Transient Analysis of First Order RC and RL circuits

Transcription

1 Transien Analysis of Firs Order and iruis The irui shown on Figure 1 wih he swih open is haraerized by a pariular operaing ondiion. Sine he swih is open, no urren flows in he irui (i=0) and v=0. The volage aross he apaior, v, is no known and mus be defined. I ould be ha v=0 or ha he apaior has been harged o a erain volage v = V0. i v v Figure 1 e us assume he nonrivial iniial equilibrium or iniial seady sae ondiion for he apaior volage v = V 0 and le s lose he swih a ime = 0, resuling in he irui shown on Figure 2. i =0 v v Figure 2 Afer losing he swih, urren will begin o flow in he irui. Energy will be dissipaed in he resisor and evenually all energy iniially sored in he apaior, 1 2 E = v, will be dissipaed as hea in he resisor. Afer a long ime, he urren will 2 be zero and he irui will reah a new, albei rivial, equilibrium or seady sae ondiion (i=0, v=0, v=0). The ransien haraerisis of he irui desribes he behavior of he irui during he ransiion from one seady sae ondiion o anoher. In his lass we will develop he ools for desribing and undersanding his ransien phenomena / Spring 2006, hanioakis and ory 1

2 Soure Free irui As ou firs example le s onsider he soure free irui shown on Figure 3. i =0 v v Figure 3 e s assume ha iniially he ideal apaior is harged wih a volage v V. = = 0 A ime = 0, he swih is losed, urren begins o flow in he irui and we would like o obain he form of he volage v as a funion of ime for >0. Sine he volage aross he apaior mus be oninuous he volage a = 0 is also Vo. Our firs ask is o deermine he equaion ha desribes he behavior of his irui. This is aomplished by using Kirhhoff s laws. Here we use KV whih gives, v () v () = 0 (0.1) Using he urren volage relaionship of he resisor and he apaior, Equaion (0.1) beomes dv () v () = 0 (0.2) d Noe ha he produ has he uni of ime. (Ohm)(Farad) seonds is alled he ime onsan of he irui and i is ofen assigned he variable τ =. Equaion (0.2) along wih he iniial ondiion, v = 0 = V0 desribe he behavior of he irui for >0. In fa, sine he irui is no driven by any soure he behavior is also alled he naural response of he irui. Equaion (0.2) is a firs order homogeneous differenial equaion and is soluion may be easily deermined by separaing he variables and inegraing. However we will employ a more general approah ha will also help us o solve he equaions of more ompliaed iruis laer on. Assume ha a soluion o Equaion (0.2) is of he form given by s v() = Ae (0.3) / Spring 2006, hanioakis and ory 2

3 The parameers A and s are o be deermined by he speifi haraerisis of he sysem. By subsiuing equaion (0.3) ino equaion (0.2) we obain, s s Ase Ae = 0 (0.4) Or equivalenly, s s 1 Ae = 0 (0.5) ( ) The only nonrivial soluion of Equaion (0.5) follows from ( s 1) = 0 (0.6) This is alled he haraerisi equaion of he sysem. Therefore s is And he soluion is 1 s = (0.7) v() = Ae = Ae τ (0.8) The onsan A may now be deermined by applying he iniial ondiion v = 0 = V 0 whih gives A= V 0 (0.9) And he final soluion is v() = V e (0.10) The plo of he volage v is shown on Figure 4. A =0 he volage sars a and subsequenly i exponenially deays o zero. V 0 0 Figure / Spring 2006, hanioakis and ory 3

4 Soure Free irui Now le s onsider he irui shown on Figure 5. v a b =0 i v Figure 5 Iniially he swih is a posiion a and here is a urren I 0 irulaing in he loop onaining he ideal induor. This is he iniial equilibrium sae of he irui and is shemai is shown on Figure 6(a). A ime =0 he swih is moved from posiion a o posiion b. Now he resisor is inorporaed in he irui and he urren I 0 begins o flow hrough i as shown Figure 6(b). Io i() b a v v v (a) (b) Figure 6 Our goal is o deermine he form of he urren i(). We sar by deriving he equaion ha desribes he behavior of he irui for >0. KV around he mesh of he irui on Figure 6(b) gives. v () v () = 0 (0.11) Using he urren volage relaionship of he resisor and he induor, Equaion (0.11) beomes di () i () = 0 (0.12) d 6.071/ Spring 2006, hanioakis and ory 4

