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GATE ECE · Analog Circuits

Operational Amplifiers Op Amps Practice Questions

Generate GATE-level questions on Op-Amps. Focus on: 1. Ideal Op-Amp characteristics and Virtual ground. 2. Applications: Inverting/Non-inverting amplifiers, Summing, Differentiator, Integrator. 3. Precision rectifiers, Log/Antilog amplifiers, and Active filters.

66 questions · 6 PYQs · 0 AI practice · GATE ECE 2027

Q61GATE 2003MCQ

Three identical amplifiers with each one having a voltage gain of 50, input resistance of 1 k $\Omega$ and output resistance of 250 $\Omega$ are cascaded. The opened circuit voltages gain of the combined amplifier is

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📖 Explanation

\begin{array}{l} \quad A_{v}=\frac{V_{4}}{V_{1}} \\ \Rightarrow \quad A_{v}=\frac{V_{4}}{V_{3}} \times \frac{V_{3}}{V_{2}} \times \frac{V_{2}}{V_{1}} \end{array}

Voltage across 1k after 1 st stage is,

\begin{aligned} V_{2}&=\frac{1000 \times 50 V_{1}}{1250}\\ \Rightarrow \quad \frac{V_{2}}{V_{1}}&=40\\ \text{Similarly }\frac{V_{3}}{V_{2}}&=40 \\ \therefore \quad A_{v}&=40 \times 40 \times 50=8 \times 10^{4} \\ A_{v} \text { \in }(\mathrm{dB})&=20 \log \left(8 \times 10^{4}\right)=98 \mathrm{dB} \end{aligned}

Explanation Figure 1

Q62

GATE 2002

MCQ

A 741-type op-amp has a gain-bandwidth product of 1 MHz. A non-inverting amplifier using this op-amp and having a voltage gain of 20 dB will exhibit a 3-dB bandwidth of

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📖 Explanation

\begin{aligned} 20 \log x &=20 \mathrm{dB} \\ x &=10 \\ \text{Gain} \times B W &=1 \times 10^{6} \\ B W &=\frac{1 \times 10^{6}}{\text { Gain }}=\frac{10^{6}}{10}=100 \mathrm{k}=10^{5} \end{aligned}

Q63GATE 2002MCQ

An amplifier using an opamp with a slew-rate SR = 1V/ $\mu$ sec has a gain of 40 dB. If this amplifier has to faithfully amplify sinusoidal signals from dc to 20 KHz without introducing any slew-rate induced distortion, then the input signal level must not exceed.

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📖 Explanation

\begin{aligned} \text { Slew rate } &=A \cdot 2 \pi f V_{m} \\ V &=A V_{m} \sin \omega t \\ \frac{d V}{d t} &=A V_{m} \omega \cos \omega t \\ \left.\frac{d V}{d t}\right|_{\max } &=S R=A V_{m} 2 \pi f_{m} \\ 20 \log x &=40 \\ x &=100=A \\ \Rightarrow V_{m}=\frac{S R}{A \cdot 2 \pi f_{m}} &=\frac{1}{10^{-6} \times 100 \times 2 \pi \times 20 \times 10^{3}} \\ V_{m} &=79.5 \mathrm{mV} \end{aligned}

Q64GATE 2001MCQ

The inverting OP-AMP shown in figure has an open-loop gain of 100. The closed loop gain $\frac{V_{o}}{V_{s}}$ is

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📖 Explanation

If open loop gain is not infinite, then

\begin{aligned} \frac{1}{\beta} &=-\frac{R_{f}}{R_{1}}=-\frac{10 \mathrm{k}}{1 \mathrm{k}}=-10 \\ |\beta| &=\frac{1}{10} \\ A_{f} &=\frac{-A}{1+A \beta}=\frac{-100}{1+100 \times \frac{1}{10}}=-\frac{100}{11} \\ A_{f} & \cong-9 \end{aligned}

Q65GATE 2001MCQ

In figure, assume the OP-AMPs to be ideal. The output $V_{o}$ of the circuit is:

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📖 Explanation

KCL at node 1

\begin{aligned} \frac{V_{2}}{V_{s}} &=\frac{-\omega L}{R} \\ &=\frac{-100 \times 10 \times 10^{-3}}{10}=-\frac{1}{10} \\ V_{2} &=-\frac{V_{s}}{10}=-\cos 100 t\\ \frac{V_{0}}{V_{2}} &=\frac{-1 / \omega \mathrm{c}}{100} \\ &=-\frac{1}{100 \times 10 \times 10^{-6} \times 100}=-10 \\ V_{0} &=-10 \mathrm{V}_{2}=-10(-\cos 100 \mathrm{t}) \\ V_{0} &=10 \cos 100 \mathrm{t} \end{aligned}

Explanation Figure 2

Q66GATE 2001MCQ

The ideal OP-AMP has the following characteristics.

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📖 Explanation

Currently no explanation available

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