Chapter 3: Capacitors, Inductors, and Complex Impedance


 Delphia Webster
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1 haptr 3: apacitors, Inductors, and omplx Impdanc In this chaptr w introduc th concpt of complx rsistanc, or impdanc, by studying two ractiv circuit lmnts, th capacitor and th inductor. W will study capacitors and inductors using diffrntial quations and Fourir analysis and from ths driv thir impdanc. apacitors and inductors ar usd primarily in circuits involving timdpndnt voltags and currnts, such as A circuits. I. A oltags and circuits Most lctronic circuits involv timdpndnt voltags and currnts. An important class of timdpndnt signal is th sinusoidal voltag (or currnt, also known as an A signal (Altrnating urrnt. Kirchhoff s laws and Ohm s law still apply (thy always apply, but on must b carful to diffrntiat btwn timavragd and instantanous quantitis. An A voltag (or signal is of th form: (t p cos(ωt (3. whr ω is th angular frquncy, p is th amplitud of th wavform or th pak voltag and t is th tim. Th angular frquncy is rlatd to th frguncy (f by ω πf and th priod (T is rlatd to th frquncy by T/f. Othr usful voltags ar also commonly dfind. Thy includ th paktopak voltag ( pp which is twic th amplitud and th MS voltag ( MS which is /. Avrag powr in a rsistiv A dvic is computd using MS quantitis: MS p PI MS MS I p p /. (3. This is important nough that voltmtrs and ammtrs in A mod actually rturn th MS valus for currnt and voltag. Whil most ral world signals ar not sinusoidal, A signals ar still usd xtnsivly to charactriz circuits through th tchniqu of Fourir analysis. Fourir Analysis On convnint way to charactriz th rat of chang of a function is to writ th tru function as a linar combination of a st of functions that hav particularly asy charactristics to dal with analytically. In this cas w can considr th trigonomtric functions. It turns out that w can writ any function as an intgral of th form ~ ( t cos( ωt + φ dω (3.3 whr ~ and φ ar functions of th frquncy ω. This procss is calld Fourir analysis, and it mans that any function can b writtn as an intgral of simpl sinusoidal  7 
2 functions. In th cas of a priodic wavform this intgral bcoms a sum ovr all th harmonics of th priod (i.. all th intgr multiplicativ frquncis of th priod. A cos + n ( nωt ( t (3.4 n φ n An implication of this mathmatical fact is that if w can figur out what happns whn w put pur sinusoidal voltags into a linar circuit, thn w will know vrything about its opration vn for arbitrary input voltags. omplx Notation In complx notation w rplac our sinusoidal functions by xponntials to mak th calculus and bookkping asir still. Thn w can includ both phas and magnitud information. W ll dfin whr i. iφ cos φ + isinφ, (3.5 Th gnral procdur for using this notation is:. hang your problm into complx algbra (i.. rplac cos ωt with. Solv th problm. 3. Tak th ral part of th solution as your answr at th nd. iωt II. apacitors On of th most basic ruls of lctronics is that circuits must b complt for currnts to flow. This wk, w will introduc an xcption to that rul. Th capacitor is actually a small brak in a circuit. Try masuring th rsistanc of a capacitor, you will find that it is an opn circuit. Howvr, at th insid nds of th capacitor s lad, it has littl plats that act as charg rsrvoirs whr it can stor charg. For short tims, you do not notic that th brak is thr. Ngativ charg initially flows in to on sid and out from out th othr sid just as if th two lads wr connctd. For fast signals, th capacitor looks lik a shortcircuit. But aftr a whil th capacitor s rsrvoirs fill, th currnt stops, and w notic that thr rally is a brak in th circuit. For slow signals, a capacitor looks lik an opn circuit. What is fast, and what is slow? It dpnds on th capacitor and th rst of th circuit. This wk, you will larn how to dtrmin fast and slow for yourslvs. apacitors srv thr major rols in lctrical circuits (although all thr ar just variations of on basic ida: harg intgrators; High or low frquncy filtrs; D isolators
