EE207 Electrical Power - Lecture 4

7/30/2019 EE207 Electrical Power - Lecture 4

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EE207 Electrical Power

Lecture 4

TransformersPart I



Rajparthiban Kumar EE207 Electrical Power 2

Introduction

Transformers• A transformer essentially consists of two or more windings coupled by

a mutual magnetic field.

• Ferromagnetic cores are used to provide tight magnetic coupling and

high flux densities. Two types of core construction are normally used:

– Core type: windings are wound around legs of a magnetic core of

rectangular shape formed using L shaped stack of thin laminations.

– Shell type; the windings are wound around the center leg of a three legged

magnetic core formed using E shaped stack of thin laminations.

φ

windings Flux φ

Core Type

windings

Fluxφ/2φ/2

Shell Type Electrical Schematics




• Typically a transformers have two or more windings, the one

connected to the supply is called primary, and the one at the load

side is called secondary.

• Transformers are used in many electrical application such as:

(1) To change the voltage levels (step up/ step down) in power transmission

and distribution.

Introduction

Transformers

ACGenerator

20-30 Kv

Generation stage

345-765 kv

Transmission stage

To Mediumand Heavyindustries

Small industries &Residences

240-600V

Step UpTransformer

Step DownTransformer

2.4-69 kv

Distribution stage




(2) To provide isolation in low power electronic or control circuits between

low power and high power circuits (e.g. Pulse transformers are used in

Gate drive circuits for IGBT and MOSFET power Transistors).

(3) To provide impedance matching, and thus maximum power transferbetween source and load (e.g. in communication circuits matching

transformers are used to between a transmitter as source and antenna as

load).

(4) Transformers are also used to measure voltages and currents and are

known as instrumentation transformers.

Introduction

Transformers

Equivalent cct. OfAmplifier

10 V1Ω Rsp= 25Ω

MatchingTransformer

a




• Ideal transformers (practically they do not exist) have the following

properties:

(1) Winding resistances are negligible, i.e. the have no power loss

(efficiency =100%).

(2) All fluxes are confined to the magnetic core and link both primary and

secondary windings, i.e. no leakage flux.

(3) Core losses are negligible, no Eddy current and Hysteresis losses, andalso the permeability of the core is infinite (∞). Therefore the exciting

current required to establish flux in the core is negligible and thus the net

mmf is zero.

• Consider the ideal transformer shown in the figure below, with the

primary coil (N1) is connected to a time-varying voltage source and thesecondary is open circuited.

Ideal Transformer




• Time-varying flux is established in the core, and then according to

Farady’s Law this flux will induce voltage equal to the source voltage

(ideal transformer ), and is given by:

Ideal Transformer

dt

d φ N ev 111 ==

φφφφ

v1

v 2 Load

Zl

e1

e 2

N 1 N

2

i1 i

2




• The same flux (φ) will link the secondary coil ( N 2) and induce a

voltage (e2)in the secondary coil which is similar to the terminal

voltage (v2):

• The ratio of the primary induced voltage to the secondary induced

voltage gives the transformation ratio (a) as:

• If the load Z l is connected to the secondary, then a current i2 will flow

in the secondary coil and this will in turn produce an mmf =N 2i2

Ideal Transformer

dt

d φ N ev 222 ==

)ratioturns(a N

N

v

v

2

1

2

1 ==




• Immediately, a current i1 will flow in the primary to produce an mmf which will counter the effect of the secondary mmf and maintain thenet mmf in the transformer core zero. Thus,

N 1i1= N 2i2 and,

• Similar results can be obtained if the apparent power is considered, i.e.

• One of the transformer properties, is load transfer. A load connected tothe secondary of a transformer will be seen by the primary as:

Ideal Transformer

a

1

N

N

i

i

1

2

2

1 ==

a

1

N

N

i

i

v

v

ivivS S

1

2

2

1

1

2

221121

===

=⇒=




• The load voltage is given as:

• The above equation shows that an impedance connected at the

secondary side, will appear at the primary scaled by a2. And an

impedance connected at the primary side will appear at the secondary

scaled by 1/a2.

