0:38

So, the switch is in position one for this first interval of time

DTS and it is in position two for a second interval of time D prime TS.

I've also labeled the inductor voltage in

current as well as the capacitor voltage in current.

Incidentally, here is how we in practice normally

realize the switch using power transistors and diodes.

So there's a gate drive circuit that's just shown here as

a voltage source but when we want the switch to be in position one,

this gate driver will produce a high voltage on

the gate of the MOSFET that will turn it on and when it's

on it effectively pulls this node down to

ground just like when the switch is in position one.

For the switch in position two,

what we do is the gate driver turns off the MOSFET by making its gate voltage low.

When the MOSFET is turned off,

the inductor will forward bias the diode turning

it on and putting the switch effectively in position two.

Now, one thing you might wonder is why can we

put a diode here and why does it automatically come on?

Now, one way to look at this is to go back to

the formula for the inductor that the inductor voltage is

L DIL DT which is the voltage from plus to minus across here.

So, when the MOSFET is on we build up

some positive current flowing through the inductor and through the MOSFET.

When we turn the MOSFET off,

one way to think about it is we're effectively trying to interrupt the inductor current.

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 can think of it as heading towards minus infinity

for slope that maybe in practice it isn't.

DIT isn't that large but it is negative and large in magnitude.

So, if the IL DT is negative,

say let's call it minus infinity,

what that does is it will make VL become big and negative.

Well VL is plus to minus in this direction that's the reference direction.

So, if VL is a negative number effectively it means that

the negative terminal is becoming positive and the positive terminal is going down.

So this node here where the MOSFET is connected will head to

a large positive voltage from the voltage developed in the inductor.

So, it heads towards plus infinity but

long before it gets there it forward biases the diode.

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 then 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 freewheeling diode.

4:32

First of all on the original circuit,

it's important to clearly label the reference polarities of the 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 wave forms or the variables directly on the original converter.

So, I've defined the inductor voltage and

arbitrarily chosen its reference direction is plus on

the left and minus on the right which means that if the variable VLT is positive,

the voltage is in the direction shown and if VLT is

a negative number then the actual voltage will have the other polarity.

That's an arbitrary choice of the direction but we pick some direction for now.

Okay. Once we've defined a direction for the voltage,

the inductor current must be defined in a direction that flows

through the inductor from the plus terminal to the minus terminal of the voltage.

So ILT 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 as being plus on top and minus on

the bottom and so the capacitor 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.

6:12

Okay. Now, we draw the circuit with a switch in the two positions.

So, on the bottom left 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 directions of

the voltages and currents that were labeled on the original circuit.

So, VLT is shown here ILT is shown there,

et cetera and you have to keep the same polarities.

Likewise, for the switch in position two the right side of

the inductor is connected to the output like this and I again

could copy the voltage and

current reference directions for the switch in position two

and use the same reference directions as in the first position.

Okay. Having done this,

now we can write the expressions for the voltage and current in each case.

7:14

So, with the switch in position one we got this circuit,

the MOSFET is on and the right side of the inductor is connected to ground.

So, the inductor voltage here we can see is equal to the input voltage VG like

this and the capacitor current

we can see flows this way and it's equal to minus the load current.

So, the load current here is V over R and from the node equation at the output node,

the capacitor current equals minus V over R like that.

These are the two things we need to know for this interval in order to apply

volt second balance to the inductor and charge balance to the capacitor.

Okay, next we make the small ripple approximation and in particular we have

the output capacitor voltage right here V of T which

we will approximate with its DC component capital V like that.

So, for the first interval,

we have an inductor voltage VL that's equal to

VG and we have a capacitor current that's equal to minus V over R.

10:32

Okay, that's supposed to be a straight line equal to capital VG.

During the second interval, the inductor voltage,

we found was VG minus V. Now,

where should I draw VG minus V?

Is that a positive number or a negative number and why?

Well, if the volt seconds are going to balance so the average VL is

zero and if we have a positive voltage during

the first interval then the voltage had better be negative during the second interval.

If we were to have any chance of making the average come out to be zero.

So, I'm going to draw this voltage during the second interval as

a negative quantity and you

can see that since the voltage has to be negative to get the volt seconds to balance,

V needs to be bigger than VG.

