Outline. Systems and Signals 214 / 244 & Energy Systems 244 / 344. Ideal Inductor. Ideal Inductor (cont... )

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1 Outline Systems and Signals 214 / 244 & Energy Systems 244 / 344 Inductance, Leakage Inductance, Mutual Inductance & Transformers 1 Inductor revision Ideal Inductor Non-Ideal Inductor Dr. P.J. Randewijk Stellenbosch University Dep. of Electrical & Electronic Engineering Copyright c 2015 Stellenbosch University All rights reserved 2 Transformers Ideal Transformers non-ideal Transformers Nilsson & Riedel Section 6.5 & Appendix C 1 / 58 2 / 58 Ideal Inductor The typical layout of an ideal inductor, where we assume that all the magnetic flux produced by the coil is confined only to the core, is shown below. Ideal Inductor (cont... ) With the flux in the core, φ c = Ni (1) i φ c l c and the reluctance in the core, = l c µ 0 µ r A c (2) e N A c 3 / 58 4 / 58

2 Ideal Inductor (cont... ) The inductance of the core can thus be calculated as follows: L c = λ i (3) non-ideal Inductor In a non-ideal inductor, some of the flux will leak through the air, i.e. it will not be confined to the core alone. = Nφ c i = N i (4) Ni (5) = N2 (6) i e N φ φl φ c 5 / 58 6 / 58 non-ideal Inductor (cont... ) The total magnetic flux linking the coil is therefore equal to the sum of the flux confined to the core plus the leakage flux through the air. φ = φ c φ l (7) = Ni Ni R l (8) Nφ = N2 i N2 i R l (9) λ = N2 N2 i R l (10) = L c L l (11) non-ideal Inductor (cont... ) The total or self inductance of the coils is thus equal to the sum of the inductance due to the flux confined to the core, plus the leakage inductance associated with the leakage flux. L = L c L l (12) 7 / 58 8 / 58

3 Ideal Transformer The typical layout of an ideal transformer, where all the flux produced by the coils are confined only to the core, is shown below. i 1 e 1 φ c =φ m i 2 The flux in the core, φ c, is also referred to as the mutual flux, φ m, because it has is mutual to both coils. If we were to connect an ideal voltage source across coil and we assume that the coil itself does not have any resistance, we will be able to calculate the flux in the core by making use of Faraday s law, with the voltage of the voltage source known. Thus, for a purely sinusoidal voltage, the flux will also be sinusoidal. With the flux being mutual to coil as well, we will be able to calculate the voltage induced in by also making use of Faraday s law. 9 / / 58 Thus from the sketch below, i s (t) φ m v s (t) = 2V s cos(ωt) (13) = e 1 (t) (14) v s (t) e 1 and if the applied voltage is sinusoidal, we can do the following calculations. dφ m = (15) 2Vs φ m (t) = sin(ωt) (16) ω dφ m (t) = ( ) d 2Vs = sin(ωt) ω (17) (18) = e 1 (t) (19) 11 / / 58

4 The current drawn from the supply voltage source can be calculated as follows: φ m (t) = i s (t) (20) i s (t) = φ m (t) (21) = R c 2Vs sin(ωt) (22) ω 2 2Vs = sin(ωt) ωl m (23) = i m (t) (24) The mutual flux, φ m, is sometimes also called the magnetising flux because it magnetises the core and induce the secondary voltage,. Therefore, the current component which is responsible for the magnetising flux is also known as the magnetising current, i m. We would also be able to calculate the magnetising current directly from the inductance of the primary winding. This inductance is therefore also known as the magnetising inductance, L m. L m = 2 (25) 13 / / 58 The magnetising current can thus be calculated directly from the magnetising inductance by making use of phasor. v s (t) = 2V s cos(ωt) (26) V s = V s 0 (27) I s = I m Φ m I m = V s jx m (28) = V s 90 (29) ωl m V s E 1 E 2 i m (t) = = 2Vs cos(ωt 90 ) (30) ωl m 2Vs sin(ωt) (31) ωl m With X m = ωl m known as the magnetising reactance. 15 / / 58

