Let's consider now a two-winding transformer. Here we have a core that has two windings on it. We have polarity marks shown here, the dots, and what these denote is the relative directions of the currents in the two windings. Here we have current i_1 that flows into the window or the hole in the middle of the core, and likewise, for the second winding, we have a current i_2 that flows in the same direction, downward through the window of the core. So if we have flux Phi that flows around the inside of the core, then we can use Ampere's law on this flux to say that the integral of H.dl around the core, which is the magnetomotive force in the core, is equal to the sum of the currents passing through the interior of the path, or in this case, the sum of the amp turns of the two windings, n_1, i_1 plus n_2, i_2. The magnetomotive force around the core or around the path, can be written as the flux times the reluctance of the core, where the reluctance of the core, as usual, is the path length of the core material, the distance all the way around, divided by the Mu of the core and the cross-sectional area of the core. Then Ampere's law gives us this basic equation, and we can write an equivalent circuit model that goes along with it, where the flux times the reluctance is the magnetomotive force across the core, and that's equal to the sum of the amp turns of the two windings. So we can write a total equivalent circuit like this, and note the polarities of the MMF sources they add as we go around the loop. Let's construct an electrical equivalent circuit model now, to go along with our magnetic circuit model. The simplest version gives us the ideal transformer, and in the ideal transformer, the reluctance of the core tends to 0. So in this case, if this reluctance goes to 0, then the magnetomotive force also goes to 0, and what we have is that the sum of n_1, i_1 plus n_2, i_2, the two sources add up to 0 like this. This is the equation, the end of the terminal currents of the ideal transformer, and it says that the currents are related by the turns ratio. We can also model the voltages, and to do that we use Faraday's law. So we have a flux Phi that passes through the interior of the windings and in the ideal transformer we assume that it's the same flux in both windings. So it's the flux Phi which induces a voltage by Faraday's law in the first winding of n_1 d Phi, dt, and in the second winding, it will likewise induce a voltage of n_2 d Phi, dt, and the key point here in the ideal transformer is that the Phis are the same. So we have the same flux linking both windings. So we can actually then eliminate d Phi, dt. So d Phi, dt from the first winding would be equal to v_1 over n_1, and in the second winding, d Phi, dt is equal to v_2 over n_2. Since it's the same flux, then Faraday's Law tells us that the voltages are related according to the turns ratio as well. Then we get these equations for the ideal transformer, and they come out of the magnetic circuit model. For the case when the reluctance of the core goes to 0, these equations we commonly relate with the ideal transformer, or model electrically with the ideal transformer. Next, let's add back in the reluctance of the core. So let's suppose the core reluctance is not 0. In that case, we can't leave this term flux times reluctance out of the equation. So what we can do instead is use the Faraday's law equation that we just derived and solve this first one for flux and plug it into the second one to eliminate Phi. So the flux then would be equal to the sum of the amp turns divided by the reluctance. We can plug that into here, and we would get that v_1 is n_1 times the derivative of this flux. That would give us n_1 di_1 dt plus n_2 di_2 dt. What I've done in the next equation on the slide is simply to pull out the n_1, pull this n_1 out, and what we get is this. The voltage v_1 is related to this effective inductance, n_1 squared over the core reluctance, times the derivative of the sum of two current terms. We can write this in the form that v_1 is an effective inductance that we call the magnetizing inductance, L_M times the derivative of this effective current which we call the magnetizing current, i_M. Where are those inductances and currents in the equivalent circuit? Here is the ideal transformer from the previous slide. If you look at it, we have current i_2 flowing into the secondary dot. We'll have flowing out of the secondary dot, i_2 times the turns ratio, so n_2 over n_1 i_2. That current is this current. The other current, i_1, is the primary current, and the sum of the two currents actually looks like a node equation. It's this current that we define as i_M, the magnetizing current, and it's flowing through a magnetizing inductance, L_M, that is given by n_1 squared over the core reluctance. When we include the core reluctance, as a non-zero quantity, what we get is an effective inductor called the magnetizing inductance that is in parallel with the winding. You can actually draw the inductor on either side of the ideal transformer. We could push L_M through the transformer turns ratio squared, and on the secondary side write this as n_2 squared over the reluctance. The magnetizing inductance is a real inductor. It models the magnetization of the core. For example, we could do the thought experiment where we just disconnect the secondary and connect the primary up into our circuit. What would we expect to see? Well, with just the primary, we have some turns of wire on a core, and that makes an inductor, and the inductor is in fact the magnetizing inductance. The equivalent circuit reflects that if you disconnect the secondary of the ideal transformer, then we have an open circuit on the