Welcome back to the course on Magnetics for Power Electronic Converters. In the previous lesson, we saw how currents in a wire, redistribute due to time varying magnetic fields produced by the current itself. In this lesson, we will see how current distribution further changes when there are other conductors nearby. Especially, when the other conductors are also getting current and producing time varying magnetic fields. We will begin by considering a case similar to the one we saw in the last lesson. Except this time in addition to the one conductor that was carrying current, we will add additional conductors in close proximity of it. To keep the problem simple in the beginning, we will assume that other conductor are not getting any current. These are essentially dead conductors just placed in the proximity of the current carrying conductors. Again, we will assume that the current through the first conductor is time varying. And since it's in close proximity on the left side to the core there is no magnetic field on its left side. And the only magnetic field that's impinging on it is on its right side. As a result, the current will converge to the right side and be flowing in a skin depth from the surface on the right side of this conductor. Note that while we are showing is singular rectangular conductor each of these conductors could in turn be modeling multiple round wires that are on the same layer as shown by this conductor. We label these conductors as 1, 2, 3, and 4. These numbers essentially represent the layer number for each of these conductors from one through four. Now with the direction of the current in the first conductor as shown, by the head of the arrow. Which means that it is coming out of the screen. Using the right hand rule pointing our thumb in the direction of the current indicates the direction of the magnetic field. Which in this case will be pointing upwards. Now, since the second conducting layer is in close proximity of the first conductor, the H field produced by the first conductor will impinge on the left side of the second conductor. Since the H field is time baring any currents will be produced in the second conductor in order to try and expel it from the center of the second conductor. Hence the N occurrence will flow as shown. Note that I'm only showing currents along one plane, the similar thing happens along the entire length of this conductor. But for simplicity, I'm only showing whats happening in the middle. On the left side of the second conductor; the eddy current flows into the screen, and then loops around and comes back. And flows out the screen on the right side of the conductor; and then loops back again. With the direction shown this eddy current will produce a magnetic field in the center of the conductor which is pointing in the opposite direction of the impinging field. And will tend to cancel out the field that's impinging on the conductor from the center of the conductor. And confine that field to simply the surface of the conductor. However, the presence of the eddy current on the right surface of the second conductor in turn produces fields outside that conductor. Which are also pointing in the upward direction. Note that in order to cancel the original H field, these eddy currents will also have current strengths of i on both sides of the conductor. So the field that is generated by the eddy currents between conductors two and three, is also of strength h. This H field, now impinges on the third conductor, which in turn, produces eddy currents on its surface, to counteract that field and try and cancel it from entering its center. These eddy currents are also of strength, i, and produce field that tends to cancel the impinging field. This process, now repeats as the eddy current on the right side of the third conductor, now further produces an H field between the third and the fourth conductor. And then, that field impinges on the fourth conductor leading to eddy currents flowing in the fourth conductor. Which try and expel this impinging field. Now let's try and figure out, what is the power loss in this winding structure. If we did not have these extra conductors, then the only power loss would have been in the conductor that's carrying the original current i. If we assume that the width of the conductor H is much, much greater than delta. Then we can simply assume that all of the current is only flowing in the skin depth of this conductor. And we can approximate the power loss in the first conductor as simply equal to P1. Which would be equal to square of the RMS current through the first conductor times the AC resistance of that conductor. Which would, in this approximation, would H be much bigger than delta. Simply be H over delta times the dc resistance of the conductor. Recall that the dc resistance of the conductor is simply equal to the resistivity of the conductor, times the length of the conductor, divided by the cross-sectional area of the conductor. How we will know that we have additional conductors placed in proximity of the original current carrying conductor. And we have eddy currents flowing in these conductors as well. We have additional losses. Since the current in the second layer is also flowing through a width equal to the skin depth delta. And the current has a strength equal to i, which is the same as the current that was flowing in the first conductor. The loss on the left