# Oscillator Circuits

 Question 1
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 A introduce amplitude stabilization by preventing the op amp from saturating and thus producing sinusoidal oscillations of fixed amplitude B introduce amplitude stabilization by forcing the opamp to swing between positive and negative saturation and thus producing square wave oscillations of fixed amplitude C introduce frequency stabilization by forcing the circuit to oscillate at a single frequency D enable the loop gain to take on a value that produces square wave oscillations
GATE EC 2016-SET-1   Analog Circuits
Question 1 Explanation:
The given circuit is Wein-bridge oscillator which produced sinusoidal oscillations and the amplitude of output wave is decided by feedback through inverting input terminal of op-amp.
 Question 2
The circuit shown in the figure has an ideal opamp. The oscillation frequency and the condition to sustain the oscillations, respectively, are A $\frac{1}{CR} \; and \; R_{1}=R_{2}$ B $\frac{1}{CR} \; and \; R_{1}=4R_{2}$ C $\frac{1}{2CR} \; and \; R_{1}=R_{2}$ D $\frac{1}{2CR} \; and \; R_{1}=4R_{2}$
GATE EC 2015-SET-1   Analog Circuits
Question 2 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}
 Question 3
The value of C required for sinusoidal oscillations of frequency 1 kHz in the circuit of the figure is A $\frac{1}{2\pi }\mu F$ B $2\pi \mu F$ C $\frac{1}{2\pi\sqrt{6} }\mu F$ D $2\pi \sqrt{6}\mu F$
GATE EC 2004   Analog Circuits
Question 3 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}
 Question 4
The oscillator circuit shown in the figure has an ideal inverting amplifier. Its frequency of oscillation (in Hz) is A $\frac{1}{(2\pi \sqrt{6}RC)}$ B $\frac{1}{(2\pi RC)}$ C $\frac{1}{(\sqrt{6}RC)}$ D $\frac{\sqrt{6}}{(2\pi RC)}$
GATE EC 2003   Analog Circuits
Question 4 Explanation:
Frequency of oscillation for RC phase shift
oscilla tor is $\frac{1}{2 \pi \sqrt{6} R C}$
 Question 5
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 A $R_{2}=5R_{1}$ B $R_{2}=6R_{1}$ C $R_{2}=R_{1}/6$ D $R_{2}=R_{1}/5$
GATE EC 2002   Analog Circuits
Question 5 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}
 Question 6
The oscillator circuit shown in figure is A Hartley oscillator with $f_{oscillation}=79.6MHz$ B Colpitts oscillator with $f_{oscillation}=50.3 MHz$ C Hartley oscillator with $f_{oscillation}=159.2MHz$ D Colpitts oscillator with $f_{oscillation}=159.2MHz$
GATE EC 2001   Analog Circuits
Question 6 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}
There are 6 questions to complete.