0:00

In this lecture we will develop the

basic analytical techniques that

allow us to solve for the voltages

and currents of any arbitrary

switching converter. Specifically we will employ

the small ripple approximation to simplify the equations, and we'll

derive the principles of inductor volt second balance and capacitor

charge balance that give us the DC equations of the converter circuit.

0:41

So if you recall from the previous lectures we discussed

the buck converter having a switch and having a low-pass

filter that removes the switch in harmonics but allows the DC

components or the DC component of the waveform through to the output.

Now, in a practical converter, we can't build a, a perfect low-pass filter.

The low-pass filter will have continuation but

it won't completely remove the switching harmonics

of the waveform.

[COUGH]

And therefore, the output voltage of a practical converter might look

like this where there's a DC component capital V that as we previously

found, is equal to the duty cycle times the input voltage vg.

And then in addition to that, we have some small

ripple that is at the switching frequency and its harmonics that,

where a small amount of the switching ripple

gets through the filter and appears in the output.

1:58

So, in a practical circuit, we'll have some kind of specification on how

large this switching ripple can be, and generally it's a very small number.

So in an output voltage of say a volt

or two for a computer power supply,

this ripple might be ten millivolts

or, or some very small number.

so therefore we can write the equation of the

output voltage VOT, as being equal to capital V which

[COUGH]

here I'm using the capitals letters to denote the DC components.

So capital V is the desired DC component equal to DVG.

And then plus some AC variation, that's called V ripple of T here, that

is the undesired, switching harmonics that make

it through the filter to the output voltage.

[COUGH]

So in a well designed converter, the cell c filter will have lots of attenuation.

And the ripple will be very small compared to the DC component.

3:06

So we call this the small ripple approximation, where the,

the ripple is small compared to the desired DC component.

And under certain circumstances we will neglect the ripple and

simply approximate the output voltage by its desired DC component.

Okay, this has the effect of decoupling the

equations, the differential equations of the circuit, and making

them very simple to solve. Okay, so this is called the small ripple

approximation to ignore this component and simply approximate V of T by capital V.

3:55

[COUGH]

as we're

going to

see in a

minute,

the small

ripple

approximation

formally

can be

applied

only to

continuous

wave forms

that have

small

ripple.

So, we don't apply the small ripple

approximation to switched wave forms in

the circuit such as the switch output VS of T

[NOISE].

And with these being a second order circuit, in

general, the kinds of wave forms that we get are

sinusoidal or decaying exponential in nature.

So, the actual IL of T, when we solve

this circuit might have some wave form that looks like

this, for example, with some initial value IL of zero

and then some solution, of the second order differential equation.

[COUGH]

Now in a well-designed convertor, this low pass filter has a cutoff frequency

that is very low in frequency relative to the switching frequency.

What that means is that the time constants, or the time that it takes

this response, this ringing response, to happen,

is long compared to the switching period.

So one switching period might be just this long, for example.

And in the small ripple approximation, what we're actually doing is

approximating this solution for a short time with a straight line.

And for that reason we also often will call

the small ripple approximation instead, the linear ripple approximation.

And any continuous wave form

for short enough time can be approximated by a straight line.

And so that is what we're in fact doing.

Of course if it's a discontinuous wave form such as a switched

wave form then we cannot in general approximate it with a straight line.

[COUGH]

So we apply the small ripple approximation as appropriate to inductor

currents and capacitor voltages that have responses such as this one.

That we can approximate over the switching period or

a fraction of the switching period for the straight line.

7:35

Okay, we can do a similar thing with the switch in the second

position as well were the inductor left side of the inductor then is connected

to ground instead of VG We get another circuit to solve, and we can

find the, the wave forms for that interval as well, in a similar manner.

So, let's look at, for example, for the

first interval, finding the inductor current wave form.

8:04

So from solving this circuit, we have a

loop right here that the inductor is connected in.

And the loop equation for this circuit is the inductor voltage BL

is equal to the input voltage VG minus the capacitor voltage.

Here it's called V.

So we get this equation for the loop the the inductor is connected in.

Okay.

Now, V of T is the output capacitor voltage, and as we've discussed

so far, we want this capacitor voltage to have small ripple.

And so V of T again can be expressed as its DC component, capital

V, plus the small ripple, and here is a case where we can replace

[INAUDIBLE]

by its DC component capital V to find the inductor voltage for this interval.

And so this is a good approximation over the

short time of when the switch is in this position.

