I. INTRODUCTION. There are two other circuit elements that we will use and are special cases of the above elements. They are:

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1 I. INTRODUCTION 1.1 Crcut Theory Fundamentals In ths course we study crcuts wth non-lnear elements or deces (dodes and transstors). We wll use crcut theory tools to analyze these crcuts. Snce some of tools deeloped n crcut theory apply only to lnear elements (thus, lnear crcut theory), let s frst examne what we can use to analyze non-lnear elements. Crcut theory s an approxmaton to Maxwell s electromagnetc equatons to smplfy analyss of complcated crcut. There are nne fundamental crcut elements n crcut theory, denoted by ther characterstcs: Resstor: Capactor: = R = C d dt or V = 1 jωc I Inductor: = L d or V = jωl I dt Independent oltage source: = s = const. for any current Independent current source: = s = const. for any oltage and four controlled sources: oltage-controlled oltage source, (smlar to an ndependent oltage source but wth source strength dependng on oltage on another element n the crcut), current-controlled oltage source, oltage-controlled current source, current-controlled current source. There are two other crcut elements that we wll use and are specal cases of the aboe elements. They are: Short Crcut: = 0 for any current Open Crcut: = 0 for any oltage As can be seen, short crcut s a specal case of a resstor (wth R = 0) or a specal case of a oltage source (wth s = 0) whle open crcut s a specal case of a resstor (wth R ) or a specal case of a current source (wth s = 0). It s essental to remember that the aboe crcut elements do NOT represent physcal deces, rather they are dealzed elements, cooked up to smplfy the analyss (because ther oltage-current relatonshp s lnear). ECE65 Lecture Notes (F. Najmabad), Sprng

2 Any two-termnal network (a box/dece wth two wres comng out of t) whose oltage s drectly proportonal to the current flowng through t,.e., the element characterstcs equaton s = R, can be modeled as an deal resstor. Ths means, that we can take ths box/dece out of the crcut and a replace t wth a resstor and the response of the crcut does not change. Most mportantly, we do NOT need to know what s nsde the box, the only parameters we need to now s ts resstance alue. Smlarly, f we hae a black box whose oltage s a constant for all currents, t can be modeled as an ndependent oltage source (wthout any knowledge of what s nsde the box). You actually hae been dong ths n the lab, modelng the power supply (whch ncludes many transstors, dodes, resstors, capactors) wth an ndependent oltage source. On the other hand, physcal elements (.e., resstors n the lab) can only be modeled wth one of these deal elements wthn a certan range of parameters and wthn a certan accuracy. For example take a resstor n the lab. At hgh enough frequences, t wll exhbt capactance (.e. ts resstance drops as frequency ncreases). At hgh enough current, when the resstor s hot enough, the rato of / s not lnear anymore. So, an deal resstor used n crcut theory s NOT a physcal dece. Rather, a resstor n the lab can be approxmated by an deal resstor only for a range of currents or oltages (typcally rated by ts maxmum power), a range of frequences, and een a range of enronmental condtons (temperature, humdty, etc.). The bottom lne s that the characterstcs of a two-termnal network or dece s the key (not what s nsde the box). If t follows the characterstcs equaton of one of the fe elements aboe, we can use the correspondng deal crcut element n our analyss. The arables n a crcut are currents and oltages n each element. The physcs of current flow s captured n the characterstcs of each element. Two general rules goerns what happens when these elements are connected to each other: Krchhoff current low, KCL, whch s conseraton of electrc charge and Krchhoff oltage law, KVL, whch s a topology-dren constrant (.e., you get to the same place f you follow a closed loop). These two rules are ndependent of nternal physcs of elements and can be appled to non-lnear elements. In each crcut wth N elements, we hae 2N unknowns ( and of each element) and we need 2N equatons: N characterstcs equatons whch descrbe the nternal physcs of each element and N KVL and KCLs whch depend only on how the elements are connected to each other. In the crcut theory, we learned that we can reduce the number of equatons to be soled by a large number usng Node-Voltage and Mesh-Current methods. As these method really are a compact form of wrtng KVL and KCLs, they equally apply to non-lnear elements. ECE65 Lecture Notes (F. Najmabad), Sprng

