In this lecture, we will discuss how to solve for the output voltage of a converter
operating in discontinuous induction mode. And I'm going to explain it for the
example of a butt converter. What we need to do is to extend the ideas
developed in chapter two, to apply to this case where we have
inductor current ripple that is large. So here's what we talked about back in chapter two at the beginning of the
course. We had this technique of inductor volts
second balance. Which said that in steady state the average voltage applied to an inductor is
zero. And that's true in any circuit, and it doesn't matter if it's a continuous mode, or discontinuous power converter, or
something else. So it still applies. And same thing with charge balance. on capacitors the, in steady state the average current through a capcitor is also
zero. And we can apply this to solve our
discontinuous mode converter as well, but where we have to be careful is the
small ripple of approximation. So, really by definition here in
discontinuous mode we have an element where the ripple is
large. And we are talking in these lectures about
the current example where the inductor current ripple is large
compared to the DC component. So we can't approximate away the inductor
current ripple now in solving our discontinuous
mode converter. And we're going to have to use some other arguments in order to, to handle the inductor current and include
its ripple. On the other hand, we still want the output capacitor voltage to have small
ripple. And generally we're going to put a
capacitor on our output that will be big enough to filter out the ripple, and have
small voltage ripple at our output. So we can apply the small ripple equation
to the capacitor voltage but not to the inductor current. Okay, so let's do this example of the buck
converter and discontinuous mode, and apply these arguments to solve
for the output voltage. Here's our buck converter then.
And we now have three sub intervals. First interval where the moss fit is on
gives us this circuit. For the second interval, the moss fit is
turned off and the diode is on so we get this circuit with the left side
of the inductor connected to ground. And then we now have a new third interval
where both transistor and diode are off. The inductor current has gone to zero. Then the diode is turned off, and the left
side of the inductor now is left hanging in space, not connected anywhere with zero current flowing through
the inductor. So, as usual, we will write the circuit
equations for each interval. So here we have for the first interval
with a MOSFET on, as usual we can write the inductor voltage is
Vg minus V, like this. and the capacitor current would be from
the node equation, it would be the inductor current minus the
load current V over R. So we get the, the capacitor current is IL
minus V over R. Next we apply the small ripple
approximation. So the output capacitor voltage v has
small ripple, and we can replace v in these two equations
with capital V. To ignore the ripple in the output.
So that gives us these terms. But the inductor current has large ripple,
and we can't apr, approximate IL of T by it's DC component.
So we're going to just IL of T alone for now and we're going to have
to handle the, the ripple in IL With some additional
arguments in a couple of minutes. For subinterval two, the diode conducts
the inductor current is positive, and we have this circuit with the left side of
the inductor connected to ground. We can write that the inductor voltage is
minus the output voltage like this. The capacitor current is again the
inductor current minus the load current, like that. And we can replace V of T by the DC
component, capital V using the small ripple
approximation to get these equations. And again the inductor current is left as
IL of T. For sub interval three, the inductor
current has gone to zero, the diode is turned off, so the right-hand side of the inductor is connected just, or not connected, it's
hanging in space. What is the inductor voltage during this
interval? Well, the inductor voltage is the
derivative of the current. We know what v l of t is L times d i l, d
t. I L is zero so we get that the inductor
voltage is 0 during this interval. So VL is zero, we'll just, we can simply
write that. IL is also zero. The capacitor current is the inductor
current minus the lode current. Although the inductor current is zero, so
really all we need to say is the capacitor current is minus
capital D over R. where we apply the small ripple again to
be of t. So here's the inductor voltage wave form. It's a positive voltage during the first
interval, a negative voltage during the second, and it's zero
during the third interval. We can apply volt second balance to this
waveform in the usual way. We have a third interval. The inductor voltage can be written as D1 times the voltage during the first
interval. Plus D2 times the voltage during the
second interval, which is minus V. Plus D3 times the voltage during the third
interval, which is 0 [COUGH] and by volts second balance this is equal
to 0. So we get this equation, you can solve it
for the output voltage, and we get this expression
for the output voltage. Now in continuous conduction mode, without
losses, this equation told us what the output voltage of the buck in
continuous mode was. But here we have an, an unknown.
The, the duty cycle D2 is not known. D2 really depends on when the inductor
current goes to zero and the diode turns off and we have to do further
analysis to find D2. So, at this point, we have an equation
with two unknowns, V and D2. And we need to get more equations. So let's keep going.
Let's apply charge balance next. here's our node where the capacitor is connected at the output terminals of the
converter. We can write the node equation at that
point which for the butt converter the inductor
is connected there, so we can write that the inductor
current is equal to the capacitor current plus the
load current. Like this. And what we want to do is apply charge
balance to the capacitor. So the hard way to apply charge balance is
to take the inductor current waveform that we already drew, and subtract the load current to draw the capacitor current
waveform. And find its average.
