0:15

Now let's start to include some non-idealities.

And see how that modifies the global and circuit model for the transformer.

The first non-ideality that we're going to include is that we're going

to stop assuming that the permeability of the core is infinite.

Instead, we will assume a finite permeability for the core.

0:39

If you stop assuming that the permeability of the core is infinite,

then the reluctance of the core will no longer be zero.

Instead, the reluctance will be some finite value greater than zero.

If we know to go back to the magnetic circuit model for

our core, we realize that no longer does N1,

I1, plus N2, I2 equal zero.

As was the case when the core reluctance was zero.

If you recall, the magnetic circuit model for

this core was simply two MMF sources.

One, N1, i1, and another pointing

in the direction such that is also

throwing flux in the same direction,

and an N of source value of N2 i2.

1:43

Now since Rc is no longer zero, we can compute

what the core flux is going to be.

And we can write good joules voltage law

equivalent equation here, which would essentially be this expression.

From this, we can again determine the total flux in the core via C,

2:31

Let's start by looking at the first winding and

find the flux that links back into winding 1, which we also call the primary winding.

In this first winding, the flux that links back is the gold flux phi C,

as given by this expression, times the number of terms N one.

So that expression is right here.

We can substitute phi C into that

expression and what we will get for

lambda one, the flux linking back

into winding one is n1 squared,

i1 plus n1 n2 i2 divided by Rc.

I can pull N1 squared over Rc come out of this expression.

That leaves behind i1 plus N2 over

N1 times i2.

We can now find the voltage that develops at the terminals of the first winding

by substituting this flux linkage back into the expression for the voltage,

since the voltage is simply the time derivative of the flux linking back

into the first winding.

We substitute that,

we get essentially a similar expression,

N1 squared over Rc, but now with

the time derivative of the currents

i1 plus N2 over N1 times i2.

So what does this expression for

the voltage across the first winding mean in terms of the equilibrium circuit model.

For that let's define this term, i1 plus N2 over N1 i2,

as another current which we will call i mu1.

4:51

Let's also define this term N1 squared over

Rc as Lm1, which we will soon see,

is the magnetizing inductance of this transformer reflected to the primary site.

If you make these definitions, then real one simply becomes

l new one times the time derivative of I new one.

6:48

The term's ratio N1 is to N2, and

with the dots shown.

And then ask the question,

what current here would actually give

me the current n2 over n1 times i2.

Well the answer's quite simple.

Since the currents transform by the inverse of the turns ratio,

this current will simply be i2.

7:21

This, essentially, is the equivalent circuit model for

our transformer With the magnetizing inductance.

It matches the equation we have since it simply says that the voltage

we want across terminal 1 is equal to this inductance,

which we would call the magnetizing inductance, L mu 1.

And the subscript one is used because this inductance is shown

on the primary side of the transformer.

Note that we can move the magnetizing inductance off

the transformer On the secondary side of the transformer.

We can do that in one of two ways,

we can simply use the method we know of taking an impedance and

transforming it through an ideal transformer in which case Lμ1,

when it moves to second reside would simply get multiplied by

N2 over N1 squared, and that would be what we call L mu 2.

We could have also directly derived the value for

L mu 2 by doing this entire analysis for

winding 2 and getting an expression for V2 instead of V1.

8:38

In either case, we would have gotten the same answer for L Mu 2.

Remember that there is only one magnetizing inductance for a transformer.

The magnetizing inductance essentially models

the energy that is being stored in the core of the transformer.

And there is only one core and hence only one magnetizing inductance.

You can choose to show it on either side of the transformer.

You could also choose to split it up and put part of it on one side and

part of it on the other side, but the sum of the magnetizing inductance.

Once it is reflected to one side,

must match the expression that we've developed here.

So for the case of the primary side, the magnetizing inductors

would simply be N1 squared over the reluctance of the core.

Note that this also makes intuitive sense.

Since the magnetizing inductance would be, essentially,

the inductance of the structure,

if we looked into the primary side of this, and asked the question.

What is the inductance that we see through this winding,

when we completely ignore the second winding,

effectively leaving the second winding open or moving it all together.

In that case, if you remember from our lecture on inductance,

the inductance is simply given by the square of the tons ratio

which in this case will be N one squared divided by their

of the through which the flux flows, which in this case is simply Rc.

This is exactly the expression we have here.

On the other hand, if you wanted to reflect this magnetizing inductance to

the secondary side, then you could look into the secondary side and

ask what is the magnetizing inductance looking from that side?

And in that case we would get L mu 2 which would simply be

N2 squared over Rc.

