13) Example design: 6-14) Example design: Other values of L and C are valid if the inductor current is continuous with margin. 6-15) The output voltage is mainly the dc term and the first ac term. 6-16) 6-17) 6-18) Inductor current: (see Example 2-8) Capacitor current: (define t=0 at peak current) 6-19) � 6-20) Example design: 6-21) � 6-22) 6-23) � 6-24) Inductor current: (see Example 2-8) Capacitor current: For convenience, redefine t = 0 at the peak current. The current is then expressed as 6-25) 6-26) Example design: � 6-27) Example design: � 6-28) Using the equations and using f = 100 kHz (designer’s choice), results are shown in the table. Vs, (V) P (W) D R (Ω) Lmin (µH) IL (A) C (µF) 10 10 0.545 14.4 14.9 1.83 37.9 10 15 0.545 9.6 9.9 2.75 56.8 14 10 0.462 14.4 20.9 1.55 32.1 14 15 0.462 9.6 13.9 2.32 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) 6-30) � 6-31) Example design: � 6-32) � 6-33) 6-34) Equation (6-69) for the average voltage across the capacitor C1 applies: When the switch is closed, the voltage across L2 for the interval DT is Assuming that the voltage across C1 remains constant at its average value of Vs When the switch is open in the interval (1 - D)T, Since the average voltage across an inductor is zero for periodic operation, resulting in 6-35) � 6-36) 6-37) 6-38) � 6-39) 6-40) � 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. 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) � 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) From the averaged circuit of Fig. 6.33b, Boost Converter: Show from Eqs. (6-27) and (6-28) that From the averaged circuit of Fig. 6.33c, Buck-Boost Converter: Show from Eqs. (6-47) and (6-49) and preceding equations that From the averaged circuit of Fig. 6.33d, Ćuk Converter: Show from Eqs. (6-59) and (6-61) that From the averaged circuit, _1315073940.unknown _1328971718.unknown _1328989740.unknown _1329593943.unknown _1335535403.unknown _1335550266.unknown _1335592678.unknown _1335592804.unknown _1335592927.unknown _1335551331.unknown _1335536445.unknown _1335545432.unknown _1329594701.unknown _1329594818.unknown _1329594403.unknown _1328989862.unknown _1329040221.unknown _1328989839.unknown _1328987180.unknown _1328987546.unknown _1328988460.unknown _1328988025.unknown _1328987503.unknown _1328985597.unknown _1328987024.unknown _1328987030.unknown _1328971816.unknown _1315897469.unknown _1328967756.unknown _1328967757.unknown _1328967755.unknown _1315074085.unknown _1315075882.unknown _1315201462.unknown _1315074052.unknown _1313782041.unknown _1314688808.unknown _1314771953.unknown _1315070778.unknown _1315071273.unknown _1314777164.unknown _1315070536.unknown _1314774088.unknown _1314769519.unknown _1314770356.unknown _1314692975.unknown _1314293331.unknown _1314523793.unknown _1314688542.unknown _1314293800.unknown _1314296003.unknown _1313953194.unknown _1314214407.unknown _1313950345.unknown _1313952412.unknown _1313764518.unknown _1313766250.unknown _1313778412.unknown _1313779687.unknown _1313765264.unknown _1313760538.unknown _1313763484.unknown _1313759691.unknown _1313759690.unknown Eletronica_Potencia_-_Daniel_Hart_-_Solutions_Manual/HartChapter7solutions.doc CHAPTER 7 SOLUTIONS 4/03/10 7-1) 7-2) 7-3) 7-4) Example design 7-5) For continuous current 7-6) Switch is closed for DT, current returns to zero at t = tx: 7-7) 7-8) 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. 7-9) 7-10) 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. 7-11) 7-12) 7-13) 7-14) The current in the physical primary winding is the sum of iL1 and iLm in the model. The physical currents in windings 2 and 3 are the same as in the model. 7-15) 7-16) 7-17) 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 Lx is therefore 7-19) 7-20) � 7-21) 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. 7-23) Using a nonideal switch and diode produces lower values for the currents. For iLx, 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 iLx in PSpice matches that of the ideal circuit (0.55 A). 7-24) Design for θco= -210° and a gain of 20 dB for a cross over frequency of 12 kHz. � 7-25) 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. 7-27) a) A frequency response of the circuit yields Vo ≈ -2.5 dB and θ ≈ 103° at 10 kHz. 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 From 7-75, 7-85, 7-86, and 7-87, 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. � 7-28) a) The gain at 8 kHz is approximately -2.44 dB, and the phase angle is -100°. 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 From 7-75, 7-85, 7-86, and 7-87, 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. 