A little bit easier way to do it, is to look at the original node equation at this
node, and compute the DC components of each term. So we can say that the DC current flowing
out of the inductor into this node equals the DC component of the capacitor
current, plus the DC component of the lode current. And the capacitor current by charge
balance has no DC, it's zero, so you get that the average inductor current, or DC component of inductor current, equals the
lode current. So all we really have to do is take the
inductor current wave form that we've already drawn, find it's
average value, and equate that to the lode
current. I should caution you at this point this
expression of finding the average inductor current and equating it to the lode is
really only true in the butt converter. And it's because the inductor is what's
connected to the output node where the capacitor is. We'll do another example next time with a
boost converter where it's the diode that's connected to the output
node instead of the inductor. And in that case we have to draw the diode current wave form and find it's DC
component. Okay so here is the inductor current
waveform. We'll find its DC component and, and
equate it to the load. Here's the calculation for that. The DC component of inductor current is
found, as usual, by averaging the inductor current
over one period. So, the average is the integral divided by
the period. And the integral is the area under the
curve. So we need to find this area. Okay, the area in this, for this wave
form, we have the area of a triangle, so we have to use the triangle
formula, which would be one half the base. And what the base is D1 plus D2 times Ts. Times the height, I've labeled this p
current i peak. Okay, so that's the area or the integral. If you divide by Ts we get the average
value. [COUGH] Okay, and so here, here it is divided by
Ts, and the other thing I've done here is, is subtr, is substitute
in an expression for i peak. Here's how we find i peak. The inductor current starts the period at
zero. And it starts at zero because we're in
discontinuous mode, and we ended at zero in the last
period. We then, we go up during the first
interval with a slope given by the applied voltage over L. So Vg minus V over L was the slope of the inductor current during the
first interval. Here, multiply that by the length of the first interval to get the, the change in
current. And that's i peak. So we can plug that expression for i peak
into here, and finally we, we get this expression for
the average inductor current. and then lastly, we equate the average
inductor current to the lode current, and we get this expression that comes from
charge balance on the capacitor. So we have two equations. The first one was from volt second
balance. The second one is from capacitor charge
balance. We have two unknowns, the output voltage
and D2, so we can solve. So I'm not going to go through the algebra
in this lecture, I'll actually illustrate the algebra in the
next example of the next lecture. But basically you solve one of these
equations for D2, plug it into the other equation. You then get a quadratic expression that
we have to find the roots of, and finally from doing that we can solve, and get this
equation for the conversion ratio V over VG of the butt
converter in discontinuous mode. This equation is a function of D1, the
duty cycle of the first interval. It's also a function of k, which is the
same 2L over RTs that came out of the mode boundary equation in
the last lecture. Here's a plot. So this is in, which is V over VG. Versus duty cycle, and we have now two
choices for in, if we're in continuous mode, when K is
greater than K crit, then M is equal to D. Where, whereas if we're in discontinuous
mode with K less than K crit In is given by the expression from
the last slide. So we can plot this for different values
of load resistance, or different values of K,
and here are some different choices.
So example for K equals 0.1. To plot this we first have to plot our,
our K versus K crit formula. Remember K crit for the butt converter was
D prime, so it looked like this. And if say we do the K equals 0.1 curve so
here's 0.1 for K K is less than k crit for D from 0 all the
way up to 0.9 right there. So we use the discontinuous formula for D
between 0 and 0.9. And then for D between 9 and 1 we're in continuous mode, and we have to use the
continuous formula. So here is the plot, computer drawn plot
of the discontinuous formula going all the way upto D as. 9 after that we follow the continuous
formula n equals D. And that's what happens over the whole
range of, from 0 to 1 in duty cycle, for the case where K is
equal to 1. One easy way to plot this is to recognize that this continuous mode makes
the voltage rise. So you can actually just. Plot both of these expressions and then
take the largest, and you'll get the right
answer. Another thing I should point out is that
what happens when you disconnect the lode? Which is, say, let R go to infinity.
Okay, so k is 2L over RTs. So here with no load power or R is an open
circuit or infinity. If R goes to infinity K goes to zero, and
what this curve does in that case, is it actually just rises all the way to
1. Basically you get the output voltage equal
to the input voltage regardless of the duty
cycle. It's actually, the solution goes to 1
except right at the axis where it's, actually turns into a
squared off function here. So, if say the customer disconnects the
lode you will lose control of our output
voltage. And we have to do something about that, generally our build a feedback loop that
stops switching. Turns the converter off to keep the output
voltage from rising beyond the value that we want, so that we could
maintain control of the output. Generally this is a problem in discontinuous mode when k goes to zero,
we tend to lose control of the output.