GATE 2023 IN Q.59: solution and answer

GATE 2023 · INQ.59 · 2 marks · NAT

The opamps in the circuit are ideal. The input signals are

$V_{S 1} = 3 + 0.10 sin (300 t) V$ and $V_{S 2} = - 2 + 0.11 sin (300 t) V$ .

The average value of the voltage $V_{O}$ is ______ V (rounded off to two decimal places).

GATE 2023 IN question 59: Two ideal op-amps, one above the other. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

ANSWER

The average value of $V_{O}$ is $0.50 V$ .

Official answer: 0.49 to 0.51Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Assumptions: the op-amps are ideal, so no current flows into either input terminal, each op-amp's two inputs sit at the same voltage (virtual short), and each output behaves as an ideal voltage source (zero output resistance). Each op-amp's inverting input connects back to its own output through a resistor of the left chain, so both have negative feedback and the virtual short holds. The $V_{O}$ terminal is open-circuited (no load).
Virtual short at each op-amp: node A (the top op-amp's inverting input, the junction of the upper and middle resistors of the left chain) is at $V_{A} = V_{S 1}$ . Node B (the bottom op-amp's inverting input, the junction of the middle and lower resistors) is at $V_{B} = V_{S 2}$ .
Ohm's law on the middle resistor: a current $I = \frac{V _{S 1} - V _{S 2}}{R}$ flows from A to B. No current enters the op-amp inputs, so the same $I$ also flows down through the upper and lower resistors of the left chain.
Outputs of the two op-amps: $V_{o 1} = V_{A} + I R = V_{S 1} + (V_{S 1} - V_{S 2}) = 2 V_{S 1} - V_{S 2}$ $V_{o 2} = V_{B} - I R = V_{S 2} - (V_{S 1} - V_{S 2}) = 2 V_{S 2} - V_{S 1}$
Right chain: two equal resistors $R$ in series between $V_{o 1}$ and $V_{o 2}$ . The op-amp outputs are ideal voltage sources, so this divider does not change $V_{o 1}$ or $V_{o 2}$ , and with no load on the output $V_{O}$ sits at their midpoint: $V_{O} = \frac{V _{o 1} + V _{o 2}}{2} = \frac{( 2 V _{S 1} - V _{S 2} ) + ( 2 V _{S 2} - V _{S 1} )}{2} = \frac{V _{S 1} + V _{S 2}}{2}$
Substitute the inputs: $V_{O} = \frac{( 3 - 2 ) + ( 0.10 + 0.11 ) sin ( 300 t )}{2} = 0.5 + 0.105 sin (300 t) V$
Average value: over one period $T = \frac{2 π}{300} s$ the sinusoid averages to zero, so the average of $V_{O}$ is $0.50 V$ . Check with the DC levels alone: $V_{o 1} = 2 (3) - (- 2) = 8 V$ and $V_{o 2} = 2 (- 2) - 3 = - 7 V$ , and their midpoint is $\frac{8 + ( - 7 )}{2} = 0.5 V$ .

KEY IDEA

This is the input stage of a three-op-amp instrumentation amplifier. The virtual shorts put $V_{S 1}$ and $V_{S 2}$ across the middle resistor, and the same current flows through all three resistors of the left chain, so $V_{o 1} = 2 V_{S 1} - V_{S 2}$ and $V_{o 2} = 2 V_{S 2} - V_{S 1}$ . The difference between the outputs is amplified, but their average stays equal to the input common-mode voltage. The equal-resistor divider on the right picks out that average: $V_{O} = (V_{S 1} + V_{S 2}) /2$ . Only the DC parts of the inputs survive averaging, so the answer is $(3 - 2) /2 = 0.5 V$ .

WATCH OUT FOR

A common error is to treat the circuit as a differential amplifier and compute $V_{o 1} - V_{o 2} = 3 (V_{S 1} - V_{S 2})$ , which averages to $15 V$ . But $V_{O}$ is measured from the midpoint of the right divider to ground, not across the two outputs, so it is the common-mode average $(V_{o 1} + V_{o 2}) /2$ . Other slips: getting the sign of $I R$ wrong for the bottom op-amp ( $V_{o 2} = V_{S 2} - I R$ , because the current flows from B towards the bottom output), and averaging the sine amplitudes into the answer when a sinusoid averages to zero.

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.59: solution and answer

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