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-

Power Circuits and Electromechanics

up or step-down voltages at a fixed frequency (50/60 Hz), at power of

hundreds of watts to hundreds of megawatts.

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.

http://www4.hcmut.edu.vn/~nqnam/lecture.php

This course concerns only power transformers.

Lecture 4

1

Ideal transformer

i1

wound as shown. Ignore losses, stray

+

v1

–

capacitance, and leakage flux.

N1

i2

+

v2

–

N2

Magnetic permeability is infinite or zero

reluctance.

d

dt

2

Ideal transformer (cont.)

Consider a magnetic core with two coils

v1 t N 1

Lecture 4

v 2 t N 2

d

dt

v1 t N 1

a

v2 t N 2

v1 N1

a

v2 N2

i1

i1

N

1

2

i2

N1

a

i2

+

v1

v2

v1 t i1 t v2 t i2 t 0

–

v1 N1

a

v2 N 2

+

+

v1

v2

–

N1:N 2

i1

i1 N 2 1

i2 N 1 a

v1 t i1 t v 2 t i2 t

Ideal

i2

–

–

N1:N 2

a is called turns ratio.

Total mmf is given by

Ideal

+

It can be shown, for an ideal transformer

mmf N1i1 N 2 i2 R 0

i1 t

N

1

2

i2 t

N1

a

k 1

Lecture 4

L2

i1

v

1

2

i2

v1

a

L1

3

Impedance-changing property of ideal transformer

L1 N 22 L2 N 12

Lecture 4

4

Impedance matching

Consider an ideal transformer with resistive load accross winding 2

The impedance-changing property can be used for maximizing power

v2

RL

By Ohm’s law,

i2

Subsituting v2 v1 a and i2 ai1

transferring between to windings, or matching impedances.

i1

Ideal

+

2

i2

+

v1

v2

–

–

N

v1

a 2 R L 2 R L

i1

N1

RL

An ideal transformer is placed between power source (impedance Zo )

and load (impedance ZL). Turns ratio is so selected that

N1:N 2

Z o N 1 N 2 Z L

2

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

V1 N 2

I 1 N 1

2

Load resistance 4 together with R referred to the input side is (R +

2

V2 N 2

Z L a 2 Z L

I 2 N1

Lecture 4

4(2)2)(2)2. For maximum power transfer,

10 4 R 64

5

R 13.5

Lecture 4

6

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

Small power

turns) connected to load circuit.

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

Vm1 2fN 1 max

or

V1 4.44 fN 1 max

Lecture 4

Example

110 kV, oil

10 kV, oil

7

500 kV, oil

Lecture 4

8

Equivalent circuit of transformer with linear core

Ex. 3.8: Given N1, N2, core cross-sectional area, mean core length, B-

Consider now a transformer with leakage flux and winding resistances.

H curve, and the applied voltage. Find maximum flux density, and

Equivalent directly derived from physical model is simple but somewhat useless.

required magnetizing current.

The equations on the secondary side is mulplied by a (= N1/N2) and i2 is replaced

V1 4.44 fN 1 max

max

Hence,

where

by i2/a, to derive a more useful equivalent circuit.

V1 230 V, f 60 Hz, N1 200

i1

230

4.32 10 3 webers

4.44 60 200

+

v1

L1 – aM

R1

i2

+

2

a2R2 a L2 – aM

+

RL

v2

+

i1

v1

aM

i2/a

av2

a2RL

4.32 10 3

Therefore,

Bm

0.864 webers/m 2

0.005

The required H m 0.864 300 259 At/m , the peak value of

L1 – aM is termed the leakage inductance of winding 1, a2L 2 – aM is termed

magnetizing current is (259)(0.5)/200 = 0.6475 A. Hence, Irms = 0.46 A is

“referred” (to side 1) leakage inductance of winding 2. aM is the magnetizing

–

–

–

N1:N 2

–

inductance, and its associated current is called magnetizing current.

the magnetizing on the primary side.

Lecture 4

9

Equivalent circuit of transformer with linear core (cont.)

Lecture 4

10

Transformer under sinusoidal steady-state conditions

There are losses in the magnetic core due to hysteresis and eddy current.