3 Other mportant crcut theory tools nclude: 1) Theenn and Norton Equalent crcuts, 2) Proportonalty and Superposton, and 3) Frequency doman (phasors) or s-doman analyss of crcut wth snusodal sources. These tools, howeer, only apply to lnear elements and we cannot use them for crcut wth non-lnear elements. Lnear crcut analyss (ncludng these tools) are so conenent that n a lot of cases we buld approxmate lnear models for dodes and transstors so that we can apply the aboe rules n specal cases. As such, short descrptons of frequency doman and Theenn equalent crcuts are gen below. Whch Soluton Method to Use? By lookng at the crcut you should be able to decde the best method to sole the crcut. Bascally, one wants to hae the smallest number of equatons. Assumng that we hae reduced the crcut (.e., replaced parallel and seres elements): KVL & KCL: 2N element equatons Node-oltage Method: N nodes 1 N IV S equatons Mesh-Current Method: N loops N ICS equatons (IVS: ndependent oltage source, ICS: ndependent current source). Obously, one should use KVL & KCL only f there are only a few elements. Furthermore, we are mostly nterested n oltages n the crcuts. As such, usually node-oltage method s preferred as we wll hae a far number of oltage sources and the answer s also a oltage. 1. You CANNOT mx and match the three methods aboe! 2. Apply the technques consstently e.g., for example, always wrte KCL as sum of currents flowng out of a node = 0. Ths mnmzes the chance for error. 1.2 Frequency Doman In prncple, the oltages and currents n analog crcuts are arbtrary functons of tme (we call them sgnals or waeforms). Analytcal analyss of the crcut response to an arbtrary nput waeform s dffcult and requres soluton to a set of dfferental equatons. Een numercal analyss becomes dffcult when there are a lot of crcut elements. Fortunately, there are ways to fnd the response of a lnear crcut to tme-dependent sgnal. These approaches are based on the followng obseratons: 1. For crcuts dren by snusodal sources, the forced response of the state arables (currents and oltages) are all snusodal functons wth the same frequency as the source. ECE65 Lecture Notes (F. Najmabad), Sprng

4 Ths s dered from the mathematcal propertes of snusodal functons. Forced response of a set of lnear dfferental equatons (crcut equatons) to a snusodal functon s a snusodal functon. Ths property leads to specal analyss tools for AC crcuts usng phasors, or usng Fourer transform. AC steady-state analyss of lnear crcuts are coered n ECE35/45. When we use phasors, the crcut equaton do not contan tme anymore, but they nclude frequency ω. As such, ths s usually called analyss n frequency-doman to dfferentate that from tme-doman analyss where we sole the dfferental equaton to fnd the crcut response. 2. Any arbtrary but perodc sgnal can be wrtten as a sum of snusodal functons usng Fourer seres expanson. For example, a square wae wth perod T or frequency ω 0 = 2πf = (2π)/T and ampltude V m can be wrtten as: (t) = 4V m π [ sn(ω 0 t) 1 3 sn(3ω 0t) 1 ] 5 sn(5ω 0t)... Sgnals wth frequences nω 0 (n nteger) are called harmoncs of the fundamental frequency, ω 0. In general the ampltude of hgher harmoncs become smaller as n become larger. The dea of decomposton of a perodc functon to a sum of snusodal functons can be extended to an arbtrary temporal functon by usng Fourer ntegrals. As such, n prncple, any functon of tme can be wrtten as a sum of (or an ntegral of) snusodal functons. 3. Proportonalty and superposton prncples state that the response of a lnear crcut to a lnear combnaton of sources s equalent to the lnear combnaton of the crcut response to each nddual source. Bascally, n a crcut wth seeral ndependent sources, the alue of any state arable equals to the algebrac sum of the nddual contrbutons from each ndependent source. So, n a crcut wth a tme-dependent source, we can use Fourer seres decomposton and replace the source wth a lnear combnaton of seeral snusodal sources. We can then fnd the response of the crcut to each snusodal source and then use proportonalty and superposton to fnd the response to the tme-dependent source. For example, suppose we want hae a crcut dren by a source that can be decomposed nto (t) = A cos(100t)b cos(300t). We want to know the oltage across an element, o (t). We sole the crcut wth the source cos(100t) and fnd the oltage across the element nterest, suppose α cos(100t φ α ). We then repeat the analyss wth a source cos(300t) and fnd the oltage across the element nterest, suppose β cos(300t φ β ). The response of the crcut to (t) = A cos(100t) B cos(300t), then s o (t) = Aα cos(100t φ α ) Bβ cos(300t φ β ). The problem s actually much smpler than the example aboe. In prncple, soluton of AC steady-state crcut s smple and we typcally fnd the response the crcut wth frequency, ECE65 Lecture Notes (F. Najmabad), Sprng

