This week we will cover Chapter 3, concerned with the steady-state equivalent circuit modeling and switching converters. The object here is to develop equivalent circuits that model the important functions that the switching converter performs, while ignoring the unimportant things, such as the switching ripple. So what we want to do is model the important low frequency and in this case dc components of the voltages and currents of the converter. And we'll use these models to gain insight and understanding of how the circuit and converter works, how it performs the important functions associated with these dc components of the wave forms. And we will use it to model the losses in particular and the efficiency of the converter. We'll find that by use of equivalent circuits, we can gain a lot of insight as to, for example, how to develop approximations for what is important and what is not, and how to handle more complex circuits. For example, if we want to include all of the sources of loss in our model, we can refine the model and add these loss elements. Which could become fairly complicated, but we can handle them in a way by manipulating the circuit that lets us easily get the answers and the insight that we need. Later in the course, we're going to refine these models to include the phenomenon known as switching loss. And even later, we're going to extend the models to model converter dynamics, be able to find things like small signal transfer functions and, say, output in penances, and to design the feedback loops of these converters. Okay, so in this lecture I want to talk first about the equivalent circuit model of an ideal dc-dc converter. Okay, what are the basic properties of an ideal dc-dc converter? Well, first of all, we're going to say that an ideal converter is 100% efficient. And to begin with we won't model the losses. Later, in the upcoming lectures, we'll refine our model and add losses in, but we're going to start with the ideal case. So if we have a switching converter with an input voltage, Vg, applied to the input port of the converter, and what we'll call the input current, Ig. And if we have an output port of our convertor with the voltage V, with a current that we'll label I, then we can write that the input power equals the output power ideally, so VgIg =VI. I should add that in addition to this 100% efficiency requirement, these equations also implicitly assume that the converter operates in steady state. So for example, if there are inductors and compacitors inside the converter that can store energy, then during transients those reactive elements may store or release energy. And during the time that they're doing that, the input and output powers could differ. But in steady state, there's no net change in energy or stored energy in the reactive elements, and then we get this first basic equation. Okay, the second thing we found in the last chapter was that we could solve that converters and find that the output voltage is some function of duty cycle times the input voltage. And this function of duty cycle we called the conversion ratio, M, and M is a function of D. So for the buck converter, M was equal to D, and for the boost it was equal to 1 over D prime, and so on. Okay, there's a third equation that really comes from the first two. That if you take V from equation 2 and substitute it into equation 1, then, well, let's do it. We get VgIg on the left side of equation 1 is equal to V, which is MVg times the output current I, okay? We can cancel the Vgs, and what we find is that the input current, Ig, equals the same conversion ratio times the output current, I. And that's our third equation. So it says that the currents are related by the same conversion ratio, except going in the opposite direction. Okay, so those are three basic equations for the converter. Now what kind of equivalent circuit could we construct to go with those three equations? Okay, we'll I'm going to actually draw that here. So we had what, VgIg = VI is the first one, second one was V =M(D)Vg, and the third was Ig = M(D)I, okay? Now, there's actually two different ways that we commonly model these equations with equivalent circuits. The first one uses dependent sources. So let's consider our converter with an input port having Vg and Ig. And its output port has voltage V and current I. And one way to model these equations is with a dependent voltage source for equation 2, which says that the output voltage V is dependent on Vg with a constant or gain of M(D). So I can draw a dependent voltage source, And label its value M times Vg. Here in my notation a square source is a dependent source. Okay, the third equation we could similarly model with a dependent source. The current Ig is equal to this gain M times the current I. So I could write a dependent current source, That produces Ig and label its value M times I. So this is one possible equivalent circuit that models the important low frequency or dc voltages and currents at the terminals of the converter. One can verify that equation 1 is also satisfied, in fact we derived 3 using 1, but Vg times MI is equal to MVg times I. [COUGH] And we can verify from these equations that that's so. This is one way to model a switching converter, and in fact we often use this model in circuit simulators such as SPICE. Which have dependent sources in them that can be easily programmed to do this. So that's approach number I. A second approach is to recognize that these equations are actually the equations of an ideal transformer. So let's draw, Our terminal quantities, and put a transformer between them. So a transformer has input power equals output power, at least ideally. And it also has the voltages related by a factor, which in the transformer is called the turns ratio. So we could write a transformer symbol, or an ideal transformer symbol, and the turns ratio is the ratio of the voltages. So here V is equal to the turns ratio times Vg, and the turns ratio apparently is the conversion ratio M. So I can label the turns ratio