Note that L mu 1 and L mu 2 do not exist at the same time,

they're essentially the same thing Just depends where you're looking from.

And if you're looking from the first winding that's the value

of the magnetizing inductance.

If you're looking from the secondary side

that's the value of the magnetizing inductance.

11:16

the energy stored in the transformer is modeled by the magnetizing inductance, and

if you ever need to calculate the value of this energy stored in the transformer,

you can use the standard formula for energy in an inductor to calculate it.

Which is simply one-half Li squared, so in the case of the magnetizing inductance

reflected on the primary side it would simply be one-half L mu1 i mu 1 squared.

11:55

Note that the magnetizing inductance is across the terminals of the transformer.

Therefore, if a DC voltage is applied to any terminal of the transformer

this magnetizing inductance is essentially going to have zero impedance and

will short out the terminal.

Hence the transformer will not work at dc.

As we all know, transformers don't work at dc.

But our ideal transformer model did not differentiate between dc and ac.

12:30

In reality, the transformer will have issues even at low frequencies,

ACs because of the presence of this magnetizing inductance.

Let's see how.

In any real transformer we need to make sure that the magnetic flux

density in the core does not exceed its saturation level.

Because once the magnetic flux density reaches saturation,

the permeability of the core drops from its high value of mu c

to a very small value of mu naught.

And once that happens the magnetizing inductance itself

will drop by the same order of magnitude.

And if the value of the magnetizing inductance drops,

then the impedance of this inductance will drop and so

it will start to act more like a sharp circuit at the given frequency.

So under all circumstances,

we would like to avoid saturating the core of this transformer

which means avoiding the magnetic flux density from exceeding B sat.

What that translates to in terms of the total flux is that we would like to keep

the total flux phi c to be less than B sat

times Ac, where Ac is the cross-sectional of this core.

We can figure out what this constraint implies in terms of

the terminal voltage of the transformer.

If we consider the voltage across the first winding

of the transformer, v1, which is equal to d lambda 1 dt,

the lambda 1 itself is simply equal to n1 times phi c.

We can then invert this expression to find phi C in terms of V1, and

phi C is simply equal to one over N1 times the integral of v1 over time.

By substituting this expression for phi c in our constraint for

phi c we can get a constraint on the integral of the voltage

that must be satisfied in order to avoid the transformer core from saturating.

14:51

What this is saying is that the peak value of the integral of V1 over any time

must be less than the product of V sat times Ac times N1.

Let's see what this means in a more intuitive way.

Presuming that v1 is a sinusoidal voltage, then the peak value

of the integral of v1 will simply be the area of the shaded region.

This is because if we integrate for a period beyond this, then V1

starts to become negative and the area under the curve will start to decrease.

So the worst case for the integral of V1 is if we integrate

a half line of the voltage V1 starting from 0 and ending when it comes back to 0.

The value of this integral will depend on

the peak value of v1 as well as the frequency or

the time period associated with v1.

16:05

If the peak value of the voltage and its frequency are fixed, and

we are saturating our transformer.

Then from this expression, our only recourse is to either find another core

material with a higher saturation flux density if such a material is available.

Or to increase the size of the core by increasing the core

cross-sectional area or by adding modems on the core.

16:42

Say like this, then the total area under this curve would be smaller.

And therefore, a core with either a smaller cross-sectional

area could be used, or a transformer with fewer number of windings could be used.

In either case the size of the transformer would reduce as we increase frequency.

This is one of the major incentives for

trying to increase the switching frequency of power converters.

As it allows the size of the transformers to reduce.

17:16

Also note how the problem of saturation needs to be

dealt with differently in transformers versus inductors.

Recall that in the case of an inductor, if you increase

the current through the inductor then the inductor will saturate more quickly.

In the case of the transformer, the current that determines the saturation of

the core is the current going through the magnetizing inductors.

And not the current which is flowing in the terminals of the transformer.

We could arbitrarily increase the current through the terminals as long as

the current through the magnetizing inductance does not increase.

18:00

And in the case of the transformer shown here, if we simply increase the current,

I1, that current would simply flow out of the secondary terminals.

And would not increase the magnetizing inductance's current.

The current of the magnetizing inductance is really determined

by the voltage across it.

Since we could find that current by inverting v = L di/dt and

that current i u1 is simply 1/ L u1 times the integral of v1.

Hence, in the case of the transformer,

it's really the terminal voltage that determines its saturation.

While in the case of the inductor,

it was the terminal current of the inductor that determined its saturation.