5 ω, as a parameter. We can then construct the response by replacng ω wth frequences of nterest n the response equaton (e.g., set ω = 100 and 300 n the aboe example). Another major smplfcaton arses when the crcut response s frequency ndependent. In that case, the crcut response can be drectly appled to any tme-dependent functon. For example, n the aboe example, f the crcut response to cos(100t) and cos(300t) sources were, respectely, α cos(100t) and α cos(300t) (frequency ndependent), then the crcut response s smply: o (t) = α (t). Therefore, n many crcut applcatons we focus on crcuts dren by snusodal sources. We sole these crcuts n frequency doman. We try to fnd crcut parameters wth frequency ω as a parameter to facltate constructon of response to an arbtrary functon of tme. There are seeral ways to sole the crcut n frequency doman, all hang the same mathematcal foundaton. We can use phasors (whch are really Fourer Transforms). Or, we can use complex frequency doman whch s sometmes called s-doman (s = σjω). In junor leel courses and beyond, you wll probably use complex frequency doman manly. Crcut analyss wth phasors s suffcent for the work we do n ths class (to conert from phasors to s-doman), smply replace jω wth s and ω 2 wth s 2. Analyss n frequency doman s straght-forward. Resstors, capactors, and nductors are replaced by mpedances, Z: Z = R for a resstor, Z = 1/(jωC) for a capactor and Z = jωl for an nductor. Impedances obey Ohm s Law: V = ZI. Thus, wth mpedances the crcut reduces to a resste crcut and all analyss technques of resste crcuts (node-oltage method, mesh-current method, Theenn Theorem, etc.) apply. The only dfference s that analyss s performed usng complex arables. 1.3 Theenn Theorem and Theenn or Norton Equalents T R T N R N We know from lnear crcut theory that the characterstcs of a two-termnal network s n the form of (usng acte sgn conenton): = T R T ; R T = R N ; N R N = T (n frequency or s doman, we should replace R T wth Z T ). ECE65 Lecture Notes (F. Najmabad), Sprng

6 Ths means that we only need to sole and/or measure the Theenn equalent of a twoport termnal once. From then one, the two-port termnal can be replaced wth ether of ts Theenn or Norton equalents wthout affectng the response of the rest of the crcut. An mportant corollary to the Theenn Theorem s that f a two-termnal network does not nclude an ndependent source t can be reduced to a sngle resstance (een f t ncludes dependent sources). How to calculate the Theenn equalent You hae seen a detaled dscusson of Theenn/Norton forms n your crcut theory course(s). In summary, the best method s to calculate two of the the followng three parameters: (1) Open-crcut oltage, oc (found by settng = 0), (2) Short-crcut current, sc (found by shortng the termnals of the two-termnal network,.e., settng = 0), and (3) Drect calculaton of R T whch s the resstance seen at the termnals wth the ndependent sources zeroed out (.e., ther strengths set equal to zero). Remember, you should NOT zero out dependent sources. Example 1: Fnd the Theenn and Norton Equalent of ths crcut: 1. oc : Usng node-oltage method and notng that snce = 0, by KVL, 1 = oc = = 0 T = oc = 32V 25 V V A 4 3 A = 0 oc 2. R T (zerong the ndependent sources): From the crcut, we hae R T = 4 (5 20) = 4 4 = 8 Ω sc : Note that sc = 1 / = = R T 1 = 16V sc = N = = 4A So, the Theenn/Norton parameters are: T = 32 V, N = 4 A, and R T = 8 Ω. (note T = N R T.) 25 V 20 3 A sc ECE65 Lecture Notes (F. Najmabad), Sprng

