Mutual Inductance and Transformers F3 3. r L = ω o


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1 utual Inductance and Transformers F3 1 utual Inductance & Transformers If a current, i 1, flows in a coil or circuit then it produces a magnetic field. Some of the magnetic flux may link a second coil of circuit. That flux linkage, Ψ 1, will be proportional to the current i 1 providing that any magnetisable material in the neighbourhood is not out of its linear range. We then write Ψ 1 = 1 i 1 where 1 < the constant of proportionality, is the mutual inductance. If the current, i 1 is changing the voltage induced in there is a voltage induced in the second coil, e = dψ 1 = 1 di 1 This is analogous to the induced voltage in the first coil due to the same changing current. e 1 = di 1 where L is the self inductance If the second coil carries current i then we may write the flux linkage in the first coil as Ψ 1 = 1 i It can be shown from energy considerations ( see Appendix) that 1 = 1 = The mutual inductance,, between two coils is = Flux linkage in one coil current in the other coil i 1 i The unit of mutual inductance is the same as that of self inductance, namely the henry, H. Thus if the currents in the two coils are and I the induced voltages are e 1 = L di 1 1 di ; e = di 1 L di NOTE: L is always positive: may have either sign; may be zero even for coils in close proximity. Think of two partly overlapping circular coils. utual Inductance & Transformers: wtnorris: 11 July 006
2 utual Inductance and Transformers F3 Dot notation Consider a source feeding a load through a mutual inductance. I V V 1 R The dots at the end of the coil symbol signify the 'positive ' end of each coil. If both coils were wound on the same former they would each be wound in the same direction if their helices advanced along the former in the same direction. In the circuit shown we have for KVL in the frequencydomain on the supply side V 1 = jω jω I 1 The negative sign for the coefficient of the second term is because I is flowing out of the Dot or positive end. KVL for the load circuit is 0 = jω + jω I + I R Below we show the development of these equations to transformer theory. eanwhile here is an example in signal theory. Problem V i I R R V o L L The filter above is to be a band pass filter. There is no load on the output. Where should the second dot be placed  and why? Solution Suppose the second dot is at the top. KVL for input circuit: V i = R + sl + 1 sc I KVL for output circuit: V o = s + 1 sc I Combine these: V o = s + 1 sc R + sl + 1 sc so the transfer function H(s) = V o V i = = 1 + s C 1 + src + s LC V i 1 + s C 1 + src + s LC utual Inductance & Transformers: wtnorris: 11 July 006
3 utual Inductance and Transformers F s The transfer function, C 1 + src + s LC, has zeros at s= ± j 1 C which lie on the imaginary axis of the splane. So the filter is a notch filter; i.e., also a band stop filter. Since < L the two poles will be at rather lower frequencies where there may be amplification. At zero frequency and infinite frequency V o = V 1 : there is no attenuation. Numerical example. Rewrite the transfer function putting ω m = 1 C : ω o = 1 LC so H(s) = ω o ω s + ω m m s + s ω o Q + ω o r L = ω o Q Then set ω o = 1 khz, ω m = 1 1 khz and Q = 5. The Bode plot becomes that shown opposite. Note that because we made Q as high as 5 there is a slight amplification just before the notch. If the second dot were at the bottom KVL for the output circuit would be: V o = s + 1 sc and the transfer function would be 1 s H(s) = C 1 + src + s LC This has zeros at s = ± 1 C These are both on the real axis and will have only a modifying effect with no pronounced character. There is amplification near the poles but still at the extremes of frequency there is no attenuation/ agnitude HdBL Phase HdegreesL agnitude HdBL Phase HdegreesL Frequency,Hz. Log Scale Frequency,Hz. Log Scale Frequency,Hz. Log Scale Frequency,Hz. Log Scale utual Inductance & Transformers: wtnorris: 11 July 006
