Answered: v(t) R (a) vo(t) = v,(0) (b) Rc 2p(1)'a…

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 single-pole op amp has an open-loop gain of 100 dB and a unity-gain frequency of 3 MHz. Find an expression for the transfer function of the integrator shown if R = 10 kΩ and C = 0.05μF. What is the integrator’s phase margin?

v(t)
R
(a)
vo(t) = v,(0)
(b)
Rc
2p(1)'a
Figure 10.34 (a) The integrator circuit; (b) output voltage for a step-function input with vc (0) = 0.
1,

Expert Solution

Step 1

Given $f_{T} = 3 M H z$

Calculating angular frequency

$ω_{T} = 2 π f_{T} ω_{T} = 2 π (3 M H z) ω_{T} = 6 \times 10^{6} π$

Converting gain decibel to normal value

$A_{0} = 10^{\frac{A_{0} (d B)}{20}} A_{0} = 10^{\frac{100}{20}} A_{0} = 10^{5}$

Step 2

Calculating feedback factor

$β = \frac{R}{R + \frac{1}{s C}} β = \frac{s R C}{s C R + 1} β = \frac{s}{s + \frac{1}{C R}}$

Step 3

Comparing it with standard form

$β = \frac{s}{s + ω_{L}} ω_{L} = \frac{1}{C R} ω_{L} = \frac{1}{(0.05 μ F) (10 K Ω)} ω_{L} = 2000$

Calculating loop gain T

$T = A β A = \frac{ω_{T}}{s + ω_{B}} ω_{T} = A_{0} ω_{B}$

Step 4

Comparing it

$\begin{array}{rcl} A_{0} ω_{B} & = & 6 \times 10^{6} π \\ (10^{5}) ω_{B} & = & 6 \times 10^{6} π \\ ω_{B} & = & \frac{6 \times 10^{6} π}{(10^{5})} \\ ω_{B} & = & 60 π \end{array}$

substituting these values in A

$A = \frac{ω_{T}}{s + ω_{B}} A = \frac{6 \times 10^{6} π}{s + 60 π}$

Step 5

loop gain

$T = A β T = (\frac{6 \times 10^{6} π}{s + 60 π}) (\frac{s}{s + ω_{L}})$

Calculating closed loop gain

$A_{V} (s) = {A_{V}}^{i d e a l} (\frac{T}{1 + T}) {A_{V}}^{i d e a l} = \frac{- ω_{L}}{s} {A_{V}}^{i d e a l} = \frac{- 2000}{s}$

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