ECE 325 – Electric Energy System Components
4- Transformers
Instructor:
Kai Sun
Fall 2015
1
Content
• Ideal Transformer (Ch. 9)
• Practical Transformers (Ch. 10)
• Special Transformers (Ch. 11)
• Three-Phase Transformers (Ch. 12)
2
Power Transformer
• Ideal transformers
– Winding resistance is negligible
– No leakage flux
– Permeability of the core is infinite
(zero magnetizing current Ip is
needed to produce flux)
– No core loss
• Real transformers:
– Windings have resistance
– Windings do not link the same
flux (having leakage flux)
– Permeability of the core is finite
– Have core losses (hysteresis
losses and eddy current losses
due to time varying flux)
3
Ideal Transformer
• Primary winding (assuming sinusoidal flux):
Because there is no flux leakage,
1=2=m (mutual flux)
I1
I2
E1
m   max cos t
d
  N1 max sin t
dt
 E1max cos t  90 
eg  e1  N1
E1max  2 fN1 max  6.28 fN1 max  2  4.44 fN1 max
• Secondary winding:
Because there is no flux leakage
d
e2  N 2 m  E2max cos t  90 
E2max
dt
 2 fN 2  max  2  4.44 fN 2  max
• Voltage/turns ratio
Eg  E1  4.44 fN1 max 90
E2  4.44 fN 2  max 90
a=E1/E2=N1/N2
4
No-load and Load Conditions
I1
• Under no-load conditions, because of
infinitely permeable core, no magnetizing
current is required to produce flux m.
There are
I2
E1
I1=I2=0
• Under load conditions (I10, I20,)
because of the infinitely permeable core,
there is an exact mmf balance
E2
Eg, E1
I1
F1=I1N1=I2N2=F2
E1 I 2 N1
a
 
E2 I1 N 2
E2
Eg, E1
• Phasor diagram:
– Assume power factor cos
I1
I2
5
Circuit Symbol for an Ideal Transformer
E1  aE2
I1 
I2
a
Eab  aEcd
I1 
I2
a
6
Impedance Ratio
E2
Z
I2
E1  aE2
I1  I 2 / a
E1 aE2
ZX 

 a2Z
I1 I 2 / a
• Impedance transformation
– The impedance seen by the source (primary side) is a2 times
the real impedance (secondary side)
– Thus, an ideal transformer has the amazing ability to increase
or decrease the value of an impedance
7
Shifting Impedances (SP)
Zp
a2
Zs
Ep
a
Es
Ip
1/a
Is
2
3
4
1
8
Shifting Impedances (PS)
Zp
1/a2
Zs
Ep
1/a
Es
Ip
a
Is
1
2
aI1
9
Examples 9-4&9-5
10
Considering an imperfect core
If
Im
Im If
• Pm+jQm and Io are the complex power and current outputs of source Eg
under no-loading conditions. Pm is the iron loss
• Rm is a resistance causing the iron loss and resulting heat
Rm=E12/Pm
If=E1/Rm
• Xm is the magnetizing reactance that measures permeability of the core.
A smaller Xm means lower permeability and needs a higher magnetizing
current Im (i.e. bigger reactive power Qm) to set up mutual flux m
Xm=E12/Qm
Im=E1/jXm
11
Considering loose coupling and resistances of windings
f1 f2
E1 =4.44fN1m
E2 =4.44fN2m
Ef1 =4.44fN1f1 Ef2 =4.44fN2f2
E1+Ef1=Ep=Eg
E2+Ef2=Es
• Define winding leakage reactances
Xf1=Ef1/I1
Xf2=Ef2/I2
• Add winding resistances R1 and R2 in series
with Xf1 and Xf2, respectively
f1
f2
Ef1
Ef2
12
Equivalent Circuit of a Practical Transformer
• No-load conditions
– I1=I2=0
– Ignore R1, Xf1, R2 and Xf2
– EpE1=aE2=aEs
• 10%-100% load conditions
– Ip>>Io,
– Ignore Io
13
Simplified Equivalent Circuits
• No-load conditions
– Ignoring winding
leakage fluxes and losses
• 10%-100% load conditions
– Ignoring core reluctance
and losses
14
Equivalent Circuits Referred to One Side
a2
/a
a2
a
a
/a2
a2
/a2
aIo
/a
Xm/a2
Rm/a2
15
Simplified Equivalent Circuits Referred to One Side
• Under load conditions
a
Zp 
N1
N2
Rp
 j Xp
 R1  a 2 R2  j ( X f 1  a 2 X f 2 )
16
Voltage Regulation
• With the primary voltage held constant at its rated value, the
voltage regulation is % change of the secondary voltage from
no-load to full load (rated)
ENL  EFL
Voltage Regulation =
100
EFL
ENL: secondary voltage at no-load
EFL: secondary voltage at full-load
• The voltage regulation depends on the power factor of the load
on the secondary side
– If the load is capacitive, the full-load voltage may exceed the
no-load voltage, in which case the voltage regulation
becomes negative
17
Measuring Transformer Impedances
• Open-circuit (no-load) test
– Neglect R1, Xf1, R2, Xf2
– Measure Pm (core loss), |Ep|,
|Io| and |Es|
Rm 
| E p |2
Pm
| I f |
E1=aEs
| Ep |
Rm
| I m | | I o |2  | I f |2
Xm 
| Ep |
| Im |
N1 | E p |
a

