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Power transformers #1
1.0 Introduction
You have been introduced to transformers in
EE 303. In these notes, I will review some
of the main ideas related to transformers, but
I do expect that you have come across these
ideas previously.
2.0 Ideal Transformer and referring
quantities (Section 5.1)
The circuit model for the ideal transformer
is given Fig. 1 below, with turns ratio
n=N2/N1.
I1
V1
I2
E1
N1
E2
Z2
V2
N2
Fig. 1
The dots mean that the currents flowing into
the dotted terminal produces fluxes φm1 and
φm2 in the core which add, based on the
right-hand rule (see Fig. 5.1 of text).
1
A more “circuits” way of understanding the
dots is [1, pg. 215]
 If the two currents I1 and I2 are either
both entering or both leaving the dotted
terminals of the coils, then the signs of the
mutual terms are the same as the signs of
the self-terms.
 If one current is entering and the other is
leaving the dotted terminals, then the signs
of the mutual terms are opposite to the
signs of the self-terms.
One other essential idea is to remember that
a self-induced voltage across an inductor is
positive at the terminal into which the
current enters.
The voltages E1 and E2 result from the
composition of self-induced voltage and
mutual induced voltage in each coil. That is,
(a)
E1  L1 I1  MI 2
(b)
E2   L2 I 2  MI1
2
Let’s eliminate I2 from the above equations.
Solving the (b) for I2 results in:
M
1
L2 I 2  MI1  E2  I 2  I1  E2
(c)
L2
L2
Substituting this expression for I2 back into
(a) results in:
M2
M
E1  L1 I1  MI 2  L1 I1 
I 1  E2
L2
L2
(d)
2

M 
M
  L1 
 I1  E2
L2 
L2

Most circuits textbooks (in the chapter on
mutual inductance) will use Faraday’s law
and magnetic circuit theory to derive the
relationship between mutual inductance and
self inductance for two coupled coils,
resulting in the definition of the coefficient
of coupling k:
M
k
L1 L2
(e)
Two coils that are not coupled at all have
k=0. If two coils are perfectly coupled
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(which means no leakage flux, i.e., the flux
linking one coil completely links with the
other coil), then k=1. This of course is
impossible, but the practice reality is that if
you wind the coils on some ferromagnetic
material such as iron, k will be almost 1.
So let’s consider the “ideal” case, when k=1,
which results in what we have called the
ideal transformer.
M
k 1
 M 2  L1 L2
L1 L2
(f)
Substitute (f) into (d) to get:
LL
LL

LL 
E1   L1  1 2  I 1  1 2 E2  1 2 E2
L2 
L2
L2

(g)
Multiply top and bottom of (g) by L2 :
E1 
L2
L2
L1 L2
L L
L1
E2  2 1 E21 
E2
L2
L2 L2
L2
Dividing both sides by E2, we obtain:
E1

E2
L1
L2
(h)
4
Using L1 
N12 A ,
l
N 22 A for the two coils
L2 
l
of equal cross section and equal length, we
see that eq. (h) becomes:
E1 N1
(i)

E2
N2
The text derives this relation and also one
for current, resulting in:
E2 N 2
1

n
E1 N 1
a
I2 N1 1

 a
I1 N 2 n
(1)
(2)
(Same as eq. 5.2 and 5.6 in text)
Example 5.1 uses ABCD parameters to
determine the relation for Z1 and Z2 that
make the two circuits of Fig. 2 have
identical input/output characteristics.
I1
Ideal network
I2
Z1
V1
E1
Network
Network
2 1
E2
N1
N2
Fig. 2a
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V2
I1
V1
Ideal network
E1
N1
I2
E2
N2
Z2
V2
Network 2
Fig. 2b
I think it is not so clear. Let’s go through it.
Recall that the ABCD parameters are
coefficients of the terms in the equations
that express input voltage and current as a
function of output voltage and current as
follows:
V1   A B  V2 
 I   C D   I 
(3)
 2 
 1 
For each circuit, we will obtain two sets of
ABCD parameters:
 The set for the ideal transformer
 The set for the impedance (network 1, 2).
Then we will combine them together by
recalling the great thing about ABCD
parameters for cascaded circuits – they can
just be multiplied together!
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Ideal Network:
The ABCD parameters for the ideal
transformer are obtained by assuming E1 and
E2 are the input and output voltages,
respectively. We simply manipulate eqs. (1)
and (2) to get:
E1  aE2
(4)
I1 
1
I2
a
(5)
Therefore:
 E1   a 0   E 2 
 I   0 1   I 
(6)
 1  
a   2 
So we see that A=a, B=0, C=0, D=1/a, and
a 0 
1
Tideal  
0
(7)

a 
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Network 1 (Fig 2a):
We get the ABCD matrix for the network of
the left-hand dotted box in Fig. 2a. Here, the
input quantities are V1 and I1, and the output
quantities are E1 and I1. In this case:
V1  E1  Z1 I1
(8)
I1  I1
(9)
Therefore:
V1  1 Z 1   E1 
 I   0 1   I 
(10)
 1 
 1 
and
Tnetwork1
1 Z 1 


