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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 · 20 PYQs · 0 AI practice · GATE ECE 2027

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Q1GATE 2025MSQ

Which of the following statements is/are TRUE with respect to an ideal opamp?

🚀 Solve in Practice Mode

📖 Explanation

For an ideal operational amplifier, the following properties are defined:Infinite input resistance This ensures that no current flows into the input terminals. $\Rightarrow$ Statement (A) is TRUE Zero output resistance This ensures the output can drive any load without voltage drop. $\Rightarrow$ Statement (B) is FALSE (since it says infinite output resistance) Infinite open-loop differential gain The opamp amplifies the difference between the inputs with infinite gain in the ideal case. $\Rightarrow$ Statement (C) is TRUE Zero open-loop common-mode gain Ideal opamps reject any signal common to both inputs perfectly. Hence, common-mode gain is zero, not infinite. $\Rightarrow$ Statement (D) is FALSE

Q2GATE 2024MCQ

The opamps in the circuit shown are ideal, but have saturation voltages of $\pm 10 \mathrm{~V}$ . Assume that the initial inductor current is $0 \mathrm{~A}$ . The input voltage $\left(V_{i}\right)$ is a triangular signal with peak voltages of $\pm 2 \mathrm{~V}$ and time period of $8 \mu \mathrm{s}$ . Which one of the following statements is true?

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📖 Explanation

Opamp $A_{1}$ is non-inverting type Schmitt Trigger.

\begin{aligned} & V_{U T}=\frac{R_{1}}{R_{2}} \times V_{S a t}=\frac{10}{100} \times 10 \mathrm{~V}=1 \mathrm{~V} \\ & V_{L T}=-\frac{R_{1}}{R_{2}} \times V_{\text {sat }}=\frac{-10}{100} \times 10=-1 \mathrm{~V} \end{aligned}

$V_{01}$ is square wave. It is delayed by $1 \mu$ s wrt input $V_{i}$ . Opamp $A_{2}$ is integrator. $V_{02}$ is gettng saturated in less than half time period. Hence $V_{02}$ is a combination of ramp and constant portion. Thus $V_{02}$ resembles trapezoidal wave.

Explanation Figure 2

Q3GATE 2024MCQ

In the circuit below, the opamp is ideal. If the circuit is to show sustained oscillations, the respective values of $R_{1}$ and the corresponding frequency of oscillation are ______

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📖 Explanation

For given circuit, frequency of oscillations is $f_{o}=\frac{1}{2 \pi R C}$ . For sustained oscillations, minimum gain should be 3

\begin{aligned} \Rightarrow \quad 1+\frac{R_{1}}{R} & =3 \\ R_{1} & =2 R\end{aligned}

Q4GATE 2023MCQ

The $\frac{V_{\text {OUT }}}{V_{\text {IN }}}$ of the circuit shown below is

🚀 Solve in Practice Mode

📖 Explanation

Here, $A_{1}$ is an inverting amplifier and $A_{2}$ is a non-inverting amplifier.

\begin{aligned} V_{01}&=\frac{-R_{2}}{R_{1}} V_{i n} \\ V_{02}&=\left(1+\frac{R_{2}}{R_{1}}\right) V_{i n} \end{aligned}

Also, $A_{3}$ is an inverting summing amplifier,

\begin{aligned} V_{\text {out }} & =\frac{-R_{4}}{R_{3}} V_{01}-\frac{R_{4}}{R_{3}} V_{02}=\frac{-R_{4}}{R_{3}}\left[\frac{R_{2}}{R_{1}} V_{\text {\in }}+\left(1+\frac{R_{2}}{R_{1}}\right) V_{\text {\in }}\right] \\ V_{\text {out }} & =\frac{-R_{4}}{R_{3}} V_{\text {\in }} \\ \text { Gain, } \frac{V_{\text {out }}}{V_{\text {\in }}} & =\frac{-R_{4}}{R_{3}} \end{aligned}

Explanation Figure 2

Q5GATE 2022MCQ

A circuit with an ideal OPAMP is shown. The Bode plot for the magnitude (in dB) of the gain transfer function $(A_V(j\omega )=V_{out}(j\omega )/V_{\in}(j\omega ))$ of the circuit is also provided (here, $\omega$ is the angular frequency in rad/s). The values of $R$ and $C$ are

