This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses on “Multiplier and Divider – 1”.

1. Determine output voltage of analog multiplier provided with two input signal 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_{ref}^{2}

View Answer

Explanation: 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.

=> 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

View Answer

Explanation: 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

View Answer

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

a) V

_{o}= [(V

_{x}×V

_{y}) /V

_{ref}

^{2}] × [1-cos2ωt/2] b) V

_{o}= [(V

_{x}

^{2}×V

_{y}

^{2}) /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] View Answer

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

∴ V

_{x}= V

_{x}sinωt and V

_{y}= V

_{y}sinωt

=> V

_{o}= (V

_{x}×V

_{y}× sin

^{2}ωt) / 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×10^{4}t)

b) V_{o} = 2.75-(2.75×cos4π×10^{4}t)

c) V_{o} = 1.25-(1.25×cos4π×10^{4}t)

d) None of the mentioned

View Answer

Explanation: Output voltage of frequency V

_{o}=V

_{i}

^{2}/ V

_{ref}

=> V

_{i}= 5sinωt = 5sin2π×10

^{4}t

V

_{o}= [5×(sin2π×10

^{4}t)

^{2}]/10 = 2.5×[1/2-(1/2cos2π ×2×10

^{4}t)] = 1.25-(1.25×cos4π×10

^{4}t).

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

View Answer

Explanation: 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

^{-1}30 =1.019.

7. Express the output voltage equation of 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_{ref}^{2}×(V_{z}/V_{x})

View Answer

Explanation: 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 divider circuit configuration given below

View Answer

Explanation: Division is the complement of multiplication. So, the divider can be accomplished by placing the multiplier circuit element in the op-amp feedback loop.

a) I

_{Z}= 0.5372mA

b) I

_{Z}= 1.581mA

c) I

_{Z}= 2.436mA

d) I

_{Z}=9.347mA

View Answer

Explanation: Input current, I

_{Z}= -(V

_{x}×V

_{o})/(V

_{ref}×R) = -(4.79v×16.5v)/(10×5kΩ) = 1.581mA.

10. 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
[expand title="View Answer"] Answer: c
Explanation: 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 multiplier will not saturate. [/expand]

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