surface of the second layer is going to equal the loss that we had in the first conductor. So, the second conductor has a loss equal to P1 on its left surface, and then an identical loss on its right surface. The same situation being valid there so, the loss in the second conductor is simply going to equal. We call that P2, is going to equal P1 from the left surface and then another P1 from the right surface. And so this will equal 2 times P1. The same situation exists in the third layer. Again, it will have a loss P1 on its left surface. And it will have a loss P1 on its right surface. And so, the loss in the third layer P3 is also going to equal 2P1. Similarly, the fourth layer also has loss 2P1. So, P4 also equals 2P1. So the total loss in this winding structure is P1 from the first conductor, Plus, 2P1 from the second conductor, again 2P1 from the third conductor, and 2P1 from the fourth conductor, which all add up to 6P1 for total lost of 7P1. This loss now includes both skin and proximity effect. If he had only considered skin effect. A loss would have been simply equal to P1. So, proximity effect increased that loss relative to skin effect, by a factor of 7 in this example. Recall that the loss due to skin effect is already substantially larger than the loss at DC, since we have this multiplier h / 𝛿, so P1 can be written as h / 𝛿 * the DC loss. So if we were only estimating DC loss and h was greater than 𝛿 by, say, a factor of 5, then the skin effect loss along would have been greater than DC loss by a factor of 5. And now you multiply that by another factor of 7 to get the actual loss in your conductor. I should point out that this is a worst case condition and of course, when h is much, much greater than 𝛿. We will see later what happens when we pick other values of h relative to 𝛿. Also recall that while I've shown these currents as essentially flowing just in the skin depth of the conductor. The real current distribution is going to be more often exponential ledger, and the exact current distribution depends on the relative value of h versus 𝛿. In the case where h is much, much greater than 𝛿, the current decays well before it gets through to the end of the conductor width, and you can approximate it as if it's only flowing in a skin that which is what we have done in this analysis. In the previous example, we assumed there were noncurrent carrying conductors or dead conductors in proximity with our current carrying conductor inside the core. That is an unlikely situation but provided a good platform to understand proximity effect. To see a more realistic example let's consider what happens in the case of a two winding transformer. Shown here is a two winding transformer again, we only show one window of the core with the windings though it. We'll assume that this transformer is a one to one transformer and it has three primary turns or layers and three secondary turns or layers. We'll label these layers as follows. The primary layers we'll label them starting from 1, 2 and 3. And the secondary layers we'll label from the reverse side 1, 2, and 3. We'll assume that the current in the primary is some current, i, and is flowing in the direction as shown by the arrow head. Since this is a one-toone transformer, the current in the secondary will also be i except it'll be flowing into the screen as shown by the tail of the arrow. The direction of current shown for the secondary is consistent with having an existed load connected to the secondary as then the secondary currents will flow in such a way. So as to counteract the flux that the primary is throwing in the core. At low frequencies or when h is much, much less than 𝛿 the current distribution in the wires would essentially look the same as it would at DC and the currents would just flow uniformly through all of these conductors and we would have no additional loss due to skin affect or proximity affect. However as we crank up the frequency the situation is going to change. So let's see what happens again in the very high frequency regime where we assume that each is much, much, greater than 𝛿. So essentially we've increased the frequency high enough that 𝛿 became small relative to h. Before we start looking at the current distribution in this case, let's label our conductors. So we'll label 1, 2, and 3 primary windings. And then 3, 2, and 1 as our secondary windings. Now in this high frequency case, where the skin depth 𝛿 is much less than h, the current in the first layer. Our first conductor is going to flow only on the right surface of this conductor, and so relative to the DC situation, the current distribution shifts and the current instead of being uniform distributed is now flowing only along this surface. The magnitude of this current is still i and it produces an h field that is to the right of this conductor and it's pointing upwards. Now this time varying h field impinges on the second conductor which produces 80 currents to try and expel this field from its center, and these eddy currents would then again look as shown. So that in the middle, they trying field to cancel this h field. However, since these eddy currents are oppositely directed, there is no net current going through this conductor two. But we rarely have a current i, flowing in this conductor coming out of the screen. So the original current i, must still continue to flow. These [INAUDIBLE] currents also have