9:09

Okay, so this is a small ripple approximation.

Knowing the inductor voltage we can now find what the inductor current does.

Using the well known defining equation of an

inductor, that V is LDIDT in the inductor.

9:27

So if we plug this expression for the approximate inductor

voltage into there, and solve for DIDT, we get this

equation Which says that the inductor current changes with a

slope that is given by the voltage over the inductance,

with the voltage being equal to a constant value, Vg minus capital V.

9:53

Okay? So, for this first interval, The

inductor current will start at some initial value,

IL of zero. And it will increase with the, this slope.

10:12

And it will increase with that constant slope.

Until this first interval is over, and the, the switch

is changed to position 2, which happens at time DTS.

[COUGH]

' Kay.

Then, during the second interval, we get this circuit,

with the left side of the inductor connected to ground.

And the loop equation now around this loop where the inductor is

connected, says that the inductor voltage VL is equal to minus V.

10:49

We can again use the small ripple approximation to replace V of T

with its DC component capital V, and ignore the ripple during this interval.

And we can plug that voltage into the defining equation of the

inductor again to find the slope of IL during the second interval.

And what we find then is that the inductor current for

this interval changes with a slope equal to minus V over L.

[SOUND]

So we started at some initial value, we went up

during the first interval and ended at some point.

[SOUND]

And then during the second interval we now change with a different slope.

And the slope is negative, assuming the output voltage is positive.

So we get a slope for this interval of minus V over L, and the current

goes down, until the switching period is over, and we end at, at time

TS we end with some final value.

[COUGH]

' Kay, so the small ripple approximation makes it simple

to, to sketch what the inductor current wave form looks like.

We have straight lines that have constant slopes and are easy to solve.

And this greatly simplifies the equations so that

we don't have to write decaying exponential type functions.

And have pages of algebra.

12:26

So then here's a summary of what the inductor

current wave voltage and current wave forms look like.

We found the inductor voltage during each interval, it was positive

and equal to an approximately constant value during the first interval.

And during the second interval, it was

equal to an approximately constant negative value.

13:05

Okay, one thing we can do with this right off the, right

away is, get an equation that helps us choose the value of inductance.

You can see that the actual inductor current wave form has switching ripple.

The current goes up and down with a

traingle, triangular wave form at the switching frequency.

And on top of that it has an

average value or dc component, which is the dc value of inductor current.

That we'll call capital I.

13:51

How much ripple we allow is a design choise, and at least for the

next several weeks we will. Discuss converters where this ripple is

limited in value to maybe ten or twenty percent of capital I at full power.

' Kay, so the ripple, we commonly define the peak to average ripple as delta IL.

And so two delta il is the peak to peak ripple, which is from here to here.

14:22

And from this waveform and knowing the slope, we can

easily write an equation for how large the ripple is.

So for example, during the first sub-interval, with the switch

in position one, the inductor current changes from here to here.

And the net change is two delta I.

14:42

So we can write two delta IL is the

change in the inductor current, and this is simply equal

to the slope during the first interval. Vg minus V capital over

L times the length of the first interval with is DTS.

15:47

Okay, there's another thing we can do besides finding the value of L.

We would like to find the steady state voltage and current in the converter.

So, let's consider now what the inductor current

does during a turn on trangant of the converter.

[SOUND].

So, for example,

[COUGH].

Let's suppose that we, start the circuit with the inductor

current equal to zero. And the output voltage is zero also.

16:31

So we'll start at zero with the circuit turned off.

And at t equals zero, we'll start switching with So we'll

switch between positions one and two with some fixed duty cycle.

Okay?

We've already found what happens, the inductor current

will change with some slope during each interval.

And so during

the first interval, the very first switching

period, the inductor current will go up.

17:10

Okay, well initially what v is the capacitor voltage, vc in fact let's

just call this v, instead of vc. So the voltage is initially

zero at the output, and our slope during the first interval is V G over L.

18:17

' Kay, so we have a slope here, of

minus V over L. Well V, the output voltage,

is nearly zero. so minus V over L is pretty close to zero.

And there's hardly any change in inductor current during

this second interval because the slope is nearly zero.

After one switching

period, we end up with the inductor current at what IL of TS.

18:53

A little bit positive. It's a little greater than 0.

Now we repeat the process.

So, for the second switching period, we'll go up with a slope of Vg minus v over L.

