Welcome back to the course on magnetics for power electronic converters. So far we have looked at an ideal transformer and developed a circuit model for it. Now let's start to include some non-idealities. And see how that modifies the global and circuit model for the transformer. The first non-ideality that we're going to include is that we're going to stop assuming that the permeability of the core is infinite. Instead, we will assume a finite permeability for the core. If you stop assuming that the permeability of the core is infinite, then the reluctance of the core will no longer be zero. Instead, the reluctance will be some finite value greater than zero. If we know to go back to the magnetic circuit model for our core, we realize that no longer does N1, I1, plus N2, I2 equal zero. As was the case when the core reluctance was zero. If you recall, the magnetic circuit model for this core was simply two MMF sources. One, N1, i1, and another pointing in the direction such that is also throwing flux in the same direction, and an N of source value of N2 i2. And these were connected through reluctance of the core Rc. Now since Rc is no longer zero, we can compute what the core flux is going to be. And we can write good joules voltage law equivalent equation here, which would essentially be this expression. From this, we can again determine the total flux in the core via C, which is given by this expression here. Let's see what this means in terms of the relationships at the terminals, to find the terminal relationships, we must first find the flux linking back into the winding. Let's start by looking at the first winding and find the flux that links back into winding 1, which we also call the primary winding. In this first winding, the flux that links back is the gold flux phi C, as given by this expression, times the number of terms N one. So that expression is right here. We can substitute phi C into that expression and what we will get for lambda one, the flux linking back into winding one is n1 squared, i1 plus n1 n2 i2 divided by Rc. I can pull N1 squared over Rc come out of this expression. That leaves behind i1 plus N2 over N1 times i2. We can now find the voltage that develops at the terminals of the first winding by substituting this flux linkage back into the expression for the voltage, since the voltage is simply the time derivative of the flux linking back into the first winding. We substitute that, we get essentially a similar expression, N1 squared over Rc, but now with the time derivative of the currents i1 plus N2 over N1 times i2. So what does this expression for the voltage across the first winding mean in terms of the equilibrium circuit model. For that let's define this term, i1 plus N2 over N1 i2, as another current which we will call i mu1. As we will soon see, this will be the magnetizing current off this transformer, reflected onto the primary site. Let's also define this term N1 squared over Rc as Lm1, which we will soon see, is the magnetizing inductance of this transformer reflected to the primary site. If you make these definitions, then real one simply becomes l new one times the time derivative of I new one. Well what does that say? Well this is clearly just an inductor of some value. L mu one, with a current i mu one going through it. And v1 is just the voltage across this inductor. To complete our equivalent circuit model, let's see what i mu 1 really is. Well i mu 1 is the sum of two currents. A current i1 which is simply the current flowing into winding one. And another current, N2 over N1 times I2. Which adds to I1 to give the current a new one. Now if i, look at this current. It seems like it's a transformed version of the current that's flowing into the second winding. What if I go ahead and add a transformer, an ideal transformer on this side. The term's ratio N1 is to N2, and with the dots shown. And then ask the question, what current here would actually give me the current n2 over n1 times i2. Well the answer's quite simple. Since the currents transform by the inverse of the turns ratio, this current will simply be i2. This, essentially, is the equivalent circuit model for our transformer With the magnetizing inductance. It matches the equation we have since it simply says that the voltage we want across terminal 1 is equal to this inductance, which we would call the magnetizing inductance, L mu 1. And the subscript one is used because this inductance is shown on the primary side of the transformer. Note that we can move the magnetizing inductance off the transformer On the secondary side of the transformer. We can do that in one of two ways, we can simply use the method we know of taking an impedance and transforming it through an ideal transformer in which case Lμ1, when it moves to second reside would simply get multiplied by N2 over N1 squared, and that would be what we call L mu 2. We could have also directly derived the value for L mu 2 by doing this entire analysis for winding 2 and getting an expression for V2 instead of V1. In either case, we would have gotten the same answer for L Mu 2. Remember that there is only one magnetizing inductance for a transformer. The magnetizing inductance essentially models the energy that is being stored in the core of the transformer. And there is only one core and hence only one magnetizing