GATE 2022 EE Q.61: solution and answer

Official answer: 165 to 171Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

1. Read the waveform: the inductor current rises from $12\text{A}$ to $16\text{A}$ in $DT=20\mu \text{s}$ (switch on) and falls back in $(1-D)T=30\mu \text{s}$ (diode on). So $T=50\mu \text{s}$ and the duty ratio is $D=\frac{20}{50}=0.4$.

2. The inductor current is a triangular wave between $12\text{A}$ and $16\text{A}$, so its average is $I_L=\frac{12+16}{2}=14\text{A}$.

3. In a buck-boost converter the inductor current reaches the output only through the diode during the off time. The average output current is therefore $I_o=(1-D)I_L=0.6\times 14=8.4\text{A}$.

4. While the switch is on, the diode is off and the capacitor alone supplies the load current $I_o$ for $DT$. The charge it loses is $\Delta Q=I_{o}DT=8.4\times 20\times 10^{-6}=168\mu \text{C}$.

5. The peak-to-peak output ripple equals this charge divided by the capacitance: $\Delta V_o=\frac{\Delta Q}{C}=\frac{I_{o}DT}{C}$.

6. Setting $\Delta V_o=1\text{V}$ gives $C=\frac{168\mu \text{C}}{1\text{V}}=168\mu \text{F}$.

7. Check over the off interval: the capacitor charges with $(I_D-I_o)$, whose average is $14-8.4=5.6\text{A}$ over $30\mu \text{s}$. The charge gained is $5.6\times 30=168\mu \text{C}$, equal to the charge lost, as required in steady state.