Multiplier and Divider Using Op-Amp MCQ [Free PDF] – Objective Question Answer for Multiplier and Divider Using Op-Amp Quiz

11. Which circuit allows doubling the frequency?

A. Frequency doubler
B. Square doubler
C. Double multiplier
D. All of the mentioned

Answer: A

The multiplication of two sine waves of the same frequency with different amplitude and phases allows for doubling a frequency.

 

12. Compute the output of the frequency doubler. If the inputs Vx = Vsinωt and Vy = Vision(ωt+θ) are applied to a four-quadrant multiplier?

A. Vo= { Vx×Vy× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/ Vref
B. Vo= { Vx×Vy× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/ 2
C. Vo= { Vx×Vy× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×Vref)
D. Vo= – { Vx×Vy× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×Vref)

Answer: C

The output of frequency doubler

Vo= (Vx×Vy)/Vref = [(Vsinωt +Vsin(ωt+θ)]/ Vref

= {Vx×Vy×[(sin2ωt×cosθ)+(sinθ×sinωt×cosωt)]}/ Vref

=> Vo= { Vx×Vy× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×Vref).

 

13. How to remove the dc term produced along with the output in the frequency doubler?

A. Use a capacitor between load and output terminal
B. Use a resistor between load and output terminal
C. Use an Inductor between load and output terminal
D. Use a potentiometer between load and output terminal

Answer: A

The output of the frequency doubler contains a dc term and wave of double frequency. The dc term can be easily removed by using a 1µF coupling capacitor between the load and output terminal of the doubler circuit.

 

14. Find the voltage range at which the multiplier can be used as a squarer circuit?

A. 0 – Vin
B. Vref – Vin
C. 0 – Vref
D. All of the mentioned

Answer: C

The basic multiplier can be used to square any positive or negative number provided the number can be represented by a voltage between 0 to Vref.

 

15. Calculate the output voltage of a squarer circuit, if its input voltage is 3.5v. Assume Vref = 9.67v.

A. 2.86v
B. 1.27v
C. 10v
D. 4.3v

Answer: B

The output voltage of a squarer circuit

Vo = Vi2 / Vref

= (3.5)2 / 9.67v = 1.27v.

 

16. Which circuit can be used to take the square root of a signal?

A. Divider circuit
B. Multiplier circuit
C. Squarer circuit
D. None of the mentioned

Answer: A

A divider circuit can be used to find square roots by connecting both the inputs of the multiplier (in the divider) to the output of an op-amp.

 

17. Find the output voltage of squarer circuit?

A. Vo = √( Vref/|Vin|)
B. Vo = √( Vref×|Vin|)
C. Vo = √( Vref2×|Vin|)
D. Vo = √( Vref×|Vin|2)

Answer: B

The output of the square root circuit is proportional to the square root of the magnitude of the input voltage.

 

18. Find the current, IL flowing in the circuit given below

Find the current, IL flowing in the circuit given below


A. IL = 1.777mA
B. IL = 1.048mA
C. IL = 1.871mA
D. None of the mentioned
Answer: D

Voltage, VL = Vo2/ Vref =132/10 = 16.9v.

∴ IL = VL/R = 16.9/12kΩ = 1.408mA.

 

19. A square root circuit built from the multiplier is given an input voltage of 11.5v. Find its corresponding output voltage?

A. 11v
B. 15v
C. 13v
D. Cannot be determined

Answer: D

The input voltage must be negative or else the op-amp will saturate and lie between the ranges of -1 to -10v.

 

20. The circuit in which the output voltage waveform is the integral of the input voltage waveform is called

A. Integrator
B. Differentiator
C. Phase shift oscillator
D. Square wave generator

Answer: A

The integrator circuit produces the output voltage waveform as the integral of the input voltage waveform.

 

21. Find the output voltage of the integrator

A. Vo = (1/R×CFt0 Vindt+C
B. Vo = (R/CFt0 Vindt+C
C. Vo = (CF/R)×t0 Vindt+C
D. Vo = (R×CFt0 Vindt+C

Answer: A

The output voltage is directly proportional to the negative integral of the input voltage and inversely proportional to the time constant RCF.

Vo = (1/R×CFt0 Vindt+C

Where C-> Integration constant and CF-> Feedback capacitor.

 

22. Why an integrator cannot be made using a low pass RC circuit?

A. It requires a large value of R and a small value of C
B. It requires a large value of C and a small value of R
C. It requires a large value of R and C
D. It requires a small value of R and C

Answer: C

A simple low pass RC circuit can work as an integrator when the time constant is very large, which requires the large value of R and C. Due to practical limitations, the R and C cannot be made infinitely large.

 

23. How a perfect integration is achieved in op-amp?

A. Infinite gain
B. Low input impedance
C. Low output impedance
D. High CMRR

Answer: A

In an op-amp integrator, the effective input capacitance becomes CF×(1-Av). Where Av is the gain of op-amp. The gain is infinite for the ideal op-amp. So, the effective time constant of the op-amp integrator becomes very large which results in perfect integration.

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