Having discuessed how to realize switches using semi conducter devices and to implement things such as single quadrant or two quadrant switches in a switching converter We're now in a position to discuss what, what is known as the discontinuous conduction mode. And chapter five covers the first the origin of the discontinuous conduction mode and then how to solve it, and we'll do some examples of solving Bach and boost converters that operate in this discontinuous conduction mode. The converters that we've discussed so far in this course operate in what is called the continuous conduction mode by contrast to the mode of operation we're going to discuss, now. the discontinuous conduction mode occurs because there is a ripple in the current or voltage waveform that is applied to one or more of the switches inside the converter, and this ripple is larger than the DC component. and as a result, the, the current or the voltage attempts to reverse through the switch. And now, we've realized the switch to be a unidirectional switch, and we try then to reverse the polarity. the switch doesn't behave in a way that it was intended, and so it will switch on or off at a time that does not coincide with the driver signal switching, and so we get additional intervals or sub-intervals within the switching period as a result. And this completely changes the characteristics of the converter. The most traditional, discontinuous mode of operation occurs because the inductor current ripple is greater than the DC component of current, and this causes the current through the switches to try to reverse direction. the diode won't allow it. The diode will turn off, instead. And, we get a third period during the switching period, and this changes the characteristics of the converter substantially, so when the discontinuous mode happens, what we're going to find is that the output voltage now becomes load-dependent. In continuous mode, say for a buck converter, the output voltage is the duty cycle times the input voltage. Any dependence on load is small and only comes from loss at the effects of loss in the converters. in the discontinuous mode though, we have a first order dependence of the output voltage on the load current. so the properties of the converter change, drastically. We then all of a sudden, we have a high output impedance. We, we're going to find that the, or it turns out that the dynamics of the converter that we're going to discuss in several weeks, those dynamics change very substantially. And in fact, they're simpler in discontinuous mode. we also find that when you remove the load, the discontinuous mode causes the output voltage to not be controllable, and it can do bad things, so we have to account for that. Some converters are designed on purpose to always work in discontinuous mode. One of the good things about it is that the current goes to zero before the end of the switching period, the diode turns off, and then there's no reverse recovery when the MOSFET is next turn on. So there can be less switching loss. But on the other hand, there are higher p currents, and so we can get more conduction loss, and there's a trade off then. even if we don't intentionally design a converter to work in this mode many converters will operate in discontinuous mode at low output power, and so we have to be able to analyze and know what's going to happen then at those operating points. So what we're going to do in this lecture is discuss the origins of the mode and find the mode boundaries, the conditions under which we operate and continue as our discontinuous mode, and then in the next several lectures we will analyze converters to find their output voltages and other things when they operate in the discontinuous mode. So, here's an example. We have a, a basic buck converter, and we've realized the switches with conventional single-quadrant switches of a transistor and a diode. here is what the inductor current wave form looks like in the continuous mode. And it's the wave part we've been drawing all along this in this class, so it has a DC component that we're labeling capital I plus it has some switching ripple that has a peak magnitude of delta I. Kay? We previously analyzed this circuit, and we know how to calculate capital I and delta I now. here are the expressions. So the DC component, capital I, is the load current, V over R. And delta I, the, the, peak to average ripple is the slope times the time, which we can write, in terms of VG, like this. and the ripple depends on the duty cycle, VG, the inductance and the switching period. Okay, the interesting thing about this is that the DC component depends on the load resistance, or on the, the current that the load decides to draw, but the ripple doesn't. Ripple depends on all kinds of things, but the one thing it doesn't depend on is the low resistance or the load current. So what happens if we say increase the value of' R or our all of a sudden decides it doesn't need much current? Well, that will make the dc component capital I go down, but it won't change the ripple. what's important here is actually the effect of that on a diode, so when the diode is on it, conducts the inductor current, and the diode current wave form looks like this, so when the diode is on, it has the same capital I plus the ripple. And the minimum diode current happens right there at the end of the switching period. And that minimum value is the DC component