GATE 2025 IN Q.43: solution and answer

GATE 2025 · INQ.43 · 2 marks · MCQ

A 3-bit DAC is implemented using ideal opamp and switches as shown in the figure. Each of the switches gets closed when its corresponding digital input is at logic 1. For a digital input of 110, the resistance $R_{in}$ seen from the reference source and the current $I$ , are

GATE 2025 IN question 43: Binary-weighted-resistor DAC with an inverting op-amp summer. — Figure from the GATE 2025 IN paper. It is part of the question.

(A)
$R_{in} = 2 k Ω$ and $I = 3 mA$
(B)
$R_{in} = 12 k Ω$ and $I = 0.5 mA$
(C)
$R_{in} = \infty Ω$ and $I = 1 mA$
(D)
$R_{in} = \infty Ω$ and $I = 3 mA$

Answer and solution

ANSWER

$R_{in} = 2 k Ω$ and $I = 3 mA$ , option (A).

Official answer: ASource: Official answer key, GATE 2025 IN (IIT Roorkee)

Decode the input $D_{2} D_{1} D_{0} = 110$ : the $D_{2}$ switch (3 k $Ω$ ) and the $D_{1}$ switch (6 k $Ω$ ) close, and the $D_{0}$ switch (12 k $Ω$ ) stays open.
Virtual ground: the + input is grounded and the ideal op-amp has negative feedback through the 10 k $Ω$ resistor, so the − input sits at $0 V$ . The right-hand ends of all three resistors are therefore at $0 V$ .
An open switch leaves its branch floating, so the 12 k $Ω$ branch carries no current and plays no part. In this circuit it is not switched to ground.
The source therefore sees only the two closed branches in parallel, each running from $6 V$ to virtual ground: $R_{in} = 3 k Ω ∥ 6 k Ω = \frac{3 \times 6}{3 + 6} k Ω = 2 k Ω$
Branch currents by Ohm's law: $I_{3 k} = \frac{6 V}{3 k Ω} = 2 mA$ and $I_{6 k} = \frac{6 V}{6 k Ω} = 1 mA$ .
KCL at the inverting node: an ideal op-amp input draws no current, so the whole branch current goes through the 10 k $Ω$ feedback resistor: $I = 2 + 1 = 3 mA = \frac{V _{r e f}}{R _{in}} = \frac{6 V}{2 k Ω}$
(Not needed for the answer: an ideal op-amp output would be $V_{o} = - I \times 10 k Ω = - 30 V$ .) So $R_{in} = 2 k Ω$ and $I = 3 mA$ , which is option (A).

KEY IDEA

In a binary-weighted-resistor DAC, the op-amp's inverting input is a virtual ground. Each closed branch therefore runs from $V_{r e f}$ to $0 V$ and draws $V_{r e f} / R$ on its own. The reference source sees the closed branches in parallel, and their currents add in the feedback resistor, so $I = V_{r e f} / R_{in}$ .

WATCH OUT FOR

There are three common mistakes. (1) Reading the bits in the wrong order, as if $D_{0} = 1$ , $D_{1} = 1$ , $D_{2} = 0$ . That closes the 12 k $Ω$ and 6 k $Ω$ branches and gives 4 k $Ω$ and 1.5 mA, which is not an option. (2) Taking $I$ as the current in the top 12 k $Ω$ branch only, because the arrow is drawn at that node. This leads towards option (B), yet that branch is open and carries nothing. $I$ is the total current into the summing node. (3) Treating the op-amp input as an open circuit, which gives $R_{in} = \infty$ (options C and D). The inverting input draws no current, but it is held at 0 V, so each closed branch still draws current from the source.

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 18 ADC & DAC questions, with notes

GATE 2025 IN Q.43: solution and answer

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