GATE 2023 EE Q.58: solution and answer

Official answer: 39 to 40.5Source: Official answer key, GATE 2023 EE (IIT Kanpur)

1. Let $A$ be the junction of $T_1$ and $T_2$ (the source $+$ terminal) and $B$ the junction of $D_1$ and $T_3$. Then $v_{in}=v_A-v_B=V_m\sin \theta$ with $V_m=100\text{ V}$. The load is a constant current $I_o=10\text{ A}$, so only the voltage $V_o$ across it changes. The top rail is fed through $T_1$ from $A$ or through $D_1$ from $B$. The bottom rail returns through $T_2$ to $A$ or through $T_3$ to $B$.

2. $60^{\circ }$ to $180^{\circ }$ ($T_1$ and $T_3$ fired at $\alpha =60^{\circ }$): $T_1$ and $T_3$ conduct, so $V_o=v_{in}=V_m\sin \theta$.

3. $180^{\circ }$ to $240^{\circ }$: $v_{in}$ goes negative, so $T_1$ is reverse biased and turns off, and $D_1$ turns on. The constant current free-wheels through $D_1$ and $T_3$, so $V_o=0$.

4. From $240^{\circ }$ ($T_2$ fired at $\alpha =240^{\circ }$): $T_2$ takes over from $T_3$, so the current path is $B\to D_1\to \text{load}\to T_2\to A$ and $V_o=v_B-v_A=-V_m\sin \theta$, which is positive for $240^{\circ }$ to $360^{\circ }$. The load current is constant, so $T_2$ keeps conducting past $360^{\circ }$ until $T_1$ and $T_3$ are fired again at $360^{\circ }+60^{\circ }$. For $0^{\circ }$ to $60^{\circ }$ the output is still $-V_m\sin \theta$, which is now negative.

5. The average over one period is $$V_o=\frac{V_m}{2\pi }\left[{\int }_{60^{\circ }}^{180^{\circ }}\sin \theta d\theta -{\int }_{240^{\circ }}^{360^{\circ }}\sin \theta d\theta -{\int }_{0^{\circ }}^{60^{\circ }}\sin \theta d\theta \right]$$

6. The three integrals are $\cos 60^{\circ }-\cos 180^{\circ }=1.5$, $\cos 240^{\circ }-\cos 360^{\circ }=-1.5$ and $\cos 0^{\circ }-\cos 60^{\circ }=0.5$. The bracket is therefore $1.5+1.5-0.5=2.5$.

7. $$V_o=\frac{100\times 2.5}{2\pi }=39.79\text{ V}$$