First Order RC and RL Transient Circuits

Transcription

1 Firs Order R and RL Transien ircuis

2 Objecives To inroduce he ransiens phenomena. To analyze sep and naural responses of firs order R circuis. To analyze sep and naural responses of firs order RL circuis. Professor: Yasser. G. Hegazy

3 Transiens In circuis wih inducors and capaciors volage and curren canno change insananeously. The applicaion or removal of sources or circui elemens creaes a ransien behavior. Transien is he process of going from one seady sae o anoher seady sae following a sudden change in he circui configuraion. Sudden changes are mainly due o swiching process or fauls.

4 FIRST ORDER IRUITS ircuis ha conain a single energy soring elemens. Eiher a capacior or an inducor. SEOND ORDER IRUITS ircuis wih wo independen energy soring elemens in any combinaion

5 Transien Analysis The circui is modeled in ime domain using differenial equaions. The order of he differenial equaion equals he number of independen energy soring elemens in he circuis. urrens and Volages of circuis wih jus one or One L can be obained using firs order differenial equaions.

6 Firs Order ODF The firs order ordinary differenial equaion in he form dy d Ay B y( ) 0 Y 0 Has a soluion Seady sae Transien par y() where, Y F ( Y 0 - Y F )e - T Y F B A, T 1 A

7 Iniial ondiions Iniial condiions are he values of he capaciors volage or he inducor curren a saring insan of he ransien period. = 0 - is he insan jus before swiching. = 0 + is he insan jus afer swiching. In apaciors V (0 - ) = V (0 + ). Where V is he capacior volage. In inducors i L (0 - ) = i L (0 + ). Where i L is he capacior volage.

8 Firs Order R circui A. Sep Response The response of he circui o sudden applicaion of an energy supply. V TH Then, R TH dv d R TH v =0 v TH + v c _ Since, For = 0 + o inf. V TH i c i R dv d R TH TH i R V dv 1 v 1 d R R c TH 0 v TH

9 Firs Order R circui o The following is he firs order differenial equaion describing he hange in he capacior volage during he ransien period. dv 1 v 1 d R R TH o The soluion of his equaion requires he iniial condiions of he capacior volage V(0). Noe ha V(0+) = V(0-) = Vo o The general soluion of his equaion is given by: TH v TH V () V F (V 0 - V F )e - Where, - V F is he final value of V - is he circui ime consan

10 Final Value of V The final value of V occurs when he capacior is fully charge i.e. he rae of change of V c = 0 R TH Using he differenial equaion, v F R Then, V F = V TH V F is usually refers o as V ( ) TH 1 R TH V F can be obained by replacing he capacior wih an open circui since i F = 0 v TH V TH = v cf _

11 Time consan The ime consan is a measure of how fas is he charging process of he apacior. Mahemaically i s he ime required o reach 63% of he final value. R TH

12 Example Find i O (), 0 6 k m F k For 0 i O v 6k 12 V + i O ( ) _ 6 k 0 + _ 4V Hence, if he capacior volage is known he problem is solved 1) Iniial ondiions V 2 ( 0 ) V (0 ) V V 12V 2) Time onsan 6k a b v i O () TH k 6k + _ 6k 4V R TH 6k 6k 3k R TH 3) Final Value v 6V 6 TH V F 0.3 4) Diff. equaion V i 0 () 6 (2-6)e () e ma

13 Example FIND i( ), 0 STEP 2: Deermine V ( ) + V F _ STEP1: Iniial v c volage acrosscapacior USE IRUIT IN STEADY STATE PRIOR TO THE SWITHING v ( 0) vc (0) 24V i 2mA 12k ( 0) 36V (2mA)(2k) 32[ V ] i( ) mA VF STEP 3:Time consan apaciive circui : R TH R TH 2k 6k 1. 5k 100m F ( V V i( ) 3 ( ) 6 )( () 27 (32-27)e e V F) 0. 15s STEP 4:The differeni al equaion

14 B. Naural Response of R circuis The response of he circui due o he energy sored in he capacior is known as naural response (no sources in he circui). =0 i For >0 KVL equaion around he circui yields, ir v R dv d 0 v 0 V i () 0 (V dv () d 0-0)e V R o - e -

15 Naural Response apacior Volage Time consan : R

16 Example In he circui shown in he figure, he swich opens a = 0. Find he numerical expression for i().

17 Soluion Before = 0, he circui has reached seady sae so ha he capacior acs like an open circui. The circui is equivalen o ha shown in Fig. (a) afer ransforming he volage source. Iniial ondiions

18 Time onsan

19 Soluion Final Value = 0 KL yields

20 Firs Order R-L circuis 0 ) (0 ) 0 ( i L i L The differenial equaion is Solving he equaion yields A = 0 he swich closes, The iniial condiions L R s L L R S S L R LF L LF e V d di L v e R V R V i I I I i ) / ( ) / ( ) / ( ) ( ) ( ) (0 ) ( ) (0) ( ) ( The final condiions I LF = Vs/R

21 Firs Order R-L circuis sep response L = R EQ R-L Sep Response

22 Equivalen Resisance seen by an Inducor For he RL circui in he previous example, i was deermined ha = L/R. As wih he R circui, he value of R should acually be he equivalen (or Thevenin) resisance seen by he inducor. In general, a firs-order RL circui has he following ime consan: L = R EQ R = R EQ seen from he erminals of he inducor for > 0 wih independen sources killed ircui L ircui > 0 independen R EQ sources killed

23 Naural Response of R-L ircui When he swich is closed (ON) he inducor will sore he energy. As a resul, he inducor is said o be charged. When he swich is opened (OFF), his will resul in he insananeous change in he circui. The inducor will supply he energy sored o he resisive.

24 Naural Response of R-L ircui The circui when he swich is opened Using he Kirchhoff s volage law (KVL), o find he differenial equaion of he loop: L ir 0 By using differenial and inegral equaions echnique: di d i( ) I(0) e / v L ( ) RI (0) e / where = L/R is a ime consan.

25 Naural Response of R-L ircui The curren generaed agains ime shows he inducor loses is energy exponenially i( ) I(0) e /

26 Example Assuming ha i(0) = 10 A, calculae i() and i x () in he circui. The equivalen resisance is he same as he Thevenin resisance a he inducor erminals. Because of he dependen source, we inser a volage source wih v o = 1 V

27 Soluion Since here is no energy source in he circui, he final value of he inducor curren is zero i.e. I LF = 0 The ime consan = L/R = 1.5 sec. The naural response equaion yields i( ) i( ) I(0) e 10e / /1.5 The inducor volage v v L L di 1 ( ) L 0.5( ) I(0) e d 10 /1.5 ( ) e 3 I x ()= V L ()/2 /

28 The swich in he circui shown below has been closed for a long ime. I opens a = 0. Find i() for > 0. Example Answer: i( ) 2 e 10 28

29 For he circui, find i() for > Answer: i() = 2e 2 A