CAPACITANCE AND INDUCTANCE

CHAPTER 6 CAPACITANCE AND INDUCTANCE THE LEARNING GOALS FOR THIS CHAPTER ARE: Know how o use circui models for inducors and capaciors o calculae volage, curren, and power Be able o calculae sored energy for capaciors and inducors Undersand he conceps of coninuiy of curren for an inducor and coninuiy of volage for a capacior Be able o calculae volages and currens for capaciors and inducors in elecric circuis wih dc sources AAirpor Scanners Couresy of UPI/Brian Kersey/NewsCom To be searched or no o be searched is never he quesion. Air ravelers demand securiy in he skies and oday s echnology makes i possible wih jus a 5o3 second body scan insead of an inrusive padown ha can ake wo o four minues. Over 99% of airline passengers in major airpors across he naion choose o use body scanners when faced wih he opion. Scanners can spo plasic and ceramic weapons and explosives ha evade meal deecors and could evenually replace meal deecors a he naion s 2, airpor checkpoins. Mos ravelers say hey welcome any measure ha enhances safey, even if i means giving up some privacy. Today s new body scanners depend on millimeer wave echnology or backscaer xray echnology. The firs produces an image ha resembles a fuzzy phoo negaive; he second a Know how o combine capaciors and inducors in series and parallel chalk eching. Millimeer wave echnology emis, imes less radio frequency han a cell phone. Backscaer echnology uses highenergy xrays as i moves hrough clohing and oher maerials. In boh cases, images used for securiy are no reained bu desroyed immediaely. This chaper inroduces wo new circui elemens: capaciors and inducors ha sore energy in elecric and magneic fields. Volage and curren relaionships for hese componens do no follow Ohm s law bu insead connec volages and currens o heir derivaives and inegrals. Capaciors and inducors are cenral o he sudy of alernaing curren circuis. They are also key componens in making body scanners work. The circui designs behind effecive fullbody scanners help make everyone safer in he skies. Wha a feeling. 245

246 CHAPTER 6 CAPACITANCE AND INDUCTANCE 6. Capaciors A capacior is a circui elemen ha consiss of wo conducing surfaces separaed by a nonconducing, or dielecric, maerial. A simplified capacior and is elecrical symbol are shown in Fig. 6.. There are many differen kinds of capaciors, and hey are caegorized by he ype of dielecric maerial used beween he conducing plaes. Alhough any good insulaor can serve as a dielecric, each ype has characerisics ha make i more suiable for paricular applicaions. For general applicaions in elecronic circuis (e.g., coupling beween sages of amplificaion), he dielecric maerial may be paper impregnaed wih oil or wax, mylar, polysyrene, mica, glass, or ceramic. Ceramic dielecric capaciors consruced of barium ianaes have a large capacianceovolume raio because of heir high dielecric consan. Mica, glass, and ceramic dielecric capaciors will operae saisfacorily a high frequencies. Aluminum elecrolyic capaciors, which consis of a pair of aluminum plaes separaed by a moisened borax pase elecrolye, can provide high values of capaciance in small volumes. They are ypically used for filering, bypassing, and coupling, and in power supplies and moorsaring applicaions. Tanalum elecrolyic capaciors have lower losses and more sable characerisics han hose of aluminum elecrolyic capaciors. Fig. 6.2 shows a variey of ypical discree capaciors. In addiion o hese capaciors, which we deliberaely inser in a nework for specific applicaions, sray capaciance is presen any ime here is a difference in poenial beween wo conducing maerials separaed by a dielecric. Because his sray capaciance can cause Figure 6. A capacior and is elecrical symbol. [hin] A (a) Dielecric d v() i= dq (b) q() C Noe he use of he passive sign convenion. Figure 6.2 Some ypical capaciors. (Couresy of Mark Nelms and Jo Ann Loden)

SECTION 6. CAPACITORS 247 unwaned coupling beween circuis, exreme care mus be exercised in he layou of elecronic sysems on prined circui boards. Capaciance is measured in coulombs per vol or farads. The uni farad (F) is named afer Michael Faraday, a famous English physicis. Capaciors may be fixed or variable and ypically range from housands of microfarads ( F) o a few picofarads (pf). Capacior echnology, iniially driven by he modern ineres in elecric vehicles, is rapidly changing, however. For example, he capacior on he lef in he phoograph in Fig. 6.3 is a doublelayer capacior, which is raed a 2.5 V and F. An aluminum elecrolyic capacior, raed a 25 V and 68, F, is shown on he righ in his phoograph. The elecrolyic capacior can sore.5 * 6.8 * 2 * 25 2 = 2.25 joules (J). The Figure 6.3 doublelayer capacior can sore.5 * * 2.5 2 = 32.5 J. Le s connec en of he F capaciors in series for an equivalen 25V capacior. The energy sored in his equivalen capacior is 325 J. We would need o connec 47 elecrolyic capaciors in parallel o sore ha much energy. I is ineresing o calculae he dimensions of a simple equivalen capacior consising of wo parallel plaes each of area A, separaed by a disance d as shown in Fig. 6.. We learned in basic physics ha he capaciance of wo parallel plaes of area A, separaed by disance d, is C = e o A d where o, he permiiviy of free space, is 8.85 * 2 F/m. If we assume he plaes are separaed by a disance in air of he hickness of one shee of oilimpregnaed paper, which is abou.6 * 4 m, hen F = A8.85 * 2 BA.6 * 4 A =.48 * 9 m 2 and since square mile is equal o 2.59 * 6 square meers, he area is A 443 square miles which is he area of a mediumsized ciy! I would now seem ha he doublelayer capacior in he phoograph is much more impressive han i originally appeared. This capacior is acually consruced using a high surface area maerial such as powdered carbon which is adhered o a meal foil. There are lierally millions of pieces of carbon employed o obain he required surface area. Suppose now ha a source is conneced o he capacior shown in Fig. 6.; hen posiive charges will be ransferred o one plae and negaive charges o he oher. The charge on he capacior is proporional o he volage across i such ha q=cv 6. where C is he proporionaliy facor known as he capaciance of he elemen in farads. The charge differenial beween he plaes creaes an elecric field ha sores energy. Because of he presence of he dielecric, he conducion curren ha flows in he wires ha connec he capacior o he remainder of he circui canno flow inernally beween he plaes. However, via elecromagneic field heory i can be shown ha his conducion curren is equal o he displacemen curren ha flows beween he plaes of he capacior and is presen any ime ha an elecric field or volage varies wih ime. Our primary ineres is in he curren volage erminal characerisics of he capacior. Since he curren is i = dq A F doublelayer capacior and a 68, F elecrolyic capacior. (Couresy of Mark Nelms and Jo Ann Loden)

248 CHAPTER 6 CAPACITANCE AND INDUCTANCE hen for a capacior i = d (Cv) which for consan capaciance is i = C dv 6.2 Eq. (6.2) can be rewrien as dv = C i Now inegraing his expression from = q o some ime and assuming v( q)= yields v() = C 3 q i(x) dx 6.3 where v() indicaes he ime dependence of he volage. Eq. (6.3) can be expressed as wo inegrals, so ha v() = dx C 3qi(x) i(x) dx C 3 = v( ) i(x) dx C 3 6.4 where va B is he volage due o he charge ha accumulaes on he capacior from ime = q o ime =. The energy sored in he capacior can be derived from he power ha is delivered o he elemen. This power is given by he expression p() = v()i() = Cv() dv() 6.5 and hence he energy sored in he elecric field is w C () = Cv(x) dv(x) dx = C v(x) dv(x) dx 3 dx 3 dx q = C 3 v() v(q) q v(x) dv(x) = v() 2 Cv2 (x)2 v(q) = 2 Cv2 () J 6.6 since v(= q)=. The expression for he energy can also be wrien using Eq. (6.) as w C () = q 2 () 2 C 6.7 Eqs. (6.6) and (6.7) represen he energy sored by he capacior, which, in urn, is equal o he work done by he source o charge he capacior. Now le s consider he case of a dc volage applied across a capacior. From Eq. (6.2), we see ha he curren flowing hrough he capacior is direcly proporional o he ime rae of change of he volage across he capacior. A dc volage does no vary wih ime, so he curren flowing hrough he capacior is zero. We can say ha a capacior is an open circui o

SECTION 6. CAPACITORS 249 dc or blocks dc. Capaciors are ofen uilized o remove or filer ou an unwaned dc volage. In analyzing a circui conaining dc volage sources and capaciors, we can replace he capaciors wih an open circui and calculae volages and currens in he circui using our many analysis ools. Noe ha he power absorbed by a capacior, given by Eq. (6.5), is direcly proporional o he ime rae of change of he volage across he capacior. Wha if we had an insananeous change in he capacior volage? This would correspond o dv =qand infinie power. In Chaper, we ruled ou he possibiliy of any sources of infinie power. Since we only have finie power sources, he volage across a capacior canno change insananeously. This will be a paricularly helpful idea in he nex chaper when we encouner circuis conaining swiches. This idea of coninuiy of volage for a capacior ells us ha he volage across he capacior jus afer a swich moves is he same as he volage across he capacior jus before ha swich moves. The polariy of he volage across a capacior being charged is shown in Fig. 6.b. In he ideal case, he capacior will hold he charge for an indefinie period of ime, if he source is removed. If a some laer ime an energyabsorbing device (e.g., a flash bulb) is conneced across he capacior, a discharge curren will flow from he capacior and, herefore, he capacior will supply is sored energy o he device. If he charge accumulaed on wo parallel conducors charged o 2 V is 6 pc, wha is he capaciance of he parallel conducors? Using Eq. (6.), we find ha C = Q V = (6)A2 B 2 = 5 pf SOLUTION EXAMPLE 6. The volage across a 5 F capacior has he waveform shown in Fig. 6.4a. Deermine he curren waveform. Noe ha v() = = 24 3 6 * 6 ms 24 96 6 6 8 ms 3 2 * = 8 ms SOLUTION EXAMPLE 6.2 v() (V) 24 V i() (ma) 2 6 8 Figure 6.4 Volage and curren waveforms for a 5 F capacior. 6 8 6 (a) (b)

