Last time we talked about how to model a transformer, and the first order model that included an ideal transformer with a magnetizing inductance in parallel with one of the windings.
The magnetizing inductance models magnetization of the core, and it
operates like a real inductor in that we have to have volt second balance
on this magnetizing inductance. So today we're going to talk about how to put a transformer into a buck
converter, and one of the common well-known and
commercially important transformer isolated versions of the buck
converter that is called the forward converter. Now putting an isolation transformer into
a buck converter, and getting the volt seconds to balance on the magnetizing
inductance, is not so easy to do. Here's a buck converter, I've just drawn
one, and let's think about, you know, where,
where could we perhaps break the circuit and
insert an ideal, or insert an actual isolation
transformer. So we want to put a transformer somewhere in the
circuit. So perhaps we could break these wires and
put the transformer here, or we might try putting it over here
perhaps, breaking the wires right there, or perhaps we could try
breaking the wires right here. And the question is, do any of these work? And could you put the transformer in any
of these points and get the volt seconds to
balance? Okay, well if we put the transformer in
position one, we're connecting the primary of the transformer straight up
to Vg, which is DC. So we would be applying a DC voltage to
the primary winding of the transformer and the volt seconds
won't balance, so that's no good. Likewise, at position two, we would be
connecting the second, second area of our transformer up
to the output voltage, V, which is DC if the converter
works, and so we're putting DC on our secondary.
and so that also applies a DC voltage across the transformer, the volt seconds
won't balance, and that's no good. Okay, how about position three, now this
point three? Well, that's a little trickier, but that
doesn't work either, because in fact if volt seconds balance on
this inductor, then the average voltage across the, this filter inductor is zero.
And so the DC at this point, at point three,
is the same as the DC at point two. And in fact, the voltage, the switch voltage at this point switches from zero
to Vg and back, but it never goes negative,
so the average is positive, and the volt seconds don't balance there either. So none of these places work, and it's
just not so easy to do. So we have to go to more work, and we
actually have to add extra circuitry to make the volt
seconds balance and reset the transformer in any of these buck derived transformer
isolated converters. And so there are quite a few different
schemes and different ways to do that. What we're going to talk about now, is one
of the more popular ways that is called the
forward converter. Here's a schematic of the forward
converter. We have the transformer placed in series with Q1 and it's output or secondary
winding, which is actually winding three in this
diagram goes through some diodes to the output
terminals. And the voltage right here is the same
voltage driving the, the LC output filter that we have in
the regular non-isolated buck converter. So Vs of T, actually switches high and low
with some duty cycle. It turns out to switch high to n3 over n1
Vg with a swit, with a transistor on, and it's
zero with a transistor off. and so we get a DC, a voltage that has a,
an, a DC average that depends on the duty
cycle, that gets filtered to, to produce the output voltage, that is very similar to the original non-isolated buck
converter. But then we have an extra winding here, and there's some extra means for resetting
the transformer and applying negative voltage across the
magnetizing inductance during the D prime interval to get the average
value of the transformer voltage to be zero, and
get the volt seconds to balance on the
magnetizing inductance. So we need to talk about how that works in
order to understand this converter. And there's actually different versions of
the forward converter that have different circuits to
reset the transformer. This is one of the classic ones that has a
winding, this n2 winding, is here only for resetting the transformer, but
the power actually flows through the first and third
windings. So to approach the analysis of a
transformer isolated converter, the first thing we do is replace the transformer
with the equivalent circuit model. So here is the equivalent circuit model we
discussed in the last lecture that contains a, an ideal transformer with
three windings in this case. And in parallel with the first winding
we're going to put the magnetizing inductance of value
LM, and it has a magnetizing current IM.
Okay, having that that model in the circuit, now we can proceed
and analyze the circuit in the usual way. So we will work out the voltage on the
inductor, we'll work out the voltage on the magnetizing inductance, and we'll
apply a volt second balance separately to each of
these. We also will work out the capacitor
current and apply charge balance. And in general, we will also work out the
input current, IG, in a, find its average value
as well. Okay, here are the waveforms. and it turns out that with this version of
the forward converter, the magnetizing current operates in the discontinuous conduction
mode. And it has a discontinuous mode type
waveform that starts at zero, rises to a peak value while the transistor is on. When the transistor is off, the magnetizing current decreases down to
zero, and then remains at zero for the rest of
the switching period. And in the process of going back to zero,
we say that the transformer has been reset, and it's magnetizing
current has gone back to the starting point. And the voltages that are applied to the
induct, to the magnetizing inductance that make that his happen will then have
volt second balance. Okay, so for this version of the
converter, the magnetizing inductance must operate in
discontinuous mode. The output filter inductor on the other
hand, can operate either in continuous or discontinuous mode just
like any other buck filter inductor. So, we need to write the write the circuit
for the different switching intervals and work out
the waveforms.
