fundamentals of industrial electronics

The Industrial Electronics HandbookS E c o n d E d I T I o n

Fundamentals oF IndustrIal electronIcs

2011 by Taylor and Francis Group, LLC



Power electronIcs and motor drIves

control and mechatronIcs

IndustrIal communIcatIon systems

IntellIgent systems


The Electrical Engineering Handbook Series

Series EditorRichard C. DorfUniversity of California, Davis

Titles Included in the Series

The Avionics Handbook, Second Edition, Cary R. SpitzerThe Biomedical Engineering Handbook, Third Edition, Joseph D. BronzinoThe Circuits and Filters Handbook, Third Edition, Wai-Kai ChenThe Communications Handbook, Second Edition, Jerry GibsonThe Computer Engineering Handbook, Vojin G. OklobdzijaThe Control Handbook, Second Edition, William S. Levine CRC Handbook of Engineering Tables, Richard C. DorfDigital Avionics Handbook, Second Edition, Cary R. SpitzerThe Digital Signal Processing Handbook, Vijay K. Madisetti and Douglas WilliamsThe Electric Power Engineering Handbook, Second Edition, Leonard L. GrigsbyThe Electrical Engineering Handbook, Third Edition, Richard C. DorfThe Electronics Handbook, Second Edition, Jerry C. WhitakerThe Engineering Handbook, Third Edition, Richard C. DorfThe Handbook of Ad Hoc Wireless Networks, Mohammad IlyasThe Handbook of Formulas and Tables for Signal Processing, Alexander D. PoularikasHandbook of Nanoscience, Engineering, and Technology, Second Edition, William A. Goddard, III, Donald W. Brenner, Sergey E. Lyshevski, and Gerald J. IafrateThe Handbook of Optical Communication Networks, Mohammad Ilyas and Hussein T. MouftahThe Industrial Electronics Handbook, Second Edition, Bogdan M. Wilamowski and J. David IrwinThe Measurement, Instrumentation, and Sensors Handbook, John G. WebsterThe Mechanical Systems Design Handbook, Osita D.I. Nwokah and Yidirim HurmuzluThe Mechatronics Handbook, Second Edition, Robert H. BishopThe Mobile Communications Handbook, Second Edition, Jerry D. GibsonThe Ocean Engineering Handbook, Ferial El-HawaryThe RF and Microwave Handbook, Second Edition, Mike GolioThe Technology Management Handbook, Richard C. DorfTransforms and Applications Handbook, Third Edition, Alexander D. PoularikasThe VLSI Handbook, Second Edition, Wai-Kai Chen




Edited by

Bogdan M. WilamowskiJ. david Irwin


MATLAB is a trademark of The MathWorks, Inc. and is used with permission. The MathWorks does not warrant the accuracy of the text or exercises in this book. This books use or discussion of MATLAB software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB software.

CRC PressTaylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742

2011 by Taylor and Francis Group, LLCCRC Press is an imprint of Taylor & Francis Group, an Informa business

No claim to original U.S. Government works

Printed in the United States of America on acid-free paper10 9 8 7 6 5 4 3 2 1

International Standard Book Number: 978-1-4398-0279-3 (Hardback)

This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the valid-ity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint.

Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or uti-lized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopy-ing, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers.

For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged.

Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe.

Library of Congress CataloginginPublication Data

Fundamentals of industrial electronics / editors, Bogdan M. Wilamowski and J. David Irwin.p. cm.

A CRC title.Includes bibliographical references and index.ISBN 978-1-4398-0279-3 (alk. paper)1. Industrial electronics. I. Wilamowski, Bogdan M. II. Irwin, J. David. III. Title.

TK7881.F86 2010621.381--dc22 2010019980

Visit the Taylor & Francis Web site athttp://www.taylorandfrancis.comand the CRC Press Web site athttp://www.crcpress.com


vii

Contents

Preface....................................................................................................................... xiAcknowledgments................................................................................................... xiiiEditorial.Board..........................................................................................................xvEditors..................................................................................................................... xviiContributors............................................................................................................ xxi

Part I Circuits and Signals

. 1. DC.and.Transient.Circuit.Analysis.................................................................1-1Carlotta A. Berry and Deborah J. Walter

. 2. AC.Circuit.Analysis........................................................................................ 2-1Carlotta A. Berry and Deborah J. Walter

. 3. ComputationalMethods.in.Node.and.Loop.Analyses................................... 3-1Stephen M. Haddock and J. David Irwin

. 4. Transistor.Operation.and.Modeling............................................................... 4-1Tina Hudson

. 5. Application.of.Operational.Amplifiers.......................................................... 5-1Carlotta A. Berry and Deborah J. Walter

. 6. Frequency.Response.and.Bode.Diagrams....................................................... 6-1Thomas F. Schubert, Jr. and Ernest M. Kim

. 7. Laplace.Transforms..........................................................................................7-1Dalton S. Nelson

Part II Devices

. 8. Semiconductor.Diode...................................................................................... 8-1Bogdan M. Wilamowski

. 9. Bipolar.Junction.Transistor............................................................................ 9-1Bogdan M. Wilamowski and Guofu Niu


viii Contents

.10. Field.Effect.Transistors..................................................................................10-1Bogdan M. Wilamowski and J. David Irwin

.11. Noise.in.Semiconductor.Devices................................................................... 11-1Alicja Konczakowska and Bogdan M. Wilamowski

.12. Physical.Phenomena.Used.in.Sensors............................................................12-1Tiantian Xie and Bogdan M. Wilamowski

.13. MEMS.Devices...............................................................................................13-1Jos M. Quero, Antonio Luque, Luis Castaer, Angel Rodrguez, AdrianIonescu,Montserrat Fernndez-Bolaos, Lorenzo Faraone, andJohnM. Dell

.14. MEMS.Technologies.......................................................................................14-1Antonio Luque, Jos M. Quero, and Carles Can

.15. Applications.of.MEMS...................................................................................15-1Antonio Luque, Jos M. Quero, Robert Lempkowski, and Francisco Ibez

.16. Transistors.in.Switching.Circuits..................................................................16-1Tina Hudson

.17. Transistors.in.Amplifier.Circuits.................................................................. 17-1Tina Hudson

.18. A.Simplistic.Approach.to.the.Analysis.of.Transistor.Amplifiers.................18-1Bogdan M. Wilamowski and J. David Irwin

.19. Analog.and.Digital.VLSI.Design...................................................................19-1Vishal Saxena and R. Jacob Baker

Part III Digital Circuits

.20. Digital.DesignCombinational.Logic..........................................................20-1Buren Earl Wells and Sin Ming Loo

.21. Digital.DesignSequential.Logic.................................................................. 21-1Sin Ming Loo and Arlen Planting

.22. Soft-Core.Processors......................................................................................22-1Arlen Planting and Sin Ming Loo

.23. Computer.Architecture..................................................................................23-1Victor P. Nelson

.24. FPGAs.and.Reconfigurable.Systems..............................................................24-1Juan J. Rodriguez-Andina and Eduardo de la Torre


Contents ix

Part IV Digital and analog Signal Processing

.25. Signal.Processing............................................................................................25-1James A. Heinen and Russell J. Niederjohn

.26. Analog.Filter.Synthesis..................................................................................26-1Nam Pham and Bogdan M. Wilamowski

.27. Active.Filter.Implementation......................................................................... 27-1Nam Pham, Bogdan M. Wilamowski, and John W. Steadman

.28. Designing.Passive.Filters.with.Lossy.Elements.............................................28-1Marcin Jagiela and Bogdan M. Wilamowski

Part V Electromagnetics

.29. Electromagnetic.Fields.I................................................................................29-1Sadasiva M. Rao, Tyler N. Killian, and Michael E. Baginski

.30. Propagating.Electromagnetic.Fields..............................................................30-1Michael E. Baginski, Sadasiva M. Rao, and Tyler N. Killian

.31. Transmission.Line.Time-Domain.Analysis.and.Signal.Integrity................. 31-1Edward Wheeler, Jianjian Song, and David R. Voltmer

Index.................................................................................................................. Index-1


xi

Preface

The.field.of.industrial.electronics.covers.a.plethora.of.problems.that.must.be.solved.in.industrial.practice..Electronic.systems.control.many.processes.that.begin.with.the.control.of.relatively.simple.devices.like.electric.motors,.through.more.complicated.devices.such.as.robots,.to.the.control.of.entire.fabrication.processes..An.industrial.electronics.engineer.deals.with.many.physical.phenomena.as.well.as.the.sensors.that.are.used.to.measure.them..Thus,.the.knowledge.required.by.this.type.of.engineer.is.not.only.tra-ditional.electronics.but.also.specialized.electronics,.for.example,.that.required.for.high-power.applica-tions..The.importance.of.electronic.circuits.extends.well.beyond.their.use.as.a.final.product.in.that.they.are.also.important.building.blocks.in.large.systems,.and.thus.the.industrial.electronics.engineer.must.also.possess.knowledge.of.the.areas.of.control.and.mechatronics..Since.most.fabrication.processes.are.relatively.complex,.there.is.an.inherent.requirement.for.the.use.of.communication.systems.that.not.only.link.the.various.elements.of.the.industrial.process.but.are.also.tailor-made.for.the.specific.industrial.environment..Finally,.the.efficient.control.and.supervision.of.factories.require.the.application.of.intelli-gent.systems.in.a.hierarchical.structure.to.address.the.needs.of.all.components.employed.in.the.produc-tion.process..This.is.accomplished.through.the.use.of.intelligent.systems.such.as.neural.networks,.fuzzy.systems,.and.evolutionary.methods..The.Industrial.Electronics.Handbook.addresses.all.these.issues.and.does.so.in.five.books.outlined.as.follows:

. 1.. Fundamentals of Industrial Electronics

. 2.. Power Electronics and Motor Drives

. 3.. Control and Mechatronics

. 4.. Industrial Communication Systems

. 5.. Intelligent Systems

The.editors.have.gone.to.great.lengths.to.ensure.that.this.handbook.is.as.current.and.up.to.date.as.pos-sible..Thus,.this.book.closely.follows.the.current.research.and.trends.in.applications.that.can.be.found.in.IEEE Transactions on Industrial Electronics..This.journal.is.not.only.one.of.the.largest.engineering.publications.of.its.type.in.the.world,.but.also.one.of.the.most.respected..In.all.technical.categories.in.which.this.journal.is.evaluated,.its.worldwide.ranking.is.either.number.1.or.number.2.depending.on.category..As.a.result,.we.believe.that.this.handbook,.which.is.written.by.the.worlds.leading.researchers.in.the.field,.presents.the.global.trends.in.the.ubiquitous.area.commonly.known.as.industrial.electronics.

