250+ TOP MCQs on Multiplier and Divider – 2 & Answers

Tricky Linear Integrated Circuit Questions and Answerson “Multiplier and Divider – 2”.

1. Determine the relationship between log-antilog method.
A. lnV_x×lnV_y = ln(V_x+V_y)
B. lnV_x / lnV_y = ln(V_x-V_y)
C. lnV_x -lnV_y = ln(V_x/V_y)
D. lnV_x+ lnV_y = ln(V_x×V_y)
Answer: D
Clarification: The log-anilog method relies on the mathematical relationship that is the sum of the logarithm of the product of those numbers.
=> lnV_x+lnV_y = ln(V_x×V_y).

2. Which circuit allows to double the frequency?
A. Frequency doubler
B. Square doubler
C. Double multiplier
D. All of the mentioned
Answer: A
Clarification: The multiplication of two sine wave of same frequency with different amplitude and phase allows in doubling a frequency.

3. Compute the output of frequency doubler. If the inputs V_x = Vsinωt and V_y = Vsin(ωt+θ) are applied to a four quadrant multiplier?
A. V_o= { V_x×V_y× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/ V_ref
B. V_o= { V_x×V_y× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/ 2
C. V_o= { V_x×V_y× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×V_ref)
D. V_o= – { V_x×V_y× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×V_ref)
Answer: C
Clarification: The output of frequency doubler, V_o= (V_x×V_y)/V_ref = [(Vsinωt +Vsin(ωt+θ)]/ V_ref
= {V_x×V_y×[(sin²ωt×cosθ)+(sinθ×sinωt×cosωt)]}/ V_ref
=> V_o= { V_x×V_y× [cosθ-(cosθ×cos2ωt)+(sinθ×sin2ωt)]}/(2×V_ref).

4. How to remove the dc term produced along with the output in frequency doubler?
A. Use a capacitor between load and output terminal
B. Use a resistor between load and output terminal
C. Use a Inductor between load and output terminal
D. Use a potentiometer between load and output terminal
Answer: A
Clarification: 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 load and output terminal of the doubler circuit.

5. Find the voltage range at which the multiplier can be used as a squarer circuit?
A. 0 – V_in
B. V_ref – V_in
C. 0 – V_ref
D. All of the mentioned
Answer: C
Clarification: 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 V_ref.

6. Calculate the output voltage of a squarer circuit, if it input voltage is 3.5v. Assume V_ref = 9.67v.
A. 2.86v
B. 1.27v
C. 10v
D. 4.3v
Answer: B
Clarification: The output voltage of a squarer circuit, V_o = V_i² / V_ref = (3.5)² / 9.67v = 1.27v.

7. Which circuit can be used to take square root of a signal?
A. Divider circuit
B. Multiplier circuit
C. Squarer circuit
D. None of the mentioned
Answer: A
Clarification: 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.

8. Find the output voltage of squarer circuit?
A. V_o = √( V_ref/|V_in|)
B. V_o = √( V_ref×|V_in|)
C. V_o = √( V_ref²×|V_in|)
D. V_o = √( V_ref×|V_in|²)
Answer: B
Clarification: The output of the square root circuit is proportional to the square root of the magnitude of the input voltage.

9. Find the current, I_L flowing in the circuit given below

A. I_L = 1.777mA
B. I_L = 1.048mA
C. I_L = 1.871mA
D. None of the mentioned
Answer: D
Clarification: Voltage, V_L = V_o²/ V_ref =13²/10 = 16.9v.
∴ I_L = V_L/R = 16.9/12kΩ = 1.408mA.

10. A square root circuit build from 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
Clarification: The input voltage must be negative or else the op-amp will saturate and lie between the ranges of -1 to -10v.

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