Welcome back to the course on Magnetics for Power Electronic Converters. In the last lesson, we learned how to determine the inductance of a magnetic structure from first principles. In this lesson we'll learn easy ways to find inductance. But first, I must introduce you to the concept of reluctance and magnetic circuits. Recall from the last lesson our expression for the total flux in an inductor with an air gap. We had the following expression where the total flux was proportional to Ni, where i was the current going through the winding, N was the number of turns in the winding, divided by lc over mu c Ac plus lg over mu0 Ag. Does this expression remind you of something? What does it resemble? Think about it for a second. Does it not look like Ohm's Law? Indeed it does. Here's Ohm's Law. i is some voltage divided by the resistance in the path. Well, if we look at our expression for total flux, it looks like some quantity Ni divided by two other quantities. If we define this quantity lc over mu c Ac as an effective magnetic resistance, then this expression looks like some kind of a term NI divided by two resistances, in this case, reluctances. We will define this quantity lc over mu c Ac as the reluctance of the core. And in general, the reluctance of a magnetic circuit element which we will denote as script R will be given by the length of the magnetic element divided by the permeability of the magnetic material and divided by the cross-sectional area through which the flux passes. Note that Ni is like a driving force, quite similar to the voltage v. In fact, Ni is equal to the magneto-motive force, or MMF. Recall that MMF was defined as the line integral of the magnetic field H over some path length l. And by Ampere's law, that over a closed loop would equate to the total current passing through that loop, and that was, in our case, equal to N times i. So this expression for flux is quite synonymous to the expression for Ohm's law if we make the following mapping. If we map Ni, the MMF, to voltage, or EMF as it's also known, we map the total flux in the magnetic circuit to the total current i, and we map the reluctance of the magnetic circuit element to the resistance R. In fact, you'll see that the expression for the reluctance Is quite similar to the expression for the resistance. They're both proportional to the length of the element and inversely proportional to the cross-sectional area. In the case of the reluctance, it's inversely proportional to the permeability, which acts sort of like the conductivity in the case of the electrical resistance. If we do this mapping, then we can simply model our magnetic structure in a form similar to the electrical circuit by simply modeling the force that's driving, which is essentially coming from the winding, as Ni R, which is the MMF, which we will denote by script F. And then the path reluctance is essentially modeled by the radius reluctances that are in the path of that magnetic flux. Once we've done this mapping, we can use all the tricks we know from circuit analysis to solve this magnetic circuit. For example, we can go and solve for the total flux, which, if you go back and check, will exactly be what we got in the first place, when we used our basic principles and Maxwell's equations. However, this method of converting a magnetic structure into an equivalent magnetic circuit is going to be extremely helpful to simplify all the work we have to do in order to determine inductances. Let's apply the technique we have just learned to a simple magnetic structure. In fact, it's very similar to the toroidal inductor we considered, except that instead of a circular structure, we now have this squarish or rectangular structure. But we still have a core and we have an air gap. But instead of using the first principles in Maxwell's equations to determine the inductance, let's apply magnetic circuits to figure that out and see how simple that is. So first let's map this into a magnetic circuit. We have a winding in this magnetic structure with a current i. And the total number of turns is N. So the MMF that is being applied is Ni. We have to be careful about the polarity of this MMF source. Again, we use the right-hand rule. We use our right hand and curl our fingers in the direction of the current, and our thumb points in the direction in which the flux will flow, which then tells us that the positive side of the MMF force must be on the top so that it drives a flux in the same direction as it would flow in the original structure. We can further model this structure using two reluctances, a reluctance Rc which models the reluctance of the core and a reluctance Rg which models the reluctance of the air gap. And these reluctances are in series. Now that we have this reluctance model, it is very easy for us to essentially use circuit techniques to solve for the total flux that it flows. Again, just using the analogy from Ohm's Law, the flux here is simply going to be the force that's causing the flux Ni divided by the reluctance. The reluctance is simply the resistance that this path imposes through the flow of flux. So in this case, it's just going to be equal to Ni over Rc + Rg. And what are the values of Rc and Rg? Well, those we can simply determine from the geometry of the structure. For the core, Rc, remember, the reluctance is proportional to the length. We'll assume that the length that the flux has to travel through the core is lC. So Rc is simply lc over mu c times