In this lecture we'll discuss what is known as the
canonical circuit model.
All PWM, continuous conduction mode DCDC converters
perform the same basic functions, and as a result, their equivalent circuit models have the same
general form. And the canonical circuit model is a way
to represent that general form. So basically you can use the canonical
circuit model. For any converter, and then we just plug
in the specific element values for which ever converter you're using.
one of the uses of the canonical model is that we can solve larger system
problems in a generic sense for, for any converter.
And then plug in the detail parameter values for whichever converter we're using
to apply it to our particular situation. The canonical circuit model also is
actually a way to solve the circuit for the transfer functions through circuit
manipulations rather then doing algebra on paper, and so I'm
going to illustrate how to do that also. So these basic functions are this. first of all the, the DC-DC converter
transforms the voltage in current levels, ideally
with 100% efficiency. And therefore we expect the model to contain DC transformers that represent
that. We have low-pass filtering of the waveforms from the inductors and
capacitors that are really there to filter the ripple, but they also filter
signals at other frequencies as well. So the, the v hat and d hat variations get filtered by the L's and C's in the
converter, also. we can control the, the circuit by
variation of the duty cycle. And so the canonical model has sources
that depend on the duty cycle variations, D-hat. So we have these independent sources in
the model of any converter. And therefore all of these elements should
be present in any of our converter equivalent
circuits. So here is a, a development of the canonical model, based on simple
physical principles. So again, we said this first order
property that we're trying to obtain is the conversion ratio limb of d, with
this effective dc transformer function. And so we will start our converter model by building it up beginning with this DC
transformer. So the effective turns ratio is the converter conversion
ratio M of D this conversion ratio is a function of the
quiesce and duty cycle capital D, so that it c-, will vary if the steady
state duty cycle varies. Here I've also illustrated the DC input
voltage, capital v g, and the DC output voltage, capital v. So, solution of this model at this point says that capital V equals M of
capital D times capital VG. The next thing we'll do is, is introduce
variations in VG. So our power input source now, we will
write as VG plus VG hat. So we can have variations in VG. And to first order what happens is that we
would expect those variations to pass through the same conversion ratio and
cause a vhat variations in the output. At this point what I'm doing is drawing
the ideal transformer symbol containing both a DC
line so that it passes DC. And an AC sig-, line in it so that we are
now modeling how it passes the AC variations as well.
So effectively, what we've done here is
combine both the DC and AC quantities into a single equivalent circuit model, which
really combines the, the DC models we talked about in previous
weeks. With this week's AC models. The third step is introdu-, introduction
of an effective low pass filter, so the, the inductors and
capacitors that filter the ripple [COUGH] also filter the variations that come from
VG HAT. In the canonical model what we do is we
push all the inductors and capacitors to the output
side of the equivalent circuit. And so we have some effective low pass
filter there. It may turn out that the, the element
values in a low-pass filter are not the same as the actual values in a converter.
We're going to see in the example in a couple of minutes that we push an
inductor through the transformer turns ratio and the inductor then becomes a function of D in this effective low-pass
filter. So it is an effective low-pass filter that
does depend on the values of L and C but it's not. the values may not be exactly L and C. The next step is to introduce variations
in the du-, duty cycle, or the control input. So now we will represent the duty cycle as
capital D plus D hat. Capital D effects the conversion ratio or
transformer turns ratio, whereas the d hat causes introduction of
control sources. So we have some independent sources of d
hat sources, in the canonical model there are two
sources one is a voltage source. A value E of SD hat and the other is a
current source of value J of SD hat. In the canonical model we push all of the independent sources to the left side of
the model. So we have both the VG sources and the
duty cycle sources on the left. Okay we can talk about transfer functions
that are predicted by the canonical model. If we look at this circuit as its drawn
right now we have an output voltage, the AC
output V hat. And it is the function of the two
independent inputs, the VG hat input and the D hat
input. So in general, we can express the output
as a superposition of terms that come from the
two different inputs. So there are some coefficients when we
apply superposition. The coefficient of VG hat we call the
transfer function gvg of s. And the coefficient of D hat we're going
to call the transfer function gvd of s. And to find one of these transfer functions you set the opposite source to
zero. So for example to find gvg what we do is this is the transfer
function from vg hat to v hat. And it's found under the conditions that
the other input or d hat is set to zero. Okay? So in order to calculate that we can just
look at the circuit and see what the canonical model predicts.
If we set d hat to zero that will make this d hat voltage source into a short,
setting d hat to zero makes the current source into
an open. And then we have the vg hat source
connected up to the primary of the transformer. I would also note that to find AC transfer functions, we set DC quantities to zero,
or DC biases. So the capital VG will be zero. Capital V will therefore be zero also.
