Q1GATE 2016MCQ

Consider the oscillator circuit shown in the figure. The function of the network (shown in dotted lines) consisting of the 100 k $\Omega$ resistor in series with the two diodes connected

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Q2GATE 2015MCQ

The circuit shown in the figure has an ideal opamp. The oscillation frequency and the condition to sustain the oscillations, respectively, are

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📖 Explanation

Frequency of Wein bridge oscillator is $\omega_{0} =\frac{1}{R C}$ but in the question time constant is double so frequency becomes half.

\begin{aligned} \omega_{0} &=\frac{1}{2 R C} \\ Z_{1} &=2 R+\frac{1}{j \omega C}=2(R-j R) \\ Z_{2} &=\frac{R \times \frac{1}{2 j \omega C}}{R+\frac{1}{2 j \omega C}}=\frac{\frac{R^{2}}{j}}{R-j R} \\ \beta &=\frac{Z_{2}}{Z_{1}+Z_{2}}=\frac{1}{5} \\ 1+\frac{R_{1}}{R_{2}}&=5 \quad \Rightarrow R_{1}=4 R_{2} \end{aligned}

Q3GATE 2004MCQ

The value of C required for sinusoidal oscillations of frequency 1 kHz in the circuit of the figure is

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📖 Explanation

\begin{aligned} R &=1 \mathrm{k} \\ \frac{V_{o}-V_{1}}{X_{C}+R} &=\frac{V_{1}}{X_{C}}+\frac{V_{1}}{R} \\ \frac{V_{o}-V_{1}}{X_{C}+R} &=V_{1}\left(\frac{X_{C}+R}{X_{C} \cdot R}\right) \\ \frac{V_{o}}{V_{1}} &=\frac{\left(X_{C}+R\right)^{2}}{X_{C} \cdot R}+1 \quad\left[X_{C}=-\frac{j}{\omega C}\right] \\ \frac{V_{o}}{V_{1}} &=\frac{-\frac{1}{\omega^{2} C^{2}}+R^{2}-\frac{2 j R}{\omega C}}{\frac{R}{j \omega C}}+1 \\ \frac{V_{0}}{V_{1}} &=+\frac{j \omega C}{R}\left(-\frac{1}{\omega^{2} C^{2}}+R^{2}-\frac{2 j R}{\omega C}\right)+1 \end{aligned}

For oscillation imaginary part is zero. i.e. $j\left[\frac{\omega C}{R} \times \frac{-1}{\omega^{2} C^{2}}+\frac{\omega C}{R} \times R^{2}\right]\$ =0$

\begin{aligned} \frac{-1}{\omega C A}+\omega C R&=0\\ \omega^{2} C^{2} R^{2}-1&=0 \\ \omega^{2}&=\frac{1}{R^{2} C^{2}} \\ C&=\frac{1}{\omega R}=\frac{1}{2 \pi \times 10^{3} \times 10^{3}} \\ C&=\frac{1}{2 \pi} \mu \mathrm{F} \end{aligned}

Q4GATE 2003MCQ

The oscillator circuit shown in the figure has an ideal inverting amplifier. Its frequency of oscillation (in Hz) is

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📖 Explanation

Frequency of oscillation for RC phase shift oscilla tor is $\frac{1}{2 \pi \sqrt{6} R C}$

Q5GATE 2002MCQ

The circuit in figure employs positive feedback and is intended to generate sinusoidal oscillation. If at a frequency $f_{0}B(f)=\Delta \frac{V_{f}}{V_{0}}=\frac{1}{6}\angle 0^{\circ}$ , then to sustain oscillation at this frequency

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📖 Explanation

KCL at node 1

\begin{aligned} \frac{\beta V_{o}-0}{R_{1}}+\frac{\beta V_{o}-V_{o}}{R_{2}} &=0 \\ \beta\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right) &=\frac{1}{R_{2}} \\ \Rightarrow \quad \quad \beta \cdot \frac{R_{1}+R_{2}}{R_{1}} &=1 \\ \Rightarrow \quad \quad\frac{R_{1}\left(1+\frac{R_{2}}{R_{1}}\right)}{R_{1}} &=\frac{1}{\beta}=6 \\ \frac{R_{2}}{R_{1}} &=5 \\ \Rightarrow \quad \quad R_{2} &=5 R_{1} \end{aligned}

Q6GATE 2001MCQ

The oscillator circuit shown in figure is

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📖 Explanation

Fig. shown is Colpitts oscillator.

\begin{aligned} f &=\frac{1}{2 \pi \sqrt{L C_{e q}}} \\ C_{\mathrm{eq}} &=\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{2 \times 2}{4}=1 \mathrm{PF} \\ &=\frac{1}{2 \pi \sqrt{10 \times 10^{-6} \times 10^{-12}}} \\ &=\frac{1 \times 10^{9}}{2 \pi \sqrt{10}}=50.3 \mathrm{MHz} \end{aligned}

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