inductance. You can choose to show it on either side of the transformer. You could also choose to split it up and put part of it on one side and part of it on the other side, but the sum of the magnetizing inductance. Once it is reflected to one side, must match the expression that we've developed here. So for the case of the primary side, the magnetizing inductors would simply be N1 squared over the reluctance of the core. Note that this also makes intuitive sense. Since the magnetizing inductance would be, essentially, the inductance of the structure, if we looked into the primary side of this, and asked the question. What is the inductance that we see through this winding, when we completely ignore the second winding, effectively leaving the second winding open or moving it all together. In that case, if you remember from our lecture on inductance, the inductance is simply given by the square of the tons ratio which in this case will be N one squared divided by their of the through which the flux flows, which in this case is simply Rc. This is exactly the expression we have here. On the other hand, if you wanted to reflect this magnetizing inductance to the secondary side, then you could look into the secondary side and ask what is the magnetizing inductance looking from that side? And in that case we would get L mu 2 which would simply be N2 squared over Rc. Note that L mu 1 and L mu 2 do not exist at the same time, they're essentially the same thing Just depends where you're looking from. And if you're looking from the first winding that's the value of the magnetizing inductance. If you're looking from the secondary side that's the value of the magnetizing inductance. Note that unlike an ideal transformer a real transformer stores energy the energy stored in the transformer is modeled by the magnetizing inductance, and if you ever need to calculate the value of this energy stored in the transformer, you can use the standard formula for energy in an inductor to calculate it. Which is simply one-half Li squared, so in the case of the magnetizing inductance reflected on the primary side it would simply be one-half L mu1 i mu 1 squared. Now let's look at the impact of the magnetizing inductance on the performance of the transformer. Note that the magnetizing inductance is across the terminals of the transformer. Therefore, if a DC voltage is applied to any terminal of the transformer this magnetizing inductance is essentially going to have zero impedance and will short out the terminal. Hence the transformer will not work at dc. As we all know, transformers don't work at dc. But our ideal transformer model did not differentiate between dc and ac. In reality, the transformer will have issues even at low frequencies, ACs because of the presence of this magnetizing inductance. Let's see how. In any real transformer we need to make sure that the magnetic flux density in the core does not exceed its saturation level. Because once the magnetic flux density reaches saturation, the permeability of the core drops from its high value of mu c to a very small value of mu naught. And once that happens the magnetizing inductance itself will drop by the same order of magnitude. And if the value of the magnetizing inductance drops, then the impedance of this inductance will drop and so it will start to act more like a sharp circuit at the given frequency. So under all circumstances, we would like to avoid saturating the core of this transformer which means avoiding the magnetic flux density from exceeding B sat. What that translates to in terms of the total flux is that we would like to keep the total flux phi c to be less than B sat times Ac, where Ac is the cross-sectional of this core. We can figure out what this constraint implies in terms of the terminal voltage of the transformer. If we consider the voltage across the first winding of the transformer, v1, which is equal to d lambda 1 dt, the lambda 1 itself is simply equal to n1 times phi c. We can then invert this expression to find phi C in terms of V1, and phi C is simply equal to one over N1 times the integral of v1 over time. By substituting this expression for phi c in our constraint for phi c we can get a constraint on the integral of the voltage that must be satisfied in order to avoid the transformer core from saturating. What this is saying is that the peak value of the integral of V1 over any time must be less than the product of V sat times Ac times N1. Let's see what this means in a more intuitive way. Presuming that v1 is a sinusoidal voltage, then the peak value of the integral of v1 will simply be the area of the shaded region. This is because if we integrate for a period beyond this, then V1 starts to become negative and the area under the curve will start to decrease. So the worst case for the integral of V1 is if we integrate a half line of the voltage V1 starting from 0 and ending when it comes back to 0. The value of this integral will depend on the peak value of v1 as well as the frequency or the time period associated with v1. If the peak value of the voltage and its frequency are fixed, and we are saturating our transformer. Then from this expression, our only recourse is to either find another core material with a higher saturation flux density if such a material is available. Or to increase the size