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \text{maximum gain}&=12dB\\ 20 \times \log A_{max}&=12\\ A_{max}&=4\\ 1+\frac{R_2}{R_1}&=4\\ R_2&=3R_1\\ R&=3 \times 1=3k\Omega \\ \log _{10}\omega _c&=3\\ \omega _c&=1000rad/sec\\ \omega _c&=\frac{1}{R_3C}\\ C&=\frac{1}{R_3 \times \omega _c}\\ C&=\frac{1}{1000 \times 1000}=1\mu F \end{aligned}

Explanation Figure 2

Q6GATE 2022MSQ

An ideal OPAMP circuit with a sinusoidal input is shown in the figure. The 3 dB frequency is the frequency at which the magnitude of the voltage gain decreases by 3 dB from the maximum value. Which of the options is/are correct?

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📖 Explanation

\begin{aligned} \frac{V_{out}}{V_{\in}}&=\frac{-2000}{1000+\frac{1}{j\omega \times 10^{-6}}}\\ Gain&=\frac{V_{out}}{V_{\in}}\frac{-2}{1+\frac{1}{\left (\frac{j\omega }{1000} \right ) }} \end{aligned}

$\omega \rightarrow \infty \Rightarrow gain=-2$ $\omega \rightarrow 0 \Rightarrow gain=0$ $\omega _c=1000$ rad/sec = cutoff frequency Hence, it is HPF.

Q7GATE 2021MCQ

For the circuit with an ideal OPAMP shown in the figure. $V_{\text{REF}}$ is fixed. If $V_{\text{OUT}}=1$ volt for $V_{\text{IN}}-0.1$ volt and $V_{\text{OUT}}=6$ volt for $V_{\text{IN}}=1$ volt, where $V_{\text{OUT}}$ is measured across $R_{L}$ connected at the output of this OPAMP, the value of $R_{F}/R_{\text{IN}}$ is

🚀 Solve in Practice Mode

📖 Explanation

MARKS TO ALL AS PER IIT ANSWER KEY

\begin{aligned} V &=V^{+} \\ \frac{V_{\text {out }} R_{\text {\in }}+V_{\text {\in }} R_{F}}{R_{\text {\in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} \\ \frac{1 \times R_{\text {\in }}+0.1 \times R_{F}}{R_{\text {\in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} &\ldots(i)\\ \frac{6 R_{\text {\in }}+1 \times R_{F}}{R_{\text {\in }}+R_{F}} &=\frac{V_{\text {ref }} R_{2}}{R_{1}+R_{2}} &\ldots(ii) \end{aligned}

Equate equation (i) and (ii),

\begin{aligned} 1 \times R_{\text {\in }}+0.1 \times R_{F} &=6 \times R_{\text {\in }}+1 \times R_{F} \\ -5 R_{\text {\in }} &=0.9 R_{F} \\ \therefore \quad \frac{R_{F}}{R_{\text {\in }}} &=\frac{-5}{0.9}=-5.55 \end{aligned}

(According to the given data magnitude is taken)

Q8GATE 2021NAT

A circuit with an ideal $\text{OPAMP}$ is shown in the figure. A pulse $V_{\text{IN}}$ of $20\:ms$ duration is applied to the input. The capacitors are initially uncharged. The output voltage $V_{\text{OUT}}$ of this circuit at $t=0^{+}$ (in integer) is _______ V.