Okay, so let's take our equivalent circuit for the transformer and the converter, and
look at what happens during each interval. So, this is a circuit with three diodes.
Whenever we have a lot of diodes it can be tricky to figure out
which diodes conduct when. so here's the explanation.
During the first interval, the D interval, we turn the MOSFET on.
Okay, so Q1 is on, and that makes Vg appear across the
primary winding.
Okay, so here's the dot of the winding is, is at the plus of Vg, so the windings are
positive with respect to the dot.
We'll have Vg times the turns ratio appear across this n2 winding,
that's plus at the dot. And you can see then that the voltage
across this diode D1 will be Vg plus this positive trans, or winding two voltage, which is a positive voltage,
and it reverse biases D1.
So D1 is off and blocking voltage. On the third winding, we have Vg times the
turns ratio that is plus at the dot, so there's positive voltage
coming out the dot, which will turn on D2. And then we'll get positive voltage [COUGH] after D2, and that positive voltage will
turn D3 off. So, that's the situation during the first
interval, with Q1 on, D2 on, and the other two diodes are
off. Here's the circuit then. D1 is an open circuit, D3 is an open
circuit, Q1 and D2 are short circuits. Okay, what happens during this interval? Okay, first of all, we have the output
inductor current, which I'm going to call i, and it'll be approximately capital
I if we have small ripple. That current flows out of the dot of the
winding three of the transformer. There's no current in winding two because
D1 is off, and so this output current, which follows the the equations
of the ideal transformer, if we have current coming out
at the dot on the, the third winding, we must
have current going into the dot on the first
winding. Okay, re, recall that the sum of the
currents flowing into the dots times their respective turns, adds up
to zero in a transformer. So, n1i1 prime is this current flowing
into the dot, plus n2i2, which is 0. And then i3 times n3 is flowing into the dot, but the
inductor current is flowing the other direction, so we could subtract n3I, is the current flowing into the dot
of the third lining, those add up to. And so that tells us that i1 prime, the
current flowing into the ideal transformer dot primary winding, is equal to n3 over n1 I.
So basically this is the reflected output, inductor filter current is flowing this way.
Okay. The current ig during this interval is
that current in that ideal line, ideal transformer primary winding n3
over n1, I plus the magnetizing current. One other thing I'll say, is for forward converters we generally design the
transformer with no air gap and the magnetizing inductance is
very large and the magnetizing current is very
small. So in a typical forward converter, the magnetizing
converter im is much smaller than, than the reflected lode current capital I, and this ig is
approximately just n3 over n1 times I. Okay, what happens during the next
interval? In the second interval we turn Q2 off.
Okay. what happens when Q2 is off? Well the ma, magnetizing current has built up to some positive current, and even
though it's small it's, it's not zero, it's more it's positive, and it has to flow
somewhere. Before it was flowing through the, the
MOSFET, but with the MOSFET turned off it has to actually flow out of one of
the other windings. What our model here predicts is that IM will flow this
way. And it will flow through the ideal
transformer part of our primary winding model, and flow out at
the dot. So if we have positive current flowing out
of the dot on the primary, then there must be current flowing into
the dot of one of the other windings. Well it can't flow into the dot of the third winding, because that would flow
backwards through D2. So in fact, D2 gets reverse biased and it is turned off, it won't allow reverse
current. So, the only other place it can flow is
in, into the dot on the secondary winding, and so it flows
like this, and it forward biases D1. The output inductor current has to flow
somewhere also. If D2 is off, the output inductor currrent
will flow through D3. So I have D3 on. So for the second interval here's the
circuit, we have Q1 off, D2 off, D1 and D3 are on.
