EE385 & EE369
Transformers
1
References
P.C. Sen, "Principles of Electrical Machines
and Power Electronics", John Wiely & Sons,
2nd edition, 1997.
2
Transformers
Contents
Introduction
Construction and Principle of Operation
1.
Basic Components
2.
Ideal transformer
Practical Transformer
1.
Equivalent Circuit
2.
Referred & Approximate Equivalent Circuit
Voltage Regulation
Efficiency
3
Power Transformer
13.8 KV – 132 KV
4
Distribution Transformer
132 KV - 11 KV
5
Service Transformer
Circuit Transformer
11 KV – 415V
6
Introduction
A transformer is an electrical device that transfers energy from one
electrical circuit to another by magnetic coupling but without any
moving parts.

Its action is based on the laws of electromagnetic induction.

There is no electrical connection between primary and
secondary.

There is no change in frequency.

The ac power is transferred from primary to secondary through
magnetic flux.

Transformer has no moving parts.

Rugged and durable in construction.

High efficiency as well as 99%.
7
Basic Components
Insulated Copper Wire
Iron Core
Primary
winding
connected to
the source
Secondary
winding
connected to
the load
Both coils are electrically separated but magnetically linked through a
low reluctance path (Iron Core)
8
Basic Components
Laminated
iron core
Insulated
copper wire
9
Principle of operation
Ideal Transformer
• r1  r2  0
• No leakage flux
• m   (high permeability)
exciting current negligible
f
i1
v1
+
e1
_
N1
N2
c
+
e_2
S
v2
Load
d
Let the mutual flux linking both windings be sinusoidal
fm   P sin t
According to faraday’s law the induced voltage is
dfm
  p N 1cos t
dt
df
e2  N 2
  p N 2cos t
dt
e1  N1
10
Principle of operation
The rms value of the induced voltages are
E1 
N1 p 
2

2f
2
 P N1  4.44 f  P N1
E 2  4.44 f  P N 2
The ratio of the induced voltages are
E1 N1

 a  turns ratio
E2 N 2
Since the transformer is ideal E1=V1 and E2= V2
V1 E1

a
V2 E2
11
Principle of operation
Based on the assumption that μ is ∞
N1 I 1  N 2 I 2
I1 N 2 1


I 2 N1 a
• Currents are in phase.
• Current ratio is opposite to the voltage ratio
12
Principle of operation
The primary voltage may be expressed in terms of their secondary
or vice versa
V1  a V2
1
I1  I 2
a
1
V2  V1
a
I2  a I2
The power in ideal transformer
V1 I1  V2 I 2
13
Impedance Transfer
V2
Secondary impedance  Z 2 
I2
V
aV2
V
Primary impedance  Z1  1 
 a2 2  a2Z2
I1 I 2 / a
I2
I1
V1
I1
I2
V2
Z2

V1
Z’2
N1 N2
Secondary impedance is transferred to the primary side
Primary impedance is transferred to the secondary side
Z 2'  Z1  a 2 Z 2
1
Z  Z 2  2 Z1
a
'
1
14
Rated Values
• Rated voltage:
The device can
continuously operate at the rated voltage
without being damaged due to insulation
failure
• Rated current:
The device can
continuously operate at the rated current
without being damaged due to thermal
destruction
15
Transformer Rating and Name Plate
• The transformer has two windings one rated
for 1100V and the other one for 110V
a = 1100/110 = 10 = turns ratio
10kVA
1100/110 Volts
• Each winding is designed for 10 kVA
• The current rating for high-voltage winding
is 10000/1100 = 9.09 A
• The current rating for lower-voltage winding
is 10000/110 = 90.9 A
16
Example
I1
I2
A 60 Hz ideal transformer is rated 220/110
an inductive load is Z2=10+j10 Ω is
connected across the low voltage side at
V1
V2
rated secondary voltage. Calculate The
primary and secondary current and the load
impedance referred to the primary
N1 N2
Z2
V1 220

