In this lecture, we will discuss how to solve for the output voltage of a converter operating in discontinuous induction mode. And I'm going to explain it for the example of a butt converter. What we need to do is to extend the ideas developed in chapter two, to apply to this case where we have inductor current ripple that is large. So here's what we talked about back in chapter two at the beginning of the course. We had this technique of inductor volts second balance. Which said that in steady state the average voltage applied to an inductor is zero. And that's true in any circuit, and it doesn't matter if it's a continuous mode, or discontinuous power converter, or something else. So it still applies. And same thing with charge balance. on capacitors the, in steady state the average current through a capcitor is also zero. And we can apply this to solve our discontinuous mode converter as well, but where we have to be careful is the small ripple of approximation. So, really by definition here in discontinuous mode we have an element where the ripple is large. And we are talking in these lectures about the current example where the inductor current ripple is large compared to the DC component. So we can't approximate away the inductor current ripple now in solving our discontinuous mode converter. And we're going to have to use some other arguments in order to, to handle the inductor current and include its ripple. On the other hand, we still want the output capacitor voltage to have small ripple. And generally we're going to put a capacitor on our output that will be big enough to filter out the ripple, and have small voltage ripple at our output. So we can apply the small ripple equation to the capacitor voltage but not to the inductor current. Okay, so let's do this example of the buck converter and discontinuous mode, and apply these arguments to solve for the output voltage. Here's our buck converter then. And we now have three sub intervals. First interval where the moss fit is on gives us this circuit. For the second interval, the moss fit is turned off and the diode is on so we get this circuit with the left side of the inductor connected to ground. And then we now have a new third interval where both transistor and diode are off. The inductor current has gone to zero. Then the diode is turned off, and the left side of the inductor now is left hanging in space, not connected anywhere with zero current flowing through the inductor. So, as usual, we will write the circuit equations for each interval. So here we have for the first interval with a MOSFET on, as usual we can write the inductor voltage is Vg minus V, like this. and the capacitor current would be from the node equation, it would be the inductor current minus the load current V over R. So we get the, the capacitor current is IL minus V over R. Next we apply the small ripple approximation. So the output capacitor voltage v has small ripple, and we can replace v in these two equations with capital V. To ignore the ripple in the output. So that gives us these terms. But the inductor current has large ripple, and we can't apr, approximate IL of T by it's DC component. So we're going to just IL of T alone for now and we're going to have to handle the, the ripple in IL With some additional arguments in a couple of minutes. For subinterval two, the diode conducts the inductor current is positive, and we have this circuit with the left side of the inductor connected to ground. We can write that the inductor voltage is minus the output voltage like this. The capacitor current is again the inductor current minus the load current, like that. And we can replace V of T by the DC component, capital V using the small ripple approximation to get these equations. And again the inductor current is left as IL of T. For sub interval three, the inductor current has gone to zero, the diode is turned off, so the right-hand side of the inductor is connected just, or not connected, it's hanging in space. What is the inductor voltage during this interval? Well, the inductor voltage is the derivative of the current. We know what v l of t is L times d i l, d t. I L is zero so we get that the inductor voltage is 0 during this interval. So VL is zero, we'll just, we can simply write that. IL is also zero. The capacitor current is the inductor current minus the lode current. Although the inductor current is zero, so really all we need to say is the capacitor current is minus capital D over R. where we apply the small ripple again to be of t. So here's the inductor voltage wave form. It's a positive voltage during the first interval, a negative voltage during the second, and it's zero during the third interval. We can apply volt second balance to this waveform in the usual way. We have a third interval. The inductor voltage can be written as D1 times the voltage during the first interval. Plus D2 times the voltage during the second interval, which is minus V. Plus D3 times the voltage during the third interval, which is 0 [COUGH] and by volts second balance this is equal to 0. So we get this equation, you can solve it for the output voltage, and we get this expression for the output voltage. Now in continuous conduction mode, without losses, this equation told us what the output voltage of the buck in continuous mode was. But here we have an, an unknown. The, the duty cycle D2 is not known. D2 really depends on when the inductor current goes to zero and the diode turns off and we have to do further analysis to find D2. So, at this point, we have an equation with two unknowns, V and D2. And we need to get more equations. So let's keep going. Let's apply charge balance next. here's our node where the capacitor is connected at the output terminals of the converter. We can write the node equation at that point which for the butt converter the inductor is connected there, so we can write that the inductor current is equal to the capacitor current plus the load current. Like this. And what we want to do is apply charge balance to the capacitor. So