7 Theenn Equalent of two-termnal networks wth controlled sources Example: Fnd the Theenn equalent of ths twotermnal network. Fndng oc : Snce the crcut s smple, we proceed to sole t wth KVL and KCL (notng = 0): 2k 32 V 1.2k 4 KCL: 1 4 = 0 1 = 0 KCL: = 0 2 = 0 KVL: oc = 0 T = oc = 32 V 2 2k 32 V 1.2k 4 1 = 0 oc Fndng sc : Usng KVL and KCL: 2 2k 4 = sc KCL: 1 4 = 0 1 = 5 sc KCL: = 0 2 = sc KVL: = 0 32 V 1.2k sc sc = 0 N = sc = A = 4 ma Fndng R T : We zero out all ndependent sources n the crcut. The resultng crcut cannot be reduced to a smple resstor by seres/parallel formulas. We can fnd R T, howeer, by attachng a test source, x to the termnals and calculate current x (see fgure). Snce the two-termnal network should be reduced to a resstor (R T ), we should get R T = x / x. Snce the crcut s smple, we proceed to sole t wth KVL and KCL (note x = ): 2 2k 1.2k 4 1 x x KCL: 1 4 = 0 1 = 5 = 5 x KCL: = 0 2 = = x KVL: x = x x x = 0 R T = x X = = 8 kω ECE65 Lecture Notes (F. Najmabad), Sprng

8 How to measure the Theenn equalent Suppose we hae gen a box wth two termnals and want to measure the Theenn equalent of the crcut nsde the box. In prncple, we cannot use the aboe technque and try to measure oc, sc, and R T. We cannot turn off the nput sgnal and use a ohm-meter to measure R T. Nor can we short the termnals and measure sc (there s a good chance that we are gong to run the crcut f we do that). In prncple, we can use a olt-meter (or scope) to measure oc but care should be taken as t s not known a pror f the nternal resstance of the olt-meter (or scope) s large enough to act as an open crcut (there are other complcatons). There s also the ssue of measurement error that one should consder. The best way to measure the Theenn Equalent parameters (works for Resste R T ) s to measure the characterstcs of the two-termnal network. We can do ths by attachng a arable load (a resstance) to the box, ary the load whch changes the output oltage and currents, and measure seeral par of and (here we do not use the alue of R L ). Typcally ths s done wth startng from a large R L and gradually reducng the load. These data pont should le on the lne of the twotermnal network. Values of T, N, and R T can be read drectly from the graph as shown. Ths method s specally accurate as one can use a best-ft lne to the data n order to mnmze random measurement errors. A smpler, but less accurate erson of the aboe method s to measure the output oltage for TWO dfferent alues of R L (.e., R L1 and R L2 wth 1 and 2, respectely). From the crcut: = T 1 = T R L R T R L R L1 R T R L1 and Ddng the two equatons ge: 2 T = R L2 R T R L2 N slope of 1/R T RT T T R L R R L L 1 2 = R L1 R T R L1 R T R L2 R L2 whch can be soled to fnd R T. Then, one of the aboe equatons for 1 or 2 can then be used to fnd T. Typcally, we choose R L2 to be ery large, R L2 (e.g., nternal ECE65 Lecture Notes (F. Najmabad), Sprng

