GATE 2020 EC Q.8: solution and answer

GATE 2020 · ECQ.8 · 1 mark · MCQ

The components in the circuit shown below are ideal. If the op-amp is in positive feedback and the input voltage $V_{i}$ is a sine wave of amplitude 1 V, the output voltage $V_{o}$ is

GATE 2020 EC question 8: An op-amp with supply rails +5 V (top) and −5 V (bottom). — Figure from the GATE 2020 EC paper. It is part of the question.

(A)
a non-inverted sine wave of 2 V amplitude.
(B)
an inverted sine wave of 1 V amplitude.
(C)
a square wave of 5 V amplitude.
(D)
a constant of either +5 V or −5 V.

Answer and solution

ANSWER

(D) The output is a constant of either +5 V or −5 V. The Schmitt trigger's thresholds are ±5 V, and a 1 V sine never reaches them, so the output never switches.

Official answer: DSource: Official answer key, GATE 2020 EC (IIT Delhi)

Identify the circuit. The 1 kΩ feedback resistor returns the output to the $+$ input and the $-$ input is grounded, so this is a non-inverting Schmitt trigger (a comparator with hysteresis). With positive feedback there is no virtual short: $V_{+} \neq = V_{-}$ in general, and the ideal op-amp output sits at one supply rail, $V_{o} = + 5 V$ or $V_{o} = - 5 V$ (ideal op-amp, output saturates at the supply, no current into the input terminals).
Find the voltage at the $+$ input by superposition on the two equal 1 kΩ resistors (no current flows into the ideal op-amp input): $V_{+} = V_{i} \frac{1 k Ω}{2 k Ω} + V_{o} \frac{1 k Ω}{2 k Ω} = \frac{V _{i} + V _{o}}{2}$
Switching condition: the output changes state only when $V_{+}$ crosses $V_{-} = 0$ , that is when $V_{i} + V_{o} = 0$ , or $V_{i} = - V_{o}$ .
Thresholds: while $V_{o} = + 5 V$ , $V_{+} = \frac{V _{i} + 5}{2}$ goes negative only if $V_{i} < - 5 V$ , so the lower threshold is $V_{T L} = - 5 V$ . While $V_{o} = - 5 V$ , $V_{+} = \frac{V _{i} - 5}{2}$ goes positive only if $V_{i} > + 5 V$ , so the upper threshold is $V_{T H} = + 5 V$ . The hysteresis width is $V_{T H} - V_{T L} = 10 V$ .
Compare with the input: $V_{i}$ swings only between $- 1 V$ and $+ 1 V$ , well inside the band from $- 5 V$ to $+ 5 V$ , so it never reaches either threshold. (Check: with $V_{o} = + 5 V$ , $V_{+}$ stays between $2 V$ and $3 V$ , always positive; with $V_{o} = - 5 V$ , $V_{+}$ stays between $- 3 V$ and $- 2 V$ , always negative.)
So the output never switches. It stays at whichever rail it started on: a constant $+ 5 V$ or $- 5 V$ . Option (D).

KEY IDEA

With positive feedback the op-amp acts as a comparator with hysteresis: the output is always at a rail, and it switches only when the input crosses a threshold. For this non-inverting Schmitt trigger, $V_{+} = \frac{V _{i} + V _{o}}{2}$ gives thresholds $V_{i} = \mp 5 V$ . A $\pm 1 V$ input stays inside the hysteresis band, so the output is latched at $+ 5 V$ or $- 5 V$ .

WATCH OUT FOR

Treating the circuit as a linear amplifier. Assuming a virtual short ( $V_{+} = V_{-}$ ) gives a gain of $- 1$ and leads to option (B), and misreading it as a non-inverting amplifier gives option (A); neither is valid under positive feedback. A second trap is picking (C) because "a comparator turns a sine into a square wave". That only happens if the input amplitude exceeds the thresholds (here $\pm 5 V$ ), and a 1 V sine does not.

From the GATE 2020 EC paper. Every question of this paper · Download the full paper · All 13 Filters & waveforms questions, with notes

GATE 2020 EC Q.8: solution and answer

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