Operational Amplifier Application MCQ with Explanation- Objective Question Answer for Operational Amplifier Application Quiz

101. Find the output voltage of the log-amplifier

A. VO  = -(kT)×ln(Vi/Vref)
B. VO  = -(kT/q)×ln(Vi/Vref)
C. VO  = -(kT/q)×ln(Vref/Vi)
D. VO  = (kT/q)×ln(Vi/Vref)

Answer: B

the output voltage is proportional to the logarithm of the input voltage.

VO  = -(kT/q)×ln(Vi / Vref).

 

102. How to provide saturation current and temperature compensation in log-amp?

A. Applying reference voltage alone to two different log-amps
B. Applying input and reference voltage to same log-amps
C. Applying input and reference voltage to separate log-amps
D. None of the mentioned

Answer: C

The emitter saturation current varies from transistor to transistor with temperature. Therefore, the input and reference voltage are applied to separate log-amps and two transistors are integrated close together in the same silicon wafer. This provides a close match of the saturation currents and ensures good thermal tracking.

 

103. The input voltage, 6v, and reference voltage, 4 v are applied to a log-amp with saturation current and temperature compensation. Find the output voltage of the log-amp?

A. 6.314(kT/q)v
B. 0.597(kT/q)v
C. 0.405(kT/q)v
D. 1.214(kT/q)v

Answer: C

The output voltage of saturation current and temperature compensation log-amp

VO  = (kT/q)×ln(Vi / Vref)

= (kT/q)×ln(6v/4v)

= (kT/q)×ln(1.5)

VO  = 0.405(kT/q)v.

 

104. Determine the output voltage for the given circuit

Determine the output voltage for the given circuit

A. VO  = Vref/(10-k’vi)
B. VO  = Vref+(10-k’vi)
C. VO  = Vref×(10-k’vi)
D. VO  = Vref-(10-k’vi)

Answer: C

The output voltage of an antilog amp is given as

VO  = Vref (10-k’vi)

Where k’ = 0.4343 (q/kt)×[(RTC/ (R2 +RTC)].

 

105. Calculate the base voltage of the Q2 transistor in the log-amp using two op-amps?

Calculate the base voltage of Q2 transistor in the log-amp using two op-amps?

A. 8.7v
B. 5.3v
C. 3.3v
D. 6.2v

Answer: C

The base voltage of Q2 transistor,

VB  = [RTC / (R2 +RTC)]×(Vi)

= [10kΩ/(5kΩ+10kΩ)]×5v = 3.33v.

 

106. Determine output voltage of analog multiplier provided with two input signals 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

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.

= > Vo  = Vx ×Vy / Vref.

 

107. Match the multiplier with List-I and 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.

 

108. 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.

 

109. 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

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].

 

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

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

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).

 

111. 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

Vo = [Vmx×Vmy /(2×Vref)] ×cosθ

= > (Vo×2×Vref)/ (Vmx × Vmy) = cosθ

= > cosθ = (10×2×12)/(2×4) = 30.

= > θ = cos-130 = 1.019.

 

112. Express the output voltage equation of the 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

The output voltage of the divider circuit

Vo = -Vref×(Vz/Vx).

Where Vz –> dividend and Vx –> divisor.

 

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

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

Input current, IZ  = -(Vx×Vo)/(Vref×R)

= −(4.79v×16.5v)/(10×5kΩ)

= 1.581mA.

 

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

In an analog multiplier, Vref is internally set to 10v. As long as Vx < Vref and Vy < Vref, the output of the multiplier will not saturate.

 

115. Determine the relationship between the log-antilog method.

A. lnVx×lnVy  = ln(Vx+Vy)
B. lnVx / lnVy  = ln(Vx-Vy)
C. lnVx -lnVy  = ln(Vx/Vy)
D. lnVx+ lnVy  = ln(Vx×Vy)

Answer: D

The log-analog method relies on the mathematical relationship that is the sum of the logarithm of the product of those numbers.

= > lnVx+lnVy  = ln(Vx×Vy).

 

116. 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.

 

117. 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).

 

118. 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.

 

119. 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.

 

120. 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.

Scroll to Top