15-Non-Ideal Single Phase Transformers
ECEGR 450
Electromechanical Energy Conversion
Overview
•
•
•
•
Magnetizing Reactance
Core Resistance
Leakage Reactance
Winding Resistance
Dr. Louie
2
Questions
• What are the losses in real transformers
attributed to?
• How do we model these losses?
Dr. Louie
3
Magnetizing Reactance
• Non-ideal transformers do not have near infinite
permeability

A
0
 =N1I1 -N2I2 =Φm  0
• Add shunt magnetizing reactance (Xm) to ideal
transformer model
I0
I1
I2
+
Im
+
Xm e 1
e2
-
-
Note: I1 is now defined as current into ideal xfmr,
I0 is current into xfmr primary terminal.
Dr. Louie
4
Magnetizing Reactance
without load
e2
I0, Im
e1
I0
I1
I2
+
Im
I2, I1 = 0
Fm
+
Xm e 1
-
e2
-
N1I0 = Φm +N2I2
with load
(vector sum)
e2
Im
e1
f
I1
Fm
I0
I1
I2
+
I2
Current into and out
of xfmr terminals no
longer in phase
Im
Xm e 1
-
+
e2
-
I0
Dr. Louie
5
Example
• A transformer has 450 turns on the primary and
50 turns on the secondary. The primary voltage is
6000V. If the magnetizing reactance is j500W
compute:
 The no-load primary current and real power loss of
the transformer
 The primary current if a load impedance of Z = 10
+ j15 is applied to the secondary.
Dr. Louie
6
Example
I0  Im  I1 
V1
50090
 0  12  90 A
Pin  Re{V1I*0 }  0W
Pout  Re{V2 I2* }  0W
PLoss  Pin  Pout  0W
I0
V1
Im
I1
I2
+
+
Xm e 1
e2
-
No load: open circuit
-
Dr. Louie
7
Example
V1
 0  12  90 A
50090
N
e1  V1  1 e2
N2
Im 
e2  6670 V
I2 
e2
 36.98  56.3 A
10  j15
I0  Im  I1  Im 
50
I2  15.59  81.6 A
450
I0
V1
Im
I1
Note: real power in =
real power delivered to
load
I2
+
+
Xm e 1
e2
-
-
Dr. Louie
8
Core Resistance
• Non-ideal transformers have eddy current loss
 Real power loss
 occurs even with no secondary load
• Model as shunt resistance
• Rc >> Xm
I1
I0
I2
If
+
Im
X m Ic
Rc e1
-
+
e2
-
Note: xfmr are
designed to have large
Xm, Rm values
Dr. Louie
9
Leakage Flux
• Non-ideal transformers have leakage flux
• Leakage flux: flux in primary(secondary) coil that
is not linked to secondary (primary) coil
λ1=λl1+N1Φm
λ2=-λl2+N2Φm
I0
Fm
I2
+
+
dI0
dλ1
dΦm
V1=
=Ll1
+N1
dt
dt
dt
dλ
dI
dΦm
V2= 2 =-Ll2 2 +N2
dt
dt
dt
V1
-
N1
N2
V2
-
See lectures on magnetic circuits
and mutual inductance
Dr. Louie
10
Leakage Flux
• Model as series reactances on primary and
secondary
• Xmfrs are generally designed to have low leakage
reactance
 X1 << Xm
I0
I2
I1
X1
V1
X2
+
Xm
Rc e1
-
Dr. Louie
+
e2
V2
-
11
Winding Resistance
• Include winding resistance
• R1 < X1, R2 < X2
dλ1
dI
dΦm
=R1I0 +Ll1 1 +N1
dt
dt
dt
dλ
dI
dΦm
V2=-R 2I2 + 2 =-R 2I2 -Ll2 2 +N2
dt
dt
dt
V1=R1I0 +
I0
R1
V1
I2
I1
X1
X2
+
Xm
Rc e1
-
Dr. Louie
+
e2
R2
V2
-
12
Winding Resistance
If you were designing a transformer with the shown
number of turns, would you rather:
A. use the same gauge wire on the primary and
secondary
B. use larger diameter on the primary
C. use larger diameter wire on the secondary
I0
I2
+
+
V1
100
10
V2
-
-
Dr. Louie
13
Winding Resistance
More current is flowing through the secondary, so
it requires lower resistance to dissipate the same
heat. You should use larger diameter wire.
Side with fewer turns (lower voltage, higher current)
has lower resistance wire
I0
I2
+
+
V1
100
10
V2
-
-
Dr. Louie
14
Approximate Circuit
• Often desirable to simplify the transformer model
• More accurate than ideal, less accurate than
exact
• Voltage drop across Z1 is designed to be small
R1
V1
-
Z1
+
X1
X2
+
Xm
Rc e1
-
Dr. Louie
+
e2
R2
ZL
V2
-
15
Approximate Circuit
• Move Z1 to other side of shunt elements
• Next, eliminate the ideal transformer by referring
the secondary to the primary
I1
I0
R1
V1
X1
X2
+
Xm
Rc
e1
-
Dr. Louie
+
e2
R2
ZL
V2
-
16
Approximate Circuit
• Move Z1 to other side of shunt elements
• Next, eliminate the ideal transformer by referring
the secondary elements to the primary
I2
R1
V1
a
I1
I0
Xm
X1
a2X2
a2R2
a2 Z L
Rc
Dr. Louie
aV2
17
Approximate Circuit
• Combine series elements
R e1  R1  a2R 2
Xe1  X1  a2 X2
I1
I0
It is possible to refer to
the impedances from
the secondary side.
See text Figure 4.17.
Re1
V1
Xm
Rm
Dr. Louie
I2
Xe1
a2 Z L
a
aV2
18
Approximate Circuit
• Further approximations are possible
 Ignore shunt branch
 Ignore resistances
• Problem statement will indicate which model to
use
I1
I0
Re1
I2
Xe1
a2 Z L
V1
Dr. Louie
a
aV2
19
Example
• Consider a single-phase xfmr with the following
specifications:





primary turns: 10
secondary turns: 5
winding resistance: 0.2 Ohms
leakage impedance: 0.6 Ohms
infinite permeability
• If the primary is connected to a 1000 V source
and the secondary to a 5 Ohm load, find the
power supplied to the load
• Assume the xfmr impedances are referred from
the primary and include the secondary
impedances
Dr. Louie
20
Example
Xe1=j0.6
I2
I1
Re1 = 0.2 +
e1
V1=1000
N1=10
First calculate the ratio:
N
a 1 2
N2
transform the impedance
+
e2
V2
5
N2=5
Xe1 =j0.6
I1
Re1 = 0.2
V1
20
Z1  a2Z2  20 W
redraw the circuit
Dr. Louie
21
Example
Xe1 =j0.6
I1
Re1 = 0.2
V1
I1 
20
10000
 49.48  1.7 A
20.2  j0.6
e1  989.66  1.7
V
1
e2   989.66  1.7     494  1.7 V
 a
I2   49.48  1.7
 2  98.96  1.7
A
P2  V2 I2 cos  0   48.9 kW (this could have been solved in fewer steps)
Dr. Louie
22
Summary
• Non-ideal xfmrs include: magnetization
reactance, leakage reactance, winding resistance
and core loss
• Approximations can be made to simplify circuit
analysis (series impedances are small, shunt
impedances are large)
Dr. Louie
23