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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'll 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.

So, you recall from the previous lectures we discussed the buck converter,

having a switch and having a low-pass filter that removes the switching 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 perfect low-pass filter,

the low-pass filter will have attenuation,

but it won't completely remove the switching harmonics of the waveform, 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.

Then in addition to that,

we have some small ripple that is at the switching frequency and it's harmonics,

that where small amount of

the switching ripple gets through the filter and appears in the output.

As sketched here, this ripple

is actually probably a lot larger than it would be in practice.

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 10 millivolts or some very small number.

So therefore, we can write the equation of

the output voltage V of T as being equal to capital V,

which here, I'm using the capital letters to denote the DC components.

So, capital V is the desired DC component equal to DVg,

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.

So in a well-designed convertor,

this LC filter will have lots of attenuation and

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

So, we call this the small ripple approximation,

where 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 it's desired DC component.

This has the effect of decoupling 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. Okay.

Brief discussion here, 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 waveforms in the circuits,

such as the switch output voltage,

Vs of t. Instead,

we apply the small ripple approximation to continuous waveforms and

specifically the inductor currents and capacitor voltages of the circuit.

4:23

Okay. So, let's take an example here of a buck converter,

and here's our original circuit,

and we would like to solve for the wave forms such as the inductor current waveform IL

of t and the output voltage waveform V of t. Okay.

So, with the switch in each position,

the circuit reduces to a linear circuit that we can solve

by standard sophomore circuit analysis.

So, for example with the switch in position 1,

the circuit reduces to this one in which

the left side of the inductor is connected to Vg.

Okay. This is a second order RLC circuit and we could solve it,

in fact in principle,

we can simply solve the second order differential equations of the circuit,

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and with this being a second order circuit, in general,

the kinds of waveforms 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.

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.

For that reason, we also often will call

the small ripple approximation instead the linear ripple approximation.

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

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.

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 with a straight line.

7:34

Okay. We can do a similar thing with the switch in a second position as well,

where the left side of the inductor then is connected to ground instead of Vg.

We get another circuit to solve and we can find

the waveforms for that interval as well in a similar manner.

So, let's look at,

for example, for the first interval,

finding the inductor current waveform using the small ripple approximation.

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

VL is equal to the input voltage Vg minus the capacitor voltage here,

it's called V. So,

we get this equation for the loop that 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 it's DC component capital V plus the small ripple,

and here is a case where we can replace V of T by it's DC component, capital V,

to find the inductor voltage for this interval.

So, this is a good approximation over

the short time of when the switch is in this position.

Okay. So, this is the 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.

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. Okay.

So, for this first interval,

the inductor current will start at some initial value IL

of 0 and it will increase with this slope.

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Okay. 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 like this.

We can again use the small ripple approximation to

replace V of T with it's 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.

What we find then is that the inductor current for this interval

changes with a slope equal to minus V over L.

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To design the inductor and choose an inductor value,

we generally will try to limit the amount of switching ripple in the inductor current.

How much ripple we allow is a design choice and at least for the next several weeks,

we will discuss converters where this ripple is limited in value to

maybe 10 or 20 percent of capital I at full power.

Okay. 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.

Okay. From this waveform and knowing the slope,

we can easily write an equation that for how large the ripple is.

So, for example during the first subinterval with the switch in position one,

the inductor current changes from here to here and the net change is two delta i.

So, we can write two delta iL is the change in inductor current,

and this is simply equal to the slope during

the first interval vg minus capital V over L,

times the length of the first interval.

Which is DTS.

Okay. So, knowing the slope and the time,

we can find the change and we can solve then for delta iL,

just divide both sides by two and we get this expression.

Okay. This is the equation that is commonly used

to calculate the value of inductance and choose an inductor for the converter.

So, we can solve this equation for L. Push L onto that side,

and get this equation.

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With the circuit turned off,

and at t equals zero will start switching.

So, we'll switch our 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.

So, during the first interval,

the very first switching period,

the inductor current will go up.

With a slope that we've already found,

is equal to Vg minus V over L. 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 Vg over L. Okay.

So, we go up until at the end of the first interval at time DTS,

we've reached some positive current.

Okay. What happens to the output voltage?

Well, this inductor current is small but it's a little more than zero.

That inductor current flows to the output and it

will slightly charge the capacitor voltage up a little.

So, v is still small but it's a tiny bit more than zero.

Okay. During the second interval with the switch in position two,

we found that the inductor current changed with a slope of

minus V over L. Okay.

So, we have a slope here,

a minus V over L. Well,

v that 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

the second interval because the slope is nearly zero.

Okay. After one switching period,

we end up with the inductor current at what iL of Ts,

a little bit positive,

it's a little greater than zero.

Now, we repeat the process.

So, for the second switching period,

will 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 increases much.

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this is maybe at time nTS,

and here's n plus one Ts.

So, after in 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 waveforms that are the same as the previous switching period,

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

So, you can see that that will

happen 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 iL of n plus one T S equals iL of nTs.

At that point, we say that we are in steady state.

So, the waveforms from then on,

have the correct DC values where the converter is supposed to operate.

We've gone through some transient that eventually

has settled out and we've reached steady state.

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So, we'll integrate say from zero to T_s over one switching period,

and here, we'll integrate from i_L of zero to i_L of T_s.

When we evaluate these integrals,

the right-hand side gives us i_L of T_s 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 one over

L times the integral of the voltage over one period.

So, in steady-state, then the net change in inductor current must be zero.

Therefore, this integral of the voltage must be zero.

Okay? 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.

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 that period.

Another way to write this equation is simply to divide

both sides by the switching period, T_s.

In that case, we recognize that one over

T_s times the integral of the voltage is the DC component.

So, another way to say this is that the DC component of the voltage applied to

the inductor is zero when the circuit operates in steady-state.

We're going to use this notation of angle brackets

around a variable to mean the average value.

So, the average value of the inductor voltage is zero in steady state.

24:58

Let's try that.

Try finding the applied volt-seconds to the inductor in our buck converter.

We previously found this waveform was the voltage waveform applied to the inductor.

The inductor saw a voltage of Vg minus V during

the first interval and minus V during the second.

So, we have to integrate the inductor voltage,

and the easy way to do that is simply to find the area under the curve.

So, during this first interval,

we have an area that is the base DT_s.

times the height.

The height is Vg minus V. During the second interval,

we have an area that is the base here, D prime T_s.

This is the length of the second interval times the height,

which is minus V. So,

if we add these two areas,

they have to add up to zero.

Okay? So, here is that area.

If we equate it to zero like this,

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So, volt-second balance plus

the small ripple approximation then gives us a way to solve for the output voltage.

Okay. Capacitor charge balance or capacitor Amp

second second is the dual of inductor volt-second balance.

We can apply a similar principle to a capacitor in the circuit.

What we do is, we start with the defining equation of the capacitor,

i is C dv/dt in the capacitor.

We can integrate both sides of this.

So, integral of i_C dt is the C times the integral of dv_C.

If you integrate over one period, from say,

zero to T_s,

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then this part here gives us the change in capacitor voltage over one period.

Then net change is v_C of T_s minus v_C of zero.

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

If we operate in 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.

This says the integral of the capacitor current must be

zero when the converter operates in steady-state.

Again, another way to write this,

we can divide by T_s multiply by C and find that

the DC component of capacitor current must be zero when steady state.

28:58

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 second or perhaps more intuitive.

If you put a DC component of current into a capacitor,

then the capacitor will continue to charge.

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.

If 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.