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.