4 utual Inductance and Transformers F3 4 Transformer I V V 1 R We showed earlier that KVL in the frequencydomain on the supply side V 1 = jω jω I 1 KVL for the load circuit is 0 = jω + jω I + I R It is convenient to introduce V = I R, the voltage across the load and write eq. as 0 = jω + jω I + V so V = jω jω I a On noload when I = 0 V V oc = jω = V 1 3 where V oc is the open circuit secondary voltage. The load terminals are called the secondary side or secondary terminals: the supply side are the primary terminals. Also on noload when I = 0 we get from eq. 1 where I is called the magnetising current. From eq. 1 we have in general I = V 1 jω 4 = V 1 jω + I = I + I 5 Combine eq. 5 with eq. a so that on load V = jω V 1 jωl + 1 L I jω I = V 1 1 L jω 1 L 1 L I = V oc jω X I 6 with X = ω 1 L 1 L 7 X is the leakage reactance referred to the load or secondary side. utual Inductance & Transformers: wtnorris: 11 July 006
5 utual Inductance and Transformers F3 5 Another important feature of the mutual inductance between two circuits is, as shown in the Appendix, that We define a coupling coefficient, k, such that = k For two coils wound close to each other the coupling coefficient, k, will be close to unity. The factor 1 L 1 L = (1 k ) in eq. 7 will be small. Anyway we may write eq. 7 X = ω (1 k ) 7a A mutual inductor is a transformer though the latter term, transformer, is usually reserved for the arrangement where there is a magnetic circuit completely formed of iron or other ferromagnetic material and the two coils are wound onto this common magnetic circuit usually called the core. Now from electromagnetic theory we have for coils wound on magnetic circuits R N R 8 where R is the reluctance of the magnetic circuit. For ferromagnetic circuits R = l A µ r µ o 9 where l is the effective (usually average) length of the magnetic circuit, A is its crosssectional area, µ o is the magnetic constant ( 4π 10 7 H m 1 ) and µ r is the relative permeability of the ferromagnetic material from which the core is made. Now = k = k N R so k N R 10 We found the open circuit secondary voltage: V oc = V 1 3 Thus taking k as close to unity from eqq. 8 and 10, N. Thus from eq. 3 V oc = V 1 V 1 N 11 which is the simple relationship normally taken for a transformer. Although the theory is approximate the relationship is quite close in practice. V oc V 1 = N 10a utual Inductance & Transformers: wtnorris: 11 July 006
6 utual Inductance and Transformers F3 6 Similarly from eq. 5 = I + I 5 = I + N N I 1 1 = I + load 1a with load I = N 1b In summary these following relationships for a transformer are widely used V oc V 1 = N 10a load I = N 1b V = V oc jω X I 6 I = V 1 jω 4 = I + N I 1a The leakage reactance referred to the secondary; X = ω (1 k) 7a The I is the easy to compute at least to a rough order using the formulae for reactance given above; eqq. 7 & 8. The computation of X is fairly complicated since k must be determined quite accurately; finding X does not figure in nd year exam papers depending as it does on detailed geometry of the core and windings. Equations 10a, 11b and 6 allow the construction of a simple equivalent circuit: X I V V 1 turns N turns Load Perfect transformer utual Inductance & Transformers: wtnorris: 11 July 006
7 utual Inductance and Transformers F3 7 Appendix : Energy Theorems Energy stored in a coil e e i L I i t o t Look at the charging process. When the generator voltage is positive and the current through the inductor increases e = L di t o t When the voltage falls to zero the inductor is charged and the current remains constant. Note that there is no resistance in this circuit. The power, p = ei, and therefore the energy input during charging is t o t op t oe W = = 0 i = 0 i L di = I 1 i L di = 0 L I 0 This is the stored energy W = 1 L i 13 utual Inductance & Transformers: wtnorris: 11 July 006
8 utual Inductance and Transformers F3 8 To show that 1 = 1 by an energy method Consider a pair of coils of self inductance and. E 1 1 Stage 1 Charge up the first coil with a generator to a current. The stored energy is W 1 = I E 1 E 1 Stage. Keeping the current in coil 1 constant charge up the second coil to a current I. The only voltages induced in coil in this process are those due to the fields caused by current I. Thus the energy supplied by the generator E is W = 1 I 15 However there will be voltages induced in coil 1 by the changing current in coil of V 1 = 1 di where i is a value of current during the charging process. To keep at a constant value the generator E 1 must counteract this voltage and in so doing will exchange energy at a rate dw 1 = V 1 The total energy exchanged with generator E 1 during the raising of current in coil is W 1 = dw 1 = dw 1 = V 1 = di 1 = I I1 0 1 di = 1 I 16 The energy exchange may be positive or negative depending on whether 1 is positive or negative, that is on whether the field due to I strengthens (when E 1 supplies energy) or weakens the field due to. Take the positive value now. The total energy supplied by the generators is the sum of eqq. 14, 15 & 16 W 1 + W + W 1 = I + 1 I utual Inductance & Transformers: wtnorris: 11 July 006
