GATE 2024 IN Q.41: solution and answer

GATE 2024 · INQ.41 · 2 marks · MCQ

In the figure shown, both the opamps $A_{1}$ and $A_{2}$ are ideal, except that the opamp $A_{1}$ has an offset voltage ( $V_{os}$ ) of 1 mV. For $V_{in} = 0$ V, the values of the output voltages $V_{o u t 1}$ and $V_{o u t 2}$ , respectively, are

GATE 2024 IN question 41: Two op-amps, A1 (upper) and A2 (lower), each powered by +V_CC and -V_EE. — Figure from the GATE 2024 IN paper. It is part of the question.

(A)
3 mV and −1 mV
(B)
1 mV and 0 mV
(C)
1 mV and −1 mV
(D)
2 mV and 0 mV

Answer and solution

ANSWER

$V_{o u t 1} = 3 mV$ and $V_{o u t 2} = - 1 mV$ . The answer is option (A).

Official answer: ASource: Official answer key, GATE 2024 IN (IISc Bengaluru)

Assumptions: both op-amps are ideal (infinite gain, zero input current) and each has negative feedback, so each holds a virtual short between its inputs. The offset of $A_{1}$ is drawn as a battery $V_{os} = 1 mV$ in series with its inverting input, with the long plate (positive terminal) on the side of the junction node. Call the junction of the three resistors node $X$ .
Virtual short at $A_{1}$ : its $+$ input is grounded, so its inverting input sits at $0 V$ . Going through the battery from its $-$ terminal to its $+$ terminal gives $V_{X} = 0 + V_{os} = + 1 mV$ . So node $X$ is not a virtual ground.
Virtual short at $A_{2}$ : its $+$ input is grounded, so its inverting input is at $0 V$ . The resistor $R$ from node $X$ therefore carries a current $V_{X} / R$ from node $X$ towards $A_{2}$ .
KCL at node $X$ with $V_{in} = 0$ . The battery branch carries no current, because the input current of $A_{1}$ is zero. Summing the currents entering $X$ : $\frac{0 - V _{X}}{R} + \frac{V _{o u t 1} - V _{X}}{R} - \frac{V _{X} - 0}{R} = 0 \Rightarrow V_{o u t 1} = 3 V_{X}$
So $V_{o u t 1} = 3 \times 1 mV = 3 mV$ . Another way to see it: node $X$ has two resistors to $0 V$ (the source resistor, since $V_{in} = 0$ , and the resistor to the virtual ground of $A_{2}$ ) and the feedback resistor to $V_{o u t 1}$ . This gives a gain of $1 + \frac{R _{f}}{R ∥ R} = 1 + \frac{R}{R /2} = 3$ from $V_{X}$ to $V_{o u t 1}$ .
$A_{2}$ is an inverting amplifier with gain $- \frac{R}{R} = - 1$ , driven by $V_{X}$ : $V_{o u t 2} = - V_{X} = - 1 mV$
Hence $V_{o u t 1} = 3 mV$ and $V_{o u t 2} = - 1 mV$ , which is option (A).

KEY IDEA

A series offset source moves the node it is connected to. The virtual short pins the op-amp's inverting input at 0 V, so the junction node sits at $+ V_{os} = 1 mV$ instead of at virtual ground. That node sees two resistors to 0 V and one feedback resistor, so it is amplified by the noise gain $1 + R / (R ∥ R) = 3$ . The second op-amp simply inverts $V_{X}$ with gain $- 1$ .

WATCH OUT FOR

A common mistake is to treat node $X$ as a virtual ground, which gives $V_{o u t 2} = 0$ (options B or D). Another is to count only the source resistor when working out the gain, giving $1 + R / R = 2$ and $V_{o u t 1} = 2 mV$ (option D). The resistor to the virtual ground of $A_{2}$ also loads node $X$ , so the gain is 3. Reading the battery polarity backwards flips both signs.

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2024 IN Q.41: solution and answer

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