Linear Integrated Circuit Multiple Choice Questions on “Multiplier and Divider – 1”.
1. Determine output voltage of analog multiplier provided with two input signal Vx and Vy.
A. Vo = (Vx ×Vx) / Vy
B. Vo = (Vx ×Vy / Vref
C. Vo = (Vy ×Vy) / Vx
D. Vo = (Vx ×Vy) / Vref2
Answer: B
Clarification: The output is the product of two inputs divided by a reference voltage in analog multiplier. Thus, the output voltage is a scaled version of x and y inputs.
=> Vo =Vx ×Vy / Vref.
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
Clarification: 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 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
Clarification: Log amplifier requires the input and reference voltage to be of the same polarity. This restricts log-antilog multiplier to one quadrant operation.
4. An input of Vsinωt is applied to an ideal frequency doubler. Compute its output voltage?
A. Vo = [(Vx×Vy) /Vref2] × [1-cos2ωt/2].
B. Vo = [(Vx2×Vy2) /Vref] × [1-cos2ωt/2].
C. Vo = [(Vx×Vy)2 /Vref] × [1-cos2ωt/2].
D. Vo = [(Vx×Vy) /( Vref] × [1-cos2ωt/2].
Answer: D
Clarification: In an ideal frequency doubler, same frequency is applied to both inputs.
∴ Vx = Vxsinωt and Vy = Vysinωt
=> Vo = (Vx×Vy × sin2ωt) / Vref = [(Vx×Vy) / Vref] × [1-cos2ωt/2].
5. Find the output voltage for the squarer circuit given below, choose input frequency as 10kHz and Vref =10v
A. Vo = 5.0-(5.0×cos4π×104t)
B. Vo = 2.75-(2.75×cos4π×104t)
C. Vo = 1.25-(1.25×cos4π×104t)
D. None of the mentioned
Answer: C
Clarification: Output voltage of frequency Vo =Vi2 / Vref
=> Vi = 5sinωt = 5sin2π×104t
Vo = [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 Vx= 2sinωt and Vy= 4sin(ωt+θ). (Take Vref= 12v).
A. θ = 1.019
B. θ = 30.626
C. θ = 13.87
D. θ = 45.667
Answer: A
Clarification: Vo= [Vmx×Vmy /(2×Vref)] ×cosθ
=> (Vo×2×Vref)/ (Vmx × Vmy) = cosθ
=> cosθ = (10×2×12)/(2×4) = 30.
=> θ = cos-130 =1.019.
7. Express the output voltage equation of divider circuit
A. Vo= -(Vref/2)×(Vz/Vx)
B. Vo= -(2×Vref)×(Vz/Vx)
C. Vo= -(Vref)×(Vz/Vx)
D. Vo= -Vref2×(Vz/Vx)
Answer: C
Clarification: The output voltage of the divider, Vo= -Vref×(Vz/Vx).
Where Vz –> dividend and Vx –> divisor.
8. Find the divider circuit configuration given below
Answer: A
Clarification: Division is the complement of multiplication. So, the divider can be accomplished by placing the multiplier circuit element in the op-amp feedback loop.
9. Find the input current for the circuit given below.
A. IZ = 0.5372mA
B. IZ = 1.581mA
C. IZ = 2.436mA
D. IZ =9.347mA
Answer: B
Clarification: Input current, IZ = -(Vx×Vo)/(Vref×R) = -(4.79v×16.5v)/(10×5kΩ) = 1.581mA.
10. Find the condition at which the output will not saturate?
A. Vx > 10v ; Vy > 10v
B. Vx < 10v ; Vy > 10v
C. Vx < 10v ; Vy < 10v
D. Vx > 10v ; Vy < 10v
Answer: C
Clarification: In an analog multiplier, Vref is internally set to 10v. As long as Vx < Vref and Vy < Vref, the output of multiplier will not saturate.
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