I. INTRODUCTION. 1.1 Circuit Theory Fundamentals

Transcription

1 I. INTRODUCTION 1.1 Crcut Theory Fundamentals Crcut theory s an approxmaton to Maxwell s electromagnetc equatons n order to smplfy analyss of complcated crcuts. A crcut s made of seeral elements (boxes or deces) whch are connected to each other wth deal wres. The arables n a crcut are currents and oltages n all elements. The physcs of current flow nsde each element s captured by ts characterstcs. Snce crcut elements communcate wth each other only through current and oltages, elements wth smlar characterstcs are ewed by the rest of the crcut as beng dentcal regardless of what s nsde the element/box. Two general rules goern what happens when these elements are connected to each other: Krchhoff current law, 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 a crcut wth N elements, we hae 2N unknowns ( and of each element) and we need 2N equatons: N characterstcs equatons for elements and N KVLs and/or 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-oltage and mesh-current methods. As these method really are compact forms of wrtng KVL and KCLs, they equally apply to crcut wth non-lnear elements. A lnear crcut s made of lnear elements(.e., characterstcs of eery element s lnear). A lnear crcut has many desrable propertes such as proportonalty and superposton whch are essental for many practcal crcuts (e.g., an amplfer). Because of the desrablty of lnear crcuts, crcut theory ncludes symbols for fe fundamental two-termnal elements whch are lnear: Resstor: Capactor: Inductor: Independent oltage source: Independent current source: = R = C d dt or V = 1 jωc I = L d or V = jωli dt = s = const. for any current = s = const. for any oltage ECE65 Lecture Notes (F. Najmabad), Wnter

2 and four two-port networks (boxes wth two-wres gong n and two-wre comng out): 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 two-termnal elements that we wll use and are specal cases of the aboe elements. They are: Short Crcut: Open Crcut: = 0 for any current = 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 are NOT physcal deces. The dealzed resstor n crcut theory s a symbol for any element whose characterstcs equaton s = R. Ths means that f we hae a two-termnal network (box or dece) whch has a = R characterstcs, we can represent the network n a crcut wth an deal resstor. Smlarly, f we hae a box whose oltage s constant for all currents, t can be represented 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. Physcal elements (.e., resstors n the lab) can only be modeled wth one of these dealzed crcut elements wthn a certan accuracy (dscussed n Sec. 1.4) and/or wthn a certan range of parameters: For example take a resstor n the lab. At hgh enough current, when the resstor s hot enough, the rato of / s not lnear anymore. So, a resstor n the lab can be approxmated by an deal crcut-theory resstor only for a range of currents or oltages (typcally rated by ts maxmum power). The bottom lne s that the characterstcs of a two-termnal network s the key. For example, we wll make a two-termnal network wth transstors whch has a lnear characterstcs. From the pont ew of the rest crcut ths two-termnal network s dentcal to an deal resstor. ECE65 Lecture Notes (F. Najmabad), Wnter

3 Whch Soluton Method to Use? Snce we wll sole a lot of crcuts n ths course, t s essental that one should be able to choose the best method to sole a crcut. Bascally, one wants to hae the smallest number of equatons. Assumng that we hae reduced the crcut (.e., replaced parallel and seres elements), one can readly estmate the number of crcut equatons for each method: KVL & KCL: 2N element equatons Node-oltage method: N nodes 1N IVS equatons Mesh-Current method: N loops N ICS equatons where N IVS denotes the number of ndependent oltage sources and N ICS the number of ndependent current sources. 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. (PSpce uses node-oltage method but wth full non-lnear characterstcs equatons for elements). Note: 1. You CANNOT mx and match the three methods aboe! 2. Apply the technques consstently e.g., always wrte KCL as sum of currents flowng out of a node = 0. Ths mnmzes the chance for error. Functonal Crcuts: In ths course, we wll use dodes and transstors to buld crcuts whch do specfc functons (and learn to desgn smple crcuts). These crcuts are ether: 1. Two-termnal networks,.e., a box wth two wre comng out of t, such as a power supply, output termnals of an Pod, etc. (Sec. 1.2) or 2. Two-port networks,.e., a box wth two wres gong n and two wres comng out of t, such as an amplfer, rectfer, etc. (Sec. 1.3) Note that prncples of two-port networks (dscussed later) can be easly extended to crcuts wth multple nputs. ECE65 Lecture Notes (F. Najmabad), Wnter

