GATE 2022 EE Q.36: solution and answer

GATE 2022 · EEQ.36 · 2 marks · MCQ

The output impedance of a non-ideal operational amplifier is denoted by $Z_{out}$ . The variation in the magnitude of $Z_{out}$ with increasing frequency, $f$ , in the circuit shown below, is best represented by

GATE 2022 EE question 36: Op-amp voltage follower (unity-gain buffer). — Figure from the GATE 2022 EE paper. It is part of the question.

(A)
Figure — $lo g (∣ Z_{out} ∣)$ vs $lo g (f)$ : a horizontal line, constant at all frequencies.
(B)
Figure — $lo g (∣ Z_{out} ∣)$ vs $lo g (f)$ : a straight line rising from the origin, increasing at all frequencies.
(C)
Figure — $lo g (∣ Z_{out} ∣)$ vs $lo g (f)$ : flat at a low level, then a rising straight segment, then flat again at a higher level.
(D)
Figure — $lo g (∣ Z_{out} ∣)$ vs $lo g (f)$ : flat at a high level, then a falling straight segment, then flat again at a lower level.

Answer and solution

ANSWER

(C): $∣ Z_{out} ∣$ is flat at $r_{o} / (1 + A_{0})$ at low frequencies, rises at $+ 20 dB/decade$ between $f_{b}$ and $f_{T}$ , and levels off at $r_{o}$ above $f_{T}$ .

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

Model. The circuit is a voltage follower: the output is wired straight back to the inverting input, so the feedback factor is $β = 1$ . Model the non-ideal op-amp as an infinite input impedance, an open-loop output resistance $r_{o}$ and a single-pole open-loop gain $A (f) = \frac{A _{0}}{1 + j f / f _{b}}$ , where $f_{b}$ is the open-loop break frequency.
Output impedance by a test source. Set $V_{in} = 0$ , so the non-inverting input is grounded, and apply a test voltage $V_{x}$ at the output. The inverting input sits at $V_{x}$ , so the op-amp's internal source is $A (0 - V_{x}) = - A V_{x}$ . The current into the output terminal is $I_{x} = \frac{V _{x} - ( - A V _{x} )}{r _{o}} = \frac{( 1 + A ) V _{x}}{r _{o}}$ . Therefore $Z_{out} = \frac{V _{x}}{I _{x}} = \frac{r _{o}}{1 + A ( f ) β} = \frac{r _{o}}{1 + A ( f )}$ This is the standard result for voltage-sampling (shunt-output) negative feedback: the open-loop output resistance is divided by $1 + loop gain$ .
Substitute $A (f)$ : $Z_{out} = \frac{r _{o}}{1 + \frac{A _{0}}{1 + j f / f _{b}}} = r_{o} \frac{1 + j f / f _{b}}{( 1 + A _{0} ) + j f / f _{b}}$
Read off the zero and the pole. The zero is at $f = f_{b}$ . The pole is at $f = (1 + A_{0}) f_{b} \approx A_{0} f_{b} = f_{T}$ , the unity-gain frequency. Since $A_{0} ≫ 1$ , the pole lies far above the zero.
Low frequencies, $f ≪ f_{b}$ : $∣ Z_{out} ∣ \approx \frac{r _{o}}{1 + A _{0}}$ . This is very small and does not depend on $f$ , so the plot is flat at a low level.
Mid band, $f_{b} ≪ f ≪ f_{T}$ : the numerator is about $j f / f_{b}$ and the denominator is about $A_{0}$ , so $∣ Z_{out} ∣ \approx \frac{r _{o} f}{A _{0} f _{b}} = \frac{r _{o} f}{f _{T}}$ . It is proportional to $f$ and rises at $+ 20 dB/decade$ , a straight rising segment on the log-log axes. In this band the output behaves like an inductor, because the loop gain that was holding $Z_{out}$ down is falling at $- 20 dB/decade$ .
High frequencies, $f ≫ f_{T}$ : both $j f / f_{b}$ terms dominate, so $Z_{out} \to r_{o}$ . The loop gain is gone and the plot is flat again, at the higher level $r_{o}$ .
Shape: low and flat, then rising, then high and flat. Option (A) is the ideal-feedback picture with no frequency dependence, (B) never levels off, and (D) is the reverse trend. The answer is (C).

KEY IDEA

Voltage-sampling negative feedback divides the op-amp's open-loop output resistance by $1 + A (f) β$ . For a follower $β = 1$ , so $Z_{out} = r_{o} / (1 + A (f))$ . The open-loop gain rolls off above $f_{b}$ , so $∣ Z_{out} ∣$ rises from $r_{o} / (1 + A_{0})$ to $r_{o}$ between $f_{b}$ and $f_{T}$ : a zero at $f_{b}$ and a pole at about $f_{T}$ .

WATCH OUT FOR

Treating the op-amp as ideal, so that $Z_{out} = 0$ at every frequency, and picking the flat plot (A). The question says the op-amp is non-ideal. Another mistake is sketching the shape of the open-loop gain curve (flat, falling, flat) and picking (D). $Z_{out}$ varies inversely with the loop gain, so it rises where the gain falls.

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.36: solution and answer

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