Answered: A noninverting ampliﬁer is being…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Question

A noninverting ampliﬁer is being designed to have a closed-loop gain of 36 dB. What op-amp gain is required to have the gain error less than 0.15 percent?

Expert Solution

Step 1

Given:

$A_{v}^{I d e a l} = 36 d B$

Step 2

$A_{v}^{I d e a l} = 36 d B$

To convert the value from decibel to gain is shown as:

$A_{v}^{I d e a l} = 10^{\frac{G a i n i n d B}{20}}$

Substitute the value of gain value in the following equation is shown as:

$A_{v}^{I d e a l} = 10^{\frac{36}{20}} A_{v}^{I d e a l} = 63.1$

Step 3

The fractional gain error can be written as:

$F G E \leq \frac{1}{1 + T}$

Substitute the value of FGE=0.15% in the following equation is shown as:

$0.15 % \leq \frac{1}{1 + T} \frac{0.15}{100} \leq \frac{1}{1 + T}$

Here,

$T \geq 666$

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