5 The raio has he unis of ime as an be seen by simple dimensional analysis. By assuming a soluion of he form, s i () = Be (0.13) Equaion (0.12) beomes afer subsiuion s 1 Be s = 0 (0.14) The nonrivial soluion of Equaion (0.14) is s = (0.15) And he soluion beomes / i () = Be (0.16) The onsan B may now be deermined by onsidering he iniial ondiion of he irui i = 0 = I 0, whih gives B = I0. And he ompleed soluion is 0 / i () = Ie (0.17) The raio is he haraerisi ime onsan of he irui. Figure 7 shows he normalized plo of i(). Figure / Spring 2006, hanioakis and ory 5

6 and iruis wih muliple resisors. The apaior of he irui on Figure 8 is iniially harged o a volage Vo. A ime =0 he swih is losed and urren flows in he irui. The apaior sees a Thevenin equivalen resisane whih is eq = ( 2 3) (0.18) 2 =0 i 3 1 v Figure 8 Therefore one he swih is losed, he equivalen irui beomes eq v The haraerisi ime is now given by And he evoluion of he volage v is Figure 9 τ = eq (0.19) eq () V0e v = (0.20) 6.071/ Spring 2006, hanioakis and ory 6

7 irui wih muliple resisors and induors. e s onsider he irui shown on Figure 10 whih onains muliple induors and resisors. Iniially he swih is losed and has been losed for a long ime. A ime =0 he swih opens and we would like o obain he ransien behavior of he irui for >0. In pariular we are ineresed in deermining he urren i() as indiaed on he irui of Figure 11 4 Io 0.5 < Figure 10 4 i() 0.5 > Figure 11 In order o find he iniial ( = 0 ) urren flowing in he irui we onsider he irui on Figure 10. The irui may be simplified by ombining he resisors and aking ino aoun he operaional haraerisis of he induor a equilibrium. Sine under D ondiions he induors a as shor iruis he orresponding irui beomes 4 Io 0.5 <0 2 2 Figure / Spring 2006, hanioakis and ory 7

8 And hus he urren Io = 2. Therefore a he momen ha he swih is opened, he urren is known. This is he iniial ondiion for our problem. Afer he swih is opened, he irui beomes 4 i() Figure 13 By ombining he resisors and he induors he irui redues o 2 5/3 i() Figure 14 Wih he iniial ondiion for he urren i = 0 = I0 = 2 he soluion for he urren i() beomes 6 5 i () = 2 e (0.21) For his example we have been able o ombine he induanes ino an equivalen induane and hus derive he firs order differenial equaion for he behavior of he irui. However, his reduion is no possible in general. If afer he reduion more han one reaive elemen remains in he irui he order of he sysem differenial equaion is equal o he number of reaive elemens. We will disuss he ransien behavior of hese higher order sysems nex lass / Spring 2006, hanioakis and ory 8

9 Fored esponse of iruis For he irui shown on Figure 15 he swih is losed a =0. This orresponds o a sep funion for he soure volage as shown on Figure 16.We would like o obain he apaior volage v as a funion of ime. The volage aross he apaior a =0 (he iniial volage) is Vo. i =0 v v Figure 15 Figure 16 The equaion ha desribes he sysem is obained by applying KV around he mesh. Whih by using he urrenvolage relaionships beomes v() v () = (0.22) dv () v () = (0.23) d By seing τ =, he ime onsan of he irui, Equaion (0.23) beomes dv () τ v () = (0.24) d The soluion of his equaion is he ombinaion (superposiion) of he homogeneous soluion v ( h ) and he pariular soluion vp (). v = vh v p (0.25) 6.071/ Spring 2006, hanioakis and ory 9

