DC Amplifier MCQ [Free PDF] – Objective Question Answer for DC Amplifier Quiz

11. 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).

 

12. 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.

 

13. 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.

 

14. 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).

 

15. 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.

 

16. In which amplifier the output voltage is equal to the negative sum of all the inputs?

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

Answer: B

In the summing amplifier, the output voltage is equal to the sum of all input. Since the total input is a sum of negative input, the amplifier is an inverting summing amplifier.

 

17. Determine the expression of output voltage for inverting summing amplifier consisting of four internal resistors? (Assume the value of internal resistors to be equal)

A. Vo = -(Rf/R )×(Va +Vb+Vc+Vd)
B. Vo = (RF/R)×(Va +Vb+Vc+Vd)
C. Vo = (R/ RF)×(Va +Vb+Vc+Vd)
D. None of the mentioned

Answer: A

If the internal resistors of the circuit are the same i.e

Ra=Rb=Rc=Rd=R (since there are four internal resistors)

Then, the output voltage for inverting amplifier is given as

Vo= -(Rf/R)×(Va +Vb+Vc+Vd).

 

18. An inverting amplifier with gain 1 has different input voltage: 1.2v,3.2v, and 4.2v. Find the output voltage?

A. 4.2v
B. 8.6v
C. -4.2v
D. -8.6v

Answer: D

When the gain of the inverting summing amplifier gain is 1 then, the internal resistors and feedback resistors have the same value. So, the output is equal to the negative sum of all input voltages.

VO= -(Va+Vb+Vc)

=-(1.2+3.2+4.2)= -8.6v.

 

19. In which type of amplifier, the input voltage is amplified by a scaling factor

A. Summing amplifier
B. Averaging amplifier
C. Weighted amplifier
D. Differential amplifier

Answer: C

The weighted amplifier is also called a scaling amplifier. Here each input voltage is amplified by a different factor i.e. Ra, Rb and Rc are different in values ( which are the input resistors at each input voltage).

 

20. An inverting scaling amplifier has three input voltages Va, Vb and Vc. Find it output voltage?

A. VO= – {[(RF/Ra)×Va] +[(RF/Rb)×Vb]+[(RF/Rc)×Vc]}
B. VO= – [(RF/Ra)+(RF/Rb)+(RF/Rc)]×[( Va +Vb+Vc)].
C. VO = – {[(Ra/RF)×Va] +[(Rb/RF)×Vb]+[(Rc/RF)×Vc]}
D. None of the mentioned

Answer: A

Since three input voltages are given assume the input resistors to be Ra, Rb, and Rc. In a scaling amplifier, the input voltages are amplified by a different factor

=> ∴ RF/Ra ≠ RF/Rb ≠ RF/Rc

Therefore, the output voltage

Vo = -{[(RF/Ra) Va] +[(RF/Rb) Vb]+[(RF/Rc) Vc]}.

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