Bài giảng Power Circuits and Electromechanics - Lecture 4 - Nguyễn Quang Nam
Transformers – Introduction
Transferring electrical energy from one circuit to another through
time-varying magnetic field.
ECE430
Applications: both power and communications fields.
In transmission, distribution, and utilization of electrical energy: step-
up or step-down voltages at a fixed frequency (50/60 Hz), at power of
hundreds of watts to hundreds of megawatts.
Power Circuits and Electromechanics
Dr. Nam Nguyen-Quang
Fall 2009
In communications, transformers can be used for impedance
matching, DC isolation, and changing voltage levels at power of a few
watts over a very wide frequency range.
This course concerns only power transformers.
Lecture 4
1
Lecture 4
2
4
6
Ideal transformer
Ideal transformer (cont.)
i1
i2
Consider a magnetic core with two coils
wound as shown. Ignore losses, stray
capacitance, and leakage flux.
Ideal
v1 N1
i1
N2
N1
1
+
+
v2
–
a
i1
i2
v2 N2
i2
a
+
+
v1
–
v1
N1
N2
v2
–
–
v1
t
i1
t
v2
t
i2
t 0
N1:N2
Ideal
Magnetic permeability is infinite or zero
i1
i2
v1 N1
i1 N2
1
reluctance.
+
+
v2
–
a
v1
v2
t
N1
v2 N2
i2 N1
a
d
dt
d
dt
v1
–
v1
t
N1
v2
t N2
a
t
N2
v1
t
i1
t
v2
t
i2
t
N1:N2
a is called turns ratio.
It can be shown, for an ideal transformer
Total mmf is given by mmf N1i1 N2i2 R 0
L2
L1
i1
v2
v1
1
i1
t
N2
N1
L1 N22 L2 N12
1
k 1
i2
a
i2
t
a
Lecture 4
3
Lecture 4
Impedance-changing property of ideal transformer
Impedance matching
The impedance-changing property can be used for maximizing power
Consider an ideal transformer with resistive load accross winding 2
v2
i2
i1
i2
+
transferring between to windings, or matching impedances.
Ideal
RL
By Ohm’s law,
+
An ideal transformer is placed between power source (impedance Zo)
RL
v1
–
v2
–
Subsituting
and
i2 ai1
v2 v1
a
and load (impedance ZL). Turns ratio is so selected that
N1:N2
2
v1
i1
N
2
a2 RL
RL
Zo
N1 N2
ZL
2
N1
Ex. 3.7: Two ideal transformer (each of ratio 2:1) and one resistor R
are used to maximize power transfer. Find R.
The discussion can easily be extended to systems with complex load.
It can be verified that
Load resistance 4 together with R referred to the input side is (R +
4(2)2)(2)2. For maximum power transfer,
2
2
V1
I1
N2 V2
N
ZL a2 ZL
2
N1 I2
N1
R 13.5
10 4R 64
Lecture 4
5
Lecture 4
1
Power transformer
More pictures
Two windings mounted on a magnetic
core, minimizing leakage flux.
“Primary” winding (N1 turns) connected
to power supply, “secondary” winding (N2
turns) connected to load circuit.
Small power
Small 3-phase
Control
Cast resin
Assuming an ideal transformer: no leakage flux, winding resistances
are neglected, magnetic core has infinite permeability, and is lossless.
Let v1(t) = Vm1cost is the voltage applied to the primary winding, it can
be shown that
or
Vm1 2fN1max
V1 4.44 fN1max
110 kV, oil
10 kV, oil
500 kV, oil
Lecture 4
7
Lecture 4
8
10
12
Example
Equivalent circuit of transformer with linear core
Ex. 3.8: Given N1, N2, core cross-sectional area, mean core length, B-
H curve, and the applied voltage. Find maximum flux density, and
required magnetizing current.
Consider now a transformer with leakage flux and winding resistances.
Equivalent directly derived from physical model is simple but somewhat useless.
The equations on the secondary side is mulplied by a (= N1/N2) and i2 is replaced
by i2/a, to derive a more useful equivalent circuit.
where
V1 230V, f 60 Hz, N1 200
4.32103 webers
V1 4.44 fN1max
L1 – aM
a2L2 – aM
R1
a2R2
i1
i2
+
230
max
Hence,
+
+
+
4.44 60 200
4.32103
i1
a2RL
i2/a
av2
RL
v1
–
v2
–
v1
–
aM
–
Therefore,
Bm
0.864 webers/m2
N1:N2
0.005
L1 – aM is termed the leakage inductance of winding 1, a2L2 – aM is termed
“referred” (to side 1) leakage inductance of winding 2. aM is the magnetizing
inductance, and its associated current is called magnetizing current.
