GATE 2025 EE Q.56: solution and answer

GATE 2025 · EEQ.56 · 2 marks · NAT

The op-amps in the following circuit are ideal. The voltage gain of the circuit is ______ (round off to the nearest integer).

GATE 2025 EE question 56: Two ideal op-amps in cascade (a two-op-amp instrumentation/difference amplifier). — Figure from the GATE 2025 EE paper. It is part of the question.

Answer and solution

ANSWER

The voltage gain of the circuit is 2.

Official answer: 2Source: Official answer key, GATE 2025 EE (IIT Roorkee)

Assume ideal op-amps: infinite open-loop gain and zero input current. Both op-amps have negative feedback, so each has a virtual short between its two inputs. Call the voltage at the input terminal marked '−' $V_{a}$ and the voltage at the terminal marked '+' $V_{b}$ , so that $V_{in} = V_{b} - V_{a}$ .
Op-amp 1 is a non-inverting amplifier. $V_{a}$ is on its + input, so its − input also sits at $V_{a}$ . The two 10 kΩ resistors form a divider from its output $V_{1}$ to ground, with the − input at the midpoint. So $V_{1} = (1 + \frac{10 k Ω}{10 k Ω}) V_{a} = 2 V_{a}$
Op-amp 2: by the virtual short, its − input sits at $V_{b}$ . No current flows into the op-amp, so KCL at that node gives $\frac{V _{1} - V _{b}}{10 k Ω} = \frac{V _{b} - V _{o u t}}{10 k Ω}$
Solving for the output: $V_{o u t} = 2 V_{b} - V_{1}$ .
Substitute $V_{1} = 2 V_{a}$ : $V_{o u t} = 2 V_{b} - 2 V_{a} = 2 (V_{b} - V_{a}) = 2 V_{in}$
Voltage gain $= \frac{V _{o u t}}{V _{in}} = 2$ .

KEY IDEA

This is the two-op-amp instrumentation amplifier. The first stage scales $V_{a}$ by $(1 + R_{2} / R_{1})$ . The second stage outputs $(1 + R_{4} / R_{3}) V_{b} - (R_{4} / R_{3}) V_{1}$ . When $R_{4} / R_{3} = R_{1} / R_{2}$ (here all resistors are equal), the common-mode part cancels and $V_{o u t} = (1 + R_{4} / R_{3}) (V_{b} - V_{a}) = 2 V_{in}$ .

WATCH OUT FOR

Treating the second stage as a plain inverting amplifier, $V_{o u t} = - V_{1} = - 2 V_{a}$ , and forgetting that its + input carries $V_{b}$ . That gives a gain of −2 or 1 instead of 2. A second mistake is taking op-amp 1's gain as 1 (a follower) when its − input actually sits at the midpoint of a 10 kΩ/10 kΩ divider.

From the GATE 2025 EE paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2025 EE Q.56: solution and answer

Answer and solution

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