Microelectronics: Circuit Analysis and Design
Microelectronics: Circuit Analysis and Design
4th Edition
ISBN: 9780073380643
Author: Donald A. Neamen
Publisher: McGraw-Hill Companies, The
bartleby

Concept explainers

Question
Book Icon
Chapter 15, Problem 15.31P

a.

To determine

The expression for the frequency of the oscillation in the terms of C, R and Rv.

a.

Expert Solution
Check Mark

Answer to Problem 15.31P

The frequency of oscillation:

  f0=12πRC2+4RRRV

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

  Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.31P , additional homework tip  1

Redrawing the given circuit

  Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.31P , additional homework tip  2

Nodal analysis at the node v1 :

  v1v0RV+v1v2R+v11 sC=0v1(1 R V +1R+sC)v2R=v0RV...........(1)

Nodal analysis at the node v2 :

  v2v1RV+v2v3R+v21 sC=0v2+v2v3+v2sRC=v1v1=v2(2+sRC)v3..........(2)

Nodal analysis at the node v3 :

  v3v2RV+v30R+v31 sC=0v2+v2v3+v3(sRC)=0v3(2+sRC)=v2v3=v22+sRC........(5)

Now, solving the equation 2 and 3:

  v1=v2(2+sRC)v22+sRC=v2(2+sRC1 2+sRC)v1=v2( ( 2+sRC ) 2 1 2+sRC)v2=v1( 2+sRC ( 2+sRC ) 2 1).........(4)

Nodal analysis at the inverting terminal:

  0v0RF+0v3R=0v0RF=v3Rv0=RFRv3

From equation (3):

  v0=( R F R)(1 2+sRC)v2( R F R)(1 2+sRC)v2

From equation 4:

  v0=( R F R)(1 2+sRC)( 2+sRC ( 2+sRC ) 2 1)v1( R F R)(1 ( 2+sRC ) 2 1)v1

Substituting values of v0 and v2 in equation (1):

  { v 1( 1 R V + 1 R +sC) 1 R( 2+sRC ( 2+sRC ) 2 1 ) v 1}=(1 R V )( R F R)(1 ( 2+sRC ) 2 1)v11RV+1R+sC2+sRCR[ ( 2+sRC ) 21]=(1 R V )( R F R)(1 ( 2+sRC ) 2 1)

Now simplifying more:

  1[ R VR ( 2+sRC ) 2R R V]{R( 2+sRC)2RRV( 2+sRC)2RV+sRRVC( 2+sRC)2sRRVC2RVsRRVC}={( 1 R V )( R F R)( 1 ( 2+sRC ) 2 1)}

  {R(4+4sRC+ s 2 R 2 C 2)R+RV(4+4sRC+ s 2 R 2 C 2)3RV+sRVRC(4+4sRC+ s 2 R 2 C 2)2sRRVC}=RF4R+4sR2C+s2R2C2R+4RV+4sRRVC+s2R2RVC23RV+4sRRVC+4s2R2RVC2+s3R3RVC32sRRVC=RF{3R+4sR2C+s2R3C2+RV+5s2R2RVC2+s3R3RVC3+6sRRVC}=RF Evaluating the frequency of oscillation, considering s=jω :

  {3R+j4ωR2C+ω2R3C2+RV+5ω2R2RVC2+jω3R3RVC3+j6ωRRVC}=RF{3R+ω2R3C2+RV5ω2R2RVC2+j(4ω R 2C ω 3 R 3 R V C 3+6ωR R VC)}=RF

Now equating imaginary part to 0:

  4ωR2Cω3R3RVC3+6ωRRVC=0ωRC(4Rω2R2RVC2+6RV)=04Rω2R2RVC2+6RV=06RV+4R=ω2R2RVC2

Now simplifying:

  ω2=1R2C2( 2 R V +4 R V +4R R V )ω2=1R2C2(2+4R[ R V +R R R V ])ω2=1R2C2(2+ 4R R R V )

Squaring on the both sides:

  ω=1 R 2 C 2 ( 2+ 4R R R V )ω=1RC2+ 4R R R V

Since, the frequency is given as:

  2πf=1RC2+4RRRV

Hence, the required frequency of oscillation is

  f0=12πRC2+4RRRV

b.

To determine

The range in the frequency of the oscillation.

b.

Expert Solution
Check Mark

Answer to Problem 15.31P

The range of frequency of oscillation:

  19.45f022.66kHz

Explanation of Solution

Given:

The circuit is given for the phase shift oscillator:

  Microelectronics: Circuit Analysis and Design, Chapter 15, Problem 15.31P , additional homework tip  3

  R=25,C=0.001μFand15RV30

The expression for the frequency of the oscillation is given as:

  f0=12πRC2+4RRRV

Substitute the known values in the above expression:

  f0=12π( 25× 10 3 )( 0.001× 10 6 )2+4( ( 25× 10 3 25× 10 3 ×15× 10 3 25× 10 3 +15× 10 3 ))=12π( 25× 10 6 )2+4( 25× 10 3 9.375× 10 3 )=1157.08× 10 62+4( 2.667)

More simplifying:

  f0=1157.08× 10 612.667=3.599157.08× 10 6=22657.464=22.66kHz

If, RV=30 , the frequency is given as:

  f0=12πRC2+4RRRV

Substitute the known values:

  f0=12π( 25× 10 3 )( 0.001× 10 6 )2+4( ( 25× 10 3 25× 10 3 ×30× 10 3 25× 10 3 +30× 10 3 ))=12π( 25× 10 6 )2+4( 25× 10 3 13.636× 10 3 )=1157.08× 10 62+4( 1.833)

