Chapter 7 Response of First-order RL and RC Circuits

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1 Chaper 7 Response of Firs-order RL and RC Circuis 7.- The Naural Response of RL and RC Circuis 7.3 The Sep Response of RL and RC Circuis 7.4 A General Soluion for Sep and Naural Responses 7.5 Sequenial Swiching 7.6 Unbounded Response

2 Overview Ch9- discuss seady-sae response of linear circuis o sinusoidal sources. The mah reamen is he same as he dc response excep for inroducing phasors and impedances in he algebraic equaions. From now on, we will discuss ransien response of linear circuis o sep sources (Ch7-8 and general ime-varying sources (Ch-3. The mah reamen involves wih differenial equaions and Laplace ransform.

3 Firs-order circui A circui ha can be simplified o a Thévenin (or Noron equivalen conneced o eiher a single equivalen inducor or capacior. In Ch7, he source is eiher none (naural response or sep source. 3

4 Key poins Why an RC or RL circui is charged or discharged as an exponenial funcion of ime? Why he charging and discharging speed of an RC or RL circui is deermined by RC or L/R? Wha could happen when an energy-soring elemen (C or L is conneced o a circui wih dependen source? 4

5 Secion 7., 7. The Naural Response of RL and RC Circuis. Differenial equaion & soluion of a discharging RL circui. Time consan 3. Discharging RC circui 5

6 Wha is naural response? I describes he discharging of inducors or capaciors via a circui of no dependen source. No exernal source is involved, hus ermed as naural response. The effec will vanish as. The inerval wihin which he naural response maers depends on he elemen parameers. 6

7 Circui model of a discharging RL circui Consider he following circui model: For <, he inducor L is shor and carries a curren I s, while R and R carry no curren. For >, he inducor curren decreases and he energy is dissipaed via R. 7

8 Ordinary differenial equaion (ODE, iniial condiion (IC For >, he circui reduces o: IC depends on iniial energy of he inducor: i( i( I I s By KVL, we go a firs-order ODE for i(: d L i( Ri(. d where L, R are independen of boh i, and. 8

9 Solving he loop curren d ODE : L i( Ri(, IC : i( I I s; d di R L( di Ri( d, d, i L i( i( di i ln i( i( R L ln i( I e d, ln i i( ln I i( i( R L, R L L, where ime consan R, 9

10 Time consan describes he discharging speed The loop curren i( will drop o e - (37% of is iniial value I wihin one ime consan. I will be <.I afer elapsing 5. If i( is approximaed by a linear funcion, i will vanish in one ime consan..37i

11 Soluions of he volage, power, and energy. for,, for, ( ( e RI Ri v abrup change a =. The volage across R (or L is:. for, ( ( Re I R i p τ The insananeous power dissipaed in R is:. for,, ( LI w e w d e R I d p w τ τ The energy dissipaed in he resisor R is: iniial energy sored in L

12 Example 7.: Discharge of parallel inducors ( Q: Find i (, i (, i 3 (, and he energies w, w sored in L, L in seady sae (. For <: ( L, L are shor, and ( no curren flows hrough any of he 4 resisors, i ( 8 A, i ( 4 A, i 3 (, w ( (5 H(8 A 6 J, w ( ((4 6 J.

13 Example 7.: Solving he equivalen RL circui ( For >, swich is open, he iniial energy sored in he inducors is dissipaed via he 4 resisors. The equivalen circui becomes: I =(8+4= A L eq =(5H//H=4 H v( - = R eq = 8 The soluions o i(, v( are: L R eq eq s, i( Ie e v( Ri( 96e V. A, 3

14 Example 7.: Solving he inducor currens (3 The wo inducor currens i (, i ( can be calculaed by v(: 96e V i ( i ( L v( d e d.6 9.6e A. i ( i ( L v( d 4 96e d.6.4e A. 4

15 Example 7.: Soluions in seady sae (4 Since i i (.6 9.6e (.6.4e.6 A,.6 A, he wo inducors form a closed curren loop! The energies sored in he wo inducors are: w w ( (5 H(.6 A 6.4 J, ( (( J, which is ~% of he iniial energy in oal. 5

16 Example 7.: Solving he resisor curren (5 By curren division, i 3 ( =.6i 4 (, while i 4 ( can be calculaed by v(: i 4 i 4 ( 4 v( (5 // 96e 9.6e A. i 3 ( 3 5 i 4 ( 5.76e A. 6

17 Circui model of a discharging RC circui Consider he following circui model: For <, C is open and biased by a volage V g, while R and R carry no curren. For >, he capacior volage decreases and he energy is dissipaed via R. 7

18 ODE & IC For >, he circui reduces o: IC depends on iniial energy of he capacior: v( v( V V g By KCL, we go a firs-order ODE for v(: d v( C v(. d R where C, R are independen of boh v, and. 8

19 Solving he parallel volage d v( ODE : C v(, IC : v( V Vg ; d R v( V e, where RC ime consan Reducing R (loss and parasiic C is criical for high- speed circuis. 9

20 Soluions of he curren, power, and energy. for,, for, ( ( e R V R v i abrup change a =. The loop curren is:. for, ( ( e R V R v p τ The insananeous power dissipaed in R is:. for,, ( CV w e w d p w τ The energy dissipaed in he resisor R is: iniial energy sored in C