of the core by increasing the core cross-sectional area or by adding modems on the core. Note however, that if the frequency of the voltage waveform was higher. Say like this, then the total area under this curve would be smaller. And therefore, a core with either a smaller cross-sectional area could be used, or a transformer with fewer number of windings could be used. In either case the size of the transformer would reduce as we increase frequency. This is one of the major incentives for trying to increase the switching frequency of power converters. As it allows the size of the transformers to reduce. Also note how the problem of saturation needs to be dealt with differently in transformers versus inductors. Recall that in the case of an inductor, if you increase the current through the inductor then the inductor will saturate more quickly. In the case of the transformer, the current that determines the saturation of the core is the current going through the magnetizing inductors. And not the current which is flowing in the terminals of the transformer. We could arbitrarily increase the current through the terminals as long as the current through the magnetizing inductance does not increase. And in the case of the transformer shown here, if we simply increase the current, I1, that current would simply flow out of the secondary terminals. And would not increase the magnetizing inductance's current. The current of the magnetizing inductance is really determined by the voltage across it. Since we could find that current by inverting v = L di/dt and that current i u1 is simply 1/ L u1 times the integral of v1. Hence, in the case of the transformer, it's really the terminal voltage that determines its saturation. While in the case of the inductor, it was the terminal current of the inductor that determined its saturation. Also, because of that, the way the turns ratio play a roll in saturation is quite different. In the case of the transformer, we can avoid saturation by increasing the number of turns of the winding. Because in the case of the transformer, it's really the total flux through the core that we need to keep below a certain limit. And since the flux is related to 1 over n1 times the integral of the voltage, assuming that the integral of the voltage is constant. Then by increasing the number of turns, we actually introduce the total flux through the core, and hence we can avoid saturation. Also recall that in the case of the inductor we were able to avoid saturation by adding an air gap. That effectively reduced the inductance value, but it increased the current level at which the inductor saturated. In the case of the transformer, adding an air gap has no effect on saturation. As you can see from the terms that appear in the expression that determines saturation. Adding an air gap in the case of a transformer does reduce the value of l u1 and increases the value of i u1 at which the core would saturate. But because the voltage and the integral of the voltage is the same. By decreasing the value of L u1, we simply increase the current that flows through this inductor. Therefore overall, adding or not adding an air gap has no effect on the saturation of the transformer. Now let's consider a second non-ideality in the transformer. Up to now, we have been assuming that all of the flux that is generated by either of these windings, flows in the core. This made sense when we assumed that the permeability of the core was infinite. And in that case, the reluctance provided by the core was zero, and so the flux had no reason to try and find a part of lesser reluctance. When the permeability of the core is finite, and in real core materials the permeability of the core material Is only three to five orders of magnitude higher than the permeability of air or free space. And in that case, some of the flux that's generated by one winding will flow through the air surrounding the core. And return back to the winding through a part that does not take it through the second winding. Similarly, there'll be some flux generated by the second winding that will also travel in the air. And return back to the winding and lock couple the first winding. How does this impact the electrical circuit model for our transformer? To figure that out, let's first develop a magnetic circuit model that takes into account these linkage parts. We do that here, again, in the central portion we have our original magnetic circuit model that we used to develop the model for our ideal transformer. As well as the model for the transformer with just the magnetizing inductance. However, now we've added two more parts to this model. One part modeled by a reluctance Rl1, models the reluctance of the part that is seen by the flux. That flows through the air on the side of winding one. This is essentially. The reluctance seen by the flux generated by the first winding that travels through the air and does not couple to winding two. Similarly, we include a reluctance RL2 to model the reluctance of the part taken by the flux generated by the second winding. Which travels through the air, and comes back through the second winding. We can go back and look at this magnetic circuit model, to assure that this well matches the physics of what's going on in the actual magnetic circuit. Note that Rc, is the core reluctance