25 CHAPTER 6 CAPACITANCE AND INDUCTANCE Using Eq. (6.2), we find ha dv () i() = C = 5 * 6 A4 * 3 B 6 ms = 2 ma 6 ms i() = 5 * 6 A2 * 3 B 6 8 ms =6 ma 6 6 8 ms and i() = 8 ms Therefore, he curren waveform is as shown in Fig. 6.4b and i()= for >8 ms. EXAMPLE 6.3 SOLUTION Deermine he energy sored in he elecric field of he capacior in Example 6.2 a =6 ms. Using Eq. (6.6), we have w() = 2 Cv2 () A =6 ms, w(6 ms) = 2 A5 * 6 B(24) 2 = 44 J Learning Assessmen E6. A µf capacior has an accumulaed charge of 5 nc. Deermine he volage across he capacior. ANSWER:.5 V. EXAMPLE 6.4 SOLUTION The curren in an iniially uncharged 4 F capacior is shown in Fig. 6.5a. Le us derive he waveforms for he volage, power, and energy and compue he energy sored in he elecric field of he capacior a =2 ms. The equaions for he curren waveform in he specific ime inervals are Since v()=, he equaion for v() in he ime inerval 2 ms is and hence, i() = 6 * 6 2 * 3 2 ms v() = =8 * 6 2 ms 4 ms = 4 ms 6 8A 3 Bx dx = 3 2 (4)A 6 B 3 v(2 ms) = 3 A2 * 3 B 2 = 4 mv

SECTION 6. CAPACITORS 25 Volage (mv) Curren (A) 5 4 3.5 3 2.5 5 2 5 2 4.5.5 2.5 3 3.5 Time (ms).5.5.5.5 2 2.5 3 3.5 4 Time (ms) (a) (b) Energy (pj) Power (nw) 35 3 6 5 4 3 2 2 3.5.5 2 3.5 2.5 3 4 (c) Time (ms) 25 2 5 5.5.5 2 2.5 3 3.5 4 (d) Time (ms) In he ime inerval 2 ms 4 ms, v() = (4)A 6 B 3 =2 8 * 3 2A 3 B A8BA 6 Bdx A4BA 3 B The waveform for he volage is shown in Fig. 6.5b. Since he power is p()=v()i(), he expression for he power in he ime inerval 2 ms is p()=8 3. In he ime inerval 2 ms 4 ms, he equaion for he power is p() =(8)A 6 BA2 8 * 3 B Figure 6.5 Waveforms used in Example 6.4. = 6A 6 B 64A 9 B The power waveform is shown in Fig. 6.5c. Noe ha during he ime inerval 2 ms, he capacior is absorbing energy and during he inerval 2 ms 4 ms, i is delivering energy. The energy is given by he expression w() = 3 p(x) dx wa B

252 CHAPTER 6 CAPACITANCE AND INDUCTANCE In he ime inerval 2 ms, Hence, In he ime inerval 2 4 ms, w() = 3 w() = 8x 3 dx = 2 4 3 w(2 ms)=32 pj 2 * 3 CA6 * 6 Bx A64 * 9 BD dx 32 * 2 = CA8 * 6 Bx 2 A64 * 9 BxD 3 2 * 32 * 2 = A8 * 6 B 2 A64 * 9 B 28 * 2 From his expression we find ha w(2 ms)=32 pj and w(4 ms)=. The energy waveform is shown in Fig. 6.5d. Learning Assessmens E6.2 The volage across a 2 F capacior is shown in Fig. E6.2. Deermine he waveform for he capacior curren. ANSWER: v() (V) i() (ma) 2 2 Figure E6.2 2 3 4 5 6 6 6 2 3 4 5 E6.3 Compue he energy sored in he elecric field of he capacior in Learning Assessmen E6.2 a =2 ms. ANSWER: w=44 J. E6.4 The volage across a 5 F capacior is shown in Fig. E6.4. Find he waveform for he curren in he capacior. How much energy is sored in he capacior a =4 ms. v() (V) 5 2 3 4 5 6 7 8 9 5 Figure E6.4

SECTION 6. CAPACITORS 253 ANSWER: 25 J. i() (ma) 25 2 3 4 5 6 7 8 9 25 E6.5 The waveform for he curren in a nf capacior is Fig. E6.5. If he capacior has an iniial volage of 5V, deermine he waveform for he capacior volage. How much energy is sored in he capacior a =6 ms? i() ( A) 2 3 4 5 6 7 8 9 Figure E6.5 ANSWER: 32.5 nj. v() (V) 5 5 2 3 4 5 6 7 8 9 5 35

254 CHAPTER 6 CAPACITANCE AND INDUCTANCE 6.2 Inducors An inducor is a circui elemen ha consiss of a conducing wire usually in he form of a coil. Two ypical inducors and heir elecrical symbol are shown in Fig. 6.6. Inducors are ypically caegorized by he ype of core on which hey are wound. For example, he core maerial may be air or any nonmagneic maerial, iron, or ferrie. Inducors made wih air or nonmagneic maerials are widely used in radio, elevision, and filer circuis. Ironcore inducors are used in elecrical power supplies and filers. Ferriecore inducors are widely used in highfrequency applicaions. Noe ha in conras o he magneic core ha confines he flux, as shown in Fig. 6.6b, he flux lines for nonmagneic inducors exend beyond he inducor iself, as illusraed in Fig. 6.6a. Like sray capaciance, sray inducance can resul from any elemen carrying curren surrounded by flux linkages. Fig. 6.7 shows a variey of ypical inducors. From a hisorical sandpoin, developmens ha led o he mahemaical model we employ o represen he inducor are as follows. I was firs shown ha a currencarrying conducor would produce a magneic field. I was laer found ha he magneic field and he curren ha produced i were linearly relaed. Finally, i was shown ha a changing magneic field produced a volage ha was proporional o he ime rae of change of he curren ha produced he magneic field; ha is, v() = L di() 6.8 The consan of proporionaliy L is called he inducance and is measured in he uni henry, named afer he American invenor Joseph Henry, who discovered he relaionship. As seen in Eq. (6.8), henry (H) is dimensionally equal o volsecond per ampere. Following he developmen of he mahemaical equaions for he capacior, we find ha he expression for he curren in an inducor is i() = L 3 q v(x) dx 6.9 Flux lines i() Flux lines i() v() v() L i() (a) Figure 6.6 Two inducors and heir elecrical symbol (b) (c) Figure 6.7 Some ypical inducors. (Couresy of Mark Nelms and Jo Ann Loden)

SECTION 6.2 INDUCTORS 255 which can also be wrien as i() = ia B v(x) dx L 3 6. The power delivered o he inducor can be used o derive he energy sored in he elemen. This power is equal o p() = v()i() = c L di() d i() 6. Therefore, he energy sored in he magneic field is w L () = 3 Following he developmen of Eq. (6.6), we obain q c L di(x) d i(x) dx dx w L () = 2 Li2 () J 6.2 Now le s consider he case of a dc curren flowing hrough an inducor. From Eq. (6.8), we see ha he volage across he inducor is direcly proporional o he ime rae of change of he curren flowing hrough he inducor. A dc curren does no vary wih ime, so he volage across he inducor is zero. We can say ha an inducor is a shor circui o dc. In analyzing a circui conaining dc sources and inducors, we can replace any inducors wih shor circuis and calculae volages and currens in he circui using our many analysis ools. Noe from Eq. (6.) ha an insananeous change in inducor curren would require infinie power. Since we don have any infinie power sources, he curren flowing hrough an inducor canno change insananeously. This will be a paricularly helpful idea in he nex chaper when we encouner circuis conaining swiches. This idea of coninuiy of curren for an inducor ells us ha he curren flowing hrough an inducor jus afer a swich moves is he same as he curren flowing hrough an inducor jus before ha swich moves. Find he oal energy sored in he circui of Fig. 6.8a. 6 L =2 mh 3 L 2 =4 mh EXAMPLE 6.5 C 9 V ± =2 F C 2 =5 F 6 3 A (a) 6 I L A 3 I L2 9 V ± V C 3 A V C2 6 Figure 6.8 Circuis used in Example 6.5. (b)

256 CHAPTER 6 CAPACITANCE AND INDUCTANCE SOLUTION This circui has only dc sources. Based on our earlier discussions abou capaciors and inducors and consan sources, we can replace he capaciors wih open circuis and he inducors wih shor circuis. The resuling circui is shown in Fig. 6.8b. This resisive circui can now be solved using any of he echniques we have learned in earlier chapers. If we apply KCL a node A, we ge I L2 = I L 3 Applying KVL around he ouside of he circui yields 6I L 3I L2 6I L2 = 9 Solving hese equaions yields I L =.2 A and I L2 =.8 A. The volages V C and V C2 can be calculaed from he currens: V C =6I L 9 = 6.2 V V C2 = 6I L2 = 6(.8) =.8 V The oal energy sored in he circui is he sum of he energy sored in he wo inducors and wo capaciors: The oal sored energy is 3.46 mj. w L = 2 A2 * 3 B(.2) 2 =.44 mj w L2 = 2 A4 * 3 B(.8) 2 = 6.48 mj w C = 2 A2 * 6 B(6.2) 2 = 2.62 mj w C2 = 2 A5 * 6 B(.8) 2 = 2.92 mj The inducor, like he resisor and capacior, is a passive elemen. The polariy of he volage across he inducor is shown in Fig. 6.6. Pracical inducors ypically range from a few microhenrys o ens of henrys. From a circui design sandpoin i is imporan o noe ha inducors canno be easily fabricaed on an inegraed circui chip, and herefore chip designs ypically employ only acive elecronic devices, resisors, and capaciors ha can be easily fabricaed in microcircui form. EXAMPLE 6.6 SOLUTION i() (ma) 2 The curren in a mh inducor has he waveform shown in Fig. 6.9a. Deermine he volage waveform. Using Eq. (6.8) and noing ha and i() = 2 * 3 2 * 3 2 ms i() = 2 * 3 2 * 3 4 * 3 2 4 ms v() (mv) i()= 4ms< Figure 6.9 Curren and volage waveforms for a mh inducor. 2 4 (a) 2 4 (b)

