Chapter 7. Response of FirstOrder RL and RC Circuits


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1 Chaper 7. esponse of FirsOrder L and C Circuis 7.1. The Naural esponse of an L Circui 7.2. The Naural esponse of an C Circui 7.3. The ep esponse of L and C Circuis 7.4. A General oluion for ep and Naural esponses 7.5. equenial wiching 7.6. Unbounded esponse 7.7. The Inegraing Amplifier 1
2 7.3 The ep esponse of L and C Circuis Finding he currens and volages in firsorder L or C circuis when eiher dc volage or curren sources are suddenly applied. The ep esponse of an L Circui The circui is shown in Fig Energy sored in he inducor a he ime he swich is closed is given in erms of a nonzero iniial curren i(). The ask is o find he expressions for he curren in he circui and for he volage across he inducor afer he swich has been closed. We derive he differenial equaion ha describes he circui and we solve he equaion. 2
3 Afer he swich has been closed, Kirchhoff s volage law requires ha di = i + L (7.29) d which can be solved for he curren by separaing he variables i and, hen inegraing. The firs sep is o solve Eq. (7.29) for di d : di d = i + = i (7.3) L L Nex, we muliply boh sides by d. di i = d (7.31) L We now separae he variables in Eq. (7.32) o ge di = d i ( / ) L and hen inegrae boh sides. Using variables for he inegraion, we obain x and (7.32) y as 3
4 = ) ( ) / ( i I dy L x dx (7.33) where I is he curren a = and ) ( i is he curren a any >. Therefore L I i = ) / ( ) / ( ) ( ln (7.34) from which L e I i ) / ( ) / ( ) / ( ) ( = or L e I i ) / ( ) ( + = (7.35) When he iniial energy in he inducor is zero, I is zero. Thus eq. (7.35) reduces o 4
5 i( ) ( / L) = e (7.36) Eq. (7.36) indicaes ha afer he swich has been closed, he curren increases exponenially from zero o a final value /. The ime consan of he circui, L /, deermines he rae of increase. One ime consan afer he swich has been closed, he curren will have reached approximaely 63% of is final value, or i( τ ) = e (7.37) If he curren were o coninue o increase a is iniial rae, i would reach is final value a = τ ; ha is because di 1 / τ / τ = e = e (7.38) d τ L he iniial rae a which i () increases is di d ( ) = (7.39) L 5
6 If he curren were o coninue o increase a his rae, he expression for i would be i = (7.4) L from which, a = τ, i L = = (7.41) L Equaions (7.36) and (7.4) are ploed in Fig The values given by Eqs. (7.37) and (7.41) are also shown in his figure. The volage across an inducor is + (7.35), for, L di d, so from Eq. v L ( / L) ( I ) e ( / L) = L I e = (7.42) The volage across he inducor is zero before he swich is closed. Eq. (7.42) indicaes ha he inducor volage jumps o I a he insan he swich is closed and hen decays exponenially o zero. 6
7 Does he value of v a + = makes sense? Because he iniial curren is I and he inducor prevens an insananeous change in curren, he curren is I in he insan afer he swich has been closed. The volage drop across he resisor is I, and he volage impressed across he inducor is he source volage minus he volage drop, ha is, I. When he iniial inducor curren is zero, Eq. (7.42) simplifies o v ( / L) = e (7.43) If he iniial curren is zero, he volage across he inducor jumps o. We also expec he inducor volage o approach zero as increases, because he curren in he circui is approaching he consan value of. 7
8 Fig shows he plo of Eq. (7.43) and he relaionship beween he ime consan and he iniial rae a which he inducor volage is decreasing. If here is an iniial curren in he inducor, Eq. (7.35) gives he soluion for i. The algebraic sign of I is posiive if he iniial curren is in he same direcion as ; oherwise, I carries a negaive sign. i Example 7.5 The swich shown in Fig has been in posiion a long ime. A =, he swich moves from a o b. The swich is a makebeforebreak ype; so, here is no inerrupion of curren hrough he inducor. a) Find he expression i () for b) Wha is he iniial volage across he inducor jus afer he swich has been moved o posiion b? c) Does he iniial volage make sense in erms of circui behavior? d) How many milliseconds afer he swich has been moved does he inducor volage equal 24? e) Plo boh i () and v () versus. 8
9 We can also describe he volage v () across he inducor direcly, no jus in erms of he circui curren. We begin by noing ha he volage across he resisor is he difference beween he source volage and he inducor volage. We wrie v( ) i( ) = (7.44) where is a consan. Differeniaing boh sides wih respec o ime yields di d dv = 1 (7.45) d Muliply each side of Eq. (7.45) by he inducance L. v L dv = (7.46) d Puing Eq. (7.46) ino sandard form yields 9
10 dv d + v L = (7.47) erify ha he soluion o Eq. (7.47) is idenical o ha given in Eq. (7.42). v L ( / L) ( I ) e ( / L) = L I e = (7.42) A his poin, a general observaion abou he sep response of an L circui is perinen. When we derived he differenial equaion for he inducor curren, we obained Eq. (7.29). We now di rewrie Eq. (7.29) = i + L as d di + i = (7.48) d L L Observe ha Eqs. (7.47) and (7.48) have he same form. pecifically, each equaes he sum of he firs derivaive of he variable and a consan imes he variable o a consan value. 1
