Sect. 3.2 Inductor Copper Loss

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En provenance du cours de University of Colorado Boulder

Introduction to Power Electronics

1390 notes

University of Colorado Boulder

Introduction to Power Electronics

1390 notes

Cours 1 sur 6 dans Specialization Power Electronics

This course introduces the basic concepts of switched-mode converter circuits for controlling and converting electrical power with high efficiency. Principles of converter circuit analysis are introduced, and are developed for finding the steady state voltages, current, and efficiency of power converters. Assignments include simulation of a dc-dc converter, analysis of an inverting dc-dc converter, and modeling and efficiency analysis of an electric vehicle system and of a USB power regulator. After completing this course, you will: ● Understand what a switched-mode converter is and its basic operating principles ● Be able to solve for the steady-state voltages and currents of step-down, step-up, inverting, and other power converters ● Know how to derive an averaged equivalent circuit model and solve for the converter efficiency A basic understanding of electrical circuit analysis is an assumed prerequisite for this course.

À partir de la leçon

Ch 3: Steady-State Equivalent Circuit Modeling, Losses, and Efficiency

Equivalent circuit modeling of switching converters, to predict their power conversion functions and efficiency under steady-state conditions

Sect. 3.2 Inductor Copper Loss12:32

Sect. 3.3 Construction of Equivalent Circuit Model19:24

Rencontrer les enseignants

Dr. Robert Erickson
Professor
Electrical, Computer, and Energy Engineering

As a first

example of

modeling loss in a converter,

let's model the resistance of the wire in the inductor of a boost converter.

The resistance of the winding, or DC copper loss

of an inductor, is important. because it has a first

order effect on the cost and efficiency of, of the inductor and the converter.

If you don't care how large this resistance is.

I can wind you an inductor that is arbitrarily small and inexpensive.

That can, give you the desired inductance, and handle the desired

current without saturating.

but it will use many turns of very

thin wire that will have a very high resistance.

So the inductor then will actually have very large loss and will burn up.

So, the resistance of the wire is important.

And one of the important things we have to specify and design,

[COUGH].

In order to optimize the cost and efficiency of our converter design.

So, to model this, The common way to do

it is with a lumped element model, as shown here.

Where we replace the physical inductor with

a, a model, consisting of an ideal inductor.

Having a voltage VL.

In series with a resistor, that is the resistance of the winding.

' Kay?

So between these terminals, this is a model of the physical inductor.

And what we do is we insert this model into the converter circuit like this,

and then analyze the circuit.

so here is our inductor model. Okay?

The ideal part of the inductor.

with its voltage V sub L, still must obey inductor volt-second balance.

But now we have an added resistor and series in the circuit, that

has a voltage drop equal to the inductor current multiplied by, the resistance.

[COUGH]

So, we apply a volt-second balance to VL of t in this model.

And, as usually, we plot, apply charge balance to the capacitor current.

and we can get a set of equations that we can

then solve for the output voltage and other quantities of the converter.

Okay?

So let's do it.

Here's our, here's our circuit now to be analyzed.

Here's the inductor

voltage, and the voltage of the ideal part of the inductor model.

And we have a capacitor current there.

And as usual we, we now write the circuit with the switch in a two positions.

And proceed to work out the waveforms of vl of t and ic of t.

So with a switch in the first position we get this circuit.

We can identify what the inductor voltage is from this loop equation.

The inductor voltage Vl will be equal to the input voltage Vg.

Minus the voltage drop on the resistor RL, which is iRL like this.

the capacitor current for the switch in the first position is

found from the node equation at the output, and as usual.

It is equal to the load current. Actually, minus the load current.

so ic is minus V over R.

We now apply the small ripple approximation, as usual.

So the output voltage, v of t, we replace with its dc component, capital B.

And additionally, on the inductor.

We will assume that the inductor has small ripple as well.

So we

will replace i of t with its DC component, capital I.

Similar for the switch in the second position we get this circuit.

The inductor voltage this time is the input voltage VG Minus

the drop on the resistor iRL minus the output voltage V.

And the capacitor current is the inductor current i minus the load current V over R.

So we get these equations and we

again apply the small ripple approximation as appropriate.

Okay?

So here are the waveforms then, this is

the inductor voltage waveform, and the capacitor current waveform.

Both of these must have zero average when the converter works in steady state.

the inductor voltage waveform now has some additional terms that

come from the voltage drop on the, the winding resistance.

How we can apply volt-second balance to this as usual.

So, zero the average inductor

voltage is zero, and it would be D times the value during the first interval.

