Let's consider now a
two-winding transformer. Here we have a core that
has two windings on it. We have polarity marks
shown here, the dots, and what these denote is the relative directions of the currents in the two windings. Here we have current
i_1 that flows into the window or the
hole in the middle of the core, and likewise, for the second winding, we have a current i_2 that
flows in the same direction, downward through the
window of the core. So if we have flux Phi that flows around the
inside of the core, then we can use Ampere's law on this flux to say
that the integral of H.dl around the core, which is the magnetomotive
force in the core, is equal to the sum of the
currents passing through the interior of the
path, or in this case, the sum of the amp turns
of the two windings, n_1, i_1 plus n_2, i_2. The magnetomotive force around the core or around the path, can be written as the flux times the
reluctance of the core, where the reluctance
of the core, as usual, is the path length of
the core material, the distance all the way around, divided by the Mu of the core and the cross-sectional
area of the core. Then Ampere's law gives
us this basic equation, and we can write an
equivalent circuit model that goes along with it, where the flux times
the reluctance is the magnetomotive
force across the core, and that's equal to
the sum of the amp turns of the two windings. So we can write a total
equivalent circuit like this, and note the polarities of the MMF sources they add
as we go around the loop. Let's construct an electrical equivalent circuit model now, to go along with our
magnetic circuit model. The simplest version gives
us the ideal transformer, and in the ideal transformer, the reluctance of
the core tends to 0. So in this case, if this reluctance goes to 0, then the magnetomotive
force also goes to 0, and what we have is that the
sum of n_1, i_1 plus n_2, i_2, the two sources
add up to 0 like this. This is the equation, the end of the terminal currents of the ideal transformer, and it says that the currents are related by the turns ratio. We can also model the voltages, and to do that we
use Faraday's law. So we have a flux Phi that
passes through the interior of the windings and in the
ideal transformer we assume that it's the same
flux in both windings. So it's the flux Phi which
induces a voltage by Faraday's law in the first
winding of n_1 d Phi, dt, and in the second winding, it will likewise induce
a voltage of n_2 d Phi, dt, and the key point here in the ideal transformer is that the Phis are the same. So we have the same flux
linking both windings. So we can actually then
eliminate d Phi, dt. So d Phi, dt from the first winding would
be equal to v_1 over n_1, and in the second winding, d Phi, dt is equal
to v_2 over n_2. Since it's the same flux, then Faraday's Law tells
us that the voltages are related according to the
turns ratio as well. Then we get these equations
for the ideal transformer, and they come out of the
magnetic circuit model. For the case when the reluctance
of the core goes to 0, these equations we commonly relate with the
ideal transformer, or model electrically with
the ideal transformer. Next, let's add back in the
reluctance of the core. So let's suppose the core
reluctance is not 0. In that case, we can't leave this term flux times reluctance
out of the equation. So what we can do instead is use the Faraday's law
equation that we just derived and solve this first one for flux and plug it into the
second one to eliminate Phi. So the flux then would be equal to the sum of the amp turns divided by the reluctance. We can plug that into here, and we would get that v_1 is n_1 times the derivative
of this flux. That would give us n_1 di_1 dt plus n_2 di_2 dt. What I've done in
the next equation on the slide is simply
to pull out the n_1, pull this n_1 out, and what we get is this. The voltage v_1 is related to
this effective inductance, n_1 squared over the
core reluctance, times the derivative of the
sum of two current terms. We can write this in
the form that v_1 is an effective inductance that we call the magnetizing inductance, L_M times the derivative
of this effective current which we call the
magnetizing current, i_M. Where are those inductances and currents in the
equivalent circuit? Here is the ideal transformer
from the previous slide. If you look at it, we have current i_2 flowing
into the secondary dot. We'll have flowing out
of the secondary dot, i_2 times the turns ratio, so n_2 over n_1 i_2. That current is this current. The other current, i_1, is the primary current, and the sum of the two currents actually looks like
a node equation. It's this current that we define as i_M, the
magnetizing current, and it's flowing through
a magnetizing inductance, L_M, that is given by n_1 squared over the
core reluctance. When we include the
core reluctance, as a non-zero quantity, what we get is an
effective inductor called the magnetizing inductance that is in parallel with the winding. You can actually draw the inductor on either side
of the ideal transformer. We could push L_M through the transformer
turns ratio squared, and on the secondary
side write this as n_2 squared over
the reluctance. The magnetizing inductance
is a real inductor. It models the
magnetization of the core. For example, we could do the thought experiment where
we just disconnect the secondary and connect the
primary up into our circuit. What would we expect to see? Well, with just the primary, we have some turns
of wire on a core, and that makes an inductor, and the inductor is in fact
the magnetizing inductance. The equivalent
circuit reflects that if you disconnect the secondary
of the ideal transformer, then we have an open circuit
