In the last lecture, we modeled the copper loss of the inductor of a boost converter by inserting an extra resistor in series effectively with the inductor in a lumped element model, and with that model we then calculated the effect of this resistor on the output voltage of the boost converter. In this lecture, we're going to extend that result and construct an equivalent circuit model to go along with the equations. So, here are the equations that were found in the last lecture. For this example, the first equation was from inductor volt second balance and the second equation we got from capacitor charge balance. The object now is to find an equivalent circuit to go along with these equations. The approach that we're going to take is to view these equations as the loop and node equations of our equivalent circuit model. Indeed, the first equation was found by finding the average voltage around the loop where the inductor is connected. So, this effectively is a loop equation in our model. The second equation was found by finding the average current flowing into the capacitor from the node where the capacitor is connected, namely the output node. So, we can view this equation as a node equation in our model as well. So, what we're going to do is construct equivalent circuits to go with each of these equations. Now in basic circuits classes, I hope you were taught how to take a circuit and apply Kirchhoff's laws to find the equations. Here we're going backwards, we're given the equations and we want to construct a circuit. Well, this process of going backwards turns out to not be unique. There's more than one way to construct a circuit that satisfies these equations. So, we have to be careful how we do it to make sure that our resulting model has physical significance, and my advice here is to use the equations in the form that they come without further manipulations. So, the first equation was found from inductor volt second balance, we should keep it as a sum of voltages around a loop corresponding to a Kirchhoff voltage law equation and don't manipulate this. If you for example divided through by say R, then this first equation would have terms that have dimensions of current rather than voltage and you might think then that this is a node equation with currents flowing into a node. While that equation might be mathematically correct, the resulting model is not physically connected to the actual converter. So, we don't manipulate these equations or do any kind of algebra on them, we construct circuits to go along with the equations right away. So, let's do it. I'm going to copy this equation that the average inductor voltage is Vg minus IRL minus D prime V, and we will construct a circuit to go with it. It's on the next slide, but I'm going to do it actually on some graph paper here. So, our equation was that zero is the average inductor voltage which is Vg minus IRL minus D prime V. Okay. So, this is an equation describing the loop where the inductor is connected and it's describing actually the DC components of the voltages around that loop. So, I'm going to draw a little dashed inductor in that loop that has an average inductor voltage equal to zero and there is a current flowing through this inductor that is the inductor current DC component that we've been calling capital I. Now, the average inductor voltage is zero and in fact the impedance of the inductor at DC goes to a short circuit. So, in fact this is a short circuit in our DC model but it's a good starting place to have a physical connection to the actual model and the actual circuit to recognize that the current in this loop is the inductor current I, and the voltage is the average inductor voltage that is defined with a reference direction or polarity that is consistent with the direction of the current. So, that is the voltage on the left hand side of the equation. On the right hand side of the equation, we have the input voltage Vg. So, what should I draw in our loop to represent that? Well, we'll draw an independent voltage source of value Vg and since it's on the other side of the equation, then as our current goes around the loop and it's going from plus to minus through VL because Vg is on the other side of the equation, our current will go from minus to plus or with the other direction through Vg. Okay, the next term is I times RL, that term sounds like the voltage across a resistor having current I which is the current in this loop multiplied by a resistance RL. So, we'll draw a resistor here and label its value RL and the voltage across it with I going from left to right would be a voltage from left to right a value IRL. That polarity has the opposite polarity going around the loop as the Vg term had, which is consistent with a minus sign. The last term is minus D prime V, but how do we handle this term? Well, here's how V is the output voltage. This isn't exactly the output voltage, but it's D prime times the output voltage. So, I'm going to draw for now a dependent source of value D prime V. So, it's a voltage source that depends on the output voltage V, but it's multiplied by D prime. Let's see the minus sign would mean that this voltage is plus on top and minus on the bottom, it has the same polarity going around the loop as the RL of the IRL voltage. So, here is an equivalent circuit whose loop equation is the same as the equation that we got from volt second balance and it is the inductor loop part of our equivalent circuit. Let's do the other equation. So, the other equation was from capacitor charge balance which said that the average current through the capacitor or DC component of capacitor current is zero, and what we found there was that that current was equal to D prime I minus the load current capital V over R. This equation was found by charge balance on the capacitor, and again basically it was found by computing the DC components of the currents going into the node where the capacitor is connected. So, we can view this as a sum of currents equaling zero which sounds like a node equation. So, I'm going to draw a node. This is the node where the capacitor is connected. So, I'll draw a dashed capacitor. So, the current through this capacitor, our average iC is zero, and we know the voltage across the capacitor is V, capital V actually in our DC model, the DC component of the output voltage. In our model, our DC model, the impedance of our capacitor at DC is an open circuit, so this will actually be an open circuit. But for reference, we know that V is the voltage at the node, and iC is coming out of the node and flowing into the capacitor. So, what our equation says, our charge balance equation says is that, there are two other currents flowing into the node. The first one is D prime I. So, this is a current that is dependent on the inductor current, which is in our other