Q41GATE 2011MCQ

The circuit below implements a filter between the input current $I_{i}$ and the output voltage $V_{o}$ . Assume that the opamp is ideal. The filter implemented is a

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📖 Explanation

At $\omega=0 ; X_{L}=\omega L=0$ Hence circuit can be redrawn as below: $\therefore v_{0}-0$ At $\omega=\infty ; X_{L}=\infty$ Hence, circuit can be redrawn as below: $\therefore v_{0}=R_{1}i_{L}$ Hence, given circuit is a high pass filter.

Q43GATE 2010MCQ

The transfer characteristic for the precision rectifier circuit shown below is (assume ideal OP-AMP and practical diodes)

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📖 Explanation

\begin{aligned} \mathrm{At}, \quad V_{I}&=-10 \mathrm{V}\\ \frac{V_{0}-0}{R}&=\frac{0-20}{4 R}+\frac{0(-10)}{R}\\ V_{0}&=-5+10=5 \mathrm{V}\\ A t, \quad V_{I}&=-5 V\\ \frac{V_{0}-0}{R} &=\frac{0-20}{4 R}+\frac{0-(-5)}{R} \\ V_{0} &=-5+5 \\ &=0 \mathrm{V} \end{aligned}

For $V_{I}>-5 \mathrm{V}$ , both diodes are conducting. So, $\quad V_{0}=0 \mathrm{V}$

Q44GATE 2009MCQ

In the circuit shown below, the op-amp is ideal, the transistor has VBE = 0.6 V and $\beta$ = 150. Decide whether the feedback in the circuit is positive or negative and determine the voltage V at the output of the op-amp.

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📖 Explanation

\begin{aligned} I&=\frac{10-5}{5}=1 \mathrm{mA} \\ V^{\prime}&=1.4 \times 1=1.4 \mathrm{Volt} \\ \therefore \quad V&=V^{\prime}+V_{B E} \\ &=1.4+0.6=2 \mathrm{Volts} \end{aligned}

Q45GATE 2008MCQ

Consider the following circuit using an ideal OPAMP. The I-V characteristic of the diode is described by the relation $I=I_{0}(e^{\frac{V}{V_{T}}}-1)$ where $V_{T}$ = 25 mV, $I_{O}$ = 1m A and V is the voltage across the diode (taken as positive for forward bias). For an input voltage $V_{i}$ =- 1V, the output voltage $V_{O}$ is

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📖 Explanation

\begin{aligned} I &=I_{0}\left(e^{V / V_{T}}-1\right) \\ &=\frac{0-(-1)}{100 \mathrm{k} \Omega}\\ \Rightarrow \quad 10^{-6}\left(\frac{V}{25 \times 10^{-3}}-1\right)&=\frac{1}{10^{5}} \\ \Rightarrow \quad V&=0.06 \mathrm{V} \\ \Rightarrow \quad \frac{V_{0}-V}{4 \mathrm{k} \Omega}&=\frac{1}{100 \mathrm{k} \Omega} \\ \Rightarrow \quad \frac{V_{0}-0.06}{4 \mathrm{k} \Omega}&=\frac{1}{100 \mathrm{k} \Omega} \\ \Rightarrow \quad V_{0}&=0.1 \mathrm{V} \end{aligned}

Q46GATE 2008MCQ

The OPAMP circuit shown above represents a

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📖 Explanation

\begin{aligned} \frac{V_{i}}{R_{1}+L s}&=\frac{-V_{0}}{\frac{R_{2}}{R_{2} C_{2} s+1}} \\ \frac{V_{0}}{V_{i}}&=-\frac{R_{2}}{\left(R_{1}+L s\right)\left(R_{2} C_{2} s+1\right)} \end{aligned}

which is equivalent to standard form of transfer function of low pass filter, i.e. $H(s)=\frac{a}{p s^{2}+q s+r}$

Q47GATE 2007MCQ

Consider the Op-Amp circuit shown in the figure. If $V_{i} = V_{1} sin(\omega t)$ and $V_{0} = V_{2} sin(\omega t + \phi)$ , then the minimum and maximum values of $\phi$ (in radians) are respectively

