In this lecture, we'll talk about the flyback converter, which is another very popular form of transformer-isolated converter. The flyback converter is a transformer isolated version of the buck boost converter. Now, it turns out that in the buck boost converter, it is actually easy to add a transformer. Here's a diagram of the buck boost. And have a point in the middle of the buck boost converter where there is already an inductor. And you could simply place a transformer at this point, and the volt seconds will balance, because they already balance on the inductor. In fact, what we do in the flyback is, is we turn the inductor into a transformer. it's actually, we often call it a flyback transformer. And it functions, in fact, more like a two winding inductor. So here is a way to think about it. suppose we're in the lab winding an inductor with, say, some length of wire that we wind some number of turns on a magnetic cord to make an inductor. And suppose instead of just wounding one wire, we wound two. So perhaps it's like we needed a fatter wire or a thicker wire to have lower resistance. And we decided to just take two wire and put them together as if they were one and wind them together. The same number of turns Okay. So I can dry, draw that this way, as is shown in this lower diagram, and it actually looks like a one to one transformer. The inductor though, what, what we were calling the inductor in our original buck boost converter We might call now as the magnetizing inductance of this transformer. So it's like we have a trans, we have an inductor and then we have two, a two winding ideal transformer. Now so far we really haven't changed anything, it looks more complicated, but it's, it's really just functioning as an inductor. With the inductor current splitting in some way between the two windings. What we can do, though, in this case now, is to now break this connection. This changes how the current flows through the two windings, although it doesn't change the sum of the current. So if we have current in this winding, call it i 1. And a current in this winding, we call i 2. The sum of those currents adds up to the original i, of the buck-boost converter. But the difference is, that now we have i 1 flowing when q 1 is on N i 2 flows when d 1 is on. So i is distributing between the two windings in a different way. Now, the way we normally draw the flyback converter is, is the bottom diagram, and it's nearly the same as the one above it To get a positive output voltage, what I've done is reverse the polarity. I rev, reversed the dots of the winding. Recall that the buck boost converter inverts the polarity. If we want our flyback converter to make a positive voltage, then we must reverse the dots, and we reverse the dio-direction also so we get a positive output. And then the other we can do if we like is now apply a turns ratio. And the turns ratio will let us scale the voltage up and down. And this is the conventional flyback converter. There's one other small change. The border of key 1 and the primary winding are reversed here. We usually do that to put the source of the transistor at ground and make it easy to drive, with just a regular ground reference driver. So the flyback converter is a buck boost converter, in which we realize the inductor using two windings, that effectively give us isolation and a turns ratio. The flyback transformer then, which is this two winding magnetic device, can be modeled in the same way as the usual transformer, with an ideal transformer and an magnetizing inductance put in parallel. But now we're using the magnetizing inductance as the actual physical inductor in our converter. And this inductor is designed to store energy. Not only that but the winding currents look very different than the way they look in an ideal transformer. We never actually have current flowing in both windings. There's either current flowing in the primary when the MOSFET is on, or in the secondary when the diode is on. But we don't have the case that the secondary current equals the turns ratio times the primary current. So the magnetizing inductance is very significant. Its current is large and as a result we don't see what we would normally expect in an ideal transformer. Far from it. we won't talk about magnetics design in this Coursera course, but I will say that the design of the flyback transformer is very different from the design of a more ideal transformer such as the one that is in the forward converter. We put an air gap in this transformer that's, lets our inductance store substantial energy. And its behavior, as I mentioned, is different. Okay, [COUGH] having said that, we can still model our converter in the usual way. We replace the transformer with our transformer model, having a magnetizing inductance. We then solve the converter, apply volt second balance to the magnetizing inductance, charge balance to the capacitor, and so on. And we can get the equations of the converter. So here's what happens during subinterval 1. In subinterval 1, the MOSFET is on The diode is off as is usual in the buck boost converter. And with the MOSFET turned on, then we have the magnetizing inductance connected to the g. And so, we will have what vl is equal to vg, and we have current flowing around this path. The capacitor current is simply minus the load current or minus v over r, and the input current ig is equal to the magnetizing current i. We'll assume we're in