250+ TOP MCQs on Peaking Amplifier & Answers

Linear Integrated Circuit Multiple Choice Questions 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
Answer: C
Clarification: 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.
Answer: A
Clarification: The resonant frequency is also called as peak frequency, which is determined by the combination of L and C.
f_p = 1/(2π√LC..

3. 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
Clarification: Frequency, f_p= 1/[2π×√(LC.] =1/[2π√(0.1µF×8mH)]=1/1.776×10_-4= 5.63kHz.
=> X_L = 2πf_pL = 2π×5.63kHz×8mH = 282.85.
The figure of merit of coil, Q_coil= X_L/R₁= 282.85/100Ω = 2.8285.
∴ R_p = (Q_coil)² ×R₁ = (2.8285^2)×100Ω= 800Ω.
The gain of the amplifier at resonance is maximum and given by
A_F =-(R_F||R_p)/R₁ = -(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
Answer: B
Clarification: The gain of the amplifier at resonance is the maximum and given by,
A_F =-(R_F||R_p)/ R₁ =-[(R_p×R_F)/ (R_F+R_p)] /R₁ = -[ (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
Answer: B
Clarification: 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).

6. 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
Clarification: Given L=10mH and the internal resistance of the inductor R=30Ω. Assume R₁=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_pL)/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)/[R₁×(R_F+R_p)]
=>R_F = (A_f ×R_p ×R₁)/ (R_p -A_F ×R₁)
R_F = (-10×13.5×10⁶×100) / (13.5×10⁶-(10×100))=1000Ω
=> R_F = 1kΩ.
Thus the component values are R₁=100Ω, R_F= 1kΩ, L=10mH at R=30Ω and C = 4µF.

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