of current minus the ripple. Kay? So, the diode current has to stay positive for the diode to work. And, when we account for the ripple, the, the, minimum diode current is actually less than capital I. So, let's consider increasing the value of R to the point where the ripple equals capital I. Actually, we reduce capital I, so it's equal to the ripple. And the inductor current looks like this, and the diode current follows the inductor current for the second half of the period. So you can see that the diode's current starts out at I plus delta i, and it ends I minus delta i which for this particular load current is equal to zero. Okay, what happens if we increase the load resistance even more? Well, here is that case. DC component of lo, load current and inductor current, capital I, is now less than delta i. So even though the average current and the load current are positive, they're less than delta i, and so the diode current and inductor current go to zero before the end of the switching period. Okay, once that happens, the diode will turn off, and it will not allow the inductor current to continue in the same direction and go negative. So we get a third interval now this, from the point the diode turns off until the end of the switching period where, the diode and the MOSFET are off and the inductor current just sits at zero current. So now, our switching period is divided into 3 intervals. There's the D1 interval, the first interval, which we're also going to call DTS. It's the, D is the tran-, the MOSFET duty cycle, and D1 is the duty cycle of the first interval. Then they're the same here. we have a D2 interval now, when the diode conducts. And we have a D3 interval when nobody conducts. So, this is the discontinuous mode of operation, and this extra interval changes all the equations of the converter. So first of all, we need to be able to write the equations of the mode boundary. And we now, we have a we have a way to do that, that the mode boundary happens when delta i is equal to the DC component capital I. Kay? and in fact, the conditions then are that we're in DCM, or discontinuous mode, when, when capital I is less than delta I. Now, at the boundary, the continuous mode equations are still valid, so we can plug the continuous mode equations that we had for capital I and Delta I into these equations to find the boundary. So here's, here's the expression for capital I in terms of VG. Here's the expressions for the delta I. And so we can plug them into this. And we can solve, so what, the three VG's cancel out. we can move the two l and t s over to the left hand side. And cancel the d's it looks like. And we'll get this expression. So this is the expression or the condition for operation in the discontinuous mode. Kay, the quantity on the left-hand side is a function of element values in the switching period. it's traditional to call this quantity K, and so in the power electronics business, we generally call capital K is two L over RTS. And this two L over RTS factor keeps coming up in many of our equations, over and over from here on. the right hand side is the critical value of K at the boundary between modes, so we call that K crit. so the right hand side is a function of duty cycle. It will vary with duty cycle. but, when K is equal to K crit, then we're at the mode boundary, and if K is less than K crit, then we operate in the discontinuous conduction mode. For the buck converter, we, we find here that K crit is equal to D prime. For different converters, they'll have, K crit will be a different function of duty cycle. But in general, we can write this expression for the mode boundary, in some equation of the form K less than K crit of D for discontinuous mode. So here's a plot of that. K crit for the buck converter is one minus D or D prime. So here's a plot of K crit versus D. So K crit is one at D of zero. K crit is zero at D of one, and it looks like that. Okay, so we compare that to K. So suppose K is less than one. But the, the maximum value of K crit is one, so K is less than one, says some value like here. Then we'll be in discontinuous mode over this range where K is less than K crit, and then at higher duty cycles over this range we'll be in continuous mode. On the right side here is another example where K is bigger than one, and in that case, K is always greater than K crit, so we're always in continuous mode then. For the different converters Buck, Boost and Buck-Boost it with their K crit expressions labeled so, K crit for the Buck is D prime. We're going to show in an upcoming lecture, that K crit for the Boost is D times D prime squared, and on the homework, you're going to work out for the buck boost, that K crit is D prime squared. for the buck, the maximum value of K crit for any D between zero and one is one. So if K is greater than one, then we're always in continuous mode. But if K is less than one, then we have to worry about discontinuous mode. For the boost, this, this maximum value turns out to be four twenty-sevenths. We'll discuss that. it's an interesting number that comes out of nature here. the boost is less likely to run in discontinuous mode, but if K is smaller than four twenty-sevenths, then it can. for the buck boost, the maximum value of K crit is one. So again, the K less than one, we may run into discontinuous mode. Okay, so that's the mode boundary in the next lecture. We're going to solve for the output voltage.
