In this lecture, we'll do a, a design example where we're
going to design a compensator. Of the feedback loop for a buck converter
with output voltage regulation. So the object in this system is to produce an output
voltage of 15 volts. So the first thing we should do is decide
what the sensor or feedback gain, h of s should
be. In this example, we have a reference
voltage that is a well-regulated and accurate five
volts. So as we've discussed previously, in a
well designed feedback loop, the error signal will be zero, or at least very,
very small, and so the fedback. Output voltage, h times v, should be equal
to the reference when the, the circuit is working
in steady state. So we have five volts here, and that means
that h, therefore, should be 1 3rd. So we'll have gain of a 3rd and our 15
volts will turn into 5. So H equals 1 3rd. With this choice of values, we can work
out the quiescent operating point of the converter, define, for
example, the steady state duty cycle. We're not going to model losses, so the
steady state duty cycle would be 15 volts divided by 28 volts for
the buck converter. And we can work out the other quantities,
such as the control voltage and, and other
things from there. So, here is a list of the voltages and duty cycle at this quiescent operating
point. Knowing that, we can now plug in the small
signal model for our buck converter, and here is the buck converter
AC model that we have previously derived. I also have shown a block diagram then for
the rest of the feedback loop. We can solve this ac model for the buck converter then to find the important
transfer functions. So the control-to-output transfer function
Gvd of s for this buck converter, we've found previously for
the buck is this expression. We'll plug in the numbers from the
previous slides to these component values, and what we find
is that we have a dc gain in gvd, of 28 volts, which
corresponds to 29 db volts.
And we have a resonant frequency, f0. That turns out to be 1 kilohertz, so we'll
have a two pole slope beyond 1 kilohertz.
The q factor works out to be 9 and a half, which corresponds to 19.5 db, so we
can sketch in the resonance from that. And we can also sketch the phase
asymptotes of gbd. By using our phase asymptote formulas, we
find that the phase asymptotes have break frequencies at 900 hertz and 1.1
kilohertz, so the phase changes very sharply over that narrow frequency range.
From 0 degrees down to minus 180 degrees. So, there's our control-to-output transfer
function. We could also work out the line-to-output
transfer function and the output impedance of the converter power stage as usual and
here are the expressions for those. So given all of those things, then, we have this complete system block diagram
with the converter power stage model, as we previously
discussed, and with the, the transfer functions of the buck
converter included. And we have our feedback loop. As well. From this block diagram, we can work out
the loop gain. So the loop gain T will be the compensator
gain G c times the modulator gain, 1 over V M, times the control to output transfer
function G v d, and then times the H of s. So, that's, that's these quantities. And when we plug in our GVD of s with its
two poles, then we get this. So, let's first plot the uncompensated
loop gain. This is the loop gain t of s when the
compensator gain is 1. So we haven't designed a compensator yet. Let's see what our starting point is. So the uncompensated loop gain then has
the two poles from the control to output
transfer function, and then it has a dc gain that is the
product of these dc gains around the loop. This DC gain works out then to be 2.33 or
7.4 dB. So we have a fairly low DC gain in our
uncompensated loop. And then we have our same two poles at 1
KHz with a Q of 9.5. So there's the complete loop gain. Without a compensator, you can see that the crossover frequency looks like it's
right there, from the drawing.
Which is 1.8 kilohertz, and the phase at that point.
Is almost minus 180 degrees. There's actually a little deviation from
the asymptote and if you work out the exact phase at that point, we find
that it's minus 175 degrees. So, our uncompensated loop would have a
phase margin of five degrees. Well, so this is our starting point. I would say, some defects in the
uncompensated loop gain are, first of all, we don't have much
phase margin. Second of all, we could probably push this
crossover frequency up to a higher frequency and get some more
bandwidth. And third, we don't have much low
frequency loop gain, so our loop won't do much of a job of
regulating. First thing we'll do then, is design a
lead compensator to improve the phase margin.
So we discussed already how to do that. What we're going to do is I'm going to try
to design for a crossover frequency of five
kilohertz Which is here. It's, it's 1 20th of the switching
frequency. So we'll need to increase our loop gain
by, this much, to push our, our gain up at 5 kilohertz so that our crossover frequency
ends up being there. We'll also need to add some phase, so
we'll need to improve our phase margin, and I'm going to design for
a closed loop q of 1. If you recall our plot of q of the closed loop transfer function versus phase
margin of t. If we want a Q, closed-loop Q of 1, this means we need a phase margin of 52
degrees. Looking at this plot for the uncompensated
loop gain, at 5 kilohertz there really is basically
no phase margin. So, we need to design a compensator, that
will add 52 degrees at 5 KHz. And we'll add this much gain at 5 KHz.
From working out the asymptote of T, the expression for the asymptote of the
uncompensated T. One finds that the existing loop gain at 5
KHz is minus 20.6 dB. So, we need to add 20.6 dB more of gain at
5 kHz to, to get the crossover frequency to be there.
Here were are the asymptotes of the lead compensator from the previous
lecture. We'll then want this frequency here to be
the crossover frequency, fc or 5 kHz. And we'll want this phase right at this
point. We were calling theta to be plus 52
degrees. This slide is also from the previous
lecture. These are the formulas for designing the
lead compensator. So, again, f c is 5 KHz.
