A buck converter is operating at 1-kHz switching frequency from a 120-V dc source. The inductance is 50 mH. If the output voltage is 60 V and the load resistance is 12 Ω, we determine the following:

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Duty ratio D = 60 V ÷ 120 V = 0.50 period T = 1/1000 = 0.001 s = 1 ms. Ton = 0.50 × 1 = 0.5 ms Toff=1-0.5 = 0.5 ms. Average load current = 60 V ÷ 12 02 = 5 A Output power = 60 × 5 = 300 W. Average source current = 5 x 0.50 = 2.5 A Input power = 120 x 2.5 = 300 W. Equation 13.4 gives peak-to-peak ripple current: ALL Vout ToffL = 60 x 0.0005 ÷ 0.050 = 0.6 A. =