on the primary side, and effectively the model reduces down to the
magnetizing inductance. This is a real inductor also. It has saturation and hysteresis which is in fact coming from
the actual core material. You can saturate the
transformer then by saturating the
magnetizing inductance. If the current and the
magnetizing inductance becomes too large, it will saturate the transformer, the magnetizing inductance will turn into a short-circuit, and that'll short out the
windings of your transformer. The one other thing the
magnetizing inductance does, is it causes the currents to
differ from the turns ratio. As I mentioned a minute ago, the transformer will
saturate when the magnetizing current
becomes too large. However, this is not
the same as saying the transformer saturates when the winding current is too large, because the winding currents can differ from the
magnetizing current. In fact, if we put
a large current in the primary and it all comes out the secondary with no current flowing through
the magnetizing inductance, the transformer doesn't saturate. We generally don't think
of transformers as saturating because the
winding current is too large. Rather, we think of them as saturating when the applied
volt seconds are too large. In fact, we can bypass the magnetizing current
altogether and simply go to Faraday's law. Where if we integrate
Faraday's law to get the flux density in the core, it's given by one over
the turns times the cross sectional area times
the integral of the voltage, and the current doesn't
really come into play. If you apply too many volt
seconds to the winding, then you will make your
transformer saturate. We think of transformers
as saturating based on applied voltage rather than
the applied winding current. We saw previously
in the last lecture that if your inductor
is saturating, you can fix that by removing
turns or adding an air gap. In the case of the transformer, neither of those work. In fact, what we need to do is reduce the flux density
in this equation. To do that, you add turns. So make n_1 larger
rather than smaller, and that will tend to make your transformer stop saturating. So we saw that the
magnetizing inductance causes the winding currents to differ from the turns ratio. The voltages can also differ
from the turns ratio, and this happens from a phenomenon called
leakage inductances. So to relate the voltages
in the ideal transformer, we assume that we
had the same flux passing through the
interiors of both windings. Here in this drawing, that's called the
mutual influx piece sub M. But in practice, there can be fluxes that pass through one winding
and not the other, and these are called
leakage inductances. Here I've drawn some fluxes that we say leak out of
the core into the air, and the bottom line
here is that this is flux that passes through one
winding and not the other. So in this case, when we write Faraday's Law, V_1 is n_1 times
the derivative of the flux passing through the
interior of the winding. That flux in this
case is drawn here, is the a sum of
leakage flux VL_1 plus the the mutual flux V_M. V_2 likewise is n_2 times the derivative
of a leakage flux VL_2 plus the mutual flux V_M. The point here now is that the leakage flux has
caused the voltages to defer and no longer scale
according to the turns ratio. Really, this case as drawn
here is topologically equivalent to having
separate little inductors in series with each winding, and those inductors
carry the flux. So this term here
effectively looks like an inductor with n_1
turns carrying flux VL_1, which I've just drawn
as a separate core, and similar for winding 2. So these things look then effectively like
series inductors. The voltage drops across those
series inductors then will cause the terminal
voltages to differ. So here is an equivalent circuit that includes the leakage fluxes. So I have a
magnetizing inductance and then we have two
leakage inductances. This is the well-known T
model of the transformer. It contains four
parameters: there is a magnetizing inductance
or a mutual inductance, there are two
leakage inductances, and there's a turns ratio. Now we know from basic circuit
theory that you can write the terminal equations of a two-winding transformer
with this matrix form, where this is called
the inductance matrix. The inductance matrix
turns out to have three independent
parameters, L_11 and L_22, which are called the
self-inductances of the primary and
secondary windings, and then L_12, which is the same, it's the off-diagonal
term n. It turns out by reciprocity that it's the
same term in both cases here. So the inductance
matrix in general has three independent parameters
rather than four, and in the equivalent
circuit model of a two-winding transformer, there are three
independent parameters, and the fourth can be
chosen arbitrarily. A common case is to
choose the turns ratio, here is the physical turns ratio of the transformer winding, and then solve with measurements
to find the inductances, the three inductances
of the T model. The T model parameters
can be related to the inductance matrix
with these relationships. It's common in circuit
theory to also define an effective turns ratio, which is defined to
be the square root of the ratio of the
self-inductances. This is a parameter
that may not be the same as the actual
physical turns ratio of the transformer. This effective turns
ratio is based on the elements in the
transformer model. We also can define a
coupling coefficient. The coupling coefficient
is a measure of how closely or tightly coupled
the two windings are. As the coupling
coefficient tends to one, the leakage inductances
go away and tend to zero. Good transformers will have coupling coefficients
slightly less than one.