Linear Integrated Circuit Questions and Answers – Peaking Amplifier

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

Answer: c
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) fp = 1/[2π×√(LC)].
b) fp = (2π×√L)/C
c) fp = 2π×√(LC)
d) fp = 2π/√(LC)
View Answer

Answer: a
Explanation: The resonant frequency is also called as peak frequency, which is determined by the combination of L and C.
fp = 1/(2π√LC).

3. From the circuit given below find the gain of the amplifier
Find the gain of the amplifier for the given circuit
a) 1.432
b) 9.342
c) 5.768
d) 7.407
View Answer

Answer: d
Explanation: 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 = (2.8285^2)×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.
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4. The parallel resistance of tank circuit and for the circuit is given below.Find the gain of the amplifier?
Find the gain of the amplifier which has parallel resistance of tank circuit
a) -778
b) -7.78
c) -72.8
d) None of the mentioned
View Answer

Answer: b
Explanation: 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.

5. 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
View Answer

Answer: b
Explanation: 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).
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6. Design a peaking amplifier circuit to provide a gain of 10 at a peak frequency of 32khz given L=10mH having 30Ω resistance.
a)
Find the peaking amplifier circuit from the given diagram
b)
Find the peak frequency from the given diagram
c)
Find the internal resistance of the inductor from the given diagram
d) All of the mentioned
View Answer

Answer: a
Explanation: 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.

Sanfoundry Global Education & Learning Series – Linear Integrated Circuits.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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