21 Procedures o ge naural response of RL, RC circuis. Find he equivalen circui.. Find he iniial condiions: iniial curren I hrough he equivalen inducor, or iniial volage V across he equivalen capacior. 3. Find he ime consan of he circui by he values of he equivalen R, L, C: L R, or RC; 4. Direcly wrie down he soluions:, v( V e. i( Ie

22 Secion 7.3 The Sep Response of RL and RC Circuis. Charging an RC circui. Charging an RL circui

23 Wha is sep response? The response of a circui o he sudden applicaion of a consan volage or curren source, describing he charging behavior of he circui. Sep (charging response and naural (discharging response show how he signal in a digial circui swiches beween Low and High wih ime. 3

24 ODE and IC of a charging RC circui IC depends on iniial energy of he capacior: v( v( V Derive he governing ODE by KCL: I v( R where C d d v(, dv d v, s V f RC, V f are he ime consan and final (seady-sae parallel volage. I s R 4

25 5 Solving he parallel volage, branch currens. ( f f e V V V v. for, ( (, ( ( R v i e R V I v d d C i R s C, (, ( ln,,, ( f f f f v V s f f e V V V v V V V v d R I v dv d V v dv V v d dv The charging and discharging processes have he same speed (same ime consan =RC. The branch currens hrough C and R are:

26 Example 7.6 ( Q: Find v o (, i o ( for. For <, he swich is conneced o Terminal for long, he capacior is an open circui: 6 k v o ( (4 V 3 V, i o (. ( 6 k 6

27 Example 7.6 ( A +, he charging circui wih wo erminals and G can be reduced o a Noron equivalen: G v o ( V v o 3 V ( + v I s.5 ma 7

28 Example 7.6 (3 The ime consan and he final capacior volage of he charging circui are: RC ( 4 k(.5 F ms, V f I R s (.5 ma(4 k 6 V. v o ( V f V V e 6 9e V. f i o ( I V R e.5e ma. s A he ime of swiching, he capacior volage is coninuous: ( 3 V v( ( 6 9 V, v o while he curren i o jumps from o -.5 ma. 8

29 Charging an RL circui d Vs Ri( L i(, i( I f d L Vs where, I f. R R IC depends on iniial energy of he inducor: i( i( I I I e, The charging and discharging processes have he same speed (same ime consan =L/R. f 9

30 Secion 7.5 Sequenial Swiching 3

31 Wha is and how o solve sequenial swiching? Sequenial swiching means swiching occurs n ( imes. I is analyzed by dividing he process ino n+ ime inervals. Each of hem corresponds o a specific circui. Iniial condiions of a paricular inerval are deermined from he soluion of he preceding inerval. Inducive currens and capaciive volages are paricularly imporan for hey canno change abruply. Laplace ransform (Ch can solve i easily. 3

32 Example 7.: Charging and discharging a capacior ( Q: v (=? for. =5 ms = IC: V v( For 5 ms, he 4-V source charges he capacior via he -k resisor, V f = 4 V, = RC = ms, and he capacior volage is: v ( V f ( V V f e 4 4e V. 3

33 Example 7. ( For >5 ms, he capacior is disconneced from he 4-V source and is discharged via he 5- k resisor, V = v (5 ms = 3.75 V, V f =, = RC = 5 ms. The capacior volage is: v ( V e 3.75e ( (.5 V. 33

34 Secion 7.6 Unbounded Response 34

35 Definiion and reason of unbounded response An unbounded response means he volages or currens increase wih ime wihou limi. I occurs when he Thévenin resisance is negaive (R Th <, which is possible when he firs-order circui conains dependen sources. 35

36 Example 7.3: ( Q: v o (=? for. For >, he capacior seems o discharge (no really, o be discussed via a circui wih a curren-conrolled curren source, which can be represened by a Thévenin equivalen. 36

37 Example 7.3 ( Since here is no independen source, V Th =, while R Th can be deermined by he es source mehod: vt it k R v Th T vt vt 7 k k i -5 k. T, 37

38 Example 7.3 (3 For, he equivalen circui and governing differenial equaion become: V vo( V, v o ( V e e 4 5 ms RC, V...grow wihou limi. 38

39 Why he volage is unbounded? Since -k, -k resisors are in parallel, i k =i, he capacior is acually charged (no discharged by a curren of 4i! Charging effec will increase v o, which will in urn increase he charging curren (i = v o / k and v o iself. The posiive feedback makes v o soaring. 6i i 39

40 Lesson for circui designers & device fabricaion engineers Undesired inerconnecion beween a capacior and a sub-circui wih dependen source (e.g. ransisor could be caasrophic! 4

41 Key poins Why an RC or RL circui is charged or discharged as an exponenial funcion of ime? Why he charging and discharging speed of an RC or RL circui is deermined by RC or L/R? Wha could happen when an energy-soring elemen (C or L is conneced o a circui wih dependen source? 4

First Order RC and RL Transient Circuits

First Order RC and RL Transient Circuits Firs Order R and RL Transien ircuis 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.