# 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

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

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

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

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

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.

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 A. 1.432
B. 9.342
C. 5.768
D. 7.407

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

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

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.

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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