Let's work an example, and go through the analysis of a boost converter in discontinuous conduction mode. Here is our boost converter. we know from the discussions of the previous lectures, that the mode boundary
between continuous and discontinuous mode occurs, when the DC component of inductor current equals the
the ripple. since the diode in this converter conducts
the inductor current when it's on, that is, in fact, the
condition, for whether the diode is turned off by the ripple. We can use our continuous mode solutions
for the inductor current and its ripple. from, from previous analysis, here's what
we had before that the DC component of the inductor current was
VG over D prime squared R. And the ripple was given by this
expression. So, we can plug these two into here to get
the expression for the mode boundary. So, here it is again. This is the DC component and this is the
ripple. So, for continuous mode, the DC component
is greater than the ripple, we can simplify
this expression. Let's see, the VG's cancel out. We can put the terms that depend on duty
cycle on the right side of the equation. And the other terms on the left hand side. And we get this expression. We recognize that this is our usual K 2L
over RTS. And this is the critical value at the boundary.
K crit of D. So, this is what comes out of the
equations for the boost converter. K crit is equal to D times D prime
squared. Here's a plot of that function. the function D times D prime squared has a
root at D equals 0, right here. And it actually has two roots of D prime
equal 0, or D equals 1, right there. So, it looks parabolic about D of 1.
and so, K crit is 0, at both D of 0 and D of 1.
And in between, it rises to a maximum. This is qualitatively different than what
happened in the buck. And actually, it's different than what
happens in many other converters, as well. In the buck, K crit was maximum at D of 0. So, the converter was most likely to run
in discontinuous mode at low duty cycle. That's not the case here.
here at K crit goes to 0 at both extremes of duty cycle. For high duty cycle, most converters will
go continuous. And the reason for that is that it high,
as the duty cycle approaches 1, the ripple goes to 0, while
the DC component does not. So, the converters generally run in
continuous mode at high duty cycle. Here the boost is going continuous, also,
at low duty cycle.
Why is that? Well, let's look back at the original
boost converter. The at low duty cycle, for example, in the
extreme as D goes to 0, you still have a path for current to flow
this way through the converter. If we never turn the MOSFET on, current
still flow out of VG through the inductor and through
the load. At D of 0, the load voltage is VG.
We have load current equal to VG over R. So, there's substantial current flowing
through our inductor. However, if we don't switch the MOSFET,
then, there's no ripple. So, the, the current ripple goes to 0, but
the DC current does not. And we find that the converter must run in
continuous conduction mode. So, the boost tends to not want to run around discontinuous, but
certainly, if K is small enough, it can. And one can work out the maximum of this K
crit function. It turns out to be at, D of 1 3rd, or the
value is 4 27ths. And that's roughly. 015 .
So if we have k less than. 015, then the converter will run into
discontinuous mode over some intermediate range of duty
cycles. So, here's an example with k just less
than. 1 . And we run in discontinuous mode from here
to here. Okay that's the analysis of the mode boundary, now let's work out the output
voltage. So, as usual here we will draw what the
circuit reduces to. With the switches in the different
positions. So, for subinterval one, the MOSFET is on
and right side of the inductor is connected to
ground. In position 2, the diode conducts. And in position 3, the diode and MOSFET are both off, and the inductor current is
0. So, we, we'll go through volt second
balance and charge balance for the two two
reactive elements. So, at subinterval one, the inductor
voltage is Vg and the load current, or the capacitor current, is
minus the load current. We make the small ripple approximation for
the output voltage as usual and replace v of t with
V. In position two, the inductor's connected
to the output. [COUGH] So the inductor voltage is Vg minus V. Well, we can again make the small ripple approximation, and replace v of t with
capital V. The capacitor current in this center bowl
is equal to the inductor current minus the
load current. We do not make the small ripple
approximation on I, because the ripple is large in the
inductor current. But, we can make the small ripple
approximation in the voltage. For the third interval, the inductor
current is 0. And, the capacitor current is minus the
load current. We'll make the small ripple approximation
in the load current, as well. So, we apply volt-second balance. Here are the inductor voltages, that we
have found for the three intervals, so we can draw the V L of
t waveform. To apply Volt-second balance, we apply the
average inductor voltage, which will be D times the, or D1 times the
value in the first interval. Vg plus D2 times the value in the second