circuit, our earlier circuit. But it's not just I, it's D prime times I. So, I'm going to draw a dependent source, where the current D prime I that flows into the node. Then, the second term is minus V over R. So, this is the load current, and it is flowing out of the node because of the minus sign. We can model that as a resistor of value R. It has a voltage V across it. So, the resistors are in parallel with the capacitor, and that way, we get a current capital V over R flowing through our load. So, this is the equivalent circuit that comes from the capacitor charge balance equation. The last thing to do is to combine the two circuits together. So, let's go back to the slides. When we draw the two circuits together, we get this, this top circuit. So, the left hand side was the inductor loop equation, where the inductor went right here, and the right hand side was the capacitor node equation, where the capacitor went right there. The last thing to say about this is that we recognize we have two dependent sources as discussed in Section 3.1 are equivalent to a DC transformer. So, we have one dependent source, this D prime V source, depends on the voltage V across the other source with a factor of D prime, and the dependent current source depends on the current I through the voltage source with the same factor of D prime. So, since the factors are the same, and they depend on the voltages and currents of the corresponding sources, we can combine these into the DC transformer, as we discussed previously. I think the easiest way to understand what the turns ratio and polarity mark should be is to look at the voltages. So, we have a voltage V on the output side, and D prime V on the input side of our transformer. So, the turns ratio is D prime to one, which will get us the correct polarities of voltages. To look at the polarity marks, we have look at the polarities of the voltages. So, we have V is positive on the top, on the secondary side and D prime V is positive on the top on the primary side. So, the polarity marks are of the same phase, we have the dots on the top in both cases. As a check, we should look at whether the currents work. So, here in our model, we have a current, I, flowing into the dot on the primary side and coming out of the dot on the secondary side, which should be D prime I, and that's indeed what the model says. So, the currents are consistent with the transformer model as well. So, we have an equivalent circuit now for our boost converter that has the ideal transformer, in this case D prime the one is its turns ratio, or if you want to call it one to M, M is one over D prime. Then, it has an added loss resistor that comes from the inductor loss resistance. An easy way to solve this I think, is to push the elements through the transformer. We've actually already done this exercise. We pushed Vg and RL through to the secondary side if we want to solve for the output voltage. So, when we do that, Vg will be multiplied by the by M, or divide it by D prime, RL will get multiplied by M squared or in fact divided by D prime squared. So, that's the effective circuit between the secondary terminals of the transformer. We connect that to our load resistor, and recognize that the secondary voltage is V, and we can then just solve this circuit using the voltage divider formula, V would be the voltage source Vg over D prime multiplied by the divider ratio of these two resistors here. If you like, you can divide top and bottom by R to write the expression in this form, which is the way we wrote the expression in the last lecture. So, we found the same result as in the previous lecture, except instead of using algebra. We found it using circuits, but we can do more with the equivalent circuit than just solve for the output voltage. We could also for example solve for the current I on the primary side. So, if you say push the load resistor through the transformer to this side, will get effectively between these terminals a resistor that would be D prime squared R and we can then find the current I would be Vg divided by the sum of the two resistors like this. Another thing we can do is finding the efficiency. To find the efficiency, we find the input power and the output power. So, the input power is the power going into the converter terminals here, and the output power is the power coming out this the output side, and the efficiency is the output power over the input power. So, from the circuit we can easily write expressions for the input and output power. So, the input power we can write is the input voltage Vg multiplied by the input current I and the output power would be the output voltage V multiplied by the output current. What is the output current or the current flowing here? We'll, use the transformer. We have current I flowing in the dot on the primary. So, the current coming out of the secondary will be D prime times I. So, we can write that the output power is V times D prime I. That's shown here, and now we can cancel the I's, and we get that the efficiency is V over Vg times D prime. We previously found that V over Vg for this converter is one over D prime times one over one plus RL over D prime squared R. So, we have to take that V over Vg and multiply it by D prime. So, if you multiply it by D prime, the one over D prime factor goes away and then this added multiplying factor is in fact the efficiency. Here's a plot of the efficiency. So, for different values of RL, actually for different values of RL over R, we can plot a series of curves of efficiency versus duty cycle. What you can see is that for low duty cycles, the efficiency is a lot higher whereas for high duty cycles, the efficiency comes to some shoulder and then drops off quickly. So, there's a range of duty cycles from zero up to some value depending on RL, where we have high efficiencies. Beyond that, the curve, we get low efficiencies very quickly. Efficiency drops off quickly. This drop off coincides with the output voltage deviating substantially from the ideal case, the one over D prime curve. So, we can take our volt second balance and charge balance equations and view them as the loop and node equations of our equivalent circuit model. Therefore, we can reconstruct loop equations that go with our volt second balance equations and reconstruct equivalent circuits that go with those. We can also reconstruct equivalent circuits based on node equations that correspond to the capacitor charge balance. Finally, we can combine these using ideal DC transformers to get a final model. This final model has a physical interpretation. We can plug this into a larger system, and then model how the converter works in the larger system, and we can solve it for things such as the efficiency or do manipulations on the circuit to find things like the voltages and currents of the converter.