4 1.2 Two-termnal networks and Theenn Theorem The smplest functonal crcut s a two-termnal network. Snce the rest of the crcut communcates wth our network through oltages and currents, we only need to calculate(or measure) ts characterstcs. In general, we sole the crcut and fnd the output oltage as a functon of output current (.e., assume output oltage s and compute ). Note that the conenton s to use acte sgn conenton for our two-termnal networks. = f() or = g() Our calculaton would be much smpler f the characterstcs equaton of a two-termnal network s lnear. Theenn theorem states that f a two-termnal network contans only lnear elements, ts characterstcs equaton would be lnear. Furthermore, f the characterstcs equaton of a two-termnal network s lnear, we can represent the network wth a combnaton of two of deal crcut-theory elements. Proof of the Theenn Theorem s straght forward. A lnear characterstcs equaton,.e., A B C = 0 can be wrtten as = T R T (wth T = C/A and R T = B/A). A crcut contanng a oltage source T and a resstor R T n seres would hae the same characterstcs equaton and, thus, can used nstead of the two-termnal network n crcut analyss. Such a smple crcut s called the Theenn equalent crcut. Smlarly, the lnear characterstcs equaton can be wrtten as = N /R N (wth N = C/B and R N = B/A) and the Norton equalent crcut contanng a current source N and a resstor R N n parallel can be constructed. Note that R T = R N and N R N = T. T R T Theenn Equalent N R N Norton Equalent 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 resstor (een f t ncludes dependent sources). Not only Theenn Theorem smplfes calculaton of the characterstcs of a two-termnal network (as s dscussed below), t allows us to dentfy a two-termnal network wth only two numbers (nstead of an equaton) How to calculate characterstcs and the Theenn parameters 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 followng three parameters: ECE65 Lecture Notes (F. Najmabad), Wnter

5 1. Open-crcut oltage, oc (found by settng = 0) = T R T = oc = T R T 0 = T 2. Short-crcut current, sc (found by shortng the termnals of the two-termnal network,.e., settng = 0), = N /R N = sc = N 0 = N 3. DrectcalculatonofR T whchstheresstanceseenatthetermnalswththendependent 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: 5 4 Usng Node-oltage-method: 1. oc : Note that by Ohm s law across 4-Ω resstor oc = = 0 1 = 32V T = oc = 1 = 32V 25 V 25 V 20 3 A = A oc 2. sc : Note that by Ohm s law across 4-Ω resstor sc = 1 / = 0 1 = 16V N = sc = = 4A V A sc R T (zerong the ndependent sources): From the crcut, we hae R T = 4(5 20) = 44 = 8 Ω. 20 R T ECE65 Lecture Notes (F. Najmabad), Wnter

6 Whle Theenn Theorem leads to a smpler analyss, t cannot be used for crcuts wth non-lnear elements. In general, we need to calculate the characterstcs of the network drectly. The way to do ths s to assume the output oltage s known and compute n terms of. If we use the node-oltage method, frst we hae to compute all node oltages n terms of. Computng n terms of node oltages would then ge n terms of. In general, computng characterstcs drectly s qute tme-consumng. We wll sole the crcut of Example 1 to demonstrate how the method works (for ths crcut, the soluton s rather smple because there s only one unknown node, 1 ). Usng Node-oltage method, we frst fnd all node oltages n terms of : = = V A We now calculate from node oltages. Here, can be found from Ohm s law across the 4-Ω resstor: 4 = 1 = (0.5 16) = = 328 Note that snce the characterstcs s lnear, ths crcut can be reduced to ts Theenn equalent: = 328 T R T whch ges T = 32 V and R T = 8 Ω. Example2 : Fnd the Theenn equalent of ths two-termnal network. 2k 4 32V 1.2k Fndng oc and sc are left as an exercse (answer: T = oc = 32 V and N = sc = 4 ma). Fndng R T : We zero out all ndependent sources n the crcut. Howeer, the resultng crcut cannot be reduced to a smple resstor by seres/parallel formulas. To fnd R T, we attach a test source, x to the termnals and calculate current x (see crcut). Snce the two-termnal network should be reduced to a resstor (R T ), we should get R T = x / x. ECE65 Lecture Notes (F. Najmabad), Wnter

