2:26

So, the inductor current that was positive is now trying to be

forced by the MOSFET turning off to go immediately to zero.

So what is dIL/dT?

Well, dIL/dT is big, and negative, if we're shutting it off.

You could think of it as going, heading towards minus infinity for slope.

but maybe in practice it isn't, the idt isn't

that large, but it is negative, and, and, and sub,

large in magnitude.

So, if the iLdt is negative, say let's call it minus infinity.

What that does is it will make vl become big and negative.

3:36

So it heads towards plus infinity, but long

before it gets there, it forward-biases the diode.

And so, in fact, when this voltage

rises above the output voltage, the diode will

turn on and clamp this voltage to the output.

And once it does turn on, it provides a path for the inductor current

to flow, the inductor is happy and the current no longer tries to change.

So, we can put a MOSFET and a

diode that will operate automatically with the MOSFET.

Sometimes in the business, a diode that operates in this

fashion is called a free-wheeling diode.

Okay, so what we would like to do now is apply voltsec and balance

to the inductor and charge balance to the capacitor To solve for

the voltages and currents of the converter in steady state.

So here's how we do it.

First of all, on the original curcuit, it's important

to clearly label the reference polarities of voltages and currents.

And if you're not very careful about this it's very

easy to make minus sign errors and get the wrong answer.

So to head off a lot of problems coming up soon, let's

label the, the waveforms or the variables directly on the original converter.

So I've defined the inductor

voltage and arbitrarily chosen its reference direction as

plus on the left and minus on the right.

Which means that if the variable VL of T

is positive, the voltage is in the direction shown,

and if VL of T is a negative number,

then the actual voltage will have the other polarity.

That's an arbitrary choice of the direction,

but we picked some direction for now.

Okay? Once we've

defined a direction for the voltage, the induct

or current must be defined with in a

direction that flows through the inductor from the

plus terminal to the minus terminal of the voltage.

So I L of T is a quantity having a

reference direction that goes from left to right through the inductor.

Likewise here I've drawn the output voltage which coincides with

the capacitor voltage the plus on top and minus on the bottom.

And so the capicitor current must be defined as flowing through the

capacitor from the plus terminal to the minus terminal in this direction

so positive and capacitor current is flowing in this direction.

Okay? Now we draw the circuit with

a switch in the two positions.

So on the left, bottom here is the circuit that

we get with the switch in position one, with the

right hand side of the inductor connected to ground like

this, and for this circuit, I now carefully copy over.

The, the directions of the voltages and

currents that were labeled on the original circuit.

So VL of T is shown here. IL of T is shown there, etcetera.

And you have to keep the same polarities.

15:05

At D of zero we have D prime as one and the

conversion ratio was one meaning that the output voltage equals the input voltage.

As we increase the duty cycle this function gets larger and

larger and in fact as D goes to one, D prime goes to zero and the

ideal conversion ratio function heads to infinity at D of one.

Now of course, real converters

won't do that. They won't make infinite voltage.

And we'll find next week that the maximum output

voltage of the boost convertor is limited by losses.

But we haven't modeled any losses yet, and so to the extent that

the convertor is ideal, we can get

conversion ratios that approach very large numbers.

16:30

So we can find the average capacitor current or the DC component

And by charge balance, set that average current to zero.

It's, in steady state.

[COUGH]

So, the average capacitor current will be D

times the value during the first interval,

minus V over R. Plus D prime times the value

during the second interval, I minus V over R, and that's equal to 0

and steady state. So given that, that's this

equation, we can solve this equation for capital I.

19:04

and we get this expression.

We can divide by two and find the peak to average

ripple delta I L and we get this equation which we

can use to select the inductor to get a given ripple if we like.

I would also note that the peak current in our inductor, which is also the

peak current in our transistor and diode, is

the DC component capital I, plus the ripple.

[NOISE]

Generally, when we design a power convertor, we want to limit

the peak current to a value that our components can handle.

And so we'll want to control the value of delta IL.

Capital I is determined by the load current, but

delta IL is determined by the choice of inductor value.

So, we can

choose the inductance here to make the peak current be a desired value.

21:09

according to whatever specification we have.

And the only thing I would add to this is that Often times, in practice, capacitors

have significant series resistance, it's called equivalent series resistance.

So the model for a practical capacitor, especially, say, an

electrolytic capacitor, is this equivalent series resistance or

ESR, in series with an ideal capacitor, C.

22:09

So we've done a boost converter example and

we've used the principles of conductor volts second

balance and capacitor charge balance to find the

DC voltage and DC currents of the converter.

And we've been able to derive the equation for example

for the output voltage as a function of input

voltage we've also been able to derive equations for

the ripples in the inductor current and capacitor voltage

and use them to choose the values of the capacitance.