Chapter 15, Problem 15.1EP

Design a two−pole low−pass Butterworth filter with a bandwidth of2.5 kHz. The largest capacitor value to be used is 50 pF. (Ans. Set $C_3=50\text{pF}$ ,then $C_4=25\text{pF}$ , $R=180\text{k}\Omega$ )

To determine

The design parameters of a low-pass Butterworth filter.

Answer to Problem 15.1EP

The design parameters are:

  $C_3=50\text{pF}$

  $C_4=25\text{pF}$

  $R=180\text{k}\Omega$

Explanation of Solution

Given Information:

Bandwidth of low-pass Butterworth filter is $f_{3\text{dB}}=25\text{kHz}$ .

Value of largest capacitor is 50 pF.

Calculation:

A general low-pass Butterworth circuit is shown below.

  

Consider $C_3=50\text{pF}$ .

The relation between $C_3$ and $C_4$ is

  $C_3=2C_4$

So, the value of capacitor $C_4$ is,

  $\begin{array}{c}{C}_4=\frac{C_3}{2}\\ =\frac{50\text{pF}}{2}\\ =25\text{pF}\end{array}$

dB frequency or bandwidth of the circuit is

  $f_{3\text{dB}}=\frac{1}{2\pi CR}$

Substituting the values,

  $\begin{array}{c}RC=\frac{1}{2\times 3.14\times 25\times {10}^3}\\ =6.369\times {10}^{-6}\\ =6.63\text{kHz}\end{array}$ (1)

The value of capacitor C is

  $\begin{array}{c}{C}_3=1.414C\\ C=\frac{C_3}{1.414}\\ =\frac{50\times {10}^{-12}}{1.414}\\ =35.36\times {10}^{-12}\\ =35.36\text{pF}\end{array}$

So, the value of resistance R, from equation (1) is

  $\begin{array}{c}R=\frac{1}{35.36\times {10}^{-12}}\times 6.369\times {10}^{-6}\\ =180.118\times {10}^3\\ \approx 180\text{k}\Omega \end{array}$