🚀 Solve in Practice Mode

📖 Explanation

At, $t=0^{+}:$ Capacitor is short circuit

\begin{aligned} \therefore\quad V^{-}=V_{\text {\in }}=5 \mathrm{~V} \\ V^{+}=0 \mathrm{~V} \end{aligned}

\begin{aligned} \text{If}\qquad V^- &>V^{+} \\ V_{\text {out }} &=-V_{\text {sat }}=-12 \mathrm{~V} \end{aligned}

Explanation Figure 2

Q9GATE 2021MCQ

Consider the circuit with an ideal OPAMP shown in the figure. Assuming $\left | V_{\text{IN}} \right |\ll \left | V_{\text{CC}} \right |$ and $\left | V_{\text{REF}} \right |\ll \left | V_{\text{CC}} \right |$ , the condition at which $V_{\text{OUT}}$ equals to zero is

🚀 Solve in Practice Mode

📖 Explanation

For ideal op-amp, $V^{\prime}=V^{+}=0$ KCL at node $\mathrm{V}^{-}:$

\begin{aligned} \frac{V_{\text {IN}}-0}{R}+\frac{\left(-V_{\text {REF }}-0\right)}{R}+\frac{V_{\text {OUT }}-0}{R_{F}} &=0 \\ \frac{V_{\text {OUT }}}{R_{F}} &=\frac{1}{R}\left(V_{\text {REF }}-V_{\text {IN }}\right) \\ V_{\text {OUT }} &=\frac{R_{F}}{R}\left(V_{\text {REF }}-V_{\text {IN }}\right)\\ \text { We want, }\qquad V_{\text {OUT }}&=0 \\ \Rightarrow\qquad V_{\text {REF }}-V_{\text {IN }}&=0 \\ \Rightarrow\qquad V_{\text {IN }}&=V_{\text {REF }} \end{aligned}

Q10GATE 2020MCQ

The components in the circuit shown below are ideal. If the op-amp is in positive feedback and the input voltage $V_i$ is a sine wave of amplitude 1 V, the output voltage $V_o$ is

🚀 Solve in Practice Mode

📖 Explanation

Given circuit is a Schmitt trigger of non-inverting type. $V_{o}=\pm 5\, V$ $V^{+}=\frac{V_{o}\times 1+V_{i}\times 1}{1+1}=\frac{V_{o}+V}{2}$ let, $V_{o}=-5\, V,\, \, \, \,V^{+}=\frac{-5+V_{i}}{2}$ $V_{o}$ can change from -5 V to +5 V if $V^{+} \gt 0$ i.e. $\frac{-5+V_{i}}{2} \gt 0\Rightarrow V_{i} \gt 5\, V$ similarly, $V_{o}$ can change from -5 V to +5 V if $V_{i} \lt -5\, V$ But given input has peak value 1 V. Hence output cannot change from +5 V to -5 V or -5 V to +5 V. Output remain constant at +5 V or -5 V.

Explanation Figure 2

Q11GATE 2020NAT

In the circuit shown below, all the components are ideal. If $V_i$ is +2 V, the current $I_o$ sourced by the op-amp is __________ mA.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} V_o=(1+1) \times 2&=4V \\ \text{KCL at node } V_o & \\ \frac{2-4}{1k\Omega }I_o+\frac{0-4}{1k\Omega }&=0 \\ -2+I_o-4 &=0 \\ I_o&=6mA \end{aligned}

Q12GATE 2020MCQ

The components in the circuit given below are ideal. If R = 2 $k\Omega$ and C = 1 $\mu F$ , the -3 dB cut-off frequency of the circuit in Hz is

🚀 Solve in Practice Mode

📖 Explanation

Op-amp active filter (LPF) inverting type 3 dB cut-off frequency, $f_{c}=\frac{1}{2\pi RC}=\frac{1}{2\pi \times 2\times 10^{3}\times 10^{-6}}=\frac{500}{2\pi }=79.58Hz$

Q13GATE 2018MCQ

In the circuit shown below, the op-amp is ideal and Zener voltage of the diode is 2.5 volts. At the input, unit step voltage is applied, i.e. $v_{IN}(t)= u(t)$ volts. Also, at t= 0, the voltage across each of the capacitors is zero. The time t, in milliseconds, at which the output voltage $v_{OUT}$ crosses -10 V is

🚀 Solve in Practice Mode

📖 Explanation

$\text{For} \quad t \gt 0,$ $I=\frac{1 V}{1 \mathrm{k} \Omega}=1 \mathrm{mA}$ Till $t=2.5 \mathrm{msec}$ , both $V_{1}$ and $V_{2}$ will increase and after $t=2.5 \mathrm{msec}$ , $V_{2}=2.5 \mathrm{V}$ and $V_{1}$ increases with time.

