电路基础chapter3

1Chapter3MethodsofAnalysis3.1Motivation3.2Nodalanalysis.3.3Nodalanalysiswithvoltagesources.3.4Loop(Mesh)analysis.3.5Loop(Mesh)analysiswithcurrentsources.3.6Nodalandmeshanalysisbyinspection.3.7Nodalversusmeshanalysis.3.8Applications:DCTransistorCircuits3.9Summary2Whatarethethingswhichweneedtoknowinordertodeterminetheanswers?3.1Motivation(1)Ifyouaregiventhefollowingcircuit,howcanwedetermine(1)thevoltageacrosseachresistor,(2)currentthrougheachresistor.(3)powergeneratedbyeachcurrentsource,etc.3Thingsweneedtoknowinsolvinganyresistivecircuitwithcurrentandvoltagesourcesonly:Howshouldweapplytheselawstodeterminetheanswers?•Kirchhoff’sCurrentLaws(KCL)•Kirchhoff’sVoltageLaws(KVL)•Ohm’sLaw3.1Motivation(2)4Example13.2NodalAnalysis(1)Itprovidesageneralprocedureforanalyzingcircuitsusingnodevoltagesasthecircuitvariables.353.2NodalAnalysis(2)Stepstodeterminethenodevoltages:1.Selectanodeasthereferencenode.2.Assignvoltagesv1,v2,…,vn-1totheremainingn-1nodes.Thevoltagesarereferencedwithrespecttothereferencenode.3.ApplyKCLtoeachofthen-1non-referencenodes.UseOhm’slawtoexpressthebranchcurrentsintermsofnodevoltages.4.Solvetheresultingsimultaneousequationstoobtaintheunknownnodevoltages.6ConsiderthecircuitinFig.3.1(a)Fig.3.17Selectnode0asreferencenode,whilenodes1and2areassignedvoltagesv1andv2,respectively.ApplyingKCLtoeachnonreferencenodeinthecircuit.Forconvenience,weredrawFig.3.1(a)inFig.3.1(b),wherewenowlabeli1,i2andi3asthecurrentsthroughresistorsR1,R2andR3,respectively.Atnode1,applyingKCLgives1212IIii(3.2.1)Atnode2223Iii(3.2.2)8ApplyingOhm’slawtoexpresstheunknowncurrentsi1,i2,andi3intermsofnodevoltage.Thekeyideatobearinmindisthat,bythepassivesignconvention,currentmustalwaysflowfromahigherpotentialtolowerpotential.Withthisinmind,weobtainfromFig.3.2.1(b),111111vioriGvR12222122()vvioriGvvR232323vioriGvR(3.2.3)SubstitutingEq(3.2.3)intoEqs(3.2.1)and(3.2.2)andrearrangingterms,weobtain121221222322()()GGvGvIIGGGvI(3.2.4)9mutualconductance(互电导)ConsiderthecircuitinFig.3.21.SelectnodeDasreferencenode2.Togeneratethenode-voltageequations,weapplyKCLandOhm’slaw3.Thenode-voltageequationderivedatnodeAisFig.3.2011664RuRuRsAsCABA66116464111)111(RuRuRRRRRssCBAself-conductance(自电导)5522554241)111(1RuRuRRRRRssCBAAtnodeBandCweobtain556665356)111(11RuRuRRRRRssCBAThealgebraicsumofthevaluesofallcurrentsourcesconnectedtothenode(与节点相连的电流源电流的代数和)10Self-conductanceisequaltothesumofallconductancesatthenode.Mutualconductanceisequaltothesumofconductancesdirectlyconnectingtwonodes.Thevaluesofthosesourceswhosecurrentflowstowardsthenodearetakenpositivewhile,intheoppositecase,theyaretakenasnegative.11Review:Stepsinnode-voltageanalysisStep1:Onenodeofthecircuitisdefinedasthereferencenode.Usuallychoosethenodetowhichmostbranchesareconnectedasthereferencenode.Step2:FortheremainingN-1nodes,afternumberthem,thenodevoltageswithrespecttothereferencenodearedefined.Step3:Thenode-voltageequationsarewrittenintheformasinthepreviouscase.Step4:Theequationsaresolvedandthenodevoltageareknown.Step5:Thevoltageofallbranchesarecalculatedcombiningthenodevoltageandasaconsequencethecurrentsofallcircuitelementsareknown.12observationWhenapplyingnodalvoltageanalysistosolveelectricanalysisproblems,forconveniencewecantransfertheseriescombinationofanindependentvoltagesourceandaresistorintotheparallelcombinationofanindependentcurrentsourceandaresistor.iiFig.3.3(a)Fig.3.3(b)AnequivalentconditionisssssviorvRiR13Example1.Findiinthefollowingcircuit.iSolution:Thereferencenodeisdefinedandtheremainingnodesarelabeled.ABCThenodeequationsareC=-4.21416VHencei=-0.527ASolvingforCgives24021101)21101101(CBA42081)8141101(101CBA240)818121(8121CBA14I1I3I2Example2.DetermineI1,I2andI3inthefollowingcircuit.Solution:Thereferencenodeisdefinedandtheremainingnodesarelabeled.Thenodeequationsare25.06.0BA17.02.25.0CBA27.07.0CBThusVA864.3VB615.0VC242.2SolvingcurrentyieldsI1=0.3864AI2=0.615AI3=1.4285AABC153.2NodalAnalysis(3)v1v2Example3–circuitindependentcurrentsourceonly3ApplyKCLatnode1and216SolutionApplyingKCLtonode1,weget1121026vvv(3.2.5)TheKCLequationnode2is2124067vvv(3.2.6)RearrangeEqs(3.2.5)and(3.2.6),weobtain12111()1266vv12111()4667vv(3.2.7)(3.2.8)Answer:v1=-2V,v2=-14V173.2NodalAnalysis(4)Example3–currentwithdependantcurrentsourceAnswerv1=4.8V,v2=2.4V,v3=-2.4V183.3NodalAnalysiswithVoltageSource(1)Approach1AvoltagesourceistheonlyelementbetweentwonodesABCDFirst,thereferencenodeisdefinedatthenegativeterminalofthevoltagesource.Then,thevoltageofitspositiveterminalisknown.Finally,othernode-voltageequationsarewrittenaspreviousway.11021)31211(CBA8B110)51411(41CBA19First,IntroducethecurrentofthevoltagesourcebecausewecannotexpressthecurrentwithnodeApproach2:1103121)31211(DBAiDA)5131(31Then,theequationsaresupplementedbythevoltageacrosstheconsideredvoltagesource.8DBABCDiThen,othernode-voltageequationsarewrittenaspreviousway.iBA)4121(2120Approach3: