Electronic Circuits I
Electronic Circuits I EECS 312
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This 39 page Class Notes was uploaded by Melissa Metz on Monday September 7, 2015. The Class Notes belongs to EECS 312 at Kansas taught by James Stiles in Fall. Since its upload, it has received 14 views. For similar materials see /class/186777/eecs-312-kansas in Elect Engr & Computer Science at Kansas.
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Introduction Analysis of Electronic Circuits Reading Assignment KlL and KCL fexf from EECS 211 Just like EECS 211 the majority of problems hw and exam in EE 312 will be circuit analysis problems Thus a key to doing well in 312 is to thorough lyknow the mata ial from 211 So before we get started with 312 let39s review 211 and see how it applies to electronic circui Q IacedEECS 211as semester can I just s z ill395 re Wen 9 A Even if you did extremely well in 211 you will want to pay attention to this review You will see that the concepts of 211 are applied a little differently when we analyze electronic circuits Both the conventions and the approach used for analyzing electronic circuits will perhaps be unfamiliar to you at first I thus imagine that everyone I hope will find this review to be helpful Electronic Circuit Notation KVL and Electronic Circuit Notation Analysis of Electronic Circuits Even the quantities of current and resistance are a little different for electronic circuits Q you mean we don t use Ampere and Oil715 A Not exactly VoltsI MilliAmpsI KiloOhms Now let39s try an example Exam le Circuit Anal sis usin Electronic Circuit Notation Electronic Circuit Notation The standard electronic circuit notation may be a little different that what you became used to Seeing in in EECS 211 The electronic circuit notation has a few quotshorthandquot standards that can simplify circuit Schematics Consider the circuit below 1 R11K V1 39 fgl Isl l55V R24K 3V2 pg4ilt V3 0 Note the voltage values in this circuit ie lSv1v2v3 provide values of potential difference between two points in the circuit For example from the voltage source we can conclude The electric potential at this point in the circuit is 5 volts greater than the electric potential at this point in the circuit l55V Or The resisTor volTage V3 means 3 394 The elecTric poTenTial aT This poinT in The circuiT is V3 volTs greaTer Than The elecTric poTenTial aT This poinT in The circuiT BuT remember V3 could be a negaTive value Thus The values of volTages are comparaTive They Tell us The difference in elecTric poTenTial beTween Two poinTs wiTh in The circuiT As an analogy Say John Sally and Joe work in a very full building Our circuiT volTages are liTTle like saying quotfam is 5 oors above foe quot50y is 2 oors above foe From This comparaTive informaTion we can deduce ThaT John is 3 floors above Sally WhaT we cannoT deTermine is on whaT floor John Sally or Joe are acTually locaTed They could be locaTed aT The highes l39 floors of The building or aT The lowesT or anywhere in beTween Similarly we canno l39 deduce from The values l5v1v2v3 The elecTric poTenTial aT each poinT in The circuiT only The relaTive values relaTive To oTher poinTs in The circuiT EGZ quot Pom R has an eecfrc pofem l39a 5V7g75r fhan pom B quot Pom 6 has an eecfrc pofem l39a v3 higher fhan pom B Q 50 how do we defermne fhe vaue of eecfrc p0 fen a of a speci c pom in a circuit A Recall ThaT elecTric poTenTial aT some 7 poinT is equal To The poTenTial energy possessed by 1 Coulomb of charge if 39 i locaTed aT ThaT poinT Thus To deTermine The absoluTe as opposed To relaTive value of The elecTric poTenTial we firsT mu deTermine where ThaT elecTric poTenTial is zero The problem is similar To ThaT of The poTenTial energy possessed by 10 kg of mass in a graviTaTional field We ask ourselves Where does This poTenTial energy equal zero The answer of course is when The mass is locaTed on The ground BUT This answer is a biT subjecTive is The quotgroundquot A where The car peT is locaTed B where The sidewalk is locaTed C The basemenT floor D Sea level E The cenTer of The Ear Th The answer is iT can be any of These Things We can raTher arbitrarily seT some poinT as The locaTion of ground The poTenTial energy is Therefore described in reference To This ground poinT i For Tall buildings The ground u floor is usually defined as The floor conTaining The fron l39 door I 7 Le The sidewalk buT iT doesn39T 7 quot have To be jusT look aT Ea l39on j Hall Now having defined a ground reference if we add To our earlier sTaTemenTs quotJoe is 32 floors above ground We can deduce quot John is 5 oors above foe fherefore John is on 75 3739 floor quot 5ay is 2 oors above foe fherefore 5ay is on 75 3439 floor Q 50 can we de ne aground pa fen a for our39 circuit A AbsoluTer We jusT pick a poinT on The r cir39cuiT and call iT The ground poTenTial We ever39y poinT in The cir39cuiT wiTh respecT To This ground poTenTiaI Consider now The circuiT 1 R11K o V1 39 vs5v R24K V2 R34K V3 o Look aT This NoTe we have added an upsidedown Trianglequot To The cir39cuiT This denoTes The locaTion we define as our39 ground poTenTial Now if we add The sTaTemenT quotPairf B 2539 afar eecfr39ic pa fenfb of zero V0fs wfh respecf f0 ground We can conclude quotPom R 39539 07 an eecfrc p0 fem ia of 5 Volf39s wfh respecf 7 0 ground quotPom 6 39539 07 an eecfrc p0 fen 7 39a of V3 Volf39s wfh respecf 7 0 ground NoTe ThoT all The poinTs wiThin The cir39cuiT ThoT r39eside oT ground poTenTiol form a r39oTher39 large node 1 R11K o v oi oi l55V R24K V2 R34K V3 0 f Look IT This STondor d elecTronic noToTion simplifies The SchemoTic by placing The ground symbol oT each device Terminal 1 R11K T V1 39 fgl l55V R24K V2 pg4lt NoTe ThaT all Terminals connecTed To ground are likewiSe connecTed To each oTher Now in The C036 where one Terminal of a device is connecTed To ground poTenTial The eecTric poTenTiaI wiTh respecT To ground of The oTher Terminal is easily deTermined The eecTric poTenTial aT This poinT in The circuiT is 5 volTs greaTer Than ground ie 5 l55V volTs This poinT is aT ground poTenTial ie zero volTs For This example iT is apparenT ThaT The volTage source simply enforces The condiTion ThaT The Terminal is aT 50 VolTs wiTh respecT To ground Thus we ofTen simplify our eecTronic circuiT SchemaTics as ll R11K lg5v a O T V1 39 Igl I l R24K V2 R34K Finally we find Thai The elecTr ic poTenTial of This poimL in The cir cuif is V2 vol39l39s greater Than ground poTenTial R34K V3 ie V3 This poimL is of ground poTenTial ie zero volTs Thus we can simplify our39 cir cuiT fur Ther as ll R11K ls5v V gt 0 V3 V1 igl igl R24K V2 R34K This cir cuif schemaTic is precisely The same as our39 original schematic I 1 R1 1 K o V1 39 igl Igl l55V R24K 3V2 pg4lt 3 o KVL and Electronic Circuit Notation Consider this circuit We can apply Kirchoff39s Voltage Low KVL to relate the voltages in this circuit in any number of ways For example The KVL around This loop is39 4VR1VR2VR3720 We could mulTiply boTh sides of The equaTion by 1 and likewise geT a valid 3 equaTion 47Vprvpzivp320 Q Buf which equafian is carrecf Which one do we use Which one is fhe KVL resuf A Each r39esulT is equally valid both will provide The same cor39r39ecT answer39s EssenTially The first KVL equaTion is consTr ucTed using The convention ThaT we add The cir cuiT elemenT voITage if we firsT encounTer39 a plus sign as we move along The loop and subfr acT The circuiT elemenT volTage if we firsT encounTer39 a minus sign as we move along The loop For example 5V1 BLIT we could also use The convenTion ThaT we subtract The circLIiT elemenT volTage if we firsT encoLInTer a plus sign as we move along The loop and add The circLIiT elemenT volTage if we firsT encoLInTer a minus sign as we move along The loop For example 5iV1 This convention would provide us wiTh The second of The Two KVL equaTions for our original circLIiT 4rvmrvpzrvpz20 Q HM Whar kind of sense does This conven on make We sub Irac when encounferng a 9 We add when encounferng a 9 A AcTually This second convenTion is more logical Than The firsT if