# Linear Integrated Circuit MCQ [Free PDF] – Objective Question Answer for Linear Integrated Circuit 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)

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

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

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

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

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?

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

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

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 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

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?

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

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].

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

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

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

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)

The output voltage of the divider,

Vo= -Vref×(Vz/Vx).

Where Vz –> dividend and Vx –> divisor.

113. 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

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?

A. Vx > 10v ; Vy > 10v
B. Vx < 10v ; Vy > 10v
C. Vx < 10v ; Vy < 10v
D. Vx > 10v ; Vy < 10v

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 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)

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

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)

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

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

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