10 The homogeneous soluion saisfies he equaion And he pariular soluion he equaion dvh() τ vh() = 0 (0.26) d dvp () τ vp () = (0.27) d The homogeneous equaion orresponds o he soure free problem already invesigaed and is soluion is v () = Ae τ (0.28) h The onsan A is undefined a his poin bu any value will saisfy he differenial equaion. The pariular soluion is found by inspeion o be And hus he oal soluion beomes vp = (0.29) v () = Ae τ (0.30) The onsan A may now be deermined by onsidering he iniial ondiion of he apaior volage. The iniial apaior volage is Vo and hus A=Vo. And he omplee soluion is ( ) v () = Vo e τ (0.31) Figure 17 shows he plo of v() for Vo=1 Vol, =5 Vol as a funion of he normalized quaniy / τ. Noe ha afer 5 ime onsans he volage v is wihin 99% of he volage / Spring 2006, hanioakis and ory 10

11 Figure 17 Now le s onsider he irui shown on Figure 18. The swih has been a posiion a for a log ime and hus here is no volage aross he apaior plaes a ime =0. A ime =0 he wih is moved from posiion a o posiion b where i says for ime 1 and subsequenly reurned o posiion a. This swih aion orresponds o he reangular pulse shown on Figure 19. b a v v Figure Figure / Spring 2006, hanioakis and ory 11

12 We would like o obain he volage v(). Firs we know ha he iniial ondiion is v = 0 = 0. We also know ha afer a long ime (>>1) he volage will go bak o zero. The soluion of he sysem will ell us he evoluion of he volage v from ime =0 o =1 and for >1. The soluion for 1>>0 is v () = (1 e τ ) (0.32) For >1 he soluion is deermined by onsidering as he iniial ondiion, he volage aross he apaior a =1. 1 v (1) = (1 e τ ) (0.33) And he soluion for >1 is 1 τ τ v () = (1 e ) e (0.34) Figure 20 shows he omplee evoluion of he volage v where we have aken 1=2τ. Figure / Spring 2006, hanioakis and ory 12

13 Proedure for ransien analysis of and iruis. 1. Deermine he equivalen induane/apaiane (, ) 2. Deermine he Thevenin equivalen resisane, eq, seen by ( eq, eq ) eq 3. The haraerisi ime is now known τ = eqeq or τ = 4. alulae he iniial value for he volage/urren flowing in he irui a. apaior as as an open irui under d ondiions i. For a ransiion happening a = 0, v( = 0 ) = v( = 0 ) b. Induor as as a shor irui under d ondiions i. For a ransiion happening a = 0, i ( = 0 ) = i ( = 0 ) 5. Esimae he value of v, i as (final value) 6. The omplee soluion is: soluion = final value [ iniial value final value] e τ eq eq eq v() = v( ) v v ( 0 ) ( ) e τ = i() = i( ) i i ( 0 ) ( ) e τ = 6.071/ Spring 2006, hanioakis and ory 13

14 The operaion of he irui shown on Figure 21 is similar o he one disussed above. The swih has been a posiion a for a log ime and hus here is no volage aross he apaior plaes a ime =0. A ime =0 he wih is moved from posiion a o posiion b where i says for ime 1 and subsequenly reurned o posiion a. This swih aion reaes he reangular pulse shown on Figure b 2 v2 a v1 v Figure 21 The irui shown on has wo ime onsans. For 0<<1 he ime onsan is τ 1 = ( 1 2). For >1 he ime onsan is τ 2= 2. The soluion now is τ1 ( ) = (1 ) For 1 v e 1 τ1 τ2 v ( ) = (1 e ) e For > 1 (0.35) The plo of v() is shown on Figure 22 for 1 = 2, τ1 = 2( τ 2) and 1= 2( τ1) Time onsan τ 1 1 Time onsan τ 2 = ( τ1) 2 Figure / Spring 2006, hanioakis and ory 14

15 Problems. The fuse elemen is a resisor of resisane f whih is desroyed when he urren hrough i exeeds a erain value. The swih in he irui has been in he losed posiion for a long ime. A ime =0 he swih is opened. If he maximum urren ha an flow hrough he fuse is Im, alulae he minimum resisane of he fuse (f) as a funion of Im and he oher irui parameers. 2 =0 1 f Deermine 1 and 2 so ha v = 2 Vols and v ( = o ) ( = 1 ms) = 1 Vols 2 v Is 1 A =0 1 50mH 6.071/ Spring 2006, hanioakis and ory 15