SECTION 6.2 INDUCTORS 257 we find ha and v() = A * 3 B = mv v() = A * 3 B = mv 2 * 3 2 * 3 2 ms 2 * 3 2 * 3 2 4 ms and v()= for >4 ms. Therefore, he volage waveform is shown in Fig. 6.9b. The curren in a 2mH inducor is i()=2 sin 377 A Deermine he volage across he inducor and he energy sored in he inducor. SOLUTION EXAMPLE 6.7 From Eq. (6.8), we have and from Eq. (6.2), v() = L di() = A2 * 3 B d (2 sin 377) =.58 cos 377 V w L () = 2 Li2 () = 2 A2 * 3 B(2 sin 377) 2 =.4 sin 2 377 J The volage across a 2mH inducor is given by he expression v() = ( 3)e 3 mv = 6 Le us derive he waveforms for he curren, energy, and power. SOLUTION EXAMPLE 6.8 The waveform for he volage is shown in Fig. 6.a. The curren is derived from Eq. (6.) as i() = 3 ( 3x)e 3x dx 2 3 = 5 e e 3x dx 3 xe 3x dx f 3 3 = 5 e e3x 3 2 = 5e 3 ma = 6 3c e3x 9 (3x )d f

258 CHAPTER 6 CAPACITANCE AND INDUCTANCE A plo of he curren waveform is shown in Fig. 6.b. The power is given by he expression p() = v()i() The equaion for he power is ploed in Fig. 6.c. The expression for he energy is This equaion is ploed in Fig. 6.d. = 5( 3)e 6 W = 6 w() = 2 Li2 () = 2.5 2 e 6 J = 6 Volage (mv) Curren (ma)..7.8.6.4.6.5.4.3.2.2.5.5 2 2.5 3 3.5 Time (s) (a).2..5.5 2 2.5 3 3.5 (b) Time (s) Power (W) Energy (nj) 4.2 35.5 3..5.5.5.5 2 2.5 Time (s) 25 2 5 5..5.5 2 2.5 Time (s) Figure 6. (c) (d) Waveforms used in Example 6.8.

Learning Assessmens E6.6 The curren in a 5mH inducor has he waveform shown in Fig. E6.6. Compue he waveform for he inducor volage. 2 i() (ma) SECTION 6.2 INDUCTORS 259 ANSWER: v() (mv) 2 3 4 5 2 3 4 Figure E6.6 E6.7 Compue he energy sored in he magneic field of he inducor in Learning Assesmen E6.6 a =.5 ms. ANSWER: W=562.5 nj. E6.8 The curren in a 2H inducor is shown in Fig. E6.8. Find he waveform for he inducor volage. How much energy is sored in he inducor a =3 ms? i() (ma) 5 2 3 4 5 6 7 8 9 2 5 Figure E6.8 ANSWER: 25 J. v() (V) 2 4 6 3 5 9 7 8 2 3.33 5

26 CHAPTER 6 CAPACITANCE AND INDUCTANCE E6.9 The volage across a.h inducor is shown in Fig. E6.9. Compue he waveform for he curren in he inducor if i()=.a. How much energy is sored in he inducor a =7 ms? v() (V) 5 2 3 4 5 6 7 8 9 5 Figure E6.9 i() (A) ANSWER:.25 mj..2.. 2 3 4 5 6 7 8 9.25 E6. Find he energy sored in he capacior and inducor in Fig. E6.. ANSWER:.72 J,.5 J. 2 ma nf H 6 k 8 ma 3 k 2 V 2 k Figure E6.

SECTION 6.2 INDUCTORS 26 CAPACITOR AND INDUCTOR SPECIFICATIONS There are a couple of imporan parameers ha are used o specify capaciors and inducors. In he case of capaciors, he capaciance value, working volage, and olerance are issues ha mus be considered in heir applicaion. Sandard capacior values range from a few pf o abou 5 mf. Capaciors larger han F are available bu will no be discussed here. Table 6. is a lis of sandard capacior values, which are ypically given in picofarads or microfarads. Alhough boh smaller and larger raings are available, he sandard working volage, or dc volage raing, is ypically beween 6.3 V and 5 V. Manufacurers specify his working volage since i is criical o keep he applied volage below he breakdown poin of he dielecric. Tolerance is an adjunc o he capaciance value and is usually lised as a percenage of he nominal value. Sandard olerance values are ; 5%, ; %, and ; 2%. Occasionally, olerances for singledigi pf capaciors are lised in pf. For example, 5 pf ;.25 pf. TABLE 6. Sandard capacior values pf pf pf pf F F F F F F F..., 2 2 2.2.2.2 2 2 2 2,.5 5 5 5.5.5.5 5 5 5 5, 8 8 8.8.8.8 8 8 8 8, 2 2 2 2.2.2 2. 2 2 2 2, 22 22 22.22.22 2.2 22 22 22 22, 27 27 27.27.27 2.7 27 27 27 27, 3 33 33 33.33.33 3.3 33 33 33 33, 4 39 39 39.39.39 3.9 39 39 39 39, 5 47 47 47.47.47 4.7 47 47 47 47, 6 5 5 5.5.5 5. 5 5 5 5, 7 56 56 56.56.56 5.6 56 56 56 56, 8 68 68 68.68.68 6.8 68 68 68 68, 9 82 82 82.82.82 8.2 82 82 82 82, The wo principal inducor specificaions are inducance and resisance. Sandard commercial inducances range from abou nh o around mh. Larger inducances can, of course, be cusom buil for a price. Table 6.2 liss he sandard inducor values. The curren raing for inducors ypically exends from a few dozen ma s o abou A. Tolerances are ypically 5% or % of he specified value. TABLE 6.2 Sandard inducor values nh nh nh H H H mh mh mh...2 2 2.2 2 2.2 2.5 5 5.5 5 5.5 5.8 8 8.8 8 8.8 8 2 2 2 2. 2 2 2. 2 2.2 22 22 2.2 22 22 2.2 22 2.7 27 27 2.7 27 27 2.7 27 3 33 33 3.3 33 33 3.3 33 4 39 39 3.9 39 39 3.9 39 5 47 47 4.7 47 47 4.7 47 6 5 5 5. 5 5 5. 5 7 56 56 5.6 56 56 5.6 56 8 68 68 6.8 68 68 6.8 68 9 82 82 8.2 82 82 8.2 82

262 CHAPTER 6 CAPACITANCE AND INDUCTANCE As indicaed in Chaper 2, wirewound resisors are simply coils of wire, and herefore i is only logical ha inducors will have some resisance. The major difference beween wirewound resisors and inducors is he wire maerial. Highresisance maerials such as Nichrome are used in resisors, and lowresisance copper is used in inducors. The resisance of he copper wire is dependen on he lengh and diameer of he wire. Table 6.3 liss he American Wire Gauge (AWG) sandard wire diameers and he resuling resisance per foo for copper wire. TABLE 6.3 Resisance per foo of solid copper wire AWG No. Diameer (in.) m /f 2.88.59 4.64 2.54 6.58 4.6 8.4 6.5 2.32.4 22.253 6.5 24.2 26.2 26.59 4.6 28.26 66.2 3. 5 32.8 67 34.63 267 36.49 428 38.39 684 4.3 94 EXAMPLE 6.9 SOLUTION We wish o find he possible range of capaciance values for a 5mF capacior ha has a olerance of 2%. The minimum capacior value is.8c=4.8 mf, and he maximum capacior value is.2c=6.2 mf. EXAMPLE 6. SOLUTION The capacior in Fig. 6.a is a nf capacior wih a olerance of 2%. If he volage waveform is as shown in Fig. 6.b, le us graph he curren waveform for he minimum and maximum capacior values. The maximum capacior value is.2c=2 nf, and he minimum capacior value is.8c=8 nf. The maximum and minimum capacior currens, obained from he equaion are shown in Fig. 6.c. i() = C dv()

SECTION 6.2 INDUCTORS 263 4 3 2 i() v() (V) 2 v() ± C 3 4 2 3 4 5 6 7 Time ( s) (a) (b) 4 3 2 v() i() a C min i() a C max 8 6 4 Volage (V) 2 2 Curren (ma) 2 4 3 6 4 8 2 3 4 5 6 7 Time ( s) (c) Figure 6. Circui and graphs used in Example 6.. The inducor in Fig. 6.2a is a H inducor wih a olerance of %. If he curren waveform is as shown in Fig. 6.2b, le us graph he volage waveform for he minimum and maximum inducor values. The maximum inducor value is.l= H, and he minimum inducor value is.9l = 9 H. The maximum and minimum inducor volages, obained from he equaion are shown in Fig. 6.2c. v() = L di() SOLUTION EXAMPLE 6.

