GATE 2020 IN Q.49: solution and answer

GATE 2020 · INQ.49 · 2 marks · NAT

A circuit consisting of capacitors, DC voltage source and an amplifier having a voltage gain G = −5 is shown in the figure. The effective capacitance across the nodes A and B (in µF, rounded off to one decimal place) is ____

GATE 2020 IN question 49: Ladder circuit with an amplifier. — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

ANSWER

The effective capacitance across A and B is $C_{A B} = \frac{40}{14} + 2 (1 - (- 5)) = 2.857 + 12 = 14.857 μ F \approx 14.9 μ F$ .

Official answer: 14.5 to 15.0Source: Official answer key, GATE 2020 IN (IIT Delhi)

Deactivate the source. The capacitance seen between A and B is found with the independent source switched off, so the DC source $V_{s}$ is replaced by a short circuit. The left end of the 7 µF capacitor is then tied to B (ground). The amplifier is taken as ideal: its input draws no current and its output is a stiff voltage $V_{o} = G V_{A}$ .
Passive branch into A. With the source shorted, the 7 µF and 3 µF capacitors both run from node X to ground, so they are in parallel: $7 + 3 = 10 μ F$ . That combination is in series with the 4 µF capacitor: $C_{1} = \frac{4 \times 10}{4 + 10} = \frac{40}{14} = 2.857 μ F$
Miller's theorem for the 2 µF feedback capacitor. The capacitor sits between the amplifier input (A) and its output, where $V_{o} = G V_{A}$ . Its current, flowing out of A, is $i = C \frac{d}{d t} (V_{A} - G V_{A}) = C (1 - G) \frac{d V _{A}}{d t}$ Seen from A, it therefore behaves as a capacitance to ground of $C_{M} = C (1 - G) = 2 (1 - (- 5)) = 12 μ F$
Combine at node A. Both $C_{1}$ and $C_{M}$ run from A to ground (B), so they are in parallel. The ideal amplifier input draws no current, so nothing else loads node A: $C_{A B} = C_{1} + C_{M} = 2.857 + 12 = 14.857 μ F$
Check with a test voltage $V$ applied at A. The charge drawn through the 2 µF capacitor is $2 (V - G V) = 2 (1 + 5) V = 12 V$ , and through the passive branch it is $\frac{40}{14} V$ . So $Q = (12 + 2.857) V \Rightarrow C_{A B} = \frac{Q}{V} = 14.857 μ F \approx 14.9 μ F$

KEY IDEA

Short the independent DC source, reduce the passive ladder (7 µF parallel with 3 µF, then in series with 4 µF, giving 40/14 µF), and use Miller's theorem on the feedback capacitor: at the input side it looks like $C (1 - G)$ to ground, which is 12 µF for G = −5. The two capacitances at A are in parallel.

WATCH OUT FOR

Three slips are common. (1) Getting the Miller sign wrong. Writing $C (1 + G)$ with G = −5 gives a negative capacitance, and using the output-side value $C (1 - 1/ G) = 2.4 μ F$ gives about 5.3 µF. (2) Treating the source as an open circuit, so the 7 µF branch drops out and the passive part becomes $4 \times 3/7 = 1.71 μ F$ , for a total of about 13.7 µF. (3) Putting the 7 µF and 3 µF in series instead of in parallel once the source is shorted.

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 16 Theorems questions, with notes

GATE 2020 IN Q.49: solution and answer

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