Welcome back to the course on Magnetics for Power Electronic Converters. Our objective in this course is to learn how to design good inductors and transformers for our power converters. By good inductors and transformers, we mean inductors and transformers. That have low losses and are relatively small. Before we can learn how to design inductors and transformers, let's figure out how to analyze them. Let's first see how the inductance of a magnetic structure is calculated. In this lesson, we will learn how to determine the inductance Of a magnetic structure from it's basic physical properties by applying Maxwell's equations and then we'll learn how to do it much faster by simplifying the method. Let's consider the simple inductor form by coiling a wire In order to determine the terminal relationship or the VI relationship for this coil, we will need to apply Faraday's law. Let's recall what Faraday's law states. Well, Faraday's law states, that the integral law states that the integral of the electric field around a close path or a loop. Let's say we have a loop L like this. And we'll defined the direction of this loop to be in the clockwise direction. Note that keeping directions straight is important, because these are all integrals of vector quantities. Then if we integrate the electric field Along this direction dl that integral according to Faraday's law is given by the negative of the rate of change of the surface integral of the magnetic flux density through the surface bounded by this loop l. So we have this loop L here. And then the surface that it is bounding is a surface that I'm shading. And the flux density that is crossing the surface. Let's say it's going into the screen, so it's going in that direction, which we will many times for clarity simply show as the tail of an arrow by marking a circle with an X. That's when the field is going into the screen, in this case, so that's our magnetic flux density direction. Now, again, to keep all of the signs correct, we much define the positive direction of the surface As we're integrating B with respect to that surface. The positive direction of that surface has to be related to the direction that we've chosen for the loop and that's going to be defined by the right-hand rule which states that if you take right hand and curl our fingers in the direction of the loop. And are thumb points in the direction of the positive surface. So, in that case that's also going into the screen. So, that would be the direction of a ds as well. Now, we can take this Faraday's Law and apply this to our coil structure here First let's identify the loop in this case. Here the loop will be along the length of the wire. And it will return back through the external circuit. So that entire thing is a loop. And it's positive direction again we have selected. If we select this to be the positive direction of the loop then again curling our fingers in the direction of the loop, with our right hand our thumb points in the direction which is the positive direction associated with the surface s. The magnetic field that will be generated by this coil, with a currant i flowing also in the direction that we chose for the loop, will also point in the direction as shown, which is pointing downwards, using the right hand rule. And the actual value of this magnetic field would be determined by Ampere's law. Now, in order to compute Faraday's law, let's make an assumption that the conductor that you're using to form this coil, Is a perfect conductor. So there is no electric field inside the conductor itself, and the electric field only exists between the terminals of this conductor in the outside circuit. So it, the electric field really exists Between the bottom terminal of this coil and its top terminal. So when we do the line integral of the electric field using Law and integrate along this coil the integral along the path of the coil itself will be zero. And the only integral value that we will get Will be the contribution from outside the coil. Notice that this is simply equal to the negative of the voltage that is being applied to this coil just by the definition of voltage. Remember that voltage is defined in our case here From the top terminal to the bottom terminal while we're integrating from the bottom terminal to the top terminal. Hence there is a negative sign. So, the left hand side of Faraday's Law simply reduces To minus v, where v is the voltage across the terminal of this coil.For the right hand side, let's make a new definition. We'll define flux linkage as the integral of the minetic flux density Over the surface that's suspended by this loop. But what does flux linkage physically mean? Well, if we have a simple loop, just a single loop, the flux linkage is simply equal to the total flux passing through this loop. However, in a mu complicated structure, where we have multiple loops, such as the case of this coil, where the magnetic flux crosses the surface boundaried by that loop multiple times, we have to be more careful in determining flux linkage. In this example, Your phi assume that there are N turns, then the flux linkage lambda, will equal N times the flux, or the total flux phi that we had defined earlier. This is because this coil suspends it's surface, which is maybe a little hard to visualise in 2D but if you think hard about it, it's really a surface that is bounded by this turns. And is similar to if you took a coil like this and you dip it in soap water and then pull it out, you would see that the soap fill actually forms that surface s. And if you visualize the magnetic field according to that you can easily see that it is cutting through this surface n times if there are n turns. Hence, in this case the flux linkage is really n times three. For now, let's simply use this definition for flex linkage and substitute this back into Faraday's law. If you do that, we see that Faraday's Law simplifies to simply minus v where again v is the voltage across the terminals of this coil equal to minus the time derivative of flux linkage lambda. Since lambda has no spacial dependence and only time dependence, we can simplify this partial derivative to an ordinary derivative of Lambda and the terminal relationship for this coil can then simply be written as the voltage v equal to t lambda dt, where lambda is the flux linkage. We can argue that