7 Snce the crcut s smple, we proceed to sole t wth KVL and KCL. We note that = x and current x flows n the 2 k resstor (see crcut): x 2k 1 1.2k 4 x x KCL: 1 4 x = 0 1 4( x ) x = 0 1 = 5 x KVL: x (5 x ) x = 0 x = x R T = x x = = 8 kω How to measure the characterstcs and Theenn parameters Suppose we hae a two-termnal network and want to measure ts Theenn equalent crcut. In prncple, we cannot use the aboe technque and try to measure oc, sc, and R T. We cannot use an ohm-meter to measure R T. It s not adsable to 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 scope (or olt-meter) 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. The best way to measure the Theenn Equalent parameters s to measure the characterstcs of the twotermnal network. We can do ths by attachng a arable load (a arable resstance) to the box and ary the load (R L ). For each alue of R L, we measure the par of and (hghlghted wth a n the fgure). We can fnd the characterstcs of the dece by repeatng ths measurement wth seeral alues of R L as s shown. Typcally ths s done wth startng from a large R L and gradually reducng the load. If the crcut s lnear, these data pont should le on a lne (see the fgure). Values of T, N, and R T can be read drectly from the graph as s shown. Ths method s specally accurate as one can use a best-ft lne to the data n order to mnmze random measurement errors. N slope of 1/R T T R L ECE65 Lecture Notes (F. Najmabad), Wnter

8 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 measured oltages of 1 and 2, respectely). From the crcut: T R T R L 1 T = R L1 R T R L1 and 2 T = R L2 R T R L2 Ddng the two equatons ges: 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 be used to fnd T. Typcally, we choose R L2 to be ery large, R L2 (e.g., nternal 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 In general, 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 0.3 to 0.7). Note that the aboe method has to be modfed f the box ncludes tme-dependent sources as R T may be an mpedance. (Exercse: How?) In addton, ths method would not ge the Theenn equalent crcut f T = 0 (.e., the crcut reduces to a resstor) How to fnd the characterstcs and Theenn parameters usng PSpce You can use the same technque descrbed aboe for measurng the Theenn parameters wth PSpce. Attach a arable load (usng Parameter settng n PSpce) to the crcut. Ask PSpce to compute output oltage for dfferent alues of the parameter 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!). ECE65 Lecture Notes (F. Najmabad), Wnter

9 1.3 Two-port networks 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 (amplfers, flters, etc.) that can be used n a arety of deces. A typcal analog crcut s composed of a source, and a load both of whch are two-termnal networks dscussed before. A typcal crcut also contans seeral two-port networks as s shown below. Source 2port Network 2port Network Load In a two-port network 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 n order to mnmze power dsspaton n the crcut. As such the relatonshp between the output oltage and the nput oltage dctates the response of the two-port network. Ths relatonshp s called the oltage transfer functon of the network. For two-port networks wth non-lnear elements, transfer functon s non-lnear (e.g., dode waeform shapng crcuts of Sec. 2). Howeer, If a two-port network ncludes only lnear elements, ts transfer functons s lnear (.e., rato of o / does not depend on ). In addton, such a lnear two-port network can be modeled by four lnear crcut elements (often 3) as we dscuss n the transstor amplfer secton 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 ECE65 Lecture Notes (F. Najmabad), Wnter

10 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. sg Rsg 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 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). sg Rsg 2port Network o o R L The aboe obseratons ndcate that we do not need to sole a complete crcut. For a two-termnal network lke the source, we only need to fnd ts characterstcs (or T and R T for a lnear network) 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 sg, R sg, and R L as parameters. We wll dscuss ths n depth n transstor amplfer secton. Note that the concept of two-port networks can be easly extended to crcut wth multple nputs and one output as we wll see n our dscuss of dgtal gates. 1.4 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. Secondly, we can buld components only wthn a certan accuracy(or tolerance). As a general rule, the hgher the accuracy n a component, the more expense t s. Therefore, we always desgn and buld component based on the accuracy that s needed (to reduce the cost). Lastly, we use dealzed model to descrbe physcal phenomena n practcal systems by makng approxmatons. We also gnore terms n our analyss by argung that they are small or large. Some of these ssues are dscussed below. Approxmatons and gnorng terms n analyss are also related to the accuracy. ECE65 Lecture Notes (F. Najmabad), Wnter

