Power Transformers 1. Download/read notes on transformers from website. 2. Download HW4 on website; I will give due-date next week. 3.

Transcription

1 Power Transformers. Download/read notes on transformers from website.. Download HW4 on website; I will give due-date next week.. Read Chapters 5 and 6 in Kirtley s text

2 Power Industry Uses of Transformers Instrument Transformers: Current (CT) and potential (PT) transformers: Step down quantity from power system level (gen, trans, st) so that quantity is compatible at the instrument level, in order to perform protective relaying (you need to take EE 457 to learn about these). Power Transformers:. Step up voltage from generator to transmission (GSU). Step down voltage from transmission to stribution primary levels. Step down voltage from stribution primary to stribution secondary 4. Interconnecting fferent system voltage levels in HV and EHV systems

3 Ampere s Law: Magnetic circuits Line integral of mag fld intensity about a closed path equals current enclosed. Apply it here: Make the path the dotted line. H is along rection of ϕ, which is same rection as dl Let l be length of dotted path, therefore LHS is Hl RHS is current enclosed: i. Hl i

4 Magnetic circuits Hl i Recall B H BA l A i A l i 4

5 Magnetic circuits A l i Define: magnetomotive force (MMF): reluctance: F This looks suspiciously familiar. What does this R remind you of? Write down the relations for F, R, and ϕ. F R l A i 5

6 Magnetic circuits I V R F R Units: ϕ webers (kg-m /sec /Amp) B webers/m =tesla F ampere-turns R amperes/weber μ = μ μ =μ (4π 0-7 )tn/amp 6

7 Example Compute ϕ. 7

8 Example F R F I 00* ampere-turns lg 0.5 / 000 R gap = 99, 47amperes/Weber 7 Ag (4 0 )(40 / 00 ) lc. Rcore = 95, 49amperes/Weber 7 Ac 500(4 0 )(40 /00 ) F 780 B A 40 / (00) R 99, 47 95, Wb Weber/m =Tesla 8

9 Recall: Inductance A i l l A i Define: Flux Linkage: Self Inductance: Li L A l R 9

10 Inductance Some notation that will be useful later: L i The self inductance L is the ratio of the flux from coil linking with coil, λ to the current in coil, i Linking? Passing through the coil interior 0

11 Inductance Flux not linking Flux linking

12 Inductance Some notation that will be useful later: L i The self inductance L is the ratio of the flux from coil linking with coil, λ to the current in coil, i Inductance L? The ability of a current in coil, i, to create flux φ that links with coil.

13 L Inductance A l To make large self-inductance, we need to make, μ, and A large; make l small And so a large L results from many turns ( ) large μ (core made of iron) large cross section (A) compact construction (small l)

14 And so a large L results from many turns ( ) large μ (core made of iron) large cross section (A) compact construction (small l) Recalling l R A L L R A l we see that a magnetic circuit characterized by a large self-inductance will have a small magnetic path reluctance. 4

15 Example Example : Compute the self-inductance of the magnetic circuit given in Ex. We just found that L L R Alternative derivation: Rc i i i c R R g i R g 5

16 Example Example : Compute the self-inductance of the magnetic circuit given in Ex. L R R R c g From Ex, =00 and R 95,49amperes/Weber c R g 99,47amperes/Weber 00 L henries 6

17 Mutual inductance i φ i From our self-inductance work, we express for each coil L L i i where we recall the self inductance L jj is the ratio of the flux from coil j linking with coil j, λ jj to the current in coil j, i j 7

18 Mutual inductance λ jj The self inductance L jj is the ratio of the flux from coil j linking with coil j, λ jj to the current in coil j, i j L L i i λ ij Likewise, mutual inductance L ij is the ratio of the flux from coil j linking with coil i, λ ij to the current in coil j, i j L ij i ij j 8

19 Mutual inductance λ Mutual inductance L is the ratio of the flux from coil linking with coil, λ to the current in coil, i L i λ Mutual inductance L is the ratio of the flux from coil linking with coil, λ to the current in coil, i L i 9

20 Recall that Mutual inductance Likewise L L And the mutual inductances become i i i i 0

21 Mutual inductance Assumption: All flux produced by each coil links with the other coil. This implies there is no leakage flux. This leakage flux is assumed to be zero. In reality there is some leakage flux, it is quite small because the iron φ has much less i reluctance than the air i With no leakage flux, it must be the case that all flux developed by one coil must completely link with the other coil.

