25 200 - Today'sOutline 7 FirstOrderCircuits ECSE200 1 =...

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Unformatted text preview: Today'sOutline 7. FirstOrderCircuits responsetoaconstantinput ECSE200 1 responsetoaconstantinput ConsideranRLcircuitbeingswitchedatt=0betweentwodifferentopen circuitvoltages.Assumesteadystatehasbeenreachedfort<0. t=0ms V1 + vR(t) - R iL(t) + - + - V2 L ECSE200 2 responsetoaconstantinput ConsiderfirstthesteadystatecondiJonsfort<0. t<0 + vR(0-) - V1 + - KVL: 0 = -V1 + vR (0-) + vL (0-) = -V1 + iL (0-)R + 0 V1 R + 0 L + - V2 iL(0-) iL (0-) = R - di Steadystate: vL = L L = 0 dt ECSE200 3 responsetoaconstantinput ConsiderthecircuitequaJonsfort>0. t>0 KVL: + vR(t) - V1 + - 0 = -V2 + vR + vL = -V2 + R iL + L diL dt R + vL(t) L + - V2 iL(t) - V2 R + iL = L dt L diL t >0 conJnuityofinductorcurrent: iL (0+) = iL (0-) = R V1 WehaveafirstorderlineardifferenJal equaJonwithiniJalcondiJons. ECSE200 4 responsetoaconstantinput Considersteadystateast>. t> KVL: 0 = -V2 + vR () + vL () = -V2 + R iL () + 0 V2 + vR() - V1 + - R + 0 L iL () = R + - V2 iL() - Thiscanalsobeconcludedfrom thecircuitequaJonfort>0: diL dt t V2 R + iL = L L ( ) ( ) ( ) di Steadystate: vL = L L = 0 dt ECSE200 0 V2 R + iL = L L V iL = 2 R 5 responsetoaconstantinput SolvethedifferenJalequaJon. diL R V + iL = 2 t >0 L dt L Recall: dx + kx = G dt iL (0+) = R t TheformofthesoluJonis: il (t) = c1 + c2 exp - L/R V1 x(t) = c1 + c2 exp(-kt) t UseouriniJalandfinalcondiJons: iL () = lim c1 + c2 exp - = c1 t RC V2 c1 = iL () = R iL (0+) = c1 + c2 exp 0 = c1 + c2 () c2 = iL (0+) - c1 = ECSE200 V1 -V2 R 6 t=0ms V1 + - + vR(t) - R responsetoaconstantinput Theresponseisagainasumofanatural/transient responseandaforced/steadystateresponse. iL(t) + - V2 L iL(t) iL(0+)=V1/R iniJalsteadystate V1 -V2 t soluJonforiL(t)fort>0: iL (t) = + exp - R R L/R V2 V1 -V2 t exp - natural/transientresponse L/R R iL()=V2/R finalsteadystate forced/steadystateresponse t *WeassumeV1>V2inthisgraph. ECSE200 7 responsetoaconstantinput TheJmeconstantofanRLcircuitisthuseasilyidenJfied: t t inatural (t) = i exp - = i exp - L/R iL(t) iL(0+)=V1/R =L/R Checkunits: H Vs/A = = s V/A 0.368i iL()=V2/R *WeassumeV1>V2inthisgraph. ECSE200 t 8 theJmeconstant Ques0on:TheJmeconstant=RCforanRCcircuitand=L/RforanRL circuit,suggesJngthatalargerRlengthensthetransientresponseforanRC circuit,whilealargerresistanceshortensthetransientresponseofanRL circuit.Why?Let'sconsidertwocircuitsconverJngstoredenergytoheat. A A +q -q R i + vC(t) + R v C B L B iL(t) - - v dq = -i = - c dt R dvC -i v = =- c C RC dt d = -v = -iLR dt diL -v iR i = =- L =- L L L L/R dt increasingRdecreases therateatwhichcharge separaJonisreduced ECSE200 increasingRincreases therateatwhichflux linkageisreduced 9 constantinput:generalprocedure