For steady-state operation, impedances and phasors can be used in

These losses are very difficult to calculate analytically. The sum of these losses

the equivalent circuit.

represents the total loss in the magnetic circuit of the transformer, and depends

+

be placed in parallel with the magnetizing inductance aM to account for them.

i1

R1

L1 – aM

2

a2R2 a L2 – aM

+

v1

V1

Ideal

+

Rc1

(aM)1

–

v2

–

–

I1

ja2xl2

I2 a

Rc1

jXm1

+

I2

+

aV 2

V2

–

–

Ideal

ZL

N1:N 2

where

RL

N1:N 2

Real load RL and its associated voltage and current can be retained by

L1 aM xl1 Leakage reactance of winding 1

aM X m1 Magnetizing reactance referred to winding 1

L2 M a xl 2 Leakage reactance of winding 2

a 2 L2 aM a 2 xl 2 Leakage reactance of winding 2 referred to side 1

referring them back to the secondary side, using an ideal transformer.

Lecture 4

a2R2

–

i2

+

av2

jxl1

R1

only upon the value of B m. They are called core or iron losses. A resistance can

11

Lecture 4

12

2

Transformer under steady-state (cont.)

Approximate equivalent circuit

All quantities can be referred to winding 1

jxl1

R1

+

Magnetizing branch makes computation somewhat difficult, hence this

a2R2

I1

branch is moved to the terminals of winding 1, yielding an approximate

ja2xl2

V1

Rc1

equivalent circuit, with no serious numerical error introduced.

+

I2 a

a2Z

jXm1

aV2

L

–

R1

+

–

ZL

jXm1/a 2

+

V2

–

I1

V1 Rc1

–

I2 a

jXm1

R1eq R1 a 2 R2

aV2

x1eq xl1 a 2 xl 2

–

13

Open- and short-circuit tests of transformers

+

a2ZL

–

Lecture 4

aV2

jx1eq

R1eq

+

I2

Rc1/a2

a2ZL

–

jxl2

R2

aI1

V1 a

+

I2 a

jXm1

–

jxl1/a2

R1/a2

+

ja2xl2

a2R2

I1

V1 Rc1

Or they can be referred to winding 2

jxl1

Lecture 4

14

Open-circuit test

Parameters in equivalent circuit can be determined by two simple

The test is performed with all instrumentation on the LV side with the

tests: open-circuit test and short-circuit test.

HV side being open-circuited. Rated voltage is applied to LV side. Voc,

In power transformers, the windings are called high-voltage (HV) and

Ioc, and Poc are measured with the meters.

low-voltage (LV) windings.

A

Rc

I oc

W

Voc

V oc

Hence,

IR

V

Rc

LV

IX

Voc2

Poc

IR

Voc

Rc

I oc I R I X

I oc

I X I oc2 I R2

Voc

V

X m oc

IX

Xm

HV

IR

IX

Rc

Rc and Xm are the values referred to the LV side.

Open-circuit test

Equivalent circuit

Lecture 4

15

Short-circuit test

Lecture 4

16

Example

All the instrumentation is on the HV side. Rated current is supplied to

Ex. 3.9: Given OC and SC tests’ readings. Find equivalent circuit

HV side. Vsc, Isc, and Psc are measured with the meters.

parameters referred to the HV side.

A

Xm

I sc

W

Req

From OC test

Xeq

Vsc

Vsc

Rc

220 2

50

2

I X 12 0.227 0.974 A

V

IR

968

Xm

220

0.227 A

968

220

225.9

0.974

From SC test

HV

Req

Psc

I sc2

Z eq

LV

V sc

I sc

Req

X eq Z eq2 Req2

Z eq

Req and Xeq are referred to the HV side.

Lecture 4

60

17 2

0.2076

15

0.882

17

X eq 0.882 2 0.2076 2 0.8576

17

Lecture 4

18

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.

Pout

Pout

Pout

100%

100%

Pin Pout losses

Pout Pc Pi

Losses are copper loss Pc and iron losses Pi.

Alternatively, if input power is known,

Pin Pc Pi

100%

Pin

Voltage regulation is defined as

% voltage regulation

Vno load Vload

100%

Vload

Lecture 4

19

Lecture 4

20

4

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-

Power Circuits and Electromechanics

up or step-down voltages at a fixed frequency (50/60 Hz), at power of

hundreds of watts to hundreds of megawatts.

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.

http://www4.hcmut.edu.vn/~nqnam/lecture.php

This course concerns only power transformers.

Lecture 4

1

Ideal transformer

i1

wound as shown. Ignore losses, stray

+

v1

–

capacitance, and leakage flux.

N1

i2

+

v2

–

N2

Magnetic permeability is infinite or zero

reluctance.

d

dt

2

Ideal transformer (cont.)

Consider a magnetic core with two coils

v1 t N 1

Lecture 4

v 2 t N 2

d

dt

v1 t N 1

a

v2 t N 2

v1 N1

a

v2 N2

i1

i1

N

1

2

i2

N1

a

i2

+

v1

v2

v1 t i1 t v2 t i2 t 0

–

v1 N1

a

v2 N 2

+

+

v1

v2

–

N1:N 2

i1

i1 N 2 1

i2 N 1 a

v1 t i1 t v 2 t i2 t

Ideal

i2

–

–

N1:N 2

a is called turns ratio.

Total mmf is given by

Ideal

+

It can be shown, for an ideal transformer

mmf N1i1 N 2 i2 R 0

i1 t

N

1

2

i2 t

N1

a

k 1

Lecture 4

L2

i1

v

1

2

i2

v1

a

L1

3

Impedance-changing property of ideal transformer

L1 N 22 L2 N 12

Lecture 4

4

Impedance matching

Consider an ideal transformer with resistive load accross winding 2

The impedance-changing property can be used for maximizing power

v2

RL

By Ohm’s law,

i2

Subsituting v2 v1 a and i2 ai1

transferring between to windings, or matching impedances.

i1

Ideal

+

2

i2

+

v1

v2

–

–

N

v1

a 2 R L 2 R L

i1

N1

RL

An ideal transformer is placed between power source (impedance Zo )

and load (impedance ZL). Turns ratio is so selected that

N1:N 2

Z o N 1 N 2 Z L

2

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

V1 N 2

I 1 N 1

2

Load resistance 4 together with R referred to the input side is (R +

2

V2 N 2

Z L a 2 Z L

I 2 N1

Lecture 4

4(2)2)(2)2. For maximum power transfer,

10 4 R 64

5

R 13.5

Lecture 4

6

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

Small power

turns) connected to load circuit.

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

Vm1 2fN 1 max

or

V1 4.44 fN 1 max

Lecture 4

Example

110 kV, oil

10 kV, oil

7

500 kV, oil

Lecture 4

8

Equivalent circuit of transformer with linear core

Ex. 3.8: Given N1, N2, core cross-sectional area, mean core length, B-

Consider now a transformer with leakage flux and winding resistances.

H curve, and the applied voltage. Find maximum flux density, and

Equivalent directly derived from physical model is simple but somewhat useless.

required magnetizing current.

The equations on the secondary side is mulplied by a (= N1/N2) and i2 is replaced

V1 4.44 fN 1 max

max

Hence,

where

by i2/a, to derive a more useful equivalent circuit.

V1 230 V, f 60 Hz, N1 200

i1

230

4.32 10 3 webers

4.44 60 200

+

v1

L1 – aM

R1

i2

+

2

a2R2 a L2 – aM

+

RL

v2

+

i1

v1

aM

i2/a

av2

a2RL

4.32 10 3

Therefore,

Bm

0.864 webers/m 2

0.005

The required H m 0.864 300 259 At/m , the peak value of

L1 – aM is termed the leakage inductance of winding 1, a2L 2 – aM is termed

magnetizing current is (259)(0.5)/200 = 0.6475 A. Hence, Irms = 0.46 A is

“referred” (to side 1) leakage inductance of winding 2. aM is the magnetizing

–

–

–

N1:N 2

–

inductance, and its associated current is called magnetizing current.

the magnetizing on the primary side.

Lecture 4

9

Equivalent circuit of transformer with linear core (cont.)

Lecture 4

10

Transformer under sinusoidal steady-state conditions

There are losses in the magnetic core due to hysteresis and eddy current.