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📖 Explanation

\begin{aligned} \angle \frac{V_{0}}{V_{i}} &=\angle \frac{1-R_{S C}}{1+R_{S C}} \\ \Rightarrow & \angle(1-s R C)-\angle 1+s C R \\ \theta &=-\tan ^{-1} \omega R C-\tan ^{-1} \omega R C \\ \theta &=-2 \tan ^{-1} \omega R C \end{aligned}

minimum value of $\theta=-\pi \quad(\text{at} \;\omega \rightarrow \infty )$ maximum value of $\theta=0 \quad(at\; \omega=0)$

Q48GATE 2007MCQ

For the Op-Amp circuit shown in the figure, $V_{0}$ is

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📖 Explanation

\begin{aligned} X&=1 \mathrm{volt}\\ y&=0.5 \mathrm{V} \\ &\text{(using voltage division rule)}\\ \text{So }\quad V_{0}&=X\left(-\frac{2 \mathrm{K} \Omega}{1 \mathrm{K} \Omega}\right)+y\left(1+\frac{2 \mathrm{K} \Omega}{1 \mathrm{K} \Omega}\right) \\ &=1 \times (-2)+0.5 \times (3)=-0.5 \mathrm{V} \end{aligned}

Q49GATE 2007MCQ

Consider the Op-Amp circuit shown in the figure. The transfer function $V_{0}(s)/V_{i}(s)$ is

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📖 Explanation

From the figure given in the question

\begin{aligned} V_{0}&=V_{i}\left(-\frac{R_{1}}{R_{1}}\right)+V_{i} \frac{\frac{1}{S C}}{\frac{1}{S C}+R}\left(1+\frac{R_{1}}{R_{1}}\right)\\ \frac{V_{0}}{V_{i}}&=-1+\frac{1}{1+R S C} \times 2 \\ \frac{V_{0}}{V_{i}}&=\frac{2-(1+R s C)}{1+R s C} \\ \frac{V_{0}}{V_{i}}&=\frac{1-R s C}{1+R s C} \end{aligned}

Q50GATE 2007MCQ

In the Op-Amp circuit shown, assume that the diode current follows the equation $I = I_{s} exp(V/ V_{T})$ . For $V_{i} = 2 V, \; V_{o} = V_{o1}$ , and for $V_{i} = 4 V,V_{o} = V_{o2}$ . The relationship between $V_{o1}$ and $V_{o2}$ is:

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📖 Explanation

\begin{aligned} V_{o} &=-V_{T} \ln \left(\frac{V_{I}}{R_{1} I_{s}}\right) \\ V_{o_{1}}-V_{o_{2}} &=-V_{T} \ln \frac{2}{R_{1} I_{s}}+V_{T} \ln \frac{4}{R_{1} I_{s}} \\ &=V_{T} \ln \left(\frac{4}{2}\right)=V_{T} \ln 2 \end{aligned}

Q51GATE 2006MCQ

For the circuit shown in the following figure, the capacitor C is initially uncharged. At t=0, the switch S is closed. The $V_c$ across the capacitor at t = 1 millisecond is In the figure shown above, the OP-AMP is supplied with $\pm$ 15V and the ground has been shown by the symbol $\bigtriangledown$ .

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📖 Explanation

\begin{aligned} V_{C} &=15\left(1-e^{-t / R C}\right) \\ &=15\left(1-e^{-1}\right)=9.45 \mathrm{V} \end{aligned}

Q52GATE 2005MCQ

The voltage $e_{0}$ is indicated in the figure has been measured by an ideal voltmeter. Which of the following can be calculated ?

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📖 Explanation

$V_{1}=-I_{B_{1}} \times 1 \mathrm{M} \Omega, V_{2}=V_{1}=-I_{B_{1}} \times 1 \mathrm{M} \Omega$ (due to virtual ground) Drop in feedback resistor $1 \mathrm{M} \Omega=I_{B_{2}} \times 1 \mathrm{M} \Omega$ $e_{0}=V_{2}+I_{B_{2}} \times 1 \mathrm{M} \Omega$ $e_{0}=-I_{B_{1}} \times 1 \mathrm{M} \Omega+I_{B_{2}} \times 1 \mathrm{M} \Omega$ $e_{0}=\left(I_{B_{2}}-I_{B_{1}}\right) \times 1 \mathrm{M} \Omega$ where $\left(I_{B_{2}}-I_{B_{1}}\right)$ is offset current

