# Log and Antilog Amplifier MCQ [Free PDF] – Objective Question Answer for Log and Antilog Amplifier Quiz

1. Which of the following functions does the antilog computation required to perform continuously with log-amps?

A. In(x)
B. log(x)
C. Sinh(x)
D. All of the mentioned

Log-amp can easily perform function such as In(x), Log(x), Sinh(x) to have direct dB display on digital voltmeter and spectrum analyser.

2. 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).

3. 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.

4. 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.

5. 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)].

6. 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.

7. 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.

8. 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.

9. 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.

10. 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].