264 CHAPTER 6 CAPACITANCE AND INDUCTANCE Figure 6.2 Circui and graphs used in Example 6.. i() L v() (a) i() 5 5 v() a L min v() a L max 2 i() (ma) 5 5 i() (ma) 5 5 2 v() (V) 3 5 2 3 4 5 6 Time ( s) 5 2 3 4 5 4 6 Time ( s) (b) (c) 6.3 Capacior and Inducor Combinaions SERIES CAPACITORS If a number of capaciors are conneced in series, heir equivalen capaciance can be calculaed using KVL. Consider he circui shown in Fig. 6.3a. For his circui v()=v ()v 2 ()v 3 () p v N () 6.3 bu v i () = i() v C i A B i 3 6.4 Figure 6.3 Equivalen circui for N seriesconneced capaciors. v() v () v i() 2 () v 3 () C C 2 C 3 v N () C N (a) i() v() (b) C S

SECTION 6.3 CAPACITOR AND INDUCTOR COMBINATIONS 265 Therefore, Eq. (6.3) can be wrien as follows using Eq. (6.4): v() = a a N i = C i b 3 i() a N i = v i A B 6.5 where = i() va B C S 3 6.6 va B = a N i = v i A B and C S = a N i = = C i C C p 2 C N 6.7 [hin] Capaciors in series combine like resisors in parallel. Thus, he circui in Fig. 6.3b is equivalen o ha in Fig. 6.3a under he condiions saed previously. I is also imporan o noe ha since he same curren flows in each of he series capaciors, each capacior gains he same charge in he same ime period. The volage across each capacior will depend on his charge and he capaciance of he elemen. Deermine he equivalen capaciance and he iniial volage for he circui shown in Fig. 6.4. Noe ha hese capaciors mus have been charged before hey were conneced in series or else he charge of each would be equal and he volages would be in he same direcion. The equivalen capaciance is C S = 2 3 6 where all capaciance values are in microfarads. Therefore, C S = F and, as seen from he figure, va B =3 V. Noe ha he oal energy sored in he circui is SOLUTION EXAMPLE 6.2 w A B = 2 C2 * 6 (2) 2 3 * 6 (4) 2 6 * 6 () 2 D = 3 J However, he energy recoverable a he erminals is w C A B = 2 C S v 2 () = 2 C * 6 (3) 2 D = 4.5 J v() 2 V 2 F 6 F 3 F V 4 V Figure 6.4 Circui conaining muliple capaciors wih iniial volages.

266 CHAPTER 6 CAPACITANCE AND INDUCTANCE EXAMPLE 6.3 SOLUTION Two previously uncharged capaciors are conneced in series and hen charged wih a 2V source. One capacior is 3 F and he oher is unknown. If he volage across he 3 F capacior is 8 V, find he capaciance of he unknown capacior. The charge on he 3 F capacior is Q=CV=(3 F)(8 V)=24 C Since he same curren flows in each of he series capaciors, each capacior gains he same charge in he same ime period: C = Q V = 24 C 4V = 6 F [hin] Capaciors in parallel combine like resisors in series. PARALLEL CAPACITORS To deermine he equivalen capaciance of N capaciors conneced in parallel, we employ KCL. As can be seen from Fig. 6.5a, where i() = i () i 2 () i 3 () p i N () = C dv() = a a N i = = C p dv() C 2 dv() C i b dv() C 3 dv() p C N dv() C p =C C 2 C 3 p C N 6.2 6.8 6.9 Figure 6.5 Equivalen circui for N capaciors conneced in parallel. v() i() i () i 2 () i 3 () i N () C C 2 C 3 C N v() i() C p (a) (b) EXAMPLE 6.4 SOLUTION Deermine he equivalen capaciance a erminals AB of he circui shown in Fig. 6.6. C p = 5 F Figure 6.6 Circui conaining muliple capaciors in parallel. A v() B 4 F 6 F 2 F 3 F

SECTION 6.3 CAPACITOR AND INDUCTOR COMBINATIONS 267 Learning Assessmens E6. Two iniially uncharged capaciors are conneced as shown in Fig. E6.. Afer a period of ime, he volage reaches he value shown. Deermine he value of C. ANSWER: C =4 F. C Figure E6. 24 V 6 V 2 F E6.2 Compue he equivalen capaciance of he nework in Fig. E6.2. ANSWER: C eq =.5 F. 3 F 2 F 4 F C eq 2 F 3 F Figure E6.2 2 F E6.3 Deermine C T in Fig. E6.3. ANSWER:.667 F. A 6 F 4 F 6 F 8 F 5 F 2 F C T 6 F Figure E6.3 B 3 F F 6 F SERIES INDUCTORS If N inducors are conneced in series, he equivalen inducance of he combinaion can be deermined as follows. Referring o Fig. 6.7a and using KVL, we see ha v()=v ()v 2 ()v 3 () p v N () 6.2 and herefore, v() = L di() = a a N i = = L S di() L 2 di() L i b di() L 3 di() p L N di() 6.22 6.23

268 CHAPTER 6 CAPACITANCE AND INDUCTANCE [hin] Inducors in series combine like resisors in series. where L S = a N i = L i = L L 2 p L N 6.24 Therefore, under his condiion he nework in Fig. 6.7b is equivalen o ha in Fig. 6.7a. Figure 6.7 Equivalen circui for N seriesconneced inducors. v() i() v () L v 2 () v 3 () L 2 L 3 v() i() L S L N v N () (a) (b) EXAMPLE 6.5 SOLUTION Find he equivalen inducance of he circui shown in Fig. 6.8. The equivalen inducance of he circui shown in Fig. 6.8 is L S =H2H4H =7H Figure 6.8 Circui conaining muliple inducors. v() H 2 H 4 H PARALLEL INDUCTORS Consider he circui shown in Fig. 6.9a, which conains N parallel inducors. Using KCL, we can wrie However, Subsiuing his expression ino Eq. (6.25) yields i()=i ()i 2 ()i 3 () p i N () 6.25 i() = a a N j = L j b 3 v(x) dx a N = v(x) dx ia L B p 3 i j () = v(x) dx i L j A B j 3 j = i j A B 6.26 6.27 6.28

SECTION 6.3 CAPACITOR AND INDUCTOR COMBINATIONS 269 where = L p L L 2 L p 3 L N 6.29 [hin] Inducors in parallel combine like resisors in parallel. and ia B is equal o he curren in L p a =. Thus, he circui in Fig. 6.9b is equivalen o ha in Fig. 6.9a under he condiions saed previously. i() i() Figure 6.9 v() i () L i 2 () i 3 () i N () L 2 L 3 L N v() L p Equivalen circuis for N inducors conneced in parallel. (a) (b) Deermine he equivalen inducance and he iniial curren for he circui shown in Fig. 6.2. The equivalen inducance is = L p 2 6 4 SOLUTION EXAMPLE 6.6 where all inducance values are in millihenrys: and he iniial curren is ia B = A. L p =2 mh v() i() 3 A 2 mh 6 A 6 mh 2 A 4 mh Figure 6.2 Circui conaining muliple inducors wih iniial currens. The previous maerial indicaes ha capaciors combine like conducances, whereas inducances combine like resisances. Learning Assessmen E6.4 Deermine he equivalen inducance of he nework in Fig. E6.4 if all inducors are 6 mh. ANSWER: 9.429 mh. L eq Figure E6.4

27 CHAPTER 6 CAPACITANCE AND INDUCTANCE E6.5 Find L T in Fig. E6.5. ANSWER: 5 mh. 2 mh 4 mh 5 mh A 2 mh 6 mh L T 2 mh 4 mh B Figure E6.5 2 mh 3 mh 2 mh CHIP CAPACITORS In Chaper 2, we briefly discussed he resisors ha are used in modern elecronic manufacuring. An example of hese surface moun devices was shown in Fig. 2.4, ogeher wih some ypical chip capaciors. As we will indicae in he maerial ha follows, modern elecronics employs primarily resisors and capaciors and avoids he use of inducors when possible. Surfacemouned chip capaciors accoun for he majoriy of capaciors used in elecronics assembly oday. These capaciors have a large range of sizes, from as small as mils on a side up o 25 mils on a side. All ceramic chip capaciors consis of a ceramic dielecric layer beween meal plaes. The properies of he ceramic and meal layers deermine he ype of capacior, is capaciance, and reliabiliy. A cuaway view of a sandard chip capacior is shown in Fig. 6.2. The inner meal elecrodes are alernaely conneced o he opposing sides of he chip where meal erminaors are added. These erminaors no only make connecion o he inner elecrodes, bu also provide a solder base for aaching hese chips o prined Figure 6.2 Cross secion of a mulilayer ceramic chip capacior. Ceramic dielecric Tin Nickel Copper Inner elecrodes (Ni/Cu) circui boards. The number of alernaing layers, he spacing beween hem, along wih he dielecric consan of he ceramic maerial, will deermine he capaciance value. We indicaed earlier ha resisors are normally manufacured in sandard sizes wih specific power raings. Chip capaciors are also manufacured in his manner, and Table 6.4 provides a parial lising of hese devices. The sandard sizes of chip capaciors are shown in Table 6.4.