we consider The physical meaning of volTage Remember The volTagequot is simply a meaSUre of poTenTial energy The poTenTial energy of 1 Coulomb of charge If 1 C of charge were To be TransporTed around The circuiT following The paTh defined by our KVL loop Then The poTenTial energy of This charge would change as is moved Through each circuiT elemenT In oTher words iTs poTenTial energy would go up or iT would go down The second convenTion describes This increasedecreaSe For example as our 16 charge moves Through 5V The volTage source iTs poTenTial energy is increases by 5 Joules The poTenTial is 5 V higher aT The Terminal Than iT was aT The minus Terminal BuT when iT moves Through The resisTor iTs 5V poTenTial energy drops by vl Joules The poTenTial aT The minus Terminal is vl VolTs less Than ThaT aT The plus Terminal Thus The second convenTion is a more accuraTe accounfing of The change in poTenTial A This convenTion is The one Typically used for elecfronic circuiTs You of course will geT The correcT answer eiTher way buT The second convenTion allows us To easily deTermine The absolufe poTenTial ie wiTh respecT To ground aT each individual poinT in a circuiT To see This leT39s reTurn To our original circuiT The KVL from TheSe loops are Thus 4 121 v4 O v4 v22 v5 0 2 v5 123 0 4 vkl vk2 v5 0 v4 vlt2 VR320 Q I a on f See how fhs new con ven an heps u539 defer11076 fhe 39bbsoufe pafem a 07 each pom m 16 arm397 A That39s becau3e we have not defined a ground potential Let39s do that now 144v We can thus rewrite this CIrCUIt Schematlc as RFZK Remember that all ground terminals are connected to each other so we can perform KVL by starting and ending at a ground node 4 VR1 VR2 VR32O 4 vk1 v420 V4 VR2 V520 The same resulTs as before Now we can further simplify The SchemaTiC R31K NoTe ThaT we were able To replace The volTage sources wiTh a direcT simple sTaTemenT abouT The elecTric poTenTial aT Two poinTs wiThin The circuiT 4V fl The elecTric 144v R13K VP poTenTial here musT be 4 V V4 RZZZK V92 The elecTric poTenTial here N 0 V5 musT be 2 V Vszv R31K V93 NoTe The KCL equaTion we deTermined earlier 4 ilt1 v22 VR320 LeT39s subTracT 20 from boTh sides 4 121 VRZ 123 2 This is The same equaTion as before a valid resulT from KVL YeT This resulT has a very inTeresTing inTerpreTaTion The value 40 V is The iniTiaI elecTric poTenTial The poTenTial aT beginning node of The loop The values VR1VR2 and VR3 describe The volTage drop as we move Through each resisTor The poTenTial is Thus decreased by These values and Thus They are subTracTed from The iniTiaI poTenTial of 40 l l When we reach The boTTom of The circuiT The poTenTial aT ThaT poinT erg wiTh respecT To ground musT be equal To 4 VRl sz VR3 BuT we also know ThaT The poTenTial aT The boTTom of The circuiT is equal To 20 V Thus we conclude 4 VR1 sz VR3 2 Our KVL equaTion In general we can move Through a circuiT wriTTen wiTh or elecTronic circuiT noTaTion wiTh This law The eecTrc p0 TehTa aT The im39TI39a node erg ml39huscbus The voTage drop hcrease of each circuiT eemehT encoun Terea w be equa T0 The eecTrc p0 TehTa aT The final node erg For example eT39s analyze our cir cuiT in The opposiTe dir39ecTion fl R13K R22K R31K 4V VV W ltO Here The elecTr39ic poTenTial aT The firsT node is 20 voTs erg and The poTenTial aT The lasT is 40 NoTe as we move Through The r39esisTor s we find ThaT The poTenTial increases by 14 2vR3vR2vR14 NoTe This is The effecTively The same equaTion as before 4 VR1 V22 V23 2 2 BoTh equaTions accuraTely sTaTe KVL and eiTher will The same cor r ecT answer m M Mr My Hand knowgag Na in cum in 51 lnvtl EIMIUSI39IHI 5 1 12 mm mm sans n m 2VR3V5 39 m g 44 7V v m m 5 R22K W 72vmvmv 39 v m Analysis of ElecTronic CircuiTs In EECS 211 you acquired The Tools necessary for circuiT analysis ForTunaTely al Those Tools are sTill applicable and useful when analyzing elecTronic circuiTs Ohm39s Law KVL and KCL are all sTill valid buT isn39T There always a buT The complicafing facTor in elecTronic circuiT analysis is The new devices we will inTroduce