Fundamentals of Industrial Electronics. deals. with. the. fundamental. areas. that. form. the. basis. for.the.field.of.industrial.electronics..Because.of.the.breadth.of.this.field,.the.knowledge.required.spans.a.wide.spectrum.of.technology,.which.includes.analog.and.digital.circuits,.electronics,.electromagnetic.machines,.and.signal.processing..The.knowledge.gained.here.is.then.applied.in.Power Electronics and Motor Drives,.Control and Mechatronics,.Industrial Communication Systems,.and.Intelligent Systems,.and.in.total.form.the.Industrial.Electronics.Handbook.


xii Preface

For.MATLAB.and.Simulink.product.information,.please.contact

The.MathWorks,.Inc.3.Apple.Hill.DriveNatick,.MA,.01760-2098.USATel:.508-647-7000Fax:.508-647-7001E-mail:[email protected]:.www.mathworks.com


xiii

Acknowledgments

The.editors.wish.to.express.their.heartfelt.thanks.to.their.wives.Barbara.Wilamowski.and.Edie.Irwin.for.their.help.and.support.during.the.execution.of.this.project.


xv

Editorial Board

Jake BakerBoise.State.UniversityBoise,.Idaho

Alicja KonczakowskaGdansk.University.of.TechnologyGdansk,.Poland

Victor P. NelsonAuburn.UniversityAuburn,.Alabama

Guofu NiuAuburn.UniversityAuburn,.Alabama

John W. SteadmanUniversity.of.South.AlabamaMobile,.Alabama


xvii

Editors

Bogdan M. Wilamowski.received.his.MS.in.computer.engineering.in.1966,.his.PhD.in.neural.computing.in.1970,.and.Dr..habil..in.integrated.circuit.design.in.1977..He.received.the.title.of.full.professor.from.the.president.of.Poland. in.1987..He.was. the.director.of. the.Institute.of.Electronics. (19791981). and. the. chair. of. the. solid. state. electronics.department. (19871989). at. the. Technical. University. of. Gdansk,.Poland..He.was.a.professor.at.the.University.of.Wyoming,.Laramie,.from. 1989. to. 2000.. From. 2000. to. 2003,. he. served. as. an. associate.director. at. the. Microelectronics. Research. and. Telecommunication.Institute,.University.of.Idaho,.Moscow,.and.as.a.professor.in.the.elec-trical.and.computer.engineering.department.and.in.the.computer.sci-ence.department.at.the.same.university..Currently,.he.is.the.director.

of.ANMSTCAlabama.Nano/Micro.Science.and.Technology.Center,.Auburn,.and.an.alumna.professor.in.the.electrical.and.computer.engineering.department.at.Auburn.University,.Alabama..Dr.Wilamowski.was.with.the.Communication.Institute.at.Tohoku.University,.Japan.(19681970),.and.spent.one.year.at.the.Semiconductor.Research.Institute,.Sendai,.Japan,.as.a.JSPS.fellow.(19751976)..He.was.also.a.visiting.scholar.at.Auburn.University.(19811982.and.19951996).and.a.visiting.professor.at.the.University.of.Arizona,.Tucson.(19821984)..He.is.the.author.of.4.textbooks,.more.than.300.refereed.publications,.and.has.27.patents..He.was.the.principal.professor.for.about.130.graduate.students..His.main.areas.of.interest.include.semiconductor.devices.and.sensors,.mixed.signal.and.analog.signal.processing,.and.computa-tional.intelligence.

Dr..Wilamowski.was.the.vice.president.of.the.IEEE.Computational.Intelligence.Society.(20002004).and.the.president.of.the.IEEE.Industrial.Electronics.Society.(20042005)..He.served.as.an.associate.edi-tor.of.IEEE Transactions on Neural Networks,.IEEE Transactions on Education,.IEEE Transactions on Industrial Electronics,.the.Journal of Intelligent and Fuzzy Systems,.the.Journal of Computing,.and.the.International Journal of Circuit Systems and IES Newsletter..He.is.currently.serving.as.the.editor.in.chief.of.IEEE Transactions on Industrial Electronics.

Professor.Wilamowski. is.an.IEEE.fellow.and.an.honorary.member.of. the.Hungarian.Academy.of.Science..In.2008,.he.was.awarded.the.Commander.Cross.of.the.Order.of.Merit.of.the.Republic.of.Poland.for.outstanding.service. in. the.proliferation.of. international. scientific.collaborations.and. for.achieve-ments.in.the.areas.of.microelectronics.and.computer.science.by.the.president.of.Poland.


xviii Editors

J. David Irwin.received.his.BEE.from.Auburn.University,.Alabama,.in. 1961,. and. his. MS. and. PhD. from. the. University. of. Tennessee,.Knoxville,.in.1962.and.1967,.respectively.

In.1967,.he.joined.Bell.Telephone.Laboratories,.Inc.,.Holmdel,.New.Jersey,.as.a.member.of.the.technical.staff.and.was.made.a.supervisor.in.1968..He. then. joined.Auburn.University. in.1969.as.an.assistant.professor.of.electrical.engineering..He.was.made.an.associate.profes-sor.in.1972,.associate.professor.and.head.of.department.in.1973,.and.professor.and.head.in.1976..He.served.as.head.of.the.Department.of.Electrical.and.Computer.Engineering.from.1973.to.2009..In1993,.he.was.named.Earle.C..Williams.Eminent.Scholar.and.Head..From.

1982.to.1984,.he.was.also.head.of.the.Department.of.Computer.Science.and.Engineering..He.is.currently.theEarle.C..Williams.Eminent.Scholar.in.Electrical.and.Computer.Engineering.at.Auburn.

Dr.. Irwin. has. served. the. Institute. of. Electrical. and. Electronic. Engineers,. Inc.. (IEEE). Computer.Society.as.a.member.of.the.Education.Committee.and.as.education.editor.of.Computer..He.has.served.as. chairman. of. the. Southeastern. Association. of. Electrical. Engineering. Department. Heads. and. the.National. Association. of. Electrical. Engineering. Department. Heads. and. is. past. president. of. both. the.IEEE.Industrial.Electronics.Society.and.the.IEEE.Education.Society..He.is.a.life.member.of.the.IEEE.Industrial.Electronics.Society.AdCom.and.has.served.as.a.member.of.the.Oceanic.Engineering.Society.AdCom..He.served.for.two.years.as.editor.of.IEEE Transactions on Industrial Electronics..He.has.served.on. the. Executive. Committee. of. the. Southeastern. Center. for. Electrical. Engineering. Education,. Inc.,.and.was.president.of.the.organization.in.19831984..He.has.served.as.an.IEEE.Adhoc.Visitor.for.ABET.Accreditation. teams..He.has.also.served.as.a.member.of.the.IEEE.Educational.Activities.Board,.and.was.the.accreditation.coordinator.for.IEEE.in.1989..He.has.served.as.a.member.of.numerous.IEEE.com-mittees,.including.the.Lamme.Medal.Award.Committee,.the.Fellow.Committee,.the.Nominations.and.Appointments.Committee,.and.the.Admission.and.Advancement.Committee..He.has.served.as.a.mem-ber.of.the.board.of.directors.of.IEEE.Press..He.has.also.served.as.a.member.of.the.Secretary.of.the.Armys.Advisory.Panel.for.ROTC.Affairs,.as.a.nominations.chairman.for.the.National.Electrical.Engineering.Department.Heads.Association,.and.as.a.member.of.the.IEEE.Education.Societys.McGraw-Hill/Jacob.Millman. Award. Committee.. He. has. also. served. as. chair. of. the. IEEE. Undergraduate. and. Graduate.Teaching.Award.Committee..He.is.a.member.of.the.board.of.governors.and.past.president.of.Eta.Kappa.Nu,.the.ECE.Honor.Society..He.has.been.and.continues.to.be.involved.in.the.management.of.several.international.conferences.sponsored.by.the.IEEE.Industrial.Electronics.Society,.and.served.as.general.cochair.for.IECON05.

Dr.. Irwin. is. the. author. and. coauthor. of. numerous. publications,. papers,. patent. applications,. and.presentations,. including. Basic Engineering Circuit Analysis,. 9th. edition,. published. by. John. Wiley. &.Sons,.which.is.one.among.his.16.textbooks..His.textbooks,.which.span.a.wide.spectrum.of.engineering.subjects,.have.been.published.by.Macmillan.Publishing.Company,.Prentice.Hall.Book.Company,.John.Wiley.&.Sons.Book.Company,.and.IEEE.Press..He.is.also.the.editor.in.chief.of.a.large.handbook.pub-lished.by.CRC.Press,.and.is.the.series.editor.for.Industrial.Electronics.Handbook.for.CRC.Press.

Dr..Irwin.is.a.fellow.of.the.American.Association.for.the.Advancement.of.Science,.the.American.Society. for. Engineering. Education,. and. the. Institute. of. Electrical. and. Electronic. Engineers.. He.received. an. IEEE. Centennial. Medal. in. 1984,. and. was. awarded. the. Bliss. Medal. by. the. Society. of.American.Military.Engineers.in.1985..He.received.the.IEEE.Industrial.Electronics.Societys.Anthony.J..Hornfeck.Outstanding.Service.Award.in.1986,.and.was.named.IEEE.Region.III.(U.S..Southeastern.Region). Outstanding. Engineering. Educator. in. 1989.. In. 1991,. he. received. a. Meritorious. Service.Citation. from. the. IEEE. Educational. Activities. Board,. the. 1991. Eugene. Mittelmann. Achievement.Award.from.the.IEEE.Industrial.Electronics.Society,.and.the.1991.Achievement.Award.from.the.IEEE.Education.Society..In.1992,.he.was.named.a.Distinguished.Auburn.Engineer..In.1993,.he.received.the.IEEE.Education.Societys.McGraw-Hill/Jacob.Millman.Award,.and.in.1998.he.was.the.recipient.of.the.


