10-Source Transformations Text: ECEGR 210 Electric Circuits I

Transcription

1 10Source Trnsformtions Text: ECEGR 210 Electric Circuits I

2 Introduction Source Trnsformtion Thevenin s Theorem Norton s Theorem Overview Dr. Louie 2

3 Introduction Consider the two circuits (A nd B) shown elow Compute the voltge cross the 2W resistor in ech circuit Circuit A 4V Circuit B 0.5W 4V 1A 0.5W 2W 0.5V 2W Dr. Louie 3

4 Introduction Circuit A (superposition): V R1 = 2I R1 = 2 x 1(0.5/2.5)= 0.4V (current source) V R2 = 4(2/2.5) = 3.2V (voltge source) V R = V R1 V R2 = = 3.6V Circuit B: V R = 4.5(2/2.5)=3.6V (voltge divider) Circuit A 4V Circuit B 0.5W 4V 1A 0.5W 2W 0.5V 2W Dr. Louie 4

5 Introduction Solving Circuit B ws much esier Sme voltge cross (current through) the resistor Circuits re equivlent looking into the terminls Circuit A 4V Circuit B 0.5W 4V 1A 0.5W 2W 0.5V 2W Dr. Louie 5

6 Introduction Clerly there cn e dvntgeous of trnsforming sources Source trnsformtions nd equivlence re the focus of this lecture source source R V s I s R Dr. Louie 6

7 Introduction Two wys of modeling rel (nonidel) voltge nd current sources re shown R s : smll vlue, prevents infinite current from flowing if terminls (,) re shorted R : lrge vlue, prevents infinite voltge t the terminls (,) under open circuit conditions Genericlly: cn e ny voltge source in series with resistnce, or ny current source in prllel with resistnce V s R s I s R Dr. Louie 7

8 Source Trnsformtion Not possile to trnsform current (voltge) sources to voltge (current) sources directly V s I s But we cn trnsform sources with series or prllel resistnces s seen y terminls R s V s I s R Dr. Louie 8

9 Source Trnsformtions For the two circuits to e equivlent, they must hve the sme iv chrcteristics t their terminls under ll externl circuit connections Due to linerity, only need to verify iv chrcteristics under two different externl connections (short, open circuit) How re V s, I s, R s nd R relted? R s V s I s R Dr. Louie 9

10 Source Trnsformtion Consider when the externl circuit is short Both circuits must hve sme short current out of their terminls I sc = V s /R s I sc = I s Therefore: I s = V s /R s V s R s I I s R I Dr. Louie 10

11 Source Trnsformtions Consider n open circuit Both circuits must sme open circuit voltge V oc V oc = V s V oc =R I s Therefore: I s = V s /R V s R s V oc I s R V oc Dr. Louie 11

12 Reltionships: I s = V s /R Source Trnsformtions I s = V s /R s Therefore R s = R = R I s = V s /R V s = I s R Source trnsformtion equtions R V s I s R Dr. Louie 12

13 Source Trnsformtions Verify the results hold for Circuit A nd Circuit B Circuit A 4V Circuit B 0.5W 4V 1A 0.5W 2W 0.5V 2W Dr. Louie 13

14 Exmple Use source trnsformtions to find V 0 Consider the current source first. Which resistor cn we ssocite it with? 4W (they re in prllel) Should we trnsform them? Yes, the resistor will e in series with the 2W resistor, nd we cn comine the two 2W 3W 4W 3A 8W V 0 12V Dr. Louie 14

15 Trnsform the source: R = 4W V s = I s R = 3 x 4 = 12V Exmple Py creful ttention to the polrity 2W 3W 4W 2W 3W 4W 3A 8W V 0 12V 12V 8W V 0 12V Dr. Louie 15

16 Exmple Trnsform the 12V source (on the right), if it is eneficil 6W 3W 12V 8W V 0 After comining series resistnce 12V Dr. Louie 16

17 Exmple Yes, eneficil (results in prllel comintion with V 0 ). Comine with 3W resistor to get: R = 3W I s = V s /R = 12/3 = 4A 6W 3W 6W 12V 8W V 0 12V 12V 8W V 0 3W 4A Dr. Louie 17

18 Exmple Now trnsform voltge source with 6W resistor 6W 12V 8W V 0 3W 4A Dr. Louie 18

19 Result: R = 6W I s = V s /R = 12/6 = 2A Exmple 6W 12V 8W V 0 3W 4A 2A 6W 8W V 0 3W 4A Dr. Louie 19

20 Exmple The rest is esy. Current division: I 0 = (42)x(2/10) = 0.4A V 0 = 3.2V 2A 6W 8W V 0 3W 4A 8W V 0 2W 2A Dr. Louie 20

21 Source Trnsformtions Dependent sources re hndled using the sme procedure Be creful Dr. Louie 21

22 Thevenin s Theorem Often, most elements of circuit re fixed nd only one element (the lod) chnges Do not wnt to resolve the entire circuit every time the lod chnges Better pproch: represent unchnging prt of circuit with voltge source with series resistnce (Vrile Resistor) V V TH R TH Dr. Louie 22