strength i, to create the h field necessary to counteract the j field impending on them. This picture is consistent with the fields as we have an h field of strength h on the left side of the conductor, and therefore we have a single current, i, going into the screen on this side. And on the other conductor, conductor one, we have a current of value, again, I, which is flowing out of the screen. Now let's consider what happens on the right side of conductor two. On the right side of conductor two, we have an effective current of strength 2i coming out of the screen, so we will have a field to its right which is of strength 2h. Now when this time varying field 2h impinges on the third conductor, it will produce 80 currents of strength to I in order to counteract this field. We'll show those as separate dots and crosses. In general, when I draw one dot and one cross, it is of strength i. So each of these have strength I. Now again, these eddy currents, are not causing any net current to flow in the third layer. But we do have a net current i, that is supposed to flow in the third layer. So we must still have another current i flowing in this third conductor. And that will flow again on the right side of this conductor in order to keep the field consistent with the currents. Now that we have three currents on the right surface of the third layer. We will have a field to the right of this conductor that is of strength 3H. So now we have a time varying field of strength 3H that is impinging on layer 3 of the secondary winding. As a result, there must be three currents going into the screen on the left side of layer 3. However since layer 3 is suppose to carry a net current into the screen. Only two of these currents are really associated with eddy currents. And so, on the right side of layer 3 of the secondary. There are only two currents that are coming out off the screen. Again, although I don't show the strength of these currents, all of these currents have strength i. So we have 3i current that's flowing on the left side of layer 3 and 2i current flowing on the right side of layer 3. Now with only 2i current on the right side of layer 3, there will be a field on the right side of layer 3, that will have strength equal to 2H. And this time varying field 2H, when it impinges on layer 2 of the secondary, now produces two currents. That are going into the screen on the left side of layer 2. And again, because we need to have a net of one current going into the screen into layer 2. There's only one of these that is the eddy current, so we only have one current on the right side of layer 2. This current then produces a field of strength, simply H, in the space between layer 2 and layer 1 of the secondary winding. And finally, to prevent this time varying EH field from entering the center of the first layer of the secondary winding. It is a current going into the screen on the left side of layer 1. So this completes a picture of how the currents redistribute in all of these conductors due to proximity effect. Again, we assume that the currents only flow in a skin-depth form the surface of each of these layers. In reality, those current distributions are often exponential in nature. And the exact current distribution depends on the relative value of the weird H relative to the skin depth,delta. But note that in the case where age is much greater than delta, these currents are essentially flowing only on the surface of the conductor. And the relative values are as indicated by these circles with either the dots or the crosses in them. You can convince yourself that our model of just using crosses and dots is a rough approximation of the actual current distribution. Now that we have the current distribution, let's figure out what is the power loss in this winding. Shown here again, is the current distribution that we caught in the primary and the secondary winding's of our one is to one transformer. Again, this is for the case where we've assumed H is much greater than Delta. And we're approximating the exponential drop in current with a model. In which we assume that the current solely flows in a skin depth from the surface. In this case we can model the ac resistance of our conductor as simply equal to the dc resistance multiplied by H over delta. This essentially is the skin effect in the case H much greater than delta. Now let's go ahead and calculate the losses in our winding's. We'll focus on the primary winding, but the losses in the secondary winding will be identical to that in the primary winding. Given the symmetry of the current distributions in the two places. The loss in the first layer, which we'll call P1, can simply be calculated by again taking the square of the rms current and multiplying that by the ac resistance. Since the ac resistance is simply H over delta times the dc resistance of the winding, the actual power loss. Which we also call the ac loss, is simply equal to H over delta times the dc power loss. These essentially is just the skin effect loss in layer 1, as layer 1 has no additional loss due to proximity effect. Layer 2 on the other hand, whose loss we will call P2 has additional loss due to proximity effect. This is because it is being affected by the field that's produced by layer 1. On the left side of layer 2, we have a loss equal to P1, since we have a current i flowing again through the skin depth of that layer. So that gives us