In the previous lecture, we perturbed and linearized the converter equations and we derived this set of three small-signal equations for the, the buck-boost converter. We're now in a position then, to construct an equivalent circuit model for the converter. And the process for doing this is very similar to the way we've been constructing the DC equivalent circuit models. We just have some different terms now. but in these equations we have signals and a converter that are the hat quantities. Again, the capital values are taken as constants or parameters in the model that are known. So they're known given quantities and it's the hat quantities that are signals in our equivalent circuit. So as, as in the DC case, what we're going to do is construct an equivalent circuit to go with each of these equations. so the equations are treated as loop or node equations of the converter equivalent circuit model. And we will take then the three circuits that we get and combine them with DC transformers, actually, in this case, they're AC transformers to get the final equivalent circuit model for the converter. So again, the lecture slides have the material that I'm about to present, but I'm going to actually draw them by hand on the screen to illustrate the process. So here is the first equation which we got from the the averaged inductor equation that was then perturbed and linearized. So this is, in fact, the small-signal AC equation that describes the the loop In the small-signal AC model that contains the inductor. So we're going to start on the left side of the equation with an inductor, and we re, we recognize that L times di hat dt, is the voltage across the inductor in our small-signal model. So I'm going to just put an inductor in the middle of the circuit here having value L with a current i hat flowing through it. And, therefore, the voltage across this inductor with this polarity would be L times di hat dt. 'Kay? And this equation says that that in, inductor voltage, L di hat dt, is equal to a sum of three terms, which we represe, which we recognize as voltages, and these are the voltages around the loop then that is the Kirchhoff Voltage Law or loop equation for this part of the model. D times vg hat is a dependent source in our model. It depends on the input voltage, vg hat. This is actually the input AC variation in vg that is applied to our small-signal AC model. And so that gets multiplied by the constant or the value D, and we're going to, represent this with a dependent source of value D times vg hat. The next term D prime times v hat, this is another dependent source that depends on the output voltage, v hat, in our AC model. And so I'll draw a dependent source on the output side here of value D prime v hat. 'Kay? going around the loop this term has the same polarity as the vg hat term, so as we go around the loop, we have to go through plus to minus in the same direction. So this is actually minus on top and plus on the bottom. And, finally, we have another source, capital Vg minus V, then all times d hat. Now, in this equivalent circuit or the AC model, d hat is an independent input to this circuit. It's actually the control input of the system. And so I'm going to draw all the d hat sources as independent sources. So round sources. And this is d hat with a gain of capital Vg minus V. 'Kay? This one also is a plus sign and so, as we go around the loop, it's minus on the left and plus on the right. So this is the equivalent circuit then to go with our inductor equation. 'Kay? Lets do the capacitor. Here's the equation we got from, from this small-signal capacitor equation. C dv hat dt we'll recognize as the current in the capacitor in our model. So I'm going to draw a capacitor here, of value C with a voltage v hat across it, so that it has a current of C dv hat dt. 'Kay. This equation says that that current is a sum of three other currents. So this is a node equation and we'll draw a node here where the capacitor's connected. And we have three currents of the size of the capacitor current flowing in that node. The first term, minus D prime times i hat, is a current that depends on the inductor current, i hat, from the previous circuit we just drew. So this is a dependent source, and I'll put a dependent current source, of value D prime i hat. 'Kay? This current, because of the minus sign, goes out of the node, not into the node and so its current points down. The second term is the load current. We have a, actually it's the AC variation in the load current. And so we can recognize v hat over R is the current through a resistor of value R, having a voltage, v hat. And because of the minus sign, this also flows out of the nodes, so that's consistent with the resistor. And then the last term is another d hat source. This is a current that depends on the control input, d hat. So we will draw an independent source for the control input. with the plus sign, this flows into the node, and it's of value capital I times d hat. [COUGH]. So here is the equivalent circuit to go with the capacitor equation. Finally, our last equation is for ig hat. ig hat, we recognize as the current coming out of the vg hat source at the input of our converter. So I'm going to draw a voltage source of value vg hat. So this is the independent input to the, to our small-signal AC model that has vg variations. And the current coming out of that source is ig hat. And this equation says that that current is equal to a sum of two other currents. So this is also a node equation. Here's the node. And what? We have the current D times i hat. This is a dependent source. It depends on the i hat inductor current elsewhere in the model. So we'll draw a dependent source for, for it. And with a plus sign, it goes out of the node. And then, I times d hat. This is another control source, d hat or d hat source. So it's an independent input that comes from the control. From the plus sign, it goes out of the node and its value is capital I times d hat. So here is the model for the input cord of the converter that we get from the ig hat equation. Okay. Now we can put all of the circuits together. So here are the three circuits. In the middle is the first inductor loop equation, on the right is the capacitor equation and on the left is the ig hat equation that we constructed. And they're placed together, and, just like in the DC model, now we can recognize that the dependent sources behave as effective transformers. And so these were DC transformers in our DC model. I'm going to actually represent them as AC transformers here, put a squiggle, little AC symbol through the transformer. This denotes, really, that it's an ideal transformer model and not a physical transformer. It's really the same as the DC transformer. The little AC squiggle helps us keep track of the fact that this the small-signal AC model. So quantities like capital Vg or capital V are simply constants in the model and not actual signals, and it's the hat quantities that are the AC signals in this circuit. 'Kay? So we have a one to D transformer from these two sources that's shown here and we have a D prime to one transformer here from these sources. because the this D prime v hat source is minus on the top, whereas V, the reference polarity of V is plus on the top the dots are reversed on this second transformer as shown here. 'Kay? So construction of the equivalent circuit model for the small-signal AC equations proceeds in a very similar manner to what you've already learned for the DC model. And once we have the small-signal AC equations, then, I think, we, we can proceed in the usual manner to construct the small-signal AC model. Just for reference here are the results for the three basic converters. The first shown here is the buck converter and the AC model contains a buck type transformer in the AC equivalent circuit, this one to D, that has some d hat sources in the output LC filter. For the boost converter, we get the inductor on the left side. There's a D prime to one boost type transformer in the middle of this AC model and there are d hat sources around it. And, finally, the buck-boost converter has both a buck and a boost transformer, very similar to the DC model. We have three d hat sources and this is, in fact, the model that we've just derived. So we can derive the basic models or the models of all the basic converters. you will do homework where you will derive one of these and we're in a position now to be able to use these models and solve for things like the small-signal transfer functions of the converter.