# Comparators and Waveform Generator MCQ [Free PDF] – Objective Question Answer for Comparators and Waveform Generator Quiz

1. What is Barkhausen criterion for oscillation?

A. Aß > 1
B. Aß < 1
C. Aß = 1
D. Aß ≠ 1

The Barkhausen criterion for oscillation is Aß = 1.
Where A-> gain of amplifier and ß-> transfer ratio.

2. At what condition the output signal can be continuously obtained from the input signal?

A. When the product of input voltage and the feedback voltage is equal to 1
B. When the product of amplifier gain and transfer ratio is equal to 1
C. When the product of feedback voltage and transfer ratio is equal to 1
D. When the product of amplifier gain and the input voltage is equal to 1

When Aß=1, the feedback signal will be equal to the input signal. At this condition, the circuit will continue to provide output, even if the external signal is disconnected. This is because the amplifier cannot distinguish between external signal and signal from the feedback circuit. Thus, the output signal is continuously obtained.

3. An oscillator is a type of

A. Feedforward amplifier
B. Feedback amplifier
C. Waveform amplifier
D. RC amplifier

An oscillation is a type of feedback amplifier in which a part of the output is fed back to the input via a feedback circuit.

5. What is the condition to achieve oscillations?

A. |Aß|=1
B. ∠Aß=0o
C. ∠Aß=multiples of 2π
D. All the mentioned

All the conditions should be simultaneously satisfied to achieve oscillations.

6. What happens if |Aß|<1

A. Oscillation will die down
B. Oscillation will keep on increasing
C. Oscillation remains constant
D. Oscillation fluctuates

If |Aß| becomes less than unity, the feedback signal goes on reducing in each feedback cycle and oscillation will die down eventually.

7. How sustained oscillation can be achieved?

A. Maintaining |Aß| slightly greater than unity
B. Maintaining |Aß| equal to unity
C. Due to the non-linearity of transistor
D. Due to using the feedback network

When |Aß| is kept slightly greater than unity the signal, however, cannot go on increasing and gets limited due to the non-linearity of the device (that is transistor enters into saturation). Thus, it is the non-linearity of the transistor because sustained oscillation can be achieved.

8. Why it is difficult to maintain Barkhausen’s condition for oscillation?

A. Due to variation in temperature
B. Due to variation in supply voltage
C. Due to variation in a components lifetime
D. All of the mentioned

The Barkhausen condition |Aß|=1 is usually difficult to maintain in the circuit as the value of A and ß vary due to temperature variations, aging of components, change of supply voltage, etc.

9. Name the type of noise signal present in the oscillation?

A. Schmitt noise
B. Schottky noise
C. Saturation noise
D. None of the mentioned

Schottky noise is the noise signal always present at the input of the transistor due to variation in the carrier concentration.

10. A basic feedback oscillator is satisfying the Barkhausen criterion. If the ß value is given as 0.7072, find the gain of the basic amplifier?

A. 2.1216
B. 0.7072
C. 1
D. 1.414

Barkhausen criterion for oscillation is given as

Aß=1

=> A=1/ ß

= 1/0.7072 = 1.414.

11. The feedback signal of the basic sine wave oscillator is given as

A. Vf = Aß ×Vo
B. Vf = Aß ×Vi
C. Vf = Aß × (Vo/ Vi)
D. Vf = Aß × (Vi/ Vo)

The feedback signal of an oscillator is given as the product of the externally applied signal & the loop gain of the system.

=> Vf= Aß ×Vi.

12. Express the requirement for oscillation in polar form

A. Aß =1∠360o
B. Aß =1∠90o
C. Aß =1∠πo
D. Aß =1∠270o

There are two requirements for oscillation

1. The magnitude of Aß=1
2. The total phase shift of Aß=0o or 360o.

13. What will be the phase shift of the feedback circuit in the RC phase shift oscillator?

A. 360o phase shift
B. 180o phase shift
C. 90o phase shift
D. 60o phase shift

The RC feedback network provides a 180o phase shift and the amplifier used in the RC phase shift oscillator provides a 180o phase shift (op-amp is used in the inverting mode) to obtain a total phase shift of 360o.

14. How many RC stages are used in the RC phase shift oscillator?

A. Six
B. Two
C. Four
D. Three

The RC stage forms the feedback network of the oscillator. It consists of three identical RC stages, each of 60o phase shifts to provide a total phase shift of 180o.

15. Calculate the frequency of oscillation for the RC phase shift oscillator having the value of R and C as 35Ω and 3.7µF respectively.

A. 1230 Hz
B. 204 Hz
C. 502Hz
D. 673 Hz

The frequency of oscillation of RC phase shift oscillator is,

fo=1/(2πRC√6)

= 1/(2×3.14×√6×3.7µF×35Ω)

=> fo= 1/ 1.9921×10-3 = 502Hz.

16. What must be done to ensure that oscillation will not die out in the RC phase shift oscillator?

A. Gain of the amplifier is kept greater than 29
B. Gain of the amplifier is kept greater than 1
C. Gain of the amplifier is kept less than 29
D. Gain of the amplifier is kept less than 1

For a sustained oscillation in the RC phase shift oscillator, the gain of the inverting op-amp should be at least 29. Therefore, the gain is kept greater than 29 to ensure the variation in circuit parameter will not make |Aß|<1, otherwise, oscillation will die out.

17. Calculate the feedback voltage from phase shit network. A. Vf = ( VoR3S3C3) / (1+ 6SRC+5S2C2R2+S3R3C3)
B. Vf = ( VoR3S3C3) / (1+ 5SRC+6S2C2R2+S3R3C3)
C. Vf = ( VoR3S3C3) / (1+ 5SRC+5S2C2R2+S3R3C3)
D. Vf = ( VoR2S3C3) / (1+ 6SRC+5S2C2R2+S3R3C3)

Applying KVL equation to the circuit, we get

=> I1(R+(1/SC.)-I2=Vo -> Equ1

=> -I1R+ I2(2R+(1/SC.)- I3R=0 -> Equ2

=> 0- I2R+ I3(2R+(1/SC.=6 -> Equ3

Vf = I3× 2R,

Solving Equ 1, 2, and 3 for I3

=> I3= ( VoR2S3C3)/ (1+ 5SRC+6S2C2R2+S3R3C3)

=> Vf = I3× 2R

=( VoR3S3C3) / (1+ 5SRC+6S2C2R2+S3R3C3).

18. Which type of op-amp is avoided for high frequencies?

A. LM318
B. Op-amp 741
C. LF 351
D. None of the mentioned

Op-amp741 is generally used for low frequencies < 1 kHz.

19. Find out the constant values of α and ß in phase shift oscillator.

A. α = √6, ß = -1/29
B. α = 6, ß = -1/29
C. α = √6, ß = 1/29
D. α = 6, ß = -1/29

From phase shift network, we obtain

ß= 1/(1-5 α2)+j α(6- α2) -> Equ1

For Aß=1, ß should be real and the imaginary terms must be zero

α(6- α2) =0

=> α=√6

Now substituting α2=6 in Equ 1, we get

ß=-1/29 (Negative sign indicates that the feedback network produces a phase shift of 180o).

20. A phase shift oscillator is designed to oscillate at 155Hz. Determine the value of Rf. (Take C=0.30µF)

A. 399Ω
B. 3.98MΩ
C. 13.9kΩ
D. 403kΩ