EEE5: CI CUI T THEOY CHAPTE 6: FIST-ODE CICUITS 6. Inroducion This chaper considers L and C circuis. Applying he Kirshoff s law o C and L circuis produces differenial equaions. The differenial equaions resuling from analyzing he C and L circuis are of he firs order. Hence, he circuis are known as firs-order circuis. A firs-order circui is characerized by a firs-order differenial equaion. Two ways o excie he firs-order circui: (i) source-free circui The energy is iniially sored in he capaciie of inducie elemens. The energy couses he curren o flow in he circui and gradually dissipaed in he resisors. (ii) Exciing by independen sources 6. The Source-Free C Circui A source-free C circui occurs when is dc source is suddenly disconneced. 8
EEE5: CI CUI T THEOY The energy already sored in he capacior is released o he resisors. Consider he circui in Figure 6.: Figure 6. Assume olage () across he capacior. Since he capacior is iniially charged, a ime, he iniial olage is ( ) V wih he corresponding of he energy sored as w ( ) CV applying KCL a he op node of he circui, i C By definiion, + i d i C C and d i. 9
EEE5: CI CUI T THEOY Thus, or C d d d d + + C This is he firs-order differenial equaion. earrange he equaion, d C Inergraing boh sides, d ln + ln C where ln A is he inegraion consan. Thus, ln A C Taking powers of e produces, ( ) Ae Bu from he iniial condiion, Thus, C A ( ) A V C ( ) V e (6.) 3
EEE5: CI CUI T THEOY This shows ha he olage response of he C circui is an exponenial decay of he iniial olage. Since he response is due o he iniial energy sored and he physical characerisics of he circui and no due o some exernal olage or curren source i is called he naural response of he circui. The olage response of he C circui: Figure 6. As increases, he olage decreases oward zero. The rapidiy wih which he olage decreases is expressed in erms of he ime consan, τ. The ime consan of a circui is he ime required for he response o decay by a facor of /e or 36.8% of is iniial alue. This implies a τ, τ C ( ) Ve Ve. 368V 3
EEE5: CI CUI T THEOY or τ C (6.) The olage is less han % afer 5 ime consan he circui reaches is final sae or saedy sae. The curren i () is gien by i ( ) V τ ( ) e (6.3) The power dissipaed in he resisor is p( ) V τ i e (6.4) The energy absorbed by he resisor up o ime is w ( ) pd τv e V e τ τ τ w ( ) CV ( e ) (6.5) 3
EEE5: CI CUI T THEOY Noe: as, w ( ) CV w ( C, which is he same as ), he energy iniially sored in he capacior. The key o working wih a source-free C circui: (i) Find he inial olage capacior. (ii) Find he ime consan τ. ( ) V across he (iii) Obain he capacior olage C ( ) ( ) () e /τ τ C Noe: In finding he ime consan, is ofen he Theenin equialen resisance a he erminals of he capacior (ake ou he capacior and find TH a is erminal). Example : In Figure 6.3, le ( C ) 5V. Find c x for >., and i x Figure 6.3 33
EEE5: CI CUI T THEOY Change he circui o he sandard C circui as shown in Figure 6.4 Thus, eq Figure 6.4 (8 + ) 5 4Ω τ C ( 4)(.) eq () e C / τ 5e 5e.5 V Using olage diision, /.4.4 V s Finally, x i.6(5e ) 9e + 8.5.5 V e x.5 x.75 A 34
EEE5: CI CUI T THEOY Example : The swich in he circui in Figure 6.5 has been closed for a long ime and i is opened a. Find Calculae he iniial energy sored in () for. he capacior. Figure 6.5 For <, he swich is closed; he capacior is an open circui o dc. Figure 6.6 Using olage diision, 9 ( ) () 5 9 + 3 C V, <. 35