\begin{aligned} \text { when } v_{\text {out }}(t) &=-10 \mathrm{V} \\ & V_{1}=7.5 \mathrm{V}\\ \text{So,}\\ \frac{1}{1 \mu F} \int_{0}^{t}(1 \mathrm{m} \mathrm{A}) d t &=7.5 \mathrm{V} \\ 10^{3} t &=7.5 \\ t &=7.5 \mathrm{msec} \end{aligned}

Explanation Figure 2

Q14GATE 2018NAT

An op-amp based circuit is implemented as shown below. In the above circuit, assume the op-amp to be ideal. The voltage (in volts, correct to one decimal place) at node A, connected to the negative input of the op-amp as indicated in the figure is _________.

🚀 Solve in Practice Mode

📖 Explanation

Applying the concept of virtual ground, we get

\begin{aligned} V_{o}&=-\frac{R_{2}}{R_{1}} \cdot V_{i n}\\ [\therefore & \text{ non-inverting amplifiel}]\\ \therefore \quad V_{o}&=-\frac{31 \mathrm{k} \Omega}{1 \mathrm{k} \Omega} \times 1 \mathrm{V}\\ V_{0}&=-31 \mathrm{V} \lt -15 \mathrm{V} \end{aligned}

which is not possible Hence, the output voltage of the op-amp is equal to $-15 \mathrm{V}$ Now applying KCL of node 'A', we get.

\begin{aligned} \frac{V_{A}-(-15)}{31 \mathrm{k} \Omega}+\frac{V_{A}-1}{1 \mathrm{k} \Omega} &=0 \\ \frac{V_{A}}{31 \mathrm{k} \Omega}+\frac{V_{A}}{1 \mathrm{k} \Omega} &=\frac{-15}{31 \mathrm{k} \Omega}+\frac{1}{1 \mathrm{k} \Omega} \\ V_{A}\left[\frac{1}{31}+\frac{1}{1}\right] &=-\frac{15}{31}+1 \\ V_{A} &=0.5 \mathrm{V} \end{aligned}

Explanation Figure 2

Q15GATE 2017NAT

In the voltage reference circuit shown in the figure, the op-amp is ideal and the transistors $Q_{1},Q_{2}...,Q_{32}$ are identical in all respects and have infinitely large values of common-emitter current the relation $I_{C}=I_{S}\; exp \;(V_{BE}/V_{T})$ , where $I_{s}$ is the saturation current. Assume that the voltage $V_{P}$ shown in the figure is 0.7 V and the thermal voltage $V_{T}$ =26mV. The output voltage $V_{out}$ (in volts) is__________.

🚀 Solve in Practice Mode

📖 Explanation

KCL at a

\begin{aligned} \frac{V_{0}-V_{x}}{20} &=\frac{V_{x}-0.7}{5} \\ V_{0}-V_{x} &=4 V_{x}-2.8 \\ V_{0} &=5 V_{x}-2.8 \quad\ldots(i)\\ \text{Now,}\quad I_{1} &=31 I \\ I_{s} e^{V_{x} / V_{I}} &=31 I_{s} e^{v_{p} / v_{T}} \\ \frac{V_{x}}{V_{T}} &=\ln 31+\frac{V_{P}}{V_{T}} \\ \frac{V_{x}-V_{P}}{V_{T}} &=\ln 31 \\ \Rightarrow\qquad V_{x} &=0.789 \mathrm{V} \end{aligned}

rom equation (i) $V_{0}=5 \times 0.789-2.8=1.145 \mathrm{V}$

Explanation Figure 2

Q16GATE 2017NAT

The amplifier circuit shown in the figure is implemented using a compensated operational amplifier (op-amp), and has an open-loop voltage gain, $A_{o}=10^{5}V/V$ and an open-loop cut-off frequency $f_{c}=8Hz.$ The voltage gain of the amplifier at 15 kHz, in V/V is __________.