3 In ordr to prform ths functions analytically, w will nd to introduc a numbr of nw concpts and som significant mathmatical formalism. In this procss w will also dvlop a numbr of nw concpts in analyzing lctronic circuits. apacitanc A capacitor is a dvic for storing charg and lctrical nrgy. It consists of two paralll conducting plats and som nonconducting matrial btwn th plats, as shown in figur 3. on th right. Whn voltag is applid positiv charg collcts on on plat and ngativ charg collcts on th othr plan. Sinc thy ar attractd to ach othr this is a stabl stat until th voltag is changd again. A capacitor s charg capacity or capacitanc ( is dfind as: Q (3.6 which rlats th charg stord in th capacitor (Q to th voltag across its lads (. apacitanc is masurd in Farads (F. A Farad is a vry larg unit and most applications us µf, nf, or pf sizd dvics. Many lctronics componnts hav small parasitic capacitancs du to thir lads and dsign. Th capacitor also stors nrgy in th lctric fild gnratd by th chargs on its two plats. Th potntial nrgy stord in a capacitor, with voltag on it, is E (3.7 W usually spak in trms of currnt whn w analyz a circuit. By noting that th currnt is th rat of chang of charg, w can rwrit th dfinition of capacitanc in trms of th currnt as: or Q Idt Figur 3.: A capacitor consist of two paralll plats which stor qual and opposit amounts of charg (3.8 d I & (3.9 dt This shows that w can intgrat a function I(t just by monitoring th voltag as th currnt chargs up a capacitor, or w can diffrntiat a function (t by putting it across a capacitor, and monitoring th currnt flow whn th voltag changs
4 A Simpl ircuit W will start by looking in dtail at th simplst capacitiv circuit, which is shown in figur 3. on th right. An circuit is mad by simply putting a rsistor and a capacitor togthr as a voltag dividr. W will put th rsistor in first, so w can connct th capacitor to ground. By applying Kirchhoff s aws to this circuit, w can s that:. Th sam currnt flows through both th rsistor and th capacitor, and. Th sum of th voltag drops across th two lmnts qual th input voltag. This can b put into a formula in th following quation: IN I + Idt. (3. which can also b writtn as IN I + Idt. (3. W can also put this into th form of a diffrntial quation in th following way: or d di I IN + (3.a dt dt & IN I& + I. (3.b Ths quations show that tims ar masurd in units of, and that what you s dpnds on how quickly things chang during on tim intrval. If th currnt changs quickly, thn most of th voltag will show up across th rsistor, whil th voltag across th capacitor slowly chargs up as it intgrats th currnt. If th voltag changs slowly, thn most of th voltag shows up across th capacitor as it chargs. Sinc this usually rquirs a small currnt, th voltag across th rsistor stays small. But, what happns at intrmdiat tims? To dtrmin this quantitativly w will hav to dvlop som mor sophisticatd mathmatical tchniqus. IN OUT Figur 3.: A simpl circuit which intgrats currnt. Solutions to ircuit athr than produc th gnral solution, w will concntrat on two spcial cass that ar particularly usful. Th first will b for a constant voltag and th scond will b a sinusoidal input.  