5 The polarity of can be determined by making use of Lenz s law. The polarity of will be such that, if a load were connected over, a current, i 2, will flow so as to oppose the cause of (i.e. will try and oppose φ m ). By looking at the way in which and are wound around the core and using the right hand grip rule, the polarity of w.r.t. e 1 can be determined. v s (t) i s (t) e 1 φ m φ2 i 2 The polarity relationship between e 1 and is usually indicated by dots if the winding configuration self is not shown. With φ 2 which now tries to oppose φ m, it implies that the total flux that sees will now become less / / 58 But from Faraday s law: φ m v s (t) = e 1 (t) = dφ m Thus with v s (t) constant (i.t.o. amplitude and frequency), it implies that the flux which sees must also remain constant. (32) i m (t) i 1 (t) v s (t) e 1 φ1 φ2 i 2 (t) The transformer will now draw an additional current component, i 1, from the supply which will produce an additional flux, φ 1, which will cancel out the effect of φ 2, i.e. φ 1 = φ 2. The nett flux in the core will thus remain constant and equal to φ m. 19 / / 58

6 The relationship between i 1 and i 2 can now be calculated as follow: φ 2 = i 2 (t) (33) φ 1 = i 1 (t) (34) φ 1 = φ 2 (35) i 1 (t) = i 2 (t) (36) i 1 (t) = i 2 (t) (37) i 1 (t) = i 2 (t) (38) The transformer can also be represented completely electrical, with out the need to refer to its magnetic circuit, by making use of the electrical symbol for an ideal transformer as shown below, i s i 1 i 2 i m L m :N2 v s e 1 21 / / 58 with the following relationship between the primary and the secondary s currents and voltages: i 1 = i 2 (39) = e 1 (40) For a transformer to really be considered ideal, the relative permeability of, µ r, of the magnetic core of the transformer must tend to infinity. And that would imply that the reluctance of the core will tend to zero. This in turn would imply that the magnetising inductance of the transformer will tend to infinity. As a result, the magnetising current will tend to zero. The ideal transformer in the time domain, can thus electrically be shown as only: i s = i 1 i 2 :N2 v s = e 1 23 / / 58

7 This is also true for the phasor domain : to rather work with the winding ration 1 : a, so that: V s I s = I 1 I 2 1:a E 1 E 2 Z load a = N2 (41) E 2 = ae 1 (42) I 2 = I 1 a (43) The load can thus be defined as: Another simplification is to, instead of using the winding ratio : N2, Z load = E 2 I 2 (44) = a 2 E 1 I 1 (45) 25 / / 58 The load, as referred to the supply side of the transformer, Z load, can be defined as follows: non-ideal Transformers For a non-ideal transformer, not all the flux from coil 1 couples with coil 2 and reverse. I s = I 1 φ m i 1 φ1 i 2 V s E 1 Z load a 2 e 1 φl1 φl2 φ2 Z load = Z load a 2 = Z load ( ) 2 (46) 27 / / 58

8 N.B. φ 1 and φ 2 are now defined differently as in the previous section and are now directly associated with and respectively. φ 1 is the total magnetic flux that links coil 1 due to the current in coil1, i 1, and the current in coil 2, i 2. Similarly is φ 2 the total magnetic flux that links coil 2 due to the current in coil 1, i 1, and the current in coil 2, i 2. Not all the flux that is linking coil 1, links with coil 2 and reverse. The portion of the flux from coil 1 that also links with coil 2 (i.e. which is common to both coils) is known as the mutual flux, φ m. The portion of the flux from coil 1 which does not link with coil 2, is known as the leakage flux, φ l1. Similarly, φ l2 is the leakage flux of coil 2, i.e. the portion of the flux that links coil 2, but which does not link with coil 1. The total flux of coils 1 & 2 can thus mathematically be expressed as: φ 1 = φ l1 φ m (47) φ 2 = φ l2 φ m (48) 29 / / 58 Or in terms of the different reluctance values: φ 1 = i 1 R l1 i 1 i 2 (49) φ 2 = i 2 R l2 i 1 i 2 (50) In order to obtain the flux linkages of coils 1 & 2, we multiply (49) with and (50) with. φ 1 = 2 i 1 2 i 1 i 2 (51) R l1 2 i 2 φ 2 = 2 i 2 i 1 (52) R l2 R ( ) c 2 λ 1 = ( ) 1 i 1 i 2 (53) R l1 ( ) ( ) 2 N2 λ 2 = i 1 2 i 2 (54) R l2 31 / / 58

