GATE 2020 IN Q.39: solution and answer

GATE 2020 · INQ.39 · 2 marks · NAT

If the opamps in the circuit shown are ideal and $V_{X} = 0.5 mV$ , the steady state value of $V_{O}$ (in volts, rounded off to two decimal places) is ____

GATE 2020 IN question 39: Two ideal op-amps in cascade. — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

ANSWER

$V_{O} = 1000 V_{X} = 0.50 V$ .

Official answer: 0.45 to 0.55Source: Official answer key, GATE 2020 IN (IIT Delhi)

Assumptions. Both op-amps are ideal: open-loop gain $A \to \infty$ and zero input current. $V_{X}$ is an ideal DC source. Op-amp 1's inverting input draws no current, so the left $100 Ω$ resistor carries no current and $V_{-} = 0$ .
$V_{X}$ connects only to op-amp 1's non-inverting input, which draws no current, so no current flows through $V_{X}$ . The $100 Ω$ and $99.9 k Ω$ resistors are therefore an unloaded divider from $V_{O}$ to node N: $V_{N} = V_{O} \frac{100}{100 + 99 900} = \frac{V _{O}}{1000}$
The $+$ terminal of $V_{X}$ is at node N and its $-$ terminal is at op-amp 1's $+$ input, so going through the source gives $V_{+} = V_{N} - V_{X} = \frac{V _{O}}{1000} - V_{X}$
Op-amp 2 is an inverting integrator: $V_{O} (s) = - \frac{V _{1} ( s )}{s R C}$ with $R C = 20 k Ω \times 100 nF = 2 ms$ . Here $V_{1} = A (V_{+} - V_{-})$ is op-amp 1's output.
Check that the loop settles. Combining steps 1 to 4: $s R C V_{O} (s) = - A (\frac{V _{O} ( s )}{1000} - V_{X} (s)) \Rightarrow \frac{V _{O} ( s )}{V _{X} ( s )} = \frac{A}{s R C + A /1000}$ This has a single pole at $s = - \frac{A}{1000 R C}$ , which is in the left half-plane for any $A > 0$ . The loop is negative feedback: if $V_{O}$ rises, $V_{+}$ rises, $V_{1}$ rises, and the inverting integrator pulls $V_{O}$ back down. So a steady state exists.
Steady state. Any non-zero constant $V_{1}$ at the integrator's input would make $V_{O}$ ramp without limit, so in steady state $V_{1} = 0$ . Since $V_{1} = A (V_{+} - V_{-})$ , this means $V_{+} = V_{-} = 0$ .
So $V_{N} = V_{X}$ , which gives $\frac{V _{O}}{1000} = 0.5 mV$ and $V_{O} = 1000 \times 0.5 mV = 0.50 V$ . Check with the final value theorem for a DC input $V_{X} (s) = V_{X} / s$ : $V_{O} (\infty) = s \to 0 lim s \cdot \frac{A}{s R C + A /1000} \cdot \frac{V _{X}}{s} = 1000 V_{X} = 0.50 V$

KEY IDEA

An integrator inside a negative-feedback loop drives its own input to zero at DC. That input is op-amp 1's output, so op-amp 1's inputs end up at the same voltage: $V_{+} = V_{-} = 0$ . The divider voltage $V_{O} /1000$ must therefore exactly cancel $V_{X}$ , so $V_{O} = 1000 V_{X}$ . The integrator enforces this even when $A$ is finite, so the result does not depend on $A$ .

WATCH OUT FOR

One mistake is reading the source polarity backwards. Its $-$ sign is drawn as a short vertical bar because the source is drawn horizontally, and a backwards reading gives $V_{O} = - 0.50 V$ . Another is treating op-amp 1 as an open-loop comparator that saturates. A third is assuming a DC input makes the integrator output ramp forever, when in fact the loop drives the integrator's input to zero. Students also sometimes load the divider with a current through $V_{X}$ , but that current is zero because $V_{X}$ feeds an op-amp input. Always check that the overall loop is negative feedback before using the virtual short.

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2020 IN Q.39: solution and answer

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