GATE EE 2023 Set 1 — Question 37

The non-inverting input is grounded, so the inverting node is a virtual ground, $V^- = 0$, and the input resistor and the feedback network carry the same current $i = V_{in}/1\text{ k}\Omega$.The feedback network between the inverting node and the output has two parallel paths:

• the $1\text{ k}\Omega$ resistor in series with $D_1$;
• the Zener $D_Z$ in series with $D_2$, oriented opposite to $D_1$.

Positive half cycle ($V_{in} > 0$): the current flows into the inverting node and must leave it towards the output. Only the branch containing $D_1$ is oriented to carry current in that direction, so $D_1$ conducts and the feedback element is the $1\text{ k}\Omega$ resistor:$$V_O = -\frac{1\text{ k}\Omega}{1\text{ k}\Omega}V_{in} = -V_{in}$$With $V_{in}$ peaking at $10$ V, the output swings down to $-10$ V, which the $\pm 15$ V supplies permit.Negative half cycle ($V_{in} < 0$): the current must now flow from the output into the inverting node, which is the direction allowed by $D_2$. With $D_2$ forward biased, the Zener is reverse biased; as soon as the output reaches the breakdown voltage the Zener clamps it (diode conduction drops are ignored):$$V_O = +V_Z = +5\text{ V}$$So the positive excursion is clipped at $+5$ V instead of following $-V_{in}$ up to $+10$ V.Hence $V_{O,\max} = +5$ V and $V_{O,\min} = -10$ V.Answer: (D) The maximum and minimum values of the output voltage $V_O$ are +5 V and -10 V, respectively.