Very High Input Impedance OP-Amp Circuit MCQ [Free PDF] - Objective Question Answer for Very High Input Impedance OP-Amp Circuit Quiz

11. 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

V_O = (kT/q)×ln(V_i / V_ref)

=(kT/q)×ln(6v/4v)

=(kT/q)×ln(1.5)

V_O = 0.405(kT/q)v.

12. Determine the output voltage for the given circuit

A. V_O = V_ref/(10^-k’v_i)
B. V_O = V_ref+(10^-k’v_i)
C. V_O = V_ref×(10^-k’v_i)
D. V_O = V_ref-(10^-k’v_i)

Answer: C

The output voltage of an antilog amp is given as

V_O = V_ref (10^-k’v_i)

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

13. 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

Answer: C

The base voltage of Q₂ transistor,

V_B = [R_TC / (R₂ +R_TC)]×(V_i)

= [10kΩ/(5kΩ+10kΩ)]×5v =3.33v.

14. Determine output voltage of analog multiplier provided with two input signals 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_ref²

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.

=> V_o =V_x ×V_y / V_ref.

15. 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

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.

16. 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.

17. An input of Vsinωt is applied to an ideal frequency doubler. Compute its output voltage?

A. V_o = [(V_x×V_y) /V_ref²] × [1-cos2ωt/2].
B. V_o = [(V_x²×V_y²) /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].

Answer: D

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

∴ V_x = V_xsinωt and V_y = V_ysinωt

=> V_o = (V_x×V_y × sin²ωt) / V_ref

= [(V_x×V_y) / V_ref] × [1-cos2ωt/2].

18. 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×10⁴t)
B. V_o = 2.75-(2.75×cos4π×10⁴t)
C. V_o = 1.25-(1.25×cos4π×10⁴t)
D. None of the mentioned

Answer: C

Output voltage of frequency V_o =V_i² / V_ref

=> V_i = 5sinωt = 5sin2π×10⁴t

V_o = [5×(sin2π×10⁴t)² ]/10

= 2.5×[1/2-(1/2cos2π ×2×10⁴t)] = 1.25-

(1.25×cos4π×10⁴t).

19. 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

Answer: A

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^-130 =1.019.

20. Express the output voltage equation of the 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_ref²×(V_z/V_x)

Answer: C

The output voltage of the divider,

V_o= -V_ref×(V_z/V_x).

Where V_z –> dividend and V_x –> divisor.

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