GATE 2023 EC Q.39: solution and answer

GATE 2023 · ECQ.39 · 2 marks · MCQ

The $\frac{V _{O U T}}{V _{I N}}$ of the circuit shown below is

GATE 2023 EC question 39: Three op-amps, each powered from V_CC (top rail) and -V_EE (bottom rail). — Figure from the GATE 2023 EC paper. It is part of the question.

(A)
$- \frac{R _{4}}{R _{3}}$
(B)
$\frac{R _{4}}{R _{3}}$
(C)
$1 + \frac{R _{4}}{R _{3}}$
(D)
$1 - \frac{R _{4}}{R _{3}}$

Answer and solution

ANSWER

$\frac{V _{O U T}}{V _{I N}} = - \frac{R _{4}}{R _{3}}$ , option (A).

Official answer: ASource: Official answer key, GATE 2023 EC (IIT Kanpur)

Assume ideal op-amps (infinite input resistance, zero output resistance, infinite open-loop gain) that stay in their linear region, well inside the $V_{C C}$ and $- V_{E E}$ rails. Each stage has negative feedback from its output to its inverting input, so each has a virtual short across its inputs and draws no input current.
Top op-amp (inverting amplifier): its + input is grounded, so its − input is at $0 V$ . KCL at the − node (current in through $R_{1}$ equals current out through $R_{2}$ ): $\frac{V _{I N} - 0}{R _{1}} = \frac{0 - V _{1}}{R _{2}}$ , which gives $V_{1} = - \frac{R _{2}}{R _{1}} V_{I N}$
Bottom op-amp (non-inverting amplifier): $V_{I N}$ is wired straight to its + input, so its − input is also at $V_{I N}$ . KCL at the − node: $\frac{V _{I N} - 0}{R _{1}} = \frac{V _{2} - V _{I N}}{R _{2}}$ , which gives $V_{2} = (1 + \frac{R _{2}}{R _{1}}) V_{I N}$
Right op-amp (inverting summer): both $R_{3}$ resistors meet at its − input. That input is a virtual ground because the + input is grounded. KCL at the − node: $\frac{V _{1}}{R _{3}} + \frac{V _{2}}{R _{3}} = - \frac{V _{O U T}}{R _{4}}$ , so $V_{O U T} = - \frac{R _{4}}{R _{3}} (V_{1} + V_{2})$
Add the two first-stage outputs. The $R_{2} / R_{1}$ terms cancel: $V_{1} + V_{2} = (- \frac{R _{2}}{R _{1}} + 1 + \frac{R _{2}}{R _{1}}) V_{I N} = V_{I N}$
So $\frac{V _{O U T}}{V _{I N}} = - \frac{R _{4}}{R _{3}}$ which is option (A). The result does not depend on $R_{1}$ or $R_{2}$ .

KEY IDEA

With the same $R_{1}$ and $R_{2}$ , the inverting gain $- R_{2} / R_{1}$ and the non-inverting gain $1 + R_{2} / R_{1}$ always add to exactly 1. An equal-weight summer that adds the two outputs therefore sees just $V_{I N}$ , and the whole circuit reduces to the summer's inverting gain $- R_{4} / R_{3}$ .

WATCH OUT FOR

One mistake is to read the bottom op-amp as a second inverting stage because it also has an $R_{1}$ and an $R_{2}$ . In fact $V_{I N}$ drives its + input, so it is non-inverting, and $R_{1}$ goes to ground. Misread this way, the two outputs add to $- 2 (R_{2} / R_{1}) V_{I N}$ and $R_{2} / R_{1}$ does not cancel. Another mistake is to forget that the output stage is an inverting summer (its + input is grounded) and choose (B) $+ R_{4} / R_{3}$ . Treating it as a non-inverting stage gives (C).

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

GATE 2023 EC Q.39: solution and answer

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