Question:
Consider the buck-boost converter shown. Switch Q operates at 25 kHz with a duty-cycle of 0.75. Assume diode and switch are ideal. Under steady-state, the average current flowing through the inductor is \(\underline{\hspace{1cm}}\) A.
Consider the buck-boost converter shown. Switch Q operates at 25 kHz with a duty-cycle of 0.75. Assume diode and switch are ideal. Under steady-state, the average current flowing through the inductor is \(\underline{\hspace{1cm}}\) A.
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In a buck-boost converter, the inductor supplies the load only during the OFF time, so the average inductor current is amplified by \(\frac{1}{1-D}\).
Updated On: Feb 3, 2026
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Correct Answer: 24
Solution and Explanation
For a buck-boost converter, steady-state output voltage is:
\[
V_o = \frac{D}{1-D} V_{in}
\]
Given:
\[
V_{in} = 20V, D = 0.75
\]
\[
V_o = \frac{0.75}{0.25} \cdot 20 = 3 \cdot 20 = 60V
\]
Load resistance:
\[
R = 10\Omega
\]
Load current:
\[
I_o = \frac{V_o}{R} = \frac{60}{10} = 6A
\]
Relation between inductor current and output current in buck-boost:
\[
I_L = \frac{I_o}{1 - D}
\]
\[
I_L = \frac{6}{0.25} = 24A
\]
Thus the average inductor current is 24 A, within the given range 24 to 24.
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