5 To study a constant supply voltag on an circuit, w st th lft sid of quation 3. qual to a constant voltag. Thn w hav a simpl homognous diffrntial quation with th simpl solution for th currnt of a dcaying xponntial, I ( t / I, (3.3 which will account for any initial conditions. Aftr a tim of a fw tim priods, this solution will hav dcayd away to th supply voltag. And now lt us considr th othr solution. In th prior sction, w argud that if w can undrstand th circuit s bhavior for sinusoidal input w can dal with any arbitrary input. Thrfor, this is th important on. t s look at our simpl circuit and suppos that w apply (or driv a simpl sin wav into th input: IN ( ωt cos. (3.4 In complx notation, this mans that w will st th driv voltag to IN ( iωt xp, (3.5 and w just hav to rmmbr to tak th al part at th nd of our calculation. If w put this driv voltag into th diffrntial quation (quation 3., thn it bcoms a rlativly simpl inhomognous diffrntial quation: din di iω xp( iωt + I. (3.6 dt dt This is rlativly simpl bcaus it shows up so oftn in physics that you might as wll mmoriz th solution or at last th way to gt th solution. Not that mathmatically it looks just lik a drivn harmonic oscillator. W can obtain th solution by using th standard rcip for first ordr linar diffrntial quations. W start by rwriting quation as di dt iω + I xp( iωt, (3.7 which w thn multiply by ( t / xp to obtain di xp( t / iω xp( t / + I xp[( iω + t]. (3.8 dt Th lft handsid of this quality can b rwrittn undr th form of a total drivativ (multiplication rul so that w now hav d dt I( txp( t iω This quation is asily intgrabl and can b rwrittn as xp[( iω + t]. (3.9 t iω I( txp( i t dt xp[( ω + ]. (3.  
6 Th intgral is straightforward and yilds th following xprssion: iω I( t xp( iωt + st xp( + iω t (3. Th first trm rprsnt th stady stat oscillatory bhavior of th drivn circuit, whil th scond trm dscribs th transint bhavior of th currnt aftr switching on th driving voltag. Sinc w ar only intrstd in th longtrm bhavior of th circuit, w nglct th scond trm and concntrat on th first. Aftr a littl bit of algbra, w can rwrit th stadystat currnt as iω ω ω + i I( t xp( iωt xp( iωt (3. i + ω + ( ω + ( ω Th scond fraction can b intrprtd as a phas trm with xprssion for th currnt bcoms with tan φ, so that th ω I ( t I xp( iω t + φ (3.3 ω I cos( φ (3.4 + ( ω Th ral solution of this simpl circuit can b obtaind by taking th ral part of quation 3, and is lft as an xrcis to th radr. Th solution of th simpl circuit appars to b rathr complicatd and involvd, howvr it simplifis considrably whn w plug quation 3 back in to th original intgral quation from Kirchhoff s loop law (quation. Aftr intgrating th xponntial and a littl bit of algbra, w obtain in ( t; ω I( t + I( t (3.5 i ω This rmarkably simpl xprssion looks a lot lik th standard Kirchhoff s loop law for rsistors, xcpt that th capacitor trm bhavs with a frquncy dpndnt imaginary rsistanc. Impdanc W will obtain th sam solution as th on w obtaind for th original voltag dividr, as long as w assign an imaginary, frquncy dpndnt, rsistanc to th capacitor. Th imaginary part just mans that it will produc a π phas shift btwn th voltag and th currnt for a sinusoidal input. W will call this impdanc Z. (3.6 iω  
7 Now, th solution for an dividr bcoms somwhat simplifid. W can comput th total currnt flowing through th circuit as I Z in tot in + Z + / iωt iω iωt iω iωt ( iω + iω ( + iω cos( φ ( ωt+ φ i (3.7 Th voltag across an lmnt is just this currnt tims th lmnt s impdanc. For th voltag drop across th rsistor it is largly th sam as bfor: ( ωt+ φ i I cos( φ. (3.8 For th capacitor, w gt th following voltag drop: IZ I iω cos( φ iω cos( φ i ω i( ωt+ φ i sin( φ ( ωt+ φ ( ωt+ φ π / i( ωt+ φ π / i sin( φ (3.9 If vrything is corrctly calculatd thn th sum of th voltag drops across th two lmnts should b qual to th input voltag. t s try it: + ( i( ω