Editors xix

IEEE.Undergraduate.Teaching.Award.. In.2000,.he.received.an.IEEE.Third.Millennium.Medal.and.the.IEEE.Richard.M..Emberson.Award..In.2001,.he.received.the.American.Society.for.Engineering.Educations. (ASEE). ECE. Distinguished.Educator. Award.. Dr.. Irwin.was.made. an.honorary.profes-sor,.Institute.for.Semiconductors,.Chinese.Academy.of.Science,.Beijing,.China,.in.2004..In.2005,.he.received.the.IEEE.Education.Societys.Meritorious.Service.Award,.and.in.2006,.he.received.the.IEEE.Educational.Activities.Board.Vice.Presidents.Recognition.Award..He.received.the.Diplome.of.Honor.from.the.University.of.Patras,.Greece,.in.2007,.and.in.2008.he.was.awarded.the.IEEE.IES.Technical.Committee.on.Factory.Automations.Lifetime.Achievement.Award..In.2010,.he.was.awarded.the.elec-trical.and.computer.engineering.department.heads.Robert.M..Janowiak.Outstanding.Leadership.and.Service.Award..In.addition,.he.is.a.member.of.the.following.honor.societies:.Sigma.Xi,.Phi.Kappa.Phi,.Tau.Beta.Pi,.Eta.Kappa.Nu,.Pi.Mu.Epsilon,.and.Omicron.Delta.Kappa.


xxi

Contributors

Michael E. BaginskiDepartment.of.Electrical.and.Computer.

EngineeringAuburn.UniversityAuburn,.Alabama

R. Jacob BakerDepartment.of.Electrical.and.Computer.

EngineeringBoise.State.UniversityBoise,.Idaho

Carlotta A. BerryDepartment.of.Electrical.and.Computer.

EngineeringRose-Hulman.Institute.of.TechnologyTerre.Haute,.Indiana

Carles CanNational.Microelectronics.CenterBarcelona,.Spain

Luis CastaerDepartment.of.Electronic.EngineeringPolytechnic.University.of.CataloniaCatalonia,.Spain

John M. DellMicroelectronics.Research.GroupUniversity.of.Western.AustraliaPerth,.Western.Australia,.Australia

Lorenzo FaraoneMicroelectronics.Research.GroupUniversity.of.Western.AustraliaPerth,.Western.Australia,.Australia

Montserrat Fernndez-BolaosEcole.Polytechnique.Fdrale.de.LausanneLausanne,.Switzerland

Stephen M. HaddockDepartment.of.Electrical.and.Computer.


James A. HeinenDepartment.of.Electrical.and.Computer.

EngineeringMarquette.UniversityMilwaukee,.Wisconsin

Tina HudsonDepartment.of.Electrical.and.Computer.


Francisco IbezEuropean.CommissionBrussels,.Belgium


xxii Contributors

Adrian IonescuEcole.Polytechnique.Fdrale.de.LausanneLausanne,.Switzerland

J. David IrwinDepartment.of.Electrical.and.Computer.


Marcin JagielaFaculty.of.Applied.InformaticsUniversity.of.Information.Technology.

andManagement.in.RzeszwRzeszw,.Poland

Tyler N. KillianDepartment.of.Electrical.and.Computer.


Ernest M. KimDepartment.of.EngineeringUniversity.of.San.DiegoSan.Diego,.California

Alicja KonczakowskaFaculty.of.Electronics,.Telecommunications.

andInformaticsGdansk.University.of.TechnologyGdansk,.Poland

Robert LempkowskiMotorola.Applied.Research.and.Technology.CenterSchaumburg,.Illinois

Sin Ming LooDepartment.of.Electrical.and.Computer.


Antonio LuqueDepartment.of.Electronic.EngineeringUniversity.of.SevilleSevilla,.Spain

Dalton S. NelsonDepartment.of.Electrical.and.Computer.

EngineeringThe.University.of.Alabama.at.BirminghamBirmingham,.Alabama

Victor P. NelsonDepartment.of.Electrical.and.Computer.


Russell J. Niederjohn (deceased)Department.of.Electrical.and.Computer.

EngineeringMarquette.UniversityMilwaukee,.Wisconsin

Guofu NiuDepartment.of.Electrical.and.Computer.


Nam PhamDepartment.of.Electrical.and.Computer.


Arlen PlantingDepartment.of.Electrical.and.Computer.


Jos M. QueroDepartment.of.Electronic.EngineeringUniversity.of.SevilleSevilla,.Spain

Sadasiva M. RaoDepartment.of.Electrical.and.Computer.



Contributors xxiii

Angel RodrguezDepartment.of.Electronic.EngineeringPolytechnic.University.of.CataloniaCatalonia,.Spain

Juan J. Rodriguez-AndinaDepartment.of.Electronic.TechnologyUniversity.of.VigoVigo,.Spain

Vishal SaxenaDepartment.of.Electrical.and.Computer.


Thomas F. Schubert, Jr.Department.of.EngineeringUniversity.of.San.DiegoSan.Diego,.California

Jianjian SongDepartment.of.Electrical.and.Computer.


John W. SteadmanCollege.of.EngineeringUniversity.of.South.AlabamaMobile,.Alabama

Eduardo de la TorreCenter.of.Industrial.ElectronicsPolytechnic.University.of.MadridMadrid,.Spain

David R. VoltmerDepartment.of.Electrical.and.Computer.


Deborah J. WalterDepartment.of.Electrical.and.Computer.


Buren Earl WellsDepartment.of.Electrical.and.Computer.

EngineeringThe.University.of.Alabama.in.HuntsvilleHuntsville,.Alabama

Edward WheelerDepartment.of.Electrical.and.Computer.


Bogdan M. WilamowskiDepartment.of.Electrical.and.Computer.


Tiantian XieDepartment.of.Electrical.and.Computer.



1-1

1.1 Introduction

Direct.current.(DC).circuit.analysis.is.the.study.of.circuits.with.a.constant.voltage.or.current.source..The.most.popular.example.of.a.DC.circuit.is.a.battery.and.a.light.bulb..A.DC.circuit.contains.an.active circuit.element.(i.e.,.battery).capable.of.generating.electric.energy..These.electric.sources.convert.nonelectric.energy. to.electric.energy.(i.e.,.a.voltage.or.current)..Independent.electric.sources.produce.a.constant.voltage.or.current.in.the.circuit.regardless.of.the.current.through.or.voltage.across.the.source..The.sym-bols.for.an.ideal.DC.voltage.and.current.source.are.shown.in.Figure.1.1..It.should.be.noted.that.an.ideal.voltage.and.current.source.can.deliver.or.absorb.power.to.an.electric.circuit..An.example.of.an.ideal.voltage.source.absorbing.power.is.a.rechargeable.battery.

Dependent.sources.establish.a.voltage.or.current.in.a.circuit.that.is.based.upon.the.value.of.a.voltage.or.current.elsewhere.in.the.circuit..One.use.of.dependent.sources.is.to.model.operational.amplifiers.and.transistors..Table.1.1.presents.a.summary.of.the.four.types.of.dependent.sources.

A.passive.circuit.element.models.devices.that.cannot.generate.electric.energy.such.as.a. light.bulb..The.most.common.passive.circuit.elements.are.inductors,.capacitors,.and.resistors..The.voltagecurrent.relationships.for.these.devices.will.be.described.in.the.subsequent.section.

1.1.1 Ohms Law

Ohms. law.states. that. the.voltage.(V).difference.across.a.resistor. is. linearly.related. to. the.current. (I).through.the.resistor.(see.Equation.1.1):

. V IR= . (1.1)

1DC and Transient

Circuit Analysis

1.1. Introduction....................................................................................... 1-1Ohms.Law. . Inductors.and.Capacitors. . Kirchhoffs.Current.Law. . Kirchhoffs.Voltage.Law. . Series.and.Parallel.Relationships. . Voltage.and.Current.Divider.Rule. . DeltaWye.(Y).Transformations

1.2. Systematic.Circuit.Analysis.Techniques........................................ 1-7Node-Voltage.Method. . Mesh-Current.Method. . Superposition

1.3. Circuit.Modeling.Techniques.........................................................1-16Source.Transformations. . Thevenin.and.Norton.Equivalent.Circuits. . Maximum.Power.Transfer

1.4. Transient.Analysis........................................................................... 1-19First-Order.Circuits. . Second-Order.Circuits

1.5. Conclusions...................................................................................... 1-36Bibliography................................................................................................. 1-36

Carlotta A. BerryRose-HulmanInstituteofTechnology

Deborah J. WalterRose-HulmanInstituteofTechnology


1-2 FundamentalsofIndustrialElectronics

where. R. is. the. resistance. of. the. resistor. in. Ohms. ().. The. conductance.(G) of. a. resistor. is. the. inverse. of. the. resistance. (1/R). and. is. in. units. of.Siemens.(S)..Resistors.always.absorb.power,. so. the. standard.way. to.rep-resent.a.resistive.element. is. to.draw.the.resistor. in. the.passive.sign.con-vention. (see. Figure. 1.2).. If. the. resistor. is. not. drawn. in. the. passive. sign.convention,.then.V.=.IR.

1.1.2 Inductors and Capacitors

As. previously. stated,. the. other. two. passive. circuit. elements. are. inductors. and. capacitors.. Both. the.inductor.and.the.capacitor.have.the.ability.to.store.energy..Inductors.store.energy.in.the.form.of.current.and.capacitors.store.energy.in.the.form.of.voltage..The.energy.stored.in.these.elements.is.released.back.into.the.circuit.when.a.DC.source.is.removed..Therefore,.these.two.elements.exhibit.behavior.that.is.a.function.of.time..The.analysis.of.these.types.of.circuits.is.transient.analysis.that.will.be.addressed.later.in.this.chapter..Table.1.2.describes.the.currentvoltage.relationship.for.inductors.and.capacitors.where.the.inductance.(L).is.in.henrys.(H),.capacitance.(C).is.in.farads.(F),.and.time.(t).is.in.seconds.(s).

1.1.3 Kirchhoffs Current Law

The.law.of.conservation.of.energy.states.that.energy.can.neither.be.created.nor.destroyed,.only.trans-ferred..Another.way.to.state.this.law.is.for.any.electric.circuit,.the.total.power.delivered.by.the.elements.must.be.equal.to.the.total.power.absorbed.by.the.elements..Kirchhoffs.current.law.(KCL).is.based.upon.the.law.of.conservation.of.energy..A.node.in.a.circuit.is.any.point.at.which.two.or.more.circuit.elements.are.connected..KCL.states.that.the.sum.of.currents.entering.a.node.is.zero.(i.e.,.current.in.=.current.out)..KCL.can.be.applied.to.any.node.in.a.closed.circuit..The.circuit.in.Figure.1.3.has.three.branch.currents:.I1,.I2,.and.I3..Since.all.of.these.currents.are.leaving.Node.A,.KCL.at.this.node.yields.Equation.1.6:

. I I I1 2 3 0+ + = . (1.2)

TABLE 1.1 Summary.of.Dependent.Sources

Element Description Symbols

Current-controlled.current.source.(CCCS)

Establishes.a.current.in.the.circuit.based.upon.the.value.of.controlling.variable,.Ix ,.and.the.gain.

