This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses on “Peaking Amplifier”.

1. 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

View Answer

Explanation: 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.

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

a) f_{p} = 1/[2π×√(LC)]
b) f_{p} = (2π×√L)/C

c) f_{p} = 2π×√(LC)

d) f_{p} = 2π/√(LC)

View Answer

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

f

_{p}= 1/(2π√LC).

a) 1.432

b) 9.342

c) 5.768

d) 7.407

View Answer

Explanation: Frequency, f

_{p}= 1/[2π×√(LC)] =1/[2π√(0.1µF×8mH)]=1/1.776×10

_{-4}= 5.63kHz.

=> X

_{L}= 2πf

_{p}L = 2π×5.63kHz×8mH = 282.85.

The figure of merit of coil, Q

_{coil}= X

_{L}/R

_{1}= 282.85/100Ω = 2.8285.

∴ R

_{p}= (Q

_{coil})

^{2}×R

_{1}= (2.8285^2)×100Ω= 800Ω.

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

A

_{F}=-(R

_{F}||R

_{p})/R

_{1}= -(10kΩ||800)/100Ω =-740.740/100 = -7.407.

4. The parallel resistance of 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

View Answer

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

A

_{F}=-(R

_{F}||R

_{p})/ R

_{1}=-[(R

_{p}×R

_{F})/ (R

_{F}+R

_{p})] /R

_{1}= -[ (10kΩ×35kΩ)/ (10kΩ+35kΩ)] /1kΩ

=> A

_{F}=- 7.78kΩ/1kΩ= -7.78.

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

a) BW = (f_{p}× X_{L})/ (R_{F}+R_{p})

b) BW =[ f_{p}×(R_{F}+R_{p})× X_{L} ] / (R_{F}×R_{p})

c) BW =[ f_{p}×(R_{F}+R_{p})] / (R_{F}×R_{p})

d) BW = [f_{p}×(R_{F}+R_{p}) ]/ X_{L}

View Answer

Explanation: The bandwidth of the peaking amplifier,

BW = f

_{p}/Q

_{p}, where Q

_{p}– figure of merit of the parallel resonant circuit = (R

_{f}||R

_{p})/X

_{l}= (R

_{f}×R

_{p})/[(R

_{f}+R

_{p})× X

_{l}] => BW = [f

_{p}×(R

_{f}+R

_{p})× X

_{l}] / (R

_{f}×R

_{p}).

View Answer

Explanation: Given L=10mH and the internal resistance of the inductor R=30Ω. Assume R

_{1}=100Ω. The gain times peak frequency= 10×32kHz=320kHz

f

_{p}= 1/2π√LC

=> C = 1/[(2π)

^{2}× (f

_{p})

^{2}×L]= 1/ [(2π)

^{2}×(320)

^{2}×10mH] = 1/252405.76 = 3.96µF ≅4µF.

Q

_{coil}= x

_{L}/R =(2πf

_{p}L)/R =(2π×320kHz×10mH)/30 = 20096/30 =669.87

=> R

_{p}= (Q

_{coil})

^{2}×R = (669.88)

^{2}×30 = 13.5MΩ

To find R

_{f},

A

_{F}= (R

_{F}×R

_{p})/[R

_{1}×(R

_{F}+R

_{p})] =>R

_{F}= (A

_{f}×R

_{p}×R

_{1})/ (R

_{p}-A

_{F}×R

_{1})

R

_{F}= (-10×13.5×10

^{6}×100) / (13.5×10

^{6}-(10×100))=1000Ω

=> R

_{F}= 1kΩ.

Thus the component values are R

_{1}=100Ω, R

_{F}= 1kΩ, L=10mH at R=30Ω and C = 4µF.

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