22 Mutual inductance If all flux developed by one coil completely links with the other coil, then the flux from coil linking with coil is equal to the flux from coil linking with coil, i.e., A i l the flux from coil linking with coil is equal to the flux from coil linking with coil, A i l Substitute above into slide 0 inductance expressions: L A A i i l A l A L R i i l R i i l

23 L i i l Mutual inductance A A i i l A l A L R i i l R Comparing the above leads to L L R This says that the mutual inductances are reciprocal: The ratio of flux from coil linking with coil, λ, to i L i i is the same as the ratio of flux from coil linking with coil, λ, to i L i i

24 L R Mutual inductance We showed on slide 4 that Solve for and : L R L R L R Substitute into slide mutual inductance expression: L L L L L L R R R R 4

25 Mutual inductance It is conventional to denote mutual inductance as M: L R L R M L L L L R R Mutual inductance gives the ratio of: flux from coil k linking with coil j, λ jk to the current in coil k, i k, M i i 5

26 Polarity & dot convention for coupled ccts i φ φ i v e e Open cct DC voltage We have a al Coil : very small resistance to increase v so in steady-state, i Increase voltage v to some higher value current i increases with time flux from coil, φ, increases with time flux linkages λ increases with time. Who speaks when you have dλ/ (=d(ϕ)/) 0? 6

27 Polarity & dot convention for coupled ccts i φ φ i v e e Who speaks when you have dλ/ (=d(ϕ)/) 0? d d( ) d Faraday! e L But what about the sign of the right-hand-side (RHS)? Is it positive or negative? 7

28 Polarity & dot convention for coupled ccts i φ φ i v e e But what about the sign of the right-hand-side (RHS)? Is it positive or negative? The sign of RHS is positive because self-induced voltage across a coil is always positive at terminal the current enters, & e is defined positive at this terminal. e L If e would have been defined negative at terminal in which the current entered, then sign of RHS would be negative. 8

29 Polarity & dot convention for coupled ccts i φ φ i v e e If e would have been defined negative at terminal in which the current entered, then sign of RHS would be negative. e L 9

30 Polarity & dot convention for coupled ccts i φ φ i v e e ow consider coil it sees same flux that coil sees which we denote by φ (and corresponngly, the flux linkages are denoted as λ ). Considering our action of using the al to increase v, we again have, by Faraday s Law, d d( ) d e L M Is the sign of RHS positive or negative? 0

31 Polarity & dot convention for coupled ccts i φ φ i v e e d d( ) d e L M Is the sign of RHS positive or negative? That is, how do we know which of below are correct? e M e M

32 Polarity & dot convention for coupled ccts i φ φ i v e e That is, how do we know which of below are correct? e M e M Alternatively: Does assumed e polarity match actual polarity of voltage induced by changing current i? If yes, we choose positive sign. If not, we choose negative sign.

33 Polarity & dot convention for coupled ccts i φ φ i v e e That is, how do we know which of below are correct? e M e M Use Lenz s Law: induced voltage e must be in a rection so as to establish a current in a rection to produce a flux opposing the change in flux that produced e. See

34 Polarity & dot convention for coupled ccts i φ φ i v e e When e increases, i increases, and by the right-hand-rule (RHR), φ increases. Assumed polarity of e causes current to flow into the load in rection shown. How do we know e polarity is correct? Use Lenz s Law: induced voltage e must be in a rection so as to establish a current in a rection to produce a flux opposing the change in flux that produced e. 4

35 Polarity & dot convention for coupled ccts i φ φ i v e e We know e polarity is correct because RHR says that a current in rection of i causes flux in rection opposite to the rection of the φ increase. This is the φ increase, i.e., it is the change in flux that produced e and not necessarily the rection of φ itself (in this particular case, the φ increase is the same as the rection of φ itself). 5

36 Polarity & dot convention for coupled ccts i φ φ i v e e We know e polarity is correct because RHR says that a current in rection of i causes flux in rection opposite to the rection of the φ increase. e M Question: How might we obtain a fferent answer? 6

37 Polarity & dot convention for coupled ccts i φ φ v e e i There are two ways. First way: Switch sign of e, as above. Here, we also must switch current i rection, because, in using Lenz s Law, the i rection must be consistent with the e rection. Here, the current i, by RHR, produces a flux in the same rection as the φ increase, in violation of Lenz s Law: e M 7

38 Polarity & dot convention for coupled ccts i φ φ i v e e There are two ways. Second way: Switch the sense of the coil wrapping, while keeping the rections of e and i as they were originally. The current i, by RHR, produces flux in same rection as the φ increase. Therefore e M 8

39 Polarity & dot convention for coupled ccts i φ φ i v e e Let s articulate what we are trying to do: We want to know which secondary terminal, when defined with positive voltage polarity, results in using Faraday s Law with a positive sign. e M On paper, there are approaches for doing this.. Draw the physical winng go through Lenz s Law analysis as we have done in previous slides. 9

40 Polarity & dot convention for coupled ccts i φ φ i v e e On paper, there are approaches for doing this.. Draw the physical winng; go through Lenz s Law analysis as we have done in previous slides.. Use the dot convention. In dot convention, we mark terminal on each coil so that when e is defined positive at dotted terminal of coil and i is into the dotted terminal of coil, then e M 40