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) 7-30) 7-31) _1332262374.unknown _1332262560.unknown _1332262831.unknown _1332263111.unknown _1332263155.unknown _1332262861.unknown _1332262645.unknown _1332262490.unknown _1332262522.unknown _1332262436.unknown _1331324598.unknown _1331741220.unknown _1332262236.unknown _1332262333.unknown _1331741221.unknown _1331741222.unknown _1331710868.unknown _1331741219.unknown _1331579822.unknown _1331580795.unknown _1331581597.unknown _1331490501.unknown _1330694593.unknown _1331322390.unknown _1331323274.unknown _1330694757.unknown _1330694508.unknown _1330694551.unknown _1330632186.unknown _1330694461.unknown _1330629161.unknown Eletronica_Potencia_-_Daniel_Hart_-_Solutions_Manual/HartChapter8solutions.doc CHAPTER 8 SOLUTIONS 4/24/10 8-1) 8-2) 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. 8-3) b) 8-4) 8-5) n Vn Zn In,rms 1 331 29.3 8.0 3 110 77 1.02 5 66 127 0.37 8-6) .37ms________ ���������������������������������������������������������������������������������������������������������������� n Vn Zn In,rms 1 88.6 31.3 2.0 3 29.5 61.8 0.34 5 17.7 97.5 0.13 Using PSpice, 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 1.200E+02 2.830E+00 1.000E+00 -3.716E+01 0.000E+00 2 2.400E+02 5.377E-06 1.900E-06 -1.203E+02 -4.594E+01 3 3.600E+02 4.778E-01 1.688E-01 -6.658E+01 4.490E+01 4 4.800E+02 3.589E-06 1.268E-06 -1.223E+02 2.629E+01 5 6.000E+02 1.818E-01 6.422E-02 -7.587E+01 1.099E+02 6 7.200E+02 2.858E-06 1.010E-06 -1.162E+02 1.068E+02 7 8.400E+02 9.427E-02 3.331E-02 -8.028E+01 1.798E+02 8 9.600E+02 2.523E-06 8.913E-07 -1.095E+02 1.878E+02 9 1.080E+03 5.743E-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%. 8-8) n |Vn| Zn In,rms 1 90 12.5 5.08 3 51.6 24.7 1.5 5 4.43 39 0.08 8-9) 8-10) 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), 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, 8-13) 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 0 30 0 30.01 1 54 1 54.01 0 66 0 66.01 1 114 1 114.01 0 126 0 126.01 1 150 1 150.01 0 210 0 210.01 -1 234 -1 234.01 0 246 0 246.01 -1 294 -1 294.01 0 306 0 306.01 -1 330 -1 330 0 360 0 8-14) a) n 1 3 5 7 9 Vn 149.5 0 -2.79 -3.04 -14.4 8-15) α1 α2 α3 Mi 15 25 55 0.815 20 30 40 0.857 10 30 50 0.831 10 30 70 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) From Table 8-3, n Vn/Vdc Vn Zn In=Vn/Zn 1 0.8 76.8 33.3 2.30 mf 17 0.82 78.7 157 0.50 mf - 2 15 0.22 21.1 139 0.151 mf + 2 19 0.22 21.1 175 0.121 8-19) From Table 8-3, n Vn/Vdc Vn Zn In=Vn/Zn 1 0.9 225 27.5 8.18 mf 31 0.71 178 585 0.305 mf - 2 29 0.27 67 547 0.122 mf + 2 33 0.27 67 622 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: 8-22) Example solution: 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) 8-25) For f = 25 Hz: n VnL-N Zn In In,rms 1 255 11.1 23.0 16.3 5 50.9 25.6 2.0 1.41 7 36.4 34.5 1.06 0.75 11 23.1 52.8 0.44 0.31 13 19.6 62.0 0.32 0.22 For f = 100 Hz, n VnL-N Zn In In,rms 1 255 21.3 11.9 8.43 5 50.9 94.8 0.54 0.38 7 36.4 132 0.27 0.19 11 23.1 208 0.12 0.08 13 19.6 245 0.08 0.06 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) _1334392276.unknown _1334392284.unknown _1334392288.unknown _1334392292.unknown _1334392294.unknown _1334392296.unknown _1334392297.unknown _1334392295.unknown _1334392293.unknown _1334392290.unknown _1334392291.unknown _1334392289.unknown _1334392286.unknown _1334392287.unknown _1334392285.unknown _1334392280.unknown _1334392282.unknown _1334392283.unknown _1334392281.unknown _1334392278.unknown _1334392279.unknown _1334392277.unknown _1334392272.unknown _1334392274.unknown _1334392275.unknown _1334392273.unknown _1334392270.unknown _1334392271.unknown _1334392269.unknown Eletronica_Potencia_-_Daniel_Hart_-_Solutions_Manual/HartChapter9solutions.doc CHAPTER 9 SOLUTIONS 3/13/10 9-1) 9-2) 9-3) 9-4) � 9-5) 9-6) 9-7) a) The circuit is shown with diode D2 added to make the switch unidirectional. (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) 9-9) � 9-10) � 9-11) � 9-12) 9-13) 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. 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) 9-16) 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) From the output file, THD = 10.8%. From Probe: VC,peak=149 V.; IL,peak=8.12 A. 9-18) 9-19) 9-20) A PSpice simulation using the circuit of Fig. 9-6(a) gives an output voltage of approximately 5.1 V. 9-21) 9-22) 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) 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) � 9-25) 9-26) � 9-27) 9-28) 9-29) 9-30) 9-31) � 9-32) 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. � 9-34) 9-35) _1318097303.unknown _1330011781.unknown _1330018511.unknown _1330020479.unknown _1330023060.unknown _1330024566.unknown _1330024933.unknown _1330023993.unknown _1330022420.unknown _1330019332.unknown _1330015377.unknown _1330016213.unknown _1330014880.unknown _1329929871.unknown _1330009920.unknown _1330010739.unknown _1329977935.unknown _1329978365.unknown _1329982405.unknown _1329976377.unknown _1318181565.unknown _1318183168.unknown _1318491329.unknown _1318492901.unknown _1318182347.unknown _1318180859.unknown _1318181002.unknown _1318097990.unknown _1317579537.unknown _1317581273.unknown _1317583846.unknown _1317580287.unknown _1317403598.unknown _1317408856.unknown _1317403595.unknown
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