GATE 2018 IN Q.42: solution and answer

GATE 2018 · INQ.42 · 2 marks · NAT

The circuit given uses ideal opamps. The current $I$ (in $μ$ A) drawn from the source $v_{s}$ is (up to two decimal places) ____.

GATE 2018 IN question 42: A source v_s = 1 V drives a node; — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

ANSWER

$I \approx 0.99 μ A$ (exactly $\frac{0.1}{101} mA = 0.990 μ A$ ).

Official answer: 0.98 to 1.00Source: Official answer key, GATE 2018 IN (IIT Guwahati)

Assume ideal op-amps (infinite gain, infinite input resistance, zero output resistance). Each op-amp has negative feedback through its $20.0 k Ω$ resistor and its (+) input is grounded, so by the virtual short both inverting inputs sit at $0 V$ . The source node is held at $v_{s} = 1 V$ by the ideal source, so the $10.1 k Ω$ path does not feed anything back into the amplifiers.
Op-amp 1 is an inverting amplifier: $V_{o 1} = - \frac{20.0}{10.0} v_{s} = - 2 \times 1 = - 2 V$ .
Op-amp 2 is an inverting amplifier with gain $- 1$ : $V_{o 2} = - \frac{20.0}{20.0} V_{o 1} = + 2 V$ . Its output resistance is zero, so the $10.1 k Ω$ resistor connected to it does not change this value.
Current from the source node through the $10.0 k Ω$ resistor into the virtual ground: $I_{1} = \frac{1 - 0}{10.0 k Ω} = 100 μ A$ .
Current from the source node through the $10.1 k Ω$ resistor towards the output of op-amp 2: $I_{2} = \frac{1 - 2}{10.1 k Ω} = - 99.01 μ A$ . The minus sign means this current actually flows from the op-amp 2 output into the source node.
KCL at the source node (the source must supply whatever leaves the node): $I = I_{1} + I_{2} = (\frac{1}{10} - \frac{1}{10.1}) mA = \frac{0.1}{101} mA \approx 0.990 μ A$
To two decimal places, $I = 0.99 μ A$ .

KEY IDEA

The two cascaded inverting stages give a net gain of $+ 2$ , so the right end of the $10.1 k Ω$ resistor sits at $2 v_{s}$ . That resistor returns almost all of the current that the $10.0 k Ω$ resistor draws into the virtual ground, and the source supplies only the small difference $v_{s} (\frac{1}{10 k Ω} - \frac{1}{10.1 k Ω})$ . The circuit makes the source see a very large input resistance of about $1.01 M Ω$ , and this is the principle behind bootstrapping and negative-impedance input stages.

WATCH OUT FOR

Stopping at $100 μ A$ (only the $10.0 k Ω$ branch), or taking $V_{o 2} = - 2 V$ by forgetting that the second inverting stage flips the sign again. The second error gives $I = 1/10 + 3/10.1 \approx 397 μ A$ . Another slip is rounding $1/10.1 mA$ to $0.099 mA$ too early. The answer is a small difference of two nearly equal currents, so keep at least five significant figures until the final subtraction.

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2018 IN Q.42: solution and answer

Answer and solution

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