GATE EC 2017 Set 2 — Question 13

Step 1: When does the diode branch conduct?$D_1$ and the Zener $D_2$ are in series across the output and face in opposite directions.

• For $V_{out} > 0$: $D_1$ is forward biased ($0.7\text{ V}$) and $D_2$ is reverse biased. The branch conducts only once $D_2$ breaks down, at

$$V_{out} = 0.7 + 6.8 = 7.5\text{ V}.$$

• For $V_{out} < 0$: $D_1$ is reverse biased, so the branch is open.

The branch therefore clamps the output at a maximum of $+7.5\text{ V}$.Step 2: Clamping action (official interpretation).Take $V_{out} = V_{in} - V_C$. During the $+14\text{ V}$ half cycle, the capacitor charges through the conducting branch until the output sits at the clamp level:$$7.5 = 14 - V_C \;\Rightarrow\; V_C = 6.5\text{ V}.$$When the input switches to $-14\text{ V}$, the branch turns off. If the capacitor holds this charge for the rest of the period, then$$V_{out,\min} = -14 - 6.5 = -20.5\text{ V}.$$So the output swings between $7.5\text{ V}$ and $-20.5\text{ V}$, which is option (A), the official answer.Step 3: About the time-constant condition.The calculation in Step 2 is the standard clamper result. It assumes the capacitor barely discharges over a period, that is, a long time constant ($\tau \gg T$). The paper prints "$10\tau \ll T$", which says the opposite. Taken literally, a very short $\tau$ would let the capacitor fully recharge in every half cycle. The negative half would then decay towards $0$ and the minimum would be the jump $-14 - 14 = -28\text{ V}$, which is not among the options. The official key (A) therefore corresponds to the long-time-constant clamper ($\tau \gg T$). Read the printed condition as that intent.

• (B) and (D) use $6.1\text{ V}$ ($6.8 - 0.7$) for the clamp level, but the two drops add because $D_1$ is forward biased and $D_2$ is in breakdown.
• (C) and (D) have minima that do not equal $-14 - V_C$ for any consistent clamp level.

Answer: (A) $7.5\text{ V}$ and $-20.5\text{ V}$