step#1:FindtheiniJalvalueofthecircuitvariableofinterest,x(0+),using circuitanalysisandconJnuityofcapacitorvoltageorinductorcurrent. step#2:Findthefinalvalueofthevariableofinterest,x(),usingdcsteady statemodelsforthecapacitororinductor. step#3:FindtheThveninequivalentresistanceRTasseenfromtheterminals ofthecapacitororinductor.TheJmeconstant=RTCor=L/RT. step#4:ConstructthesoluJon. x(t) x(0+) x() t x(0+) - x() exp - x(t) = x() + t ECSE200 10 example1 Thecircuitisindcsteadystateatt=0-.Att=0,theswitchchangesposiJon. Findi(t)fort>0. t=0ms i(t) 12V + - 6k 100F 3k ECSE200 11 example1 step#1:Att<0,thecircuitisindcsteadystate,sothecapacitoractsasan openandthecapacitorvoltageisfoundbyvoltagedivision. 3k v(0-) = 12V = 4V 3k+6k i(0+) 12V + - 6k 0 100F + v(0-) 3k - ECSE200 12 example1 step#1:Att=0+theswitchhasjustmoved,andwefindthecurrenti(0+) usingcapacitorvoltageconJnuity. conJnuity: v(0+) = v(0-) = 4V v(0+) = 1.33mA Ohm'sLaw: i(0+) = 3k i(0+) 12V + - 6k 100F + v(0+) 3k - ECSE200 13 example1 step#2:Ast>,thecircuitagainreachesdcsteadystate.Wefindthe currenti(). v() v() nodeequaJon: 0 = +0+ 3k 6k v() = 0 v() i() = =0 3k i() 12V + - 6k 0 100F + v() 3k - ECSE200 14 example1 step#3:FindRTasseenfromthecapacitorterminals(idenJfiedAandB). FindtheJmeconstant=RTC. RT = 3k||6k= 36 k=2k 3+6 = RT C = 2k100F = 200ms A 12V + - 6k 100F B ECSE200 15 3k example1 step#4:AssemblethesoluJon. i() = 0mA i(0+) = 1.333mA = 200ms t i(0+) - i() exp - i(t) = i() + t i(t) = 1.333mA exp - 200ms i(t) 1.33mA 200ms i(t) t 12V + - t=0ms 6k 100F 3k ECSE200 16 example2 Thecircuitisindcsteadystateatt=0-.Att=0,theswitchchangesposiJon. Findv(t)fort>0. t=0ms + 1k 10mA v(t) 70mH 200k - ECSE200 17 example2 step#1:Att<0,thecircuitisindcsteadystate,sotheinductoractsasa short. 0 0 0= - 10mA + i(0-) + 1k 200k i(0-) = 10mA t=0ms + 1k 10mA 0 i(0-) 70mH 200k - ECSE200 18 example2 step#1:Att=0+,theswitchmoves,andwefindv(0+)usinginductor currentconJnuity. conJnuity: i(0+) = i(0-) = 10mA v(0+) nodeeqn.: 0 = i(0+) + 200k v(0+) = -i(0+)200k = -2kV t=0+ + 1k 10mA v(0+) i(0+) 70mH 200k - ECSE200 19 example2 step#2:Ast->,thecircuitapproachesdcsteadystatesotheinductor actsasashort. di v() = L dt =0 t + 1k 10mA 0 i() 70mH 200k - ECSE200 20 example2 step#3:FindRTasseenfromtheinductorterminals(idenJfiedAandB).Find theJmeconstant=L/RT. RT = 200k = L 70mH = = 0.35s = 350ns RT 200k t>0 A 1k 10mA 70mH 200k B ECSE200 21 example2 step#4:AssemblethesoluJon. v() = 0mA v(0+) = -2kV = 350ns t v(t) = v() + v(0+) - v() exp - t v(t) = -2kV exp - 350ns v(t) 350ns t -2kV 1k 10mA t=0ms v(t) + 70mH 200k - ECSE200 22 ...
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This note was uploaded on 11/23/2010 for the course ECSE ECSE 200 taught by Professor Stevemcfee during the Fall '09 term at McGill.

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