1 /30/008 ntroducton / ntroducton: Analyss of Electronc Crcuts Readng Assgnment: KVL and KCL text from EECS Just lke EECS, the majorty of problems (hw and exam) n EECS 3 wll be crcut analyss problems. Thus, a key to dong well n 3 s to thoroughly know the materal from!! So, before we get started wth 3, let s revew and see how t apples to electronc crcuts. Q: aced EECS last semester; can just skp ths revew?? A: Even f you dd extremely well n, you wll want to pay attenton to ths revew. You wll see that the concepts of are appled a lttle dfferently when we analyze electronc crcuts. Both the conventons and the approach used for analyzng electronc crcuts wll perhaps be unfamlar to you at frst thus magne that everyone ( hope) wll fnd ths revew to be helpful.
2 /30/008 ntroducton / Electronc Crcut Notaton KVL and Electronc Crcut Notaton Analyss of Electronc Crcuts Even the quanttes of current and resstance are a lttle dfferent for electronc crcuts! Q: You mean we don t use Amperes and Ohms?? A: Not exactly! Volts, MllAmps, KloOhms Now let s try an example! Example: Crcut Analyss usng Electronc Crcut Notaton
3 /30/008 Electronc Crcut Notaton /9 Electronc Crcut Notaton The standard electronc crcut notaton may be a lttle dfferent that what you became used to seeng n n EECS. The electronc crcut notaton has a few shorthand standards that can smplfy crcut schematcs! Consder the crcut below: R =K v 3 V s =5V R =4K v R 3 =4K v 3 Note the voltage values n ths crcut (.e., V s,v,v,v 3) provde values of potental dfference between two ponts n the crcut. For example, from the voltage source we can conclude: V s =5V The electrc potental at ths pont n the crcut s 5 volts greater than: the electrc potental at ths pont n the crcut.
4 /30/008 Electronc Crcut Notaton /9 Or the resstor voltage v 3 means: R 3 =4K v 3 The electrc potental at ths pont n the crcut s v 3 volts greater than: the electrc potental at ths pont n the crcut. But remember, v 3 could be a negatve value! Thus, the values of voltages are comparatve they tell us the dfference n electrc potental between two ponts wth n the crcut. John Sally As an analogy, Say John, Sally, and Joe work n a very tall buldng. Our crcut voltages are lttle lke sayng: John s 5 floors above Joe Sally s floors above Joe Joe From ths comparatve nformaton we can deduce that John s 3 floors above Sally.
5 /30/008 Electronc Crcut Notaton 3/9 What we cannot determne s on what floor John, Sally, or Joe are actually located. They could be located at the hghest floors of the buldng, or at the lowest (or anywhere n between). Smlarly, we cannot deduce from the values V s,v,v,v 3 the electrc potental at each pont n the crcut, only the relatve values relatve to other ponts n the crcut. E.G.: Pont R has an electrc potental 5V hgher than pont B Pont G has an electrc potental v 3 hgher than pont B Q: So how do we determne the value of electrc potental at a specfc pont n a crcut? A: Recall that electrc potental at some pont s equal to the potental energy possessed by Coulomb of charge f located at that pont. Thus to determne the absolute (as opposed to relatve) value of the electrc potental, we frst must determne where that electrc potental s zero. The problem s smlar to that of the potental energy possessed by.0 kg of mass n a gravtatonal feld. We ask ourselves: Where does ths potental energy equal zero?
6 /30/008 Electronc Crcut Notaton 4/9 The answer of course s when the mass s located on the ground! But ths answer s a bt subjectve; s the ground : A. where the carpet s located? B. where the sdewalk s located? C. The basement floor? D. Sea level? E. The center of the Earth? The answer s t can be any of these thngs!
7 /30/008 Electronc Crcut Notaton 5/9 We can rather arbtrarly set some pont as the locaton of ground. The potental energy s therefore descrbed n reference to ths ground pont. For tall buldngs, the ground floor s usually defned as the floor contanng the front door (.e. the sdewalk) but t doesn t have to be (just look at Eaton Hall!). Now, havng defned a ground reference, f we add to our earler statements: Joe s 3 floors above ground We can deduce: John John s 5 floors above Joe therefore John s on the 37 th floor Sally Sally s floors above Joe therefore Sally s on the 34 th floor Joe
8 /30/008 Electronc Crcut Notaton 6/9 Q: So, can we defne a ground potental for our crcut? A: Absolutely! We just pck a pont on the crcut and call t the ground potental. We can then reference the electrc potental at every pont n the crcut wth respect to ths ground potental! Consder now the crcut: R =K v 3 V s =5V R =4K v R 3 =4K v 3 Look at ths! Note we have added an upsdedown trangle to the crcut ths denotes the locaton we defne as our ground potental! Now, f we add the statement: Pont B s at an electrc potental of zero volts (wth respect to ground).
