As a first example of modeling loss in a converter, let's model the resistance of the wire in the inductor of a boost converter. The resistance of the winding, or DC copper loss of an inductor, is important. because it has a first order effect on the cost and efficiency of, of the inductor and the converter. If you don't care how large this resistance is. I can wind you an inductor that is arbitrarily small and inexpensive. That can, give you the desired inductance, and handle the desired current without saturating. but it will use many turns of very thin wire that will have a very high resistance. So the inductor then will actually have very large loss and will burn up. So, the resistance of the wire is important. And one of the important things we have to specify and design, [COUGH]. In order to optimize the cost and efficiency of our converter design. So, to model this, The common way to do it is with a lumped element model, as shown here. Where we replace the physical inductor with a, a model, consisting of an ideal inductor. Having a voltage VL. In series with a resistor, that is the resistance of the winding. ' Kay? So between these terminals, this is a model of the physical inductor. And what we do is we insert this model into the converter circuit like this, and then analyze the circuit. so here is our inductor model. Okay? The ideal part of the inductor. with its voltage V sub L, still must obey inductor volt-second balance. But now we have an added resistor and series in the circuit, that has a voltage drop equal to the inductor current multiplied by, the resistance. [COUGH] So, we apply a volt-second balance to VL of t in this model. And, as usually, we plot, apply charge balance to the capacitor current. and we can get a set of equations that we can then solve for the output voltage and other quantities of the converter. Okay? So let's do it. Here's our, here's our circuit now to be analyzed. Here's the inductor voltage, and the voltage of the ideal part of the inductor model. And we have a capacitor current there. And as usual we, we now write the circuit with the switch in a two positions. And proceed to work out the waveforms of vl of t and ic of t. So with a switch in the first position we get this circuit. We can identify what the inductor voltage is from this loop equation. The inductor voltage Vl will be equal to the input voltage Vg. Minus the voltage drop on the resistor RL, which is iRL like this. the capacitor current for the switch in the first position is found from the node equation at the output, and as usual. It is equal to the load current. Actually, minus the load current. so ic is minus V over R. We now apply the small ripple approximation, as usual. So the output voltage, v of t, we replace with its dc component, capital B. And additionally, on the inductor. We will assume that the inductor has small ripple as well. So we will replace i of t with its DC component, capital I. Similar for the switch in the second position we get this circuit. The inductor voltage this time is the input voltage VG Minus the drop on the resistor iRL minus the output voltage V. And the capacitor current is the inductor current i minus the load current V over R. So we get these equations and we again apply the small ripple approximation as appropriate. Okay? So here are the waveforms then, this is the inductor voltage waveform, and the capacitor current waveform. Both of these must have zero average when the converter works in steady state. the inductor voltage waveform now has some additional terms that come from the voltage drop on the, the winding resistance. How we can apply volt-second balance to this as usual. So, zero the average inductor voltage is zero, and it would be D times the value during the first interval. Vg minus iRL plus D prime times the value during the second interval. which is vg minus iRL minus V. Okay? we can collect terms and equate this to zero and we get this equation for inductor volt-second balance. Okay? In the ideal case, we were able to solve this for the output voltage. And get an expression for the ideal output voltage. But now have an added term that comes from the inductor winding resistance. And there's an added variable, capital I. the inductor current at this point is an unknown. So we have one equation with two unknowns and we can't yet solve. So to get a second equation we get that from the charge balance on the capacitor. So the average capacitor current must be zero on steady state. and it's, we actually get the same waveform and same equation as in the ideal case for the capacitor current. and when we collect terms then we get this equation, which is a second equation that relates the inductor current and the output voltage. [SOUND] Okay? So, when taken together then, we have these two equations. First is from volt-second balance, the second is from charge balance and we have two unknowns, capital I and capital V. So we can solve now to find the output voltage. For example, solve the second equation for I, and plug the result into the first equation, and then solve for V. And if you do that, here is the result, V over Vg. I've written this equation in a nice form. It has a. Two terms, the first term one over D prime is the result for the ideal case with no loss. And the second term is, and we can think of as a modification of how the inductor resistance affects the output voltage. And you can see that if you let RL goes to zero, the second term goes to one and we get the expression 1 over D prime, okay? I've also written the second term in a nice form where every term is normalized and the denominator is 1 plus. A function of RL. So, as RL is increased, the denominator is increased, and the output voltage goes down. If you would like RL to have a small affect, on the output voltage then we need RL to be small compared to the term it's divided by. Which in this case, is D prime squared R. So for example, if RL was 1% of D prime squared R. Then the denominator would be 1.01, and the overall second term would be approximately 0.99. So the output voltage would decrease by 1%. It's nice to write the second term in this form because we can compare, see what to compare RL to, and how, get a good idea then of how small RL needs to be. Okay, on the right we can see A plot of this expression V over Vg, for different values of RL over R. The top curve is where the RL is equal to zero and we get the ideal case, that's 1 over D prime. This function goes to infinity as, as D goes to 1. the lower curves are for different values of RL, and you can see that as RL increases the output voltage goes down as we would expect. But not only that there's a major qualitative change, where the curve actually bends over. It reaches a maximum and then it starts decreasing. And at D of one the curve goes to zero. So this is a pretty big difference. You add a small amount of RL, anything bigger than zero, and at D of one the curve goes to zero instead of infinity. So what's exactly going on there? Well we can go back and look at the original circuit. What happens in the original circuit as you let D approach one? Well, if D goes to one and the switch is always in position one and never in position two, so you just connect the inductor across Vg, and the inductor current is large. And, in fact, with zero RL, the inductor current tends to infinity. and the volt-seconds don't even balance. if there is some D prime interval, it's greater than zero. Then, during the D prime interval, some current will go to the output and can charge up the output voltage. in the ideal case, as D approaches zero, the average current driving the output is infinite amps times something approaching 0 duty cycle. And, what is the limit? well, the ideal equation tends to, infinity, at least as approached from the left. But it's a non physical answer. And with any amount of RL at all the the inductor current is limited to a finite value. And as D goes to one, in any event we, we get essentially zero average current deriving the output, and there's no output voltage. And so The actual case with the loss resistance is the, the physical answer. In which the output voltage is limited, and doesn't go to infinity at [UNKNOWN] 1. And, in fact, it goes to zero. Moreover, we can see that, the maximum voltage that we can attain is limited by the loss resistance. So, for example, suppose we wanted our converter to boost the voltage by a factor of three. Well, if an RL over R was, if you have RL over R equal to 0.05, it's not, no good. It's not going to work, because we can't make enough output voltage. RL over R equal, equals 0.02 looks like it would work. And 0.01 would be even better. So the peak amount that we can boost by, is, is a function of RL. So, we've seen, in this first example, how we can model loss. And in this case, we modeled the inductor winding loss with a lumped element model. That, that had an ideal inductor and series with an effective resistor. The ideal inductor still obeys, volt-second balance, although the, the total inductor model does not, because of the, resistor. So, in general, we can model losses by adding elements to our converter, and then applying volt-second balance to the ideal part of the inductor model. And charge balance to the ideal part of the capacitor model. We then get a set of equations that we can solve for things like the output voltage.