primary side, and effectively the model reduces down to the magnetizing inductance. This is a real inductor also. It has saturation and hysteresis which is in fact coming from the actual core material. You can saturate the transformer then by saturating the magnetizing inductance. If the current and the magnetizing inductance becomes too large, it will saturate the transformer, the magnetizing inductance will turn into a short-circuit, and that'll short out the windings of your transformer. The one other thing the magnetizing inductance does, is it causes the currents to differ from the turns ratio. As I mentioned a minute ago, the transformer will saturate when the magnetizing current becomes too large. However, this is not the same as saying the transformer saturates when the winding current is too large, because the winding currents can differ from the magnetizing current. In fact, if we put a large current in the primary and it all comes out the secondary with no current flowing through the magnetizing inductance, the transformer doesn't saturate. We generally don't think of transformers as saturating because the winding current is too large. Rather, we think of them as saturating when the applied volt seconds are too large. In fact, we can bypass the magnetizing current altogether and simply go to Faraday's law. Where if we integrate Faraday's law to get the flux density in the core, it's given by one over the turns times the cross sectional area times the integral of the voltage, and the current doesn't really come into play. If you apply too many volt seconds to the winding, then you will make your transformer saturate. We think of transformers as saturating based on applied voltage rather than the applied winding current. We saw previously in the last lecture that if your inductor is saturating, you can fix that by removing turns or adding an air gap. In the case of the transformer, neither of those work. In fact, what we need to do is reduce the flux density in this equation. To do that, you add turns. So make n_1 larger rather than smaller, and that will tend to make your transformer stop saturating. So we saw that the magnetizing inductance causes the winding currents to differ from the turns ratio. The voltages can also differ from the turns ratio, and this happens from a phenomenon called leakage inductances. So to relate the voltages in the ideal transformer, we assume that we had the same flux passing through the interiors of both windings. Here in this drawing, that's called the mutual influx piece sub M. But in practice, there can be fluxes that pass through one winding and not the other, and these are called leakage inductances. Here I've drawn some fluxes that we say leak out of the core into the air, and the bottom line here is that this is flux that passes through one winding and not the other. So in this case, when we write Faraday's Law, V_1 is n_1 times the derivative of the flux passing through the interior of the winding. That flux in this case is drawn here, is the a sum of leakage flux VL_1 plus the the mutual flux V_M. V_2 likewise is n_2 times the derivative of a leakage flux VL_2 plus the mutual flux V_M. The point here now is that the leakage flux has caused the voltages to defer and no longer scale according to the turns ratio. Really, this case as drawn here is topologically equivalent to having separate little inductors in series with each winding, and those inductors carry the flux. So this term here effectively looks like an inductor with n_1 turns carrying flux VL_1, which I've just drawn as a separate core, and similar for winding 2. So these things look then effectively like series inductors. The voltage drops across those series inductors then will cause the terminal voltages to differ. So here is an equivalent circuit that includes the leakage fluxes. So I have a magnetizing inductance and then we have two leakage inductances. This is the well-known T model of the transformer. It contains four parameters: there is a magnetizing inductance or a mutual inductance, there are two leakage inductances, and there's a turns ratio. Now we know from basic circuit theory that you can write the terminal equations of a two-winding transformer with this matrix form, where this is called the inductance matrix. The inductance matrix turns out to have three independent parameters, L_11 and L_22, which are called the self-inductances of the primary and secondary windings, and then L_12, which is the same, it's the off-diagonal term n. It turns out by reciprocity that it's the same term in both cases here. So the inductance matrix in general has three independent parameters rather than four, and in the equivalent circuit model of a two-winding transformer, there are three independent parameters, and the fourth can be chosen arbitrarily. A common case is to choose the turns ratio, here is the physical turns ratio of the transformer winding, and then solve with measurements to find the inductances, the three inductances of the T model. The T model parameters can be related to the inductance matrix with these relationships. It's common in circuit theory to also define an effective turns ratio, which is defined to be the square root of the ratio of the self-inductances. This is a parameter that may not be the same as the actual physical turns ratio of the transformer. This effective turns ratio is based on the elements in the transformer model. We also can define a coupling coefficient. The coupling coefficient is a measure of how closely or tightly coupled the two windings are. As the coupling coefficient tends to one, the leakage inductances go away and tend to zero. Good transformers will have coupling coefficients slightly less than one.