23 Thevenin s Theorem Thevenin s Theorem: liner twoterminl circuit cn e replced y n equivlent circuit consisting of voltge source in series with resistor R TH : Thevenin Resistnce V TH : Thevenin Voltge Liner Two Terminl Circuit = V TH R TH Dr. Louie 23

24 Thevenin s Theorem How do we find V TH nd R TH? One wy: keep pplying resistnce nd source trnsformtions until there is voltge source in series with resistnce etween the terminls R TH V TH Dr. Louie 24

25 Thevenin s Theorem Better wy: recognize tht V TH = V OC nd R TH = input resistnce (looking into terminls, nd ), or R TH =V OC /I SC V OC V TH R TH Dr. Louie 25

26 Finding Thevenin Resistnce No Dependent Sources: short ll voltge sources open ll current source then find equivlent resistnce R TH = R eq Dependent Sources: short ll voltge sources open ll current source Apply test voltge V 0, compute current I 0 R TH = V 0 /I 0 Dr. Louie 26

27 Exmple Find the current through the lod resistor if R L is 6, 16 nd 36W Perfect sitution for Thevenin Equivlent Find Thevenin Equivlent, then solve equivlent circuit for vrious vlues of R L 4W 1W 32V 12W 2A R L Dr. Louie 27

28 Exmple Strt with finding the Thevenin voltge V TH = V OC By superposition (or mesh nlysis) V OC1 = 32(12/16) = 24V V OC2 = 4x2x(12/16) = 6V V TH = V OC1 V OC2 = 30V 4W 1W 32V 12W 2A Dr. Louie 28

29 Exmple Now find the Thevenin resistnce R TH = R eq dectivte ll sources R eq = 1 (4x12)/16 = 4W 4W 1W 12W R eq Dr. Louie 29

30 Exmple Thevenin equivlent circuit is shown elow Current through vrious lod resistnces cn e esily computed 4W 1W 4W 32V 12W 2A 30V Dr. Louie 30

31 Exmple Find the Thevenin Equivlent of the circuit etween the terminls, 60W 2A 30W 30V Dr. Louie 31

32 Vi superposition Exmple V OC1 = 30x(30/90) = 10V V OC2 = 30x2(60/90) = 40V V TH = V OC1 V OC2 = 50V 60W 2A 30W 30V Dr. Louie 32

33 Exmple Now find R TH R TH = (30x60)/(3060) = 20W 60W 30W Dr. Louie 33

34 Thevenin s Theorem Find the Thevenin equivlent Note the dependent source Finding V TH is the sme procedure s efore 5W I x 3W 6V 1.5I x 4W Dr. Louie 34

35 Mesh Anlysis Thevenin s Theorem 6 = 5I 1 7I x (Supermesh) 1.5I x I 1 = I x (current source constrint eqution) Solving I x = 1.33A Therefore V OC = 1.33x4 = 5.33V = V TH 5W 3W 6V I 1 I x 4W Dr. Louie 35

36 Thevenin s Theorem To find R TH pply either test voltge or current to the terminls dectivte independent sources compute either terminl current or voltge R TH = V 0 /I 0 5W I x 3W 6V 1.5I x 4W Dr. Louie 36

37 Thevenin s Theorem Apply test voltge V 0 Let V 0 = 1V Now find I 0 (note polrity if I 0 ) 5W I x 3W I 0 1.5I x 4W V 0 Dr. Louie 37

38 Thevenin s Theorem I 1 1.5I x = I x (Nodl Anlysis) I 1 0.5I x = 0 0.2V1 0.5(V1 1)/3= V =0 V 1 =5V I x = 2A (Ohm s Lw) I 2 = 0.25A (Ohm s Lw) I 0 = 2.25A (KCL) 5W I 1 I x 3W I 0 1.5I x I 2 4W V 0 Dr. Louie 38

39 Thevenin s Theorem Therefore R TH = V 0 /I 0 = 1/2.25 = 0.444W 0.444W 5.33V Dr. Louie 39

40 Norton s Theorem Thevenin equivlent circuit cn e trnsformed into current source in prllel with resistor From discussion on source trnsformtion: R N = R TH Thevenin Equivlent Norton Equivlent R TH V TH I N R N Dr. Louie 40

41 Norton s Theorem I N is found y shorting the terminls of the circuit I N = I SC Liner Two Terminl Circuit I SC I N I N Dr. Louie 41

42 Norton s Theorem Process for finding R TH is identicl to tht for Thevenin s theorem No Dependent Sources: short ll voltge sources open ll current source then find equivlent resistnce R TH = R eq Dependent Sources: short ll voltge sources open ll current source Apply test voltge V 0 (or current), compute current I 0 (or voltge) R TH = V 0 /I 0 Dr. Louie 42

43 V L (V) Prcticl Sources Idel source 12V R S V L R s = 0.25W R s = 1W R L (W) Dr. Louie 43

44 I L (A) Prcticl Sources 6 5 Idel source 4 5A R I 3 L R = 50W R = 100W R L (W) Dr. Louie 44