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ELEC2131 Machines and SensorsChapter 4: Transformers

Transformers 2

Topics• Mutually Coupled Coils

− Mutual Flux and Mutual Inductance− Coupling Coefficient

• Transformer Construction and Ideal Transformer• The Practical Transformer Model• Transformer Parameter Evaluation• Per-Unit Values

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Transformers 3

Mutually Coupled Coils

Transformers 4

Mutually Coupled Coils• Consider two coils mounted on the same magnetic structure:

• Self flux - total flux linking a coil with only that coil excited. Symbol: • Leakage flux - flux linking a coil with only that coil excited but links

no other coil in the system. Symbol: • Magnetising flux is the flux which links with the coil excited and links

other coil/s in the system. Symbol: (n is the number of the coil; l : leakage; m: magnetising)

nnφ

lnφ

mnφ

Coil 1 – N1 turns Coil 2 – N2 turns

11 φ

1 lφ

1mφ

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Transformers 5

Mutually Coupled Coils

• Self inductance,

• Leakage inductance,

• Magnetising inductance,

• Self flux,

• Self inductance,

n

nnnnn i

NL φ==

current coillinkageflux self

n

nlnnl i

NL φ==

current coillinkageflux leakage

n

nmnmn i

NL φ==

current coillinkageflux gmagnetisin

mnnlnn φφφ +=

mnnlnn LLL +=

Transformers 6

Mutual Flux• The flux that links a coil due to other sources of excitation in the

system, but with itself unexcited, is called the mutual flux. • Mutual Flux:

• (M: Mutual; k: the number of the coil)• the flux component that is produced by excitation of coil j that

links with coil k.• Mutual flux linkages with coil k:

where Nk is the number of turns of coil k• Mutual inductance between two coils is defined as the flux linkages

with one coil per unit current flowing in the second coil. • That is,

kj

nj

kjjMk φφ ∑

=

≠=

=,1

kjφ

( )j

kjkkjkj i

NLorM

φ= jkkj MM =

knkkkkk

nj

kjjkjkMkkMk NNNNN φφφφφλ +⋅⋅⋅++=== ∑

=

≠=21

,1

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Transformers 7

Mutual Flux• Total flux linking with coil k:

• Total flux linkage with coil k:

• But

• Therefore,

• Expressing in matrix for a three coil system

( ) ∑=

≠=

++=+=nj

kjnjkjmklkMkkkk

, φφφφφφ

∑=

≠=

+=+=nj

kjjkjkkkkMkkkk NN

,1 φφλλλ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

3

2

1

333231

232221

131211

3

2

1

iii

LMMMLMMML

λλλ

jkjkjkkkkkkk iMNiLN == φφ and

∑=

≠=

+=nj

kjjjkjkkkk iMiL

,1

λ

iNL φ

=

Transformers 8

Induced emf Equations• The induced emf in coil k

which for a three coil system

• Note:− In static systems, the inductance coefficients are independent of

time and the equations can be simplified. − However, in systems where motion is involved, the inductance

coefficients are often a function of position and hence time. − The equations then have to be solved as specified

dtde k

kλ

=

( )

( )

( )3332321313

3232222212

3132121111

iLiMiMdtde

iMiLiMdtde

iMiMiLdtde

++=

++=

++=

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Transformers 9

Voltage Equations• The difference between the induced emf in a coil and the applied

voltage is the resistance drop in the coils

• Thus, the voltage equations for a three coil system aredt

direirv kkkkkkk

λ+=+=

( )

( )

( )333232131333

323222121222

313212111111

iLiMiMdtdirv

iMiLiMdtdirv

iMiMiLdtdirv

+++=

+++=

+++=

Transformers 10

Coupling Coefficient• Consider a two coil

system• Let

• Now by definition

• Therefore,

• Thus,

tcoefficien coupling theasknown is and 21

2211

kkkwhere

LLkM

=

=

2211212

1111

1

22222 . LLkki

Nki

NkM ==φφ

2

11

1

221221 i

NiNMMM mm φφ

====

2222

1111 and φφφφ

kk

m

m

==

j

kjkkj i

NM

φ=

Coil 1 – N1 turns Coil 2 – N2 turns

11φ

1lφ

1mφ

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Transformers 11

Transformer ConstructionIdeal Transformer

Transformers 12

Transformer Construction• The transformer is essentially a pair of mutually coupled circuits.