And Theta is our 52 degrees. So we can simply plug the numbers in and
see what zero and pole frequency we need. Then here are those computations. It turns out our zero needs to be at 1.7kHz and the poles need to, pole needs to be at
14.5kHz. And our compensator dc gain needs to be
11.3 db, or a gain of 3.7. Here is a Bode plot of that then, so this
is our 5 kilohertz right there. Let's plot the compensated loop gain now
with this compensator. Here's the original loop gain with the,
the dc value of the uncompensated loop gain and
its two poles. And then, these three terms are the added
terms that come from our compensator with our zero pole and additional dc gain. When we add those in to our loop gain,
we'll get the zero right here and the pole right
there. The phase asymptotes then are increased,
so we have our phase margin of 52 degrees at 5 kilohertz.
And it turns out now that our DC loop gain is 8.6, or 18.7 db.
Which is better than it was before. This loop gain is not bad. We do have a good phase margin, but we
could further improve it by increasing the low frequency
or dc value of the loop gain. The reason we would want to do that is to
reduce the disturbance transfer functions. If you recall from previous lectures, the
disturbance transfer functions such as the, gvg of
transfer function, from vg hat to v hat. End up having a factor of 1 plus t in
them, like this, in the closed loop case. So our, we can use our loop to reduce this
transfer function in magnitude. A typical case for wanting to reduce this
transfer function. Is in a dc voltage regulator that operates off the power line, where g, vg includes
variations at the second harmonic of the ac line
frequency. In other words, at 100 hertz or 120 hertz. So if we have voltage ripple from
rectification That occurs down at frequencies down here.
Then our feedback loop can reduce those variations if it has more loop gain at 100
or 120 Hz. So it would be nice if we could reduce 1
over 1 plus T or increase the loop gain.
To better reject those harmonics. What we can do then, is to add a pi type
compensator or lag compensator, which is this inverted zero
form, like this. What this does then, is increase the loop
gain of frequencies below the inverted zero frequency, so that we better reject.
Variations or disturbances. Since the inverted zero term adds phase
lag at frequencies below 10 times omega L, it is a good idea to choose this omega
L to be. Perhaps no more than 1 10th of our crossover frequency, so that
it doesn't reduce our phase margin. Therefore with a crossover frequency of 5
kilohertz, we might choose this f sub l to be 500 hertz. Now, the lower in frequency we put it, the less benefit there is, say, at 100 or 120
hertz. So we might want to put this in fact at a
higher frequency, but if we do, then we're going to have to redesign our lead
compensator to correct the phase margin. So here is a design then where we choose 500 hertz, and then we obtain this loop
gain, with our additional term at 500 hertz.
And. Loop gain being improved at frequencies
below that. The 1 over 1 plus key factor then looks
like. Then let's construct the closed loop
transfer function from vg to the output. Here's our open loop gbg. And it has the same poles as in gbd, plus
it has a dc gain equal to the duty cycle. So I'm going to just sketch them on this
plot. We have a duty cycle that is slightly
greater than a half. Or slightly larger than minus 6 db so,
it's down here. Then we have the same two poles at f not. So this quantity gvg, then gets multiplied
by 1 over 1 plus t, so we have multi, multiply this
asymptote by that asymptote. Well, at high frequency above the
crossover, 1 over 1 plus d is 1 and it doesn't change anything, so we get the
close loop gvg is the same as the open loop. And that's the case all the way up to f,
or frequencies of fc and higher. When we go below fc, the 1 over 1 plus T term adds a plus 20 dB per decade slope
there. We had a minus 40 slope on Gvg, so we go
to plus, or to minus 20. Then we have our compensator zero.
Which adds a plus 40 dB per decade slope below that, which will make this
whole function flatten out below the 0.
So that is at f sub z. The next corner frequency we come to is f
not. This resonance at f not. Is contained in both GVG, and in 1 over 1 plus t.
1 over 1 plus t has resonant zeros, and GVG has resonant pulls, and they come
from exactly the same place. So, in fact, when you multiply these two quantities together, the resonance
cancels out. So it doesn't appear in our closed loop
gbg or closed loop transfer function from vg
to output. Finally we have the compensator inverted 0
at f sub l. And below that. We get this 20 dB per decade slope in 1
over 1 plus T. So, our Gvg function, or Gvg over 1 plus
T, will have a plus 20 dB per decade slope below that.
So, this is Gvg over 1 plus T. Some things to note about this. First of all, we can work out what is the gain at 100 Hz, or 120 Hz which is this
value. We can calculate how much it is, and if we know how big our input variation is nvg
at. Then we can multiply by this gain to find
the resulting output variation. Another thing that we can see from this is
that the maximum value of this transfer function happens over
this range of frequencies right here, between f sub l and f sub z. So this is where this transfer function
from the disturbance in vg to the output would be
maximum. And where most susceptible then to
variations in vg over this range of frequencies, which is around a kilohertz.
If we need to reduce variations more, we can also gain the insight from this
kind of construction to see what we need to do. For example, if we want to reduce this transfer function more, perhaps we need to
push up the crossover frequency more, which will
push down all of these asymptotes at frequencies
below the crossover. Here is a nice drawn plot. On, on this slide of the quantity that I
just computed. So, I've done an example of how to design
a PID compensator for a voltage regulator circuit.
We came up with this compensator. This gv, gc of s that has this kind of
transfer function. In the next lecture, we're going to design an op amp compensator circuit that
realizes this gain.