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
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)
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
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.
4. The parallel resistance of tank circuit and for the circuit is given below.Find the gain of the amplifier?
d) None of the mentioned
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
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).
6. Design a peaking amplifier circuit to provide a gain of 10 at a peak frequency of 32khz given L=10mH having 30Ω resistance.
Explanation: Given L=10mH and the internal resistance of the inductor R=30Ω. Assume R1=100Ω. The gain times peak frequency= 10×32kHz=320kHz
=> 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 Circuit.