GATE 2017 EE (EE2) – Question 60

Power Electronics · DC to DC conversion: Buck, Boost and Buck-Boost Converters · 2 marks · Numerical answer

All elements are ideal. Switch S runs at 10 kHz and 60% duty ratio. The large output capacitor has negligible ripple and steady voltage 75 V with the shown polarity. Find peak current drawn from the 50 V source in amperes, to one decimal place.

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Correct answer: 39 to 41

Explanation

This is an inverting buck–boost converter. The output load current is 75/5=15 A. During the off fraction 1−D=0.4, the inductor supplies the load, so its average current is $I_L=15/0.4=37.5$ A.

During the on time $DT=0.6/10000=60$ μs, inductor current rises by $\Delta I=V_{in}DT/L=50(60\ \mu s)/(0.6\mathrm{\ mH})=5$ A peak to peak.

Source current equals inductor current while on. Its peak is $37.5+5/2=$ **40.0 A**. The minimum is positive, confirming continuous conduction.

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