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GATE 2021 EE – Question 64

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

Consider the buck-boost converter shown. Switch Q is operating at 25 kHz and 0.75 duty-cycle. Assume diode and switch to be ideal. Under steady-state condition, the average current flowing through the inductor is ________ A.

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Correct answer: 24

Explanation

For the buck-boost converter $|V_o|=\frac{D}{1-D}V_{in}=3\times20=60$ V, so $I_o=\frac{60}{10}=6$ A. The diode carries the load current only during the off-time, so $I_o=(1-D)I_L$ and $I_L=\frac{6}{0.25}=24$ A.