EEE5: CI CUI T THEOY Since he olage across a capacior acnno change insananeously, he olage across he capacior a is he same a, or ( ) V 5 C V For >, Thus, eq Figure 6.7 + 9 Ω τ eq C ( ) C ()( () e / τ 5e 3 ) /. The iniial energy sored in he capacior is w () C C C () ()( 3 )(5 V ). s.5 J 36
EEE5: CI CUI T THEOY 6.3 The Source-Free L Circui Consider he circui in Figure 6.8: Figure 6.8 Goal o deermine he curren i () hrough he inducor. Why we selec he inducor curren as he response? The inducor curren canno change insananeously. A, we assume ha he inducor has an iniial curren I or i ( ) I. Energy sored in he inducor, w ( ) LI 37
EEE5: CI CUI T THEOY Applying KVL, Bu, Thus, L + di L L and i d di L + i d di + i d L earranging erms and inergraing gies, i( ) I di i ln i( ) ln i( ) ln I Taking he powers of e, L d ln i i( ) I L L I + L i( ) / L Ie (6.6) Thus, he naural (curren) response of he L circui is as shown: 38
EEE5: CI CUI T THEOY Figure 6.9 From Equaion 6.6, he ime consan is L τ (6.7) The olage across he resisor, ( ) /τ i I e (6.8) The power dissipaed in he resisor is, p /τ i I e (6.9) 39
EEE5: CI CUI T THEOY The energy absorbed by he resisor is, Noe: as w ( ) pd τi I e e τ τ τ w ( ) LI ( e ) (6.), w ( ) LI w L ( d, which is he same as ), he energy iniially sored in he inducor. The key o working wih a source-free C circui: (i) Find he inial curren capacior. () Find he ime consan τ. i ( ) I across he (i) Obain he capacior olage i L ( ) i( ) i() e /τ Noe: is ofen he Theenin equialen resisance a he erminals of he inducor (ake ou he inducor and find TH a is erminal). 4
EEE5: CI CUI T THEOY Example : Assuming ha i( ) A, calculae i() and i x () in he circui in Figure 6.. Figure 6. Two Mehods o find he i(), Mehod The equialen resisance is he same as he Theenin resisance a he inducor erminals. Exisance of dependen source inser a olage source wih V a he inducor erminals. Figure 6. 4
EEE5: CI CUI T THEOY Applying KVL, Loop : Loop : Thus, Hence, Thus, ( i i 6i i i i 5 6 i ) + i 3i i 3A and i i 3 eq τ i( ) L eq TH 3 i() e s i 3 Ω e A /τ /τ A 4
EEE5: CI CUI T THEOY Mehod : Consider he foloowing circui: Applying KVL, Loop : Loop : Thus, di d 6i i di d di d di i + ( i + 4( i i 5 6 i Figure 6. i i 3i + i 3 d 3 ) ) 43
EEE5: CI CUI T THEOY Since i, di i Inegraing gies, ln i i( ) i( ) ln i() 3 i() d Taking he power of e, i( ) i() e 3 3 ( 3) ( 3) e A which is he same as Mehod The olage across he inducor is, di L d (.5)() e 3 ( 3) V Thus, e 3 i x ( ) ( 3) V ( 3).667e A for > 44
EEE5: CI CUI T THEOY Example : In he circui shown in Figure 6.3, find i, and i for all ime, assuming ha he swich was open for a long ime. Figure 6.3 For <, he swich is opened he inducor acs like a shor circui o dc, From Figure 6.4, i A. Figure 6.4 45
EEE5: CI CUI T THEOY i ( ) A for < + 3 ( ) 3i( ) 6 V for < i( ) A Thus, For >, he swich is closed he olage source is shor-circuied. Figure 6.5 A he inducor erminal, Hence, TH τ i( ) ( 3 6 Ω L eq s /τ () ) i e e di L L d ( e ) A for > 4e V for > 46
EEE5: CI CUI T THEOY i ( ) Thus, for all ime: L e 6 3 A for > A e 3 6V 4e V A e A i ( ) A ( ) i( ) < > < > < Figure 6.6 47
EEE5: CI CUI T THEOY 6.4 Singulariy Funcions Definiion: Singulariy funcions are funcions ha eiher are diconinuous or hae disconinuous deriaies. Uni sep funcion: Figure 6.7 ( ) < > u (6.) 48
EEE5: CI CUI T THEOY If u () is delayed by seconds: If () u ( Figure 6.8 ) u is adanced by seconds: < > u ( + Figure 6.9 ) < > 49
