0:50

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.

1:37

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.

2:11

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.

2:37

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.

3:21

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.

5:24

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.

5:50

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.

6:04

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.

6:35

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,

6:50

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.

7:48

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.

9:15

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.

10:31

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.

11:43

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.

12:47

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.

13:13

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.

13:32

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.

13:57

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.

15:03

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.

15:42

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?

16:04

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.

17:22

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.

18:33

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.

18:58

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.

19:22

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.