Microelectronics: Circuit Analysis and Design
Microelectronics: Circuit Analysis and Design
4th Edition
ISBN: 9780073380643
Author: Donald A. Neamen
Publisher: McGraw-Hill Companies, The
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Textbook Question
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 pF ,then C 4 = 25 pF , R = 180 k Ω )

Expert Solution & Answer
Check Mark
To determine

The design parameters of a low-pass Butterworth filter.

Answer to Problem 15.1EP

The design parameters are:

  C3=50pF

  C4=25pF

  R=180kΩ

Explanation of Solution

Given Information:

Bandwidth of low-pass Butterworth filter is f3dB=25kHz .

Value of largest capacitor is 50 pF.

Calculation:

A general low-pass Butterworth circuit is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.1EP

Consider C3=50pF .

The relation between C3 and C4 is

  C3=2C4

So, the value of capacitor C4 is,

  C4=C32=50pF2=25pF

dB frequency or bandwidth of the circuit is

  f3dB=12πCR

Substituting the values,

  RC=12×3.14×25× 103=6.369×106=6.63kHz (1)

The value of capacitor C is

  C3=1.414CC=C31.414=50× 10 121.414=35.36×1012=35.36pF

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

  R=135.36× 10 12×6.369×106=180.118×103180kΩ

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Chapter 15 Solutions

Microelectronics: Circuit Analysis and Design

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