V is a little positive now, so the slope during the second interval,

this slope, is a little bit less than that slope was, last time.

So we don't quite increase as much.

19:22

Okay, and then during the, second, with the switch in position two, for

this position period, we'll go down, with a slope of minus v over

l, v is a little more positive than it used to be, so

this slope is a little more negative than it was during the previous period.

And we end up with some net increase in current but

it's not as much as we increased in the first period.

19:46

And we, we repeat

this process. Okay, with the output voltage

slowly charging and eventually, after many switching

periods, we, we end up at some current. Let's, let me just draw it out here.

[COUGH].

And, and some output voltage where, the amount we go up during

the, the first interval with a slope of vg minus v over l.

20:31

Capacitor voltage

does something also and it ends up at the same place it started.

Okay. And so at this point, this is maybe

at time nTS and here's n plus 1 Ts, so

after n switching periods where n is some

large number we end up in steady state.

Where the, the wave forms become periodic. Each switching

period has wave forms that are the same as the previous switching period.

And the current and voltage end up at the same place they started.

21:11

So you can see that that will happen in when the output voltage rises to a large

enough value. So V is large enough so that

these slopes have adjusted, and we our

final value of n plus one T S equals I N of N T S.

22:42

What we do is we integrate this equation to find the net change in inductor current

over one switching period. So what we can do is integrate both

sides of the equation so integrate VL of TdT over one switching period

[SOUND]

So we'll integrate

say from zero to Ts over one

switching period. And here

we'll integrate from il of zero to il of Ts.

And when we evaluate these integrals, the right hand side gives us i L of

TS minus i L of zero, which is the net change in current.

If I divide both sides by L, I can write the other side of

the equation as 1 over L times the integral of the voltage over one period.

23:40

So, in steady state, then the net change in inductor current must

be zero, and therefore this integral of the voltage must be zero.

Okay, and this

is the basic relationship that, where we say

this is volt second balance on the inductor.

That the integral of the voltage has dimensions of volt

seconds and if you integrate the inductor voltage over one

complete period, that integral or area under the curve must

be zero if the inductor current has no net change

over what that period.

24:20

Another way to write this equation is simply to

divide both sides by switching period TS and in

that case, we recognize that 1 over TS times

the integral of the voltage is the DC component.

26:15

We get an equation and can simplify this equation, D plus D prime

is, in fact, 1. We'll call d prime as 1 minus d.

So, sum of d plus d prime works, works out to be 1.

And we get dvg minus v equals zero.

You can solve this for the output voltage, v.

And we find that the output voltage is dvg.

26:42

Okay, well we already knew that for the buck convertor.

But in fact this was a way to find that the output voltage without

using the arguments of fourier series and low pass filtering.

Here we, we simply draw the inductor wave form and set it's average value to zero.

And this is something we can do for any converter.

So we could similarly

use this approach find the output voltage of

say the boost converter or the Buckman's converter.

[SOUND].

So volt second balance plus the small ripple approximation then

gives us a way to solve for the output voltage.

27:26

Okay. capacitor charge balance, or capacitor

amp second balance, is the dual of inductor volt

second balance. This is a, we can apply a similar

principle to a capacitor in the circuit. What we do is we start with a defining

equation of the capacitor, I is CDBDT in the capacitor, we can integrate both sides

of this, so integral of ICDT is the, C times

the integral of DVC. And if you integrate

over one period, from say 1 0 to

TS, then this part here gives us the

change in capacitor voltage over

one period.

The net change is B C of TS plus BC of zero

and that's equal to one over C times the integral of the current.

28:28

If we operate on steady state, then there is

also no net change in capacitor voltage over one period.

So the left hand side of this equation is zero and

this says the integral of the capacitor current must be zero when

the converter operates in steady state.

28:59

Okay, so we can use the principle of capacitor charge balance to get a similar

equation to find the conditions under which

the capacitor voltage will operate in steady state.

This might be more familiar, than the inductor volt second balance.

Or perhaps, more intuitive, if you put a DC component of current into a capacitor.

Then the

capacitor will continue to charge and if you keep putting

charge on the plates of the capacitor, just positive charge,

say, then the charge will build up and the capacitor

voltage will increase, and it won't operate in steady state.

It's for part of the time we put a positive current, say part of

the period we put a positive current

on the capacitor, making its voltage increase.

For the other part of the period we must put a negative current

to bring the charge back down to have no net change in charge.

And therefore have no net change in capacitor voltage.