Ix

Voltage-controlled.voltage.source.(VCVS)

Establishes.a.voltage.in.the.circuit.based.upon.the.value.of.controlling.variable,.Vx ,.and.the.gain.

+Vx

Voltage-controlled.current.source.(VCCS)

Establishes.a.current.in.the.circuit.based.upon.the.value.of.controlling.variable,.Vx ,.and.the.gain.

Vx

Current-controlled.voltage.source.(CCVS)

Establishes.a.voltage.in.the.circuit.based.upon.the.value.of.controlling.variable,.Ix ,.and.the.gain.

Ix

+

Vs(a)

Is(b)

FIGURE 1.1 Ideal. DC.sources.. (a). Voltage. source..(b).Current.source.


DCandTransientCircuitAnalysis 1-3

1.1.4 Kirchhoffs Voltage Law

Kirchhoffs.voltage.law.(KVL).is.also.based.upon.the.law.of.conservation.of.energy..A.loop.is.any.closed.path.in.a.circuit..KVL.states.that.the.sum.of. the.voltages.around.a. loop. is. zero. (i.e.,. sum.of. the. voltage.drops=.sum.of.the.voltage.rises)..KVL.is.applied.to.the.loop.shown.in.Figure.1.4..Note.that.the.direction.of.the.loop.goes.from.the.negative.terminal.to.the.

I + V

R

FIGURE 1.2 Resistor.

TABLE 1.2 Inductor.and.Capacitor.CurrentVoltage.Relationships

Element Circuit.Symbol Relationship

Inductor i+ v

L

v L didt

=

.(1.3)

iL

v dt it

= +1 00

( ).

(1.4)

Capacitori

+ V

C

i C dvdt

= . (1.5)

vC

idt vt

= +1 00

( ) . (1.6)

Vs+

I1

I2I3

R1

R2 R3

A

FIGURE 1.3 KCL.at.Node.A.

Vs

V1

Loop A

R1R2 R3 V2

+

+ +

FIGURE 1.4 KVL.applied.around.Loop.A.



positive.terminal.on.the.voltage.source,.which.indicates.it.is.a.voltage.rise..For.the.KVL.expression.in.Equation.1.7,.voltage.rises.are.negative.and.voltage.drops.are.positive:

. + + =V V Vs 1 2 0 . (1.7)

Example 1.1: DC Circuit Analysis with Independent Sources

For the circuit shown in Figure 1.5, apply Ohms law, KVL, and KCL to solve for the labeled voltages and currents.

The first step in the analysis is to apply KCL at Node A and KVL at the left and right loop. These equa-tions are provided in Equations 1.8 through 1.10:

KCL at Node A : + + =I I Is 2 3 0 (1.8)

KVL at left loop : + + =120 01 2V V (1.9)

KVL at right loop : + + =V V V2 3 4 0 (1.10)

Next, use Ohms law to rewrite Equations 1.9 and 1.10 in terms of the branch currents and resistor values. These equations are shown in Equations 1.11 and 1.12:

KVL at left loop: 50 100 1202I Is + = (1.11)

KVL at right loop: + + =100 20 80 02 3 3I I I (1.12)

Solving the simultaneous set of equations, (1.8), (1.11), and (1.12) yields

I I Is = = =1 2 0 6 0 62 3. , . , .A A A (1.13)

The results in (1.13) and Ohms law can be used to find the unknown voltages:

V Is1 50 60= = V (1.14)

V I2 100 60= =2 V (1.15)

V I3 V = 20 = 12 3 (1.16)

V I4 = 80 = 48 V3 (1.17)

120 V

+

+ ++ +

I2 I3

Is V1

A50 20

80 100 V3

V2 V4

FIGURE 1.5 DC.circuit.with.independent.sources.



1.1.5 Series and Parallel relationships

At.times,.it.is.useful.to.simplify.resistive.networks.by.combining.resistors.in.series.and.parallel.into.an.equivalent.resistance..Exactly.two.resistors.that.are.connected.at.a.single.node.share.the.same.current.and.are.said.to.be.connected.in.series..It.is.important.to.note.that.the.equivalent.resistance.of.series.resis-tors.is.larger.than.each.of.the.individual.resistances..Resistors.that.are.connected.together.at.a.pair.of.nodes.(single.node.pair).have.the.same.voltage.and.are.said.to.be.connected.in.parallel..The.equivalent.conductance.of.resistors.in.parallel.is.the.sum.of.the.conductances.of.the.individual.resistors..Therefore,.the. reciprocal. of. the. equivalent. resistance. is. the. sum. of. the. individual. conductances.. Note. that. the.equivalent.resistance.of.parallel.resistors.is.smaller.than.each.of.the.individual.resistances..Figure.1.6a.provides.an.example.of.a.circuit.with.series.resistors.and.the.equivalent.resistance.seen.by.the.voltage.source..Figure.1.6b.provides.an.example.of.a.circuit.with.parallel.resistors.and.the.equivalent.resistance.seen.by.the.current.source.

Example 1.2: Analysis of Example 1.1 by Combining Resistors

It is possible to analyze the circuit in Example 1.1, to find the source current, Is. The first step is to recognize that the 80 and 20 resistors are in series and combine to yield 100 . This simplified circuit is shown in Figure 1.7.

The next step is to note that the two 100 resistors are in parallel. Combine these two resistors to yield the equivalent resistance of 50 (see Figure 1.8).

The last simplification is to note that the 50 resistors in Figure 1.8 are in series and yield the equiva-lent resistance of 100 (see Figure 1.9).

Vs+

16 12

4

(a) (b)

Is 96 120 80

Req=16+12+4=32 1 +96Req=

1 (1) = 32 801 +120

FIGURE 1.6 Resistors.in.(a).series.and.(b).parallel.

120 V+

Is

50

100 100

FIGURE 1.7 Circuit.in.Example.1.1.simplified.by.putting.80.W.in.series.with.20.W.



Finally, the last step is to use Ohms law to solve Is, which yields

Is = =

120100

1 2. A

(1.18)

Note that this result is consistent with the answer to Example 1.1.

1.1.6 Voltage and Current Divider rule

Given.a.set.of.series.resistors.with.a.voltage.sourced.across.them,.the.voltage.across.each.individual.resistor.divides.in.direct.pro-portion.to.the.value.of.the.resistor..This.relationship.is.referred.to.as.the.voltage.divider.rule.and.it.can.be.derived.from.KVL..Given.a.set.of.parallel.resistors.with.a.current.sourced.through.them,.the.current. through. each. individual. resistor. divides. inversely. pro-portional.to.the.value.of.the.resistor..This.relationship.is.defined.as.the.current.divider.rule.and.it.can.be.derived.from.KCL..These.two.rules.are.shown.for.the.circuits.in.Figure.1.6.and.are.shown.in.Figure.1.10.

Example 1.3: Analysis of Example 1.1 Using Voltage and Current Divider

For the circuit in Figure 1.5, given that Is = 1.2 A, use the current divider to find I2 and the voltage divider to find V4. The first step in the analysis is to recognize that the 100 resistor is in parallel with the 80 and 20 series combination. The current divider relationship to find I2 is shown in Equation 1.19:

I Is2

100 80 20100

0 6=+

=

( ). A

(1.19)

120 V+

Is

50

50

FIGURE 1.8 Circuit.in.Figure.1.7.simpli-fied.by.putting.100.W.resistors.in.parallel.

120 V

+

Is

100

FIGURE 1.9 Circuit. in. Figure. 1.8.simplified.by.putting.50.W.resistors.in.series.

8 V+

16 12

4 48 mA 96 120 80

16(16+12+4)8

V16 = =4 V

16(16+12+4)8

V12 = =3 V

4(16+12+4)8

V4= =1 V(a)

96||120||8096I96

= 48 =16 mA

96||120||80120

= 48 =12.8 mA

96||120||8080

= 48 =19.2 mA

I120

I80(b)

FIGURE 1.10 Voltage. and. current. divider. rule. for. circuits. in. Figure. 1.6.. (a). Series. circuit. (voltage. divider)..(b)Parallel.circuit.(current.divider).



Ohms law can be used to find the voltage, V2, across the 100 resistor, V2 = 100I2 = 60 V. The voltage divider can be used to find the voltage, V4, as shown in Equation 1.20:

V V4 2

8080 20

48=+

= V

(1.20)

Note that these results are consistent with the solution to Example 1.1.

1.1.7 DeltaWye (Y) transformations

There.are.some.resistance.configurations.that.are.neither.in.series.or.parallel..These.special.configura-tions.are.referred.to.as.delta.().or.wye.(Y).interconnections..These.two.configurations.are.equiva-lent.based.upon. the. relationships. shown. in.Table.1.3..Equivalence.means. that.both.configurations.have.the.same.voltage.and.current.characteristics.at.terminals.a,.b,.however.internal.to.the.network,.the.values.may.not.be.the.same.

1.2 Systematic Circuit analysis techniques

There.are.two.general.approaches.to.solving.circuits.using.systematic.techniques..The.systematic.tech-niques.are.the.node-voltage.method.based.on.KCL.and.the.mesh-current.method.based.on.KVL..These.techniques.are.used.to.derive.the.minimum.number.of.linearly.independent.equations.necessary.to.find.the.solution.

1.2.1 Node-Voltage Method

The.node-voltage.method.is.a.general.technique.that.can.be.applied.to.any.circuit..An.independent.KCL.equation.can.be.written.at.every.essential.node.(nodes.with.three.or.more.elements.connected).except.for.one..The.standard.practice.is.to.choose.the.ground.node.as.the.reference.node.and.omit.the.ground.

TABLE 1.3 DeltaWye.(Y).Transformations

.Configuration Y.Configuration

Rc

RaRb

a b

c

R3

R2R1

a b

c

R R R R R R RRa

=

+ +1 2 2 3 3 1

1R R R

R R Rb c

a b c1 =

+ +

R R R R R R RRb

=

+ +1 2 2 3 3 1

2R R R

R R Ra c

a b c2 =

+ +

R R R R R R RRc

=

+ +1 2 2 3 3 1

3R R R

R R Ra b

a b c3 =

+ +



node.from.the.set.of.equations..Next,.each.essential.node.is.labeled.with.a.voltage.variable.(V1,.V2,.etc.)..The.node.voltage.represents.the.positive.voltage.difference.at.the.labeled.node.with.respect.to.the.refer-ence.node..A.KCL.equation.is.written.summing.the.currents.leaving.the.node.in.terms.of.the.unknown.node.voltages..Lastly,.this.set.of.linearly.independent.equations.is.solved.for.the.unknown.node.voltages..Finally,.the.node.voltages.can.be.used.to.find.any.current.in.the.circuit.