41 Example Ex : Express the voltage for each pair of coils below. From previous slide: In dot convention, we mark terminal on each coil so that when e is defined positive at dotted terminal of coil, and i is into the dotted terminal of coil, then e M () Recall the sign on the RHS is determined not by rection of flux flow (or current i flow) but by rection of change in flux flow (or current i flow). () Our above dot convention seems to depend only on rection of current (i ) flow and not on rection of change in current flow. Question: How can our dot convention give correct sign if it does not account for rection of change in current i flow? Answer: It does account for rection of change in current flow in that the above e equation implies positive rection of change ( / is positive). i i i i e e e e e e e e M M M 4 M

42 A second question So far, we have focused on answering this question: Given dotted terminals, how to determine the sign to use in Faraday s law? A second question: If you are given the physical layout, how do you obtain the dot-markings? Approach : Use Lenz s Law and the right-hand-rule (RHR) to determine if a defined voltage rection at the secondary produces a current in the secondary that generates flux opposing the flux change that caused that voltage. (This is actually a conceptual summary of Approach below.) 4

43 A second question If you are given the physical layout, how do you obtain the dot-markings? Approach : Do it by steps. (This is actually a step-bystep articulation of the first approach.). Arbitrarily pick a terminal on one side and dot it.. Assign a current into the dotted terminal.. Use RHR to determine flux rection for current assigned in step. 4. Arbitrarily pick a terminal on the other side and assign a current out of (into) it. 5. Use RHR to determine flux rection for current assigned in Step Compare the rection of the two fluxes (the one from Step and the one from Step 5). If the two flux rections are opposite (same), then the terminal chosen in Step 4 is correct. If the two flux rections are same (opposite), then the terminal chosen in Step 4 is incorrect dot the other terminal. This approach depends on the following principle (consistent with words in italics in above steps): Current entering one dotted terminal and leaving the other dotted terminal should produce fluxes inside the core that are in opposite rections. An alternative statement of this principle is as follows (consistent with words in underline bold in above steps): Currents entering the dotted terminals should produce fluxes inside the core that are in the same rection. 4

44 Example Example 4: Determine the dotted terminals for the configuration below, and then write the relation between i and e. Remember: Current entering one dotted terminal and leaving the other dotted terminal should produce fluxes inside the core that are in opposite rections. i e 44

45 Example Example 4: Determine the dotted terminals for the configuration below, and then write the relation between i and e. Solution: Steps -: φ i e Steps 4-6: φ Remember: Current entering one dotted terminal and leaving the other dotted terminal should produce fluxes inside the core that are in opposite rections. i e φ i 45

46 Example Example 4: Determine the dotted terminals for the configuration below, and then write the relation between i and e. Solution: ow write equation for the coupled circuits. Recall that in dot convention, we mark terminal on either side of transformer so that when e is defined positive at the dotted terminal of coil and i is into the dotted terminal of coil, then e M Here, however, although i is into the coil dotted φ terminal, e is defined i negative at the coil dotted e terminal. Therefore e M φ i 46

47 Example Example 4: Determine the dotted terminals for the configuration below, and then write the relation between i and e. Solution: there is another way we could have solved this problem, as follows Steps -: e i' φ Steps 4-6: i e i' φ φ 47

48 Example Example 4: Determine the dotted terminals for the configuration below, and then write the relation between i and e. Solution: There is another way we could have solved this problem! Write equation for the coupled circuits. Recall that in dot convention, we mark terminal on either side of transformer so that when e is defined positive at the dotted terminal of coil and i is into the dotted terminal of coil, then e M Here, i is into the coil dotted terminal, e is defined positive at the coil i dotted terminal. Therefore e e M If, however, we wanted to express e as a function of i (observing that i =-i' ) then we would have e i' M φ φ 48

49 Example Example 5: For the configuration below, determine the dotted terminals and write the relation between i and e. ote: Problems a,b,c,d are very similar to this one. 49

50 Example Example 5: For the configuration below, determine the dotted terminals and write the relation between i and e. Solution: Steps -: Steps 4-6: Here we arbitrarily assign dot to upper terminal of coil ; then, with i out of this dotted terminal, we use RHR to determine flux φ is in same rection as coil flux. This means our choice of coil terminal location dot is wrong. Therefore we know dot must be at other terminal, and the below shows clearly this is the case, since the flux from coil, φ, is opposite to the flux from coil, φ. 50

51 Example Example 5: For the configuration below, determine the dotted terminals and write the relation between i and e. Solution: ow we can write the equation for e. Recall that in the dot convention, we mark one terminal on either side so that when e is defined positive at the dotted terminal of coil and i is into the dotted terminal of coil, then e M e M 5

52 Writing circuit equations for coupled coils We have so-far focused on transformers or similar circuits having magnetic coupling between coils. We may also encounter other kinds of circuits having elements that are magnetically coupled. Example Example: In this paper, the feasibility of resonant electrical coupling as a wireless power transfer technique is stued. Published 05 in IEEE Transactions on Microwave Theory and We are well-positioned to handle such circuits by combining our (new) knowledge of the dot convention with our (old) knowledge of circuit analysis. Issue: When we write a voltage equation, we must account for the self-induced voltage in an inductor from its own current as well as any mutually-induced voltage in the inductor from a current in a coupled coil. 5

