GATE 2023 EE Q.23: solution and answer

Official answer: BSource: Official answer key, GATE 2023 EE (IIT Kanpur)

1. The switching frequency is $100\text{kHz}$, so the period is $T=\frac{1}{100\text{kHz}}=10\mu \text{s}$. From figure (ii), S1 is ON from $0$ to $3\mu \text{s}$ and OFF from $3$ to $10\mu \text{s}$, so the duty ratio of S1 is $D=\frac{3}{10}=0.3$. (No firing angle is involved: this is a DC chopper.)

2. The $5\text{A}$ current source is connected from the midpoint of the leg to the negative rail, and its arrow points downward. The load current therefore always leaves the midpoint and flows down into the negative rail. The voltage across the current source is the midpoint voltage $v_o$ measured with respect to the negative rail.

3. Interval $0$ to $3\mu \text{s}$: S1 is ON, so the $20\text{V}$ source drives the $5\text{A}$ through S1 into the midpoint. Hence $v_o=20\text{V}$.

4. Interval $3$ to $10\mu \text{s}$: S1 is OFF. The current source still takes $5\text{A}$ out of the midpoint, so that current must reach the midpoint from the negative rail, flowing upward. Only the anti-parallel diode D2 can carry current in that direction, so D2 conducts and clamps the midpoint to the negative rail: $v_o=0$.

5. S2 is ON from $5$ to $8\mu \text{s}$, but this changes nothing. An IGBT carries current only from collector to emitter, that is from the midpoint down to the negative rail, whereas the current needed here is upward. So D2 still carries the current and $v_o=0$ throughout $3$ to $10\mu \text{s}$ (all components are ideal).

6. The average voltage over one period is $$V_o=\frac{1}{T}{\int }_0^T{v}_{o}dt=\frac{1}{10}\left(20\times 3+0\times 7\right)=6\text{V}$$ which equals $D{V}_s=0.3\times 20\text{V}$.

7. The average voltage across the $5\text{A}$ current source is $6\text{V}$, option (B).