Vg minus iRL plus D prime times the value during the second interval.

which is vg minus iRL minus V.

Okay?

we can collect terms and equate this to zero

and we get this equation for inductor volt-second balance.

Okay?

In the ideal case, we were able to solve this for the output voltage.

And get an expression for the ideal output voltage.

But now have an added term that comes from the inductor winding resistance.

And there's an added variable, capital I.

the inductor current at this point is an unknown.

So we have one equation with two unknowns and we can't yet solve.

So to get a second equation we get that from the charge balance on the capacitor.

So the average capacitor current must be zero on steady state.

and it's, we actually get the same waveform and same

equation as in the ideal case for the capacitor current.

and when we collect terms then we get this

equation, which is a second equation that relates the inductor

current and the output voltage.

[SOUND]

Okay?

So, when taken together then, we have these two equations.

First is from volt-second balance, the second is from charge

balance and we have two unknowns, capital I and capital V.

So we can solve now to find the output voltage.

For example, solve the second equation for I, and plug

the result into the first equation, and then solve for V.

And if

you do that, here is the result, V over Vg.

I've written this equation in a nice form.

It has a.

Two terms, the first term one over D prime

is the result for the ideal case with no loss.

And the second term

is, and we can think of as a modification

of how the inductor resistance affects the output voltage.

And you can see that if you let RL goes to zero, the second

term goes to one and we get the expression 1 over D prime, okay?

I've also written the second term in a nice form

where every term is normalized and the denominator is 1 plus.

A function of RL.

So, as RL is increased, the denominator

is increased, and the output voltage goes down.

If you would like RL to have a small affect, on the output voltage

then we need RL to be small compared to the term it's divided by.

Which in this case, is D prime squared R.

So for example, if RL was 1% of D prime squared R.

Then the denominator would be 1.01, and

the overall second term would be approximately 0.99.

So the output voltage would decrease by 1%.

It's nice to write the second term in this form because we can compare, see what

to compare RL to, and how, get a good idea then of how small RL needs

to be. Okay, on the right we can see

A plot of this expression V over Vg, for different values of RL over R.

The top curve is where the RL is equal to zero

and we get the ideal case, that's 1 over D prime.

This function goes to infinity as, as D goes to 1.

the lower curves are for different values of RL, and you can see that as RL

increases the output voltage goes down as we would expect.

But not only that there's a major

qualitative change, where the curve actually bends over.

It reaches a maximum and then it starts decreasing.

And at D of one the curve goes to zero. So this is a pretty big difference.

You add a small amount

of RL, anything bigger than zero, and at D

of one the curve goes to zero instead of infinity.

So what's exactly going on there?

Well we can go back and look at the original circuit.

What happens in the original circuit as you let D approach one?

Well, if D goes to one and the switch is always in position one and never

in position two, so you just connect the

inductor across Vg, and the inductor current is large.

And, in fact, with zero RL, the inductor current tends to infinity.

and the volt-seconds don't even balance.

if there is some D prime interval, it's greater than zero.

Then, during the D prime interval, some current will go

to the output and can charge up the output voltage.

in the ideal case, as D approaches zero, the average current driving the

output is infinite amps times something approaching 0 duty cycle.

And, what is the limit?

well, the ideal equation tends to, infinity,

at least as approached from the left.

But it's a non physical answer.

And with any amount of RL at all the

the inductor current is limited to a finite value.

And as D goes to one, in any event we, we get

essentially zero average current deriving the output, and there's no output voltage.

And so The

actual case with the loss resistance is the, the physical answer.

In which the output voltage is limited, and doesn't go to infinity at

[UNKNOWN]

1. And, in fact, it goes to zero.

Moreover, we can see that, the maximum voltage that

we can attain is limited by the loss resistance.

So, for example, suppose we wanted our converter

to boost the voltage by a factor of three.

Well, if an RL over R was, if you have RL over R equal to 0.05, it's not, no good.

It's not going to work, because we can't make enough output voltage.

RL over R equal, equals 0.02 looks like it would work.

And 0.01 would be even better.

So the peak amount that we can boost by, is, is a function of RL.

So, we've seen, in this first example, how we can model loss.

And in this case, we modeled the inductor winding loss with a lumped element model.

That, that had an ideal inductor and series with an effective resistor.

The ideal inductor still obeys, volt-second balance, although the,

the total inductor model does not, because of the, resistor.

So, in general, we can model

losses by adding elements to our converter, and then applying

volt-second balance to the ideal part of the inductor model.

And charge balance to the ideal part of the capacitor model.

We then get a set of equations that we can solve for things like the output voltage.

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