So, therefore this is a boost converter where the output voltage V is bigger than

the input voltage VG and volt seconds won't balance otherwise.

So, this is during interval D second interval of length D prime TS.

That's our inductor voltage waveform.

We can now apply volt second balance to this waveform.

So the average VL.

11:57

Will be where we can find the area under the curves and divide by the period.

But an easier way to find the average is simply to

take the value during the first interval, Vg,

multiply that by the duty cycle D,

and add to that the voltage during the second interval,

Vg minus V multiplied by D prime,

and that will give us the average.

So, in steady state,

by volt-second balance, this must be equal to zero.

Now, we have an equation that we can solve to find the output voltage.

Let's in fact solve it.

Let's collect terms.

We have, Vg times D plus D prime.

Then we have minus D prime V equals zero.

Recall D plus D prime equals one.

So, we have, D prime V equals Vg and solve for V,

we get V equals one over D prime times Vg.

Okay. So, that's the equation for

the output voltage of the boost converter in steady state.

Since D prime is a number less than one,

V is greater than Vg.

13:45

So then, the slope diL_dt is the inductor voltage over L. So,

when we have positive voltage,

we have positive slope in the current.

The current goes up with a slope of the applied voltage Vg over

L. During the second interval,

we have negative voltage so we get a negative slope and the inductor current goes back

down with a slope

of Vg minus V over L. If we're in steady state,

we end up at the same current that we started at.

14:41

Okay. So, there's the inductor voltage waveform that I drew.

Here's the volt second balance that we just worked out. Here is a plot.

This is the plot of the conversion ratio of

the boost converter which is

the output voltage over the input voltage and it's the one over D prime factor.

At D of zero,

we have D prime is one and the conversion ratio is 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 converter is limited by losses.

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

we can get conversion ratios that approach very large numbers.

15:59

Okay. To find the DC component of the inductor current,

we apply charge balance to the capacitor.

So, here is the capacitor current waveform.

We found that it was minus the load current during the first interval.

During the second interval,

it is the inductor current minus the load current and to get the average to be zero,

that second interval had better come out to be a positive current,

and zero is right there.

So, we can find the average capacitor current or the DC component and by charge balance,

set that average current to zero in steady state.

16:42

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 zero in steady state.

So, given that, that's this equation,

we can solve this equation for capital I, again,

this quantity here is one and when you solve for I,

you find that the inductor current DC component,

capital I, is equal to the load current V over R times one over D prime.

So, the relationship between

the inductor current and the load current is the same one over D prime factor,

just like the conversion ratio of the converter.

18:21

Okay. We can also find the current ripple.

We, previously, drew the inductor current

waveform and we know the slopes so we can find Delta iL,

the peak to average inductor current ripple.

Here's capital I, the average current is the DC current,

we just found, the ripple Delta iL can be found from the slope.

We just simply take the slope which during the first interval is Vg over L,

multiply it by the length of the interval dtS,

that gives us the peak to peak ripple which is two Delta iL and we get this expression.

We can divide by two and find the peak to average ripple Delta iL 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.

19:35

Generally, when we design a power converter,

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.

The 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 that is controlled and not too large.

We can do a similar thing to find the capacitor voltage ripple.

We found the capacitor current and from the capacitor current,

we can, that tells us the slope and we can find the ripple on the capacitor.

Okay. So, what we found was that for the first interval,

the capacitor current was minus the load current.

So, the load current will discharge

the capacitor and the capacitor voltage will go down with

a slope that is the current divided by C, okay, from here.

20:42

So, knowing the slope,

we can multiply the slope which is this,

by the length of the first interval which is D prime Ts and

that's equal to the peak to peak ripple or two Delta V. So,

we can solve for Delta V and find the value of capacitance then,

that will make the output voltage ripple be sufficiently small,

according to whatever specification we have.

The only thing I would add to this is that often times,

in practice, capacitors have significant series resistance.

It's called the 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. The ripple that we've just

found is the voltage ripple on the ideal part of the capacitor.

But to that, we have to add voltage drop across the equivalent series resistance.

So, if we know Ic,

then you can multiply Ic by the ESR and get

the voltage wave form across the ESR and then add

that to this voltage ripple waveform that I've just

drawn which would be the voltage ripple on the ideal part of the capacitor.

So, we've done a boost converter example and we used the principles of

inductor volt-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.