t + φ i φ i( ω i t + φ i ω cos( sin( φ t φ (3.3 mmbr, you gt th actual wavforms by taking th ral parts of ths complx solutions. Thrfor cosφ cos(ωt+φ and (3.3 sinφ cos(ωt+φπ/  sinφ sin(ωt+φ (3.3 This looks complicatd, but th limits of high frquncy and low frquncy ar asy to rmmbr. At high frquncis ( φ, th capacitor is lik a short, and all th voltag shows up across th rsistor. At low frquncis ( φ π /, th capacitor is lik an opn circuit, and all th voltag shows up across th capacitor. If you considr th lading trms for th lmnts with th small voltags, you find that ( iω + ( ω ω ( i + ω + ( ω ( as ω iω as ω i ω (3.33 Thus, at high frquncy, th voltag across th capacitor is th intgral of th input voltag, whil at low frquncy th voltag across th rsistor is th drivativ of th input voltag. This says that as long as all th important frquncis ar high, th capacitor will intgrat th input voltag. If all th important frquncis ar small, th rsistor will diffrntiat th voltag. If thr ar intrmdiat frquncis, or a mixtur of som high  3 
8 and som low frquncis, th rsult will not b so simpl but it can b dtrmind from th voltag dividr algbra using complx notation. W finish by noting that th voltag on th capacitor is always π/ out of phas with th voltag on th rsistor. III. Inductors An inductors is a coil of wir, or solnoid, which can b usd to stor nrgy in th magntic fild that it gnrats (s figur 3.3 on th right. It is mathmatically similar to a capacitor, but has xactly th opposit bhavior: it bhavs as a short circuit for low frquncis and as an opn circuit for high frquncis (i.. it passs low frquncy signals and blocks high frquncy signals. Th nrgy stord in th fild of an inductor with inductanc is givn by th following formula: B I E (3.34 Th SI unit of inductanc is th Hnry (H. ommrcially availabl inductors hav inductancs that rang from nh to mh. Small millimtrsiz and cntimtr siz solnoids typically hav inductancs in th rang of µh, whil magntic fild coils can hav a inductancs in th mh rang, and can somtims hav inductancs of up to svral H. Most lctronics componnts hav small parasitic inductancs du to thir lads and dsign (for xampl, wirwound powr rsistors. Figur 3.3: An inductor consists of a coild wir, also calld a solnoid. Th dashd arrow B, rprsnt th magntic fild gnratd by th currnt in th inductor. In an lctric circuit, a voltag, or lctromotiv potntial, is gnratd across th trminals of th inductor whn th currnt changs du to Faraday s law. Th voltag drop is givn by th following simpl xprssion: di (3.35 dt From this quation, w s that th inductor oprats xactly opposit to a capacitor: an inductor diffrntiats th currnt and intgrats th voltag. Th circuit W can analyz th circuit in much th sam way that w drivd th opration of th circuit. W start by applying Kirchhoff s loop law to th circuit in figur 3.4 blow, and w find that  4 
9 di IN I +. (3.36 dt If w apply a constant voltag th solution can b calculatd using th tchniqus dvlopd for th circuit and w calculat that I( t I xp( t. (3.37 Th circuit approachs th stady stat currnt I IN / with a tim constant of /. impdanc Instad of solving th diffrntial quation for th circuit with a sinusoidal applid input voltag such as that givn by quations 4 and 5, as w did with th circuit, w will just assum that th currnt has th form I ( t I xp( iω t + φ (3.38 W plug this ansatz solution back into th diffrntial quation of quation 3 and find that IN I( t + iωi, (3.39 from which w dduc that th inductor bhavs as a rsistor with frquncy dpndnt imaginary rsistanc. Th impdanc of an inductor is thrfor Z iω (3.4 Just as with th circuit, w can apply Ohm s law to th circuit to calculat th total currnt. Sinc and ar in sris, w obtain iω iωt iωt in iωt i( ωt φ I ( t cos( φ (3.4 Ztotal + Z + iω + ω whr th phas is givn by tan( φ ω. W calculat th voltag drop across th rsistor using th xprssion for th currnt and find that i( ωt φ I( t cos( φ (3.4 Th voltag drop across th inductor is calculatd th sam way, and w find i( ωt φ i( ωt φ i( ωt φ + π / iωi ( t iω cos( φ i sin( φ cos( φ (3.43 IN OUT Figur 3.4: A simpl circuit
