We'll now begin the last chapter of the
course, chapter nine on controller design. We're going to apply the models developed
in chapter seven, and the transfer functions analysis developed in
chapter eight, to the problem of designing power converter control
systems, in this chapter. So, what I want to discuss in this lecture is, the object of our feedback control
systems that. And we're going to introduce these then
around these switching converters. And then in upcoming lectures, we're going
to talk about the design of the control
system. So, to introduce the, the topic. We have, then, our switching converter
with a, an output voltage. That, for now, we're going to treat as the
output. Although we have other examples, where we're controlling other things, like
currents. But let's take the example of a, an output
voltage controller. So, the output voltage then is a function of the different independent inputs to the
converter. We have the power input Vg.
We have the duty cycle input D. And we may have other variations, such as load current variations, that also
affect the output voltage. So, just as a functional block, we can
view the switching converter for this example; as having an output that we
identify as the output voltage. And we have inputs that affect the output
voltage from Vg, from the load current, from the duty
cycle, and possibly other quantities. In the DC voltage regulator application,
then, the object is to make the output voltage be a fixed and
well-controlled value. However, the output voltage is a function
of all of these quantities, such as Vg and the
load current. So, you can't expect to just set the duty
cycle to one value and always get the desired
output voltage. So, if there are variations, for example in Vg, they will affect the output
voltage. Or variations in the load current can
affect the output voltage. And so what we want to do is adjust the duty cycle automatically to regulate
the output voltage. And we view, then, Vg and the load load
current as being disturbances. What we really like to do is make the
transfer functions from these disturbances to the output voltage, be as small as
possible. And we would like the output voltage to
follow some controlled reference, and not be a function of Vg or
the load current. Some typical examples if Vg comes from the
rectified AC power line, then that rectified voltage
has second harmonics of the power line. So, within the 50 or 60 hertz AC system, the second harmonic will be 100 hertz or
120 hertz. And so Vg may have some Vg hat variations
at 100 or 120 hertz. And those will propagate to the switching converter, according to
the line to output transfer function at 100 or 120 hertz, to produce
variations in v. So one of the classic things we want to do
with feedback, then, is to build a feedback loop that adjusts the duty cycle to remove these variations in
v. And, or another way to look at that is to
make the transfer function from Vg to Ve at 100 and 120 hertz be as small as
possible. Likewise, if there are load variations,
they affect the output voltage through the converter
output impedance. And what we would like to do then is make the output impedance of the converter be
as small as possible. Load current variations may also have step
changes that can be very significant. And our feedback loop can affect the
output voltage response to, say, a step change in current. Circuit elements may have tolerances. There may be variations in component
values. And we would like the output voltage to
stay regulated and not be a function of those variations
or tolerances. So all of these are reasons to add
feedback to a power converter system, and we invariably have feedback in practically every power
converter. Here is a classic feedback system then. A simple negative feedback controller
where we sense the output voltage through some
sensor gain. Often this sensor is just a voltage
divider that may divide down the output voltage to an, to
value that is more convenient to work with in our
control circuitry. But we get a voltage signal, h times v then, that is proportional to the output
voltage. In with a negative feedback system, we
compare that sensed output voltage to a reference, and this reference
voltage is well controlled. We will go to some lengths to build an
accurate reference voltage that doesn't have variations due to temperatures or initial tolerances or
aging. So we have a well known reference signal. We compare the fed back output to that
reference and find what's called an Error Signal that tells
us how much the output voltage differs from
the error. The object here is to make the error
signal equal to zero. If we do that, that means the fed back output voltage exactly equals the
reference. So, we can say that the object of feedback
is to make the h times the output voltage
equals the reference. Now, a feedback loop is not perfect, but
we can approach this value and have the difference between
these two quantities be very small. The error signal goes into what's called a
compensator. Typically the compensator is simply an op
amp circuit and and analog loop. If we're building a digital controller
then we may have an a to d converter, or analog to digital
converter, and then some digital controller.