Example 1.4: Node-Voltage Method with Independent Sources

Given the circuit in Figure 1.11, use the node-voltage method to find the power delivered by eachsource.

Recall that the first step in the analysis was to label the essential nodes. The four essential nodes in Figure 1.11 have already been labeled as V1, V2, V3, and ground (0 V). Since V1 is the voltage at that node with respect to the reference node (ground node), it is tied to the 200 V source so V1 = 200 V. The node voltages V2 and V3 are unknown, thus KCL must be performed to find these values. In order to simplify analysis, the KCL equations are derived such that the current is drawn leaving the node if it is not given. The KCL equations at V2 and V3 are given in Equations 1.21 and 1.22:

KCL at V I I I

V V V V V2 500 250 400

2 1 2 2 3

500 250 4000: + + =

+ +

=

(1.21)

KCL at 100V I I

V V V V3 : + =

+

=4003 1 3 21100 400

1 0

(1.22)

By substituting V1 = 200 into Equations 1.21 and 1.22, and solving the simultaneous system of equations yields

V V V1 2 3 = 200 V, = 125 V, = 265 V (1.23)

Using the results of Equation 1.23, it is possible to find the power associated with the 1 A current source. Since the voltage across the current source is V3, and it is not in the passive sign convention, the power is

200 V

500 100

400

250 1 A

V1

V2 V3

+

FIGURE 1.11 Node-voltage.method.circuit.



P= V3 (1) = 265 W or 265 W delivered. In order to find the current through the 200 V source, it is necessary to use KCL at V1. The KCL equation at V1 is given in Equation 1.24:

KCL at V I I I I

V V V Vs s1 : + + = +

+

=500 1001 2 1 3

500 1000

(1.24)

Is = 500 mA (1.25)

Since the 200 V source obeys the passive sign convention, the power is P = 100Is =100 W absorbed. In order to check that the analysis is correct, the law of conservation can be used to verify that the sum of all of the power delivered equals the sum of all of the power absorbed.

Example 1.5: Analysis of Example 1.4 with -Y Transformations

For the circuit in Figure 1.11, use -Y transformations to find the power associated with the 200 V source. The first step in the analysis is to identify that the 500, 100, and 400 resistors form a configuration as Ra, Rb , and Rc, respectively. This circuit can be simplified by converting the configuration to a Y configu-ration. Equations 1.26 through 1.28 are used to find the resistor values in the Y configuration. The simpli-fied circuit is shown in Figure 1.12.

R

R RR R R

b c

a b c1

400 100500 100 400

40=+ +

=

+ +=

( )( )

(1.26)

R

R RR R R

c a

a b c2

400 500500 100 400

200=+ +

=

+ +=

( )( )

(1.27)

R

R RR R R

a b

a b c3

500 100500 100 400

50=+ +

=

+ +=

( )( )

(1.28)

In order to find the power associated with the 200 V source, perform KCL at essential Node A. The equa-tion and solution are shown in Equations 1.29 and 1.30:

KCL at V I I IV V V

A sA A: + =

+ =+200 2501

50 4501 0 (1.29)

VA = 225 V (1.30)

+200 V 200

250

50

40

V1

A

1 A

V2 V3

R3

R1R2

FIGURE 1.12 Circuit.in.Figure.1.11.simplified.



Using the result in Equation 1.30 to find the current through the 200 V source yields

I

Vs

A=

=

25

5 mA00

000

(1.31)

Thus, the power absorbed by the 200 V source is 100 W, consistent with the prior solution.

Example 1.6: Node-Voltage Method with Dependent Sources

The circuit in Figure 1.13 models an operational amplifier. An operational amplifier is an active circuit element used to perform mathematical operations such as addition, subtraction, multiplication, divi-sion, differentiation, and integration. This electronic unit is an integrated circuit that can be modeled as a VCVS. The gain of the op amp is the ratio of the output voltage to the input voltage, (Vo/Vs). Use KCL to determine the gain of the circuit in Figure 1.13.

The KCL equations at Nodes A and B are shown in Equations 1.32 and 1.33:

KCL at k M kV I I I

V V V V VA

A s A B A: 10 2 2010 20 2

0 + + =

+

+ =k k M

(1.32)

KCL at kV I I

V V V VB

B A B d: 50 205

202 1050

0 + =

+

=

k (1.33)

Note that the dependent source introduces a constraint equation based upon the relationship between the node voltage and the controlling voltage, Vd. This relationship is VA = Vd. This produces two equa-tions and two unknowns that can be solved for the gain shown in Equation 1.34:

VVo

s

=

2010

2kk

(1.34)

A special case of the node-voltage method is when there is a voltage source between two nonreference essential nodes (see Figure 1.14).

In this case, an additional unknown variable must be introduced to describe the current in the branch with the voltage source. To minimize the number of unknowns, an alternate method to introducing another variable is to label the voltage source and any element in parallel with it as a supernode. The supernode in Figure 1.14 is denoted by the superimposed oval. The node-voltage method with super-nodes involves deriving a KCL and KVL equation at the supernode as well as KCL equations at any other essential nodes where the voltage is unknown and solving the simultaneous system ofequations.

+

10 k

20 k

50

2 M 2105 Vd VoVdVs

A B

+

+

+

FIGURE 1.13 DC.circuit.with.dependent.sources.(operational.amplifier.model).



Example 1.7: Node-Voltage Method with Supernodes

Use the node-voltage method on the circuit in Figure 1.14 to find the current through the voltage source. The first step in the analysis is to label the node voltages and supernode. These have already been labeled in the circuit in Figure 1.14. Next, KCL at the supernode yields Equation 1.35, and KVL at the supernode yields Equation 1.36:

KCL at supernode : 2 2

500 100 1250500 100 125

1 2 2+ + + = + + + =I I IV V V

(1.35)

KVL at supernode 25 1 2: + + =V V 0 (1.36)

Solving these two equations and two unknowns yields

V1 775 V= . (1.37)

V2 1 25 V= 0 . (1.38)

To find the current through the voltage source, it is necessary to perform KCL at V1 or V2. Since the 2 A current source is connected to V1, this selection will have one less term with a voltage variable. Assuming the current through the voltage source, Is, flows from right to left and applying KCL at V1 yields the following equation:

KCL at supernode: = 2 + + = 2 155 m + 100500 250I I Is m = 1.945 A (1.39)

1.2.2 Mesh-Current Method

The.goal.of.the.mesh-current.method.is.to.determine.all.of. the.unknown.mesh.currents.in.a.circuit..A.mesh.is.a.loop.in.a.circuit.that.does.not.contain.any.other.loops..The.mesh-current.method.is.only.applicable.to.planar.circuits,.circuits.that.can.be.drawn.on.a.plane.with.no.crossing.branches..The.first.step.in.the.analysis.is.to.label.all.of.the.mesh.currents.in.a.circuit..The.mesh.currents.are.fictitious.cur-rents.that.circulate.in.a.mesh..Note.that.a.mesh.current.may.or.may.not.be.a.branch.current,.but.all.of.the.branch.currents.can.be.found.from.the.mesh.currents..The.next.step.is.to.write.KVL.equations.summing.voltage.drops.around.the.mesh.in.terms.of.the.unknown.mesh.currents..For.n.meshes,.there.will.be.n.linearly.independent.mesh-current.equations.to.solve.

+

2 A

25 VV1 V2

500 100 125

250

FIGURE 1.14 Node-voltage.method.with.a.supernode.



Example 1.8: Mesh-Current Method on Example 1.6

For the circuit in Figure 1.15, use the mesh-current method to determine the output voltage Vo if the input voltage Vs = 3 V.

The first step in the analysis is to label the two mesh currents, and this has been done in Figure 1.15. The second step is to write the KVL equations around meshes 1 and 2 in terms of the mesh currents, I1 and I2 (see Equations 1.40 and 1.41):

KVL at mesh 1: 3 + 10 + 2 ( ) = 01 1 2k MI I I (1.40)

KVL at mesh 2: 2 ( ) + 20 +50 + 2 10 = 02 1 2 25M kI I I I V d (1.41)

Similar to the prior analysis, the dependent source introduces the following constraint equation:

constraint: = 2 ( )V I Id M 2 1 (1.42)

Solving these three simultaneous equations for the mesh currents yields

I I Vd1 2A A = 299.997 , = 300.002 , = 30.075 V (1.43)

Using the mesh current value to find Vo yields

V I Vo d = 50 + 2 10 = 6 V5

2 (1.44)

The reader should verify that this gain is consistent with Example 1.6.A special case of the mesh-current method occurs when a current source is shared between two

meshes. In this case, it is necessary to introduce another variable to describe the voltage across the current source in order to write the KVL equation. An alternate approach is to define the two meshes that include the current source and anything in series with it as a supermesh. The 6 A current source in series with the 1 resistor in Figure 1.16 creates a supermesh denoted by the superimposed rectangle.

In order to analyze a supermesh, it is necessary to perform KVL and KCL at the supermesh. Lastly, write a KVL equation for any other unknown mesh currents in the circuit and solve the simultaneous system of equations. This method will be demonstrated on the circuit in Figure 1.16.

+

10 k

20 k

50

2 M 2105 Vd VoVdVs I1

I2

3 V

A B

+

+

+

FIGURE 1.15 Mesh-current.method.example.



Example 1.9: Mesh-Current Method with a Supermesh

Use the mesh-current method to find the power associated with the 6 A current source. The first step in the analysis is to label the supermesh and mesh currents. These have already been labeled in the circuit in Figure 1.16. The next step is to derive the KVL and KCL equations at the supermesh and these are shown in Equations 1.45 and 1.46:

KVL at supermesh: 2 + 3 +5 10 + 9 + 7 = 02 1 I I I I1 2 (1.45)

KCL at supermesh: = 62I I1 (1.46)

Solving this simultaneous set of equations yields

I I1 2 = 4 , = 2 A A (1.47)

In order to determine the power associated with the 6 A current source, it is necessary to perform KVL at the left or right mesh to find the voltage across the current source. Assuming the voltage across the current source, Vs, is positive on top and applying KVL at the left mesh yields

KVL at mesh 1: = 1( ) 3 + 2 7 = 44 V2 1 1 1V I I I Is (1.48)

The 6A current source is drawn in the passive sign convention and since Vs is negative, the power associ-ated with this source is P = +(44)(6) = 264 W or 264 W delivered.

1.2.3 Superposition

Superposition.applies.to.linear.circuits.that.have.multiple.independent.sources..The.principle.of.super-position.states. that. the.electrical.quantities,.voltage.or.current,.due. to.all. the.sources.acting.at. the.