this first part of loss in layer 2 which is P1. Now on the right side of layer 2, we have twice the current flowing and since power loss is proportional to the square of the current. We will have four times the power loss compared to the left side of this conductor. So the loss on the right side of layer 2 is going to be 4 times P1. And the total loss in layer 2 will be P1 + 4P1 = 5P1. We can similarly determine the loss in layer 3. The left side of layer 3 has a current 2i flowing near that surface, so the loss in the left side of layer three will also 4P1. On the right side of layer 3 we have a current 3i. And since, power loss is proportional to the square of the current the loss on the right side of layer 3 will be equal to 9 times P₁. So the total loss in layer 3 will be 4+9, equal to 13P₁. Hence the total loss in the primary winding Will equal the sum of all of these losses will be P1 + 5P1 + 13P1, which adds to 19P1. The loss in the secondary, by symmetry, Will also equal 19 P1. So the total loss In the winding, Will equal 38 P1. Again, this loss includes both skin and proximity effect. If we had only assumed that skin effect was in play, then we would simply have had a loss P1 in each of these layers. And so our loss with only skin effect Would have simply been 6P1. Therefore, proximity effect has increased the loss in the winding by more than a factor of 6, relative to when we only considered skin effect. Recall that skin effect loss itself is substantially greater than simply the dc loss. So if the width of the conductor h was greater than delta by a factor of ten, the skin effect loss is already a factor of ten greater than the dc loss. And now, by including proximity effect loss, you would have a total loss which is greater than more than 60 times the dc loss that you would have estimated. Clearly, you don't want to estimate the loss in your winding based only on the dc resistance of the winding. The example we just did had three layers of bindings in the primary and the secondary. We can generalize this analysis for a larger number of layers. Let's do that next. If you look at the current distribution pattern that we saw developing in the example with two windings, each with three layers, we can see that as the number of layers increases, the currents on the two sides of the layer also increase. So in general, in the nth layer, you will have m- 1 currents flowing on the left side of the conductor and m currents flowing on the right side of the conductor. Therefore, we can generalize the power loss in the nth layer using the formula shown here. Again, i is the outermost current through that layer, and Rdc is the dc resistance of that layer. This formula also only holds for the case where h is much greater than delta. To find the total loss in either the primary or the secondary winding, we can sum over all the layers in the transformer. Assuming that we have a total of N layers in both the primary and the secondary, then we can write the power loss in the primary and the secondary as shown here. By using formulas for cd summation, this summation can be simplified to yield the following result. Note again, that capital M here is the total number of layers in either the primary or the secondary, and this formula gives you the power loss in either the secondary or the primary. And since we have a symmetric situation, the loss in both the primary and the secondary is the same. A common way of comparing the actual loss in the winding, also known as ac loss to the dc loss in the winding, is by defining a ratio of the actual loss to the dc loss, and this is sometimes referred to as the ac loss factor or the ac resistance factor. For the case of a two winding transformer with M primary layers and M secondary layers, with each layer carrying the same current, the ac loss factor can be written as follows. Again, remember that this formula is only valid for the case where h is much greater than delta. And written in this form, you can identify the factor that is due to the skin effect. And the remaining factors Are due to the proximity effect. Now, all the examples we've looked at so far are only for the case where h is much greater than delta. So all the results we've derived are only valid for that case. This limit, where h is much greater than delta, turns out to give you the worst case losses. In reality, your losses may not be quite as bad. Next, we'll turn our attention to finding losses in the case where each takes on arbitrary value relative to delta. Before looking at the expressions for losses in windings, for the case where H takes on arbitrary values relative to delta, we need to understand one more concept. This concept is that of an MMF diagram. We recall, MMF stands for Magneto Motor Force and was simply the integral of H over some part. So far, when we were looking at current redistribution in layers or conductors, we evaluated the magnetic field that existed between those layers inside a winding. For example, with the structure shown here, where we have four separate windings, which we could again model as a single rectangular layer, we have an H-field, which is directed upwards going in this direction. Now, recall from Ampere's Law that the integral off this H-field around this loop Is equal to the total current enclosed inside that loop. So, if we have nl turns in each layer, each turn carrying a current i, then the integral of H, over