0:11

Let's work another example in which we will include

the semiconductor conduction losses in our equivalent circuit model.

The semiconductor conduction losses are the power losses that come

from the forward voltage drops of the semiconductor devices.

So, what we're going to do in this example is to model a boost converter.

We will, again, include the winding resistance of the inductor.

But now, we'll also include the on-resistance of the MOSFET,

and the forward voltage drop of the diode.

So, for the MOSFET,

a typical power MOSFET behaves as a resistance when it is in the on-state.

So, an equivalent circuit that we might use for the MOSFET is

an ideal switch in series with a resistor that we'll call Ron.

2:10

From this plot, we can fit a curve.

So, if we model the diode as forward voltage plus the resistance,

we might approximate this curve with some straight line that's

valid over the intended important operating range.

So, this intercept voltage would be VD and the slope here would give us RD.

The slope on these axis is actually one over RD.

Okay, so for the diode then,

we have this model when the diode is on.

If we put it in series with an ideal switch like the MOSFET model,

2:55

then we get a model that can be used in the switching converter for both intervals.

So, what we do is we then, as usual,

write the circuit when the switch is in the two positions.

So, what we call Position One is when the diode is conducting and the MOSFET is off,

and the circuit reduces to this,

which now includes the on-resistance of the MOSFET in the subinterval one circuit.

For subinterval two, the MOSFET is off,

or its switch is off,

the diode is on, and the circuit reduces to this,

where now we have the VD and

RD model for the diode forward drop in our equivalent circuit,

or in our subcircuit.

With these, as usual,

we can write the inductor voltage and capacitor current waveforms.

So, here, as before,

vL(t) is the voltage across the ideal inductor part of the lumped element inductor model,

and iC(t) is the current in the capacitor.

The capacitor current waveform is the same as it has always been.

So far, none of the losses that we model have affected that.

But all of the loss elements add terms to the inductor voltage waveform.

But no matter, we can work out the average inductor voltage anyway.

VL(t) is D times the voltage during the first interval

5:13

So, now, we construct the equivalent circuit,

and I'm going to do this on some graph paper also.

So, here is the inductor voltage equation that we got.

The average inductor voltage is zero and it is equal

to VG minus IRL when we collect terms

minus DIRon minus D-prime VD minus

D-prime IRD minus D-prime V. So,

there are more terms but we can write this equation and

construct an equivalent circuit that includes an element for each term as usual.

Okay, so, the first term is VG,

we'll write a VG voltage source.

6:54

If so, then that means that the current through the resistor is D times I,

or the other choice is that we associate D with the value of the resistor.

So, perhaps, this should be a resistor of value D times Ron that has a current of I.

So, what do you think is the correct answer?

Well, the correct answer is that this is

a resistor value D times Ron that has a current I flowing in it,

and the reason is that the current in this loop

for this loop equation is the inductor current.

If you recall in previous lectures,

I've drawn a little dashed inductor to model where the inductor goes.

It has an average voltage that is zero,

and it has a current that is the inductor current I.

So, this is the loop equation that contains

the voltages around the loop where

the inductor is connected and the current in that loop is,

in fact, the inductor current I.

So, we have to have a resistor with a value I,

and therefore, the value of the resistor is D times Ron.

Well, so, now, we have an element,

a resistor whose value varies with the duty cycle,

which is a strange element.

Why do we get such a thing?

Well, the reason is that the on-resistance is only in the circuit when the MOSFET is on,

which is during the first interval of length that has a duty cycle of D. So, for example,

if we let D go to zero and we never turn the MOSFET on,

then we would not expect there to be any power loss in this resistor.

The power loss that you get is a function of the duty cycle,

or the fraction of time that there is current flowing in this resistor making loss.

So, this model predicts that the power loss in this resistor is the current-squared,

or I-squared times the element value DRon.

Now, in fact, what's happening is during the first interval,

we're getting a power loss of I-squared Ron,

but we're only getting it during the DTS interval, and so,

the average power loss with zero loss in this resistor for the second D-prime interval,

the average loss will be the duty cycle times I-squared Ron.

So, by changing the effective value of this resistor,

the model correctly predicts the power loss in the MOSFET.

Okay, let's keep going.

The next element is minus D-prime VD.

Okay, so VD is the diode forward voltage drop in our diode model,

and it is a constant voltage.

I'm going to draw it as an independent source.

It's the value of D-prime VD and it's plus to minus from left to right.

So, the power loss in this element also depends on

the length of time that the diode conducts and the VD source gets multiplied by D-prime.

10:22

Finally, we have the output voltage multiplied by D-prime.

This is a voltage that depends on the output capacitor voltage,

and we'll draw it as a dependent source of value D-prime V. So,

there is our circuit model for

the inductor loop that includes all of the terms in the inductor equation.

Okay. Here is the model we just constructed.

The capacitor node equation model is the same as it's been in the previous examples,

at least in the boost examples.