Also, because of that,

the way the turns ratio play a roll in saturation is quite different.

In the case of the transformer,

we can avoid saturation by increasing the number of turns of the winding.

Because in the case of the transformer, it's really the total flux

through the core that we need to keep below a certain limit.

And since the flux is related to 1 over n1 times

the integral of the voltage, assuming that the integral of the voltage is constant.

Then by increasing the number of turns, we actually introduce the total flux through

the core, and hence we can avoid saturation.

Also recall that in the case of the inductor

we were able to avoid saturation by adding an air gap.

That effectively reduced the inductance value, but

it increased the current level at which the inductor saturated.

In the case of the transformer, adding an air gap has no effect on saturation.

As you can see from the terms that appear

in the expression that determines saturation.

20:07

Adding an air gap in the case of a transformer does reduce the value

of l u1 and increases the value of i u1 at which the core would saturate.

But because the voltage and the integral of the voltage is the same.

By decreasing the value of L u1,

we simply increase the current that flows through this inductor.

Therefore overall, adding or

not adding an air gap has no effect on the saturation of the transformer.

20:47

Up to now, we have been assuming that all of the flux

that is generated by either of these windings, flows in the core.

This made sense when we assumed that the permeability of the core was infinite.

And in that case, the reluctance provided by the core was zero, and so

the flux had no reason to try and find a part of lesser reluctance.

21:15

When the permeability of the core is finite, and

in real core materials the permeability of the core material Is only three

to five orders of magnitude higher than the permeability of air or free space.

And in that case, some of the flux that's generated by

one winding will flow through the air surrounding the core.

And return back to the winding through a part that does not take it

through the second winding.

22:17

We do that here, again, in the central portion we have our original magnetic

circuit model that we used to develop the model for our ideal transformer.

As well as the model for the transformer with just the magnetizing inductance.

However, now we've added two more parts to this model.

One part modeled by a reluctance Rl1,

23:00

The reluctance seen by the flux generated by the first winding

that travels through the air and does not couple to winding two.

Similarly, we include a reluctance RL2 to

model the reluctance of the part taken by the flux generated by the second winding.

Which travels through the air, and comes back through the second winding.

We can go back and look at this magnetic circuit model, to assure that this

well matches the physics of what's going on in the actual magnetic circuit.

23:42

Note that Rc, is the core reluctance and

is modeling the reluctance of this core.

And at both of its ends are the two nnf sources,

the nnf source N1 i1, which is modeling the first winding.

And then, an MFF source N2i2 which is modeling the second winding.

24:08

The reluctance Rl2 is simply across the second winding as it

modeling the reluctance of the part that seen by the flux going through the ear

on the second winding side, and Rl1 is the reluctance of the pot to the air.

Again, it is simply warning the reluctance seen by the flux that's generated

by winding 1, and going through air on the side of the first winding.

24:38

Now that we have this magnetic circuit model,

we can figure out the flux that flows in each of these parts.

And from that we can determine first the flux linking back to each of the windings.

And then from there we can figure out the voltages at each of the terminals.

So let's proceed by first figuring out the flux that flows through each of the parts.

The flux that goes through Rc, the reluctance of the core,

is simply, again, given by figuring out the mmf,

or essentially the voltage on one side of Rc.

25:17

And the voltage on the other side of Rc, and so

my Ohms law that's again simply given by n1, i1 + n2, i2 over Rc.

The flux that flows through the part Rl1 can simple be found,

again, by Ohms law since the MMF across Rl1 is simply n1, i1.

So the flux flows in the direction shown, and

is given by N1 i1 over the reluctance Rl1.

Similarly, the flux that flows in the reluctance Rl2 is

simply given by N2 i2 over Rl2.

Now that we have all the fluxes in each of the paths,

we can go and determine the flux linking back to each of the windings.

26:17

The flux linking back to a particular winding is positive.

When it is flowing from its positive terminal outwards and

coming back into its negative terminal.

So if you want to find the flux linking the first winding.

That in terms of polarity will them simply be the sum of the fluxes Phi C

plus the flux phi l 1, so our lambda 1.

Which is the flux linking back into the first winding,

is simply going to be the total flux that's going into that winding

times the number of terms of that winding.

Since again the flux is linking each of the terms of that winding.

We can substitute in the values for vl1 and vc into this expression

to determine the final expression for flux linking the first and

that'll be given by n1 squared i1.

Over Rl1

+ N1squared

i1 over Rc +

n1n2i2 over Rc.

27:48

We can similarly find the flux linking back into winding 2.

Here again, we have to be careful about the polarity of the fluxes.

