GATE 2015 EC – Question 51

Analog Circuits · Op-amp Circuits · 2 marks · Multiple choice

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

A Wien bridge oscillator. The positive feedback network is a series $C$ and $2R$ branch, together with a parallel $2C$ and $R$ branch. The negative feedback network has resistors $R_1$ (feedback) and $R_2$ (to ground).

$\frac{1}{CR}$ 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 $R_1 = 4R_2$

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Correct answer: (D) $\frac{1}{2CR}$ and $R_1 = 4R_2$

Explanation

For this modified Wien bridge the feedback network has a series arm $C$ and $2R$ and a parallel arm $2C$ and $R$. The oscillation condition from the loop gain gives $\omega = \frac{1}{2CR}$ and the amplifier gain must make up the attenuation of the network, which needs $R_1 = 4R_2$. This matches the official key.