The output of the compensator is a control signal that goes into the pulse-width
modulator, which we've already discussed. And causes a duty cycle or command
controls the duty cycle that the transistor, power
transistor operates at. Okay what can you do to make the error
signals small? Well, some students initially think that
there's something wrong here because if we have an error signal of zero, doesn't that
mean that the compensator output of zero? Doesn't that mean the duty cycle is zero. Which means we don't get any output
voltage. So how can that work? Well, that's the wrong way to think about
it. Here's a better way to think about it. Let's suppose we want an output voltage of
whatever, say five volts. What duty cycle does it take to get five
volts? Well that depends on Vg, but there's some
value we can find. And so we need a certain duty cycle coming
out of our pulse-width modulator. Which means we need a certain voltage
going into the pulse-width modulator. I'll call capital Vc, which maybe is a
couple of volts. Whatever the number works out to be. So given that voltage, how can we make the
error signal small? Well, the error signal Vc Ve is actually
Vc divided by the compensator gain, G sub c. If you want to make the error small, what we need to do is make the compensator gain
big. So, if Vc is say two volts, it's what it
needs to be. If we can build a compensator that has a gain of a million, then the
error signal will be two microvolts. Which means that the reference and the fed
back error, or the fed back outlet voltage, differ by
only two millivolts. So that's pretty good regulation. If you can make the compensator gain
arbitrarily large, then you can make the error signal
arbitrarily small. And, then that is the, really the thinking behind this
negative feedback system. We'll refine the block diagram now of the
negative feedback system. So we have blocks for the, the sensor,
which I said had a gain, that would be called H of
S. We have our reference input and our error signal and compensator gain, we call G sub
c. The pulse-width modulator we've already
modeled in previous lectures and previous
homework. So it has a gain that was one over V sub
m, the height of the ramp. And that produces a duty cycle that drives
the switching converter block. So there is a feedback loop then from the duty cycle to the output, going around
this, this loop. Then we also have these disturbance
signals, so Vg hat and i load hat are variations that propagate
through the converter to the output. And these disturbance transfer functions are things
that we want to minimize to make V not depend on the these
disturbances. To refine the model of the convertible
block, we use our cannonical model for the converter
which we previously solved. So we represented the output voltage V hat
as a linear combination of terms coming from the different independent inputs to the
converter. And here I represented the G hat, D hat,
and also the load current variation, i load
hat, as independent sources in this model.
We can find transfer functions from each of those disturbances to V hat by solving
the canonical model. So we previously solved GVd and GVg. We can also find the output impedance.
You actually worked that out graphically already, and include it as well.
So here is our converter equivalent circuit model with those
disturbances included. And here is our block diagram of the
feedback network. We could perturb and linearize the
elements in the feedback loop as well. And so I've explicitly written these
feedback elements and block diagrams in terms of hat quantities.
So the output of the sensor is h of s times V hat. Now, and we've removed the DC components,
or quiescent values of the, the signals in the feedback
loop. The reference I've. Then we have V ref hat. In some converters there is no variation
in the reference. Although, many times in DC regulators
nowadays, we do have variations in the reference as
well. So I'm going to leave it in as v ref hat. We have an error signal Ve hat then, that
is the AC variation in the error signal, and
so on. Then here is our block diagram, including the transfer functions of the converter
power stage. With the small signal variations VG hat
and i Ioad hat, driving their respective
transfer functions. So this is a block diagram now that we can
solve. 'Kay, the solution is given on this slide
where we've actually gone through and solved the
block diagram to find this.
I'm not going to do the details of this manipulation, but I am
going to outline them. What we can do is write an equation for
the error signal, Ve hat is given by V ref
hat, minus H times the fed back output voltage, v hat.
And, we also have an equation for v hat. V hat is, what GVg times Vg hat, minus Z out times i load hat, plus GVd, times D hat, which is in turn the error signal, Ve hat, times these gains, Gc and 1/Vm.
So what you do is you take this Ve hat expression, and you plug it in over
here to eliminate Ve hat. And then you saw for the output voltage. So, if you do the algebra for that, here
is the result. This looks fairly intimidating, but, in
fact, it can be written in a simplier form. If you recognize that all of these terms
have the same denominator. And this quantity right here is called the
Loop Gain. And we're going to call the Loop Gain T of
S in this course. So, so, the Loop Gain appears there. It appears here and it appears here. And actually, you can multiply this top by
H and then divide by H, and then the loop gain
appears there also. The loop gain is actually the gain around
the feedback loop. You start at the error signal and follow all of the gains around the
loop. G of c times 1/Vm times GVd, and then times h.
