GATE 2022 EE Q.44: solution and answer

GATE 2022 · EEQ.44 · 2 marks · MCQ

Q.44 The current gain $(I_{o u t} / I_{in})$ in the circuit with an ideal current amplifier given below is

GATE 2022 EE question 44: An ideal op-amp (gain marked ∞). — Figure from the GATE 2022 EE paper. It is part of the question.

(A)
$\frac{C _{f}}{C _{c}}$
(B)
$\frac{- C _{f}}{C _{c}}$
(C)
$\frac{C _{c}}{C _{f}}$
(D)
$\frac{- C _{c}}{C _{f}}$

Answer and solution

ANSWER

The current gain is $I_{o u t} / I_{in} = C_{c} / C_{f}$ , option (C).

Official answer: CSource: Official answer key, GATE 2022 EE (IIT Kharagpur)

Assumptions: ideal op-amp (infinite open-loop gain, zero input current, zero output resistance). Work in the Laplace (or phasor) domain, where a capacitor $C$ has impedance $\frac{1}{s C}$ . The ratio asked for is the signal (AC) current gain.
Virtual short: negative feedback through $C_{f}$ holds the two inputs at the same potential. The (+) input is grounded, so the (−) input is also at $0 V$ (virtual ground).
Read the source's direction: its arrow points down, so $I_{in}$ flows from the inverting node, through the source, into ground. It draws current out of the inverting node.
KCL at the inverting node: the op-amp input takes no current, so the current leaving through the source must arrive through $C_{f}$ from the output. The current in $C_{f}$ from the output to the node is $(V_{o} - 0) s C_{f}$ , so $V_{o} s C_{f} = I_{in} \Rightarrow V_{o} = \frac{I _{in}}{s C _{f}}$
The op-amp output is an ideal voltage source (zero output resistance), so the load $C_{c}$ does not change $V_{o}$ . The current through $C_{c}$ from the output to ground, which is the direction of the $I_{o u t}$ arrow, is $I_{o u t} = (V_{o} - 0) s C_{c} = \frac{I _{in}}{s C _{f}} s C_{c}$
Therefore $\frac{I _{o u t}}{I _{in}} = \frac{C _{c}}{C _{f}}$ This is a positive, frequency-independent ratio, which is option (C).

KEY IDEA

The "ideal current amplifier" is an ideal op-amp with capacitive feedback. Virtual ground at the inverting input forces all of $I_{in}$ through $C_{f}$ , which sets $V_{o} = I_{in} / (s C_{f})$ . The zero-resistance output then drives $C_{c}$ to ground, giving $I_{o u t} = s C_{c} V_{o}$ . The $s$ terms cancel, so the gain is the capacitor ratio $C_{c} / C_{f}$ . Its sign depends only on the reference directions drawn for $I_{in}$ and $I_{o u t}$ .

WATCH OUT FOR

Using the inverting-amplifier rule "gain $= - Z_{f} / Z_{in}$ " out of habit and picking a negative option, (B) or (D). The sign has to come from the arrows. The source draws $I_{in}$ out of the virtual-ground node, so $V_{o}$ comes out positive ( $V_{o} = + I_{in} / (s C_{f})$ ). $I_{o u t}$ is defined in the same direction as the current $V_{o}$ drives through $C_{c}$ to ground, so the ratio is $+ C_{c} / C_{f}$ . Another slip is inverting the ratio to $C_{f} / C_{c}$ , which confuses the impedance $1/ (s C)$ with the admittance $s C$ .

From the GATE 2022 EE paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2022 EE Q.44: solution and answer

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