if the material through which the magnetic field is passing is linear then lambda would be proportional to the current i, why? Well, remember from Ampere's law, that the magnetic field intensity H is proportional to the current i and B is proportional to H. And the way we define lambda, lambda is proportional to B. Therefore for at least linear materials, we can say that lambda would be proportional to i. Now, if we define the constant of proportionality between lambda and i to be the quantity L then we can write lambda is equal to Li. And if we substitute this back into our relationship v is equal to the lambda dt, we simply get v is equal to Ldi dt which is what we expect as the terminal relationship for an inductor. Hence, we can say that this constant of proportionality l is really the inductance of our coil and that's how we're going to define inductance. Inductance will be defined as the ratio of the flux linkage to the current passing through that coil. We can now use this definition to determine the inductance of a number of different structures. As a first example, let us determine the inductance of a simple toroidal inductor. Here we have a core, a submagnetic core around which I'm going to wind some turns. Let's put N turns around this toroidal core, right? Let's say this core has permeability mu c and across sectional area, AC and a mean length which is the length along the core Lc. Now to determine the inductance L, we're going to proceed as follows. First we're going to use Ampere's law to determine the magnetic field H going to the structure. Then we'll use the relationship between B and H to find the magnetic flux density B. And find the total flux and then from the total flux, we'll find the total flux linkage and then finally using the definition for inductance, we'll find the inductance of the structure. Let's assume we have a current i flowing through this wire in the direction shown. Then, to find the direction of the magnetic field that flows in this core, we simply use the right-hand rule. That is, we wrap the fingers of our right hand along the direction of the current, and our thumb then points in the direction of the magnetic field. So the magnetic field in this case will flow along the core in the direction shown. We'll assume that the magnetic field flows uniformly through this core and then we can go ahead and apply ampere's law to this case. So our integral in the case of Ampere's law, is simply integral of this H along this length, Lc, which is essentially our loop. And so we can write the left hand side of Ampere's Law as simply H, which is the magnitude of this field, times the length Lc. From the right hand side of Ampere's Law, which states that this left hand side must equal the integral of the current density through the surface that's bounded by this loop. And the surface, in this case, would simply be this shaded area which is being bounded by this loop of length Lc. The current density if integrated over this would simply be the total current that passes through the surface. Now we have current I passing through it each time the loop cuts or the wire cuts through this surface and therefore the total current that passes through the surface is simply going to equal N times i. Ad so from Ampere's Law we simply get that Hl is equal to N times i or that H is simply equal to Ni over lc. Now, since we know the permeability of this material to be mu c, b is simply given as mu c times h, which would become mu c ni over lc. And then the total flux is simply the integral of the flux density or the surface through which the flux flows, which in this case we're assuming all of the flux is contained within the core and is flowing through an area Ac. Therefore the total flux is simply B times A c which will then be mu c N A c i over l c. Once we have the total flux, we can go back and calculate the flux linkage. Recall that flux linkage has to take into account how many times the total flux cuts back through the surface or if you think of the surface as the full surface that is suspended by this coil, then Really lambda, the flux linkage is N times the flux that we just computed. And therefore lambda is then equal to Mu c, and I'm going to move the Ac first. Use the Ac N squared i over Lc. Now we can simply use our definition for inductance, L being the flux linkage lambda divided by the current, i. And this then, simply gives us that L = Mu c Ac N squared Over lc. That's a fairly simple relationship. In fact, it is quite easy to remember because it actually looks very similar to the relationship of a simple capacitor. So if you recall, the capacitance Of a parallel plate capacitor with some plate area, A, and some distance between the plates, L, that capacitance is simply given by epsilon, which is the permiativity of the material between the plates. Times are A divided by that distance L. And if compare, the formula for inductance is very similar, except epsilon is replace by mu. The area A is the cross-sectional area through were flux passes, And lc is the length that the flux has to travel. The big difference between the formula for the inductance and the formula for the capacitance, is that in the case of the inductance, you have a factor N squared, which is the squared of the number of turns. You have no such factor in the case of capacitance. Let's examine the expression for inductance that we've just determined. The inductance depends on the permeability of the core, mu c, the cross-sectional area of the core, Ac, the number of terms Let's redraw those turns here. And say we have N turns then it is inductance is proportional to N squared and it's inversely propositional to the mean length of the bas that the flux takes. As you would expect, the inductance depends on the core dimensions, the number of turns, and the permeability of the core. Now that we know how the inductance depends on these parameters, let's see if the inductive design that we've done Would be a good choice. The core dimensions would remain constant. The number of turns would remain constant, but what about the permeability of the core? If you recall, the permeability of the core is actually not the constant. The permeability depends on the strength of the magnetic field And as the magnetic field increases the permeability tends to drop. And as the material saturates the permeability