0:00

In the last

lecture,

we modeled the copper loss of the inductor of a

goose converter. by inserting

an extra resistor and series effectively with the inductor in a lumped

element model. And with that model, we then calculated

the effect of this resistor on the output voltage of the goose converter.

In this lecture we're going to extend that result

and construct an equivalent circuit model to go along with,

0:46

So, here are the equations that were found in the last lecture for this example.

the first equation was from inductor volt second balance

and the second equation we got from capacitor charge balance.

And the object now is to find an

equivalent circuit to go along with these equations.

Okay, the approach that we're going to take is to

view these equations as the loop and note equations of our equivalent circuit model.

And, indeed, the first equation was found by finding the

average voltage around the loop where the inductor is connected.

2:10

Well this process of going backwards turns out to not be unique.

There's more than one way to construct a circuit that satisfies these equations.

And so, we have to be careful how we do it to make

sure that our resulting model has physical significance.

And my advice here is to use the, the

equations in the form that they come without further manipulations.

So the first equation was found from inductor volt second balance.

We should keep it as a sum of voltages around a loop corresponding to a Kirchhoff

voltage law equation and don't manipulate this.

2:58

Then this first equation would have terms

that have dimensions of current rather than voltage.

And you might think then that this is,

a node equation with currents flowing into a node.

3:12

while that equation might be mathematically correct.

The resulting model is not physically connected to the actual converter.

So we don't manipulate these equations or do any kind of algebra on them.

We construct circuits to go along with the equations right away.

3:45

Okay, it's on the next slide but I'm going

to do it actually on some graph paper here.

So our equation was that 0 is the average inductor voltage which is

Vg minus IRL minus D prime V.

4:06

Okay.

So, this is an equation describing the loop where the inductor is connected.

And it's describing actually the DC

components of the voltages around that loop.

So I'm going to draw a little dashed

inductor in that loop. That has an average

inductor voltage equal to 0.

And there's a current flowing through this inductor that is

the inductor current DC component that we've been calling capital I.

Now, the average inductor voltage is 0 and in fact the

impedance of the inductor at DC goes to a short circuit.

So in fact, this is a short circuit in our DC model.

But it's a good starting place to have a physical

connection to the actual model and the actual circuit.

To recognize that the current in this loop is the inductor current I and the

voltage is VL, is the average inductor voltage

that is defined with a, a reference direction.

Or polarity that is consistent with the direction of the current.

5:41

And since it's on the other side of the equation then as our

current goes around the loop and it's going from plus to minus through VL.

Because Vg is on the other side of the equation our current

will go from minus to plus, so with the other direction, through Vg.

'Kay, the next term is I times RL.

6:02

That term sounds like the voltage across a resistor having

current I which is the current in this loop multiplied by a resistance RL.

So we'll draw a resistor here and label it's value RL.

And the voltage across it with I going from left to

right would be a voltage from left to right of value IRL.

And that polarity has the opposite polarity going around the loop

as the Vg term had which is consistent with a minus sign.

6:40

Well, here is how.

V is the output voltage, this isn't exactly the

output voltage but it's D prime times the output voltage.

So I am going to draw for now a dependent source of value D prime V.

So it's a voltage source that depends on the output

voltage V but it's multiplied by D prime.

7:15

Okay, so here is an equivalent circuit whose loop equation is

the same as the equation that we got from volt second balance.

And it is the inductor loop part of our equivalent circuit.

Okay, let's do the other

[COUGH],

other equation.

So the other equation was from capacitor charge balance which said that the

average current through the capacitor our DC component of capacitor current is 0.

And what we found there is that current was equal to

D prime I minus the load current capital V over R.

7:56

Okay, this equation was found, found by charge balance

on the capacitor and again, basically, it was found

by finding computing the DC components of the currents

going into the node where the capacitor is connected.

So we can view this as a sum of

currents equalling 0 which sounds like a node equation.

So, I'm going to draw a node.

8:35

so the current through this capacitor or average

ic is 0 and we know the voltage across

the capacitor is V, capital V actually in our

DC module, the DC component of the output voltage.

8:51

In our model, our DC model, the what?

The impedance of our capacitor at DC is an open circuit.

So this will actually be an open circuit but

for reference, we know that V is the voltage at the node

and ic is coming out of the node and flowing into the capacitor.