the hard way to apply charge balance is to take the inductor current waveform that we already drew, and subtract the load current to draw the capacitor current waveform. And find its average. A little bit easier way to do it, is to look at the original node equation at this node, and compute the DC components of each term. So we can say that the DC current flowing out of the inductor into this node equals the DC component of the capacitor current, plus the DC component of the lode current. And the capacitor current by charge balance has no DC, it's zero, so you get that the average inductor current, or DC component of inductor current, equals the lode current. So all we really have to do is take the inductor current wave form that we've already drawn, find it's average value, and equate that to the lode current. I should caution you at this point this expression of finding the average inductor current and equating it to the lode is really only true in the butt converter. And it's because the inductor is what's connected to the output node where the capacitor is. We'll do another example next time with a boost converter where it's the diode that's connected to the output node instead of the inductor. And in that case we have to draw the diode current wave form and find it's DC component. Okay so here is the inductor current waveform. We'll find its DC component and, and equate it to the load. Here's the calculation for that. The DC component of inductor current is found, as usual, by averaging the inductor current over one period. So, the average is the integral divided by the period. And the integral is the area under the curve. So we need to find this area. Okay, the area in this, for this wave form, we have the area of a triangle, so we have to use the triangle formula, which would be one half the base. And what the base is D1 plus D2 times Ts. Times the height, I've labeled this p current i peak. Okay, so that's the area or the integral. If you divide by Ts we get the average value. [COUGH] Okay, and so here, here it is divided by Ts, and the other thing I've done here is, is subtr, is substitute in an expression for i peak. Here's how we find i peak. The inductor current starts the period at zero. And it starts at zero because we're in discontinuous mode, and we ended at zero in the last period. We then, we go up during the first interval with a slope given by the applied voltage over L. So Vg minus V over L was the slope of the inductor current during the first interval. Here, multiply that by the length of the first interval to get the, the change in current. And that's i peak. So we can plug that expression for i peak into here, and finally we, we get this expression for the average inductor current. and then lastly, we equate the average inductor current to the lode current, and we get this expression that comes from charge balance on the capacitor. So we have two equations. The first one was from volt second balance. The second one is from capacitor charge balance. We have two unknowns, the output voltage and D2, so we can solve. So I'm not going to go through the algebra in this lecture, I'll actually illustrate the algebra in the next example of the next lecture. But basically you solve one of these equations for D2, plug it into the other equation. You then get a quadratic expression that we have to find the roots of, and finally from doing that we can solve, and get this equation for the conversion ratio V over VG of the butt converter in discontinuous mode. This equation is a function of D1, the duty cycle of the first interval. It's also a function of k, which is the same 2L over RTs that came out of the mode boundary equation in the last lecture. Here's a plot. So this is in, which is V over VG. Versus duty cycle, and we have now two choices for in, if we're in continuous mode, when K is greater than K crit, then M is equal to D. Where, whereas if we're in discontinuous mode with K less than K crit In is given by the expression from the last slide. So we can plot this for different values of load resistance, or different values of K, and here are some different choices. So example for K equals 0.1. To plot this we first have to plot our, our K versus K crit formula. Remember K crit for the butt converter was D prime, so it looked like this. And if say we do the K equals 0.1 curve so here's 0.1 for K K is less than k crit for D from 0 all the way up to 0.9 right there. So we use the discontinuous formula for D between 0 and 0.9. And then for D between 9 and 1 we're in continuous mode, and we have to use the continuous formula. So here is the plot, computer drawn plot of the discontinuous formula going all the way upto D as. 9 after that we follow the continuous formula n equals D. And that's what happens over the whole range of, from 0 to 1 in duty cycle, for the case where K is equal to 1. One easy way to plot this is to recognize that this continuous mode makes the voltage rise. So you can actually just. Plot both of these expressions and then take the largest, and you'll get the right answer. Another thing I should point out is that what happens when you disconnect the lode? Which is, say, let R go to infinity. Okay, so k is 2L over RTs. So here with no load power or R is an open circuit or infinity. If R goes to infinity K goes to zero, and what this curve does in that case, is it actually just rises all the way to 1. Basically you get the output voltage equal to the input voltage regardless of the duty cycle. It's actually, the solution goes to 1 except right at the axis where it's, actually turns into a squared off function here. So, if say the customer disconnects the lode you will lose control of our output voltage. And we have to do something about that, generally our build a feedback loop that stops switching. Turns the converter off to keep the output voltage from rising beyond the value that we want, so that we could maintain control of the output. Generally this is a problem in discontinuous mode when k goes to zero, we tend to lose control of the output.