9 utual Inductance and Transformers F3 9 This is the total stored energy in the magnetic field. It can be extracted by reversing the process. If we had begun by charging coil first and then coil 1 the total stored energy would have been W = W 1 + W + W 1 = I + 1 I Since the two final conditions are indistinguishable the final stored energy must be the same however the charging was done (else we could charge by one process and discharge by the other and thereby defeat the first law of thermodynamics, a trick which is not possible). The two expressions for stored energy must be the same. Thus W 1 = W 1 and 1 = 1 = 17 utual inductance is reciprocal. There is only one mutual inductance for a pair of coils. Thus the total stored energy in the total magnetic field due to the currents in both coils is W = I + I. 18 NOTE The mutual inductance,, may be positive or negative. If is negative then current in one coil tends to cancel the magnetic flux due to current in the other coil. utual Inductance & Transformers: wtnorris: 11 July 006
10 utual Inductance and Transformers To show that F3 10 R G i 1 i = 0 i Step 1. Charge up the magnetic field in Inductor 1 so that a current flows. Inductor is on open circuit Step Short circuit Inductor 1. The current will continue to flow there being no resistance. The stored energy is 1 Step 3 Short circuit Inductor. There is no electromagnetic change. Step 4 Open the circuit of Inductor 1. This can be done by inserting a variable resistance and increasing its value to infinity until no current flows in Inductor 1. This can be done very quickly though the rate does not matter as we shall see. The differential equation governing the current in Inductor during Step 4 is di + di 1 = 0 Integrate this with regard to time for Step 4: t t di + di 1 = or, changing variables, I L 0 di + di1 = 0 0 where I is the final value of i Thus I = 0 and the current in coil after Step 4 is I = The final stored energy is all in coil and is 1 I = 1 I 1 L = 1 L I 1 This must be less than or equal to the energy stored in Inductor 1 just after it had been charged. The first law of thermodynamics holds: all the excess energy is dissipated in the resistor or radiated. Thus 1 L I 1 1 or. We often write = k 19 where k is called the coupling coefficient. k 1. NOTE some authors define the coupling coefficient by = k utual Inductance & Transformers: wtnorris: 11 July 006
11 utual Inductance and Transformers Force between two current carrying coils F3 11 Consider again a pair of coils of self inductances and and of mutual inductance and carrying currents and I. The supplies to the two coils keep the currents constant. I We are now going to calculate the force between the coils in the xdirection. G 1 x G Suppose that the coils move apart against the force F x by a distance δx in time δt. Stage 1. The work done against the force, F x, is W x = F x δx. Stage. There is a change in flux in circuit 1, δψ = I δ = I d dx δx. The voltage induced in circuit 1 is V 1 = δψ δt = I d δx dx δt. The power provided by the supply G1, P 1 = V 1 = I d dx δx δt. Thus the energy supplied, W 1 = P 1 δt = I d dx δx = I δ. Stage 3. Similarly the energy supplied by G is the same: W = I δ. Stage 4. The increase in stored energy of the coils is: 1 W s = d dx + 1 I + I δx = I δ. Stage 5. The energy supplied by the electric power supplies, G1 and G, to sustain the current goes partly into increasing the stored energy in the magnetic field and partly into moving the constraining force: W 1 + W = W s + W x I δ = I δ + F x δx Whence F x = I d dx 0 Now the flux linking circuit due to the current in 1 is Ψ 1 = ; thus F x = I dψ 1 dx = dψ 1 dx. 1 This force is independent of the source of the flux; it could be from a permanent magnet. utual Inductance & Transformers: wtnorris: 11 July 006
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