SECTION 6.3 CAPACITOR AND INDUCTOR COMBINATIONS 27 TABLE 6.4 Ceramic chip capacior sandard sizes Size Code Size (Mils) Power Raing (Was) 2 2 /2 42 4 2 /6 63 6 3 / 85 8 5 /8 26 2 6 /4 2 2 /2 252 25 2 CHIP INDUCTORS A chip inducor consiss of a miniaure ceramic subsrae wih eiher a wire wrapped around i or a hin film deposied and paerned o form a coil. They can be encapsulaed or molded wih a maerial o proec he wire from he elemens or lef unproeced. Chip inducors are supplied in a variey of ypes and values, wih hree ypical configuraions ha conform o he sandard chip package widely uilized in he prined circui board (PCB) indusry. The firs ype is he precision chip inducor where copper is deposied ono he ceramic and paerned o form a coil, as shown in Fig. 6.22. Eched Copper Coil Alumina Subsrae Copper (Cu) Terminaion Base 2 µm Nickel (Ni) Barrier 3 µm Tin (Sn) Ouerplaing T Terminal Elecrode L Inernal Medium Ferrie W E Figure 6.22 Precision chip inducor cross secion. Figure 6.23 Ferrie chip inducor cross secion The second ype is a ferrie chip inducor, which uses a series of coil paerns sacked beween ferrie layers o form a muliplayer coil as shown in Fig. 6.23. The hird ype is a wirewound open frame in which a wire is wound around a ceramic subsrae o form he inducor coil. The compleed srucure is shown in Fig. 6.24. Each of hese configuraions displays differen characerisics, wih he wirewound ype providing he highes inducance values ( nh 4.7 uh) and reasonable olerances ( 2%). The ferrie chip inducor gives a wide range of values (47 nh 33 uh) bu has olerances in he 5% range. The precision chip inducor has low inducance values ( nh) bu very good olerances ( /. nh). B C Figure 6.24 F Wirewound chip inducor cross secion A F G E K D

272 CHAPTER 6 CAPACITANCE AND INDUCTANCE 6.4 RC Operaional Amplifier Circuis [hin] The properies of he ideal opamp are v v and i i. Two very imporan RC opamp circuis are he differeniaor and he inegraor. These circuis are derived from he circui for an invering opamp by replacing he resisors R and R 2, respecively, by a capacior. Consider, for example, he circui shown in Fig. 6.25a. The circui equaions are However, v = and i =. Therefore, C d Av v B v o v R 2 = i v o () =R 2 C dv () 6.3 R 2 C 2 v () ± C v i v ± i ± v o v () ± R v i v ± i ± v o (a) (b) Figure 6.25 Differeniaor and inegraor operaional amplifier circuis. Thus, he oupu of he opamp circui is proporional o he derivaive of he inpu. The circui equaions for he opamp configuraion in Fig. 6.25b are v v R C 2 d Av o v B = i bu since v = and i =, he equaion reduces o or v R =C 2 dv o v o () = v R C (x) dx 2 3q = v R C (x) dx v o () 2 3 6.3 If he capacior is iniially discharged, hen v o ()=; hence, v o () = v R C (x) dx 2 3 6.32 Thus, he oupu volage of he opamp circui is proporional o he inegral of he inpu volage.

SECTION 6.4 RC OPERATIONAL AMPLIFIER CIRCUITS 273 The waveform in Fig. 6.26a is applied a he inpu of he differeniaor circui shown in Fig. 6.25a. If R 2 =k and C =2 F, deermine he waveform a he oupu of he opamp. Using Eq. (6.3), we find ha he opamp oupu is v o () =R 2 C dv () dv ()/=(2) 3 for <5 ms, and herefore, v o ()= 4 V <5 ms dv ()/= (2) 3 for 5 < ms, and herefore, v o ()=4 V =(2) 3 dv () 5 < ms Hence, he oupu waveform of he differeniaor is shown in Fig. 6.26b. SOLUTION EXAMPLE 6.7 v () (V) v o () (V) 4 Figure 6.26 Inpu and oupu waveforms for a differeniaor circui. 5 5 4 (a) (b) If he inegraor shown in Fig. 6.25b has he parameers R =5k and C 2 =.2 F, deermine he waveform a he opamp oupu if he inpu waveform is given as in Fig. 6.27a and he capacior is iniially discharged. The inegraor oupu is given by he expression SOLUTION EXAMPLE 6.8 which wih he given circui parameers is v o () = v R C (x) dx 2 3 v o () = 3 v (x) dx 3 In he inerval <. s, v ()=2 mv. Hence, v o () = 3 (2) 3 6. s =2 A =. s, v o ()= 2 V. In he inerval from. o.2 s, he inegraor produces a posiive slope oupu of 2 from v o (.)= 2 V o v o (.2)= V. This waveform from = o =.2 s is repeaed in he inerval =.2 o =.4 s, and herefore, he oupu waveform is shown in Fig. 6.27b.

274 CHAPTER 6 CAPACITANCE AND INDUCTANCE v () (mv) 2 v o () (V)..2.3.4 (s)..2.3.4 (s) 2 2 (a) (b) Figure 6.27 Inpu and oupu waveforms for an inegraor circui. Learning Assessmen E6.6 The waveform in Fig. E6.6 is applied o he inpu erminals of he opamp differeniaor circui. Deermine he differeniaor oupu waveform if he opamp circui parameers are C =2 F and R 2 =2. ANSWER: v () (V) v o () (V) 6 2 3 4 (s) 24 24 2 3 4 (s) Figure E6.6 6.5 Applicaion Examples APPLICATION EXAMPLE 6.9 In inegraed circuis, wires carrying highspeed signals are closely spaced as shown by he micrograph in Fig. 6.28. As a resul, a signal on one conducor can myseriously appear on a differen conducor. This phenomenon is called crossalk. Le us examine his condiion and propose some mehods for reducing i.

SECTION 6.5 APPLICATION EXAMPLES 275 Figure 6.28 SEM Image (Tom Way/ Ginger Conly. Couresy of Inernaional Business Machines Corporaion. Unauhorized use no permied.) The origin of crossalk is capaciance. In paricular, i is undesired capaciance, ofen called parasiic capaciance, ha exiss beween wires ha are closely spaced. The simple model in Fig. 6.29 can be used o invesigae crossalk beween wo long parallel wires. A signal is applied o wire. Capaciances C and C 2 are he parasiic capaciances of he conducors wih respec o ground, while C 2 is he capaciance beween he conducors. Recall ha we inroduced he capacior as wo closely spaced conducing plaes. If we srech hose plaes ino hin wires, cerainly he geomery of he conducors would change and hus he amoun of capaciance. However, we should sill expec some capaciance beween he wires. SOLUTION v () ± Wire i 2 () C C 2 Wire Figure 6.29 2 A simple model for v 2 () invesigaing crossalk. C 2 i 2 () In order o quanify he level of crossalk, we wan o know how much of he volage on wire appears on wire 2. A nodal analysis a wire 2 yields i 2 () = C 2 c dv () Solving for dv 2 (), we find ha dv 2 () Inegraing boh sides of his equaion yields dv 2() C 2 = c d dv () C 2 C 2 C 2 v 2 () = c d v C 2 C () 2 d = i 2 () = C 2 c dv 2() d

276 CHAPTER 6 CAPACITANCE AND INDUCTANCE Figure 6.3 Use of a ground wire in he crossalk model. Noe ha i is a simple capaciance raio ha deermines how effecively v () is coupled ino wire 2. Clearly, ensuring ha C 2 is much less han C 2 is he key o conrolling crossalk. How is his done? Firs, we can make C 2 as small as possible by increasing he spacing beween wires. Second, we can increase C 2 by puing i closer o he ground wiring. Unforunaely, he firs opion akes up more real esae, and he second one slows down he volage signals in wire. A his poin, we seem o have a ypical engineering radeoff: o improve one crierion, ha is, decreased crossalk, we mus sacrifice anoher, space or speed. One way o address he space issue would be o inser a ground connecion beween he signalcarrying wires as shown in Fig. 6.3. However, any advanage achieved wih grounded wires mus be raded off agains he increase in space, since insering grounded wires beween adjacen conducors would nearly double he wih consumed wihou hem. Wire Ground wire Wire 2 v () ± C G C 2G C C 2 v 2 () Redrawing he circui in Fig. 6.3 immediaely indicaes ha wires and 2 are now elecrically isolaed and here should be no crossalk whasoever a siuaion ha is highly unlikely. Thus, we are promped o ask he quesion, Is our model accurae enough o model crossalk? A more accurae model for he crossalk reducion scheme is shown in Fig. 6.32 where he capaciance beween signal wires and 2 is no longer ignored. Once again, we will deermine he amoun of crossalk by examining he raio v 2 () v (). Employing nodal analysis a wire 2 in he circui in Fig. 6.33 yields i 2 () = C 2 c dv () dv 2() d = i 2 () = AC 2 C 2G B c dv 2() d Figure 6.3 Elecrical isolaion using a ground wire in crossalk model. v () ± Solving for dv 2 (), we obain Wire C C C C G 2G 2 Ground wire Wire 2 v 2 () dv 2 () C 2 = c d dv () C 2 C 2 C 2G Inegraing boh sides of his equaion yields C 2 v 2 () = c d v C 2 C 2 C () 2G Noe ha his resul is very similar o our earlier resul wih he addiion of he C 2G erm. Two benefis in his siuaion reduce crossalk. Firs, C 2 is smaller because adding he ground wire moves wires and 2 farher apar. Second, C 2G makes he denominaor of he crossalk equaion bigger. If we assume ha C 2G = C 2 and ha C 2 has been halved by he exra spacing, we can expec he crossalk o be reduced by a facor of roughly 4.