9 Or in matrix format: with: [ λ1 λ 2 ] [ ] [ ] L1 M i1 = M L 2 i 2 (55) L 1 = 2 1 R l1 (56) = L l1 L m1 (57) The self inductance of coil 1 is equal to the leakage inductance of coil 1 plus the mutual inductance as referred to coil 1 (i.e. the primary side) L 2 = 2 2 R l2 (58) = L l2 L m2 (59) The self inductance of coil 2 is equal to the leakage inductance of coil 2 plus the mutual inductance as referred to coil 2 (i.e. the secondary side) M = (60) The mutual inductance, as seen from the primary and secondary side, i.e. not referred to any specific side. 33 / / 58 The coupling factor, k, can be defined as the mutual flux linking coil 2, φ m, divided by the total flux linking coil 1, φ 1, due to the current in coil 1, i 1. k = φ m φ 1 (61) φ m = φ l1 φ m (62) = = i 1 i 1 R l1 N 1i 1 2 i 1 ( ) (63) 2 i 1 R l1 2 i 1 (64) = L m1 L l1 L m1 (65) = L m1 L 1 (66) Similarly, the coupling factor (due to the current in coil 2, i 2 ) can also be expressed as : Which implies that: k = L m2 L 2 (67) k 2 = L m1 L 1 Lm2 L 2 (68) M = k L 1 L 2 (69) 35 / / 58

10 with: M 2 = L m1 L m2 (70) L m1 L m2 = 2 2 (71) The equivalent circuit diagram of the transformer, shown earlier, can be drawn with dots to indicate the winding direction and relationship between the primary and secondary windings. i 1 i 2 The voltages over both coils can be obtained by differentiating both sides: d [ λ1 λ 2 [ e1 ] = d [ ] [ ] L1 M i1 M L 2 i 2 ] = d [ ] [ ] L1 M i1 M L 2 i 2 (72) (73) M e 1 L 1 L 2 37 / / 58 Dot Definition If the current flows into a dot of a coil, it will induce a voltage in the other coil which will be positive at that coil s dot. The Kirchoff voltage mesh equations for the primary and secondary side of the transformer can thus be written as: Alternative Dot Definition If a current flow out of a dot of a coil, it will induce a voltage in the other coil which will be negative at that coil s dot. L 1 di 1 L 2 di 2 M di 2 M di 1 e 1 = 0 (74) = 0 (75) If we were to write the equations in matrix format, it will be identical to (73). 39 / / 58

11 If we were to reverse the winding direction of the secondary coil of the transformer, the polarity of the induced voltage in the secondary will also reverse. The equivalent circuit diagram of the transformer can now be drawn as follows with the dots once again indicating the winding direction relationship. φ m i 1 i 2 i 1 φ1 i 2 M e 1 φl1 φl2 e 1 L 1 L 2 φ2 41 / / 58 The Kirchoff voltage mech equations for the primary and secondary side of the transformer can now be written as: L 1 di 1 L 2 di 2 Or in matrix format: [ e1 M di 2 M di 1 ] = d [ ] [ ] L1 M i1 M L 2 e 1 = 0 (76) = 0 (77) i 2 (78) Equation (73) (or (78)) can also be seen as the equivalent two-port network z-parameters matrix equations of the transform and thus also be represented as an equivalent T -circuit. i 1 L 1 M L 2 M i 2 e 1 M 43 / / 58

