ISRO EC 2023 Q.55: solution and answer

ISRO 2023 · ECQ.55 · 1 mark · MCQ

Which of the following represents the correct output Vo of circuit shown in Figure? Assume RC=1.

ISRO EC 2023 question 55: Op-amp circuit: input Vi goes through a series capacitor C to the inverting (-) terminal. — Figure from the ISRO EC 2023 paper. It is part of the question.

(A)
Figure — square wave between +1 and -1: +1 for 0<t<1, -1 for 1<t<2, +1 for 2<t<3, -1 for 3<t<4
(B)
Figure — square wave between -1 and +1: -1 for 0<t<1, +1 for 1<t<2, -1 for 2<t<3, +1 for 3<t<4
(C)
Figure — square wave between -0.1 and +0.1: -0.1 for 0<t<1, +0.1 for 1<t<2, -0.1 for 2<t<3, +0.1 for 3<t<4
(D)
Figure — triangular wave with peaks of 1 at t=1 and t=3, same as the input

Answer and solution

ANSWER

Option (B): a square wave of amplitude 1 that is $- 1$ for $0 < t < 1$ , $+ 1$ for $1 < t < 2$ , and so on.

Official answer: BSource: ISRO's revised and final answer key of 16 February 2024

The capacitor $C$ is in series with the input and the resistor $R$ is in the feedback path, so the circuit is an inverting differentiator. The non-inverting input is grounded through $R$ , which only balances the bias path and does not change the ideal result.
The inverting input is a virtual ground, so the capacitor current is $i = C \frac{d V _{i}}{d t}$ . The op-amp draws no input current, so this same current flows through the feedback resistor $R$ .
The output is $V_{o} = 0 - i R = - R C \frac{d V _{i}}{d t}$ . With $R C = 1$ , $V_{o} = - \frac{d V _{i}}{d t}$ .
For $0 < t < 1$ , $V_{i}$ rises from 0 to 1, so the slope is $+ 1$ and $V_{o} = - 1$ .
For $1 < t < 2$ , $V_{i}$ falls from 1 to 0, so the slope is $- 1$ and $V_{o} = + 1$ .
For $2 < t < 3$ the slope is $+ 1$ again, so $V_{o} = - 1$ . For $3 < t < 4$ the slope is $- 1$ , so $V_{o} = + 1$ .
The output is therefore a square wave of amplitude 1, equal to $- 1$ for $0 < t < 1$ and then alternating every unit of time. This is option (b).
Option (a) has the opposite sign, which ignores the inversion. Option (c) has amplitude 0.1, but $R C = 1$ gives a slope magnitude of 1, so the amplitude is 1. Option (d) is a triangular wave, which a differentiator cannot produce from a triangular input, because the derivative of a linear ramp is constant.

KEY IDEA

A capacitor at the input with a feedback resistor gives an inverting differentiator, $V_{o} = - R C \frac{d V _{i}}{d t}$ . A triangular input has a constant slope on each segment, so the output is a square wave of opposite sign to the slope.

WATCH OUT FOR

Forgetting the minus sign of the inverting configuration and choosing the square wave that starts at $+ 1$ , or misreading the amplitude because the slope here is 1 per unit time when $R C = 1$ .

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 4 ISRO Op-amps questions, with notes

ISRO EC 2023 Q.55: solution and answer

Answer and solution

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