in EECS 312 In EECS 211 you learned abouT devices such as volTage sources currenT sources and resisTors These devices all had very simple device equafions But that word again in EECS 312 we will learn about electronic devices such as diodes and transistors The device equations for these new circuit elements will be quite a bit more complicated all D lib DK2Vss lrVbs l025 IAI55 ll1 As a result we often find that both node and mesh analysis tools are a bit clumsy when analyzing electronic circuits This is because electronic devices are non linear and so the resulting circuit equations cannot be described by as set of linear equations 1 723139122713 12i1392 1 044 224 23 0 Not from an electronic ci rcuit Instead we find that electronic circuits are more effectively analyzed by a more precise and subllc application of Circu t Equations 1 larchquot Volhigc LawL v a f flg mhoffwlm La 7 ii 4 quotquot Device i r Equations i4 lurid this Note the fir t two to these are circuit laws they eith imuit to every other volhgc of M in Hue circuit to relate every volhgc of the c the circuit KVL or relate e ery c y other meant in the aim 39 I11o magi30 ast two items of ou list are device equations they The that same evice O currentvoltage behaviorora resistor but only the behavior r 39 I esistor V2 12 R2 50 if you mathematically state the relationship between all the currents in the circuit using kcL an 2 maThemaTically sTaTe The relaTionship beTween all The volTages of The circuiT using KVL and 3 maThemaTically sTaTe The currenT volTage relaTionship of each device in The circuiT Then you have maThemaTically described your circuiT compleTely AT This poinT you will find ThaT The P Lu number of unknown currenTs and volTages will equal The number of Lit mm l y equaTions and your circuiT analysis 1 1 mgr 39 simply becomes an algebra problem Cm b m BuT be careful In order To geT The 0 f b correcT answer from your analysis 6quot by you musT unambiguously define each and every volTage and currenT variable ELA39F in your Clr CUI39l39lllllllll We do This by defining The direcTion of a posiTive currenT wiTh and arrow and The polariTy of a posiTive volTage wiTh a and 1906779 fIS unambguous I707 a7 f07 07 your CIrcuf 5 07 absoufe reqw remenf Q An absow e requremenf 397 order 7 0 achieve whm A An absoluTe requiremenT in order To 1 deTermine The correcT answers 2 receive full crediT on examshomework Q BU why mu5397 I unambiguousy defHe each currenf and vofage var39abe in order 7 0 defernine The correcf answers A The maThemaTical expressions deSCripTions of The circuiT provided by KVL KCL and all device equaTions are direchy dependenT on The polariTy and direcTion of each volTage and currenT definiTion For example consider a Three currenT node wiTh currenTs I 1392 1393 We can of courSe USe KCL To relaTe TheSe values buT The reSUITing maThemaTical expression depends on how we define The direcTion of TheSe currenTs 4 TI Q Lg4o 3 gt 6 44 go I1 I2 1 2 Lg4 gt gt A 4 4 go 1 2 L4g Q BUT fhaf s The probem How do I know which a iquotec on The currem 39539 owHg in before I anayze The Urey7 2 Whaf if I pm The arrow in The wrong a iquotec on A Remember There is no way To incor39r39echy or39ienT The cur39r39enT arrows of volTage polar39iTy for39 KCL and KVL If The cur39r39enT or39 volTage is opposiTe ThaT of your39 convenTion Then The numeric r39e3ulT will simply be negaTive For39 example say ThaT in a 3wir39e node There is 3 mA flowing Toward The node in wire 1 2 mA flowing Toward The node in wire 2 5 mA flowing away from The node in wire 3 Depending on how you define The cur39r39enTs The numerical answers for39 I1 I2 and I3 will all be differ39enT buT Ther39e physical inTer39pr39eTaTion will all be The same a b Ii I2 T13 I1 3mA I2 2mA I3 5mA gt Ii I2 173 71 3m1I2 2mA I 5mA gt gt I1 1 2 I1 3mA I 2mA I 5mA I3 Remember a negaTive value of cur39r39enT or39 volTage means ThaT The cur39r39enT is flowing in The opposiTe dir39ecTion or39 polar39iTy of ThaT denoTed in The cir39cuiT So wiThouT cur39r39enT arrows and volTage polar39iTies