that loop, would simply equal nl times i. Now, this integral of H over this loop, lets call it L, is defined as MMF, which we also write as script F. Now if we again assume that the permeability of the core itself is much greater than the permeability of air or the space that's inside the window, then the H field in the core will essentially be equal to zero and the H field would only exist In the window itself. Therefore, essentially, there will be no MMF drop in the core and all the MMF drop will be across the window of the core. We can see this more clearly in this diagram here. So our MMF drop exists only from this point, which is at the edge of the core, to the other edge of the core. So, essentially, this MMF, which is equal to the integral of H over this loop, becomes equal to the value of H multiplied by the height of this core. Now the value of the magnetic field inside the core window will vary as you moves along the length of the core in the direction shown as x. Therefore the value of MMF will also vary as you move along this dimension x. We call the plot of MMF along this dimension x the MMF diagram. Let's sketch the MMF diagram for this example to get a better understanding. Since MMF is simply the interval of H of this loop, the value of MMF at any given point is equal to the total current that's enclosed in the loop Formed as shown. In evaluating the value of MMF at the position x, it is easiest to always consider a loop that only goes through the window at that position x, and returns back through the core. In this case, snce the H field is approximately zero inside the core, the total integral is simply equal to F(x). Therefore, F(x) can simply be evaluated by looking at the current enclosed by this loop. And, so, we can find F(x) by simply looking at what current is enclosed by this particular loop. Applying this method to our example here to find f of x, as a function of x, let's first consider the value of f of x, between the first layer and the second layer. To do that, we consider this first loop here and note that it encloses four turns each carrying the current I. Therefore the MMF at this position is simply equal to 4 I. Note that the value of MMF does not change in the space between layer one and layer two. Since whenever we did this loop integral we would have gotten the same value as we would always inclose for a current in that loop. We can repeat this process of finding the MMF value at this position, and in this case we would do a loop that encloses both of these layers. And in this case we enclose a current of value 8i and therefore the MMF at this position is equal to 8I. If you moved further along the core window and we did the MMF calculation at this position, and we now consider a loop again, this time the loop is enclosing not only 8I, but also another minus $! therefore the net current that are isn't enclosing is only 4I and therefore the MMF disposition is again back to 4I. If we went all the way to the right side of the chore window and did a loop here now we would enclose no net current and so the MMF value would be zero. Note that the MMF value would also be zero to the left side of the first layer, and so here we could also say that the MMF was zero since we do not enclose any current in a loop here. The value of MMF inside the conductor depends on the current distribution inside the conductor itself. If we assume that the conductors are getting very low frequency current or that if H is much much smaller than delta, so that the current distribution inside the conductor is uniform, then the change in MMF across the conductor will actually be linear So for the case, where h is much smaller than delta, so that the current distribution inside the conductor can resume constant. We can draw the mmf diagram, assuming that the mmf values, are connected through these linear segments. We'll consider another example in which h is not much less than delta we can see how looks like inside the conductor in that case. Here's another example which shows what the MMF diagram looks like under the two extreme cases, one in which h is much less than delta, similar to the example we just looked at, and another where h is much greater than delta. In both of these cases, the values of MMF between two layers since the values of MMF between layers do not depend on the current distribution inside the layer itself. Let's first consider the case where h is much less than delta To find the value of MMF between layer one and layer two, we select a loop that goes as shown. Since this loop encloses a current i, the MMF between layer one and layer two, will simply have a value of i. We can do the same exercise for the second case, where h is much greater than delta, and to find the value of MMF between layer one and layer two, we again do a loop as shown. And since this loop, again, encloses a current i, the value of MMF between layer one and layer two Is again, equal to i. Again note that the value of MMF to the left of layer 1, is 0, and it's also the same in the first case. We can also compare the MMF between two other layers in these structures. So, between the third layer of the primary, and the third layer of the secondary, we can again find the MMF by doing a loop, and noting that it encloses the current 3i, so the mmf in this range is equal to 3i. The same is true in this other case where we have h much greater than delta. So again if