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
introduce amplitude stabilization by preventing the op amp from saturating and thus producing sinusoidal oscillations of fixed amplitude | |
introduce amplitude stabilization by forcing the opamp to swing between positive and negative saturation and thus producing square wave oscillations of fixed amplitude | |
introduce frequency stabilization by forcing the circuit to oscillate at a single frequency | |
enable the loop gain to take on a value that produces square wave oscillations |
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
\frac{1}{CR} \; and \; R_{1}=R_{2} | |
\frac{1}{CR} \; and \; R_{1}=4R_{2} | |
\frac{1}{2CR} \; and \; R_{1}=R_{2} | |
\frac{1}{2CR} \; and \; R_{1}=4R_{2} |
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}
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
\frac{1}{2\pi }\mu F | |
2\pi \mu F | |
\frac{1}{2\pi\sqrt{6} }\mu F | |
2\pi \sqrt{6}\mu F |
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
\frac{1}{(2\pi \sqrt{6}RC)} | |
\frac{1}{(2\pi RC)} | |
\frac{1}{(\sqrt{6}RC)} | |
\frac{\sqrt{6}}{(2\pi RC)} |
Question 4 Explanation:
Frequency of oscillation for RC phase shift
oscilla tor is \frac{1}{2 \pi \sqrt{6} R C}
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
R_{2}=5R_{1} | |
R_{2}=6R_{1} | |
R_{2}=R_{1}/6 | |
R_{2}=R_{1}/5 |
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}
There are 5 questions to complete.