9 /30/008 Electronc Crcut Notaton 7/9 We can conclude: Pont R s at an electrc potental of 5 Volts (wth respect to ground). Pont G s at an electrc potental of v 3 Volts (wth respect to ground). Note that all the ponts wthn the crcut that resde at ground potental form a rather large node: R =K v 3 V s =5V R =4K v R 3 =4K v 3 Look at ths! Standard electronc notaton smplfes the schematc by placng the ground symbol at each devce termnal: R =K V s =5V v 3 v R =4K R 3 =4K v 3
10 /30/008 Electronc Crcut Notaton 8/9 Note that all termnals connected to ground are lkewse connected to each other! Now, n the case where one termnal of a devce s connected to ground potental, the electrc potental (wth respect to ground) of the other termnal s easly determned: V s =5V The electrc potental at ths pont n the crcut s 5 volts greater than ground (.e., 5 volts). Ths pont s at ground potental (.e., zero volts). For ths example, t s apparent that the voltage source smply enforces the condton that the termnal s at 5.0 Volts wth respect to ground. Thus, we often smplfy our electronc crcut schematcs as: V s =5V R =K v 3 v R =4K R 3 =4K v 3
11 /30/008 Electronc Crcut Notaton 9/9 Fnally, we fnd that: R 3 =4K v 3 The electrc potental at ths pont n the crcut s v volts greater than ground potental (.e., v 3 ). Ths pont s at ground potental (.e. zero volts). Thus, we can smplfy our crcut further as: R =K V s =5V v 3 v 3 v R =4K R 3 =4K Ths crcut schematc s precsely the same as our orgnal schematc: R =K v 3 V s =5V R =4K v R 3 =4K v 3
12 /30/008 KVL and Electronc Notaton / KVL and Electronc Consder ths crcut: Crcut Notaton V s =4V V s =V R =3K v 4 R =K v 5 R 3 =K v R v R v R3 We can apply Krchoff s Voltage Law (KVL) to relate the voltages n ths crcut n any number of ways.
13 /30/008 KVL and Electronc Notaton / V s =4V V s =V R =3K v 4 R =K v 5 R 3 =K v R v R v R3 For example, the KVL around ths loop s: 4 v v v = 0 R R R3 We could multply both sdes of the equaton by and lkewse get a vald equaton: 4 v v v = 0 R R R3 Q: But whch equaton s correct? Whch one do we use? Whch one s the KVL result? A: Each result s equally vald; both wll provde the same correct answers. Essentally, the frst KVL equaton s constructed usng the conventon that we add the crcut element voltage f we frst encounter a plus () sgn as we move along the loop, and subtract the crcut element voltage f we frst encounter a mnus () sgn as we move along the loop.
14 /30/008 KVL and Electronc Notaton 3/ For example: v 5 v 5V But, we could also use the conventon that we subtract the crcut element voltage f we frst encounter a plus () sgn as we move along the loop, and add the crcut element voltage f we frst encounter a mnus () sgn as we move along the loop! For example: v 5 v 5V Ths conventon would provde us wth the second of the two KVL equatons for our orgnal crcut: 4 v v v = 0 R R R Q: Huh?! What knd of sense does ths conventon make? We subtract when encounterng a? We add when encounterng a??
15 /30/008 KVL and Electronc Notaton 4/ A: Actually, ths second conventon s more logcal than the frst f we consder the physcal meanng of voltage! Remember, the voltage s smply a measure of potental energy the potental energy of Coulomb of charge. v 5V C f C of charge were to be transported around the crcut, followng the path defned by our KVL loop, then the potental energy of ths charge would change as s moved through each crcut element. n other words, ts potental energy would go up, or t would go down. The second conventon descrbes ths ncrease/decrease! 5V v For example as our C charge moves through the voltage source, ts potental energy s ncreases by 5 Joules (the potental s 5 V hgher at the termnal than t was at the mnus termnal)!