+ 1

7

5 3

9 6 A

10 VI1 I2

2 V

+

FIGURE 1.16 Mesh-current.method.with.a.supermesh.



same.time.is.equal.to.the.sum.of.the.same.quantity.due.to.each.source.acting.alone..The.method.to.solve.for.an.unknown.variable.in.a.circuit.involves.solving.for.the.variable.of.interest.for.one.source.acting.alone.by.deactivating.all.the.other.independent.sources,.then.sum.the.results.for.each.source.act.ing.alone..To.deactivate.an.independent.voltage.source,.replace.the.voltage.source.with.a.short.cir-cuit. (0.V).. To. deactivate. an. independent. current. source,. replace. the. current. source. with. an. open.cir.cuit. (0. A).. Dependent. sources. are. never. deactivated. (turned. off ).. The. benefit. in. applying. the.principle.of.superposition.is.that.many.times,.the.circuit.with.the.deactivated.source.is.simpler.to.solve.for.the.unknown.value.

Example 1.10: Circuit Analysis Using Superposition

For the circuit in Figure 1.16, apply the principle of linear superposition to solve for the unknown branch current I1 (see Figure 1.17).

The first step in the analysis is to disable the 6 A and 10 V sources and use KVL to calculate I1. The solu-tion to this analysis is shown in Equation 1.49. The variable of interest is given a prime to denote that it is due to one source acting alone (see Figures 1.18 through 1.20).

KVL at mesh 1: 2 + 3 + 5 +9 + 7 = 01 1 1 1 I I I I

l1 . mA= 83 33 (1.49)

+ 1

3 5

7 9

V1

I1

2 V

FIGURE 1.18 Superposition.circuit.with.2.V.source.activated.

+ 1

3 5

7 9 6 A

10 V

I1

2 V +

FIGURE 1.17 Superposition.circuit.for.Example.1.10.



In the next step, disable the 2 V and 6 A sources and use KVL to calculate I1. The solution to this analysis is shown in Equation 1.50:

KVL at mesh 1: 3 + 5 10 +9 + 7 = 01 1 1 1I I I I

l1 416 67 . mA= (1.50)

In the next step, disable the 2 and 10 V sources and use KCL at V1 to calculate I1. The solution to this analy-sis is shown in Equation 1.51:

KCL at V

V V1 :

3 71

++ +

+=6

5 901

V1 V = 35

l

V1

107 3

3 5 . A=

+=

(1.51)

Applying the principle of superposition yields

l l l l1 1 1 1 4A= + + = (1.52)

Note that the value for I1 is consistent with the solution to Example 1.9.

1

3 5

7 9

10 V

V1

I1

+

FIGURE 1.19 Superposition.circuit.with.10.V.source.activated.

1

3 5

7 6 A

9

I1

V1

FIGURE 1.20 Superposition.circuit.with.6.A.source.activated.



1.3 Circuit Modeling techniques

Just.like.it.is.possible.to.model.the.behavior.of.multiple.resistors.connected.in.parallel.and.series.with.a.single.equivalent.resistance,.it.is.also.possible.to.model.resistive.circuits.containing.sources.and.resistors.as.either.a.Thevenin.or.Norton.equivalent.model..These.models.are.useful.simplifying.techniques.when.only.the.circuit.behavior.at.a.single.port.is.of.interest.

1.3.1 Source transformations

Source.transformations.are.another.simplifying.technique.for.circuit.analysis..Source.transformations.are.based.upon.the.concept.of.equivalence.of.the.voltage.and.current.terminal.characteristics.at.a.single.port..A.voltage.source.in.series.with.a.resistor.can.be.replaced.by.a.current.source.in.parallel.with.a.resis-tor.if.they.have.the.relationships.given.in.Figure.1.21.

1.3.2 thevenin and Norton Equivalent Circuits

A.simple.resistive.circuit.can.be.simplified.to.an.independent.voltage.source.in.series.with.a.resistor.and.this.is.referred.to.as.the.Thevenin.equivalent.circuit..The.voltage.source.is.referred.to.as.the.Thevenin.voltage,.VTH,.and.the.resistor.is.the.Thevenin.resistance,.RTH..In.addition,.a.simple.resistive.circuit.can.be.simplified.to.an.independent.current.source.in.parallel.with.a.resistor.and.this.is.referred.to.as.the.Norton.equivalent.circuit..The.current.source.is.the.Norton.current,.IN,.and.the.resistance.is.the.same.as.the.Thevenin.resistance..These.are.important.simplification.techniques.when.the.values.of.interest.are.the.port.characteristics.such.as.the.voltage,.current,.or.power.delivered.to.a.load.placed.across.the.terminals..The.method.to.find.the.Thevenin.voltage.is.to.determine.the.open.circuit.voltage.across.terminals.a.and.b..The.method.to.find.the.Norton.cur-rent.is.to.find.the.short.circuit.current.between.terminals.a.and.b..There.are.several.techniques.to.find.the.Thevenin.equivalent.resistance..When.there.are.only.independent.sources,.one.of.the.more.popular.methods.is.to.deactivate.all.independent.sources.and.find.the.equivalent.resistance.of.the.network.across.terminals.a and.b..Alternately,.the.Thevenin.resistance.can.be.calculated.by.using.the.following.formula:

.R V

IVI

oc

sc

TH

NTH = =

.(1.53)

whereVoc.represents.the.open.circuit.voltage.across.terminals.a.and.bIsc.represents.the.short.circuit.current.between.terminals.a.and.b

Rs

Vs

a

b

+

(a) (b)

b

a

RpIp

Vs= IpRpRs=Rp Rp=Rs

VsRs

=Ip

FIGURE 1.21 Source. transformation. relationships.. (a). Series. circuit,. Vs=IpRp, Rs=Rp and (b). parallel. circuit,.Ip=Vs/Rs, Rp=Rs.



Note.that.when.there.are.dependent.sources.in.the.circuit,.there.is.a.third.technique.based.upon.deacti-vating.all.independent.sources.and.using.a.test.voltage.or.current.at.terminals.a.and.b.to.find.the.equiva-lent.resistance..The.reader.is.encouraged.to.review.this.technique.for.future.study.

Example 1.11: Thevenin Equivalent Resistance

For the circuit in Figure 1.22, determine the Thevenin equivalent resistance to the left of terminals a and b.

In order to find the Thevenin equivalent resistance, deactivate the two independent sources and find the equivalent resistance to the left of terminals a and b. The circuit in Figure 1.22 is shown in Figure 1.23 with the sources deactivated.

In the circuit in Figure 1.23, the 10 and 40 resistors are in parallel. This parallel combination is in series with the 8 resistor. Equation 1.54 shows the derivation of the Thevenin equivalent resis-tance, RTH:

RTH = 10 || 40 + 8 = 16 (1.54)

Example 1.12: Thevenin and Norton Equivalent Circuits

For the circuit in Figure 1.22, find the Thevenin and Norton equivalent circuit to the left of terminals a and b. The first step in the analysis is to find the open circuit voltage between terminals a and b (VTH = Voc = Va). Either the node-voltage or mesh-current method would be an acceptable technique to find this value, however the node-voltage method was used by writing the KCL equation at V1 and Va and these are shown in Equation 1.55:

KCL at V

V V V Va1

1:

+ +

=

6010 40 8

01 1

(1.55)

KCL at VV V

aa:

=1

84

4 A

a

b

60 V

V110

+

8

40

FIGURE 1.22 Circuit.for.Thevenin.equivalent.example.

a

b

10 8

40

FIGURE 1.23 Circuit. in.Figure.1.22.with.independent.sources.deactivated.



V V V VTH oc a1 = 80 V, = = = 112 V

The next step in the analysis is to find the short circuit current between terminals a and b (IN = Isc = Iab). The mesh-current method will be used to determine short circuit current, Isc, as shown in Figure 1.24. The result of the analysis is shown in Equation 1.56:

KVL at : 60 + 10( 4) + 40 ( ) = 01 1 1I I I IN

(1.56) KVL at : 40 ( ) + 8( 4) = 01I I I IN N N

I I I IN sc ab1 = 7.6 A, = = = 7 A

The Thevenin equivalent resistance can also be found from

R

VI

VI

THoc

sc

TH

N= = = =

1127

16

(1.57)

Note that this Thevenin resistance is consistent with Example 1.11. The final step in the result is to draw the Thevenin and Norton equivalent circuits to the left of terminals a and b. These are shown in Figure 1.25.

4 A

a

b

60 V

V110 8

40 + I1 IN

FIGURE 1.24 Circuit.in.Figure.1.22.with.terminals.a.and.b.short-circuited.

a

b

112 V

(a)

+

VTH

RTH

16

a

b

7 AIN RTH16

(b)

FIGURE 1.25 Thevenin. and. Norton. equivalent. of. circuit. in. Figure. 1.22.. (a). Thevenin. equivalent.. (b). Norton.equivalent.



1.3.3 Maximum Power transfer

One.benefit.and.purpose.for.determining.the.Thevenin.(Norton).equivalent.of.a.circuit.is.to.determine.the.power.delivered.to.a.load.placed.across.terminals.a.and.b..With.the.knowledge.of.the.Thevenin.and.Norton.equivalent,.it.is.possible.to.design.a.circuit.or.select.a.load.for.maximum.power.transfer.to.the.load.. It. can.be.shown. that. if. the. load. resistance. is. equal. to. the.Thevenin.equivalent. resistance,. then.maximum.power.is.transferred.to.the.load..Therefore,.the.condition.for.maximum.power.transfer.is.to.set.the.load.resistance.equal.to.the.Thevenin.equivalent.resistance..When.the.load.resistance.is.equal.to.the.Thevenin.equivalent.resistance,.the.maximum.power.delivered.to.the.load.is

.P V

RLTH

TH=

2

4 .(1.58)

Example 1.13: Maximum Power Transfer

For the circuit shown in Figure 1.22, determine the value of a load resistor placed across terminals a and b for maximum power transfer and calculate the value of the power for the load selected (see Figure 1.26).

Since the Thevenin equivalent resistance of this circuit is 16 , select RL = RTH = 16 for maximum power transfer. Finally, the value of the power delivered to the 16 is calculated as follows:

P

VR

LTH

TH= = =

2 2

41124 16

196( )

W

(1.59)

1.4 transient analysis

Transient analysis.describes.a.circuits.behavior.as.a.function.of.time..Since.capacitors.and.inductors.store.energy.in.the.electric.or.magnetic.field,.transient.analysis.focuses.on.the.current.and.voltage.values.in.circuits.where.energy.is.either.stored.by.or.released.by.an.inductor.or.a.capacitor.