and is modeling the reluctance of this core. And at both of its ends are the two nnf sources, the nnf source N1 i1, which is modeling the first winding. And then, an MFF source N2i2 which is modeling the second winding. The reluctance Rl2 is simply across the second winding as it modeling the reluctance of the part that seen by the flux going through the ear on the second winding side, and Rl1 is the reluctance of the pot to the air. Again, it is simply warning the reluctance seen by the flux that's generated by winding 1, and going through air on the side of the first winding. Now that we have this magnetic circuit model, we can figure out the flux that flows in each of these parts. And from that we can determine first the flux linking back to each of the windings. And then from there we can figure out the voltages at each of the terminals. So let's proceed by first figuring out the flux that flows through each of the parts. The flux that goes through Rc, the reluctance of the core, is simply, again, given by figuring out the mmf, or essentially the voltage on one side of Rc. And the voltage on the other side of Rc, and so my Ohms law that's again simply given by n1, i1 + n2, i2 over Rc. The flux that flows through the part Rl1 can simple be found, again, by Ohms law since the MMF across Rl1 is simply n1, i1. So the flux flows in the direction shown, and is given by N1 i1 over the reluctance Rl1. Similarly, the flux that flows in the reluctance Rl2 is simply given by N2 i2 over Rl2. Now that we have all the fluxes in each of the paths, we can go and determine the flux linking back to each of the windings. We have to be careful when we do this to make sure that we use the correct polarity. The flux linking back to a particular winding is positive. When it is flowing from its positive terminal outwards and coming back into its negative terminal. So if you want to find the flux linking the first winding. That in terms of polarity will them simply be the sum of the fluxes Phi C plus the flux phi l 1, so our lambda 1. Which is the flux linking back into the first winding, is simply going to be the total flux that's going into that winding times the number of terms of that winding. Since again the flux is linking each of the terms of that winding. We can substitute in the values for vl1 and vc into this expression to determine the final expression for flux linking the first and that'll be given by n1 squared i1. Over Rl1 + N1squared i1 over Rc + n1n2i2 over Rc. We can similarly find the flux linking back into winding 2. Here again, we have to be careful about the polarity of the fluxes. But again, the way we've defined them, both of these fluxes are entering the negative side of that winding, and coming out of the positive side. So we will use their polarity as shown and both will go in with positive signs into this expression. And again, to find the final expression, we simply substitute the value of five C from here and the expression for five L two from here. And we get this expression for lambda 2. Now that we have expressions for both lambda 1 and lambda 2, we can simply determine the expressions for the voltages at the terminals v1 and v2. By simply taking the time derivatives for the expressions for lambda 1 and lambda 2. That we've just derived. And these will look almost identical to the expressions above, except that the currents now come in with their time derivatives. Other than that, these expressions look very similar to the ones shown above. These expressions are clearly messy, and probably at this point you have no clue what they mean. Let's look at them more carefully and try and figure out what they mean in terms of an equivalent circuit model. Here again are the two expressions that we just derived for the terminal voltages for our two windings. In order to make sense of these expressions let's make some definitions. We're going to define a leakage inductance, Ll1, equal to N1 squared over the reluctance of the leakage part. L1 which we call Rl1, similarly we'll define a second leakage inductance, Ll2 as equal to N2 squared over Rl2. In addition to this definitions we'd reuse our definition for the magnetizing inductance Reflected on the primary side, which was N1 squared over Rc. Note that all of these expressions for inductances makes sense as inductances. Since if you remember, the expression for an inductance is equal to the square of the number of turns Divided by the reluctance of the part. So in the case of this inductance, Ll1, what it is really saying is that it is an inductance, Ll1, which is modeling the energy being stored in the magnetic flux v l one. Which is generated by the first winding, and this energy is being stored in the air. Similarly, L L two models the energy stored in the flux phi L two. Which is being generated by the second rewinding and this energy is also stored in air. By defining these inductances as shown, and by further defining the term N1N2 over RC in terms of L mu 1. And defining N2 squared over Rc, also in terms of L mu 1, and it essentially would be equal to L mu 2, then we can take these definitions and substitute them in our expression for V1 and V2. And that yields this slightly simpler expressions for v1 and v2, these expressions are still complicated But they can now be used to derive an equivalent circuit