16 /30/008 KVL and Electronc Notaton 5/ v 5V But when t moves through the resstor, ts potental energy drops by v Joules (the potental at the mnus termnal s v Volts less than that at the plus termnal). Thus, the second conventon s a more accurate accountng of the change n potental! v 5V 5 v Ths conventon s the one typcally used for electronc crcuts. You of course wll get the correct answer ether way, but the second conventon allows us to easly determne the absolute potental (.e., wth respect to ground) at each ndvdual pont n a crcut. To see ths, let s return to our orgnal crcut:
17 /30/008 KVL and Electronc Notaton 6/ V s =4V V s =V R =3K v 4 R =K v 5 R 3 =K v R v R v R3 The KVL from these loops are thus: 4 v v = 0 R 4 v v v = 4 R 5 0 v5 v R 3 = 0 4 v v v = 0 R R 5 v v v = 4 R R3 0 Q: don t see how ths new conventon helps us determne the absolute potenal at each pont n the crcut?
18 /30/008 KVL and Electronc Notaton 7/ A: That s because we have not defned a ground potental! Let s do that now: V s =4V V s =V R =3K v 4 R =K v 5 R 3 =K v R v R v R3 V s =4V R =3K v R We can thus rewrte ths crcut schematc as: R =K v R v 4 Remember that all ground termnals are connected to each other, so we can perform KVL by startng and endng at a ground node: v 5 R 3 =K V s =V v R3
19 /30/008 KVL and Electronc Notaton 8/ 4 v v v = 0 R R R3 4 v v = 0 R 4 V s =4V v v v = 4 R 5 0 The same results as before! R =3K v R Now, we can further smplfy the schematc: R =K v R v 4 R =3K 4V v R v 4 v 5 R 3 =K V s =V v R3 R =K v R R 3 =K v R3 v 5 V
20 /30/008 KVL and Electronc Notaton 9/ Note that we were able to replace the voltage sources wth a drect, smple statement about the electrc potental at two ponts wthn the crcut. 4V The electrc potental here must be 4 V! The electrc potental here must be V! V s =4V V s =V R =3K R =K R 3 =K v R v R v R3 v 4 v 5 Note the KCL equaton we determned earler: V 4 v v v = 0 R R R3 Let s subtract.0 from both sdes: 4 v v v = R R R3 Ths s the same equaton as before a vald result from KVL. Yet, ths result has a very nterestng nterpretaton!
21 /30/008 KVL and Electronc Notaton 0/ The value 4.0 V s the ntal electrc potental the potental at begnnng node of the loop. The values v R,vR, and v R3 descrbe the voltage drop as we move through each resstor. The potental s thus decreased by these values, and thus they are subtracted from the ntal potental of 4.0. When we reach the bottom of the crcut, the potental at that pont wrtg (wth respect to ground) must be equal to: 4 vr vr vr3 But we also know that the potental at the bottom of the crcut s equal to.0 V! Thus we conclude: 4 v v v = R R R3 Our KVL equaton! n general, we can move through a crcut wrtten wth or electronc crcut notaton wth ths law : The electrc potental at the ntal node (wrtg), mnus(plus) the voltage drop(ncrease) of each crcut element encountered, wll be equal to the electrc potental at the fnal node (wrtg).
22 /30/008 KVL and Electronc Notaton / For example, let s analyze our crcut n the opposte drecton! R =3K R =K R 3 =K 4V v R v R v R3 v 4 v 5 Here, the electrc potental at the frst node s.0 volts (wrtg) and the potental at the last s 4.0. Note as we move through the resstors, we fnd that the potental ncreases by v R : v v v = 4 R3 R R Note ths s the effectvely the same equaton as before: V 4 v v v = R R R3 Both equatons accurately state KVL, and ether wll the same correct answer!
23 /30/008 KVL and Electronc Notaton / Now, we can use our new found knowledge to come to these correct conclusons, see f these results make sense to you! 4 vr = v4 v4 vr = v5 = vr 3 v5 4 vr vr = v5 = vr3 vr v4 R =3K R =K R 3 =K 4V v R v R v R3 v 4 v 5 V
24 /30/008 Analyss of Electronc Crcuts /7 Analyss of Electronc Crcuts n EECS you acqured the tools necessary for crcut analyss. Fortunately, all those tools are stll applcable and useful when analyzng electronc crcuts! Ohm s Law, KVL and KCL are all stll vald, but (sn t there always a but?) the complcatng factor n electronc crcut analyss s the new devces we wll ntroduce n EECS 3. n EECS you learned about devces such as voltage sources, current sources, and resstors. These devces all had very smple devce equatons: v R v V s v s v = R v Vs = = s
25 /30/008 Analyss of Electronc Crcuts /7 But (that word agan!), n EECS 3 we wll learn about electronc devces such as dodes and transstors. The devce equatons for these new crcut elements wll be qute a bt more complcated! D D D G v GS v DS S v D ( GS t ) D = S e = K v V v v D DS DS vd nvt As a result, we often fnd that both node and mesh analyss tools are a bt clumsy when analyzng electronc crcuts. Ths s because electronc devces are nonlnear, and so the resultng crcut equatons cannot be descrbed by as set of lnear equatons. = 3 = 3 0 = = Not from an electronc crcut!
