As a 1st 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 the d.c copper loss of an inductor is important. Because it has a 1st order effect on the cost and efficiency of the inductor in 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 a 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. In order to optimize the cost and efficiency of our converter design. So to model this a common way to do it is with a lumped element model as shown here where we replace the physical inductor with a model consisting of an ideal inductor having a voltage v l. In series with a resistor that is the resistance of the winding. Okay, 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 of v symbol still must obey inductor volt 2nd balance. But now I have an added resistor in series in the circuit that has a voltage drop equal to the inductor current multiplied by. The resistance. So we apply a volt 2nd balance to v l of t in this model and as usual we apply charge balance to the capacitor current. And we can get a set of acquaintance that we can then solve for the output voltage and other quantities of the converter. Okay, so let's do it. Here's our circuit now to be analyzed, here's the inductor voltage or the village of the ideal part of the inductor model. And we will have a capacitor current there and as usual when we now write the circuit with the switch in the 2 positions and proceed to work out the wave forms of v l t. And i.c of t. So with the switch in the 1st position we get this circuit. We can identify what the inductor voltage is, from this loop equation the inductor will be equal to the input voltage of e.g. minus the voltage drop on the resistor r l. Which is i R l like this. The capacitor current from the switch in the 1st position is found from the node equation of the output and as usual it is equal to the load current actually minus the load current. So i.c is minus v over R. We now apply the small ripple approximation as usual so the output voltage of the of t. we replace with its d.c component V. 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 d.c component capital I. Similar for the switch in the 2nd 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 me 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 wave forms then this is the inductor voltage waveform in the capacitor current waveform. Both of these must have 0 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 2nd balance to this was usual so the zero or the average inductor voltage is 0 and it would be d times the value during the 1st interval. V g minus IRl. Plus D prime times of value during the 2nd interval which is v g minus r l minus v. Okay, we can collect terms and equate this to 0 and we get this equation for inductor volt 2nd 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 is an added variable I. The inductor current at this point is an unknown, so we have one equation with $21.00 owns and we can't yet solve. So to get a 2nd equation we get that from the charge balance on the capacitor so the average capacitor current must be 0 in steady state. And if 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 the 2nd equation that relates the inductor current in the output voltage. Okay, so when taken together then we have these two equations. The 1st is from volt 2nd balance, the 2nd is from charge balance and we have two unknowns I and V. So we can solve. Now to find the output voltage for example we can solve the 2nd equation for I unplugged the result into the 1st equation and then solve for v and if you do that here is the result v over v g. I've written this equation in a nice form that has a two terms. The 1st term one over d prime is the result for the ideal case with no loss and the 2nd term. It is you we can think of as a modification of how the. Inductor resistance affects the output voltage and you can see that if you let r l go to 0 the 2nd term goes to one. And we get the ideal expression one over D prime. Okay, I've also written the 2nd term in a nice form where every term is normalized and the denominator is one plus a function of r l. So as r l is increase the denominator is increased and the output voltage goes down. If you would like r l to have a small effect on the output voltage then we need r l to be small compared to the term it's divided by which in this case is D Prime squared R. So for example if r l was one percent of D prime squared R. Then the denominator would be $1.00 and the overall 2nd term would be approximately 0.99. So the output voltage would decrease by one percent. It's nice to write the 2nd term in this form because we can compare and see what to compare r l to and how to get a good idea of how small r l needs to be. Okay, on the right we can see. A plot of this expression v over of v g for different values of r l over r. The top curve is with r l equal to 0 and then we get the ideal case that's one over d prime and this function goes to infinity and as d goes to one. The lower curves are for different values of r l and you can see that as r l increases. The output voltage goes down as we would expect that not only that there's a major qualitative change where the per actually been over reaches a maximum and then it starts decreasing. And at the a one the curve goes to 0, so that's a pretty big difference you add a small amount of r l. Anything bigger than 0 and at the end of one the curve goes to 0 instead of infinity, so what's again exactly going on there. We can go back and look at the original circuit, what happens in the original circuit as you at d approach one. Well if the e goes to one image which is always in position one and never in position 2. So you just connect the inductor across v g and the inductor current is large and in fact with 0 r l. The inductor current tends to infinity. And the volt seconds don't even balance. If there is some deep prime interval was greater than 0 then during that D prime interval some current will go to the output and can charge up the output voltage. In the ideal case as the approaches 0, 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 up as approach from the left but it's a non physical answer and with any amount of r l at all. The inductor current is limited to a finite value and as e goes to one in any of the, we get essentially 0 average current driving the output. And there's no output voltage and so. The actual case with a loss resistance is the physical answer in which the output voltage is limited and doesn't go to infinity at the one and in fact it goes to 0. 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 move the voltage by a factor 3. Well if in r l over r was if you have r l over r, the point 05 it's not no good it's not going to work because we can't make enough output voltage. R l over r equal equals point 02 looks like it would work and point 01 would be even better. So the peak amount that we can boost by is is a function of r l. So we've seen in this 1st example how we can model loss and this case we model the inductor winding loss with a lumped element model. That had an ideal inductor in series with an effective resistor the ideal inductor still Obie's volt 2nd 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 volts 2nd 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.