11 In (7.47), he consan on he righhand side is zero; hence his equaion akes on he same form as he naural response equaions. In boh (7.47) and (7.48), he consan muliplying he dependen variable is he reciprocal of he ime consan, 1 ha is, =. L τ We encouner a similar siuaion in he derivaions for he sep response of an C circui. 11
12 The ep esponse of an C Circui We can find he sep response of a firsorder C circui by analyzing he circui shown in Fig For mahemaical convenience, we choose he Noron equivalen of he nework conneced o he equivalen capacior. umming he currens away from he op node in Fig generaes he differenial equaion Division by C gives dv d v + I (7.49) C C C = 12
13 dv d vc I + (7.5) C C C = Comparing Eq. (7.5) wih Eq. (7.48) di d + i = (7.48) L L reveals ha he form of he soluion for v C is he same as ha for he curren in he inducive circui, namely, Eq. (7.35). ( / L) i( ) = + I e (7.35) Therefore, by simply subsiuing he appropriae variables and coefficiens, we can wrie he soluion for v C direcly. The ranslaion requires ha I replace C replace L 1 replace replace I. We ge / C ( I ) e, vc = I + (7.51) 13
14 A similar derivaion for he curren in he capacior yields he differenial equaion di d 1 + i C = (7.52) Eq. (7.52) has he same form as Eq. (7.47) dv + v = (7.47) d L hence he soluion for i is obained by using he same ranslaions used for he soluion of Eq. (7.5). Thus i = I + / C e, (7.53) where is he iniial value for v C, he volage across he capacior. Le s see if he soluions for he C circui make sense in erms of known circui behavior. 14
15 From Eq. (7.51), noe ha he iniial volage across he capacior is, he final volage across he capacior is, and he ime consan of he circui is C. I Also noe ha he soluion for v C is valid for. These observaions are consisen wih he behavior of a capacior in parallel wih a resisor when driven by a consan curren source. Equaion (7.53) predics ha he curren in he capacior + a = is I. This predicion makes sense because he capacior volage canno change insananeously, and herefore he iniial curren in he resisor is. The capacior branch curren changes insananeously + from zero a = o I a =. The capacior curren is zero a =. Also noe ha he final value of v = I. Example 7.6 The swich in he circui shown in Fig has been in posiion 1 for a long ime. A =, he swich moves o posiion 2. Find a) v ( ) for + b) i ( ) for 15
16 7.4 A General oluion for ep and Naural esponse The general approach o finding eiher he naural response of he sep response of he firsorder L and C circuis shown in Fig is based on heir differenial equaions being he same. To generalize he soluion of hese four possible circuis, we le x () represen he unknown quaniy, giving x () four possible values. I can represen he curren or volage a he erminals of an inducor or he curren or volage a he erminals of a capacior. From he previous eqs. (7.47), (7.48), (7.5), and (7.52), we know ha he differenial equaion describing any one of he four circuis in Fig. (7.24) akes he form dx x x + = K (7.54) τ where he value of he consan K can be zero. 16
17 Because he sources in he circui are consan volages and/or currens, he final value of x will be consan; ha is, he final value mus saisfy (7.54), and, when x reaches is final value, he derivaive dx d mus be zero. Hence x f = Kτ (7.55) where x f represens he final value of he variable. We solve (7.54) by separaing he variables, beginning by solving for he firs derivaive: dx d ( x Kτ ) ( x x f ) x = + K = = (7.56) τ τ τ In wriing (7.56), we used (7.55) o subsiue K τ. We now muliply boh sides of (7.56) by divide by x x f o obain dx 1 = d x x τ f x f d for and (7.57) Inegrae (7.57). To obain as general a soluion as possible, we use ime as he lower limi and as he upper limi. 17
18 Time corresponds o he ime of he swiching or oher change. Previously we assumed ha =, bu his change allows he swiching o ake place a any ime. Using u and v as symbols of inegraion, we ge x( ) du x 1 = x( ) u f τ dv (7.58) Carrying ou he inegraion called for in (7.58) gives 18
19 x( ) f ( ) / τ [ x( ) X ] e = x + (7.59) f The significance of his equaion is he unknown variable as a = funcion of ime he final value of he variable he iniial [ ime of swiching] ime consan + value of he value of he variable he final variable e (7.6) In many cases, he ime of swiching   is zero. 19
20 When compuing he sep and naural response of circuis, follow hese seps: 1. Idenify he variable of ineres for he circui. For C circuis, i is mos convenien o choose he capaciive volage; for L circuis, i is bes o choose he inducive curren. 2. Deermine he iniial value of he variable, which is is value a. Noe ha if we choose capaciive volage or inducive curren as variable of ineres, i is no necessary o disinguish beween + = and =. This is because hey boh are coninuous variables. If we choose anoher variable, we need o remember ha is iniial + value is defined a. = 3. Calculae he final value of he variable, which is he value as. 4. Calculae he ime consan for he circui. Wih hese quaniies we can use Eq. (7.6) o produce an equaion describing he variable of ineres as a funcion of ime. 2
21 21
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