Daniel Hart Eletrônica de Potência (cap 6 Solucionário), Exercícios de Eletrônica de Potência

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CHAPTER 6 SOLUTIONS

5/17/ 6-1) 6-2) 6-3) 6-4)

Vo=5 V Vs, V D I, A. R, Ω Lmin, μH 10 0.5^ 0.5 10 12. 10 0.5^ 1.0 5 6. 15

16.7 (worst case, D = 1/3, R =

15 1/3^ 1.0 5 8. 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:

The output voltage is mainly the dc term and the first ac term. 6-16) 6-17)

Inductor current: (see Example 2-8) Capacitor current: (define t=0 at peak current)

6-20) Example design: 6-21)

6-26) 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. 10 15 0.545 9.6 9.9 2.75 56. 14 10 0.462 14.4 20.9 1.55 32. 14 15 0.462 9.6 13.9 2.32 48. 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.

6-34) Equation (6-69) for the average voltage across the capacitor C 1 applies: When the switch is closed, the voltage across L 2 for the interval DT is Assuming that the voltage across C 1 remains constant at its average value of Vs When the switch is open in the interval (1 - D)T,