1.4.1 First-Order Circuits

First-order.circuits.contain.resistors.and.either.capacitors.or.inductors,.but.not.both..These.configura-tions.are.either.RL.circuits.or.RC.circuits.based.upon.whether. they.have. resistors.and.capacitors.or.resistors.and.inductors,.respectively..RL.and.RC.circuits.are.known.as.first-order.circuits.because.the.

a

b

60 VRL

V1

4 A

8 10

40 +

FIGURE 1.26 Circuit.for.maximum.power.transfer.example.



equations.that.describe.these.circuits.are.first-order.ordinary.differential.equations..If.a.voltage.or.cur-rent.source.is.suddenly.applied.to.a.first-order.circuit.(i.e.,.a.switch),.then.energy.will.begin.to.store.in.the.capacitor.as.an.electric.field.or.in.the.inductor.as.a.magnetic.field..When.a.source.is.instantaneously.applied,.the.time-dependent.current.or.voltage.in.the.circuit.is.called.the.step.response..If.the.source.of.energy.is.suddenly.removed,.then.the.time-dependent.current.or.voltage.in.the.circuit.is.called.the.natu-ral.response..It.is.important.to.note.that.the.voltage.across.a.capacitor.cannot.change.instantaneously.and.the.current.through.an.inductor.cannot.change.instantaneously..The.natural.and.step.response.of.first-order.circuits.can.be.found.by.using.circuit.analysis.techniques.such.as.KVL.and.KCL.to.derive.the.first-order.differential.equation.that.describes.the.circuit..Using.the.initial.conditions.and.differential.equations,.these.equations.can.be.solved.for.voltage.and.current..In.order.to.find.the.initial.conditions.for.a.first-order.circuit,. it. is.necessary. to.draw.the.circuit.under.DC.conditions.before. the.switching.occurs..Note.that.under.DC.or.steady.state.conditions,.inductors.can.be.modeled.as.short.circuits.and.capacitors.can.be.modeled.as.open.circuits..The.general.form.of.the.solution.for.a.first-order.circuit.is.the.sum.of.the.transient.response.and.the.steady-state.response..The.transient.response.is.the.portion.of.the.response.that.decays.over.time..The.steady-state.response.is.the.portion.of.the.response.that.remains.after.a.long.time..Furthermore,.the.general.form.of.the.solution.can.be.described.as.the.sum.of.the.natu-ral.response.and.the.forced.response..The.forced.response.is.the.portion.due.to.the.independent.sources.and.the.natural.response.is.due.to.the.energy.stored.in.the.circuit..The.general.solution.for.natural.and.step.responses.for.first-order.circuits.is.given.in.Equation.1.60:

.

x t x x x t x t x tsteady-state transient t( ) ( ) [ ( 0 ) ( )] /= + = + = + e

xx t x x x t x t x tstep natural t t( ) ( ) ( )e ( )/ /= + = + = + 0 e .(1.60)

where.=.RTH.C.or..=.L/RTHRTH.is.equivalent.resistance.across.the.inductor.or.the.capacitorx(t).is.either.voltage.or.current

The.time.constant,.,.is.in.units.of.seconds.and.describes.how.fast.the.transient.signal.settles..At.a.time.equal.to.5,.the.transient.solution.will.settle.to.within.1%.of.the.steady-state.or.final.value..The.general.solution.can.be.applied.to.any.RL.or.RC.circuit.provided.that.the.initial.and.final.conditions.and.equiva-lent.resistance.seen.by.the.inductor.or.capacitor.can.be.found.

Example 1.14: Natural Response of an RL Circuit

For the circuit in Figure 1.27, assume that the switch is in position a for a long time and moves to position b at t = 0. Find the current through the inductor, i(t), and the voltage across the inductor, v(t), for t > 0.

a

b

10 V

i(t)

v(t)

t =0

4 20

1

+

+

100 mH

FIGURE 1.27 RL.circuit.for.Example.1.14.



The first step in the analysis is to find the initial conditions for the circuit in Figure 1.27. In order to find the initial conditions, redraw the circuit at t = 0 (before the switching occurs) under DC conditions. The circuit to find the initial conditions is shown in Figure 1.28. Since an inductor under DC or steady-state conditions is modeled as a short circuit, the initial voltage, v(0), is 0 V. It is modeled as a short circuit because the current is constant with time; therefore, the time rate of change of the current (diL/dt) is zero and the voltage (vL = L(diL/dt)) over the inductor is 0 V. Using Ohms law on the circuit in Figure 1.28, it is possible to find the initial current, i(0) as shown in Equation 1.61. Note that because the inductor is a short circuit, the 20 resistor is shorted out and has no affect on the circuit.

i( )0

101

10 = = A

(1.61)

Since the voltage across an inductor can change instantaneously, the circuit must also be analyzed immediately after switching occurs at t = 0+. For this analysis, model the inductor as a 10 A current source because current cannot change instantaneously so i(0+) = i(0) = 10 A. Next, use KCL to find the voltage across the inductor. The circuit is shown in Figure 1.29 and the analysis in Equation 1.62:

v (0 ) = [(4 + 1)||20]10 = 40 V+

(1.62)

The circuit in Figure 1.29 can also be used to find the time constant, = L/RTH, where RTH is the equivalent resistance seen by the inductor. Since the 4 and 1 resistors are in series and they are in parallel with the 20 resistor, RTH is given by Equation 1.63:

RTH = (4 + 1)||20 = 4 (1.63)

a

b

10 V v(0)

i(0)

1

4 20 +

+

FIGURE 1.28 RL.circuit.at.t.=.0.

b

10 Av(0+)

i(0+)

1

4 20

+

FIGURE 1.29 RL.circuit.at.t.=.0+.



The time constant is given by Equation 1.64.

= = =

LRTH

0 14

25.

ms

(1.64)

In order to find the final value for the current and voltage, analyze the circuit under steady-state conditions a long time after switching occurs. This circuit is shown in Figure 1.30; note that once again the 20 resistor is shorted out. Since, it is assumed that the circuit has been in this state for a long time, the current through the inductor and voltage across the inductor can be represented as i() and v(), respectively. Since the inductor is still modeled as a short circuit, thus v() = 0 V and since there are no sources, i() is 0 A.

These values make sense because for the natural response of a first-order circuit, the inductor has stored energy in the form of current and over time, it discharges until it is eventually 0 A. Using the gen-eral solution equation in (1.60) yields

i t i i i tt t( ) ( ) [ ( ) ( )] ,/= + = >+ 0 10 040e e A (1.65)

v t v v v tt t( ) ( ) [ ( ) ( )] ,/= + = >+ 0 40 040e e V

b

4

i()

v()

1

20

+

FIGURE 1.30 RL.circuit.at.t.=..

v(t), V

Time, ms

10

1025 25 50 75 100 125 150

20

30

40

50(a)

i(t), A

Time, ms

15

10

5

5

25 25 50 75 100 125 150

(b)

00

00

FIGURE 1.31 Graphs.of.voltage.and.current.for.the.inductor.in.Example.1.14..(a).Voltage.across.the.inductor.and.(b).current.through.the.inductor.



The graphs of v(t) ad i(t) are shown in Figure 1.31; note that these are exponentially decaying functions to represent the natural response and the fact that the inductor is discharging. Also notice that there is a discontinuous step at t = 0 for the voltage across the inductor to denote the change from storing or stored energy to releasing energy.

Example 1.15: Step response of an RC Circuit

For the circuit in Figure 1.32, the switch has been in position a for a long time, and at t = 0, it moves to position b Find the voltage across the capacitor, v(t), and the current through the capacitor, i(t), for t > 0.

The first step in the analysis is to find the initial conditions by analyzing the circuit under steady-state conditions before switching occurs to find i(0) and v(0). Since a capacitor under DC or steady-state condi-tions is modeled as an open circuit, the initial current, i(0), is 0 A. It is modeled as an open circuit because the voltage is constant with time; therefore, the time rate of change of the voltage (dvC/dt) is zero and the current (iC = CdvC/dt) over the inductor is 0 A. The circuit is shown in Figure 1.33 and the analysis using Ohms law yields Equation 1.66:

v (0 ) = ( )( ) = 1 V 1 1m k (1.66)

Since the current through a capacitor can change instantaneously, the circuit must be analyzed right after switching occurs to find i(0+). However, since voltage across a capacitor cannot change instanta-neously, after switching occurs, the capacitor can be modeled as a 1 V source (i.e., v(0) = v(0+) = 1 V). This circuit is shown in Figure 1.34. KCL can be used to analyze this circuit to find the current through the capacitor as shown in Equation 1.67:

i( )

112

. A030 16

4 75+ =

+

=

k km

(1.67)

a b

1 mA0.5 F

30 Vi(t)

v(t)

t =0

1 k 1 k 6 k

12 k1 k +

+

FIGURE 1.32 RC.circuit.for.Example.1.15.

a

1 mA

i(0)

v(0)

1 k 1 k

1 k

+

FIGURE 1.33 RC.circuit.for.initial.conditions.(t.=.0).



The circuit in Figure 1.34 can also be used to find the time constant, = RTHC. The equivalent resistance across the capacitor can be found by disabling the 30 V source. After deactivating the 30 V source, the 6 k resistor is in parallel with the 12 k resistor, thus RTH = 4 k. The time constant is = RTHC = (4 k)(0.5 ) = 2 ms. Finally, to find the steady-state voltage and current for the capacitor, analyze the circuit under steady-state conditions a long time after switching occurs. Since the capacitor is modeled as an open circuit under steady-state conditions, i() = 0 A. The circuit is shown in Figure 1.35. The final value of the capacitor voltage, v(), can be found by using the voltage divider as shown in Equation 1.68:

v( ) ( ) =

+=

1212 6

30 20 V

(1.68)

Using the general solution for a first-order circuit in Equation 1.60 yields the following equations for v(t) and i(t):

i t i i i tt t( ) ( ) [ ( ) ( )] . ,/= + = >+ 0 4 75 0500e e mA

(1.69)

v t v v v e tt t( ) ( ) [ ( ) ( )] ,/= + = >+ 0 20 19 0500 e V

The graphs of the current and voltage for the capacitor are shown in Figure 1.36. Note that since this is a step response and the capacitor voltage is charging, the graph is an exponentially increasing function. Also, since current through a capacitor can change instantaneously, there is a discontinuous step in the graph at t = 0 when the capacitor begins charging from 1 to 20 V.

b

+

+

30 V

i(0+)

v(0+)

6 k

12 k

+

1 V

FIGURE 1.34 RC.circuit.for.initial.conditions.(t.=.0+).

b

+

30 V

i()

v()

6 k

12 k

+

FIGURE 1.35 RC.circuit.for.final.conditions.(t.=.).