Ideal Transformer

l21

l2

1l

l

1l

1

12

1

1

1

1

2

2l

11

222

2ll22

Z a Z or

a

Z Z

,side primarytheat seenimpedancetheis Z i

vsetting

a

1

i

v

ia

a / v

i

v Z

:givesa& ,i ,vof termsin

them for ngsubstitutibyi&vreplacingthus ,i

v Z Z iv

==

=⇒===

=⇒=



Power & Electrical Machines II

Transformers

Part II




• Practical Transformers do not possess the three properties of ideal

transformers discussed earlier. However, they still maintain same

relationships that govern currents and voltages transformation.

• Practical Transformer have the following differences from idealtransformer:

1. Windings of the primary and secondary coils have resistance that can

be modelled in series with each coil. Thus, the equivalent circuit is

modified as:

Practical Transformers

v1

v 2

e1

e 2

i1

i 2

a

R1

R 2




2. There is a leakage flux in the transformer. i.e. the flux produced by the

primary windings consists of mutual flux links both the primary and

secondary coils, and leakage flux that links only the primary. Similarly,

the flux produced by the secondary will be the mutual flux and leakage

flux that links the secondary coil only. Thus to model this effect aninductance is added on the primary and secondary sides.

the inductances jX l1 and jX l2 are corresponding to the fluxes φ l1 and φ l2respectively.


v1

v 2e

1e 2

i1 i

2

a

R1 R

2 jX

l1jX

l2

φ φφ φ l1

φ φφ φ l2




3. The magnetic core is having a finite permeability, and thus a magnetising

current is required to establish a flux in the core. Also as the core is not

ideal there will be iron losses in the core due to the Eddy currents ( this

justifies why a practical transformer even under no load will be warm).

Both effects, the core losses and the finite permeability, can be modelledby adding a shunt branch consisting of resistance (for core losses) in

parallel with an inductance (finite permeability).

• The above equivalent circuit represents a practical transformer which

consists of an ideal transformer plus external impedances to represent

the imperfections.


v1

v 2

e1 e

2

i1

i 2

a

R1 R

2 jX l1

jX l2

R c1 jX

m1

Iφφφφ1

Im1

Ic1

i p




• To simplify the analysis of the transformer equivalent circuit, allimpedances, voltages and currents are referred either to the primary orsecondary side.

Transformer circuit referred to Primary side

• Note that when Secondary quantities are referred to the primary side:– Secondary impedances (including load) are multiplied by factor a2

– Secondary voltages are multiplied by a

– Secondary Currents are divided by a


av 2e

1

= ae 2

i 2

a

a 2 R 2

ja 2 X l2

jX m1

v1

i1

R1

jX l1

R c1

Iφφφφ1

Im1

Ic1

v 2

i 2 /a

Ideal Transformer




• Note also that the ideal transformer at the end of the previous circuit is

not having any significance in analysing the circuit, and therefore can

be removed from the circuit to become:


av 2

e1= ae 2

a 2 R 2

ja 2 X l2

jX m1v1

i1

R1

jX l1

R c1

Iφφφφ1

Im1

Ic1

i 2 /a

If Load is

connected

a 2

Zl




Transformer circuit referred to Secondary side

• Note that when Primary quantities are referred to the secondary side:

– Primary impedances are divided by factor a2

– Primary voltages are divided by a

– Primary Currents are multiplied by a


v 2=e

2

R 2

jX l2 ai

1

aIφφφφ1

aIm1

aIc1

i 2

Zl

2

1 R

2l1 jX

2 c R

2 m1 jX a

v1 e1




• An approximate model of referred transformer circuit (either to

primary or to secondary side) is possible in order to simplify the

analysis. The simplification is based on two assumptions:

1. The voltage drops I 1 R1 and I 1 X l1 are normally small when compared to v1 ,

thus v1≈ e1. This allows us to safely move the shunt branch to be

connected across the supply as below.


av 2

i1

a 2 Zl

Iφφφφ1

Im1

Ic1

a 2

R 2 ja 2 X

l2

jX m1

v1

R1

jX l1

R c1

i 2 /a




2. The excitation current Iφ1 is usually very small compared to the full load

(rated) current, thus further approximation is possible by removing the

shunt branch completely, and circuit becomes:

• Further the resistances of the primary and the referred secondary can

be lumped together giving a total resistance of Req1 . Similar treatmentfor the inductances produce X eq1. The final simplified circuit referred to

the primary side becomes:


av 2

i1

a 2 Zl

a 2 R 2

ja 2 X l2

v1

R1

jX l1

i 2 /a





• Similarly a simplified

version of the transformerequivalent circuit can be

shown referred to the

secondary side as below.