10 If vrything is corrctly calculatd thn th sum of th voltag drops across th two lmnts should b qual to th input voltag. t s try it: + ( i( ω t φ i φ i( ω i t φ i ω cos( + sin( φ t φ (3.44 You gt th actual wavforms by taking th ral parts of ths complx solutions. Thrfor cosφ cos(ωtφ and (3.45 sinφ cos(ωtφ+π/ sinφ sin(ωtφ (3.46 This looks complicatd, but th limits of high frquncy and low frquncy ar asy to rmmbr. At high frquncis ( φ π, th inductor is lik an opn circuit, and all th voltag shows up across th inductor. At low frquncis ( φ, th inductor is lik a short circuit or just a plain wir, and all th voltag shows up across th rsistor. It should also b pointd out that th voltag on th inductor is always +π/ out of phas with th voltag on th rsistor. I. Transformrs Transformrs ar an ingnious combination of two inductors. Thy ar usd to transfr powr btwn two circuits by magntic coupling. Th transformr changs an input voltag, without affcting th signal shap, similar to th voltag dividr of last wk. Howvr it has svral important diffrncs: a It can incras as wll as dcras a signal s amplitud (i.. A voltag. b It rquirs a timvarying (A input to work. c It is much hardr to fabricat. d It usually dos not work wll for vry fast signals (sinc inductors block high frquncis. Transformrs ar commonly usd as a major componnt in a D powr supplis sinc thy can convrt a A wall voltag into a smallr voltag that is closr to th dsird D voltag (.g. 5 or ±5. Th schmatic symbol for a transformr is shown in figur 3.5, abov. Transformrs ar passiv dvics that simultanously chang th voltag and currnt of a circuit. Thy hav (at last four trminals: two inputs (calld th primary and two outputs (calld th scondary. Thr is no ral diffrnc btwn th input and output for a transformr, you could simply flip it around and us th scondary as th input and th primary as th output. Howvr, for th sak of clarity, w will always assum that you us th primary for input and th scondary for output. IN OUT Figur 3.5: Th schmatic symbol for a transformr
11 Th coupling btwn th input and output is don magntically. This allows transformrs to hav a numbr of intrsting bnfits including: Thr is no D connction btwn input and output, so transformrs ar oftn usd to isolat on circuit from anothr. f Transformrs only work for tim varying signals, whn th inductiv coupling btwn th coils is gratr than th rsistiv losss. Sinc thy hav no xtrnal powr th output powr can not b gratr than th input powr P P IP S IS. (3.47 Usually, w will assum quality but thr ar small rsistancs (and hnc rsistiv losss in th coils and a poorly or chaply dsignd transformr many not hav th input and output sufficintly strongly coupld to ach othr. Dpnding on th dvic and th signal th output powr may wll b lss than th input powr. Transformrs ar most commonly usd to chang lin voltag ( MS at 6 Hz into a mor convnint voltag. High powr transmission lins us transformrs to incras th voltag and dcras th currnt. This rducs I powr losss in th transmission wirs. For our circuits w will us a transformr that rducs th voltag and incrass th currnt. Transformrs ar charactrizd by th ratio of th numbr of turns on th input and output windings. Th magntic coupling in an idal transformr will insur that th numbr of turns tims th currnt flowing is th sam for th input and output: N I I N S P P P NSI S I P N (3.48 S Sinc th voltag must chang in th opposit mannr to kp th input and output powr, th ratio of th voltags is th sam as th ratio of th turns: S P N N S (3.49 P Transformrs ar usually calld stpup or stpdown according to whthr th output voltag incrass or dcrass. A transformr also transforms th impdanc of a circuit, sinc it changs th ratio of /I. Using our ruls abov, th ratio of output impdanc to input impdanc is th squar of th ratio of turns: ZS S I P N S ZP IS P NP (3.5 So, if you us a transformr as a stpup transformr, it incrass th voltag and th impdanc at its output rlativ to its input. If you us a transformr as a stpdown transformr, it dcrass th voltag and th impdanc at its output