Q53GATE 2005MCQ

The input resistance $R_{i}$ of the amplifier shown in the figure is

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📖 Explanation

Connect a $V_{s}$ voltage source across inverting terminal of op-amp. By virtual short

\begin{aligned} V_{A}&=0 \mathrm{V} \\ I_{i}&=\frac{V_{s}-V_{A}}{10 \mathrm{k} \Omega}=\frac{V_{s}-0}{10 \mathrm{k} \Omega}=\frac{V_{s}}{10 \mathrm{k} \Omega}\\ \text{So, }R_{i}&=\frac{V_{s}}{I_{i}}=10 \mathrm{k} \Omega \end{aligned}

Q54GATE 2005MCQ

The Op-amp circuit shown in the figure is filter. The type of filter and its cut. Off frequency are respectively

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📖 Explanation

High pass filter,

\begin{aligned} \omega_{c} &=\frac{1}{R C} \\ &=\frac{1}{1 \times 10^{3} \times 1 \times 10^{-6}}=1000 \mathrm{rad} / \mathrm{sec} \end{aligned}

Q55GATE 2004MCQ

The circuit in the figure is a

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📖 Explanation

\begin{array}{l} \text { At } \omega=\infty , \frac{V_{0}}{V_{\text {\in }}}=0 \\ \text { and at } \omega=0, \quad \frac{V_{0}}{V_{\text {\in }}}=1 \end{array}

Q56GATE 2004MCQ

In the op-amp circuit given in the figure, the load current $i_{L}$ is

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📖 Explanation

\begin{aligned} \frac{V}{R_{2}}+\frac{V}{R_{L}}+\frac{V}{R_{2}}&=\frac{V_{0}}{R_{2}} \\ \frac{2 V}{R_{2}}+\frac{V}{R_{L}}&=\frac{V_{0}}{R_{2}} &\ldots(i)\\ \frac{V_{s}-V}{R_{1}}&=\frac{V-V_{0}}{R_{1}} \\ V_{s}-2 V&=-V_{0} &\ldots(ii)\\ \text{Putting } V_{o} \text{ from (i)},\\ \frac{-V_{s}+2 V}{R_{2}} &=+\left(\frac{2 V}{R_{2}}+\frac{V}{R_{L}}\right) \\ i_{L} &=\frac{V}{R_{L}}=-\frac{V_{S}}{R_{2}}=\left[\frac{V}{R_{L}}=i_{L}\right] \end{aligned}

Q57GATE 2004MCQ

An ideal op-amp is an ideal

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Q58GATE 2003MCQ

If the op-amp in the figure is ideal, the output voltage $V_{out}$ will be equal to

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📖 Explanation

\begin{aligned} V_{0} &=-\frac{R_{F}}{R_{1}} \times 2+\left(1+\frac{R_{F}}{R_{1}}\right) \cdot \frac{8}{9} \times 3 \\ &=-5 \times 2+6 \times \frac{8}{3} \\ V_{0} &=6 \mathrm{V} \end{aligned}

Q59GATE 2003MCQ

If the input to the ideal comparators shown in the figure is a sinusoidal signal of 8 V (peak to peak) without any DC component, then the output of the comparators has a duty cycle of

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📖 Explanation

\begin{aligned} V_{p-p}&=8 V\\ \text{So,} \quad V_{i}&=4 \sin \omega t \\ \text { At } \quad V_{i}&=2 \\ \sin \omega t &=\frac{1}{2} \\ \Rightarrow \quad \omega t&=\frac{\pi}{6} \end{aligned}

Another crossover at $\pi-\frac{\pi}{6}=\frac{5 \pi}{6}$ $\therefore \quad \text { Duty cycle }=\frac{T_{\text {on }}}{T}=\frac{\frac{5 \pi}{6}-\frac{\pi}{6}}{2 \pi}=\frac{\left(\frac{4 \pi}{6}\right)}{2 \pi}= \frac{1}{3}$

Q60GATE 2003MCQ

If the differential voltage gain and the common mode voltage gain of a differential amplifier are 48 dB and 2 dB respectively, then common mode rejection ratio is

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📖 Explanation

\begin{aligned} \mathrm{CMRR} &=20 \log A_{d}-20 \log A_{C} \\ &=48 \mathrm{dB}-2 \mathrm{dB}=46 \mathrm{dB} \end{aligned}