11 What does accuracy mean? When a number (or a measurement), A, has a relate accuracy (or tolerance) of ǫ, t means that ts alue s between A(1 ǫ) to A(1 ǫ). Ths means that we cannot dfferentate between any number n the range AǫA to AǫA. We would say that all numbers n ths range are approxmately equal to each other: B A AǫA B AǫA and we can use B and A nterchangeably as we cannot dstngush between them. Concepts of nfnty and zero are also meanngless n abstract. They are used n the context of much bgger and much smaller. For example, n the dscusson of measurng Theenn parameters, we arred at a equaton lke: o T = R L R T R L Mathematcally, f R L, o = T. In engneerng, f we assume R L R T, we hae R T R L R L and o T. Ths means that we hae defned nfnte R L as R L R T. Smlarly, f R T = 0, o = T. If we assume R T R L, we hae R T R L R L and o T. Ths means that we hae defned zero R T as R T R L. So, concepts of large and small (zero and nfnte) are defned as much smaller or much larger than. They also requre a frame of reference,.e., large or small compared to. In the aboe example, larger and smaller were defned wth respect to another resstor alue. For example, f R T = 1 Ω, a R L = 100 Ω resstor would be large, whle f R T = 1000 Ω, a 100 Ω load resstor would actually be small. Notons of much smaller ( ) and much greater ( ) are defned n term of a gen or needed accuracy, 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 AǫA B AǫA a A a ǫa AǫA Aa AǫA a ǫa Exercse: Show that wth a tolerance of ǫ, A a means A (1/ǫ)a. ECE65 Lecture Notes (F. Najmabad), Wnter

12 Measurement accuracy: Once calbrated, eery nstrument has a certan and fnte accuracy. As an example, the scopes n ECE65 lab are accurate wthn 2% (ǫ = 0.02). So, f a scope reads a alue of V, the real alue s anywhere between 1.352± 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 Note: 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! Component accuracy/tolerance: 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 SHOULD NOT 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 (.e., you hae wasted the second resstor). If you need to hae a 1,010 Ω resstor (.e., more precson), you should use a 1,010 Ω resstor wth a 1% tolerance (whch s more expense). Accuracy n modelng: As we dscussed before practcal elements/components are modeled (.e., approxmated) wth deal crcut-theory elements. Obously, the desred accuracy plays an mportant role n ths approxmaton. To see ths, consder the characterstc of a two-termnal network shown below. For ths element, when the current s ared between 0 and 80 ma, the oltage ares between 1.05 and 0.95 V. Also, note that the characterstc s NOT exactly lnear. Recall that a quantty, B A when AǫA B AǫA. If our desred accuracy s 5% (ǫ = 0.05) and settng A = 1 (the md alue of the oltages n the fgure), we fnd, AǫA = 0.95 and AǫA = That s, wthn an accuracy of 5%, ECE65 Lecture Notes (F. Najmabad), Wnter

13 all oltage alues between 0.95 to 1.05 V are approxmately equal to each other (hghlghted regonnfgurebelowrght). Assuch, wecanmodelourtwo-termnalnetworkwth 1.0V (dashed lne). Snce the oltage across the element s approxmately constant for dfferent alues of the current, ths two-port network can be modeled by a oltage source wth an accuracy of 5%. If our desred accuracy s 1% (ǫ = 0.01), we fnd, A ǫa = 0.99 and A ǫa = 1.01 (hghlghted regon n fgure below left). In ths case, the oltage across the element s NOT approxmately constant for dfferent alues of the current and ths two-port network CANNOT be modeled by a oltage source wth an accuracy of 1%. Howeer, f we compute and plot a best-ft lnear approxmaton to the cure (dashed lne, fgure below rght), we see that the two-port network matches ths approxmaton wthn 1% (hghlghted regon n fgure below rght). Therefore, wth an accuracy of 1%, the twoport network can be modeled wth a lnear approxmaton whch corresponds to a Theenn equalent crcut of T = 1.05 V and R T = 1.2 Ω. What accuracy do we need? For hand-calculatons and the frst-pass desgn work, an accuracy of 5% to 10% s more than suffcent. As a general rule, we wll use an accuracy of 10% n the analyss n ECE65 unless otherwse stated. ECE65 Lecture Notes (F. Najmabad), Wnter