Q2GATE 2014NAT

An oscillator circuit using ideal op-amp and diodes is shown in the figure. The time duration for +ve part of the cycle is $\Delta t_{1}$ and for -ve part is $\Delta t_{2}$ . The value of $e^{\frac{\Delta t_{1}-\Delta t_{2}}{RC}}$ will be ______.

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📖 Explanation

This circuit is a stable multivibrator (or) free running oscillator. When $V_0=+V_{sat}$ $V_{UPT}=V_{sat}\times \frac{1k\Omega }{1k\Omega +3k\Omega } = \frac{V_{sat}}{4}$ When, $V_0=-V_{sat}$ $V_{LTP}=V_{sat}\times \frac{1k\Omega }{1k\Omega +1k\Omega }=\frac{-V_{sat}}{2}$ $V_c=V_{final}+(V_{inital}-V_{final})e^{-t/RC}$ In time $t=\Delta t_1$ $V_c=V_{UTP};$ $V_{initial}=V_{LTP};$ $V_{final}=+V_{sat}$ $V_{UTP}=V_{sat}+(V_{LTP}-V_{sat})e^{-\Delta t_1/RC}$ $\frac{V_{sat}}{4}=V_{sat}+\left ( -\frac{V_{sat}}{2}-V_{sat} \right )e^{-\Delta t_1/RC}$ $V_{sat}\left ( -1+\frac{1}{4} \right )=-V_{sat}\left ( \frac{1}{2}+1 \right )e^{-\Delta t_1/RC}$ $\left (1-\frac{1}{4} \right )=\left ( \frac{1}{2}+1 \right )e^{-\Delta t_1/RC}$ $\frac{3}{4}=\frac{3}{2}e^{-\Delta t_1/RC}$ $e^{\Delta t_1/RC}=2\;\;....(i)$ In time $t=\Delta t_2$ $V_c=V_{LTP};$ $V_{initial}=V_{UTP};$ $V_{final}=-V_{sat}$ $V_{LTP}=-V_{sat}+(V_{UTP}+V_{sat})e^{-\Delta t_2/RC}$ $-\frac{V_{sat}}{2}=-V_{sat}+\left ( \frac{V_{sat}}{4}+V_{sat} \right )e^{-\Delta t_2/RC}$ $\left (1-\frac{1}{2} \right )V_{sat}=V_{sat}\left (1+ \frac{1}{4} \right )e^{-\Delta t_2/RC}$ $\frac{1}{2}= \frac{5}{4}e^{-\Delta t_2/RC}$ $e^{\Delta t_2/RC}=\frac{5}{2}\;\;....(ii)$ From equation (i) and (ii), $\frac{e^{\Delta t_1/RC}}{e^{\Delta t_2/RC}}$ $=e^{(\Delta t_1-\Delta t_2)/RC}$ $=\frac{2}{5/2}=\frac{4}{5}=0.8$

Q3GATE 2014MCQ

In the Wien Bridge oscillator circuit shown in figure, the bridge is balanced when

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📖 Explanation

$\frac{R_3}{R_4}=\frac{R_1}{R_2}+\frac{C_2}{C_1}$ $\omega =\frac{1}{\sqrt{R_1R_2C_1C_2}}$

Q4GATE 2014NAT

A hysteresis type TTL inverter is used to realize an oscillator in the circuit shown in the figure. If the lower and upper trigger level voltages are 0.9 V and 1.7 V, the period (in ms), for which output is LOW, is _____.

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📖 Explanation

Discharging curve, $V_c(t)=0-(0-1.7)e^{-t/RC}$ At, $t=T_2$ $V_c(t)=0.9V$ $0.9=1.7e^{-t/RC}$ $0.63=T_2$ $T_2=0.63\;ms$

Q5GATE 2013MCQ

In the feedback network shown below, if the feedback factore k is increased, then the

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📖 Explanation

The given configuration is a voltage-series feedback configuration. So, the input impedance increases $R_{if}=R_i(1+A_o k)$ So, the output impedance decreases $R_{of}=\frac{R_o}{1+A_o k}$

Q6GATE 2005MCQ

The typical frequency response of a two-stage direct coupled voltage amplifier is as shown in figure

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Q7GATE 2004MCQ

The feedback used in the circuit shown in figure can be classified as

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Q8GATE 2002MCQ

The output voltage ( $V_{0}$ ) of the Schmitt trigger shown in figure swings between +15V and -15V. Assume that the operational amplifier is ideal. The output will change from +15V to -15V when the instantaneous value of the input sine wave is

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Q9GATE 2001MCQ

For the oscillator circuit shown in figure, the expression for the time period of oscillation can be given by (where $\tau$ =RC)

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📖 Explanation

$T=2RC \ln 3=2\tau \ln 3$

Q10GATE 2001MCQ

An op-amp has an open-loop gain of $10^5$ and an open-loop upper cutoff frequency of 10 Hz. If this op-amp is connected as an amplifier with a closed-loop gain of 100, then the new upper cutoff frequency is

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📖 Explanation

$A_{OL}=10^5$ $f=10Hz$ $A_{CL}=\frac{A_{OL}}{1+\beta A_{OL}}$ $(1+\beta A_{OL})=\frac{10^5}{10^2}=10^3$ $f'_2=f_2(1+\beta A_{OL})$ $\;\;=10 \times 10^3 Hz$ $\;\; =10kHz$

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