But again, the way we've defined them, both of these fluxes are entering

the negative side of that winding, and coming out of the positive side.

So we will use their polarity as shown and

both will go in with positive signs into this expression.

28:18

And again, to find the final expression, we simply substitute the value of

five C from here and the expression for five L two from here.

And we get this expression for lambda 2.

Now that we have expressions for both lambda 1 and lambda 2,

we can simply determine the expressions for

the voltages at the terminals v1 and v2.

By simply taking the time derivatives for the expressions for lambda 1 and lambda 2.

That we've just derived.

29:41

We're going to define a leakage inductance, Ll1,

equal to N1 squared over the reluctance of the leakage part.

L1 which we call Rl1, similarly we'll define

a second leakage inductance, Ll2 as equal to N2 squared over Rl2.

In addition to this definitions we'd reuse our definition for

the magnetizing inductance Reflected on the primary side,

which was N1 squared over Rc.

30:38

So in the case of this inductance, Ll1, what it is really saying

is that it is an inductance, Ll1, which is modeling

the energy being stored in the magnetic flux v l one.

Which is generated by the first winding, and

this energy is being stored in the air.

Similarly, L L two models the energy stored in the flux phi L two.

31:11

Which is being generated by the second rewinding and

this energy is also stored in air.

By defining these inductances as shown, and by further defining

the term N1N2 over RC in terms of L mu 1.

And defining N2 squared over Rc, also in terms of L mu 1,

and it essentially would be equal to L mu 2,

then we can take these definitions and

substitute them in our expression for V1 and V2.

And that yields this slightly simpler expressions for

v1 and v2, these expressions are still complicated But

they can now be used to derive an equivalent circuit model for

a transformer that has both magnetizing inductance, as well as leakage.

32:35

This is nothing more than the magnetizing current, reflective on the primary side.

So this is, just simply i mu 1.

And similarly, this is just i mu 1.

With the substitution, we get a much simpler expression for v1 and v2.

And v1 is now the leakage inductance ll1

times di 1 dt plus the magnetizing

inductance conductor to the prime side times di mu 1 dt.

33:13

And then v2 Has another factor, N2 or

N1, multiplying Lu1 times Diu 3 dt,

and then a leakage term, L L two times Di2 dt.

To create the model using these equations, we can begin to see the same

wave we did for the case with just the magnetizing inductors.

When in this case, I'm actually going to just show the final model and

explain how this model is derived.

34:29

Plus another voltage which also looks like a voltage across an inductor.

In fact it is simply an inductance, the leakage inductance, LL1.

times the time derivative of the current flowing into the first winding.

So it simply can be modeled by another inductor LL1 whose current is simply I1.

And so, if I put the inductor in series as shown here, that will essentially

give me the expression E1 equal to The expression shown right here.

35:40

What this expression is again saying is that we have a voltage

that is equal to Ll2 times DI to

DT which again represents an inductor with a current I2 flowing through it.

And we can model that again, with this inductor LL2 which is modelling

the three cage inductors and the current again is I2.

36:07

This second expression V2 also has another term which is l mu

1 times di mu 1 dt times a tones ratio n2 over n1.

And this also makes sense because that is simply the voltage here,

which we just call vy, and clearly,

37:05

Hence, we now have an equivalent circuit model for

a transformer That include the known ideologies that bring about

the magnetizing inductance as well as the two leakage inductances.

To recall again that there's only one magnetizing inductance in a transformer,

and the magnetizing inductors arises because the permeability of the core

is not infinite.

On the other hand, there are two leakage inductances and

each leakage inductance is associated with the flux leaking out

from the winding associated with the side on which the leakage inductance is put.

And that flux is leaking out into the air and not connecting back.

Onto the other winding.

38:07

Part of the voltage on each winding drops across the leakage inductance.

Only a fraction of the voltage that is applied at the terminals

actually appears across the ideal part of the transformer.

The ideal part of the transformer still works using the ideal

turns ratio relationships.

But you have to account for these leakage and magnetizing inductances when computing

the actual transformer terminal relationships.

The current relationship, likewise,

is not related purely by the inverse of the turns ratio.

As part of the current that's coming into

the primary winding is flowing now to the magnetising inductance, and

only the remaining part actually gets to flow to the secondary side.

39:35

Spice also provides the coupled conductor model that you

can use with either just the ideal transformer model or

the model that includes the magnetizing inductors as well as the rigid inductors.

That's all for this lesson.

In our next lesson, we will look at other ways to model transformers.

[BLANK AUDIO]