continuous conduction mode here and apply the small ripple approximation to v and i And get these equations for the first interval. For the second interval, the moss fit is off and the diode is on, we now get this circuit, with the diode connecting the transformer secondary to the output. With the MOSFET on the magnetizing current cannot flow through Vg anymore. in fact the magnetizing current or the inductor current will flow out the secondary. In our transformer equivalent circuit model, the way this works is that the inductor current or the magnetizing current continues to flow and it flows around this way. Around this path. So it will flow through the ideal transformer and come out the dot on the primary side. So on the secondary side, we have current going into the dot, equal to this primary current or i, divided by the turns ratio n. So now the capacitor current for this interval is that secondary current i over n minus the load current V over R, and we get this equation. The voltage on the magnetizing inductance, VL is in fact the reflected secondary voltage, with the diode on, we have voltage V, applied across the secondary. And that voltage is negative at the dot of the secondary. So it will be negative at the dot on the primary, and the primary winding voltage will be V divided by the turns ratio n. So the inductor voltage is minus V over n for the first interval. With the MOSFET off, i g is equal to 0. And so these are the equations for the second interval. We can again apply sm, the small ripple approximation to replace v and i with their DC components, capital V and capital I. So then here are those waveforms. Here's our v l waveform, our i c waveform, and our i g waveform. So now we can apply volt-second balance to v l. So the average v l is d times this voltage plus d prime times that voltage, as is shown here. Volt second balances that that must be zero. We can solve this then for the output voltage, so what do we get? V would be Vg times D over D prime times the turns ratio n. 'Kay? So, the output voltage, is the input voltage times this buck-boost conversion ratio d over d prime, and then also times the turns ratio n. We can apply charge balance to the capacitor, so the average capacitor current will be D times the value during the first interval, plus D prime times this second interval value, which gives us this equation. We can solve that for the inductor current I, or magnetizing current. And what we find in that case, is that the magnetizing current, referred to the primary side, is V over D prime R, and then times the turns ratio n. Finally, we can write the equation of I G. The average I G is D times the value during the first interval, I, plus D prime times 0. So we get the average I G is D times I. So those are the equations of our DC model of the flyback. As we've done in previous chapters, we can construct equivalent circuit models to go with them. This first one, from the inductor, gives us this center loop equation. The second one from capacitor charge balance gives us the output side. And the third equation for the average I G gives us the input port model. So we get this circuit. And finally, we can replace the two, the ideal sources with transformers. And get an equivalent circuit for the flyback. The ideal flyback has a buck type transformer, and then it's followed by a boost tra, type transformer. And the boost transformer additionally has a turns ratio. So in the end, this is a buck boost converter with an added turns ratio And the flyback then has its origins as you would expect in the buck boost. So the flyback is a widely used and popular converter because it has a very low parts count. It's actually known to be a cheap and inexpensive way with a very small number of parts, to build a converter. you can get multiple outputs for very low cost as well, all we have to do is build another output. So we could, for example, have our flyback converter with What transistor, a single magnetic device, flyback transformer, and then one diode and one capacitor for each output. And get multiple output voltages, with different turns ratios, and, when very low parts count small converter get a lot of different voltages. One problem that is known with the flyback converter, is a problem called cross regulation. If we have say n 2 turns on this, this output and n 3 turns on this output With their respective voltages V2 and V3, we would hope that V3 would be the turns ratio n3 over n2 times V2. That's what we would hope, and if we have good cross regulation, then that's what happens. It turns out with the flyback converter, second order effects related to the leakage inductances, these small inductances of these transformers, cause the output voltages to differ substantially. And it can make the output voltages highly dependent on load current. So the cross-regulation between the two outputs is not very good. And we usually have to do some more work and more engineering to get a good design. So it's inexpensive but the performance in that sense is not as good. A flyback is often operated in disconinuous conduction mode. That makes the magnetizing inducatance very small and this transformer then can be very small. And so we not only have small parts count, but we have small magnetics, also. well, you can do an analysis of this converter in discontinuous mode and it turns out to be, to, to give the discontinuous mode buck-boost equations With the an added terms ratio.