26 /30/008 Analyss of Electronc Crcuts 3/7 nstead, we fnd that electronc crcuts are more effectvely analyzed by a more precse and subtle applcaton of:. Krchoff Voltage Law. Krchoff Current Law Crcut Equatons 3. Ohm s Law 4. Electronc devce equatons Devce Equatons Note the frst two of these are crcut laws they ether relate every voltage of the crcut to every other voltage of the crcut (KVL), or relate every current n the crcut to every other current n the crcut. = 0 V V V = The last two tems of our lst are devce equatons they relate the voltage(s) of a specfc devce to the current(s) of that same devce. Ohm s Law of course descrbes the currentvoltage behavor of a resstor (but only the behavor of a resstor!). V = R So, f you:. mathematcally state the relatonshp between all the currents n the crcut (usng KCL), and:
27 /30/008 Analyss of Electronc Crcuts 4/7. mathematcally state the relatonshp between all the voltages of the crcut (usng KVL), and: 3. mathematcally state the currentvoltage relatonshp of each devce n the crcut, then: you have mathematcally descrbed your crcut completely! At ths pont you wll fnd that the number of unknown currents and voltages wll equal the number of equatons, and your crcut analyss smply becomes an algebra problem! But be careful! n order to get the correct answer from your analyss, you must unambguously defne each and every voltage and current varable n your crcut!!!!!!!!! We do ths by defnng the drecton of a postve current (wth and arrow), and the polarty of a postve voltage (wth a and ). Placng ths unambguous notaton on your crcut s an absolute requrement! Q: An absolute requrement n order to acheve what?
28 /30/008 Analyss of Electronc Crcuts 5/7 A: An absolute requrement n order to:. determne the correct answers.. receve full credt on exams/homework. Q: But why must unambguously defne each current and voltage varable n order to determne the correct answers? A: The mathematcal expressons (descrptons) of the crcut provded by KVL, KCL and all devce equatons are drectly dependent on the polarty and drecton of each voltage and current defnton! For example, consder a three current node, wth currents,, 3. We can of course use KCL to relate these values, but the resultng mathematcal expresson depends on how we defne the drecton of these currents: 3 3 = 0 3 = 0 3 = 3 3 = 0 3 = 3
29 /30/008 Analyss of Electronc Crcuts 6/7 Q: But that s the problem! How do know whch drecton the current s flowng n before analyze the crcut?? What f put the arrow n the wrong drecton? A: Remember, there s no way to ncorrectly orent the current arrows of voltage polarty for KCL and KVL. f the current or voltage s opposte that of your conventon, then the numerc result wll smply be negatve. For example, say that n a 3wre node there s: 3 ma flowng toward the node n wre ma flowng toward the node n wre 5 ma flowng away from the node n wre 3 Dependng on how you defne the currents, the numercal answers for, and 3 wll all be dfferent, but there physcal nterpretaton wll all be the same! 3 = 3mA, = ma, = 5mA 3 3 = 3mA, = ma, = 5mA 3 3 = 3mA, = ma, = 5mA 3
30 /30/008 Analyss of Electronc Crcuts 7/7 Remember, a negatve value of current (or voltage) means that the current s flowng n the opposte drecton (or polarty) of that denoted n the crcut. So, wthout current arrows and voltage polartes, there s no way to physcally nterpret postve or negatve values! Now we know that wth respect to KCL or KVL, the current/voltage conventons are arbtrary (t up to you to decde!). However, we wll fnd that the voltage/current conventons of electronc devces are not generally arbtrary, but nstead have requred orentatons. G Q: Why s that? v GS D D v DS S ( ) v = K v V v v D GS t DS DS A: The conventons are coupled to electronc devce equatons these equatons are only accurate when usng the specfc voltage/current conventons! Thus, you must know both the devce equaton and the current/voltage conventon for each electronc devce. Furthermore, you must correctly label and uses these current/voltage conventons n all crcuts that contan these devces!