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📖 Explanation

In the given circuit, Feedback factor, $\beta=\frac{R_{1}}{R_{1}+R_{2}}=\frac{1}{80}$ $A_{o f}=\frac{A_{o}}{1+A_{o} \beta} \simeq 80$ $f_{c}^{\prime}=f_{c}\left(1+A_{o} \beta\right)=8\left(1+\frac{10^{5}}{80}\right) \mathrm{Hz}=10,008 \mathrm{Hz}$ Gain at $f=15 \mathrm{kHz}=15000 \mathrm{Hz}$ is, $A_{f}=\frac{A_{\mathrm{b}}}{\sqrt{1+\left(\frac{f}{f_{c}^{2}}\right)^{2}}}=\frac{80}{\sqrt{\left(1+\left(\frac{15000}{10008}\right)^{2}\right)}} \simeq 44.4$

Explanation Figure 2

Q17GATE 2017MCQ

For the operational amplifier circuit shown, the output saturation voltages are $\pm$ 15V. The upper and lower threshold voltages for the circuit are, respectively.

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} V_1&=\frac{3 \times 10+V_o \times 5}{15}=\frac{6+V_o}{3} \\ V_{UT}&=\frac{6+15}{3}=7V \\ V_{LT}&=\frac{6-15}{3}=-3V \end{aligned}

Explanation Figure 2

Q18GATE 2016MCQ

Consider the constant current source shown in the figure below. Let $\beta$ represent the current gain of the transistor. The load current $I_{O}$ through $R_{L}$ is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} V_{A} &=V_{C C}-V_{\text {ref }} \\ V_{B} &=V_{A}(\text { since virtual short }) \\ I_{C} &=\frac{V_{C C}-V_{B}}{R}=\frac{V_{C C}-\left(V_{C C}-V_{\text {Ret }}\right)}{R} \\ &=\frac{V_{\text {Ref }}}{R} \\ I_{0} &=I_{E}=\frac{I_{C}}{\alpha}=\left(\frac{\beta}{1+\beta}\right) \frac{V_{\text {Ref }}}{R} \end{aligned}

Explanation Figure 2

Q19GATE 2016NAT

The following signal $V_{i}$ of peak voltage 8 V is applied to the non-inverting terminal of an ideal opamp. The transistor has $V_{BE}$ = 0.7 V, $\beta$ =100; $V_{LED}$ = 1.5 V, $V_{CC}$ = 10 V and - $V_{CC}$ = -10 V. The number of times the LED glows is ________

🚀 Solve in Practice Mode

📖 Explanation

$V_{B}=\frac{10 \mathrm{V} \times 2 \mathrm{K}}{8 \mathrm{K}+2 \mathrm{K}}=2 \mathrm{V}$ When $V_{1}$ exceeds 2 V output of op-amp $V_{01}$ goes to $V_{C C}$ and drives BJT into saturation shorted LED will glow, In the given problem $V_{i}$ exceeds 2V three times and hence output $V_{01}$ of op-amp goes to $V_{C C}$ thrice so that LED glow three times.

Explanation Figure 2

Q20GATE 2016NAT

For the circuit shown in the figure, $R_{1}=R_{2}=R_{3}=1 \Omega , L=1 \mu H$ and $C=1 \mu F$ . If the input $V_{\in}=cos(10^{6}t)$ , then the overall voltage gain ( $V_{out}/V_{\in}$ ) of the circuit is __________

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} H_{1}(s) &=\frac{V_{01}}{V_{i}}=1+\frac{R_{1}}{s L} \\ &=1+\frac{1}{\frac{s L}{R_{1}}}=1+\frac{1}{s \times 10^{-6}} \\ H_{2}(s) &=\frac{V_{0}}{V_{01}}=-\frac{R_{3}}{\frac{R_{2} C s+1}{C s}} \\ &=\frac{-s R_{3} C}{1+s R_{2} C} \\ &=\frac{-C s}{1+C s}=\frac{-1}{1+\frac{1}{C s}} \\ H(s) &=H_{1}(s) H_{2}(s) \\ &=\left(1+\frac{1}{s \times 10^{-6}}\right)\left(\frac{-s \times 1 \times 10^{-6}}{1+s \times 10^{-6}}\right)\\ &=-1 \end{aligned}

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