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

1. Determine output voltage of analog multiplier provided with two input signals V_x and V_y.

A. V_o = (V_x ×V_x) / V_y
B. V_o = (V_x ×V_y / V_ref
C. V_o = (V_y ×V_y) / V_x
D. V_o = (V_x ×V_y) / V_ref²

Answer: B

The output is the product of two inputs divided by a reference voltage in the analog multiplier. Thus, the output voltage is a scaled version of x and y inputs.

=> V_o =V_x ×V_y / V_ref.

2. Match the List-I with list-II

List-I	List-II
1. One quadrant multiplier	i. Input 1- Positive, Input 2- Either positive or negative
2. Two quadrant multiplier	ii. Input 1- Positive, Input 2 – Positive
3. Four quadrant multiplier	iii. Input 1- Either positive or negative, Input 2- Either positive or negative

A. 1-ii, 2-i, 3-iii
B. 1-ii, 2-ii, 3-ii
C. 1-iii, 2-I, 3-ii
D. 1-I, 2-iii, 3-i

Answer: A

If both inputs are positive, the IC is said to be a one quadrant multiplier. A two-quadrant multiplier function properly, if one input is held positive and the other is allowed to swing. Similarly, for a four-quadrant multiplier both the inputs are allowed to swing.

3. What is the disadvantage of the log-antilog multiplier?

A. Provides four-quadrant multiplication only
B. Provides one quadrant multiplication only
C. Provides two and four-quadrant multiplication only
D. Provides one, two, and four-quadrant multiplication only

Answer: B

Log amplifier requires the input and reference voltage to be of the same polarity. This restricts the log-antilog multiplier to one quadrant operation.

4. An input of Vsinωt is applied to an ideal frequency doubler. Compute its output voltage?

A. V_o = [(V_x×V_y) /V_ref²] × [1-cos2ωt/2].
B. V_o = [(V_x²×V_y²) /V_ref] × [1-cos2ωt/2].
C. V_o = [(V_x×V_y)² /V_ref] × [1-cos2ωt/2].
D. V_o = [(V_x×V_y) /( V_ref] × [1-cos2ωt/2].

Answer: D

In an ideal frequency doubler, same frequency is applied to both inputs.

∴ V_x = V_xsinωt and V_y = V_ysinωt

=> V_o = (V_x×V_y × sin²ωt) / V_ref

= [(V_x×V_y) / V_ref] × [1-cos2ωt/2].

5. Find the output voltage for the squarer circuit given below, choose input frequency as 10kHz and V_ref =10v

A. V_o = 5.0-(5.0×cos4π×10⁴t)
B. V_o = 2.75-(2.75×cos4π×10⁴t)
C. V_o = 1.25-(1.25×cos4π×10⁴t)
D. None of the mentioned

Answer: C

Output voltage of frequency V_o =V_i² / V_ref

=> V_i = 5sinωt = 5sin2π×10⁴t

V_o = [5×(sin2π×10⁴t)² ]/10

= 2.5×[1/2-(1/2cos2π ×2×10⁴t)] = 1.25-

(1.25×cos4π×10⁴t).

6. Calculate the phase difference between two input signals applied to a multiplier, if the input signals are V_x= 2sinωt and V_y= 4sin(ωt+θ). (Take V_ref= 12v).

A. θ = 1.019
B. θ = 30.626
C. θ = 13.87
D. θ = 45.667

Answer: A

V_o= [V_mx×V_my /(2×V_ref)] ×cosθ

=> (V_o×2×V_ref)/ (V_mx × V_my) = cosθ

=> cosθ = (10×2×12)/(2×4) = 30.

=> θ = cos^-130 =1.019.

7. Express the output voltage equation of the divider circuit

A. V_o= -(V_ref/2)×(V_z/V_x)
B. V_o= -(2×V_ref)×(V_z/V_x)
C. V_o= -(V_ref)×(V_z/V_x)
D. V_o= -V_ref²×(V_z/V_x)

Answer: C

The output voltage of the divider,

V_o= -V_ref×(V_z/V_x).

Where V_z –> dividend and V_x –> divisor.

8. Find the input current for the circuit given below.

A. I_Z = 0.5372mA
B. I_Z = 1.581mA
C. I_Z = 2.436mA
D. I_Z =9.347mA

Answer: B

Input current, I_Z = -(V_x×V_o)/(V_ref×R)

= −(4.79v×16.5v)/(10×5kΩ)

= 1.581mA.

9. Find the condition at which the output will not saturate?

A. V_x > 10v ; V_y > 10v
B. V_x < 10v ; V_y > 10v
C. V_x < 10v ; V_y < 10v
D. V_x > 10v ; V_y < 10v

Answer: C

In an analog multiplier, V_ref is internally set to 10v. As long as V_x < V_ref and V_y < V_ref, the output of the multiplier will not saturate.

10. 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

The log-analog 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).

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