9 resstance of scope), then 2 = oc (open crcut oltage) and 1 oc = R L1 R T R L1 R T R L1 = oc 1 1 Note that we should choose R L1 such that 1 s suffcently dfferent from oc for the measurement to be accurate. Typcally, experment s repeated for seeral alues of R L1 untl 1 / oc s between 0.5 to 0.8). How to fnd the Theenn equalent usng PSpce: You can use the same technque descrbed aboe for computng the Theenn parameters wth PSpce. Attach a arable load ( Parameter n PSpce) to the crcut. Ask PSpce to compute output oltage V as a functon of load resstance R L. Plot the output current ersus the output oltage and you wll hae the characterstcs of the crcut smlar to the fgure aboe (Make sure that you hae the current drecton correctly!). 1.4 Crcut Components It s not practcal to desgn a complete crcut as a whole from scratch. It s usually much easer to break the crcut nto components and desgn and analyze each component separately. In ths manner we can desgn buldng blocks (such as amplfers, flters, etc.) that can be used n a arety of deces. A typcal analog crcut s composed of a source, a load whch s a two-termnal network (deces wth two wre comng out) and seeral two-port networks (deces wth two wres gong n and two wres comng out). Note that these components communcate wth each other only through the attachng wres,.e., through currents and oltages. Source 2port Network 2port Network Load For two-termnal networks, the relatonshp between output oltage and current dctates how ths network behae. If we dere ths relatonshp once for a gen two-termnal-network, we can sole any crcut whch ncludes that two-termnal network wthout solng for nternals of the two-termnal network. For two-termnal networks contanng only lnear elements, Theenn theorem can be used to model such a network wth two fundamental crcut elements. For two-termnal networks wth non-lnear elements, the relatonshp between output oltage and current s non-lnear (see for example Zener dode power supply of Sec. 2). ECE65 Lecture Notes (F. Najmabad), Sprng

10 In two-port networks the nput sgnal (ether nput current or nput oltage) s modfed by the crcut and an output sgnal (ether output current or output oltage) s generated. For most electronc crcuts, we keep the currents low and modfy oltages (as dscussed n Appendx). As such the relatonshp between output oltage and nput oltage dctates the response of the two-port network. Ths s called the transfer functon. For two-port networks wth non-lnear elements, ths s a non-lnear relatonshp (e.g., dode waeform shapng crcuts of Sec. 2). If a two-port network ncludes only lnear elements, the network can be modeled by four lnear crcut elements (often 3) and ts transfer functons s lnear (.e., rato of o / does not depend on ). In Sec. 5 we wll show that lnear amplfers can be characterzed by three parameters and we wll use ths technque to dde the crcut nto components and smplfy the analyss consderably. A more general erson of ths approach s gen n the Appendx How each sub-crcut sees other elements The strategy of ddng a lnear crcut nto nddual components works because of the Theenn Theorem. Recall that any two-termnal network can be replaced by ts Theenn equalent. In addton, f a two-termnal network does not nclude an ndependent source t wll be reduced to a sngle mpedance (een f t ncludes dependent sources). Source 2port Network 2port Network Load Source see ths twotermnal network Load sees ths twotermnal network What Load sees: The load sees a twotermnal network. Ths two-termnal network contans an ndependent source. So t can be reduced to ts Theenn equalent. T R T Load What Source sees: The source sees a twotermnal network. Ths two-termnal network does not contan an ndependent source. So t can be reduced to a sngle mpedance. Source R L ECE65 Lecture Notes (F. Najmabad), Sprng