EEE5: CI CUI T THEOY The sep funcion can be used o represen an abrup change in olage or curren. For example, if he olage is represened by, ( ) V can be expressed as < > ( ) Vu( ) u( The olage source of V ) and is equialen circui: Figure 6. The curren source of I u( ) and is equialen circui: Figure 6. 5
EEE5: CI CUI T THEOY Uni impulse funcion: The deriaie of he uni sep funcion u () - he uni impulse funcion δ (). < d δ ( ) u( ) Undefined (6.) d > Figure 6. Also known as dela funcion. The uni impulse funcion δ () is zero eerywhere excep a, where i is undefined. Impulsie currens or olages occur in elecric circuis as a resul of swiching operaions or impulsie sources. I may be isualized as a ery shor duraion pulse of uni area. 5
EEE5: CI CUI T THEOY In mahemaical form: + δ ( ) d where denoes he ime jus before + denoes he ime jus afer, and (refer o uniy) denoes he uni area. The uni area is he srengh of he impulse funcion. The effec of he impulse funcion o oher funcions: Le us ealuae he inegral where b a a < < b. f ) δ ( ) d ( Since δ ( ) excep, he inegrand is zero excep a. Thus, b a f ( ) δ ( ) d b a f ( f ( ) ) δ ( b a δ ( ) d ) d f ( *when a funcion is inegraed wih he impulse funcion, we obain he alue of he funcion a he poin where he impulse occur. ) 5
EEE5: CI CUI T THEOY Uni ramp funcion: Inegraing he uni sep funcion u () resuls in he uni ramp funcion r(). In mahemaical form: or r( ) u( ) d u( ), ( ), r (6.3) Figure 6.3 r is delayed by seconds: If () r ( ) < > 53
EEE5: CI CUI T THEOY Figure 6.4 If r () is adanced by seconds: r ( + ) Figure 6.5 + < > 54
EEE5: CI CUI T THEOY Example: Express he olage pulse in Figure 6.6 in erms of he uni sep. Calculae is deriaie and skech i. Figure 6.6 The pulse consiss of he sum of wo uni sep funcions Figure 6.7 Thus, ( ) u( ) u( 5) [ u( ) u( 5) ] 55
EEE5: CI CUI T THEOY Is deriaie: The pulse: d d [ δ ( ) δ ( 5) ] Figure 6.8 56
EEE5: CI CUI T THEOY 6.5 Sep esponse of an C Circui When he dc source is suddenly applied, he olage or curren can be modeled as a sep funcion. Known as a sep response The sep response of a circui is is behaior when he exciaion is he sep funcion, which may be a olage or a curren source. Consider he circui in Figure 6.9 (a) and (b): Figure 6.9 57
EEE5: CI CUI T THEOY Assume he iniial olage olage V a he capacior. Since he olage of a capacior canno change insanenously, Applying KVL, + ( ) ( ) V d C d d + d VSu( ) + VS u( ) C C where is he olage across he capacior. For >, d VS + d C C d VS d C d d V C S Inegraing boh sides and inroducing he iniial condiions, ln( V S ) ( ) V C 58
EEE5: CI CUI T THEOY Thus, ln S S + C VS ln V V C ( ( ) V ) ln( V V ) ( ) V S S V / τ + ( V VS ) e, > V ( ) / S + ( V VS ) e,, < > τ (6.4) Equaion 6.4 is known as he complee response (or oal response) of he C circui o a sudden applicaion of a dc olage source, assuming he capacior is iniially charged. Assuming ha V S > V, a plo of () is, Figure 6.3 59
EEE5: CI CUI T THEOY If he capacior is uncharged iniially, V or ( ) V, ( ) / τ S ( e V S ( e / τ ), ) u( ) he complee sep response of he C circui when he capcior is iniially uncharged. The curren is obained using < > d i ( ) C from () d equaion. () has wo componens wo ways o decompose he componens. Firs way naural response and forced response. Complee response Naural esponse (sored energy) + Forced esponse (independen source) or where and n + f n V e /τ 6