av 2

i1

a 2 Zl

v1

Req1 jX eq1

Simplified equivalent circuit referred to Primary side

Where

Req1

=R1+a 2 R

2 and X

eq1=X

l1+a 2 X

l2

Zeq1

= Req1

+ jX

eq1

v 2

ai1=i

2

Zl

Req2 jX eq2

Simplified equivalent circuit referred to Secondary side

Zeq2

= Req2

+ jX

eq2

a

v 1

2l 21l

2eq 2 21

2eq X X

X and R R

R

where

++++====++++====




Determination of Equivalent Circuit Parameters

• Equivalent circuit parameters can be determined experimentally by

performing two tests:

1. Open Circuit Test• This test is used to determine the shunt parameters ( Rcoc and X moc) of the

transformer ’s equivalent circuit. In this test the rated voltage is applied to one

side of the transformer and the measurements of input voltage, current , and

power is taken while the the other side of the transformer is kept open. The

test circuit and the transformer equivalent circuit under this condition is shownbelow.


v oc

A

V

W

Poc i

oc

v oc

Test Circuit configuration

i oc

I φ φφ φ oc

I moc

I coc

jX moc

v oc R

coc

P oc

Equivalent circuit




• Rcoc and X moc can be calculated from the test data as it follows:

1. From the input power (Poc) which is absorbed only by Rcoc

2. Now knowing Rcoc allows the calculation of I coc as,

3. Knowing I moc we can find X moc as,

• Note that all the calculated parameters are referred to the side at which

the voltage is applied (in this case it was assumed the primary side,thus parameters are automatically referred to the primary side).

Therefore: Rcoc=Rc1 and X moc=X m1


oc

2 oc

coc P

v R ====

2 coc

2 oc moc

moc

coc

oc coc

I i I

ased miner det be can I thus and R

v I

−−−−====

====

moc

oc moc

I

v X ====




2. Short Circuit Test

• This test is conducted to determine the windings impedance. The test is carried

by applying a low voltage to the one side of the transformer and monitoring

the current not to exceed the rated value (to avoid over heating of the

windings) while the other side is short-circuited. Measurement of the inputcurrent, voltage, and power at the side where the voltage is applied is taken.

The test circuit and the transformer equivalent circuit under this condition is

shown below. Note in class we did take the measurements on the secondary

side while the primary is short-circuited.


v sc

A

V

W

Secondary

is Short-circuited

P sc i

sc

v sc

Short circuit test configuration

i sc

v sc

Reqsc

jX eqsc

Zeqsc=Reqsc + jX eqsc

P sc

Equivalent circuit




• Calculation of the equivalent circuit parameters from the measurement data is done as it

follows:

1. The input power (Psc)is dissipated in the resistive element Reqsc, thus,

2. Z eqsc can be obtained as,

3. From step 2, we can determine X eqsc as,

• Again, note that all the calculated parameters are referred to the side at

which the voltage is applied (in this case it was assumed the primary

side, thus parameters are automatically referred to the primary side).

Therefore: Reqsc=Req1 and X eqsc=X eq1


2 sc

sceqsc

i P R ====

sc

sceqsc

i

v Z ====

2eqsc

2eqsceqsc R Z X −−−−====




• Also it is usually assumed that the resistances and the inductances are

distributed equally between the primary and the secondary side thus,

• In case that the measurements are taken at the secondary side and the

primary is short-circuited, then Reqsc=Req2 and X eqsc=X eq2


X X a and

X X

X a X X

and

2

R R a and

2

R R

R a R R

1eq 2l

21eq1l

2l

2

1l 1eq

1eq 2

21eq1

2 2

11eq

========

⇒⇒⇒⇒++++====

========

⇒⇒⇒⇒++++====




• Voltage Regulation: is a figure of merit used to identify how the

voltage changes in a transformer when the load changes. It is

calculated as the change in magnitude of secondary voltage as the

load current changes from the no load to load conditions.

• The value of V2L is usually taken as the rated voltage at the

secondary of the transformer.

Voltage Regulation

L2

L2 NL2

V

V V )VR(gulation ReVoltage

−=

EE207 Electrical Power - Lecture 4

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