53 Writing circuit equations for coupled coils Consider below circuit, where there may be currents in both winngs: Let s define λ as the sum of λ, the flux from coil seen by coil λ, the flux from coil seen by coil. Therefore: What does Faraday s law say about e as a function of λ? e d d d But from slides and 5: e L M L i, d Mi 5

54 Writing circuit equations for coupled coils Consider below circuit, where there may be currents in both winngs: e L M But wait we have equated the sum of the derivatives to e, where e has a certain assumed polarity. How can we be sure that the sign of both of those derivative terms is indeed positive? Until we can be sure of that, I want to write the above equation as: e L M 54

55 Writing circuit equations for coupled coils Consider below circuit, where there may be currents in both winngs: e L M Let s begin with the first ( self ) term (it is easiest!): Rule for determining the sign of the self term: The polarity of the self term is determined entirely by the rection of the current i : when this current is into the positive terminal (as defined by the polarity of e ), then the sign of the self term is positive; when this current is out of the positive terminal (as defined by the polarity of e ), then the sign of the self term is negative. 55

56 Writing circuit equations for coupled coils Consider below circuit, where there may be currents in both winngs: e L M ow we need to determine how to know whether to add or subtract the mutual term from the self term. We should not be surprised to learn that we will make this determination using the dot convention. Rule for determining the sign of the mutual term: Assume both coils correctly dotted.. Choose reference current rections for each coil (if not chosen for you).. Apply following to determine reference polarity of voltage induced by mutual effects: a. If reference current rection enters dotted terminal of a coil, the reference polarity of voltage that it induces in other coil is positive at its dotted terminal. b. If reference current rection leaves dotted terminal of a coil, the reference polarity of voltage that it induces in other coil is negative at its dotted terminal. 56

57 Writing circuit equations for coupled coils Example 6: Express voltages e and e as a function of currents i and i in the following circuit. e L M Then, what is e? e L M First, let s express e. Here, we observe two things:. i enters the positive terminal, and therefore the self term is positive.. i enters the dotted terminal of coil, therefore the reference polarity of the voltage it induces in coil is positive at its dotted terminal, and its dotted terminal is the positive terminal. e L M 57

58 Writing circuit equations for coupled coils Example 6: Express voltages e and e as a function of currents i and i in the following circuit. e L M e L M ow let s express e. Here, we observe two things:. i enters the positive terminal, and therefore the self term is positive.. i enters the dotted terminal of coil, therefore the reference polarity of the voltage it induces in coil is positive at its dotted terminal, and its dotted terminal is the positive terminal. e L M 58

59 Writing circuit equations for coupled coils Example 6: Express voltages e and e as a function of currents i and i in the following circuit. e L M e L M 59

60 Writing circuit equations for coupled coils Example 7: Express voltages e and e as a function of currents i and i in the following circuit. e L M e L M Let s express e. Here, we observe two things:. i enters the positive terminal, therefore the self term is positive.. i leaves the dotted terminal of coil, therefore the reference polarity of the voltage it induces in coil is negative at its dotted terminal, and its undotted terminal is the positive terminal. e L M 60

61 Writing circuit equations for coupled coils Example 7: Express voltages e and e as a function of currents i and i in the following circuit. e L M e L M ow let s express e. Here, we observe two things:. i enters the positive terminal, and therefore the self term is positive.. i enters the dotted terminal of coil, therefore the reference polarity of the voltage it induces in coil is positive at its dotted terminal, but its dotted terminal is the negative terminal. e L M 6

62 Writing circuit equations for coupled coils Example 7: Express voltages e and e as a function of currents i and i in the following circuit. e L M e L M 6

63 Writing circuit equations for coupled coils HW problem 4: Write a set of mesh current equations that describe the circuit below in terms of i, i, and i. umber of nodes=4; number of branches where current not known=b=6 b-(n-)=6-=. We need three mesh equations. We write these for the three windows in the cct. above. 6

64 Writing circuit equations for coupled coils HW problem 4: Write a set of mesh current equations that describe the circuit below in terms of i, i, and i. Focus on top loop. Apply KVL starting from ode moving clockwise. With i in rection shown, we assume voltage polarity of 4H inductor is defined positive at its dotted end. With i into the 4H inductor, the self term is positive. But the KVL moves across 4H inductor from positive to negative, therefore the first term in the mesh equation is negative. The mutually induced term of 4H inductor is also negated by this KVL movement. Also, observe the coupled current i is into dotted side of 9H inductor, but i is out of it. So: The rest of the top loop is easy. ode 4 i i i i