0
1


8
(11)
Network 2 (Fig. 2b):
Now let’s get the ABCD matrix for network
of the right-hand dotted box in Fig. 2b.
Here, the input quantities are E2 and I2, and
the output quantities are V2 and I2. In this
case:
E2  V2  Z2 I 2
(12)
I2  I2
(13)
Therefore:
 E 2  1 Z 2  V2 
 I   0 1   I 
(14)
 2 
 2 
and
Tnetwork 2
1 Z 2 


0
1


(15)
Composite ABCD Matrices:
The ABCD Matrix for Fig. 2a is:
TFig 2 a  Tnetwork1Tideal
1 Z 1   a



0 1   0
9

0  a
1  
a   0

Z1 
a 
1  (16)

a 
The ABCD matrix for Fig. 2b is:
TFig 2b  TidealTnetwork 2
a 0  1 Z  a aZ 2 
2
1 


1 

 0 a  0 1   0 a  (17)


If these two-port networks are to be
equivalent, then it must be the case that:
TFig 2a  TFig 2b
(18)
Or

a

0

Z1 
a aZ 2 

a 
1 

1
  0 a 
a 
(19)
Eq. (19) says that equivalence requires
Z1
 aZ 2
a
(20)
which means that:
1
Z 1  a Z 2 and Z 2  2 Z1
a
2
(21)
Equation (21) is important for transformers.
It says that you can obtain equivalent
networks by shifting impedances from one
side to another according to these relations.
10
The process of moving impedances from
one side of the transformer to another, while
retaining an equivalent network, is called
“referring” impedances.
So we can “refer” impedances from primary
to secondary.
But how do you remember whether to
multiply by a or 1/a? Remember that:
a
N1
N2
(22)
Rewriting eqs. (21) we get:
2
2
 N1 
 N2 


 Z 1
Z1  
Z2
Z 2  

and
 N2 
 N1 
(23)
we see that the side to which we refer
corresponds to the multiplying turns (with
the dividing turns being the side from which
we refer).
Looked at another way, we see:
Z1  N 1 

 
Z2  N 2 
2
(24)
11
which indicates that an impedance (in ohms)
has a larger numerical value on the side
with the larger turns.
So if you are looking to refer an impedance
from the 13.8 kV side to a 161 kV side, that
impedance will get larger, so you multiply
by (161/13.8)2.
3.0 “Exact” model
The circuits we have been addressing are
actually approximate models of single phase
(or per-phase) transformers, where we use
an ideal transformer together with a series
impedance Z1 (or Z2).
The series impedance represents two things:
 Winding resistance
 Leakage flux, i.e., flux from the coil that
“leaks” out of the transformer core.
To better understand the leakage flux, write
the total flux linkages with coil 1 as:
12
1   l 1  N 1 m
(25)
where λl1 is the leakage flux linkage from
coil 1 with coil 1, and φm is the mutual flux
that links all turns of both coils.
The voltage induced in coil 1 will be, by
Faraday’s law,
d1 dl 1
dm

 N1
dt
dt
dt
(26)
To account for the voltage drop due to
winding resistance, we insert an “r1”. Then
the total voltage across the winding
terminals is given by
dl 1
dm
v1  r1i1 
 N1
dt
dt
(27)
Note that only the third term affects the
voltage seen at the other (secondary) coil.
13
The second term has most of its flux in air, a
linear medium, and therefore we have:
l 1  Ll 1i1
(28)
So that:
di1
dm
v1  r1i1  Ll 1
 N1
dt
dt
(29)
A similar situation is observed on the
secondary side with respect to winding
resistance and leakage reactance.
The third term also is caused by a current,
but not i1.
This is evident by the fact that if we impress
a voltage across the primary winding but
leave the secondary open circuited (so that
i1=(N2/N1)i2=0 because i2=0), we will still be
able to measure a voltage across the
secondary winding.
14
In reality, we can also measure a current into
the primary winding, but a small current.
We call this current the magnetization
current, denoted im, the current necessary to
set up the magnetic field.
We model the impedance associated with
this open-circuit condition with a large shunt
inductance denoted as Lm in parallel with the
(ideal) winding, so that the corresponding
voltage in coil 1 is given by
dm
dim
N1
 Lm
dt
dt
(30)
Substituting eq. (30) into eq. (29), we
obtain:
di1
dim
v1  r1i1  Ll 1
 Lm
dt
dt
The circuit conforming to eq. (31) is:
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(31)
I1
V1
Ideal network
Z1
Lm
E1
N1
I2
Z2
E2
V2
N2
Fig. 3
Note that we have also modeled the
impedance Z2 on the secondary side.
The two impedances Z1 and Z2 are given by:
Z1  r1  jLl 1
Z 2  r2  jLl 2
(32)
We will see how to REFER the secondary
impedance to the primary side in the next set
of notes.
[1] K. Su, “Fundamentals of Circuits, Electronics, and Signal Analysis,” Houghton
Mifflin Company, 1978.
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