Okay. What happens to the transformer during
this interval is that with D, D1 turning on, the input voltage
Vg is is placed across the second winding, like this, and it is negative at the dot
instead of positive. So the voltage that we get across the magnetizing inductance refer to the
primary, is of opposite polarity, negative at the dot,
and the value will be Vg times the turns ratio. So here's what we have so far for the
magnetizing inductance, we have a positive voltage Vg applied
during the first interval by the MOSFET turning on. When D1 turns on, we have a negative voltage, Vg times the turns ratio, n1 over
n2. and with this voltage, the magnetizing
current will increase during the first interval,
up to some peak value, and then the negative
voltage during the second interval will make it
decrease. Finally, when the magnetizing current gets
to zero, D1 turns off because it won't allow
current to, to flow negative through it, and so we have a third interval where, where D1 is
off. And here is the, the cir, the schematic
during the third interval. D3 is still on, conducting the output
conductor current, but D1 now has turned off, and the
magnetizing current remains at zero for the, the rest
of the switching period. Okay, so transformer reset, again, this is the mechanism by which the volt seconds
balance. And we can analyze transformer reset by
writing the equations of volt second balance on
the transformer. So then here again was our voltage
waveform. The average primary voltage, which is the
voltage across LM will be the duty cycle D1 times positive Vg, plus the duty cycle, D2, times this
negative voltage, minus n1 over n2, Vg plus D3 times 0, for the third interval, and this must be 0 for
the converter to work. Here's that formula again.
The formula is repeated here. And we can actually solve this formula to
find how long D2 is. So if you just solve, you find that D2 is
equal to the turns ratio, n2 over n1 times D, or D1 the transistor duty
cycle. The constraint that this imposes is that
we have to reset the transformer before the end of
the switching period. D1 can't be, or D3, the length of the last
interval, can't be negative.
So we can, we can write that D3 has to be greater than or equal to 0, and D3 is
equal to 1 minus D1 minus D2. So here is that equation. And if you, if you required D3 to be
greater than 0 and you plug in our solution for D2, what
we find is that D1, or the transistor duty cycle, must be
less than or equal to this quantity that depends on the turns
ratio from n1 to n2. If we make the common, or the, the typical
choice that n1 equals n2, then this says that the duty cycle has
to be less than a half. And what's happening then is that we're
applying plus Vg during the, the first interval, we're
applying minus Vg during the second interval, so the second interval is
the same length as the first interval. And to reset before the switching period
is over, the duty cycle then has to be less than a half. If you like though, you can choose other
values of turns, and make this limit come out to a
different number. So, what happens if we let D be greater
than a half with n1 equals n2. Well, here's what it looks like. We, we, our magnetizing current won't get back to zero before the switching period
is over, so we'll start the next switching period with some positive value of
magnetizing current. And then during the next switching period,
we do it again. We don't give the transformer long enough
to reset back to zero, and we build up, we have some net increase in magnetizing current during the
next period also. And this keeps happening, every switching
period there's a net increase in magnetizing
current. And we say the transformer walks up its BH
loop, the flux in the core will increase each switching period, and eventually the transformer
will saturate. Once it saturates it turns into a short
circuit, and the next time we turn on our MOSFET with
the transformer saturated, we get a very large current run-out of Vg
that will make the converter fail. So we have to make sure to build a control
curcuit that doesn't let this happen. And it's typical to make n1 equal to n2,
and build a control circuit that limits the maximum duty cycle
to be no more than a half. Okay, we can talk about volt second
balance on the output inductor. And in fact, this is just the same analysis as in the non-isolated buck
converter. We can sketch the waveform of the voltage
across that inductor, and it looks like this.
When the MOSFET is on, we get a voltage here equal to Vg times the turns ratio
from n1 to n3 that, that comes out here. On the right hand side of the inductor,
the output voltage is V, so then the VL
voltage during the first interval will be n3 over
n1 Vg minus the output voltage. And for the remainder of the period when
the MOSFET's off, we have zero coming out of our, or across D3, and we, the voltage across VL is simply minus the
output voltage. So this is DTs here, D prime Ts is there. We can apply volt second balance in the
usual way. so D times n3 over n1, Vg minus V, plus D
prime times minus V equals 0, by volt second
balance. And you can solve for the output voltage
V, and you find that the output voltage works out to
be the duty cycle times the turns ratio times Vg.