65 Writing circuit equations for coupled coils HW problem 4: Write a set of mesh current equations that describe the circuit below in terms of i, i, and i. Focus on left loop. Apply KVL starting from ode moving clockwise. The first two parts are easy. v 8i i The next part is the self-induced term of the 9H inductor. With i in rection shown, we assume voltage polarity of 9H inductor is defined positive at its dotted end. With i into the 9H inductor, the self term is positive. But the KVL moves across 9H inductor from positive to negative, therefore the self term in the mesh equation is negative. ote it is comprised of current i into the dot (positive) and i out of the dot (negative) g vg 8i i 9 ode We still need the mutual term from the 4H inductor. It would be positive since i is into the dot of the 4 H inductor and out voltage is defined positive at the dot of the 9H inductor, but it is also influenced by the KVL rection of movement. vg i i

66 Writing circuit equations for coupled coils HW problem 4: Write a set of mesh current equations that describe the circuit below in terms of i, i, and i. i i i i Top Loop: Left Loop: vg 8 i i Right Loop: i i 0i 0 66

67 Development of turns ratio relations for ideal xfmr Dashed box: The ideal xfmr Ideal xfmr: o Losses, infinite permeability. Objective: See how i and i are related e and e are related in the steady-state. Voltage equation for left-hand-loop: e L M (*) Voltage equation for right-hand-loop: e L M iz M L iz (**) So we have equations (*) and (**): e L M (*) M L iz (**) Remember: Self term: when same-side current enters its positive terminal, self term is positive. Mutual term: when oppositeside current enters its dotted terminal, mutual term is positive at its dotted terminal. 67

68 Development of turns ratio relations for ideal xfmr Dashed box: The ideal xfmr Ideal xfmr: o Losses, infinite permeability. Objective: See how i and i are related e and e are related in the steady-state. e L M (*) M L (**) iz Important concept (see xfmr HW prob #6): Differential equations: characterize electrical relationships for any time periods under any type of excitation. Phasor equations: characterize electrical relationships for time periods where contions are in a steady-state under sinusoidal excitation We may convert fferential equations to phasor equations. How?... 68

69 Development of turns ratio relations for ideal xfmr Dashed box: The ideal xfmr Ideal xfmr: o Losses, infinite permeability. Objective: See how i and i are related e and e are related in the steady-state. e L M (*) M L (**) iz We may convert fferential equations to phasor equations. How?... Observe what we get when we fferentiate a sinusoid: Original function scaled by ω; rotated forward by 90. Time Domain i(t)= I sint /= I cost Phasor Domain I= I 0 I 90 /= I sin(t+90) = I 090=jI Differentiation in time domain is multiplication by jω in phasor (Fourier) domain! Let s use this to transform (*) and (**) 69

70 Development of turns ratio relations for ideal xfmr Dashed box: The ideal xfmr Ideal xfmr: o Losses, infinite permeability. Objective: See how i and i are related e and e are related in the steady-state. e L M (*) M L (**) iz jmi jl I Z I E jl I jmi (&) Recall: L R A l R l A I jm I jl Z Assume: μ is very large (infinite permeability). Then R is very small. Then L is very large. Then jωl >> Z. jm M So (#) becomes I I I jl L (#) 70

71 Development of turns ratio relations for ideal xfmr Dashed box: The ideal xfmr Ideal xfmr: o Losses, infinite permeability. Objective: See how i and i are related e and e are related in the steady-state. e L M (*) E jl I jmi (&) M L (**) iz I M I L Recall, slide 5: Recall, slide 4: Substitution into (!!) (!!) M R M L R I I I I I I R I L L R ratio of currents in coils on either side of an ideal transformer is in inverse proportion to the ratio of the coils turns 7

72 Development of turns ratio relations for ideal xfmr Dashed box: The ideal xfmr Ideal xfmr: o Losses, infinite permeability. Objective: See how i and i are related e and e are related in the steady-state. e L M (*) E jl I jmi (&) I M L (**) iz jm I jl Z Substitute (#) into (&); obtain common denominator; simplify: M L L jl Z I jl Z E = (#) I I (#*) Use M LL E = jl Z I jl Z 7

73 Development of turns ratio relations for ideal xfmr Dashed box: The ideal xfmr Ideal xfmr: o Losses, infinite permeability. Objective: See how i and i are related e and e are related in the steady-state. e L M (*) E jl I jmi (&) jl Z jl Z I M L (**) iz jm I jl Z E = I Use jωl >> Z. Recall, slide 4: From (#*) Substitute: I I L L R E = Z I Z I R E = (#) LZ I I L I Z E = I I I Use Z I =E : L Z R Z L E = R E (#*) ratio of voltage across coils on either side of an ideal transformer is in proportion to the ratio of the coils turns 7

74 Development of turns ratio relations for ideal xfmr Dashed box: The ideal xfmr Ideal xfmr: o Losses, infinite permeability. Objective: See how i and i are related e and e are related in the steady-state. e L M (*) E I = E I ratio of voltage across coils on either side of an ideal transformer is in proportion to the ratio of the coils turns M L (**) iz ratio of currents in coils on either side of an ideal transformer is in inverse proportion to the ratio of the coils turns 74

