GATE 2025 EE Q.60: solution and answer

Official answer: 1.7 to 1.9Source: Official answer key, GATE 2025 EE (IIT Roorkee)

1. In steady state the average capacitor current is zero, and in a buck converter the capacitor carries the ripple part of the inductor current. So the plotted triangle is the inductor ripple, and its peak-to-peak value is $\Delta I_L=0.1-(-0.1)=0.2\text{A}$.

2. From the figure, the minimum $-0.1\text{A}$ occurs at $t=0$ and again at $t=50\mu \text{s}$, so the switching period is $T=50\mu \text{s}$.

3. The inductor current rises while the switch is ON. The rising line goes from $-0.1\text{A}$ at $t=0$ to $0$ at $t=15\mu \text{s}$, so its slope is $\frac{0.1}{15\mu \text{s}}$. Reaching the peak, $+0.1\text{A}$, means a rise of $0.2\text{A}$, which takes $\frac{0.2}{0.1}\times 15=30\mu \text{s}$. Check on the falling side: from the peak at $30\mu \text{s}$ the current crosses zero at $40\mu \text{s}$ ($0.1\text{A}$ in $10\mu \text{s}$) and reaches $-0.1\text{A}$ at $50\mu \text{s}$, as plotted.

4. So $T_{on}=30\mu \text{s}$, $T_{off}=20\mu \text{s}$, and the duty ratio is $D=\frac{30}{50}=0.6$.

5. For the buck converter in CCM, $V_o=D{V}_s=0.6\times 30=18\text{V}$.

6. During the ON interval the inductor voltage is $V_s-V_o=12\text{V}$, so $\Delta I_L=\frac{(V_s-V_o)DT}{L}$.

7. $$L=\frac{(V_s-V_o)DT}{\Delta I_L}=\frac{12\times 30\times 10^{-6}}{0.2}=1.8\times 10^{-3}\text{H}=1.8\text{mH}$$

8. Check with the OFF interval, where the inductor voltage is $-V_o$: $L=\frac{V_o(1-D)T}{\Delta I_L}=\frac{18\times 20\times 10^{-6}}{0.2}=1.8\text{mH}$, the same value.