model for a transformer that has both magnetizing inductance, as well as leakage. Here again are the same expressions we just derived for v1 and v2. Now let's try and convert them into an equivalent circuit model. Let's take a look at them and see how we can do that. Note that both of these expressions have these terms i1 plus n2 over n1 times i2. This is nothing more than the magnetizing current, reflective on the primary side. So this is, just simply i mu 1. And similarly, this is just i mu 1. With the substitution, we get a much simpler expression for v1 and v2. And v1 is now the leakage inductance ll1 times di 1 dt plus the magnetizing inductance conductor to the prime side times di mu 1 dt. And then v2 Has another factor, N2 or N1, multiplying Lu1 times Diu 3 dt, and then a leakage term, L L two times Di2 dt. To create the model using these equations, we can begin to see the same wave we did for the case with just the magnetizing inductors. When in this case, I'm actually going to just show the final model and explain how this model is derived. Note that v1 is equal to the sum of two terms. One of the terms is lLu1 times dimu1 over dt. The second term, let me call that some voltage v of x. That voltage is essentially the voltage across an inductor and in this case, there's the voltage across the inductor L mu1 which is our magnetizing inductance. So it is essentially this voltage right here. What this first equation is saying is that V1 is the sum of this voltage, VX. Plus another voltage which also looks like a voltage across an inductor. In fact it is simply an inductance, the leakage inductance, LL1. times the time derivative of the current flowing into the first winding. So it simply can be modeled by another inductor LL1 whose current is simply I1. And so, if I put the inductor in series as shown here, that will essentially give me the expression E1 equal to The expression shown right here. So on the primary side, this equation matches well with the model you've done here. Note again that vx is exactly this since the current through this conductor l mu 1, is simply the current i mu 1, which is also the current in this expression. To verify that the second expression, v2, is also satisfied by this model, let's take a look at the model and compare it to this expression. What this expression is again saying is that we have a voltage that is equal to Ll2 times DI to DT which again represents an inductor with a current I2 flowing through it. And we can model that again, with this inductor LL2 which is modelling the three cage inductors and the current again is I2. This second expression V2 also has another term which is l mu 1 times di mu 1 dt times a tones ratio n2 over n1. And this also makes sense because that is simply the voltage here, which we just call vy, and clearly, VY is simply going to be N two over N one times VX and we just noted that VX was simply equal to L me one, DI me one over DT, which is exactly what this also saying. Therefore, this model that we have for a transformer, exactly matches these equations. And so this is certainly a model that captures the math that we have used so far. Hence, we now have an equivalent circuit model for a transformer That include the known ideologies that bring about the magnetizing inductance as well as the two leakage inductances. To recall again that there's only one magnetizing inductance in a transformer, and the magnetizing inductors arises because the permeability of the core is not infinite. On the other hand, there are two leakage inductances and each leakage inductance is associated with the flux leaking out from the winding associated with the side on which the leakage inductance is put. And that flux is leaking out into the air and not connecting back. Onto the other winding. Also note that the terminal variables of this more complicated transformer model are no longer simply related by the turns ratio or the inverse turns ratio. Part of the voltage on each winding drops across the leakage inductance. Only a fraction of the voltage that is applied at the terminals actually appears across the ideal part of the transformer. The ideal part of the transformer still works using the ideal turns ratio relationships. But you have to account for these leakage and magnetizing inductances when computing the actual transformer terminal relationships. The current relationship, likewise, is not related purely by the inverse of the turns ratio. As part of the current that's coming into the primary winding is flowing now to the magnetising inductance, and only the remaining part actually gets to flow to the secondary side. This transformer model can now be used either for hand calculations, or can be used in a circuit simulator like Spice. To simulate or power electronic circuits. To model this in Spice we could again use the dependent current source and dependent voltage source model for the ideal part of this transformer, and then simply add these additional inductances to that to get the complete transformer model. Spice also provides the coupled conductor model that you can use with either just the ideal transformer model or the model that includes the magnetizing inductors as well as the rigid inductors. That's all for this lesson. In our next lesson, we will look at other ways to model transformers. [BLANK AUDIO]