AC Amplifiers with Single Supply Voltage MCQ [Free PDF] – Objective Question Answer for AC Amplifiers with Single Supply Voltage Quiz

1. How an AC amplifier can be powered by a single supply voltage and produces voltage swing?

A. By inserting a voltage divider at the inverting input
B. By inserting a voltage divider at the non-inverting input
C. By inserting a voltage divider at the output
D. By inserting a voltage divider at the feedback circuit

Answer: B

A positive dc level is intentionally inserted using a voltage-divider network at the non-inverting input terminal so that output can swing both positively as well as negatively.

 

2. Find the maximum output voltage swing of an AC inverting amplifier using op-amp 741C?

A. +15Vpp
B. ±15Vpp
C. ±13Vpp
D. +13Vpp

Answer: A

The value of power supply for 741 op-amp=±15v. Therefore, the ideal maximum output voltage swing for an AC amplifier with a single power supply = +Vcc = +15v.

 

3. Determine the lower cut-off frequency of the circuit.

Determine the lower cut-off frequency of the circuit.

A. 21.3Hz
B. 12.15Hz
C. 1.35Hz
D. None of the mentioned

Answer: D

The input resistance of the amplifier is RIF =(R2 ||R3) || [ri×(1+AB.] –> equ 1

As [ri×(1+AB.] >> R2

=> Therefore, equ 1 becomes

RIF ≅ R2 || R3 = 100kΩ || 100kΩ

= (100×100)/(100+100) = 50kΩ.

=> Rin = Ro = 150Ω.

∴ Lower cut off frequency

fL= 1/[2πCi×(RIF+Ro)]

= 1/[2π×0.47µF×(50kΩ+150Ω)] = 6.75Hz.

 

4. In differential op-amp configuration a subtractor is called as

A. Summing amplifier
B. Difference amplifier
C. Scaling amplifier
D. All of the mentioned

Answer: C

In a subtractor, input signals can be scaled to the desired values by selecting appropriate values for the external resistors. Therefore, this circuit is referred to as a scaling amplifier

 

5. How the peaking response is obtained?

A. Using a series LC network with op-amp
B. Using a series RC network with op-amp
C. Using a parallel LC network with op-amp
D. Using a parallel RC network with op-amp

Answer: C

The peaking response is the frequency response that peaks at a certain frequency. This can be obtained by using a parallel LC network with the op-amp.

 

6. The expression for the resonant frequency of the op-amp

A. fp = 1/[2π×√(LC.].
B. fp = (2π×√L)/C
C. fp = 2π×√(LC.
D. fp = 2π/√(LC.

Answer: A

The resonant frequency is also called peak frequency, which is determined by the combination of L and C.

fp = 1/(2π√LC).

 

7. From the circuit given below find the gain of the amplifier

From the circuit given below find the gain of the amplifier
A. 1.432
B. 9.342
C. 5.768
D. 7.407

Answer: D

Frequency,

fp= 1/[2π×√(LC.]

=1/[2π√(0.1µF×8mH)]

=1/1.776×10-4= 5.63kHz.

=> XL = 2πfpL

= 2π×5.63kHz×8mH = 282.85.

The figure of merit of coil

Qcoil= XL/R1= 282.85/100Ω = 2.8285.

∴ Rp = (Qcoil)2 ×R1

= (2.82852)×100Ω= 800Ω.

The gain of the amplifier at resonance is maximum and given by

AF =-(RF||Rp)/R1

= -(10kΩ||800)/100Ω

=-740.740/100 = -7.407.

 

8. The parallel resistance of the tank circuit and for the circuit is given below. Find the gain of the amplifier?

parallel resistance of tank circuit and for the circuit is given below

A. -778
B. -7.78
C. -72.8
D. None of the mentioned

Answer: B

The gain of the amplifier at resonance is the maximum and given by,

AF =-(RF||Rp)/ R1

=-[(Rp×RF)/ (RF+Rp)] /R1

= -[ (10kΩ × 35kΩ)/ (10kΩ+35kΩ)] /1kΩ

=> AF =- 7.78kΩ/1kΩ= -7.78.

 

9. The band width of the peaking amplifier is expressed as

A. BW = (fp× XL)/ (RF+Rp)
B. BW =[ fp×(RF+Rp)× XL ] / (RF×Rp)
C. BW =[ fp×(RF+Rp)] / (RF×Rp)
D. BW = [fp×(RF+Rp) ]/ XL

Answer: B

The bandwidth of the peaking amplifier,

BW = fp/Qp,

where Qp – figure of merit of the parallel resonant circuit

= (Rf||Rp)/Xl = (Rf×Rp)/[(Rf+Rp)× Xl]

=> BW = [fp×(Rf+Rp)× Xl] / (Rf×Rp).

 

10. Design a peaking amplifier circuit to provide a gain of 10 at a peak frequency of 32khz given L=10mH having 30Ω resistance.

Answer: B

Given L=10mH and the internal resistance of the inductor R=30Ω. Assume R1=100Ω. The gain times peak frequency=

10 × 32kHz = 320kHz

fp= 1/2π√LC

=> C = 1/[(2π)2× (fp)2×L]

= 1/ [(2π)2×(320)2×10mH]

= 1/252405.76 = 3.96µF ≅4µF.

Qcoil = xL/R

=(2πfpL)/R

=(2π×320kHz×10mH)/30

= 20096/30 =669.87

=> Rp= (Qcoil)2×R

= (669.88)2×30 = 13.5MΩ

To find Rf,

AF= (RF×Rp)/[R1×(RF+Rp)]

=>RF = (Af ×Rp ×R1)/ (Rp -AF ×R1)

RF = (-10×13.5×106×100) / (13.5×106-(10×100))=1000Ω

=> RF = 1kΩ.

Thus the component values are R1=100Ω, RF= 1kΩ, L=10mH at R=30Ω and C = 4µF.

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