GATE 2023 IN Q.62: solution and answer

GATE 2023 · INQ.62 · 2 marks · NAT

In the circuit shown, assuming an ideal opamp, the value of the output voltage $V_{o}$ = ____ V (rounded off to one decimal place).

GATE 2023 IN question 62: Op-amp with inverting (−) input on top and non-inverting (+) input below; — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

ANSWER

$V_{o} = 2.0 V$ .

Official answer: 1.9 to 2.1Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Assume an ideal op-amp: infinite open-loop gain and zero input currents. Step 7 shows that the negative feedback outweighs the positive feedback, so the two inputs sit at the same voltage (virtual short): $V_{-} = V_{+}$ .
Inverting side. No current enters the input, so the two $3 R$ resistors form a voltage divider from $V_{o}$ to ground: $V_{-} = V_{o} \frac{3 R}{3 R + 3 R} = \frac{V _{o}}{2}$ . The virtual short then gives $V_{P} = V_{+} = \frac{V _{o}}{2}$ , where P is the non-inverting-input node.
KCL at node P (no current into the input). The current in from the 1 V source through $3 R$ equals the current out through $R$ to node X: $\frac{1 - V _{P}}{3 R} = \frac{V _{P} - V _{X}}{R} \Rightarrow 1 = 4 V_{P} - 3 V_{X}$
KCL at node X. The currents in from P and from the output equal the current out to ground: $\frac{V _{P} - V _{X}}{R} + \frac{V _{o} - V _{X}}{R} = \frac{V _{X}}{R} \Rightarrow 3 V_{X} = V_{P} + V_{o}$
Substitute $V_{P} = V_{o} /2$ into the node-X equation: $3 V_{X} = \frac{V _{o}}{2} + V_{o} = \frac{3 V _{o}}{2} \Rightarrow V_{X} = \frac{V _{o}}{2}$ . So $V_{X} = V_{P}$ , and no current flows in the $R$ between P and X.
Substitute into the node-P equation: $1 = 4 \cdot \frac{V _{o}}{2} - 3 \cdot \frac{V _{o}}{2} = \frac{V _{o}}{2} \Rightarrow V_{o} = 2 V$
Check. $V_{P} = V_{X} = 1 V$ , so no current flows in the $3 R$ from the 1 V source or in the $R$ between P and X. At X, $\frac{2 - 1}{R} = \frac{1}{R}$ flows in from the output and $\frac{1}{R}$ flows out to ground, so KCL balances. Feedback sign: with the source set to zero, node X sees $R$ to ground in parallel with $R + 3 R = 4 R$ to ground, which is $0.8 R$ . So $V_{X} = V_{o} \frac{0.8 R}{R + 0.8 R} = \frac{4}{9} V_{o}$ and $V_{+} = \frac{3 R}{4 R} V_{X} = \frac{1}{3} V_{o}$ . The positive-feedback fraction $\frac{1}{3}$ is smaller than the negative-feedback fraction $\frac{1}{2}$ . Net feedback is negative, so the virtual short is valid.
Answer: $V_{o} = 2.0 V$ .

KEY IDEA

The output feeds back to both inputs: negative feedback through the $3 R$ – $3 R$ divider (fraction $\frac{1}{2}$ ) and positive feedback through the R-network (fraction $\frac{1}{3}$ ). Net feedback is negative, so the virtual short holds. It gives $V_{+} = V_{o} /2$ , and nodal analysis of the non-inverting network then fixes $V_{o}$ .

WATCH OUT FOR

One mistake is to ignore the $R$ from node X to the output and treat the non-inverting side as a plain divider of the 1 V source. That gives $V_{+} = 1 \cdot \frac{2 R}{5 R} = 0.4 V$ and $V_{o} = 0.8 V$ . Another is to see positive feedback and assume the op-amp saturates, without checking that the negative feedback ( $\frac{1}{2}$ ) is stronger than the positive feedback ( $\frac{1}{3}$ ).

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

GATE 2023 IN Q.62: solution and answer

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