31 /5/008 Volts mllamps kloohms /4 Volts, MllAmps, and KloOhms Let s determne the voltage across a 7 KΩ resstor f a current of ma s flowng through t: ( )( ) v = = 4. V Or the resstance of a resstor f a current of ma results n a voltage drop of 0 V: 0 R = = 000 Ω 000. Or the current through a KΩ resstor f the voltage drop across t s 4.0 V: 4 = = 0. ma 000 There s just one bg problem wth ths analyss, and that problem s:
32 /5/008 Volts mllamps kloohms /4 The correct answers are 4 Volts, 0 KΩ, and.0 ma. The problem of course s all those decmal places! t s easy to get ncorrect answers when resstances are n the kloohms (or hgher) and the currents are n the mllamps (or smaller). Unfortunately, that s exactly the stuaton that we have to deal wth n electronc crcuts! Frequently, we fnd that n electronc crcuts:. Voltages are n the range of 0. to 50 Volts.. Currents are n the range of 0. to 00 ma. 3. Resstances are n the range of 0. KΩ to 50.0 KΩ. Fortunately, there s an easy soluton to ths problem. n electronc crcuts, the standard unt of voltage s volts, the standard unt of current s mllamps, and the standard unt of resstance s kloohms. Ths works well for Ohm s Law, because the product of current n mllamps and resstance n KΩ s voltage n volts:
33 /5/008 Volts mllamps kloohms 3/4 And so: v [ V ] = [ ma] R [ KΩ ] [ ] ma = [ ] [ Ω] vv R K [ ] [ ] vv R [ KΩ ] = ma The trck then s not to numercally express currents n Amps, or resstances n Ohms, but nstead to leave the values n ma and KΩ!!! For example, let s recompute our earler examples n ths way: The voltage across a 7 KΩ resstor f a current of ma s flowng through t: ( ) v = 7 = 4V Or the resstance of a resstor f a current of ma results n a voltage drop of 0 V: 0 R = = 0 KΩ Or the current through a KΩ resstor f the voltage drop across t s 4.0 V:
34 /5/008 Volts mllamps kloohms 4/4 4 = = 0. ma Not that these are all obvously the correct answers!!!!
35 /5/008 Example Crcut Analyss usng Electronc Crcut Notaton /5 Example: Crcut Analyss usng Electronc Crcut Notaton Consder the crcut below: v A 5.0 V 0.0 V R = K R = K R3 = K.0 ma Determne the voltage v A, and the current through each of the three resstors. Soluton Our frst task s to unambguously label the currents and voltages of ths crcut:
36 /5/008 Example Crcut Analyss usng Electronc Crcut Notaton /5 v A 0.0 V v K 5.0 V 4 v3 v K 3 K.0 ma Now lets relate all the currents usng KCL: v A 5.0 V 0.0 V K K 4 3 K.0 ma = = 3 4 And relate all the voltages usng KVL: v A 5.0 V 0.0 V v K v3 v K K.0 ma
37 /5/008 Example Crcut Analyss usng Electronc Crcut Notaton 3/5 0 v = v v = 0 v A A v v = 5 v = v 5 A A 5 v = 0 v = 50. V 3 3 And fnally, a devce equaton for each resstor: v v v = = = 3 3 R R R3 The equatons above provde a complete mathematcal descrpton of the crcut. Note there are eght unknown varables (,,, 3 4,v,v,v 3,v A ), and we have constructed a total of eght equatons! Thus, we smply need to solve these 8 equatons for the 8 unknown values. Frst, we nsert the KVL results nto our devce equatons:
38 /5/008 Example Crcut Analyss usng Electronc Crcut Notaton 4/5 v 0 v A = = = 0 R v A v v 5 A = = = va R 5 v 5 = = = 50mA 3 3. R3 And now nsert these results nto our KCL equatons: and: = ( v ) 0 v = 5 A A = 4 3 ( v ) A 5 = 5 Note the frst KCL equaton has a sngle unknown. Solvng ths equaton for v A : ( v ) 0 v = 5 A A v A = = = 70. V And now solvng the second KCL equaton for 4 :
39 /5/008 Example Crcut Analyss usng Electronc Crcut Notaton 5/5 ( va ) 4 5 = 5 = 5 v 5 = 0 7 = 30. ma 4 A From these results we can drectly determne the remanng voltages and currents: v = 0 v A = 0 7 = 30. V v = va 5 = 7 5 = 0. V = 0 v = 0 7 = 30. ma A = v 5 = 7 5 = 0. ma A 0.0 V 3.0 v A = V K.0 K 5.0 K.0 ma