75 Dashed box: The ideal xfmr Power for ideal xfmr Ideal xfmr: o Losses, infinite permeability. Objective: Does ideal power transformer transform power? E = E E E I I I Preliminary comment: This scussion has nothing to do with losses in a real xfmr. We are still considering an ideal xfmr (no losses). Express power on both sides of transformer: S EI * * S EI Substitute expressions for E and I into expression for S : S E I E I E I * * * S EI S * I 75

76 Dashed box: The ideal xfmr Power for ideal xfmr Ideal xfmr: o Losses, infinite permeability. Objective: Does ideal power transformer transform power? We have just proved that, for an ideal transformer, S =S, enabling us to conclude that power transformers do not transform power. It is a good thing, because doing so would result in a violation of the conservation of energy (otherwise known as the first law of thermodynamics), since power transformation would imply that we could provide one side with a certain amount of power P and get out a greater amount of power P on the other side. If we allowed, then, such a device to operate for an amount of time T, the output energy P T would be greater than the input energy P T, thus, the violation. IDEAL TRASFORMER E = E S S I I 76

77 Primary side Referring Quantities Objective: Can we somehow move Z to the primary side of the ideal xfmr and then do all analysis there, so that our analysis need not have to account for the effect of the ideal xfmr? Secondary side Closely-related question: What impedance is seen looking into the primary terminals of the ideal xfmr? In other words, what is Z =E /I? From our turns ratio relations: E = E Substitution into the expression for Z : Z E E Z I I Z I I Z Answer to closely-related question: Looking into the primary terminals of an ideal xfmr supplying Z across its secondary terminals, we see the impedance Z scaled by the turns ratio squared. 77

78 Primary side Referring Quantities Objective: Can we somehow move Z to the primary side of the ideal xfmr and then do all analysis there, so that our analysis need not have to account for the effect of the ideal xfmr? Secondary side E = E Closely-related question: What impedance is seen looking into the primary terminals of the ideal xfmr? In other words, what is Z =E /I? IDEAL TRASFORMER S Z Z S I I These equations relate currents, voltages, impedances, and powers that exist on one side of the transformer, i.e., the secondary (primary), to corresponng currents, voltages, impedances, and powers, that exist on the other side of the transformer, i.e., the primary (secondary). 78

79 Primary side Referring Quantities Objective: Can we somehow move Z to the primary side of the ideal xfmr and then do all analysis there, so that our analysis need not have to account for the effect of the ideal xfmr? Secondary side E = E Closely-related question: What impedance is seen looking into the primary terminals of the ideal xfmr? In other words, what is Z =E /I? IDEAL TRASFORMER S Z Z S I I Relating quantities on one side of the xfmr to quantities on the other side is fine, but does that accomplish our above main objective? Can we move Z to the primary side of the ideal xfmr? I think we can, based on our answer to the closely-related question if we move it scaled by the turns ratio squared. Let s see how this looks 79

80 Primary side Referring Quantities Objective: Can we somehow move Z to the primary side of the ideal xfmr and then do all analysis there, so that our analysis need not have to account for the effect of the ideal xfmr? Secondary side Closely-related question: What impedance is seen looking into the primary terminals of the ideal xfmr? In other words, what is Z =E /I? Equivalent ccts i e Z Z Write Ohm s law on primary side: i e e Z Z i e i Z e Z e Z i What is this telling us? It is taking the answer to the closely related question (which is that we see Z =( / ) Z from the primary side) and showing that it is equivalent to seeing Z looking into the load terminals from the secondary. 80

81 Primary side Referring Quantities Objective: Can we somehow move Z to the primary side of the ideal xfmr and then do all analysis there, so that our analysis need not have to account for the effect of the ideal xfmr? Secondary side Closely-related question: What impedance is seen looking into the primary terminals of the ideal xfmr? In other words, what is Z =E /I? What is this telling us? It is taking the answer to the closely related question (which is that we see Z =( / ) Z from the primary side) and showing that it is equivalent to seeing Z looking into the load terminals from the secondary. The point is that we can see Z on the primary side, but it looks like Z =( / ) Z. This means that the below circuit gives us everything we need to know about primary-side quantities when the secondary side is loaded with Z. e Z Z This is actually what we set out to achieve in our objective! The circuit to the left is the circuit we need! We have achieved our objective. 8

82 Referring Quantities Primary side e Secondary side Z Z This is actually what we set out to achieve in our objective! The circuit to the left is the circuit we need! We have achieved our objective. In fact, in aion to the impedances Z and Z, we can say similar things about the voltages and currents. Z =( / ) Z is primary side equivalent of Z. e =( / )e is the primary side equivalent of e. i =( / )i is the primary side equivalent of i. And it works the other way too Z =( / ) Z is secondary side equivalent of Z. e =( / )e is the secondary side equivalent of e. i =( / )i is the secondary side equivalent of i. You can move quantities from secondary side to primary side! You can move quantities from primary side to secondary side! 8