you look between the third layer of the primary and the third layer of the secondary, and we do a loop, then we again see are in this loop. We enclose again a net current of 3i. To see that the net current is 3i, note that these currents are just canceling each other so you essentially have this current one, this current second, and this third current. So again, you enclosed a current 3i, and so the mmf in this range, is again 3i. You can repeat this process, for all the other gaps between layers. And you can confirm to yourself, that the mmf values for these other places also match under the two conditions. And, in fact, they will match no matter what the value of h is relative to delta. The difference between these two cases will be in terms of the MMF distribution within the layer. As we saw earlier, that in the case where h is much less than delta, and the current distribution in the conductor is uniform, the value of MMF will vary linearly across the conductor. However the value of MMF through the conductor will not vary linearly in the case where h is much greater than delta. We can see this by determining the MMF inside a conductor. Let's say we want to find the MMF at the center of layer two. In this case, we will make a loop that looks as follows. Note that this loop encloses no net current, therefore the value of MMF at the center of layer 2 is equal to 0. And the same will be the case at the center of any other layers. This makes sense since we know that the age field at the center of the conductor in the case where age is much greater than delta is equal to zero. And so certainly an integral of h over length through the center of the conductor should also be zero. So in the case where h is much greater than delta, since the h fields are all confined to the surface of the conductor, the MMF value will also decay away from the surface of the conductor. Reaching zero at the center and then rising back to the other surface. And so we will have this decay like shape for the MMF diagram inside the conductor. I will remember that, in either of these extreme cases or any case in between, the values of MMF in between the layers are going to be the same. And it is the values of MMF between the layers that we will use in the expressions for power loss in a winding for the case, where H has an arbitrary value relative to delta. The power loss formula that we derived earlier for the loss in particular layer m, holds true only when h is much greater than delta. Note that although we derived this formula for the case of a transformer. This formula applies to both the transformers nth layer as well as the end layer of an inductor. In the case of the transformer, you have both the primary and the secondary windings and so you can apply to each of those. Finding a similar formula for the case where the value of h is arbitrary relative to the value of delta is not very simple. To determine such a formula for the arbitrary case, you would have to solve Maxwell's equations inside the conductor and determine the current density distribution within each layer. Then determine the loss distribution and then finally integrate the loss to find the total copper loss in the layer. If such a process is carried out for the case with the rectangular layers, then the following expression for power loss in a layer can be found. Note that this expression is fairly complicated, and involves a number of factors. Some of these factors are themselves complicated functions of a variable, that we've defined as phi. Phi is nothing more than the ratio of h to the effective skin depth, delta prime. Delta prime is equal to delta or the actual skin depth if the conductors are simply rectangular. However by using delta prime instead of delta in this formulation, allows us to apply this formulation to the case of round conductors as well. Recall that in the case of round conductors, you could use a porosity factor to model our round conductors as an equivalent rectangular conductor. The power loss formula for a layer for the case of an arbitrary h also involves the m m f on the two sides of the layer. Script f of h is on one side of the layer and script f of zero is on the other side of the layer. You can see that visually here, on one side of the layer, we have field h of h corresponding to an nmf of f of h. On the other side of the layer, we have some field H of 0 corresponding to an MMF F of 0. It turns out, because of the symmetry of this expression, it does not matter which side of the layer we define as small h and which side we define as 0, we can also eliminate f of h. And f of zero from this power law expression by defining a quantity m as shown. Since f of h minus f of zero is simply equal to the total current passing through this layer. This expression for m can also be written as f(h) over the total current to the layer, which we could write as n l times i where n l is the total number of wires or turns that go through this layer, and I is the current carried by each of those turns. In the case of simple winding geometries, M is simply equal to the layer number that we've also be considering in our earlier expression, in which case, M is always going to be an integer. However, in more complicated winding structures, m can turn out to be a non-integer. By using this definition of m in our power loss expression, we can re-write this power loss expression in the following form. Here