12 See Nilsson & Riedel Appendix C.1, Fig. C.2 For a representative model of the transformer, it should make sense that the equivalent circuit should include an ideal transformer model. i 1?? í 2 i 2 e 1? é 2 45 / / 58 Another advantage that the ideal transformer have, is that it eliminates the negative inductances in the equivalent T -model, especially for step-up transformers see Nilsson & Riedel Appendix C.2. The new unknown circuit elements in the T -equivalent circuit can be calculated as follows. With the relationship between é 2 and, and í2 and i 2 on both sides of the ideal transformer known, é 2 = (79) í 2 i 2 = (80) (73) can thus be rewritten as: [ ] ( e 1 ) é N2 = d [ ] [ ] L1 M ( i 1 ) 2 M L 2 í 2 [ ] [ ] [ ] e1 = d ( L 1 ) ( M ) ( i 1 ) é2 M L 2 í 2 ( ) = d L 1 M [ N ( ) ( 2 ) 2 i1í2] M L 2 [ ] = d [i1 ] L 1 Ḿ Ḿ Ĺ2 í 2 (81) (82) (83) (84) 47 / / 58

13 with: Ḿ = M ( = ) ( ) (85) (86) = N2 1 (87) = L m1 (88) ( ) 2 Ĺ 2 = L 2 (89) N ( 2 ) 2 = ( ) 2 2 (90) R l2 = 2 1 (91) R l2 = L l2 L m1 (92) The mutual inductance as referred to the primary side. The leakage inductance of the secondary coil, as referred to the primary side. 49 / / 58 This implies that the two-port z-parameters of (84), can be represented by the equivalent T -circuit as shown below. Which is basically the same as shown in Fig. C.13 of Appenidx C.2 from Nilsson & Riedel. i 1 L l1 L í 2 i 2 l2 e 1 é 2 L m1 with a = (93) 51 / / 58

14 In Nilsson & Riedel Section 6.5, the different magnetic flux components are defined somewhat different. The total magnetic flux that links coil 1, φ 1, is divided in the flux produced by the current in coil 1 itself, φ 11, as well as the flux produced by the current in coil 2, φ 12. Similarly, the total magnetic flux linking coil 2 can be divided into the flux produced by the current in coil 1, φ 21, as well as the flux produced by the current in coil 2 itself, φ 22. φ 1 = φ 11 φ 12 (94) φ 2 = φ 21 φ 22 (95) The mutual flux, φ m, as defined earlier, can be written as the sum of φ 12 and φ 21 (with the winding direction as shown earlier). φ m = φ 12 φ 21 (96) The leakage flux of coil 1 can thus be defined as the magnetising flux produced by the current in coil 1 minus the portion of the flux that links with coil 2. φ l1 = φ 11 φ 21 (97) Similarly, the leakage flux of coil 2 can be defined as the flux produced by the current in coil 2, minus the portion of the flux that links with coil 1. φ l2 = φ 22 φ 12 (98) 53 / / 58 The relationship between φ 11, φ 21, φ 12 & φ 22 and φ m, φ l1 & φ l2 can thus graphically be depicted as follows: This implies that both Fig & Fig in Nillson & Riedel is wrong. φ m =φ 21 φ 12 i 1 φ1 i 2 e 1 φl1=φ11φ21 φl2=φ22φ12 φ2 55 / / 58

15 The inductance associated with the different magnetic flux components can be calculated exactly the same as previously. φ 1 = φ 11 φ 12 (99) φ 2 = φ 21 φ 22 (100) [ λ1 λ 2 λ 1 = λ 11 λ 12 (101) λ 2 = λ 21 λ 22 (102) ] [ ] [ ] L11 L = 12 i1 L 21 L 22 i 2 (103) d [ λ1 λ 2 ] = d [ ] [ ] L11 L 12 i1 L 21 L 22 i 2 [ e1 ] = d [ ] [ ] L11 L 12 i1 L 21 L 22 i 2 (104) (105) The z-parameter matrix equation is basically equivalent to (73), which thus implies that: L 1 = L 11 (106) L 2 = L 22 (107) M = L 12 = L 21 (108) 57 / / 58

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