here as 1 to M(D). And we know with a transformer, if we apply a voltage VG to the primary with, say, the plus on the top. I'm going to put a polarity mark there where the plus sign of Vg is. Then the voltage coming out of the transformer here will be the turns ratio times the input voltage. And it's plus at the top also, so the polarity mark goes on the top of the secondary winding. [COUGH] We also know for transformers the currents follow the same rules, but in the opposite direction. So here the input current, Ig, flowing into the dot, is equal to the current on the secondary flowing out of the dot multiplied by the turns ratio, so M(D) times I. Okay, so this is the ideal transformer model of a switching converter. And it's justified, because it correctly describes the relationships between the terminal voltages and currents of the ideal converter. Now, I should note that real physical magnetic transformers that are constructed with turns of wire on magnetic cores can't even handle dc signals. If you put a dc voltage across a magnetic transformer winding, the volt seconds won't balance and the transformer will saturate. [COUGH] But that's perhaps one of the good reasons to built a dc-dc converter, because we want that function. And so here what we're going to do is, we're going to make sure we all understand that this is an ideal dc transformer model, by putting a straight line like this through the transformer to represent dc. [COUGH] And we are free to invent an equivalent circuit symbol that models these functions. And it's defines simply as being a symbol that satisfies equations 1 through 3. Okay, so we have two different ways to model the terminal equations of the ideal dc-dc converter, and they're completely equivalent. And in fact if you find that you construct an equivalent circuit that has one of these, you can always substitute the other for it. [COUGH] Okay, so again, the DC transformer model is justified, because it models the important physical properties of the desired waveforms of the dc-dc converter. Which is conversion of the voltages and currents according to a conversion ratio M with 100% efficiency. And this is now a time and variant model that doesn't have switching. So we can replace our switching converter with this transformer model, and we then have a circuit that can be much more easily solved, using the techniques of just basic undergraduate circuit analysis. So we can do dc analysis and solve for the important quantities, not worrying about what's going on with the switching and the switching ripple. The only slightly strange thing here is that the turns ratio, M(D), is a function of duty cycle. So if we change the duty cycle, we can change the turns ratio at any time. And if you want to turn down the output voltage, just turn down the turns ratio. But that's one of the additional important functions that the converter can do, and it is modeled. Okay, let's do an example of using the transformer model. So here's an example of a circuit, in fact, let's do this on the graph paper as well. So let's suppose we have a switching converter. It has its input terminal Vg, and output terminal V. And we model it with a transformer. So will, Draw inside the box of the switching converter a dc transformer and turns ration 1 to M. And let's suppose that at the input terminals of this converter, we have a power source that has some impedance associated with it. So we'll model the power source with a Thevenin equivalent. So I'm going to draw a V1, and it's the Thevenin equivalent voltage, and some Thevenin equivalent resistance that we'll call R1, [COUGH] that is connected to the input terminal. Let's also suppose we connect a load, let's say a resistor R, on the output. And the object now is to solve for what is the output voltage? Well, the way to do this, I think the easy way to solve this, is to manipulate the circuit by pushing all of the elements through to the output side of the transformer. So we remove the transformer, and then we can solve for the output voltage. So if you recall from basic circuit analysis, when you push a voltage source through a transformer, then effectively on the output or secondary terminals of the transformer, we see a voltage source of value V times the turns ratio. So I'm going to write a voltage source and a round source as an independent source. It's of value V1 times the turns ratio M. Then we also need to push the resistor R through the turns ratio. And when we push a resistance through a turns ratio, we must multiply by the turns ratio squared. So we have an effective resistance, as seen from a secondary terminals, of value M squared times R1. And that's what we get between the output terminals of the converter. Now this is driving a load R, so we'll draw R here, and we have output voltage V. Okay, now the circuit is simplified, and I think it's a straightforward matter now to write the expression for the output voltage. The output voltage V is equal to the effective voltage source, M times V1, times the divider ratio of these two resistors, which would be R / R + M squared R1. And that's the equation for the output voltage. So the conversion ratio M affects not only the voltage source, but also this divider ratio. And taken together we can solve for the output voltage. Okay, so we've derived an equivalent circuit for the ideal switching converter that models the important dc components of the waveforms and the basic power conversion functions that are performed by the converter. In the upcoming lectures, we're going to refine this model next to add in the different kinds of types of losses in the converter. And build up an equivalent circuit model that can be solved to find things like the efficiency. I would also note that if you found the derivation of the output voltage by manipulating the ideal transformer circuit to be unfamiliar, or you need a brush-up on how to do that. There's a supplementary lecture posted to supplement this lecture, that reviews the rules for manipulating ideal transformer circuits.