q prime Is a complicated expression, that again involves G1 and G2. By writing the power loss in this form, we can identify the factors that increasing power loss, due to skin effect. And due to proximity effect. The derivation of these expressions is beyond the scope of this course. But note that these derivations presume that the currents are sinusoidal and that the H fields on the two sides of a layer are in phase. We can plot the power law expression for a layer to gain more insight. In this plot chart, we plot the actual power loss in a layer relative to the power loss that would have occurred at d c as a function of different values of Phi. Recall that phi is simply equal to h over delta prime, where delta prime is the effect of skin depth taking into account in case the wires are round. For simplicity you can just think of it as the actual skin depth, delta. The actual power loss relative to dc is plotted here for different layer numbers. As you can see the higher the layer number, the higher the actual power loss. Also notice that for all layers, As each becomes much smaller than Δ, the power loss reduces considerably. And for each much smaller than Δ the actual power loss is the same as the loss you would expect at DC. Note that for layer one the actual loss in that layer does not diverge much from the DC loss until H starts to exceed Δ. But then it starts to increase rapidly. The situation is worse for higher layer numbers. For example, the loss in layer six is 100 times greater then at DC when h is only a factor of two greater than Δ. While this chart is useful and that it tells us how the actual loss in a layer compares to the loss we would have had at DC this chart by itself does not provide us a good way to determine the optimal value of h that we've produced. The reason is that as h is decreasing, while our AC loss is converging through the DC loss, our DC loss by itself is increasing. Since the cross sectional area of the conductor is going down. To find the optimum value of h, what we really need is to plot the AC loss normalized to a fixed value. We do that here in the chart at the bottom. In this chart the actual loss in a layer is plotted relative to a fixed loss. The fixed loss is taken to be the DC loss of that layer with Φ is equal to one which corresponds to h = Δ prime or Δ. Now we can see that there are really optimum values of h depending on the layer number. For example, layer one loss is minimized for values of Φ somewhere slightly greater than one. Therefore, if you have a single layer in your structure, you should pick h to be approximately equal to Δ, or slightly greater than Δ. For layer number three, the optimum value of h is somewhere else. And it turns out to be roughly close to 0.65 corresponding to h being 0.65 times Δ. As you go to higher and higher layer numbers, the optimum value of h becomes smaller and smaller. In practice, you are unlikely to make a magnetic structure with different layers of different thicknesses. Therefore, it is useful to look at the loss in a complete winding. Earlier, we had calculated the total losses in a winding with capital M layers. By dividing the actual loss P, which we also called the AC loss by the DC loss we defined an AC loss factor FR. However, this entire formulation was done for the case where h was much greater than delta and only holds for that case. For the case where h is arbitrary we've formulated an expression for the actual losses in an individual layer. Relative to its DC losses, the actual losses in the nth layer is given by this formulation. By summing this over the n layers we can calculate the total AC losses in the winding. This expression here holds for both the primary windings as well as the secondary windings and could also be applied to the case of an inductor. By normalizing these actual losses with respect to the DC losses in the entire winding, we can get an expression for the AC loss factor FR. Doing this, gives us the following expression for this loss factor FR and if you utilize the expression for q prime, the expression for FR, can be written as follows. Clearly, by multiplying this FR, by the DC losses in the winding, we get the actual losses in the winding. Again, it is useful to plot the AC losses, to see what happens in the case of the full winding. In this top chart, we plot the actual losses in the complete winding relative to the losses in the winding at DC. Again, these normalized losses are plotted against Φ which is equal to h over Δ prime. However, in this case, each of these plots now correspond to the total number of layers in our winding. So in general, the results will differ from the case where we were considering one layer at a time. If you look carefully, the result for the case where M is equal to one will match the result where we had M equal to one since in both cases we just have a single layer. With a larger number of layers in the winding the results are sort of averages of the individuals layer's results. Again, what this chart generally tells us, is that as the total number of layers increase, the actual power loss, relative to the DC loss, is going to increase rapidly. Also, with the large number of layers, the actual AC losses can be one or two orders of magnitude greater than the DC loses. Only in the case where age is much smaller than Δ due to AC loses, the actual losses converge to the DC losses. However again, this