9:11

So what our equation says or charge balance equation says

is that there are two other currents flowing into the node.

The first one is D prime I, okay, so this

is a current that is dependent on the inductor current.

Which was in our other circuit, our earlier circuit.

But it's not just i, it's D prime times I.

So I'm going to draw a dependent source with a current d

prime i that flows into the node.

10:33

When we draw the two circuits together, we get this, this top circuit.

So the left hand side was the inductor

loop equation where the inductor went right here.

And the right-hand side was the capacitor

node equation where the capacitor went right there.

[COUGH]

the last thing to say about this is

that we recognize we have two dependent

sources. That as discussed in section 3.1

are equivalent to a DC transformer. So we have

one dependent source as D prime V source depends on the voltage

V across the other source with the factor of D prime.

And the dependent current source depends on the current I

through the voltage source with the same factor of D prime.

So since the factors are the same and they

depend on the voltages and currents of the corresponding sources.

11:46

I think the easiest way to understand what the turns ratio

and polarity mark should be, is to look at the voltages.

So we have a voltage V on the output side

and D prime V on the input side of our transformer.

So the turns ratio is D prime to 1.

Which will get us the correct polarities of voltages.

And to the, to look at the polarity marks we have

to look at the polarities of the voltages. So, you have V is positive on the top on

the secondary side and D prime V is positive on the top on the primary side.

So the polarity marks are of the same phase.

We have the dots on the top in both cases. As a check we should look at

whether the currents work, so here in our model we have a current I flowing

into the dot on the primary side.

And coming out of the dot on the secondary side should

be D prime I, and that's indeed what the model says.

So the currents are consistent with the transformer model as well.

'Kay so we have an equivalent circuit now

for our boost converter that has the ideal transformer.

In this case D prime the 1 is its turns ratio or if

you want to call it 1 to M, M is 1 over D prime.

13:17

Okay, an easy way to solve this I

think is to push the elements through the transformer.

We have actually already done this exercise.

We pushed Vg and RL through to the secondary

side if we want to solve for the output voltage.

So, when we do that Vg would be multiplied by the by M or divided by D prime.

14:03

And recognize that the second area voltage is V and we

can then just solve this circuit using the voltage divider formula.

V would be the voltage source Vg over D prime

multiplied by the divider ratio of these two resistors here.

And, if you like you can divide top and bottom by R to write the

expression in this form which is the way

we wrote the expression in the last lecture.

14:53

so if you say, push the load resistor through the transformer to this side.

We'll get effectively between these terminals a

resistor that would be D prime squared R.

[COUGH]

And we can then find the current.

I would be Vg divided by the sum of the two resistors like this.

15:25

To find the efficiency, we find the input power and the output power.

So, the input power is the power going into the converter terminals, here.

And the output power is the power coming out the output side.

16:52

Okay, we previously found that V over Vg for this converter is 1 over D prime.

Times, 1 over 1 plus, what was it? RL over D prime squared R.

So we have to take that V over Vg and multiply it by D prime.

So if you multiply it by D prime the, the 1

over D prime factor goes away and we and then this

added multiplying factor is in fact the efficiency.

17:26

Here's the plot of the efficiency.

[COUGH]

So for different values of RL actually for different values of RL over

R, we can plot a series of curves of efficiency versus duty cycle.

And what you can, can see is that for

low duty cycles the efficiency is a lot higher.

Where as for high duty cycles the efficiency comes to some

shoulder and then drops off quickly. So, there is a range of duty cycles

from 0 up to some value depending on RL where we have high efficiencies.

And beyond that the curve will get low

efficiencies very quickly, the efficiency drops off very quickly.

And this drop off coincides with, the, the voltage,

the output voltage deviating substantially from the, from the ideal case

the 1 over D prime curve. So, by, we

can take our volts second balance and charge balance equations and view

them as the loop and node equations of our equivalent circuit model.

And, and, therefore we can reconstruct loop equations

that go with our volt second balance equations.

And reconstruct the equivalent circuits that go with those.

We can also reconstruct equivalent circuits based on

node equations that corresponds to the capacitor charge balance.

And finally we can combined these using

ideal DC transformers to get a final model.

This final model has a physical interpretation.

You plug this into a larger system and then

model how a converter works in the larger system.

And we can solve it for things such as the efficiency or do

manipulations on the circuit to find things

like voltages and currents of the converter.