For steady-state operation, impedances and phasors can be used in

These losses are very difficult to calculate analytically. The sum of these losses

the equivalent circuit.

represents the total loss in the magnetic circuit of the transformer, and depends

+

be placed in parallel with the magnetizing inductance aM to account for them.

i1

R1

L1 – aM

2

a2R2 a L2 – aM

+

v1

V1

Ideal

+

Rc1

(aM)1

–

v2

–

–

I1

ja2xl2

I2 a

Rc1

jXm1

+

I2

+

aV 2

V2

–

–

Ideal

ZL

N1:N 2

where

RL

N1:N 2

Real load RL and its associated voltage and current can be retained by

L1 aM xl1 Leakage reactance of winding 1

aM X m1 Magnetizing reactance referred to winding 1

L2 M a xl 2 Leakage reactance of winding 2

a 2 L2 aM a 2 xl 2 Leakage reactance of winding 2 referred to side 1

referring them back to the secondary side, using an ideal transformer.

Lecture 4

a2R2

–

i2

+

av2

jxl1

R1

only upon the value of B m. They are called core or iron losses. A resistance can

11

Lecture 4

12

2

Transformer under steady-state (cont.)

Approximate equivalent circuit

All quantities can be referred to winding 1

jxl1

R1

+

Magnetizing branch makes computation somewhat difficult, hence this

a2R2

I1

branch is moved to the terminals of winding 1, yielding an approximate

ja2xl2

V1

Rc1

equivalent circuit, with no serious numerical error introduced.

+

I2 a

a2Z

jXm1

aV2

L

–

R1

+

–

ZL

jXm1/a 2

+

V2

–

I1

V1 Rc1

–

I2 a

jXm1

R1eq R1 a 2 R2

aV2

x1eq xl1 a 2 xl 2

–

13

Open- and short-circuit tests of transformers

+

a2ZL

–

Lecture 4

aV2

jx1eq

R1eq

+

I2

Rc1/a2

a2ZL

–

jxl2

R2

aI1

V1 a

+

I2 a

jXm1

–

jxl1/a2

R1/a2

+

ja2xl2

a2R2

I1

V1 Rc1

Or they can be referred to winding 2

jxl1

Lecture 4

14

Open-circuit test

Parameters in equivalent circuit can be determined by two simple

The test is performed with all instrumentation on the LV side with the

tests: open-circuit test and short-circuit test.

HV side being open-circuited. Rated voltage is applied to LV side. Voc,

In power transformers, the windings are called high-voltage (HV) and

Ioc, and Poc are measured with the meters.

low-voltage (LV) windings.

A

Rc

I oc

W

Voc

V oc

Hence,

IR

V

Rc

LV

IX

Voc2

Poc

IR

Voc

Rc

I oc I R I X

I oc

I X I oc2 I R2

Voc

V

X m oc

IX

Xm

HV

IR

IX

Rc

Rc and Xm are the values referred to the LV side.

Open-circuit test

Equivalent circuit

Lecture 4

15

Short-circuit test

Lecture 4

16

Example

All the instrumentation is on the HV side. Rated current is supplied to

Ex. 3.9: Given OC and SC tests’ readings. Find equivalent circuit

HV side. Vsc, Isc, and Psc are measured with the meters.

parameters referred to the HV side.

A

Xm

I sc

W

Req

From OC test

Xeq

Vsc

Vsc

Rc

220 2

50

2

I X 12 0.227 0.974 A

V

IR

968

Xm

220

0.227 A

968

220

225.9

0.974

From SC test

HV

Req

Psc

I sc2

Z eq

LV

V sc

I sc

Req

X eq Z eq2 Req2

Z eq

Req and Xeq are referred to the HV side.

Lecture 4

60

17 2

0.2076

15

0.882

17

X eq 0.882 2 0.2076 2 0.8576

17

Lecture 4

18

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.

Pout

Pout

Pout

100%

100%

Pin Pout losses

Pout Pc Pi

Losses are copper loss Pc and iron losses Pi.

Alternatively, if input power is known,

Pin Pc Pi

100%

Pin

Voltage regulation is defined as

% voltage regulation

Vno load Vload

100%

Vload

Lecture 4

19

Lecture 4

20

4

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