Answered: An inverting ampliﬁer is built with R2…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

An inverting ampliﬁer is built with R₂ =47kΩ and R₁ = 2.4kΩ using an op amp with an open-loop gain of 100 dB, an input resistance of 300 kΩ, and an output resistance of 200 Ω. What are the closed-loop gain, input resistance, and output resistance for this ampliﬁer? (b) Repeat if the open-loop gain is changed to 94 dB.

Expert Solution

Step 1

Given

a) Open loop-gain $A = 100 dB$

$\begin{array}{rcl} R_{1} & = & 2.4 kΩ \\ R_{2} & = & 47 kΩ \\ R_{i d} & = & 300 kΩ \\ R_{o d} & = & 200 kΩ \end{array}$

convert the gain from the decibel

$\begin{array}{rcl} A & = & 10^{\frac{100}{20}} \\ = & 10^{5} \end{array}$

The expression for $β$ is given by

$\begin{array}{rcl} β & = & \frac{R_{1}}{R_{1} + R_{2}} \\ = & \frac{2.4}{2.4 + 47} \\ = & \frac{1}{20.58} \end{array}$

Step 2

The expression for ideal amplifier gain is

$\begin{array}{rcl} A_{v}^{i d e a l} & = & - \frac{R_{2}}{R_{1}} \\ = & - \frac{47}{2.4} \\ = & - 19.6 \end{array}$

The expression for gain transmission is

$\begin{array}{rcl} T & = & A β \\ = & {(10)}^{5} \frac{1}{20.58} \\ = & 4859 \end{array}$

The closed loop voltage gain is given by

$\begin{array}{rcl} A_{v} & = & A_{v}^{i d e a l} (\frac{T}{T + 1}) \\ = & (- 19.6) (\frac{4959}{4859 + 1}) \\ A_{v} & = & - 19.6 \end{array}$

Step 3

The expression fro the input resistance of the amplifier is given by

$\begin{array}{rcl} R_{i n} & = & R_{1} + R_{i d} ∥ \frac{R_{2}}{1 + A} \\ = & 2.4 + (300 ∥ \frac{47}{10^{5}}) \\ = & 2.40 kΩ \end{array}$

The output resistance is given by

$\begin{array}{rcl} R_{o u t} & = & \frac{R_{o d}}{1 + T} \\ = & \frac{200}{1 + 4859} \\ = & 41.2 mΩ \end{array}$

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