This doesn't mean that cat-, the DC component of EG is zero in
the actual converter. It simply means that in order to solve for this transfer function, we set the DC
quantities in our model to zero. Okay so with vg hat applied across the
primary of the transformer, the voltage at the secondary side would be
vg hat multiplied by the turns ratio m. That voltage gets multiplied by the
transfer function of our low-pass filter, so that the voltage at the output will be n times
that transfer function times vg hat. So then this transfer function from vg hat
to v hat is n times he of s, like this. Likewise, to find the control to output
transfer function, gbd of s, this is found as the transfer
function from d hat to v hat, under the conditions that we set the other source, vg hat, to
zero. So, let's set vg hat to zero here, and
capital VG to zero. Vg hat is set to zero, so we get a short
circuit, at the input of our converter. That shorts out the current source. And therefore, the voltage across the
input of the transformer is e of s times d hat. The voltage coming out of the transformer
then will be the turns ratio m times ed hat. And the voltage at the output of our
converter then would be this quantity multiplied by the effect
of low-pass filter transfer function. So I get m, or he of s, times m of d,
times e of s, times d hat. [COUGH]. So this is the controlled throughout the
transfer function of gvd, as shown here. We can solve other quantities here if we
want as well. For example, we could find the input
impedance or the output impedance and we'll be talking
about those later. OK, here is the buck boost small signal
model that we previously derived. Let's try to manipulate this into
canonical form. So in order to do that, we need to push
all the d hat sources or independent sources to the
left, to the input side of, of our model. And we need to push the L's and C's to the
right, and then finally in the middle we need to
combine the transformers. So lets do that. I'm going to do it one step at a time. And what we'll do is we'll take this current source and push it through the
transformer. When we do that since the dots are
reversed on this transformer, it reverses the
polarity, so it will point down. And then, we also have to divide by d
prime, this turns ratio, so that on the primary
side of this transformer, we get i d hat over d
prime. I'm also going to push this voltage source
through to here. Okay? The voltage source is in series with the
inductor. We can write them in any order so, we can just push this straight through the
inductor. And then when we push it through the turns
ratio of this transformer we'll have to what,
divide by d. So this voltage source would be vg minus v
over d times d hat. OK, that's shown on this slide, here's the current source, and here's the voltage
source. The next thing to do is we need to push
this current source past this inductor. Now that's maybe not so obvious how to do
it, but here's the trick. What we do is break this, this path and
connect the current source there. Now you can't just do that. what, in order to compensate for
connecting the current there instead of here, we need to put
another current to compensate. And basically we're taking this current
and putting into this node where it doesn't
belong. So we need to take the correct back out of
the node and put it where it does belong. And this current source then is the same
value as that one, it's capital ID hat over d
prime. So I've shown that right here. My claim then is that we're allowed to do
this because the loop and node equations of
this circuit are unchanged. Since we put the current into the node and
then take it back out again we effectively don't
change the node equation there. Now what do we do with this? We can take this second current source and
push it through the transformer to here. And what? In the process of doing that we would
multiply by d, the turns ratio. So this becomes d over d prime times
capital I times d hat. Okay, so that takes care of that one.
What do we do with this current source? Well, there's a trick for that also. What we have to do is make a thevenin
equivalent of this inductor in parallel with this current, so we have basically
this between these terminals here.
And, let's see. We have a impedance of SL, and we have a current going that way of id hat over D
prime. So this is a, we can view this as a Norton
equivalent type circuit. We need to turn it into a Thevenin
equivalent circuit. So it would be an impedance in series with
a voltage source. And the voltage source would be the open
circuit voltage between these terminals. So it would be what? Plus on the left, minus on the right, and
it will have a value equal to the current
times the impedance. So sL times capital I over D prime times d
hat. So this is a thevenin equivalent model for
this network, and we can place, put that in, in place of
this. So that's what we have here. Here's the Thevenin equivalent voltage and
inductance. And looks like I haven't pushed the current source through the transformer
yet. Next step, we'll push this current source through the transformer, as I described
previously. And once we get it here, we will further
simplify by trying to push it past this voltage source.
So that's on the next slide. Here is the current source written there. And to push it past the voltage source, we do the same trick that we did with the
inductor. We're going to break this connection.
Put the current source into this node. And then take it right back out again and put the current into the node where it
belongs. Okay, so this further simplifies the
circuit. We now have two current sources in
parallel that we can add. So we make a composite source that would
be, id hat times what, one plus d over d
prime. And then as for these, we have a voltage source in parallel with
a current source between these terminals. Okay, what, what is the IV characteristic
between these terminals of the voltage source in parallel to the
current source. Well in this case the voltage source wins and the current source makes no
difference. The independent voltage source holds the
voltage at, between these terminals to be the given value,
regardless of the current. So in fact, you can just ignore the
current source. Okay. Last step, we will push the inductor
through the transformer. We'll push this voltage source to this
transformer. When the inductor gets pushed through the transformer, we
have to divide by the terms ratio squared or D prime
squared. So we get this. Here's the effective low-pass filter, has
the inductance over D prime squared, plus the
capacitor. The two transformers are now combined. And the sources are combined as well. So we had, we got two voltage sources that I then combined and we have a current
source. Okay, an interesting thing happened when
we combined the two voltage sources. One of the voltage sources depended on the
inductor impedance and it picks up an s times L factor inside
it. So the coefficient of d hat is now
frequency dependent, and it has terms that depend on
s. In general, the e of s and j of s coefficients of the d hat sources can be
functions of s. And in this case, we picked up the factor of s by pushing the current source past
the inductor. Here is that e of s coefficient for the e
of s d hat voltage source. We can actually use the steady state
relationships for the, this converter. For example, v is, the output voltage. as a function of the input voltage, vg,
and duty cycle, and our expression for the inductor current,
capital I, can be simplified from the steady state
model. And when we do that, we can then express e
of s in this form. Okay?
This frequency dependence is actually causes control problems and we're going to
talk about those in the next chapter. So here then is the canonical model for the basic buck, boost and buck-boost
converters. The effective low-pass filter contains an
inductor and capacitor. Although the value of inductance can be a
function of duty cycle, and we can get the d hat sources and the m of
d transformer. And here's a handy table that gives the results, so the element values in the
canonical model are listed here. And if you want to apply or find the AC
model for one of these converters you can then, the work is done for you you can just simply plug into this table to get
your model.