interval Vg minus V, plus D3 times 0. And in steady state D average conductor
voltage is 0. We can solve this equation for the output
voltage. If we do this is what we get. [COUGH] The output voltage is a function of D1, D2
and BG. D2 is an unknown still at this point. So, we need another equation to solve for
V2. For D2. So, to get the second equation, we apply
capacitor charge balance. In the case of the boost converter the
elements connected to the capacitor node, besides the capacitor,
are the load and the diode. So, to work out capacitor charge balance,
we should write the node equation. So, at this node, the diode current is equal to the capacitor current plus the
load current. Like this. Capacitor charge balance tells us that the
DC component of capacitor current is 0. So, we add, we find the DC components of
these currents. Capacitor current has zero average and,
therefore, the average diode current is equal to the DC load
current. So, for the boost converter, to apply
charge balance, we need to work out the average diode
current. Here is our inductor current waveform,
that looks like the usual, discontinuous mode
inductor current. The diode current, equals the inductor
current when the diode conducts and it's 0 the rest of the time, so this is the
diode current waveform. To find the DC component of this current,
we should integrate over 1 period and divide by the period.
The integral of the diode current then, would be the area of this triangle. And that area is one half. The base of the triangle, which is for
this example, D2Ts. Times the height, which is this peak
current. We'll call i peak.
We can find i peak, knowing the slopes. So, during the first interval the inductor
current increases with a slope of Vg over L, and I peak then would
be that slope Vg over L times the length of
the first interval D1 Ts. So, we can take that i peak and plug in
into here. Okay. So, here is the area times 1 over Ts, to
get the average. We plug in our expression for i peak, and
we get this. And so, if we equate the average diode
current to the load current, finally this is the equation that
results from capacitor charge balance. So, we have two equations and two
unknowns. First equation came from volt 2nd balance. The second equation came from capacitor
charge balance, and the two unknowns are the output voltage V and the
duty cycle D2. So, we can take these two equations now, eliminate D2 and solve for the output
voltage. The steps in that analysis are outlined
here. So, what we're going to do is take the
first equation and solve it for D2. So, the little Algebra, this is what you
get. We'll then plug that expression for D2
into the 2nd equation. And then, if we rearrange terms to try to
solve for V, we get this. Okay, so this it, it turns out that this
gives us a quadratic equation in the output
voltage, that we need to solve. Here is the same equation repeated. So, we apply the quadratic formula to
solve for V and this is the answer. as usual, usual with the quadratic
equation, we get two roots. So, which root is correct? Well, if you look at the the discriminate,
or the quantity inside the radical. That quantity is greater than 1. So, 1 plus the square root of the discriminant, gives us a number bigger
than 2. Dividing by 2, gives us a number bigger
than 1. And, the conversion ratio is bigger than
one. Which is actually what we expect for the
boost converter. On the other hand, the other route, if we
take the minus sign, gives us 1 minus a number
bigger than 1. So, we get a negative answer. Well, we know the boost converter gives a
positive voltage. But even if we didn't, we could go back
and look at our volt second-balance equation and
see that a VG is positive, then V has to be positive. Because the duty cycles have to be
positive. So, a negative V is actually happening,
because our solution is giving a negative D2, which is a non
physical answer. So we select the positive sign. And then, this is our answer for the conversion ratio of the boost in
discontinuous mode. And it's valid again when K is less than K
crit. Here's a plot of what, what the function
looks like. So, we have that M. The conversion ratio is 1 over D prime in
continuous mode. And it's the function with the radical in
a discontinuous mode. So, for, for example, for K equals a
number bigger than 4 27ths will always be in
continuous mode. And we have this function. Whereas, if K is, say, 0.1, then we'll be in discontinuous mode, over some
intermediate range of duty cycles. And we’ll follow the continuous mode
result outside that range. So here’s a plot, for that, that example. K, actually for large K the discontinuous
mode equation can be approximated. 4D squared over K turns out to be much
greater than 1. You can ignore the 1, take the radical,
and you get that M is equal to one half plus D over
root K. So, it's actually approximately a linear
function of duty cycle for small K. So, for example if K is 0.01 we get this
function. Where it's nearly always in discontinuous
mode with a small k, and it's approximately
linear function of duty cycle, with a slope of
one over root K, which would be ten [COUGH]. Okay? Here's a handy table of the answers for
the basic converters. So, here's what K crit is.
