Question

Design a peaking amplifier circuit to provide a gain of 10 at a peak frequency of 32khz given L=10mH having 30Ω resistance.

I have been asked this question at a job interview.

I need to ask this question from Peaking Amplifier in chapter Operational Amplifier Applications of Linear Integrated Circuits

Answer 1

To design a peaking amplifier circuit, we need to use the general configuration of a series resonant circuit that incorporates both the inductor (L) and resistor (R). The peaking amplifier uses these components to set the peak frequency and gain characteristics.

Given the information:

• Peak frequency (fpf_pfp​) = 32 kHz
• Inductance (L) = 10 mH = 10×10−310 \times 10^{-3}10×10−3 H
• Resistance (R) = 30Ω
• Gain (A) = 10

Step 1: Determine the Peak Frequency

The peak frequency fpf_pfp​ for the resonant circuit is related to the inductance LLL and resistance RRR by the following equation:

fp=12πLCf_p = \frac{1}{2\pi \sqrt{L C}}fp​=2πLC​1​

Where CCC is the capacitance. We can solve for CCC as follows:

C=1(2πfp)2LC = \frac{1}{(2 \pi f_p)^2 L}C=(2πfp​)2L1​

Substituting the known values:

fp=32×103 Hz,L=10×10−3 Hf_p = 32 \times 10^3 \, \text{Hz}, \quad L = 10 \times 10^{-3} \, \text{H}fp​=32×103Hz,L=10×10−3H C=1(2π×32×103)2×10×10−3C = \frac{1}{(2 \pi \times 32 \times 10^3)^2 \times 10 \times 10^{-3}}C=(2π×32×103)2×10×10−31​

Calculating this gives:

C≈1.57 nFC \approx 1.57 \, \text{nF}C≈1.57nF

Step 2: Calculate the Quality Factor (Q)

The quality factor QQQ of the peaking amplifier determines the sharpness of the peak at the resonant frequency and is related to the resistance RRR, inductance LLL, and capacitance CCC. The quality factor for a series resonant circuit is given by:

Q=RL/CQ = \frac{R}{\sqrt{L / C}}Q=L/C​R​

Substitute the known values:

Q=30(10×10−3)/(1.57×10−9)≈18.9Q = \frac{30}{\sqrt{(10 \times 10^{-3}) / (1.57 \times 10^{-9})}} \approx 18.9Q=(10×10−3)/(1.57×10−9)​30​≈18.9

Step 3: Verify Gain

For a peaking amplifier, the gain at resonance is given by:

A=11−(f/fp)2A = \frac{1}{\sqrt{1 - (f / f_p)^2}}A=1−(f/fp​)2​1​

At the resonant frequency (f=fpf = f_pf=fp​), the gain AAA becomes:

A=11−1=infinityA = \frac{1}{\sqrt{1 - 1}} = \text{infinity}A=1−1​1​=infinity

But, in practice, the gain is limited by the component values. For a gain of 10, you will have to adjust the circuit design, potentially using feedback resistors to control the gain. This result suggests that the values of LLL, CCC, and RRR ensure the amplifier reaches the peak frequency with a high gain, but practical considerations will modify the feedback to achieve the desired gain.

Step 4: Circuit Design

Using an operational amplifier (op-amp) in the standard RLC resonant circuit configuration, the values of LLL, RRR, and CCC can be incorporated into the design to achieve a peaking amplifier that delivers the desired gain of 10 at the peak frequency of 32 kHz.

• Inductor (L): 10 mH
• Capacitor (C): 1.57 nF
• Resistor (R): 30Ω

The circuit should be designed with these components, ensuring proper feedback control to achieve a gain of 10.