N 2 | Es |
18
• Short-circuit test
– Apply a low voltage ESC to the
primary side to create ISC less than
the nominal value to prevent
overheating and rapid change in
winding resistance
– Neglect Rm and Xm due to low core
flux
| Esc |
| Z p |
| I sc |
Psc
Rp 
| I sc |2
X p  | Z p |2  R p2
19
Construction of a power transformer
• Core: made of iron for high permeability; laminated and high
resistive to reduce iron losses
• Windings: the primary and secondary coils are wound closely on
top of each other with careful insulation for tight coupling; the
HV winding has more turns but uses a smaller size of conductor,
so copper/aluminum of two windings are about the same
20
Standard terminal markings and polarity tests
• High-Voltage winding (H1&H2) and Low-Voltage winding (X1& X2)
EH1,H2/EX1,X2=EH/EX=NH/NX
• Polarity test:
1.
2.
3.
4.
Connect HV winding to a low voltage source Eg
Connect a jumper J between any two adjacent HV and LV terminals
Connect two voltmeters to as shown in Figure 10.11
The polarity is additive if |Ex|>|Ep| or, otherwise, is subtractive.
21
The Per-Unit System
• Quantity in Per-Unit = Actual quantity / Base or nominal value of quantity
• Why per-unit notations?
Choose 20kV,
345KV and 138kV
as base voltages
– Neglecting different voltage levels of transformers, lines and generators
– Powers, voltages, currents and impedances are expressed as decimal
fractions of respective base quantities
(Source: EPRI Dynamic Tutorial)
22
• Four base quantities are required to completely define a per-unit system
I
E
Z
S
I

E

Z

S pu 
pu
pu
pu
In
Zn
En
Sn
• We need to select two independent base quantities of the four and calculate the
other two, e.g. selecting Sn and En
E 
E
Zn  n  n
In
Sn
Sn
In 
En
2
• For a transformer:
– Usually, Sn, Epn, Esn and the total impedance Zp (in p.u. or %) referred to the
primary side are given.
I np 
Sn
Enp
Z np 
Enp
I np
I ns 

 Enp 
Sn
Sn
Ens
a
2
()
Z ns
Np
Ns

Enp
Ens
E 
E
 ns  ns
I ns
Sn
2
()
Z p ()  Z p (p.u.)  Z np
23
Examples 10-5, 10-6, 10-7, 10-8 & 10-10
24
Autotransformer
• A conventional two-winding transformer can be changed to an autotransformer by
connecting its two coils in series (note: 4 combinations).
• The connection may use a sliding contact to providing variable output voltage.
(Source: EPRI Power System Dynamic Tutorial)
20MVA (115/69kV) McGraw-Edison Substation
Auto-Transformer (Y-Y) (Source:
http://www.tucsontransformer.com)
25
N1
• The primary and secondary windings have a common terminal
A, and hence are not isolated from each other
E2/E1=ECA/EBA=N2/N1 (There is always N1>N2)
I1N1=I2N2 
I1(N1-N2)=(I2-I1)N2
E1I1=E2I2
• An autotransformer has a smaller size (or equivalently, a higher
kVA rating for the same size) but loses insulation between
primary and secondary windings
26
Conventional transformer connected as an
autotransformer (Example 11-2)
• Reconnect the transformer to obtain
– 600V / 480V
– 600V / 720V
– 120V / 480V
– 120V / 720V
• I1=15000/600=25A
• I2=15000/120=125A
When a transformer connects a source (on the primary side) with a
load (on the secondary side), it is the load (current flows into
terminal +) of the primary circuit and is the source (current flows
out of terminal +) of the secondary circuit, so two winding
currents (I1 and I2) have opposite directions.
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I2=125A
X2
I2-I1=100A
600V
120V
H2
+
120V
X1
H1
-
600V
X1