Here's the discontinuous mode M of D. Here's the completing the solution for D2. And then, the last column is the
continuous mode result. So, for all the basic converters, you can
just simply go to the table and plug in the
results. Finally. here's a plot of the discontinuous mode
functions in the D functions, or discontinuous mode for the
basic three converters. So, the boost was this function that
approached this linear function with a slope of 1 over
root k. It turns out that the buck-boost
converter, exactly has a slope of 1 over root K and it is exactly linear.
So, let's plot it here. the buck-boost is inverting, and so, the
inversion is not included in this plot. And then, the buck turn, its function
turns out to be asymptotic to the buck-boost line
into one. And so, it's this function that's below
those asymptotes, while the boost function is
above those asymptotes. Okay. So, for the boost and the buck boost, we actually get functions that are almost
linear, in discontinuous mode. One last point, I would make here is, for
both the boost and the buck-boost formulas, what happens to the output
voltage, in this case, when you remove the load? Well, if we remove the load this is the
case where r goes to infinity. And that makes k go to 0. So, what happens here, our slope is 1 over
root K. If K goes to 0, our slope goes to infinity, and the output voltage goes
to infinity. This is a real problem and it really
happens. If you remove the load from your boost,
your buck-boost converters, its output voltage tends to infinity.
So, physically, why is that? Let's go back and look at the converter
circuit. Can you explain why the output voltage of
the boost converter would tend to infinity if you
remove the load? Well, the answer is if we look at just how
the circuit works, when the MOSFET turns on.
The inductor current charge is up, to some current and some energy
from Vg is stored in the inductor. When we turn the MOSFET off, that inductor
current, and its stored energy flows through the diode
to the output. And it charges up the capacitor. So, when it char, when that energy is transferred to the capacitor, the
capacitor voltage increases and its stored energy
increases. Now, there's no mechanism for discharging
the capacitor if you remove the load. So, the only thing that can happen is,
every time we switch, every switching period the capacitor is
charged with a little more energy. And, its voltage rises more and more every
time, or during every switching period. And with no mechanism for discharging, the
load, or discharging this capacitor through the
load, the voltage can build up to an arbitrarily large
number. So, it is something to watch out for. Or, we have to be careful and limit the output voltage or, if our feedback loop
senses this happening. It needs to stop switching, or turn down
its duty cycle to [COUGH] keep this from happening. But if you run an open converter and you,
say, your load wire falls out. Then the converter will make a very large
voltage. It will probably exceed voltage ratings of
some of the components. And, and the converter will fail. So, we've seen that the discontinuous
conduction mode is a consequence of using single quadrant switches, or
unidirectional switches. In an application, where the applied
switch waveforms become bidirectional. This causes the switches to, to change
their conducting state at times that don't coincide with
the drive signal. And it adds additional intervals to the converter, which change the properties of
the converter. We found how to find the the mode
boundaries by equating the DC component of the inductor
current to the ripple. There are some sample problems that have
multiple inductors. And then, this gets a little more
complicated. But, in the case with multiple inductors,
we have to carefully draw the diode current waveform and find out under
which conditions it goes to zero. But in general we can find an equation of
the form, k is less than k crit for
discontinuous mode operation. We also found how to solve for a steady
state in the converters. We apply volt second balance and charge
balance as usual to get a set of equations. But we have to be careful with the small
ripple approximation, and in those cases where the ripple is large we
have to use additional arguments.