SECTION 6.5 APPLICATION EXAMPLES 277 C 2 Figure 6.32 Wire Ground wire Wire 2 A more accurae crossalk model. v () ± C C G C 2G C 2 v 2 () C 2 Figure 6.33 Wire i 2 () Wire 2 A redrawn version of he more accurae crossalk model. v () ± C C G C 2G i 2 () C 2 v 2 () An excellen example of capacior operaion is he memory inside a personal compuer. This memory, called dynamic random access memory (DRAM), conains as many as 4 billion daa sorage sies called cells (circa 27). Expec his number o roughly double every 2 years for he nex decade or wo. Le us examine in some deail he operaion of a single DRAM cell. Fig. 6.34a shows a simple model for a DRAM cell. Daa are sored on he cell capacior in rue/false (or /) forma, where a largecapacior volage represens a rue condiion and a APPLICATION EXAMPLE 6.2 SOLUTION V I/O V I/O To sense amps C ou 45 ff I leak 5 pa C cell 5 ff v cell () I leak 5 pa C cell 5 ff v cell ().5 V C ou 45 ff C cell 5 ff 3 V (a) (b) (c) low volage represens a false condiion. The swich closes o allow access from he processor o he DRAM cell. Curren source I leak is an uninenional, or parasiic, curren ha models charge leakage from he capacior. Anoher parasiic model elemen is he capaciance, C ou, he capaciance of he wiring conneced o he oupu side of he cell. Boh I leak and C ou have enormous impacs on DRAM performance and design. Consider soring a rue condiion in he cell. A high volage of 3. V is applied a node I/O and he swich is closed, causing he volage on C cell o quickly rise o 3. V. We open Figure 6.34 A simple circui model showing (a) he DRAM memory cell, (b) he effec of charge leakage from he cell capacior, and (c) cell condiions a he beginning of a read operaion.

278 CHAPTER 6 CAPACITANCE AND INDUCTANCE he swich and he daa are sored. During he sore operaion he charge, energy, and number of elecrons, n, used are Q=CV=A5* 5 B(3)=5 fc W = 2 CV2 = (.5)A5 * 5 BA3 2 B = 225 fj n=q/q=5* 5 /A.6* 9 B=937,5 elecrons Once daa are wrien, he swich opens and he capacior begins o discharge hrough I leak. A measure of DRAM qualiy is he ime required for he daa volage o drop by half, from 3. V o.5 V. Le us call his ime H. For he capacior, we know v cell () = C cell 3 i cell V where, from Fig. 6.34b, i cell ()= I leak. Performing he inegral yields v cell () = C cell 3 AI leakb = I leak C cell K We know ha a =, v cell =3 V. Thus, K=3 and he cell volage is v cell () = 3 I leak C cell V 6.33 Subsiuing = H and v cell A H B=.5 V ino Eq. (6.33) and solving for H yields H =5 ms. Thus, he cell daa are gone in only a few milliseconds! The soluion is rewriing he daa before i can disappear. This echnique, called refresh, is a mus for all DRAM using his oneransisor cell. To see he effec of C ou, consider reading a fully charged Av cell =3. VB rue condiion. The I/O line is usually precharged o half he daa volage. In his example, ha would be.5 V as seen in Fig. 6.34c. ATo isolae he effec of C ou, we have removed I leak.b Nex, he swich is closed. Wha happens nex is bes viewed as a conservaion of charge. Jus before he swich closes, he oal sored charge in he circui is Q T = Q ou Q cell = V I O C ou V cell C cell Q T = (.5)(45 * 5 ) (3)(5 * 5 ) = 825 fc When he swich closes, he capacior volages are he same (le us call i V o ) and he oal charge is unchanged: Q T =825 fc= V C ou C cell = A45* 5 5* 5 o V o V o B and V o =.65 V Thus, he change in volage a V I O during he read operaion is only.5 V. A very sensiive amplifier is required o quickly deec such a small change. In DRAMs, hese amplifiers are called sense amps. How can v cell change insananeously when he swich closes? I canno. In an acual DRAM cell, a ransisor, which has a small equivalen resisance, acs as he swich. The resuling RC ime consan is very small, indicaing a very fas circui. Recall ha we are no analyzing he cell s speed only he final volage value, V o. As long as he power los in he swich is small compared o he capacior energy, we can be comforable in neglecing he swich resisance. By he way, if a false condiion (zero vols) were read from he cell, hen V o would drop from is precharged value of.5 V o.35 V a negaive change of.5 V. This symmeric volage change is he reason for precharging he I/O node o half he daa volage. Review he effecs of I leak and C ou. You will find ha eliminaing hem would grealy simplify he refresh requiremen and improve he volage swing a node I/O when reading daa. DRAM designers earn a very good living rying o do jus ha.

SECTION 6.6 DESIGN EXAMPLES 279 6.6 Design Examples We have all undoubedly experienced a loss of elecrical power in our office or our home. When his happens, even for a second, we ypically find ha we have o rese all of our digial alarm clocks. Le s assume ha such a clock s inernal digial hardware requires a curren of ma a a ypical volage level of 3. V, bu he hardware will funcion properly down o 2.4 V. Under hese assumpions, we wish o design a circui ha will hold he volage level for a shor duraion, for example, second. We know ha he volage across a capacior canno change insananeously, and hence is use appears o be viable in his siuaion. Thus, we model his problem using he circui in Fig. 6.35 where he capacior is employed o hold he volage and he ma source represens he ma load. As he circui indicaes, when he power fails, he capacior mus provide all he power for he digial hardware. The load, represened by he curren source, will discharge he capacior linearly in accordance wih he expression DESIGN EXAMPLE 6.2 SOLUTION 3 V ± C Opens on power ouages ma load Figure 6.35 A simple model for a power ouage ridehrough circui. v() = 3. i() C 3 Afer second, v() should be a leas 2.4 V, ha is, he minimum funcioning volage, and hence Solving his equaion for C yields 2.4 = 3. (.) C 3o C = 67 F Thus, from he sandard capacior values in Table 6., connecing hree 56 F capaciors in parallel produces 68 F. Alhough hree 56 F capaciors in parallel will saisfy he design requiremens, his soluion may require more space han is available. An alernae soluion involves he use of doublelayer capaciors or wha are known as Supercaps. A Web search of his opic will indicae ha a company by he name of Elna America, Inc. is a major supplier of doublelayer capaciors. An invesigaion of heir produc lising indicaes ha heir DCK series of small coinshaped supercaps is a possible alernaive in his siuaion. In paricular, he DCK3R3224 supercap is a 22mF capacior raed a 3.3 V wih a diameer of 7 mm, or abou /4 inch, and a hickness of 2. mm. Since only one of hese iems is required, his is a very compac soluion from a space sandpoin. However, here is ye anoher facor of imporance and ha is cos. To minimize cos, we may need o look for ye anoher alernae soluion.

28 CHAPTER 6 CAPACITANCE AND INDUCTANCE DESIGN EXAMPLE 6.22 Le us design an opamp circui in which he relaionship beween he oupu volage and wo inpus is v o () = 5 3 v () 2v 2 () SOLUTION In order o saisfy he oupu volage equaion, we mus add wo inpus, one of which mus be inegraed. Thus, he design equaion calls for an inegraor and a summer as shown in Fig. 6.36. Using he known equaions for boh he inegraor and summer, we can express he oupu volage as v o () =v 2 ()c R 4 R 3 d c R 4 R 2 d e R C 3 v () f = R 4 R R 2 C 3 v () c R 4 R 3 d v 2 () Figure 6.36 Opamp circui wih inegraor and summer. v () C R R 2 v 2 () R 3 R 4 v o () If we now compare his equaion o our design requiremen, we find ha he following equaliies mus hold: R 4 R R 2 C = 5 R 4 = 2 R 3 Noe ha we have five variables and wo consrain equaions. Thus, we have some flexibiliy in our choice of componens. Firs, we selec C = 2 F, a value ha is neiher large nor small. If we arbirarily selec R 4 = 2 k, hen mus be k and furhermore R R 2 = 2 * 9 If our hird choice is R = k, hen R 2 = 2 k. If we employ sandard opamps wih supply volages of approximaely ; V, hen all currens will be less han ma, which are reasonable values. R 3 SUMMARY The imporan (dual) relaionships for capaciors and inducors are as follows: q = Cv i() = C dv() v() = C 3 q i(x) dx v() = L di() i() = L 3 q v(x) dx p() = Cv() dv() w C () = 2 Cv2 () p() = Li() di() w L () = 2 Li2 () The passive sign convenion is used wih capaciors and inducors. In dc seady sae, a capacior looks like an open circui and an inducor looks like a shor circui.

PROBLEMS 28 The volage across a capacior and he curren flowing hrough an inducor canno change insananeously. Leakage resisance is presen in pracical capaciors. When capaciors are inerconneced, heir equivalen capaciance is deermined as follows: capaciors in series combine like resisors in parallel, and capaciors in parallel combine like resisors in series. When inducors are inerconneced, heir equivalen inducance is deermined as follows: inducors in series combine like resisors in series, and inducors in parallel combine like resisors in parallel. RC operaional amplifier circuis can be used o differeniae or inegrae an elecrical signal. PROBLEMS 6. An uncharged F capacior is charged by a consan curren of ma. Find he volage across he capacior afer 4 s. 6.2 A 2 F capacior has an accumulaed charge of 48 C. Deermine he volage across he capacior. 6.3 A capacior has an accumulaed charge of 6 C wih 5 V across i. Wha is he value of capaciance? 6.4 A 25 F capacior iniially charged o V is charged by a consan curren of 2.5 A. Find he volage across he capacior afer min. 2 2 6.5 The energy ha is sored in a 25 F capacior is w() 2 sin 2 377. Find he curren in he capacior. 6.6 A capacior is charged by a consan curren of 2 ma and resuls in a volage increase of 2 V in a s inerval. Wha is he value of he capaciance? 6.7 The curren in a F capacior is shown in Fig. P6.7. Deermine he waveform for he volage across he capacior if i is iniially uncharged. i() (ma) 6.9 The volage across a 2 F capacior is shown in Fig. P6.9. Deermine he waveform for he curren in he capacior. v() V 5 6 2 Figure P6.9 6. Derive he waveform for he curren in a 5 F capacior in he volage across he capacior as shown in Fig. P6.. v() (V) 8 4 Figure P6.7 2 6.8 The volage across a F capacior is shown in Fig. P6.8. Deermine he waveform for he curren in he capacior. Figure P6. 6. If he volage waveform across a F capacior is shown in Fig. P6., deermine he waveform for he curren. v() (V) 4 8 2 v() V 6 4 5 2 Figure P6.8 4 8 2 6 Figure P6. 5 5 2