EEE5: CI CUI T THEOY f V S ( e /τ n is as discussed in Secion 6.. is known as he forced response because i is f produced by he circui when an exernal force (a olage or curren source) is applied. Second way ransien response and seady-sae response. or where and Complee esponse Transien esponse (emporary par) + Seady-sae esponse (permanen par) + ss ( V ss V S VS ) e ) /τ The ransien response is emporary he porion of he complee response ha decays o zero as ime approaches infiniy. The ransien response is he circui s emporary response ha will die ou wih ime. 6
EEE5: CI CUI T THEOY The seady-sae response ss is he porion of he complee response ha remains afer he ransien response has died ou. The seady-sae response is he behaior of he circui a long ime afer an exernal exciaion is applied. The complee response may be wrien as ( ) ( ) + / τ [ () ( ) ] e (6.5) where () is he iniial olage a is he final or seady-sae alue. + and ( ) To find he sep response of an C circui (i) Find he iniial capacior olage, (). - obain from he gien circui for <. (ii) Find he final capacior olage, ( ). - obain from he gien circui for >. (iii) The ime consan, τ. - obain from he gien circui for >. 6
EEE5: CI CUI T THEOY Example : The swich in Figure 6.3 has been in posiion A for a long ime. A, he swich moes o B. deermine () for >. Figure 6.3 For <, The capacior acs like an open circui o dc, bu is he same as he olage across he 5 kω resisor. Hence, he olage across he capacior jus before is obained by olage diision as 5 ( ) (4) 5 5 + 3 Since he capacior olage canno change insananeously, V + ( ) ( ) ( ) 5 V 63
EEE5: CI CUI T THEOY For >, TH 4 kω τ TH C 4 3.5 Since he capacior acs like an open circui o dc a seady sae, ( ) 3V. Thus, ( ) ( ) ( ) + [ () ( ) ] 3 + (5 3) e 3 5e.5 V / τ e 3 / τ s Example : The swich in Figure 6.3 is closed a. Find i () and () for all ime. Figure 6.3 64
EEE5: CI CUI T THEOY, < u ( ), >, i( ). 5 ( + e ) A, V,.5 ( + e ) V, < > < > 65
EEE5: CI CUI T THEOY 6.6 Sep esponse of an L Circui Consider he circui in Figure 6.33, Figure 6.33 Goal find he inducor curren i as he circui response. Decompose he response ino naural and forced curren: i i n + i f 66
EEE5: CI CUI T THEOY where in Ae /τ, τ A is he consan o be deermined The naural response dies ou afer fie ime consans he inducor becomes a shor circui and he olage across i is zero. The enire source olage V S appears across. Thus, he forced response is i f i VS Ae / τ VS + To find A, le I be he iniial curren hrough he inducor. Since he curren hrough canno change insananeously, Thus, a, + i ( ) i( ) I I A VS A + VS I L 67
EEE5: CI CUI T THEOY Thus, i( ) VS + I VS e /τ Figure 6.34 Or, he response can be wrien as, i( ) i( ) + / τ [ i() i( ) ] e (6.6) To find he sep response of an L circui (i) Find he iniial inducor curren, i (). - obain from he gien circui for <. () Find he final inducor curren, i ( ). - obain from he gien circui for >. (i) The ime consan, τ. - obain from he gien circui for >. 68
EEE5: CI CUI T THEOY Example : Find () i in he circui in Figure 6.35 for >. Assume he circui has been closed for a long ime. Figure 6.35 When <, The 3 Ω is shor-circuied. The inducor acs like a shor circui. i( ) 5 Since he inducor curren canno change insananeously, When >, The swich is open. A + i( ) i( ) i( ) 5 A 69
EEE5: CI CUI T THEOY Thus, i( ) + 3 + 3 5Ω TH τ L TH i( ) 3 5 i( ) + A 5 s [ i() i( ) ] + (5 ) e 5 e / τ + 3e 5 A 7