Ac, where Ac is the cross-sectional area of the core. Similarly Rg is simply equal to the length of the air gap, lg, over mu0 times Ag. And if we approximate Ag to be approximately equal to Ac, we could further simplify this as approximately being equal to lg over mu0 Ac. Now that we have Rc and Rg, we could go and substitute these here, and we'll find that we get exactly the same expressions that we got for the flux as before. However, if you're interested in finding the inductance L, we can proceed as follows. From the total flux, we can figure out what is the flux linkage lambda. Recall that lambda is the flux that's linking back into the coil. Since the coil has N turns, this flux linkage will be N times the total flux, phi, which is therefore simply equal to N squared i over R c + R g. And now that we have the total flux linkage, we can find inductance from its definition, which is simply the flux linkage divided by the current in the winding, which gives us n squared over Rc + Rg. And there you have it. So the expression for inductance is simply N squared over Rc + Rg, where Rc and Rg are given by the expressions that we could easily determine from the geometry of our structure. In general, we can always write inductance L as N squared over the effective reluctance of the magnetic path through which the flux has to pass. And this expression is so useful, while so simple. Easy to remember, yet extremely powerful. We can use this method for finding the inductance L for much more complicated geometries. Let's try that in an example. Let's use the method we've just learned to find the inductance of a more complicated magnetic component. We have a magnetic structure which has three legs, it has a single winding, and again, let's say it has N turns. So to find the inductance, all we have to do is figure out the effective reluctance that this coil sees. So L, remember, is simply given by N squared over R effective. And R effective is what we want to find. So let's first map this magnetic structure to its equivalent magnetic circuit model. We have an MMF source, which is N times i, i is the current that's flowing through this wire. Again, the polarity of this MMF source has to be determined using the right-hand rule. If you apply the right-hand rule to this coil, the flux will flow in the top direction, and that must also be the direction of flux in the circuit model. Now, we must also model the reluctances of this magnetic structure. We've modeled this here using R2, which models the reluctance of the leg to the right, so we have some reluctance of this piece here. We'll call that R2, and it's modeled here. Similarly, by symmetry, the reluctance of the leg on the left, we'll call that R2 as well. And that gets modeled as a parallel path to the right leg, because the flux can go through either of these paths, so they're in parallel. Now, the MMF source itself has in series with it a couple more reluctances. There's the reluctance associated with the center leg of the core, and this we will call R1. And so we model that here. And then in series with R1 is the air gap. And so we will model the reluctance of the air gap with this Rg. And so that forms the complete equivalent magnetic circuit model. Once we have this magnetic circuit model, we can easily determine the effective resistance using the standard technique we use for circuit analysis. We just have to think of the individual elements of this magnetic circuit model as if they were resistors. And this is treated just like a voltage source. And so what you have is you have R2 in parallel with R2, but then they are in series with Rg and R1. So the effective reluctance of this structure as seen by the winding is simply R effective equal to R1 plus Rg plus the parallel combination of the two R2's. Now that we have that, we can simply write the expression for L by inspection. L is simply going to be N squared over R effective, which in this case is just R1 plus Rg plus the parallel combination of R2 and R2, which is simply going to be R2 divided by 2. And there you have it, is the expression for the inductance for this structure. So even though the structure seemed complicated, we can very simply calculate its inductance. What are the values of these individual reluctances R1, Rg, and R2? Well, that we can again determine from the geometry. And in a typical case of a geometry like this, you will make some approximations, but here's what I've computed. R1 is taken to be essentially the length associated with the center post. That's l1 minus lg, where l1 is this roughly the total height of this core, where I've tried to keep it through the centers. And similarly, the reluctance of the air gap is lg over mu 0 A1. A1 is the cross-sectional area of the center post. And similarly, we can find the reluctance of the outer legs, which are identical. And that will be simply this length l1 plus 2 times length l2, because you get contribution from here as well as from there. And divided by, again, mu c over A2. As you can see, this is much simpler way of finding inductance, and works for even fairly complicated structures. Hence we don't have to go back to Maxwell's equations each time, but in fact we'll find that this magnetic circuit modeling technique is a much faster and a much more simpler way to calculate not only just inductance, but also fluxes to various parts. And this will help us not only in the design of inductors, but as we will soon see, in the design of transformers as well.