The required Hm 0.864 300 259 At/m, the peak value of
magnetizing current is (259)(0.5)/200 = 0.6475 A. Hence, Irms = 0.46 A is
the magnetizing on the primary side.
Lecture 4
9
Lecture 4
Equivalent circuit of transformer with linear core (cont.)
Transformer under sinusoidal steady-state conditions
There are losses in the magnetic core due to hysteresis and eddy current.
These losses are very difficult to calculate analytically. The sum of these losses
represents the total loss in the magnetic circuit of the transformer, and depends
only upon the value of Bm. They are called core or iron losses. A resistance can
be placed in parallel with the magnetizing inductance aM to account for them.
For steady-state operation, impedances and phasors can be used in
the equivalent circuit.
jxl1
ja2xl2
R1
a2R2
jXm1
I2
Ideal
+
+
+
I1
I2 a
ZL
Rc1
V1
aV2
V2
L1 – aM
a2L2 – aM
R1
a2R2
i1
–
–
–
i2
+
Ideal
N1:N2
+
+
where
RL
v1
–
av2
v2
–
Rc1
(aM)1
Leakage reactance of winding 1
L1 aM
xl1
–
N1:N2
Magnetizing reactance referred to winding 1
Leakage reactance of winding 2
aM X m1
L2 M a xl 2
a2 L2 aM a2 xl 2
Real load RL and its associated voltage and current can be retained by
referring them back to the secondary side, using an ideal transformer.
Leakage reactance of winding 2 referred to side 1
Lecture 4
Lecture 4
11
2
Transformer under steady-state (cont.)
Approximate equivalent circuit
All quantities can be referred to winding 1
Magnetizing branch makes computation somewhat difficult, hence this
branch is moved to the terminals of winding 1, yielding an approximate
equivalent circuit, with no serious numerical error introduced.
jxl1
ja2xl2
R1
a2R2
+
+
I1
I2 a
ja2xl2
a2ZL
jxl1
Rc1
R1
a2R2
V
aV2
jXm1
1
–
+
V1
–
+
–
I1
I2
a
a2ZL
aV2
Rc1
jXm1
Or they can be referred to winding 2
–
jxl1/a2
jxl2
R1/a2
R2
jx1eq
R1eq
+
+
aI1
I2
jXm1/a2
+
V1
–
+
aV2
R1eq R1 a2 R2
x1eq xl1 a2 xl 2
I1
I2
a
a2ZL
Rc1/a2
V2
ZL
V
a
1
Rc1
jXm1
–
–
–
Lecture 4
13
15
17
Lecture 4
14
16
18
Open- and short-circuit tests of transformers
Open-circuit test
Parameters in equivalent circuit can be determined by two simple
tests: open-circuit test and short-circuit test.
The test is performed with all instrumentation on the LV side with the
HV side being open-circuited. Rated voltage is applied to LV side. Voc
,
Ioc, and Poc are measured with the meters.
In power transformers, the windings are called high-voltage (HV) and
Vo2c
Voc
low-voltage (LV) windings.
IR
Rc
Ioc IR I X
Voc
Rc
P
oc
Ioc
Ioc
Hence, IX
Io2c IR2
Voc
IX
W
A
IX
IR
Voc
Voc
IX
IR
Xm
Rc
X m
Xm
V
Rc
LV HV
Lecture 4
Rc and Xm are the values referred to the LV side.
Open-circuit test
Equivalent circuit
Lecture 4
Short-circuit test
Example
All the instrumentation is on the HV side. Rated current is supplied to
HV side. Vsc, Isc, and Psc are measured with the meters.
Ex. 3.9: Given OC and SC tests’ readings. Find equivalent circuit
parameters referred to the HV side.
2
220
220
From OC test
Xeq
Req
Rc
968
IR
0.227 A
Isc
W
A
50
968
225.9
Vsc
Vsc
220
2
12
0.227
0.974 A
X m
IX
V
0.974
From SC test
HV LV
60
Req
0.2076
2
17
15
P
Is2c
Vsc
sc
Xeq
Ze2q Re2q
Req
Zeq
Isc
Zeq
Xeq
0.882
17
Req and Xeq are referred to the HV side.
0.8822 0.20762 0.8576
Lecture 4
Lecture 4
3
Efficiency and voltage regulation
In-class quiz
Efficiency is defined as the ratio of output power to input power.
Problem 3.22 and 3.23.
P
P
P
out
out
out
100%
100%
P
P
out losses
Pout P P
c i
in
Losses are copper loss Pc and iron losses Pi.
Alternatively, if input power is known,
P P P
i
in
c
100%
P
in
Voltage regulation is defined as
Vno load V
% voltage regulation
load 100%
Vload
Lecture 4
19
Lecture 4
20
4
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