simply becomes the permeability of free space. The permeability also depends on other factors, including temperature As well as how exactly was this core made. In fact, when you buy two cores from the same manufacturer, presumably made out of the same material, you'll find that their permeability will not exactly be the same. In fact, it could be as different as ten or twenty percent. So, would this design be a good choice for making an inductor? Note that our inductance depends directly on the permeability and if the permeability is changing then our inductance is changing. So, this is not a good design if you want to have a specific value of inductance for a converter Let us see if there is another way to design an inductor that will give us an inductance value that does not change so much. Here we have another design of an inductor in fact very similar except that I have gone ahead and I have Cut a air gap into our [INAUDIBLE] core. Essentially, I just took a saw and I chopped off a little piece to create a little gap here of length LG and this creates and air gap in out [INAUDIBLE] core. We're again going to go ahead and put our findings on the. Core and again I'll add interns. And let's go ahead and figure out what is the inductance of this structure? We'll start the same way, with Ampere's Law. Again, our h field If a current is coming into the wire in this direction, our H field will be directed according to the [INAUDIBLE] rule in direction shown. We integrate it, over the length L that it travels and that would be the left hand side of Ampere's Law. Except we have to be a little careful there. Because nowhere have we said that the H field needs to be the same in the core as well as in the air. Sop let's clarify this by marking the H field in the core as Hc. And we'll markup The H field that goes through here as Hg. Now, we can go ahead and apply the left hand side of Ampere's Law. And we do that, what we're going to get again is The magnitude of hc times the length that it travels in the core, we'll call that length, again, lc. Note that lc is now just the length that's part of the core and so it's less than the original lc but only by the amount lg. Another part of this integral will be Hg times Lg, now this becomes the left hand side of an integral. The right hand side again, Is the total current that passes through the surface bounded by the loop l. The loop l is now of length of lc + lg, and it's again the same loop that we had before and so the total current that goes through the surface bounded by this loop Is again going to be N times i. So our right hand side of this equation is again this N times i. Next I'd like to be able to calculate the actual flux density in the core but note we have an issue here because in this equation we have Two unknowns. We have Hc as unknown,as well as Hg as unknown. So, I need to bring in another equation to be able to figure out how Hc and Hg are related. And I can do that through the magnetic flux continuity condition, which says that, The total flux that is passing through our core as well as the air, that is continuous. But basically says is that while H CN HG are not equal I do have the total flux which I can write Essentially as the field that's in the core, and I'm going to call that Bc, times the cross sectional area of the core, I'll call that Ac. That must equal the flux density in the airgap, we'll call that Bg. Times the glass section area of the air gap, Ag which is not shown here. Ag is not going to be exactly equal to Ac although it will be quite close to the value of Ac. And the reason that Ag is only approximately equal to. Easy is the fact that when the flux gets to the air gap, it does not go straight through but can start to fringe. And so you will have flux lines that will sort of have more area to go through because unlike the core where the permeability was high and the flux was being guided by the core, once it gets into the air It'll find the shortest impedance path to get to the other side and it'll start to French. Now that we have this relationship between Bc and Bg, we can also utilize the relationship between Bc and Hc as well as Bg and Hg. To get and relationship between Hc and Hg. So let's just do that. From here we have that the magnetic field in the core, Hc is equal to Bc over mu c. And we also have the Hg The medic field in the air gap is equal to Bg/mu naught. Mu naught is of course the permeability of air or free space. From our magnetic flux continuity expression we have that Bg can be expressed in terms of Bc Ac and Ag. And I can use that to write Hg then also in terms of Bc which will become Bc times Ac over mu knot Ag. Now I have Hg in terms of bc as well as hc in terms of bc and I can substitute these back into Ampere's Law and get an expression which only has one on loan and that s bc. Let's do that. Substituting these in Ampere's Law we get that BcLc over mu c plus BcAcLg over mu not Ag Is simply equal to Ni, that means make this equation a little symmetric. I'm going to divide both sides of this equation by Ac so lemme do that. I'm just going to divide here by Ac, I'm going to divide here by Ac, I'm going to divide here by Ac as well. This Ac's go away and now I can write an expression for Bc which is the magnetic flux density in the core as well as So my Bc is therefore equal to Ni over Ac divided by Lc over mu c Ac + Lg over New not ad. So there we have it, an expression for the magnet flux density in the core. Once we have this expression we can easily compute the total flux going to the core which is simply going to be Bc times Ac, when we do that, the Ac actually gets cancelled, so the expression simplifies simply to Ni over Lc over myu C Ac plus Lg over myu know Ag. And we can then find the flux linkage by simply multiplying the total flux by M. And that gives us N squared I over the same denominator, Lc over Nc, Ac plus. lg over mu knot Ag. Once we have the flux linkage, we can calculate the inductance simply by dividing the flux linkage with the current i and that gives us that L is equal to N squared Over lc over mu c ac plus lg over mu 0 a g. Well, we've computed an expression for l That now seems even more complicated than when we simply had a core without an air gap. So what's the whole point? Let's ask the following question. What is the relative value between mu not and mu c? Recall