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.