GATE 2024 EE Q.64: solution and answer

Official answer: 13Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. The inductor is in series with the source, the switch $S$ connects the node after the inductor to the bottom rail, and the diode leads to the output capacitor and load. This is a boost converter, so $V_o=\frac{V_i}{1-D}$.

2. Duty ratio: $40=\frac{20}{1-D}$, so $1-D=0.5$ and $D=0.5$.

3. Load current: $I_o=\frac{V_o}{R}=\frac{40}{10}=4\text{A}$.

4. The converter is taken as lossless, so $V_i{I}_i=V_o{I}_o$. The input current is the average inductor current: $I_L=I_i=\frac{40\times 4}{20}=8\text{A}$. This is the same as $I_L=\frac{I_o}{1-D}=\frac{4}{0.5}=8\text{A}$.

5. Period: $T=\frac{1}{500}=2\text{ms}$. While $S$ is ON the inductor voltage is $V_i$, and the ON time is $DT=1\text{ms}$. The peak-to-peak ripple is $$\Delta I=\frac{V_{i}DT}{L}=\frac{20\times 0.5\times 2\times 10^{-3}}{2\times 10^{-3}}=10\text{A}.$$

6. Peak inductor current: $I_{pk}=I_L+\frac{\Delta I}{2}=8+5=13\text{A}$.

7. The minimum inductor current is $8-5=3\text{A}>0$, which confirms that conduction is continuous, as the question states.