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Transformers 13

Transformer Construction• The transformer relies for its operation on a time varying core flux

which is common to two or more windings. • Transformer core is built from grain-oriented silicon steel

laminations, which minimise Eddy current and Hysteresis losses.• In core type construction

− a single ring type core is used encircled by one or more groups of windings.

− mean length of magnetic circuit is long for this type of construction, whereas mean length of the windings is short.

• In shell type construction− magnetic circuit encloses windings− Windings are made with copper to minimise resistance loss. − high voltage (hv) and low voltage (lv) winding sections are usually

wound on top of one another to minimise leakage flux.

Transformers 14

Ideal Transformer• Primary and secondary windings would have zero resistance.• No leakage flux associated with either primary or secondary

windings.• Transformer core has infinite permeability and thus require no mmf

to establish mutual flux linking primary and secondary circuits.

• With primary winding connected to an alternating voltage source, an alternating flux is set up in the core of transformer. Alternating flux induces an emf in both primary and secondary windings.

Figure 27: Magnetic circuit of a single phase transformer

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Transformers 15

Ideal Transformer• Volts/turn and ampere-turns for each winding must be the same

• If , the transformer is described as “step-up”• If , the transformer is described as “step-down”.

• If , then the induced emf is,

• The induced emf is proportional to flux, turns and frequency and the voltage phasor leads the flux phasor by 90°

nNN

vv

==2

1

2

1

1

2

2

1

NN

ii=

( )1 121 ∝<<NN( )1 121 ∝>>NN

( ) ( )tt ωφφ sin^

=

( )

( )

( ) fNfNEtE

tfN

tNdtdNte

φφπω

ωπφ

ωωφφ

ˆ44.4ˆ2

2 where 90sin2

90sin2ˆ

cosˆ

1111

1

111

==°+=

°+=

==

ratioturnsn =

( ) ( )tINtwhere ωφ sin ^

ℜ=

Transformers 16

Ideal Transformer• The dot on respective transformer

windings indicate terminals of corresponding polarity.

• In essence the dots indicate the sensein which the windings are wound relative to the core.

• Consider Fig. 28. In this case voltages and are in phase.• Load impedance as seen from primary terminals of transformer:

• When a transformer secondary circuit impedance is multiplied by the turns ratio squared, it is said to be referred to the primary side.

21 vv

loadZNN

iv

NN

iv

2

2

1

2

2

2

2

1

1

1⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= 2

1

212

2

11 i

NNiv

NNv ==

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Transformers 17

The Practical Transformer Model

Transformers 18

The Practical Transformer Model• The practical transformer has non-ideal properties. • In particular, we need to address the following issues:

− windings have resistance− windings have leakage fluxes − requires a finite mmf to establish mutual flux in the core− core has Hysteresis and Eddy current losses

Transformer Flux Model

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Transformers 19

The Primary Circuit• There are two flux components that link the primary winding

− Mutual flux that links both primary and secondary windings. Predominantly in ferromagnetic mediumResult of both primary and secondary winding mmf’s

− Primary leakage flux component which links only primary winding. Predominantly in air and is a result of the primary mmf only.

Primary Leakage Inductance• Primary leakage flux component

− induces a voltage in the primary winding proportional to winding current, in magnitude, and leading it by 90o in phase.

• Leakage flux path is mainly air − Constant of proportionality between induced emf due to leakage

flux and primary current is not affected by any non-linearities associated with the B-H characteristic of the core.

