GATE 2023 EE Q.37: solution and answer

GATE 2023 · EEQ.37 · 2 marks · MCQ

Consider the OP AMP based circuit shown in the figure. Ignore the conduction drops of diodes $D_{1}$ and $D_{2}$ . All the components are ideal and the breakdown voltage of the Zener is 5 V. Which of the following statements is true?

GATE 2023 EE question 37: Inverting op-amp circuit. — Figure from the GATE 2023 EE paper. It is part of the question.

(A)
The maximum and minimum values of the output voltage $V_{O}$ are +15 V and -10 V, respectively.
(B)
The maximum and minimum values of the output voltage $V_{O}$ are +5 V and -15 V, respectively.
(C)
The maximum and minimum values of the output voltage $V_{O}$ are +10 V and -5 V, respectively.
(D)
The maximum and minimum values of the output voltage $V_{O}$ are +5 V and -10 V, respectively.

Answer and solution

ANSWER

(D) The maximum and minimum values of the output voltage $V_{O}$ are $+ 5 V$ and $- 10 V$ , respectively.

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

Assumptions: ideal op-amp, ideal diodes $D_{1}$ and $D_{2}$ (zero forward drop, as stated), and an ideal Zener with $V_{Z} = 5 V$ . While a feedback path conducts, the inputs are at virtual ground, so $V_{-} = V_{+} = 0$ . The input current is then $i = \frac{V _{in}}{1 k Ω}$ flowing into the $-$ node, and none of it enters the op-amp.
Check which way each feedback branch conducts. The branch of $1 k Ω$ in series with $D_{1}$ (anode on the $-$ node side) conducts only from the $-$ node to the output, which needs $V_{O} < V_{-} = 0$ . The $D_{Z}$ – $D_{2}$ branch (cathode to cathode) conducts only from the output to the $-$ node. That needs $D_{2}$ forward-biased and the Zener in breakdown, so $V_{O} - V_{-} = 0 + V_{Z} = 5 V$ . In the other direction this branch is blocked, because $D_{2}$ is then reverse-biased.
Positive half-cycle, $V_{in} > 0$ : the current $i$ has to leave the $-$ node through the feedback, from the node to the output. Only the $D_{1}$ branch allows this, so $D_{1}$ conducts and the Zener branch is off. The circuit is an inverting amplifier: $V_{O} = - \frac{1 k Ω}{1 k Ω} V_{in} = - V_{in}$ Check: $V_{O} < 0$ here, as $D_{1}$ 's conduction requires, and the Zener branch sees $V_{-} - V_{O} = V_{in} > 0$ in its blocked direction.
Minimum output: at the positive peak $V_{in} = + 10 V$ , $V_{O, m i n} = - 10 V$ . This is inside the $- 15 V$ rail, so the op-amp does not saturate.
Negative half-cycle, $V_{in} < 0$ : current now flows out of the $-$ node toward the source, so the feedback must push current into the node from the output. $D_{1}$ blocks that direction. With no feedback path the op-amp is momentarily open-loop, and its output rises until the Zener branch conducts. At that point $D_{2}$ is forward-biased and the Zener is in breakdown, which clamps the output: $V_{O} = V_{-} + 0 + V_{Z} = 0 + 0 + 5 = + 5 V$ Check: $D_{1}$ now has its cathode at $+ 5 V$ and its anode side at $0 V$ , so it is reverse-biased and stays off.
The Zener carries $∣ V_{in} ∣/1 k Ω$ , up to $10 mA$ at the negative peak, and $V_{O}$ stays at $+ 5 V$ for the whole negative half-cycle. This is below the $+ 15 V$ rail, so there is no saturation. Therefore $V_{O, m a x} = + 5 V$ and $V_{O, m i n} = - 10 V$ , which is option (D).

KEY IDEA

Each feedback branch conducts in only one direction. For $V_{in} > 0$ , the $1 k Ω$ – $D_{1}$ branch makes the circuit a unity-gain inverting amplifier, so $V_{O} = - V_{in}$ and it reaches $- 10 V$ . For $V_{in} < 0$ , the $D_{2}$ –Zener branch clamps the output at $V_{-} + V_{Z} = + 5 V$ . Neither case reaches a rail, so the $\pm 15 V$ supplies are distractors.

WATCH OUT FOR

Thinking the op-amp saturates at $+ 15 V$ in the negative half-cycle because $D_{1}$ is off, which leads to option (A). This overlooks the $D_{2}$ –Zener path, which closes the loop at $V_{O} = + 5 V$ . Two other slips give the wrong numbers: misreading the Zener's orientation, so the clamp lands on the negative half (option C), or letting the positive half-cycle run to the $- 15 V$ rail (option B) instead of following the unity-gain relation $V_{O} = - V_{in}$ .

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2023 EE Q.37: solution and answer

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