More simplifying:

  f0=1157.08× 10 69.333=3.055157.08× 10 6=0.01945×106=19.45kHz

Hence, the required range of frequency of oscillation:

  19.45f022.66kHz

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
i need the answer quickly
i need the answer quickly
i need the answer quickly

Chapter 15 Solutions

Microelectronics: Circuit Analysis and Design

Ch. 15 - Prob. 15.6TYUCh. 15 - Prob. 15.6EPCh. 15 - Redesign the street light control circuit shown in...Ch. 15 - A noninverting Schmitt trigger is shown m Figure...Ch. 15 - For the Schmitt trigger in Figure 15.30(a), the...Ch. 15 - Prob. 15.9TYUCh. 15 - Prob. 15.8EPCh. 15 - Prob. 15.9EPCh. 15 - Consider the 555 IC monostablemultivibrator. (a)...Ch. 15 - The 555 IC is connected as an...Ch. 15 - Prob. 15.10TYUCh. 15 - Prob. 15.11TYUCh. 15 - Prob. 15.12TYUCh. 15 - Prob. 15.12EPCh. 15 - Prob. 15.13EPCh. 15 - (a) Consider the bridge amplifier in Figure 15.46...Ch. 15 - Prob. 15.14EPCh. 15 - Prob. 15.15EPCh. 15 - Prob. 15.16EPCh. 15 - Prob. 1RQCh. 15 - Prob. 2RQCh. 15 - Consider a lowpass filter. What is the slope of...Ch. 15 - Prob. 4RQCh. 15 - Describe how a capacitor in conjunction with two...Ch. 15 - Sketch a onepole lowpass switchedcapacitor filter...Ch. 15 - Explain the two basic principles that must be...Ch. 15 - Prob. 8RQCh. 15 - Prob. 9RQCh. 15 - Prob. 10RQCh. 15 - Prob. 11RQCh. 15 - What is the primary advantage of a Schmitt trigger...Ch. 15 - Sketch the circuit and explain the operation of a...Ch. 15 - Prob. 14RQCh. 15 - Prob. 15RQCh. 15 - Prob. 16RQCh. 15 - Prob. 17RQCh. 15 - Prob. 18RQCh. 15 - Prob. D15.1PCh. 15 - Prob. 15.2PCh. 15 - The specification in a highpass Butterworth filter...Ch. 15 - (a) Design a twopole highpass Butterworth active...Ch. 15 - (a) Design a threepole lowpass Butterworth active...Ch. 15 - Prob. 15.6PCh. 15 - Prob. 15.7PCh. 15 - Prob. 15.8PCh. 15 - A lowpass filter is to be designed to pass...Ch. 15 - Prob. 15.10PCh. 15 - Prob. 15.11PCh. 15 - Prob. D15.12PCh. 15 - Prob. D15.13PCh. 15 - Prob. D15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - A simple bandpass filter can be designed by...Ch. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. D15.22PCh. 15 - Prob. 15.23PCh. 15 - Consider the phase shift oscillator in Figure...Ch. 15 - In the phaseshift oscillator in Figure 15.15, the...Ch. 15 - Consider the phase shift oscillator in Figure...Ch. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - A Wienbridge oscillator is shown in Figure P15.32....Ch. 15 - Prob. 15.33PCh. 15 - Prob. D15.34PCh. 15 - Prob. D15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. D15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - For the comparator in the circuit in Figure...Ch. 15 - Prob. 15.43PCh. 15 - Prob. 15.44PCh. 15 - Prob. 15.45PCh. 15 - Consider the Schmitt trigger in Figure P15.46....Ch. 15 - The saturated output voltages are VP for the...Ch. 15 - Consider the Schmitt trigger in Figure 15.30(a)....Ch. 15 - Prob. 15.50PCh. 15 - Prob. 15.52PCh. 15 - Prob. 15.53PCh. 15 - Prob. 15.54PCh. 15 - Prob. 15.55PCh. 15 - Prob. 15.56PCh. 15 - Prob. 15.57PCh. 15 - Prob. D15.58PCh. 15 - Prob. 15.59PCh. 15 - The saturated output voltages of the comparator in...Ch. 15 - (a) The monostablemultivibrator in Figure 15.37 is...Ch. 15 - A monostablemultivibrator is shown in Figure...Ch. 15 - Prob. D15.63PCh. 15 - Design a 555 monostablemultivibrator to provide a...Ch. 15 - Prob. 15.65PCh. 15 - Prob. 15.66PCh. 15 - Prob. 15.67PCh. 15 - Prob. 15.68PCh. 15 - An LM380 must deliver ac power to a 10 load. The...Ch. 15 - Prob. 15.70PCh. 15 - Prob. D15.71PCh. 15 - Prob. 15.72PCh. 15 - (a) Design the circuit shown in Figure P15.72 such...Ch. 15 - Prob. 15.74PCh. 15 - Prob. 15.75PCh. 15 - Prob. 15.76PCh. 15 - Prob. D15.77PCh. 15 - Prob. 15.78P
Knowledge Booster
Background pattern image
Electrical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:PEARSON
Text book image
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning
Text book image
Programmable Logic Controllers
Electrical Engineering
ISBN:9780073373843
Author:Frank D. Petruzella
Publisher:McGraw-Hill Education
Text book image
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:9780078028229
Author:Charles K Alexander, Matthew Sadiku
Publisher:McGraw-Hill Education
Text book image
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:9780134746968
Author:James W. Nilsson, Susan Riedel
Publisher:PEARSON
Text book image
Engineering Electromagnetics
Electrical Engineering
ISBN:9780078028151
Author:Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:Mcgraw-hill Education,