So, we get this equivalent circuit as usual,

and we can now combine the two dependent sources into a transformer.

So, here are the two models written together.

We combine these sources into a transformer.

We get a D-prime to one boost-type transformer,

and this is the equivalent circuit model then for the converter.

Here is the model that we derived,

and let's solve the model now for the output voltage.

The way to do this I think that is easiest is to push

all of the elements to the secondary side and combine them together.

So, first of all, I'm going to combine the two independent voltage sources.

Since the VD voltage source is in series with these resistors,

we can swap the order without changing

the circuit and I can combine it with VG into one source.

So I'm going to write one source that is a value,

VG minus D prime V sub D. Then I can combine all the resistors together,

14:01

And that's the expression for the output voltage.

You can see the divider ratio again is changing quite a bit with duty cycle,

all of these resistor values effectively are changing,

as the duty cycle changes,

as does the effect of the input voltage.

So here is the result that I just derived on

the previous slide and it's manipulated to make dimensionless terms.

So this is writing the result in a nice form where

the first term is the ideal conversion ratio with no losses,

the second term is

the numerator term that accounts for the losses that appear in the numerator,

and in this case the V sub D voltage drop of the diode appears in

this term and the final term is

the denominator that comes from all the resistive voltage divider terms.

We can work out the efficiency as well, as usual.

So we find the input power,

15:22

The input power is VG multiplied by the current on the input terminals,

which is the inductor current I and

the output power is the output voltage V multiplied by

this current coming out of the terminals and that current

is the current I reflected through the turns ratio,

which gives us D prime I.

So V times D prime I is the output power.

So if you divide those,

then the efficiency becomes equal

to V over VG times one over D prime and the Is cancel out.

So, if we go back to our previous expression for the V over

VG times D prime not one over D prime.

So, if we go back to our previous expression and multiply V over VG times D prime,

the one over D prime term cancels and the remaining terms are the efficiency and so,

we get this expression for the efficiency.

Basically, we have a term in the numerator from the V sub D source and we have

all the resistive terms in the denominator

and this expression again because it's written in this nice normalized form,

tells us, really the relative effects or the effect

of each of the loss elements on the efficiency.

So, here if we want the V sub D term to have a small effect on efficiency,

then V sub D should be small compared to VG over D prime.

Likewise at the on-resistance,

Ron is to have a small effect on efficiency,

then Ron should be small compared to D prime squared R over D and so on.

Okay, one last little point about prediction of

the conduction losses using this averaged model.

So we found with the MOSFET on resistance,

that the model effectively changes the value of Ron,

it gets multiplied by a factor of D. This makes the model

correctly predict the average power loss in the on-resistance of the MOSFET.

We should take this just a little bit further and recognize that we've also made

the small ripple approximation and

so these models are ignoring the effects of switching ripple.

So we would expect the accuracy of the models to be good as long as the ripple is small,

but if the ripple gets large then the model could be inaccurate.

So, let's consider for the case of the MOSFET on-resistance,

how does the ripple effect the prediction of loss?

Okay. So, here's the plot of the current and the MOSFET,

this I of t here is the MOSFET current.

Basically, we have a MOSFET that we're modeling

with a switch that

turns on and off as the MOSFET turns on and off in series with an on-resistance.

There's a current I flowing in this switch,

this is I is different than the inductor current,

this is really the MOSFET current and so,

the power loss in the on-resistance is the RMS current,

so irms squared times the on-resistance.

And here's what that I of T looks like,

it's zero when the MOSFET is off and when the MOSFET is on,

the MOSFET current follows the inductor current.

So, when we ignore the ripple,

we're really drawing I of T as case a shown here where there is no ripple.

With no ripple, I then simply is looks like this.

When it's on, we have a current capital I the DC inductor current,

when the MOSFET is off,

we have a current zero.

And in that case,

we can show that the RMS value of I, for this case,

a is the square root of the duty cycle times capital I.

20:38

is this value root D times

I and then times another factor that accounts for the ripple and

that factor turns out to be one plus one-third delta i over capital I squared.

So, this extra term accounts for ripple.

Here is the value then of the RMS current including ripple for two cases.

In case B, the ripple Delta I is 10% of the DC current I,

and we would expect those small ripple approximation to be pretty accurate there.

So, you can compute this term and see how big the RMS current is,

it turns out to be 0.167% higher than the ideal case with no ripple.

So, this factor here is this term right here and it's very close to one.

So when you square this to find the power loss,

we find that the computed average power loss is

0.33% higher than our model with the small ripple approximation predicts,

0.33% is a pretty small number,

pretty small difference and I think the model is very good here.

In fact, the tolerance on the on-resistance is a lot more than that.

Case c is the extreme case shown here where the ripple is equal to I.

So we have a very large,

we have 100% ripple relative to the DC current and

the actual current waveform starts at zero and ramps up to twice the average.

So, in that case where Delta i equals i,

we can calculate this quantity and it turns out to be 15.5% higher.

So, 1.155.