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Question

If a certain oscillator, Av = 50, the attenuation of the feedback circuit must be :

This question was previously asked in
UGC NET 2015 Paper 1 Question Paper (27-Dec-2015)
The correct answer is

0.02

 The Barkhausen criterion. For sustained oscillation the loop gain must be exactly unity in magnitude and the loop phase shift must be zero (or a multiple of 360°):

\(A_v\beta=1, \qquad \angle A_v\beta=0^{\circ}\)

Step — solve for the feedback factor.

\(\beta=\dfrac{1}{A_v}=\dfrac{1}{50}=0.02\)

which is option 4.

Why "attenuation" is the right word for β. The feedback network is passive, so it can only reduce the signal: it returns 1/50 of the output to the input. The amplifier then makes up exactly the factor of 50 that the network threw away, and the signal circulates round the loop with unchanged amplitude — which is what "sustained" oscillation means.

The distractors and the errors behind them.

OptionWhere it comes from
1Confusing the loop gain Aβ = 1 with β itself
0.01Using 1/100 — the reciprocal of twice the gain
10A β greater than 1 would require the passive network to amplify, which is impossible

What happens if the product is not exactly 1.

If Aβ < 1 the loop loses a little on each circuit and any disturbance dies away — no oscillation starts.

If Aβ > 1 the amplitude grows on every circuit until the amplifier clips, so the waveform is a distorted square rather than a sine.

In practice the loop is designed with Aβ slightly greater than 1 so that oscillation is certain to start from noise, and an amplitude-limiting element — a thermistor, a lamp, back-to-back diodes, or simply the onset of gentle saturation — then pulls the effective gain back to exactly 1 at the working amplitude. The Wien-bridge oscillator's incandescent lamp is the classic example.

The phase half of the criterion. Magnitude alone is not enough. An RC phase-shift oscillator uses three RC sections to supply 180° to add to the amplifier's own 180°, and at that frequency its attenuation happens to be 1/29 — which is exactly why such an oscillator needs an amplifier gain of at least 29.

Hence, the feedback attenuation must be 0.02.

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