83 otation: Referring Quantities First, observe that subscripts and (for turns, currents, voltages, impedances) tell us whether the quantity actually exists (physically) on the primary side (having subscript of ) or the secondary side (having subscript of ). Second, we use unprimed notation, i.e., I, E, Z, and I, E, Z, to denote the quantity represented on the side on which it actually exists. Thus, we say I, E, Z are the current, voltage, and impedance of primary side quantities referred to the primary side, and I, E, Z are the current, voltage, and impedance of secondary side quantities referred to the secondary side. Third, we use primed notation, i.e., I, E, Z, and I, E, Z, to denote the quantity represented on the opposite side from where it actually exists. Thus, we say I, E, Z are the current, voltage, and impedance of primary side quantities referred to the secondary side, and I, E, Z are the current, voltage, and impedance of secondary side quantities referred to the primary side. 8

84 Referring Quantities Referring quantities from secondary to primary I I I I E = E E E Z Z Z Z Referring secondary current to primary Referring secondary voltage to primary Referring secondary impedance to primary Referring quantities from primary to secondary I I I I E = E E E Z Z Z Z Referring secondary current to primary Referring secondary voltage to primary Referring secondary impedance to primary 84

85 Referring Quantities Example 8: Find the current I in the primary side of the below circuit. Solution: We can solve this problem in one of two ways. Approach : Refer all quantities to secondary, solve for I ; then refer this current to the primary. Approach : Refer all quantities to primary, solve for I. Approach : We begin by referring Z to the primary side via: 5 5 Z Z So we know what Z is it is the secondary impedance Z as seen from the primary side. Let s redraw the circuit accorngly. 85

86 Referring Quantities Example 8: Find the current I in the primary side of the below circuit. Solution: We can solve this problem in one of two ways. Approach : Refer all quantities to secondary, solve for I ; then refer this current to the primary. Approach : Refer all quantities to primary, solve for I. Approach : Use of Ohm s Law results in 50 0 I I 500 And we are done But wait what if we wanted to obtain secondary quantities, such as I and E? 86

87 Referring Quantities Example 8: Find the current I in the primary side of the below circuit. Solution: We can solve this problem in one of two ways. Approach : Refer all quantities to secondary, solve for I ; then refer this current to the primary. Approach : Refer all quantities to primary, solve for I. Approach : I 50 0 I 500 But wait what if we wanted to obtain secondary quantities, such as I and E? Then we refer I and E (which are quantities we obtain on the primary side) back to the secondary side. We already know I ; we obtain E by inspection of the above circuit, observing that it is the same as E, i.e., E =

88 Referring Quantities Example 8: Find the current I in the primary side of the below circuit. Solution: We can solve this problem in one of two ways. Approach : Refer all quantities to secondary, solve for I ; then refer this current to the primary. Approach : Refer all quantities to primary, solve for I. Approach : I 50 0 I 500 I I I I 0 E 0 = E E E 5 Check: Z E I

89 Referring Quantities: Example 9 I Z I =.+j4.0ω w 0.0+j0.04Ω I V = ,000Ω j,544ω E E V 4 +j Ω Find I and V. Ideal Xfmr / =0/ First, we need to refer all quantities to one side or the other. Because there are more elements on the primary side, it is easier to refer secondary quantities to primary quantities. 89

90 Referring Quantities: Example 9 Z I I s = Z =.+j4.0ω w 0.0+j0.04Ω I V = ,000Ω j,544ω E E V Z L =4 +j Ω 0 Z Z Z Ideal Xfmr / =0/ Apply to first impedance (call it Z s ): 0 Z s Zs (0.0 j0.04). j4 Apply to second impedance (call it Z L ): 0 Z L ZL (4 j) 400 j00 90

91 Referring Quantities: Example 9 Z s =.+j4ω I I I Z =.+j4.0ω w V = ,000Ω j,544ω E =E V Z L = 400 +j00 Ω ow we have a circuit to solve. Recall our goal is to find I & V. I am going to do this as follows:. Get Z eq =80,000//j,544//(40.+j04) (parallel combination). Get I =V /[Z +Z eq ].(Ohm s law). Use KVL or voltage vision to get E 4. Get I w =I =E /[Z s +Z L ] (Ohm s law) 5. Get V =I [Z L ] (Ohm s law) 6. Refer I and V back to the secondary (turns ratio) 9

92 Referring Quantities: Example 9 Z s =.+j4ω I I I Z =.+j4.0ω w V = ,000Ω j,544ω E =E V Z L = 400 +j00 Ω. Get Z eq =80,000//j,544//(40.+j04). This is a parallel combination of the 80,000Ω resistor, the j,544ω inductor, and the series combination Z a +Z L, as incated by the dotted arrows above. Recalling the parallel combination of three impedances is given by Z Z Z Z eq a b c (80, 000)( j,544)(40. j04) 85 j8 (80, 000)( j,544) ( j,544)(40. j04) (80, 000)(40. j04) 9 Z eq Z Z Z Z Z Z a b b c a c