H1
-
H2
+
X2
+
480V
I1=25A
-
|Si|=600100=60kVA
|So|=480125=60kVA
I2=125A
I2+I1=150A
+
600V

X1
X2

H1
+
+
H2
120V
-
I2-I1=100A
720V
+
120V
I1=25A
H2
-
I1=25A
|Si|=600150=90kVA
|So|=720125=90kVA
-
-
-
-
600V
H1


X1
+
I2=125A
480V
+
X2
|Si|=120100=12kVA
|So|=48025 =12kVA
The maximum apparent power |Smax|=max(|E1|,|E2|)(|I1|+|I2|)= (|E1|+|E2|) max(|I1|, |I2|)
28
Voltage Transformers and Current Transformers
• Voltage transformers (VTs),
also called potential
transformers (PTs), are
installed on HV lines to
measure line-to-neutral
voltage
• Current transformers (CTs)
are used to measure or
monitor the current in a line
for system measurement and
protection
7kVA
80.5kV
50/60Hz PT
5kVA,
100A/5A
60Hz CT
29
Three-Phase Transformers
• A bank of three single-phase transformers connected in Y or 
configurations
– Four possible combinations: Y-Y, △-△, Y-△ and △-Y
– Y: lower insulation costs, with neutral for grounding, 3rd harmonics
– △: more insulation costs, no neutral, no 3rd harmonics
30
3rd harmonics problem with three-phase transformers
• A △ configuration provides a closed path for 3rd harmonics, or
in other words, all triple harmonics are trapped in the △ loop.
 Distorted waveform
 Fundamental
 3rd harmonic
, 3, 5,
7 9
, 5, 7
, 3, 5,
7 9
3
9
, 5, 7
, 3, 5, 7, 9
, 5, 7
31
Three-phase transformers:
- connection
• Each transformer, considered
alone, acts as if it were placed in
a single-phase circuit.
• Because EH1H2 and EX1X2 are in
phase for each single
transformer, line voltages EAB
with E12, EBC with E23, and ECA
with E31 are in phase.
• A balanced three-phase load
produces balanced line currents
in line 1-2-3 and lines A-B-C
• The power rating is 3x of a
single transformer.
• One transformer may be absent
(i.e. open-delta connection)
under maintenance or
emergency conditions
3 Ip
3 Is
32
Three-phase transformers:
-Y (or Y-) connection
• X2 terminals of 3 secondary
windings are connected to a
common neutral N.
• Because EH1H2 and EX1X2 are in
phase for each single
transformer, EAB with E1N, EBC
with E2N, and ECA with E3N are
in phase.
• There is a 30o phase shift
between the line voltages of the
incoming (EAB) and outgoing
(E12) lines
• △-Y is commonly used for
voltage step-up transformers
while Y-△ is commonly used
for voltage step-down
transformers
3 Ip
E3N
EN2
30o
E1N
E2N
33
Three-phase transformers: Y-Y connection
• No phase shift between the incoming and outgoing line voltages
• Without a  configuration to trap the 3rd harmonic, it has seriously distorted voltages due to
the 3rd harmonics.
• Two solutions:
– Ground the neutrals of the primary winding and the source
– Provide each transformer with a tertiary winding, and connect 3 tertiary windings in delta
34
Phase-shifting Principle
• A 3-phase system can easily shift the phase
angle of a voltage to create 2-, 6-, and 12-phase
systems used in power electronic converters
and controllers
• Phase shifting (performed by phase-shifting
transformers) is also used in power flow
control over long transmission lines
35
Phase-shifting transformers
• A special type of 3-phase autotransformer that
shits the phase angle between the incoming and
outgoing lines without changing the voltage ratio.
• Twist all incoming line voltages by angle 
(0~20o) without changing their magnitudes
36
Calculations involving 3-phase transformers
• For any 3-phase transformer bank, assume that the primary and
secondary windings are both connected in Y configuration (if not,
transform it to Y by |ZY|=|Z|/3)
• Consider only one transformer (single phase) of the assumed Y-Y
transformer bank
– Its primary and secondary voltages are both the line-to-neutral
voltages of the incoming and outgoing lines
– Its nominal power rating and load are 1/3 of the 3-phase
transformer bank
37
Examples 12-1, 12-2, 12-3, 12-6 & 12-7
38
Homework Assignment #3
• Read Chapters 9&10
• Questions:
– Prove the polarity test on slide #21
– 9-4, 9-5, 9-6, 9-7, 10-19, 10-28, 10-29, 10-30, 10-31
• Due date:
– hand in your solution in the class on 10/5 (Mon) or
– to Denis at MK 205 or by email (dosipov@vols.utk.edu)
before the end of 10/5
39