# Analog Circuits Questions and Answers – Weinbridge Oscillator

This set of Analog Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Weinbridge Oscillator”.

1. Which of these is incorrect for a Wien Bridge oscillator?
a) Low distortion
b) Good stability at the resonant frequency
c) Difficult to tune
d) Based on frequency selective form of a Wheatstone bridge

Explanation: A Wien Bridge oscillator is a two-stage RC coupled amplifier circuit that has good stability at the resonant frequency. It has low distortion and is easy to tune. It is called so because the circuit is based on a frequency-selective form of the Wheatstone bridge.

2. At the resonant frequency, what is the phase shift for the output in a Wien Bridge oscillator?
a) 0°
b) 45°
c) 90°
d) 180°

Explanation: The Wien Bridge oscillator uses a feedback circuit which consists of a series RC circuit which is connected with a parallel RC network, producing a phase delay or phase advance circuit depending on the frequency, which is however 0° at the resonant frequency.

3. Consider the circuit shown below.

Given that R1=20kΩ, C1=2nF, R2=20kΩ, C2=2nF, find the approximate resonant frequency.
a) 4kHz
b) 3kHz
c) 25kHz
d) 15kHz

Explanation: The resonant frequency for the above Wien Bridge Oscillator circuit is fr = 1/2πRC and since R1=R2 and C1=C2 for this oscillator, for a balanced bridge, we can use either of those.
Hence fr = 3.987 kHz≈4kHz.

4. The following circuit is provided. R1=R2 and C1=C2.

What is the correct choice for sustained oscillation?
a) R1 = R2
b) R4 = 2R3
c) R4 = 3R3
d) R1 = R2 = R3 = R4

Explanation: For sustained oscillation, the gain required is given by AV = 3
Thus, in the above circuit 1+R4/R3 = 3
Thus R4 = 2R3 is the relation required.

5. Which of these is a disadvantage of the Wien Bridge oscillator?
a) It cannot fabricate a pure tune
b) Distortion observed in output is high
c) It cannot be used for high resistance values
d) There is no automatic gain control

Explanation: A Wien Bridge oscillator provides a stable low distortion output over a wide range of frequency. However, the number of components required is high and the Wheatstone bridge applied cannot be used for high resistance values.
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6. In the below circuit, the output frequency is 0.5 Mhz.

C1 = 5nF, R4 = 40kΩ, R3 = 20kΩ. Find the value of R.
a) 63Ω
b) 220Ω
c) 127Ω
d) 55Ω

Explanation: f = 1/2πRC
R = 1/2πfC = 63Ω.

7. For any Wien Bridge oscillator, R1 = R2 and C1 = C2 always in the bridge, provided the phase shift through the amplifier is zero.
a) True
b) False

Explanation: The bridge circuit is used as feedback for the oscillator, provided that phase shift through the amplifier is zero. At this point the bridge has to be balanced, however not necessarily that R1 = R2 and C1 = C2. This condition is just a case of the bridge being balanced.

8. In a Wien bridge oscillator, it is found that at the frequency ωO there is no phase shift in RF/R gain loop and the phase shift of the amplifier is also zero. Then what is the equation for the radian frequency, given R1, C1 is the series network of bridge and R2, C2 is the parallel network?
a) ωO=1/R1C1
b) ωO=1/R2C2
c) ωO=1/R1R2C1C2
d) ωO=1/RFC1RC2

Explanation: For no phase shift for the amplifier, the frequency of the output is given by ωO = 1/R1R2C1C2 and RF/R = C1/C2 + R2/R1. When R1=R2 and C1=C2 then RF = 2R.

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