93 Referring Quantities: Example 9 I Z =.+j4.0ω V = E =E Zeq 85 j8. Get I =V /[Z +Z eq ]. This is just an application of Ohm s law I V Z Z (. j4) (85 j8) eq Comment: We really don t need I, because we can get E =E (step ) by voltage vision. But if you want to perform step by KVL, we do need I. 9

94 Referring Quantities: Example 9 I Z =.+j4.0ω V = E =E Zeq 85 j8. Use KVL or voltage vision to get E I will first do it by KVL: E E V I ow do it by voltage vision: Zeq E E V Z Zeq Z 85 j (. j4) (85 j8) ( )(. j4)

95 Referring Quantities: Example 9 Z s =.+j4ω I I I Z =.+j4.0ω w V = ,000Ω j,544ω E =E V Z L = 400 +j00 Ω 4. Get I w =I =E /[Z s +Z L ] (Ohm s law). I = I' = w Z E ' s Z L (. j4) (400 j00)

96 Referring Quantities: Example 9 Z s =.+j4ω I I I Z =.+j4.0ω w V = ,000Ω j,544ω E =E V Z L = 400 +j00 Ω 5. Get V =I [Z L ] (Ohm s law) V = I' Z' L ( )(400 j00)

97 Referring Quantities: Example 9 Z s =.+j4ω I I I Z =.+j4.0ω w V = ,000Ω j,544ω E =E V Z L = 400 +j00 Ω 6. Refer I and V back to the secondary (turns ratio) V V V I' I = I'

98 Comment I strongly encourage you to read chapters 5 and 6 in Kirtley s text. Chapter 5: Magnetic circuits Chapter 6: Transformers Don t get stuck. Read through whole thing. Find where he addresses the same thing I address. Identify what he emphasizes that is fferent. Identify what he emphasizes that is similar. Be motivated to learn/understand so as to do well in this and other classes. Be motivated to learn/understand engineering to help you become a competent engineer in the workplace. 98

99 Exact & approximate transformer models I Z I =R +jx w Z =R +jx I I e V R c I c jx m I m E E V Z L Ideal Xfmr / =0/ The circuit we have been using in the last example is actually the exact equivalent model of the xfmr. (Aside: Exact and model is an oxymoron.) ote the nomenclature given to the voltages, currents, and impedances above. Let s define them. 99

100 Exact & approximate transformer models I Z I =R +jx w Z =R +jx I I e V R c I c jx m I m E E V Z L Z, Z : R, R : X, X : R c : X m : Ideal Xfmr / =0/ series impedances of each side V, V : winng resistances of each side E, E : leakage reactance of each side core loss resistance; represents I, I : losses due to eddy currents and hysteresis. I w, I : magnetizing inductance; represents current necessary to overcome the I e : core reluctance in setting up flux. I c, I m : Source, load voltages, respectively Voltages across internally-modeled prim & sec coils, respectively. Currents into and out of xfmr prim & sec terminals, respectively. Currents in internally-modeled prim & sec coils, respectively. Exciting current. 00 Core loss &magnetizing currents

101 Exact & approximate transformer models I Z I =R +jx w Z =R +jx I I e V R c I c jx m I m E E V Z L Ideal Xfmr / =0/ Let s illustrate how to refer secondary-side quantities to primary side. There are only two of them: Z, Z L. Z Z ow draw the circuit... Z L L Z 0

102 Exact & approximate transformer models I Z =R +jx I w Z Z I I e V R c I c jx m I m E =E V Z L L Z Observe that the ideal transformer is no longer used. This model is called the exact equivalent model referred to the primary. 0

103 Exact & approximate transformer models I Z =R +jx I w I e Z Z I V R c I c jx m I m E =E V Z L L Z Fact: The shunt element, R C //jx m, is actually quite large, in comparison to the series impedances Z, Z. Rc // jx m R jx This implies. Voltage drop across Z is small; most of V appears across shunt R C //jx m.. I e very small; current thru Z same as current thru Z. 0

104 Exact & approximate transformer models I I w I e Z =R +jx Z Z I V I c I m R c jx m E =E V Z L L Z Combining Z and Z, the above becomes: I Z I Z Z Z I e I w V I c I m R c jx m V Z L L Z This is called approximate circuit #. 04

105 Exact & approximate transformer models But note that with approximate circuit #, the voltage seem by the portion of the circuit with Z +Z L is V. This implies that the shunt R c //jx m does not affect I. Thus, if I am not interested in loss analysis (and therefore don t care about I c R c ), meaning I am mainly interested in voltage drop across the transformer, then the below is a good model. I Z I Z Z Z I w V V Z L L Z This is called approximate circuit #. 05

106 Exact & approximate transformer models A final model results from the fact that, for a transformer, the series reactance is significantly larger than the series impedance, i.e., with Z =R +jx, Z =R +jx : X X R R Then the following VERY simple model becomes quite reasonable. Indeed, this model, consisting of a single reactance, is often used in analysis of large power systems. I X I X X X I w V V Z L L Z This is called approximate circuit #. 06