And we stop right there. We don't include the minus sign of the
summing node. So those quantities, by convention, are
defined as the loop gain. In a negative feedback system, we leave
the minus sign out of our definition of T. And it was assumed in solving the loop
that it was there. So that is in general how we find this
quantity in the denominator. So written in, in terms of the loop gain T, this expression can be simplified into
this form. Okay, this is the expression for what we call the closed-loop transfer functions of
the converter. So, this expresses V as a function of the V ref variation times this transfer
function, plus Vg hat variations times this transfer
function, and then we have the i load variation times this
transfer function. So this is what we need to know about what happens after we close the feedback loop. Now, what are the effects of the feedback
loop on the transfer function? The key result here is that feedback
reduces the transfer functions from disturbances
to the output. So the, the transfer function that we
found for the Vg hat term was this. GVg is the original line to output
transfer function of the open loop power stage, and
that gets divided by one plus the loop gain
after we close the feedback loop. So what happens to that transfer function?
Well, if you make the loop gain big. For example, if we put this compensator in
our feedback loop that has a gain of a
million. Which makes, would make T bigger by a
factor of a million. That that reduces this transfer function.
So in fact this disturbance gets reduced through feedback
by this factor of 1 over 1 plus T. So if we can make T equal to a million then this factor will be one over a
million and one. And the variations in the output caused by
the G hat variations will be reduced by that
factor. The same arguments apply to the output impedance. And really the output current variations i
load hat is just another disturbance. So here was our open output impedance,
which was this parallel RLC circuit that we had
previously constructed. That transfer function, basically output
impedance, gets divided by one plus the loop gain
also. And so the output impedance is effectively
reduced by feedback by this factor of 1 plus T. So again, if T is a million, then the
output impedance is reduced by a factor of a
million and one. And we can get a very low output impedance
so that load current variations have very small
effect on the output voltage. Now, this business of T being a million and one is actually true only at low
frequency. All amplifiers have to roll off and at high frequency their gains are much
smaller. So, we'll find that we can reduce the output impedance by a lot at low
frequencies. But at a high enough frequency, that
doesn't work anymore. So, we're going to have to work out exactly what the transfer function of T
does. And what the effect is at all the frequencies of interest on these
close-loop disturbance transfer functions. Okay, we had one other term in our
equation, which came from d ref. And this transfer function from d ref hat to the output voltage turned out to be
this. And it has a different dependence on T
than the other transfer functions. What happens to this transfer function if
T is big? Say, take our example where T is a
million. Then, T over 1 plus T will be a million
divided by a million and one. And basically, it's equal to one.
So this term goes essentially to one. And the overall transfer function from the reference to output then becomes one over
h. Like this. This is a really nice result because it
means that the output voltage follows the reference very well and the,
the gain is just 1 over H. And it's very interesting that is doesn't
depend on anything else in the system besides H. The dependence on T cancels out. So for example, if our compensator changes
or if our, say, our control to alpha transfer function gbd
changes, it has little effect on this, this gain, because those
variations will cancel out. So we say that feedback makes the transfer function from the reference to the output
be insensitive to tolerances and variations
in the components in the forward half of the feedback loop. Things like Gc, GBm and GBd. So to accurately control the output
voltage, all we need to do is accurately control the reference
and accurately control age. Well, if age is some voltage divided with
some resistors, that means we by some accurate resisters, say, 0.1%
tolerance resistors for our voltage divider that becomes H. And we also have to build a good reference to supply the reference voltage
accurately. But you can buy integrated circuits that
have accurate references. So, we do that and then we're in business. The output can be well controlled. And it will depend only on the reference
in H and not on anything else. Okay. That's the basic idea of feedback, and these are the important transfer
functions. In the next lecture, we're going to
construct the Bode plots of the different transfer
functions.