ISRO EC 2025 Q.46: solution and answer

ISRO 2025 · ECQ.46 · 1 mark · MCQ

In the below figure, determine the signs on the OPAMP terminals A and B for negative feedback operation and gate potential of the transistor for Vref = 3V and Vdd = 5V. $μ_{n} C_{o x} (\frac{W}{L}) = \frac{1 mA}{V ^{2}}$ , $V_{T} = 0.7 V$ .

ISRO EC 2025 question 46: A supply rail Vdd at the top. — Figure from the ISRO EC 2025 paper. It is part of the question.

(A)
Terminal A is positive. Terminal B is negative
(B)
Terminal A is negative. Terminal B is positive
(C)
Both of above
(D)
None of the above

Answer and solution

ANSWER

Terminal A is positive (non-inverting) and terminal B is negative (inverting), option (a). The gate sits at $3 V$ .

Official answer: ASource: ISRO's final answer key of 15 May 2026

Read the circuit. The $0.5 mA$ current source pushes current down into a node. That node is the MOSFET drain and is also wired to terminal B. The MOSFET gate is fixed at $3 V$ . The MOSFET source is wired to the op-amp output. $V_{r e f}$ is applied to terminal A.
Find the sign of B by tracing the loop. Suppose the output rises. The source voltage rises, so $V_{GS} = 3 - V_{S}$ falls and the MOSFET can carry less current.
The current source still supplies $0.5 mA$ , but the transistor now takes less. The surplus current charges the drain node, so the drain voltage, which is $V_{B}$ , rises.
A rise in the output therefore raises $V_{B}$ . For negative feedback, a rise in the output must push the output back down, so B must be the inverting (negative) terminal. Terminal A is then the non-inverting (positive) terminal.
Check the operating point, assuming saturation: $I_{D} = \frac{1}{2} μ_{n} C_{o x} \frac{W}{L} (V_{GS} - V_{T})^{2}$ , so $0.5 = \frac{1}{2} (1) (V_{GS} - 0.7)^{2}$ . This gives $(V_{GS} - 0.7)^{2} = 1$ , so $V_{GS} = 1.7 V$ .
With the virtual short, $V_{B} = V_{A} = V_{r e f} = 3 V$ , so the drain is at $3 V$ . The source is at $3 - 1.7 = 1.3 V$ , giving $V_{D S} = 3 - 1.3 = 1.7 V$ . Since $V_{GS} - V_{T} = 1 V$ and $V_{D S} > 1 V$ , the MOSFET is in saturation, as assumed. The gate potential is $3 V$ .
So A is positive and B is negative, which is option (a). Option (b) would be the positive-feedback connection, so options (c) and (d) are also wrong.

KEY IDEA

To find which op-amp input is inverting, raise the output and follow the loop through the circuit. The input that rises with the output must be the inverting one. Here a higher output lowers $V_{GS}$ , so the transistor current falls and the drain node (terminal B) rises.

WATCH OUT FOR

Assuming the input connected to the transistor side is the non-inverting one without tracing the loop. Here the output drives the source, not the gate, so the usual common-source inversion does not apply and the sign comes out opposite to a quick guess.

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 4 ISRO Op-amps questions, with notes

ISRO EC 2025 Q.46: solution and answer

Answer and solution

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