12 Dsign Exrciss Dsign Exrcis 3: Using Kirchhoff s laws, driv a formula for th total capacitanc of two capacitors in paralll and a formula for th total capacitanc of two capacitors in sris. (Hint: prtnd that you ar working with an A signal of frquncy ω. Dsign Exrcis 3: Using Kirchhoff s laws, driv a formula for th total inductanc of two inductors in paralll and a formula for th total inductanc of two inductors in sris. (Hint: prtnd that you ar working with an A signal of frquncy ω. Dsign Exrcis 33: alculat out as a function of in in th circuit of figur 3.6 on th right, using th formulas for Z, Z, and Z (do not us Mapl / Mathmatica / MATAB / Mathad for ths calculations and show all stps. in is a prfct A voltag signal with a frquncy of ω. IN OUT Plot th magnitud and phas of out as a function of ω for kω, µf, and µh. What happns to th magnitud and th phas of out at ω? (Mapl / Mathmatica / MATAB / Mathad ar prmittd for th plots. Figur 3.6: An filtr circuit. Dsign Exrcis 34: alculat out as a function of in in th circuit dpictd on th right, using th formulas for Z, Z, and Z (do not us Mapl / Mathmatica / MATAB / Mathad for ths calculations and show all stps. in is a prfct A voltag signal with a frquncy of ω. IN OUT Plot th magnitud and phas of out as a function of ω for kω, µf, and µh. What happns to th magnitud and th phas of out at ω? (Mapl / Mathmatica / MATAB / Mathad ar prmittd for th plots. Figur 3.7: Anothr filtr circuit
13 ab 3: A signals, omplx Impdanc, and Phas Sction : Introduction to transformrs In this sction, w us a transformr to chang th impdanc of an A signal. a. Masur th input impdanc of a spakr. Masur th output impdanc of a signal gnrator with a.5 amplitud sinusoid output of khz. mmbr you ar using A signals. How do you masur currnt with an oscilloscop? What dos an A currnt rading from a DM man in trms of th wavform? hck this with th oscilloscop. If masurmnts do not match call instructor for discussion bfor you procd any furthr. b. Masur th signal gnrator output without any load. Thn connct th signal gnrator to a spakr and masur th signal amplitud. Th voltag drops so much bcaus of th impdanc mismatch. Masur th powr into th spakr. c. Us a transformr to dcras th output voltag, whil incrasing th output currnt into th spakr. Masur out and I out of th signal gnrator, and in, and I in for th spakr. How wll dos th transformr transmit powr? Dos out / in I in /I out? Estimat th ratio of primary turns to scondary turns? d. Masur th output impdanc of th signal gnrator plus transformr circuit. Dos th masurd valu agr with what you xpct thortically? What should b th ratio of th transformr for th idal impdanc matching of th signal gnrator to th spakr? Sction : Th circuit In this sction, w tak a first look at th classic circuit and th concpt of phas.. Gt two capacitors and masur thir individual capacitancs. Masur th total capacitanc with a capacitanc mtr whn thy ar in sris, and whn thy ar in paralll. Do you gt good IN agrmnt with what you xpct? f. onstruct th circuit to th right, with componnt rangs  kω and.. µf. St th function OUT gnrator at approximatly ω./ with a squar wav and dscrib what you s. Masur th tim constant of th xponntial and us it to dtrmin th capacitanc of ( should b dtrmind with a multimtr. Figur 3.8: An filtr g. (Sam stup St th function gnrator to sinusoidal circuit. output at ω/ and masur th magnitud of in and out. Do you gt what you xpct? Masur th phas of out with rspct to in and mak a issajou plot of out and in
14  3 
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