GATE 2022 IN Q.35: solution and answer

GATE 2022 · INQ.35 · 1 mark · NAT

An OPAMP has a gain of $1 0^{4}$ , an input impedance of $10 M Ω$ and an output impedance of $100 Ω$ . The OPAMP is used in unity-gain feedback configuration in a voltage buffer circuit. The closed-loop output impedance of the OPAMP (in milliohms) in the circuit is ________ (round off to one decimal place)

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Answer and solution

ANSWER

The closed-loop output impedance is $R_{o f} = \frac{R _{o}}{1 + A β} = \frac{100}{10001} Ω \approx 9.999 m Ω \approx 10.0 m Ω$ .

Official answer: 9.9 to 10.0Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Model and assumptions. Treat the op-amp as a non-ideal amplifier: an input resistance $R_{i} = 10 M Ω$ between the two inputs, and at the output a controlled voltage source $A v_{d}$ (with $v_{d} = v_{+} - v_{-}$ and $A = 1 0^{4}$ ) in series with $R_{o} = 100 Ω$ . Take $A$ as the given frequency-independent gain, and assume the signal source driving the non-inverting input is ideal (zero source resistance).
Voltage buffer. The output is wired straight back to the inverting input, and the signal goes to the non-inverting input. So the feedback factor is $β = 1$ and the loop gain is $A β = 1 0^{4}$ . The feedback samples the output voltage (series-shunt feedback), so it should lower the output resistance.
Test source. Set the signal to zero, which grounds the non-inverting input ( $v_{+} = 0$ ). Drive the output node with a test voltage $v_{x}$ and find the current $i_{x}$ flowing into that node. Because the inverting input is tied to the output, $v_{-} = v_{x}$ , so $v_{d} = 0 - v_{x} = - v_{x}$ .
Two current paths leave the output node. (i) Through $R_{o}$ into the controlled source: $\frac{v _{x} - A v _{d}}{R _{o}} = \frac{v _{x} - A ( - v _{x} )}{R _{o}} = \frac{v _{x} ( 1 + A )}{R _{o}}$ . (ii) Through $R_{i}$ to the grounded non-inverting input, since $R_{i}$ sits between $v_{-} = v_{x}$ and $v_{+} = 0$ : $\frac{v _{x}}{R _{i}}$ .
Adding the two currents: $i_{x} = v_{x} (\frac{1 + A}{R _{o}} + \frac{1}{R _{i}}) \Rightarrow R_{o f} = \frac{v _{x}}{i _{x}} = \frac{R _{o}}{1 + A β} R_{i}$
Put in the numbers: $\frac{R _{o}}{1 + A β} = \frac{100}{1 + 1 0 ^{4}} Ω = \frac{100}{10001} Ω = 9.9990 \times 1 0^{- 3} Ω$ $R_{i} = 1 0^{7} Ω$ is about $1 0^{9}$ times larger, so putting it in parallel changes the result by roughly one part in $1 0^{9}$ . It can be neglected.
$R_{o f} \approx 9.9990 m Ω \approx 10.0 m Ω (to one decimal place)$

KEY IDEA

When negative feedback samples the output voltage (shunt sampling, as in the series-shunt voltage buffer), it divides the open-loop output resistance by $(1 + A β)$ . A unity-gain buffer has $β = 1$ , so $R_{o f} = R_{o} / (1 + A)$ . The input resistance $R_{i}$ ends up in parallel with this and has no practical effect.

WATCH OUT FOR

One mistake is multiplying $R_{o}$ by $(1 + A β)$ instead of dividing, which gives about $1 M Ω$ . That factor applies to the input impedance under series mixing, $R_{i f} = R_{i} (1 + A) \approx 1 0^{11} Ω$ . Another is entering the answer in ohms (0.01) when the question asks for milliohms (10.0). A third is claiming that $R_{i}$ carries no current in the output-impedance test. With the source zeroed, $R_{i}$ connects the output to ground and does carry current; it is left out only because it is negligibly large.

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GATE 2022 IN Q.35: solution and answer

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