So this is a function that looks like a buck converter
with an added turns ratio of n3 over n1. Okay. Well, if the duty cycle is limited to say,
a half, then that limits how much output
voltage you can get. Although we can change the turns ratio to
compensate for that, and we will adjust the turns
ratio to step the voltage down, or whatever we're
trying to do. We also have this other constraint though,
that the duty cycle is limited but, to this number that depends on the turns ratio of the reset winding, or the n2 turns as
well. And so one question you might ask is, well
why don't I just make n2 be small? If I do that that makes the maximum duty cycle be larger. So if, if n2 approaches a very small
number, then the maximum duty cycle approaches one, and we can get a
full range of duty cycles. Well, the problem with that comes, [COUGH] with the voltage applied to the
transistor. During the sub interval two, when we're
resetting the transformer, diode D1 is on, transistor Q1 is off, and we're applying
voltage Vg across the second winding where it's minus at the dot and
plus at the non-dot side. And the voltage we see on the first
winding then becomes what, minus at the dot, Vg times the turns
ratio, n1 over n2. So, the voltage that is applied to the
transistor Q1 during this interval, which is the
blocking voltage of the transistor, turns out to be the input
voltage Vg, plus the voltage across the, this primary
winding, n1 over n2 Vg. So we get Vg times the quantity, one plus
n1 over n2. Okay. So, if we made n too small, made this
turns ratio n1 over n2 be big, it means we're applying a much
bigger voltage across our MOSFET. So, a small n2 will reset the transformer
quickly in a short time, but it takes a lot of voltage to do that, and
that voltage gets blocked by Q1. So we end up having a very high voltage
stress on Q1, and this can be a real problem. In a power supply, for example, that must
operate off the rectified 240 volt power line that we see in much of
the world, when you rectify 240 volts, you get a
voltage of what, 240 root 2, that's something on the order of,
or something over 300 volts. If we multiply that by 2, so n1 equals n2, then we're around 700
volts across the MOSFET, and it's hard to, to buy a
MOSFET that, that, that's that large. We may end up in order to get margin, we
may have to use say, 1000 volt MOSFET, and those
MOSFETs aren't very good. So in fact we often want to go the other direction, and make n2 be large, not
small, to reduce the voltage on our MOSFET. But then that further limits the maximum
duty cycle, so we have a trade off, and the engineer has
to decide. So for the case where n1 equals n2, and D
is less than a half, then we have this factor
of two voltage. And n1 is less than n2, then D is
restricted to less than some number less than a half, but we have reduced voltage
on the MOSFET. Here's another version of the forward converter that's also popular, and it
reduces the voltage stress on the transistors, this is called the two-transistor forward
converter. We've put a transistor on each side of the transformer and we have clamping
diodes also, as shown here, and these take the place of
that second winding that did the reset. So, what happens here is during the first
interval, Q1 and Q2 are both on, both diodes are
off. Vg is applied to the transformer, and so
nVg comes out the secondary. This turns on D3, D4 is off, and we have
the usual first interval where the reflected Vg is applied
to the output filter. During this interval, magnetizing current
builds up inside the transformer, just as in the previous
circuit. Okay, during the second interval what we
do is we turn Q1 and Q2 off, this will turn D3 off and D4 on, as in the
previous circuit. And what happens to the magnetizing
current? Well, we have magnetizing current flowing
that has built up, and it wants to flow this way. it can't flow out the, the secondary
winding, because that would mean going into the dot, and
that would reverse bias D3, so instead, it flows this
way, and it actually flows through these two
primary side diodes. So Im will flow, it will keep flowing in
this direction, like this. It turns on D1 and D2, and effectively
applies Vg in the opposite direction across the transformer,
and that resets the transformer. This is actually the same V, same primary
voltage winding we would get during the second interval if, in the
previous circuit with n1 equals n2. So, these diodes turn on to reset the
transformer, the duty cycle is limited to be less than
a half, and the circuit works fine with no
extra winding. Now, a nice thing about this circuit is
that these diodes also clamp the voltage across the transistors
to be no more than Vg. So, you can see, you know, what about say,
the drain to source voltage of Q1 is here, how large
can that be? Well, the biggest it can be is if when D2 is on, that which
puts this node at zero and this node at Vg, and so Vg is
applied across Q1. Same thing happens to Q2, the highest this
node can be is Vg before D1 will turn on and clamp this
voltage to Vg. So we have Vg is the maximum voltage on
Q2, as well. So in the previous example with 350 volts applied at the input, we can use much lower voltage
MOSFETs. In fact, if we use 600 volt MOSFETs that
are very good super junction MOSFET devices we
can buy now a days, we have plenty of voltage margin,
and so we can get a good design that's, that's
very reliable. So this is a very popular circuit that is in commercial use today. So to summarize the forward converter is a transformer isolated version of the
buck-converter. We say it's a buck-derived isolated
converter. We have to do some extra work to get the
volt seconds to balance. We saw one way to do it in which we add a
second winding on the transformer that performs reset, or
another way is to add these extra diodes across the primary to reset the, the
circuit. We have two diodes on the output instead
of one, but with this extra circuitry then we can get the volt seconds to balance on the transformer, and, and get a
functioning converter. These are very popular and widely used
converters today. [BLANK_AUDIO]