GATE 2022 EC Q.58: solution and answer

GATE 2022 · ECQ.58 · 2 marks · NAT

Consider the circuit shown with an ideal OPAMP. The output voltage $V_{0}$ is ________ V (rounded off to two decimal places).

GATE 2022 EC question 58: An R-2R resistor ladder drives an inverting op-amp amplifier, with R = 1 k Ω. — Figure from the GATE 2022 EC paper. It is part of the question.

Answer and solution

ANSWER

$V_{0} = - 0.50 V$ .

Official answer: -0.55 to -0.45Source: Official answer key, GATE 2022 EC (IIT Kharagpur)

The op-amp is ideal and its (+) input is grounded, so the (−) input is a virtual ground at $0 V$ . The ladder's final series $2 R$ therefore ends at $0 V$ . Reduce the ladder from the left, one node at a time, using Thévenin equivalents.
Node A has $2 R$ to $1.6 V$ in parallel with $2 R$ to ground. Thévenin: $V_{t h} = \frac{1.6}{2} = 0.8 V$ , $R_{t h} = 2 R ∥ 2 R = R$ . Adding the series $R$ gives $0.8 V$ behind $2 R$ .
Node B adds $2 R$ to ground: $V_{t h} = \frac{0.8 + 0}{2} = 0.4 V$ , $R_{t h} = R$ . Adding the series $R$ gives $0.4 V$ behind $2 R$ .
Node C adds $2 R$ to $1.6 V$ . Two equal resistances average their source voltages (Millman's theorem): $V_{t h} = \frac{0.4 + 1.6}{2} = 1.0 V$ , $R_{t h} = R$ . Adding the series $R$ gives $1.0 V$ behind $2 R$ .
Node D adds $2 R$ to ground: $V_{t h} = \frac{1.0 + 0}{2} = 0.5 V$ , $R_{t h} = R$ .
This equivalent source drives the final series $2 R$ into the virtual ground, so the input current is $I = \frac{0.5}{R + 2 R} = \frac{0.5 V}{3 k Ω} = \frac{1}{6} mA$
No current enters the ideal op-amp, so all of $I$ flows through the feedback $3 R$ : $V_{0} = - I \cdot 3 R = - \frac{1}{6} mA \times 3 k Ω = - 0.5 V$ This is well inside the $\pm 15 V$ rails, so the op-amp is not saturated.
Check: the voltage at node D is $V_{D} = 0.5 - I R = 0.5 - \frac{1}{6} = \frac{1}{3} V$ , and the current in the series $2 R$ is $\frac{V _{D}}{2 R} = \frac{1/3 V}{2 k Ω} = \frac{1}{6} mA$ , the same current. Full nodal analysis gives the same node voltages: $V_{A} = \frac{2}{3}$ , $V_{B} = \frac{8}{15}$ , $V_{C} = \frac{2}{3}$ , $V_{D} = \frac{1}{3} V$ . So $V_{0} = - 0.50 V$ .

KEY IDEA

This is an R-2R ladder DAC feeding an inverting amplifier. Because the (−) input is a virtual ground, reduce the ladder from its far end. At each node, two equal $2 R$ branches combine into a Thévenin source equal to the average of their two voltages, with resistance $R$ , and the next series $R$ turns that back into a source behind $2 R$ . Each stage halves the weight of every bit to its left. The final current, $V_{t h} / (R + 2 R)$ , flows entirely through $3 R$ , so $V_{0} = - 3 R \cdot I$ .

WATCH OUT FOR

The usual mistake is to treat the circuit as a standard DAC in which the last $2 R$ goes to ground and the output is taken at node D, then use $V_{0} = - \frac{R _{f}}{R} V_{D}$ with the wrong resistance. In this circuit the last $2 R$ is in series and goes to the virtual ground, so the current is $0.5/ (3 k Ω)$ , not $0.5/ (2 k Ω)$ or $0.5/ (1 k Ω)$ . A second mistake is to forget the extra $2 R$ to ground at node A. Node A has two shunt branches, so its Thévenin voltage is $0.8 V$ , not $1.6 V$ . Dropping the minus sign of the inverting amplifier gives $+ 0.5$ , which is outside the key.

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 18 ADC & DAC questions, with notes

GATE 2022 EC Q.58: solution and answer

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