250+ TOP MCQs on Weinbridge Oscillator and Answers

Analog Circuits Multiple Choice Questions 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

Answer: C
Clarification: 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°

Answer: a
Clarification: 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

Answer: A
Clarification: The resonant frequency for the above Wien Bridge Oscillator circuit is f_r = 1/2πRC and since R1=R2 and C1=C2 for this oscillator, for a balanced bridge, we can use either of those.
Hence f_r = 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

Answer: B
Clarification: For sustained oscillation, the gain required is given by A_V = 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

Answer: C
Clarification: 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.

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

Answer: A
Clarification: 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

Answer: b
Clarification: 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 ω_O there is no phase shift in R_F/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. ω_O=1/R₁C₁
B. ω_O=1/R₂C₂
C. ω_O=1/R₁R₂C₁C₂
D. ω_O=1/RFC₁RC₂

Answer: C
Clarification: For no phase shift for the amplifier, the frequency of the output is given by ω_O = 1/R₁R₂C₁C₂ and R_F/R = C₁/C₂ + R₂/R₁. When R₁=R₂ and C₁=C₂ then R_F = 2R.