2
V2 110
V
1100
 7.78  45o A
Secondary current I 2  2 
Z 2 10  j10
Turns ratio
Primary current
a
I1 
1
1
I 2   7.78  45 o  3.89  45o A
a
2
Secondary impedance is transferred to the primary side
Z2'  Z1  a 2 Z2  2 2 10  j10  40  j40 
17
Equivalent Circuit: Practical Transformer
fm
R1
I1
fl1
V1
fl2
N1
V2
N2
Xl2
Xl1
R1
R2
I2
I1
R2
I2
V1
V2
N1: N2
R1 & R2: the resistance of the primary and secondary winding
Xl1 & Xl2: the leakage reactance of the primary and secondary winding
18
Equivalent Circuit: Practical Transformer
Xl1
R1
I1
V1
Xl2
I’2
If
Ic
Rc
Im
Xm
E1
E2
R2
I2
V2
N1: N2
Ideal Transformer
The core losses due to hysteresis and eddy current losses are represented by
Rc which takes current IC. Reactance Xm takes the magnetizing current Im .
In a simplified circuit Rc and Xm are omitted since the IΦ is about 3-5% of
the full load primary current.
19
Approximate Equivalent circuit
Xl1
R1
Xl2
I’2
I1
V1
R2
I2
V2
E1=E’2
N1: N2
In transformers it is convenient to assume all of the resistances and reactances
as being in on side of the transformer
20
Referred Approximate Equivalent Circuit
X1
R1
X’2
R’2
I’2
I1
V1
V2'  aV2
I2
I 2'  I 2 / a
V’2
V2
X 2'  a 2 X 2
R2'  a 2 R2
N1: N2
Req1  R1  R2'
X eq1  X 1  X 2'
V2'
 aV1 ,
I 2'
Xeq1
I1 Req1
V1
I2
 I1 
a
Zeq1
I’2
V’2
Referred to
Primary
21
Phasor Diagram
• From the approximate equivalent circuit referred to
primary side, the voltage equation can be written as:
V1  V  I Req 1  jI X eq 1
'
2
'
2
'
2
This equation can be represented by the following phasor
diagram:
V1
V
I’
1
f2
I’2
I’2 Xeq1
V’2
I’2 Req1
Lagging power factor
I’2 Xeq1
2
f2
V’2
I’2 Req1
Leading power factor
22
Voltage Regulation
Transformers are often used to supply loads that are designed to operate at
constant voltage. The amount of load depends on the load connected to the
transformer. As this current changes, with the same applied voltage, the load
voltage changes. This change due to the voltage drop across the impedance
of the transformer. A measure of how much the voltage will change with as
the load varied is called the voltage regulation.
Zeq1
V1
V2
Referred to
Primary
Load
| V2' | NL  | V2' | L
Voltage regulation 
| V2' | L
Transformer
| V2' | NL | V1 |
| V | NL  | V | L
Voltage regulation 
| V |L
Voltage regulation 
| V1 |  | V2' | L
| V2' | L
23
Transformer Losses and Efficiency
There are two sources of losses in transformers Copper losses and Core
(Iron) Losses
Copper losses (Pcu) is the losses due to R1 and R2 i.e (I1)2 R1+ (I2)2 R2
Core losses (PC) are the hysteresis losses and eddy current losses which is
constant for a given frequency and flux density
Transformer efficiency is defined as the ratio of the output
power to the input power

Pout
Pout
Pout


Pin
Pout 
Plosses Pout  Pc  Pcu

24
Example
A 150 KVA, 2400/240 V transformer has the following parameters reffered
to the primary side. Req1=0.5Ω and Xeq1=1.5Ω. At full load the
transformer delivers rated KVA at lag 0.85 pf and the secondary voltage is
240V calculate the voltage regulation and the efficiency assuming core
losses of 600 W.
V
2400
Turns ratio a  1 
 10
V2
240
V1
V2'  a V2  10  2400 0  24000 o V
Zeq1
V2
Load
Transformer
150000
I2 
  cos 1 0.85  625  31.8o A
240
I 2 625  31.8o
Secondary current I 2  I1 

 6.25  31.8o A
a
10
'
25
The Primary voltage
V1  V2'  I 2'  Z eq1
 24000  6.25  31.8(0.5  j1.5)
 2476.81.5o V
| V1 |  | V2' |L 2476.8  2400
Voltage Re gulation 

100  3.2%
'
| V2 |L
2400
Pout  VI cos   150000  0.85  127500W
Pcu  I1  Req1  62.52  0.5  1953W
2
Pc  600W
Pout
Pout
127500



100  98%
Pin
Pout   Plosses 127500  600  1953
26
THREE-PHASE TRANSFORMER
27
THREE-PHASE TRANSFORMER (cont.)
• A set of three similar single-phase transformers may
be connected to form a three-phase transformer.
28
THREE-PHASE TRANSFORMER (cont.)
The primary and secondary windings may be connected in either star (Y) or
delta () configurations. There are therefore four possible connections for a 3phase transformer: Y-, -Y, -,Y-Y.
As a result the ratio of the 3 phase input voltage to the three phase output
line voltages depends not only on the turns ratio but also upon how they are
connected
aI
V ph1
V ph2
N
 1 a
N2
I
V
V
N1
3
N2
3 V ph1
VL1

a
VL 2
3 V ph2
V
3a
V
a
Y-Y
29
THREE-PHASE TRANSFORMER (cont.)
30
THREE-PHASE TRANSFORMER (cont.)
Delta – Way Connection  - Y
aI
3
I
N2
V
N1
I
3
V
a
3V
a
 - Y
31
THREE-PHASE TRANSFORMER (cont.)
• A three phase transformer can be constructed by having
three primary and three secondary windings on a common
magnetic core as shown in the following Figure.
fa
c
c
a
a
b
A
a
B
b
C
c
fc
fa  fb  fc  0
fb
b
Advantages
Weight less, Cost less, Required less space
Disadvantages
Magnetic current imbalance
32
THREE-PHASE TRANSFORMER (cont.)
33