[BLANK_AUDIO]. We'll now briefly discuss two more, finer points. the first one, is an example of a higher order power converter. So here in section 2.4, the slides contain the full analysis of this particular converter. It's an example of a converter that can contains two inductors and two capacitors. It's named after its inventor, Dr. Slobodan Cuk, and it's called a Cuk converter. [COUGH], and so here, here's the converter circuit, with 2 L's and 2 C's. Here is how the switches can be realized with a transistor and diode. Okay? so the same analysis approaches that we've discussed so far, can be applied to this converter as well. but you, you simply have to realize that since we have two inductors, we need to apply volt second balance to each inductor. And with two capacitors, we need to apply charge balance to each capacitor. so, this will give us four equations. Two volt second balance equations, and two charge balance equations. And there are four things to solve. The two capacitor DC voltages, and the two Inductor DC currents. So we use the small ripple approximation on each inductor current, and each capacitor voltage. We draw their waveforms and apply the volt second balance of charge balance, to get four equations. Now I'm not going to go through them all but for those of you who are interested, you can look through these slides for this section and go through the the volt second balance and charge balance equations. The result of all of that, is shown here where we have volt second balance on the two conductors, and charge balance on the two capacitors. So we have a total of four equations. We have two unknowns, or four unknowns. The two capacitor voltages, V1 and V2, and the two inductor currents I1 and I2. So we can solve this system of equations, and if we do, we get this solution for the DC voltages and currents of the converter. So the point of this example then, is that you can apply the volt second balance and charge balance techniques to more complex converters. We have to apply volt second balance on every inductor, and charge balance on every capacitor to get a system of, of equations, that can be solved then for the, the converter steady state. Here incidentally is the result. This is the conversion ratio of the converter. V2 coincides with the output voltage, and it turns out to have, that, this converter has an inverting buck boost type characteristic. We can also work out the ripples, just as in the single inductor single capacitor case. To find equations that help us chose values that the inductor, the inductances and capacitances. Okay? one other fine point, is a case such as in the output filter of the Buck converter, where we want to calculate the output voltage ripple on the output capacitor, and choose a value for that capacitance. Okay? this is the case where there is no switching associated with the output node where the capacitor is connected. Instead, we have a two pole LC filter and the current in the capacitor has ripple only. And so, what we've talked about so far to find Delta V on this capacitor, using a small ripple approximation doesn't work. To illustrate that, let's look at what is the capacitor current. Well, it's the same for both intervals. For both the first and second intervals, we have the same node equation at the output node. Which says that the capacitor current, i c, is the inductor current, i l, minus the load current, v c over R. So it doesn't switch, and so IL is the current coming out of the inductor. The inductor current, as we've drawn looks like this it has a DC component and it has some switching ripple. So first question is, where does the DC component of the inductor current go? Well, it can't go through the capacitor by charge balance. The DC component of capacitor current is 0, so all of the DC component of inductor current, in steady state at least, must flow through the load. Okay? so the DC component goes this way through the load. Now how about the ah,ripple? Well in a well design filter, where the appli capacitor does some filtering, and then it's impedance is low at high frequency. We put a large enough capacitor, so that it has a very low impedance at the switching frequency in its harmonics. And so the switching ripple coming out of the inductor, which is this triangular component of the conductor current. That switching ripple, will divide between the impedances of the two branches here, between the capacitor branch and the load branch. And in a well designed filter, the capacitor will have a low enough impedance that nearly all of the ripple will flow through the capacitor, and very little will go to the load, causing ripple in the load. So a good approximation is to say simply, that all of the inductor current ripple flows through the capacitor. So let's make that approximation. And in that case, the capacitor current looks like this. It's simply the, the same as the inductor current ripple, but with no DC component. Okay now, first of all, suppose we make the small ripple approximation on the inductor current. [COUGH] What that does is ignore the ripple, and in that case the capacitor current is zero. Which then predicts that there's no voltage ripple. So this is not a useful approximation in this case, because the ripple is the only thing, the current ripple is the only thing that makes voltage ripple in the capacitor. So we have to include the capacitor, current ripple. And calculate the resulting capacitor voltage ripple. Okay? We can to that using the defining equation of the capacitor, Ic is C times dvc dt. So, dVc/dt is ic over C, and that means that whenever the capacitor current is positive, we charge up the capacitor and increase it's voltage. So here you can see, for this time as drawn, we have positive capacitor current. Positive capacitor current, means the slope of the voltage is positive, and the capacitor voltage increases. In fact, at the 0 crossing of the current, that's where the derivative of the voltage is 0, so those are the extreme of the voltage waveform. And between the, this minimum and maximum of the voltage, we have positive current that makes the voltage increase. So, over this time the voltage increases from here to here. Okay? because of symmetry, the voltage waveform is symmetrical, and we can again define the ripple Delta V as being the peak to average, and we have one Delta V there, and another Delta V here. So over this interval, the capacitor voltage increases by a net 2 delta V. Okay. We can find the the relationship between this change. This 2 delta V. And the total charge contained in the current waveform, using the defining equation for the capacitor of q equals Cv. And in this case over this interval, where we're charging and increasing the voltage, the net change in voltage again is 2 Delta V, and that's produced by a total charge that is the area under the current curve. Total charge is the integral of the current. So this total charge q increases the voltage by 2 Delta v. And we can then apply q equals C times 2 Delta v, to find the relation, or to find the delta v. All that's left then, is to calculate the area of this triangle, or the total charge. So the area of the triangle, we can use the area formula for the triangle, one half the base times the height. The the height of the triangle is the inductor current ripple delta il. And the base is this distance, which by symmetry is half of the switching period. So the total charge then, is given by this equation from the triangle formula, the triangle area formula. We can plug that in to our q equals C v formula, and solve for Delta v and we get this expression for the the Delta v on the capacitor. And again I would point out, that if we have a capacitor with significant equivalent series resistance, we must also add the, the voltage drop and the equivalent resistance to, to this, to get the total voltage ripple on the capacitor. So, in cases where we have two pole output filters, or a higher order output filters, we can't ignore the ripple. And so we have to use arguments such as this, in order to calculate the ripple.