There is no way To physically inTerpreT posiTive or39 negaTive values Now we know ThaT wiTh r39especT To KCL or39 KVL The cur39r39enTvolTage convenTions ar39e arbiTrary iT up To you To decidel However39 we will find ThaT The volTagecur39r39enT convenTions of elecTr39onic devices are noT gener39ally ar39biTr39ar39y buT insTead have required or39ienTaTions Q Why is 7079 A The convenTions are coupled To elecTr39onic device equaTions These Di D equaTions are only accur39aTe when using The specific volTagecur39r39enT convenTions 6 105 Thus you musT know boTh The device equaTion and The cur39r39enTvolTage V95 convenTion for39 each elecTr39onic device 5 Fur39Ther39mor39e you musT correchy label and uses These cur39r39enTvolTage 7 7 7 2 VD Klalv WV V05 convenTions In all Clr39CUI l39S ThaT conTain These devices VolTs MilliAmps and KiloOhms LeT39s deTer39mine The volTage across a 7 kn r39esisTor39 if a cur39r39enT of 2 mA is flowing Thr39ough iT v 00027000 14 V Or39 The resisTance of a r39esisTor39 if a cur39r39enT of 2 mA r39e3ulTs in a voTage drop of 20 V R A 1000 0 0002 Or39 The currenT Through a 2 kn r39esisTor39 if The voTage drop across iT is 40 V 4 2 A 2000 O m Ther e39s jusT one big problem wiTh This analysis and ThaT problem is The correcT answers are 14 VolTs 10 K9 and 20 mA The problem of course is all Those decimal places IT is easy To geT incor r ecT answers when r esisTances are in The kilo ohms or39 higher and The cur r enTs are in The milliamps or smaller Unfor TunaTely ThaT39s exachy The siTuaTion ThaT we have To deal wiTh in elecTr onic cir cuiTs Frequenle we find ThaT in elecTr onic cir cuiTs 1 VolTages are in The range of 01 To 50 VolTs 2 CurrenTs are in The range of 01 To 100 mA 3 ResisTances are in The range of 01 m To 500 m For TunaTely There is an easy soluTion To This problem In elecTronic cir cuiTs The sTandar39d uniT of volTage is volTs The sTandar d uniT of cur r enT is milliamps and The sTandar39d uniT of r esisTance is kiloohms This works well for Ohm39s Law because The pr39oducT of cur r enT in milliamps and r esisTance in m is volTage in volTs vl39migtltRKQ And so vl lmAl RKQ RlKQ The Trick Then is noT To numerically express currenTs in Amps or resisTances in Ohms buT insTead To leave The values in mA and m II For example leT39s recompuTe our earlier examples in This way The volTage across a 7 kn resisTor if a currenT of 2 mA is flowing Through iT v2714l Or The resisTance of a resisTor if a currenT of 2 mA resulTs in a volTage drop of 20 V R 10KQ Or The currenT Through a 2 kn resisTor if The volTage drop across iT is 40 V 4 39 20 A 2 m NOT ThClT These are all obviously The cor r ecT answer sllll Example Circuit Analysis using Electronic Circuit Notation Consider the circuit below VA 50 V 100 V Determine the voltage VA and the current through each of the three resistors Solution Our first task is to unambiguously label the currents and voltages of this circuit li i z f OOfi 3111 E ecszonh i w 13 cn39339aon 2quot VA 50 V VA 50 V Jim S39I39Nrae The Umv ofquot Karuelrgif Dam c39u EECE 10 vlsz gt 11210 VA vA v225 gt vzsz 5 5 1320 gt V3250l And finally a device equation for each resistor The equations above provide a complete mathematical description of the circuit Note there are eight unknown variables 391392393v1v2 vanA and we have constructed a total of eight equations Thus we simply need to solve these 8 equations for the 8 unknown values First we inSert the KVL reSUlts into our device equations iii Univ Q39f Dept of 1 will 1x73quotZ 00 V 2E2 1 VA 5 3 50m1 And now inSer39T TheSe r eSulTs info our39 KCL equofions 3913921 lO VAvA 51 and Qqg VA 53945 NoTe The first KCL equofion has a single unknown Solving This equofion for39 VA lO VAvA 51 z zmy 3V 2 2 And now solving The second KCL equofion for39 a Dam 0396 15553133 Oi r 113820053 Exmawe VA 53945 gt45 VA510 730 mA Fr39om TheSe r eSulTs we can dir ecfly defermine The remaining volTages and cur39r39en rs v110 v1o 73o 1 vzsz 527 5ZO 1 41o v1o 73o mA 392vA 57 520 mA The quotu mv ofquot KCH HL AS Dap39 0396 EECE