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Oscillators Questions and Answers

101.

Which of the following devices can be used for a relaxation oscillator?

Exam Year(s)

DSSSB AE - 2023,

Answer: D

A relaxation oscillator makes a non sinusoidal wave (a sawtooth or pulse) by charging a capacitor through a resistor until a switching device fires and discharges it. The unijunction transistor (UJT) is the classic device for this: when the capacitor voltage reaches the peak point of the emitter characteristic the UJT turns on and quickly discharges the capacitor, and the cycle repeats.


An SCR is a latching switch used for power control, and a PN junction diode and a photodiode have no negative resistance region to make the circuit switch back and forth. The answer is UJT, option D.

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102.

What is the oscillator frequency of a transistor phase shift oscillator, if the value of \(R = 5\ \mathrm{k\Omega}\), \(C = 10\ \mu\mathrm{F}\), \(R_C = 10\ \mathrm{k\Omega}\) (for a single transistor stage)?

Exam Year(s)

DSSSB AE - 2023,

Answer: A

For a transistor phase shift oscillator the frequency of oscillation is


\[ f = \frac{1}{2\pi RC\sqrt{6 + 4\,(R_C/R)}} \]


With \(R = 5\ \text{k}\Omega\), \(C = 10\ \mu\text{F}\) and \(R_C = 10\ \text{k}\Omega\), so that \(R_C/R = 2\):


\[ \begin{aligned} 2\pi RC \\ &= 2\pi \times 5000 \times 10 \times 10^{-6} \\ &= 0.3142 \end{aligned} \]


\[ \begin{aligned} f = \frac{1}{0.3142 \times \sqrt{6 + 8}} \\ &= \frac{1}{0.3142 \times 3.742} = 0.85\ \text{Hz} \end{aligned} \]


The answer is 0.85 Hz, option A. The larger values would need a capacitance a hundred times or more smaller.

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103.

What is the resonant frequency of the given oscillator if \(R = 51\ \mathrm{k\Omega}\) and \(C = 0.001\ \mu\mathrm{F}\)?

Wien bridge oscillator: an op-amp with a bridge of two R1 resistors and a series RC arm and a parallel RC arm, output Vo fed back to the bridge

Exam Year(s)

DSSSB AE - 2023,

Answer: B

The circuit is a Wien bridge oscillator: the op-amp with a series RC and a parallel RC arm in the bridge. The frequency at which the bridge is balanced and the feedback has zero phase shift is


\[ \begin{aligned} f = \frac{1}{2\pi R C} \\ &= \frac{1}{2\pi \times 51 \times 10^{3} \times 0.001 \times 10^{-6}} \end{aligned} \]


\[ f = \frac{1}{3.204 \times 10^{-4}} = 3120.7\ \text{Hz} \]


The other values are factors of ten or a hundred away, which would come from a slip in the capacitance, 0.001 microfarad being \(10^{-9}\) F. The answer is 3120.7 Hz, option B.

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