282 CHAPTER 6 CAPACITANCE AND INDUCTANCE 6.2 The volage waveform across a F capacior is shown in Fig. P6.2. Derive he waveform for he curren. v() V 2 8 6.5 The volage across a 2F capacior is given by he waveform in Fig. P6.5. Find he waveform for he curren in he capacior. v C () (V) 2 4 5 2 25 3 35 4 2 3 4 5 (s) Figure P6.2 2 Figure P6.5 6.3 The curren flowing hrough a 5 F capacior is shown in Fig. P6.3. Find he energy sored in he capacior a =.4 ms, =3.3 ms, =4.3 ms, =6.7 ms, and =8.5 ms. i() (ma) 6.6 The volage across a 2 F capacior is given by he waveform in Fig. P6.6. Compue he curren waveform. v() (V) 5 2 3 6 2 5 Figure P6.6 2 3 4 5 6 7 8 6.7 Draw he waveform for he curren in a 24 F capacior when he capacior volage is as described in Fig. P6.7. 5 v() (V) i() v c () 5 F 6 4 6 6 ( s) Figure P6.3 6.4 The volage across a 25 F capacior is shown in Fig. P6.4. Deermine he curren waveform. Figure P6.7 v() (V) 2 6.8 The volage across a F capacior is given by he waveform in Fig. P6.8. Plo he waveform for he capacior curren. 2.2.4.6.8..2 v() (V) 2 2 5 Figure P6.4 Figure P6.8

PROBLEMS 283 6.9 The waveform for he curren in a 5 F capacior is shown in Fig. P6.9. Deermine he waveform for he capacior volage. 6.2 The waveform for he curren in a 5 F iniially uncharged capacior is shown in Fig. P6.2. Deermine he waveform for he capacior s volage. i() (ma) i() (ma) 2 3 4 5 Figure P6.9 2 3 4 Figure P6.2 6.2 The waveform for he volage across F capacior shown Fig. 6.2a is given in Fig. 6.2b. Deermine he following quaniies: (a) he energy sored in he capacior a = 2.5 ms, (b) he energy sored in he capacior a = 5.5 ms, (c) i c () a =.5 ms, (d) i c () a = 4.75 ms, and (e) i c () a = 7.5 ms. v() (V) 5 5 i c() v() ± µf 5 2 3 4 5 6 7 8 (a) (b) Figure P6.2 6.22 The curren in an inducor changed from o 2 ma in 4 ms and induces a volage of mv. Wha is he value of he inducor? 6.23 The curren in a mh inducor is i() = 2 sin 377 A. Find (a) he volage across he inducor and (b) he expression for he energy sored in he elemen. 6.27 The volage across a 2H inducor is given by he waveform shown in Fig. P6.27. Find he waveform for he curren in he inducor. v() (V) 5 6.24 If he curren i() =.5 A flows hrough a 2H inducor, find he energy sored a = 2s. 2 4 6 5 (s) 6.25 The curren in a 25mH inducor is given by he expressions i() = i() = ( e ) ma 6 7 Find (a) he volage across he inducor and (b) he expression for he energy sored in i. 6.26 Given he daa in he previous problem, find he volage across he inducor and he energy sored in i afer s. Figure P6.27 6.28 The volage across a 4H inducor is given by he waveform shown in Fig. P6.28. Find he waveform for he curren in he inducor. v() =, 6. v() (mv) 2.4 2 3 4 5 Figure P6.28

284 CHAPTER 6 CAPACITANCE AND INDUCTANCE 6.29 The curren in a 3mH inducor is shown in Fig. P6.29. Derive he waveform for he inducor volage. i() (ma) 2 6 6.32 The curren in a 2mH inducor is shown in Fig. P6.32. Deermine he waveform for he inducor volage. 6 4 i() (ma) Figure P6.29 2 3 4 5 6 7 2 2 2 4 6 8 2 4 6 4 6.3 The curren waveform in a 4mH inducor is shown in Fig. P6.3. Derive he waveform for he inducor volage. i() (ma) 2 5 Figure P6.32 6.33 The waveform for he curren flowing hrough a.5h inducor is shown in he plo in Fig. P6.33. Accuraely skech he inducor volage versus ime. Deermine he following quaniies: (a) he energy sored in he inducor a =.7 ms, (b) he energy sored in he inducor a =4.2 ms, and (c) he power absorbed by he inducor a =.2 ms, =2.8 ms, and =5.3 ms. i() (ma) 5 5 2 4 6 8 2 2 3 4 5 6 Figure P6.3 5 6.3 If he curren in a 6mH inducor is given by he waveform in Fig. P6.3, compue he waveform for he inducor volage. i() v L ().5 H i() (ma) Figure P6.33 8 6 4 6.34 The curren in a mh inducor is shown in Fig. P6.34. Deermine he waveform for he volage across he inducor. i() (ma) 2 2 4 6 8 2 2 3 4 5 6 Figure P6.3 Figure P6.34

PROBLEMS 285 6.35 The curren in a 5mH inducor is given in Fig. P6.35. Skech he inducor volage. i() (ma) 6.39 Find he possible capaciance range of he following capaciors. (a).68 F wih a olerance of %. (b) 2 pf wih a olerance of 2%. (c) 39 F wih a olerance of 2%. 4 2 6 8 Figure P6.35 6.36 The curren in a 5mH inducor is shown in Fig. P6.36. Find he volage across he inducor. i() (ma) 6.4 Find he possible inducance range of he following inducors. (a) mh wih a olerance of %. (b) 2. nh wih a olerance of 5%. (c) 68 H wih a olerance of %. 2 4 6 7 8 2 Figure P6.36 6.37 Draw he waveform for he volage across a 24mH inducor when he inducor curren is given by he waveform shown in Fig. P6.37. i() (A) 8 4 2 Figure P6.37 6.38 The curren in a 4mH inducor is given by he waveform in Fig. P6.38. Plo he volage across he inducor. i() (ma).2.3.6.9. (s) 6.4 The capacior in Fig. P6.4a is 5 nf wih a olerance of %. Given he volage waveform in Fig. 6.4b, graph he curren i() for he minimum and maximum capacior values. v() (V) 6 4 2 2 4 v() ± i() (a) C.5. 6 2 3 4 5 6 7 Time (ms) Figure P6.38 Figure P6.4 (b)

286 CHAPTER 6 CAPACITANCE AND INDUCTANCE 6.42 The inducor in Fig. P6.42a is 4.7 H wih a olerance of 2%. Given he curren waveform in Fig. 6.42b, graph he volage v() for he minimum and maximum inducor values. ± i() v() L 6.45 Given he nework in Fig. P6.45, find he power dissipaed in he 3 resisor and he energy sored in he capacior. 2 H 3 3 H 4 2 V ± 6 6 A 2 F 5 (a) Figure P6.45 i() (ma) 5 5 6.46 Calculae he energy sored in he inducor in he circui shown in Fig. P6.46. k 5 k 4 k 5 2 3 4 5 6 7 8 2 V 2 mh 7 kv 6 k Time (ms) Figure P6.42 (b) Figure P6.46 6.47 Calculae he energy sored in boh he inducor and he capacior shown in Fig. P6.47. 6.43 If he oal energy sored in he circui in Fig. P6.43 is 8 mj, wha is he value of L? L 5 5 25 V.5 F.5 H A 2 8 F 5 Figure P6.43 6.44 Find he value of C if he energy sored in he capacior in Fig. P6.44 equals he energy sored in he inducor. C Figure P6.47 6.48 Given a, 3, and 4 F capacior, can hey be inerconneced o obain an equivalen 2 F capacior? 6.49 Find he oal capaciance C T of he nework in Fig. P6.49. 4 F C T 2 F 2 F 2 V ±. H 2 F 3 F Figure P6.44 Figure P6.49

PROBLEMS 287 6.5 Find he oal capaciance C T of he nework in Fig. P6.5. 3 F 6.54 Find C T in he nework in Fig. P6.54. 2 F 3 F 6 F 3 F C T 4 F 6 F 3 F 4 F C T 3 F Figure P6.5 6.5 Find C T in he nework shown in Fig. P6.5. 3 F Figure P6.54 6 F 6 F 8 F 6 F 6.55 Deermine C T in he circui in Fig. P6.55 if all capaciors in he nework are 6 F. 4 F C T 2 F Figure P6.5 6.52 Find C T in he circui in Fig. P6.52. C T 5 F 4 F C T 6 F 5 F F Figure P6.55 Figure P6.52 3 F 6.53 Deermine he value of C T in he circui in Fig. P6.53. 6.56 Find C T in he circui in Fig. P6.56 if all capaciors are 6 F. 2 F 9 F 6 F C T C T F 6 F 5 F Figure P6.53 3 F Figure P6.56