1.4.2 Second-Order Circuits

Second-order.circuits.contain.resistors,.capacitors,.and.inductors..These.circuits.are.known.as.second-order.circuits.because.the.equations.that.describe.these.circuits.are.second-order.ordinary.differential.equations..Second-order.circuits.also.exhibit.a.natural.and.step.response.based.upon.whether.the.capac-itor.and.inductor.are.storing.or.releasing.energy..To.solve.these.types.of.circuits,.use.circuit.analysis.to.generate.the.second-order.differential.equation.governing.the.behavior.and.then.use.differential.equa-tions.and.initial.conditions.to.solve.for.the.solution..One.approach.to.solving.the.differential.equation.describing.the.transient.response.is.to.guess.a.solution.and.plug.into.the.differential.equation..Then.the.initial.and.final.values.can.be.used.to.determine.the.time.response..For.example,.the.general.differential.equation.for.these.types.of.circuits.is.given.by

.d xdt

dxdt

x Ko2

222+ + =

.(1.70)

wherex.represents.the.voltage.or.current.is.the.Neper.frequency.in.rad/so.is.the.resonant.frequency.in.rad/sK.is.related.to.the.steady-state.value.of.the.variable.of.interest

The.damping.factor.and.resonant.frequency.can.be.identified.from.the.second-order.equation.derived.for.the.circuit..To.solve.these.equations,.assume.that.the.solution.is.of.the.form.x(t).=.Aest..Substitute.this.value.into.Equation.1.71.to.yield.the.following:

. Ae s s Kst o( )2 22+ + = . (1.71)

Thus,.the.characteristic.equation.of.any.second-order.circuit.is. s s o2 22+ + ..The.general.form.of.the.solution.to.the.second-order.differential.equation.in.(1.72).is.given.by

. x t A A x ts t s t( ) ( )= + + 1 21 2e e . (1.72)

The.roots.of.the.characteristic.equation,.s1.and.s2,.can.be.used.to.determine.the.type.of.response..There.are.three.types.of.responses.for.second-order.circuits:.overdamped,.critically.damped,.and.underdamped..

v(t), V

Time, ms

25

20

15

10

5

5

(a)

0 5 10 15(b)

i(t), mA

Time, ms

0.51

1.52

2.5

3.5

4.5

3

4

5

5 0 5 10 150 0

FIGURE 1.36 Graphs.of.voltage.and.current.for.the.capacitor.in.Example.1.15..(a).Voltage.across.the.capacitor.and.(b).current.through.the.capacitor.



The.overdamped.response.has.a.slow.response.and.long.settling.time..The.critically.damped.response.has.a. fast. response.and.short.settling. time..The.underdamped.response.has. the. fastest. response.and.a. long.settling.time..Table.1.4.presents.the.relationship.between.the.three.responses,. the.roots.of.the.characteristics.equation,.the.Neper.frequency,.resonant.frequency,.form.of.the.solution,.and.the.graph.

Example 1.16: Natural Response of an RLC Circuit

For the circuit in Figure 1.37, assume that the switch opens instantaneously at t = 0, what is the voltage, v(t) across the capacitor and current, i(t) through the capacitor.

The first step in the analysis is to determine the initial conditions or the energy stored in the inductor and capacitor. In order to find these values, analyze the circuit under steady-state conditions right before switching occurs (t = 0). This circuit is shown in Figure 1.38. As previously stated, in this circuit, the induc-tor is modeled as a short circuit and the capacitor is modeled as an open circuit. Since the capacitor is an open circuit, i(0) is 0 V and since it is in parallel with a short circuit, v(0) is 0 V. Finally, since the inductor is a short circuit and current follows the path of least resistance, iL(0) = 2 A.

Next, the circuit must be analyzed right after switching occurs to find i(0+) (see Figure 1.39). In this circuit, the inductor is modeled as a 2 A current source and the capacitor is modeled as a 0 V voltage source or a wire. This circuit is shown in Figure 1.39. Since current is continuous for inductors and voltage is continuous for capacitors, these values do not change. However, the current through the capacitor changes to i(0+) = 2 A.

TABLE 1.4 Summary.of.Second-Order.Circuit.Responses

Overdamped.response.(.>.o)

s1.and.s2.are.realx t A A x ts t s t( ) ( )= + + 1 21 2e e

Critically.damped.(.=.(.=.o)

s1.=.s2x(t).=.A1tet.+.A2et.+.x(t..)

Underdamped.response.(.


The next step in the analysis is to analyze the circuit at some point after switching occurs to derive the second-order differential equation. This circuit is shown in Figure 1.40, and the derivation of the equation using KCL is shown in (1.73):

i i iR L C + + = 0

vvdt i

dvdt

t

20110

0 4 00

+ + + =+ ( ) m

120

110

4 02

2

dvdt

vd vdt

+ + =m

d vdt

dvdt

vd vdt

dvdt

vo2

2

2

2212 5 25 2 0+ + = + + =.

(1.73)

t=0

2 A20 10 H

4 mF v(t)

i(t) +

FIGURE 1.37 Natural.response.of.a.parallel.RLC.circuit.

2 A20 v(0)

i(0)iL(0)

+

FIGURE 1.38 Parallel.RLC.circuit.at.t.=.0.

2 A20 v(0+)

i(0+)iL(0+)+

FIGURE 1.39 Parallel.RLC.circuit.at.t.=.0+.

10 H20 4 mF v(t)

i(t) +

FIGURE 1.40 Parallel. RLC. circuit.for.Example.1.16.



From examination of Equation 1.73, it is evident that = 6.25 and o = 5 rad/s. Since > o, the voltage and current response are overdamped. The roots of the characteristic equation (s2 + 12.5s + 25) are 2.5 and 10, and the general form of the response, v(t), is given in Equation 1.74:

v t A A v t A As t ts t( ) ( ) .= + + = + 1 2 1 2 101 2 52e e e e

t

(1.74)

Note that since this is a natural response and the capacitor and inductor are discharging, v() and i() are zero. In order to find the values of A1 and A2, use the circuits initial conditions. The evaluation of Equation 1.74 and its first derivative at t = 0+ yields

v A A( )0 1 2+

= + (1.75)

dvdt

A A( )

.0

2 5 101 2+

=

Using the initial current through the inductor and the initial voltage across the capacitor with the results of (1.75), the values of A1 and A2 can be found as shown in Equation 1.76:

v A A( )0 01 2+

= + = (1.76)

i i i

vi

dvdt

R L C L( ) ( ) ( )( )

( )( )

0 0 0020

0 40

0+ + ++

++

+ + = + + =m

(1.77)

dvdt

v iA AL

( ) ( )( )( )

( ).

0 020 4

04

2 5 10 5001 2+ + +

= = =

m m

Solving the simultaneous set of Equations 1.76 and 1.77 yields A1 = 66.7 and A2 = 66.7. Finally, the solu-tions to the example are

v t tt t( ) . . ,.= + > 66 7 66 7 025 10e e V

(1.78)

i t

dvdt

tt t( ) . ,.= = > 4 667 2 667 025 10m e e mA

The reader is encouraged to verify that the solution for the transient response of the capacitors voltage and current does indeed obey the initial conditions.

Example 1.17: Natural Response of an RLC Circuit

For the circuit in Figure 1.41, assume the switch has reached steady-state before the switch moves from position a to position b. If at time t = 0 the switch moves to position b, calculate i(t) and v(t) for t > 0.

Similar to Example 1.17, the first step in the solution process is to determine the stored energy in the inductor and capacitor. The circuit in Figure 1.42 illustrates the circuit under steady-state conditions before the switch moves from position a (t = 0); the inductor is modeled as a short circuit (v(0) = 0 V) and the capacitor is modeled as an open circuit. By observation, the voltage across the capacitor is also 0 V and the current through the inductor can be found from Ohms law, i(0) = 50/10 = 5 A.



The next step in the analysis is to use the values found at t = 0 to model the initial conditions in the circuit right after switching occurs (see Figure 1.43). Since current cannot change instantaneously through the inductor, it is modeled as a 5 A current, and because voltage cannot change instantaneously across a capacitor, it is modeled as a 0 V source or wire.

Since the 20 resistor is in parallel with the 5 A current source in Figure 1.43, they have the same voltage (v(0+) = (20)(5) = 100 V). At any instant of time after the switch moves, observe that this cir-cuit is a series RLC circuit and KVL can be used to derive the second-order differential equation that describes it (see Figure 1.44 and Equation 1.79):

v v vL R C + + = 0

100 201

10000 0

0

mdidt

i idt vt

+ + + =+ ( )

100 20

11000

02

2md idt

didt

i+ + =

d idt

didt

id idt

didt

i2

2

2

22200 10 000 2 0+ + = + + =, o

(1.79)

50 V

10

20

1000 F

100 mH

i(t)

v(t)

t =0

a b

+

+

FIGURE 1.41 Natural.response.of.a.series.RLC.circuit.

50 V

10

20 i(0)

v(0)

+vc(0)a b

+

+

FIGURE 1.42 Series.RLC.circuit.at.t.=.0.

5 A

20

i(0+)

v(0+)

+ vc(0+)b

+

FIGURE 1.43 Series. RLC. circuit. at.t.=.0+.

100 mH20

i(t)

v(t)

1000 Fb

+

FIGURE 1.44 Series. RLC. circuit.at.t.=.0+.



From the examination of Equation 1.79, the Neper frequency, = 100 rad/s and the resonant fre-quency, o = 100 rad/s. By reviewing Table 1.4, since = o, this is a critically damped circuit. The characteristic equation is s2 + 200s + 10,000 and there is one repeated root, 100. The general form of the solution for the current through the inductor is given in Equation 1.80. Note that since there is no active source on the circuit after the switch moves to position b, the inductor is discharging (i(t ) = 0 A):

i t A t A i t A t At t t t( ) ( )1 2 1 2= + + = + e e e e 100 100 (1.80)

Once again, it is necessary to use the initial conditions to determine the values of A1 and A2. In order to do this, the equation in 1.80 and its first derivative must be evaluated at t = 0+. These equations are given in (1.81) and (1.82):

i A( )0 2+

= (1.81)

d

fundamentals of industrial electronics

Documents

dorfthe handbook of

sensors handbook

microwave handbook

filters handbook

applications handbook

dorfthe electronics

oklobdzijathe control

dorfdigital avionics