11 What each two-port network sees: Followng the logc aboe, ts obous that each two-port network sees a two-termnal network contanng an ndependent source n the nput sde (can be reduced to a Theenn form) and a two-termnal network that does not contan an ndependent source on the output sde (so t can be reduced to a sngle mpedance). s R s 2port Network o o R L The aboe obseratons ndcate that we do not need to sole a complete crcut. For twotermnal networks lke the source, we only need to fnd ther characterstcs (or T and R T for lnear crcuts) to be able to predct ts response when t s attached to any crcut (here modeled as R L ). For a two-port network, we only need to sole the crcut aboe wth s, R s, and R L as parameters. Then, whereer ths two-port network appears n a crcut, we can use these results. 1.5 Mathematcs ersus Engneerng You should hae learned by now that one cannot achee mathematcal accuracy n practcal systems. Frstly, our nstruments hae a fnte accuracy n measurng alues. When a number (or measurement), A, has a tolerance of ǫ, t means that ts alue s between A(1 ± ǫ) = A ± ǫa. Ths means that we cannot dfferentate between any number n the range A ǫa to A ǫa. We would say that all numbers n ths range are approxmately equal to each other: B A A ǫa B A ǫa and we can use B and A nterchangeably as we cannot dstngush between them. As an example, the scopes n ECE65 lab are accurate wthn 2%. So, f the scope (ǫ = 0.02) reads a alue of V, the real alue s anywhere between ± or n the range of to In ths context any number between and s approxmately equal to as we CANNOT dfferentate among them by our measurements: and Corollary: In the example aboe, the 4th sgnfcant dgts n s totally meanngless (see the range of numbers we cannot dstngush). It s a poor engneerng practce to een report ths 4th sgnfcant dgt! (Stll some ECE65 students report ther calculatons to 8th sgnfcant dgts, drectly wrtng the number from ther calculators!). Smlarly, t s poor engneerng practce to report numbers n whole fractons (e.g., 4/3). No measurng nstrument measure any property n whole numbers! ECE65 Lecture Notes (F. Najmabad), Sprng

12 Secondly, each element/component/system s manufactured to a certan tolerance the smaller the tolerance, the more expense s to buld that component. For example, resstors we wll use n the Lab hae a tolerance of 5%. Ths means that a 1 kω has a alue of 1, 000 ± 5% = 1, 000 ± 50 Ω or somewhere between 950 and 1,050 Ω. A corollary of ths concept s that f you desgned a crcut and found that you need a 1,010 Ω resstor, you CANNOT put a 1 kω and a 10 Ω resstor (wth 5% tolerance) n seres. The resultant combnaton would hae a alue between and 1,060.5 Ω whch s no better than a 5% 1 kω resstor. If you need to hae a 1,010 Ω resstor (.e., more precson), you should use 1% resstors (whch are more expense). Concepts of nfnty and zero are meanngless n abstract. They are used n the context of much bgger and much smaller. For example, n the dscusson of page 1-8 of measurng Theenn parameters, we noted 2 = oc = T for R L2, snce 2 T = R L2 R T R L2 Howeer, from the aboe equaton 2 oc = T when R L2 R T. Ths means that we hae defned nfnte R L2 as R L2 R T,.e., wth respect to another resstance. For example, f R T = 1 Ω, een a R L2 = 100 Ω resstance would be nfnte, whle f R T = 1000 Ω, a 100 Ω load resstance would actually be small. Smlarly, 2 = T for R T = 0, Howeer, from the aboe equaton, 2 T when R T R L2. Agan R T = 0 s defned as R T R L2. So, concepts of large and small (zero and nfnte) requre a frame of reference,.e., bg or small compared to what, and should be stated as much smaller or much bgger than. Notons of much smaller ( ) and much greater ( ) are meanngful only n term of a gen or needed relate tolerance, ǫ. Consder quantty B = Aa. We use the concept of much smaller, a A, to wrte B A. From the aboe defnton of approxmate, we should hae (assumng that a and A are poste): B A A ǫa B A ǫa a A a ǫa A ǫa A a A ǫa a ǫa Exercse: Show that wth a tolerance of ǫ, A a means A (1/ǫ)a. For most day-to-day use, a tolerance of 5% to 10% s more than suffcent. As a general rule, we wll use a tolerance of 10% n the analyss n ECE65 unless otherwse stated. ECE65 Lecture Notes (F. Najmabad), Sprng