is not the right chart to look at in order to find optimum value of h for a given number of layers in a winding. The reason again is that as h is shrinking, the DC losses themselves are going up. So what we should look at is really a plot in which the actual losses are normalized relative to a fixed loss. Again, for fixed loss we pick the DC losses with h fixed equal to Δ prime. When we plot power loss in this way, we see clearly, that there are optimum values of h, to pick. If the value of h is picked too small, then while the AC losses decrease relative to the DC losses, the DC losses themselves increase, and so, the total losses increase. On the other hand if age aspect is higher than the optimum value, than the skin effect and proximity effect loses start to play a bigger and bigger role and the total actual losses increase. The optimum value of age depends on the total number of layers in our binding structure. If a winding has only one layer, then the optimum value again is somewhere close to h being approximately equal to or slightly greater than delta prime. Note that this is the same result that we got when we looked at a similar chart plotted for each individual layer. Since the case of one total layer is the same as just looking at that one layer itself. If we have a winding structure with a total of three windings, now the optimum value is close to roughly phi is equal to 0.8, which corresponds to 8 being equal to 0.8 times the effective skin depth. Again, as the total number of layers in your winding structure increase, the optimum value of h decreases. But even at this optimum value of h, you cannot reduce the power loss to the level that you could achieve with a single layer. Therefore, if it is possible, it is always better to make a winding with a single layer. All of our power loss formulations so far has assumed that the currents in the windings are sinusoidal. However, in power electronics, we are often faced by currents that are non-sinusoidal. To deal with such situations, we can do a Fourier decomposition of our current waveform. Then we can apply the formulations we've developed already to each of the harmonics. Consider, for example, the case offered at angular current waveform. We can express this current waveform in terms of its Fourier components. Then we can calculate losses for each of these components. The Dc losses are simply equal to the square of the Dc component times the Dc resistance of the winding. The loss is due to the fundamental, and all the higher order harmonics can be determined using the formulations we developed earlier. We can then add all of these losses to get the total losses from all of these harmonics. It is also common to define a parameter called the harmonic loss factor. The harmonic loss factor is simply the sum of all the losses from the fundamental plus the higher order harmonics divided by the losses due to the fundamental. We can also express the total losses in the winding, In a form that contains this harmonic loss factor. In this formulation, FR is the ac loss factor. We can look at this example current waveform to see the kind of effect harmonics play in increasing the total losses in a winding. In this top chart, the harmonic loss factor is plotted again as a function of phi. Here, to be precise, we're calling it phi 1 since it corresponds to the fundamental, and it's equal to h/delta prime, where delta prime now corresponds to the skin depth of the fundamental in this material. You can see from this chart that the harmonic loss factor can have a substantial effect on the total losses in a winding. The effect depends on the relative value of h to delta. For large values of h, the Ac losses are already so high because of high skin and proximity effect that the additional losses, due to the harmonics, only have a modular impact. At the other extreme, when h is much smaller than delta, the harmonics again have little impact because the conductor width is so small that the current distribution in the conductor is essentially uniform, even with the harmonics. The harmonics tend to increase the loss the most when you've designed your windings well. By well, what we mean is that you've designed them well for the fundamental. Since FH represents the relative increase in loss due to the harmonics, its effect is most pronounced when the loss due to the fundamental is limited. That is exactly the case when you design your winding well for the fundamental. You can see that clearly from this chart. Suppose we had a two layer winding and we selected the conductor thickness to be less than the skin depth. Then the loss that is due to the fundamental are not much more than the Dc losses. So this point here then corresponds to our fundamental. We call that F1. Now in this chart, the higher order harmonics would show up to the right of F1 since the x-axis here, phi, is h/delta prime and the skin depth at the higher order harmonics is smaller. So the value of phi is larger. So the second harmonic, maybe here, And it clearly leads to higher losses. And the next harmonic would be further up and contribute to even higher losses. So when you've designed well for the fundamental, relatively speaking, you get hit hardest by the harmonics, as skin and proximity effect are playing a bigger and bigger role for these harmonics. In our next lesson, we will see how to mitigate the effect of proximity effect in some cases.