that new c is much much greater than u not, typically a thousand to hundred thousand times bigger, so if I I design my lg and lc to be within say an orbit of magnitude of each other, then I will still have that mu c Over lc is much much greater than mu 0 over lg. In that case, the term lc over mu c a c will be much much smaller than the term lg over mu 0 over a g. And hence, I can ignore the first term all together. Remember that ac and ad are roughly equal. In that case, my inductance simply becomes new not, ag And squared over lg. Note that none of material properties that is mu c is now present in the expression for l. So my inductance only mode depends on permeability of real space The cross-sectional area which is partly governed by the cross-sectional area of the coil itself, the number of terms squared and the length of the air gap. Hence, my inductance will now be constant. You may ask what is the point of having the core in the first place, If I'm not going to utilize any of the properties of the core. Well, think about it? Why do you think we use the core? Well, we use the core mostly because if we didn't have the core, we would not be able to specify a specific value of lg. Remember LG is the air gap in the coil and it's value is specified by the presence of the core. Also ag is specified by the cross-sectional area of the core, so having the core is important, even though we never end up using the permeability of the core as part of our design of the inductance. Most of the inductors that we will design in our electronics will utilize the airgap to set the value of inductance. And hardly ever use the core permeability to set the value of the inductance. Let's take a closer look at what the presence of an airgap. Has done to the design of our inductor. First of all, it has made the value of the inductance of our inductor less dependent on the core permeability. In fact, in the case where the permeability of the core is much greater than the permeability of air, You'll have no dependence on the permeability of the core. A negative impact, however, is that our inductance, which is now given by mu 0, ag, n squared, over lg is considerably smaller than if we had no air gap. Because now with the air gap, the permeability that is used in the expression is mu knot instead of mu c. mu c, remember was much greater than mu knot. However, this was a worthwhile tradeoff because we can get our inductance back by adding more terms. Are increasing the cross sectional area or having a smaller gap length. Another reason for adding the Airgap is that it allows us to increase the saturation current for our inductor. The saturation current is the current at which the core saturates Note that if we look at this chart, which plots the total flux going through our inductor where it says the current then until the material saturates, the flux increases linearly with the current And this makes sense since the current is proportional to the H field and from the V-H curve we know that V and H stay linear until it saturates. Now if you don't have an air gap Then our expression for the flux through the core is given by the expression shown here which has a much steeper slope because the denominator is a much smaller value given that mu c is a large number. On the other hand, once we have an air gap present, then our total flux is given by this expression, and here the dominant term in the denominator is the term which has mu nought in it and therefore This is really approximately equal to Ni over lg over mu knot Ag. And since mu knot, it's much smaller than mu c, the denominator is a much larger term and resulting in this having a much lower slope Which essentially means that the total flux reaches the saturation flux level which will be given by B sat times the cross sectional area at a larger current level. And so we've increased the saturation current of our inductor By adding the airgap. And this is an important consideration if we don't want our inductors to saturate at the current levels they're designed for. Finally, let's take a look And ask the question, where is the energy stored in an inductor with an air gap? Recall that all inductors store energy and the energy stored in an inductor from a circuit perspective would be given by one half L. I squared, when I is the current flowing through the inductor. In field terms, the energy stored in an inductor can be determined from the bought product of the magnetic flux density with the magnetic field and integrating that over the entire volume in which the fields are present. The factor of a half her indicates that we're considering these b and h fields to sinus hurdle, the half just essentially takes that factor into account. Now, in the case of our inductor with an air gap, there are two zones where the magnetic field exists that is the core And then there is the air gap and we're going to assume that the field outside can be neglected. In this case we can split this integral into two integrals, one over the volume of the core, which is this first one, and second an integral over the volume of the air gap. Now, if we substitute the fact that hc is related to bc through bc over mew c and that hg is equal to bg over mew not and we substitute that in this expression, And we assumed that the fields are uniform in the respective volumes. Then these integrals simply become what shown here, where this last 2 terms simply are the volume of the space over which we're integrating. Ac, Lc is simply the volume of the core. And AgLg is simply the volume of the air gap. Now, recall that mu c is much greater than mu knot. And if Lc and Lg are roughly of the same order of magnitude, then Lg/mu knot is a much bigger number, Then LC over mu C, therefore this first term which is the energy stored in the core is negligible, relative to the second term. And we can simply say that the energy stored in the inductor is approximately equal to, one half. Bg squared over mu naught Aglg i e most of the energy in an inductor with an air gap is stored In the air gap. This realization supports our earlier determination that the inductance of an inductor with an air gap is really determined by the properties of the air gap. Rather than the core since most of the energy is really being stored in the air gap. That's all for this lesson, we'll see you in the next session soon.