Solucionario (cap2 ) Eletrônica de Potência Daniel Hart, Exercícios de Eletrônica de Potência

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CHAPTER 3 SOLUTIONS

2/20/ 3-1) 3-2)

3-4) Using Eq. 3-15,

3-7) Using an ideal diode model, R = 48 Ω for an average current of 2 A.

Time 0s 5ms 10ms 15ms 20ms I(R1) AVG(I(L1)) 0A 4.0A 8.0A Current Average Current Iavg = 2 A for R = 48 ohms (16.700m,2.0030)

3-9) Using Eqs. 3-22 and 3-23, 3-10) Using Eq. 3-33,

Time 0s 5ms 10ms 15ms 20ms AVG(W(Vdc)) 0W 100W 200W 300W L = 0.25 H

3-13) Using Eq. 3-34, a) b) n Vn Zn In 0 54.02 12.00 4. 1 84.85 25.6 3. 2 36.01 46.8 0. 4 7.20 91.3 0. 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.)

a) b) A PSpice simulation using an ideal diode model gives 0.443 A p-p in the steady state. This compares with 2(I 1 )=2(0.199)=0.398 A p-p.

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- 3-22) Assuming Vo is constant and equal to Vm, From Eq. 3-

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).