288 CHAPTER 6 CAPACITANCE AND INDUCTANCE 6.57 Deermine he value of C T in he circui in Fig. P6.57 if all capaciors are 2 F. 6.6 If C eq = 4 F in he circui in Fig. P6.6, calculae C. 6 F C eq C 7 F C T Figure P6.6 Figure P6.57 6.58 If he oal capaciance of he nework in Fig. P6.58 is F, find he value of C T. 2 F 6 F 6.62 Find he oal capaciance C T shown in he nework in Fig. P6.62. 2 F C T F 4 F C 6 F 8 F 4 F 2 F Figure P6.58 6.59 In he nework in Fig. 6.59, if C T = 4 F, find he value of C. 4 F C T 2 F Figure P6.62 4 F C T 4 F C 2 F 2 F Figure P6.59 6.63 In he nework in Fig. P6.63, find he capaciance C T if (a) he swich is open and (b) he swich is closed. 6.6 Find he value of C in Fig. 6.6. 2 F 2 F 6 F F 2 F C T 6 F 6 F C T 4 F 3 F C 3 F 2 F Figure P6.6 Figure P6.63

PROBLEMS 289 6.64 Selec he value of C o produce he desired oal capaciance of C T = F in he circui in Fig. P6.64. 6.67 The wo capaciors shown in Fig. P6.67 have been conneced for some ime and have reached heir presen values. Find V. C V o 2 F C T = F 8 F 6 F 6 V 4 F Figure P6.64 Figure P6.67 6.65 Selec he value of C o produce he desired oal capaciance of C T = F in he circui in Fig. P6.65. 6.68 The hree capaciors shown in Fig. P6.68 have been conneced for some ime and have reached heir presen values. Find and V 2. V C C F C T F Figure P6.65 2 F F V V 2 8 F 4 F 2 V 6.66 The wo capaciors in Fig. P6.66 were charged and hen conneced as shown. Deermine he equivalen capaciance, he iniial volage a he erminals, and he oal energy sored in he nework. Figure P6.68 6.69 Deermine he inducance a erminals AB in he nework in Fig. P6.69. 6 V 2 F A mh 4 mh 2 mh 2 mh 2 V Figure P6.66 4 F 3 mh 4 mh 2 mh B Figure P6.69

29 CHAPTER 6 CAPACITANCE AND INDUCTANCE 6.7 Find he oal inducance a he erminals of he nework in Fig. P6.7. 6.73 Find L T in he circui in Fig. P6.73. All inducors are 2 H. 2 mh 2 mh 6 mh 4 mh L T 2 mh L T Figure P6.7 Figure P6.73 6.7 Find L T in he circui in Fig. P6.7. L T 6.74 Find L T in he circui in Fig. P6.74. 4 H 2 H 3 H 5 H 2 H 4 H 5 H 6 H 7 H L T 8 H 4 H 6 H 2 H 6 H H 2 H Figure P6.7 Figure P6.74 6.72 Find L T in he circui in Fig. P6.72. 2 H 6.75 Find L T in he circui in Fig. P6.75. 4 H 2 H 5 H L T 2 H 8 H 4 H 6 H 3 H L T 6 H 4 H 6 H 4 H 2 H Figure P6.72 3 H Figure P6.75 9 H 3 H 2 H

PROBLEMS 29 6.76 If he oal inducance, L T, in he nework in Fig. P6.76 is 5 H, find he value of L. 2 H H 6.79 Given he nework shown in Fig. P6.79, find (a) he equivalen inducance a erminals AB wih erminals CD shor circuied, and (b) he equivalen inducance a erminals CD wih erminals AB open circuied. L T 5 H H 3 H 6 H H L 3 H A 5 mh 2 mh 2 mh C Figure P6.76 B 6 mh D Figure P6.79 6.77 If he oal inducance, L T, of he nework in Fig. P6.77 is 6 H, find he value of L. 6.8 Find he value of L in he nework in Fig. P6.8 so ha he oal inducance, L T, will be 2 mh. 2 H 8 H 4 mh 2 mh H L T 4 H 4 H L 4 H L T L 6 mh 3 H 8 H 9 H Figure P6.77 Figure P6.8 6.78 Find L T in he nework in Fig. P6.78 (a) wih he swich open and (b) wih he swich closed. All inducors are 2 mh. L T 6.8 A 2mH inducor and a 2mH inducor are conneced in series wih a A curren source. Find (a) he equivalen inducance and (b) he oal energy sored. 6.82 If he capaciors shown in Fig. P6.82 have been conneced for some ime and have reached. Their presen values, deermine (a) he volage V and (b) he oal energy sored in he capaciors. 2 V 3 F 6 F V o Figure P6.78 Figure P6.82

292 CHAPTER 6 CAPACITANCE AND INDUCTANCE 6.83 If he capaciors in he circui in Fig. P6.83 have been conneced for some ime and have reached heir presen values, calculae (a) he volage V and (b) he oal energy sored in he capaciors. 6.86 If he capaciors in he circui in Fig. P6.86 have been conneced for some ime and have reached heir presen values, (a) calculae he volages V and V 2 and (b) deermine he oal energy sored in he capaciors. 6 F 2 F V 6 V F 36 V 9 F 3 F V V 2 Figure P6.83 Figure P6.86 6.87 For he nework in Fig. P6.87 v S () = 8 cos 377 V and v S2 () = 4 cos 377 V. Find v (). 6.84 If he capaciors shown in Fig. P6.84 have been conneced for some ime and he volage has reached is presen value, deermine (a) he volages V and V 2 and (b) he oal energy sored in he capaciors. 2 k v S () ± ± k v S2 () F v o () V 24 V V 2 Figure P6.84 2 F 4 F Figure P6.87 6.88 If he inpu o he nework shown in Fig. P6.88a is given by he waveform in Fig. P.6.88b, deermine he oupu waveform v () if v ()=. v i () 4 k k v o () 6.85 If he capaciors shown in Fig. P6.85 have been conneced for some ime and he volage has reached is presen value, find (a) he volages V and V 2 and (b) he oal energy sored in he capaciors. v i () 2 (a) 8 V V V 2 3 F 6 F 2 2 3 4 5 6 7 (s) (b) Figure P6.85 Figure P6.88

PROBLEMS 293 6.89 The inpu o he nework shown in Fig. P6.89a is shown in Fig. P6.89b. Derive he waveform for he oupu volage v () if v ()=. 2 F 6.9 Skech he oupu volage of he nework in Fig. P6.9a if he inpu is given by he waveform in Fig. 6.9b. 2 k v i () 2 k v o () v i () µf v o () (a) (a) v i () v i () (V) 3 3 2 2 3 4 (s) 2 4 6 8 (b) (b) Figure P6.9 Figure P6.89 6.92 Skech he oupu volage of he nework in Fig. P6.92a if he inpu is given by he waveform in Fig. P6.92b. 4 k 6.9 Show ha he circui shown in Fig. P6.9 acs like a differeniaor wih an oupu volage of v o () = RC dv () Assume an ideal opamp. v i () 2 µf v o () (a) C R v i () (V) 3 2 v i () v o () 2 3 4 5 6 7 (b) Figure P6.9 Figure P6.92

294 CHAPTER 6 CAPACITANCE AND INDUCTANCE 6.93 Given he nework in Fig. P6.93a, (a) Deermine he equaion for he closedloop gain v G = 2 2. v i (b) Skech he magniude of he closedloop gain as a funcion of frequency if R = k, R 2= k, and C=2 F. C 6.94 An inegraor is required ha has he following performance: where he capacior values mus be greaer han nf and he resisor values mus be greaer han k. (a) Design he inegraor. v () = 6 3 v s (b) If V supplies are used, wha are he maximum and minimum values of v? (c) Suppose v s V. Wha is he rae of change of v? v i () R R 2 v o () Figure P6.93 TYPICAL PROBLEMS FOUND ON THE FE EXAM 6PFE Given hree capaciors wih values 2 F, 4 F, and 6 F, can he capaciors be inerconneced so ha he combinaion is an equivalen 3 F capacior? a. Yes. The capaciors should be conneced as shown. 2 µf 4 µf b. Yes. The capaciors should be conneced as shown. 2 µf 4 µf 6 µf C eq 6 µf C eq 6PFE2 The curren pulse shown in Fig. 6PFE2 is applied o a F capacior. Wha is he energy sored in he elecric field of he capacior? a. b. c. d. J, w() = * 6 2 J, 6 s J, 7 s J, w() = 6 * 6 J, 6 s 6 J, 7 s J, w() = 8 * 6 2 J, 6 s 8 J, 7 s J, w() = 3 * 6 J, 6 s 3 J, 7 s c. Yes. The capaciors should be conneced as shown. i() (A) 6 2 µf 4 µf 6 µf C eq Figure 6PFE2 ( s) d. No. An equivalen capaciance of 3 F is no possible wih he given capaciors.

TYPICAL PROBLEMS FOUND ON THE FE EXAM 295 6PFE3 The wo capaciors shown in Fig. 6FE3 have been conneced for some ime and have reached heir presen values. Deermine he unknown capacior C x. a. 2 F b. 3 F 8 V 6 F c. F d. 9 F 24 V Figure 6PFE3 6PFE4 Wha is he equivalen inducance of he nework in Fig. 6PFE4? a. 9.5 mh b. 2.5 mh c. 6.5 mh d. 3.5 mh 2 mh 3 mh C x 6PFE5 The curren source in he circui in Fig. 6PFE5 has he following operaing characerisics: Wha is he volage across he mh inducor expressed as a funcion of ime? a. b. c. d. i() = e A, 6 2e 2 A, 7 v() = e V, 6.2e 2 4e 2 V, 7 v() = e V, 6 2e 2 4e 2 V, 7 v() = e V, 6.2e 2.4e 2 V, 7 v() = e V, 6 2e 2 V, 7 3 mh 6 mh 2 mh 9 mh L eq i() L v() Figure 6PFE4 Figure 6PFE5