Transformers 20

The Primary Circuit• Primary leakage inductance is defined

• Voltage induced in the primary winding as a result of primary leakage flux will be

Primary Winding Resistance• There will be a volt drop in the primary winding due to its resistance

equal to• If the voltage induced in the primary winding due to the mutual flux

is e1, then the primary circuit of the transformer may be modelled as

1

111 currentprimary

linkageflux leakageprimary iNL l

lφ

==

111 iLje ll ω=

111 irvr =

Primary Circuit Model

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Transformers 21

Magnetising Branch• Mutual flux links both primary and secondary windings and is a

result of both primary and secondary mmf’s.• Since the core is not ideal and a finite mmf is required to establish

the mutual flux, then

• Let us suppose that a current im flowing in N1 turns is required to establish the mutual flux, thus

• It is convenient to think of primary current of transformer to comprise two components− a component im required to establish the mutual flux− a component (i1 - im) required to exactly balance any secondary

winding mmf.• The current component im in the primary winding establishes the

mutual flux, which in turn induces the voltage e1 in the winding.

02211 ≠− NiNi

( ) 2211 NiNii m =−

Transformers 22

Magnetising Branch• The induced voltage e1 leads the current im by 90 deg• To represent the relationship between e1 and im, an inductance

(magnetising inductance) is introduced into the model.• Since the flux path associated with Xm is

predominantly in iron, value of Xm will depend on state of saturation of the material.

( )mm fLX π2=

Primary Circuit Model with Magnetising Inductance

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Transformers 23

Magnetising BranchCore Losses• Core material of the transformer is ferrous

− There are both Hysteresis and Eddy current losses.• Core loss is predominantly due to mutual flux.• Magnitude of mutual flux (and therefore flux density) is proportional

to induced emf e1

Primary Circuit Model with Magnetising Inductance and Core Loss Resistance

Transformers 24

Magnetising Branch• Hysteresis and Eddy current losses are proportional to flux density

squared.• Consequently, iron losses associated with transformer core is

proportional to e12

• Power dissipated in a resistor is proportional to the voltage across it squared and thus a resistor is introduced into the model to account for core losses.

Primary Circuit Model with Magnetising Inductance and Core Loss Resistance

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Transformers 25

The Secondary Circuit• Mutual flux induces an emf e2 in secondary winding. • Since mutual flux is same for both primary and secondary windings,

induced voltage ratio is same as for ideal transformer.

• Under no-load conditions, induced emf in secondary winding is also the secondary terminal voltage of transformer.

2

1

2

1

NN

ee=

Practical Transformer Model

Transformers 26

The Secondary Circuit• Under load conditions, i.e., secondary winding current flow, a

secondary winding leakage flux is established and thus a secondary winding volt-drop due to leakage inductance.

• Similarly, there would be a volt-drop due to secondary winding resistance.

Practical Transformer Model

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Transformers 27

The Secondary Circuit• The transformer that links primary and secondary windings of the

model is ideal as non-ideal nature of practical transformer has been accounted for by other model components.

• We note that

• The current is often referred to as the load component of primary current and is given the symbol .

• It is a component of current that flows in primary winding to balance secondary winding mmf caused by load (secondary) current flow.

( ) 2

1

01

2

NN

iii

=−

2i′( )01 ii −

Transformers 28

The Referred Model• Practical transformer equivalent circuit can be simplified if we refer

all parameters and variables to either primary or secondary sides.− Effectively, we can remove the ideal transformer from the

equivalent network.• For the secondary circuit

• Rearranging this equation

• Now

( )22222 jXriev +−=

22

12

2

2

12

2

2

12

1

22

2

1 vNNX

NNjr

NNi

NNe

NN

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛−

221

212

2

1 and iiNNee

NN ′==

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Transformers 29

The Referred Model• If referred parameters and variables are now introduced

22

12

2

2

2

12

2

2

2

12

vNNv

XNNX

rNNr

ll

⎟⎟⎠

⎞⎜⎜⎝

⎛=′

⎟⎟⎠

⎞⎜⎜⎝

⎛=′

⎟⎟⎠

⎞⎜⎜⎝

⎛=′ referred secondary winding resistance

referred secondary leakage inductance

referred secondary terminal voltage

Referred Practical Transformer Model

Transformers 30

Transformer Parameters• In designing transformers, need to consider transformer parameters.

Ideally:• Winding resistances should be zero.

− Any winding resistance represents power loss and reduced efficiency.

− In practice, winding resistance will be a very small impedance.• Core losses should be zero.

− Any hysteresis or eddy current effects represent power loss and reduced efficiency.

− In practice, is very small and consequently is a very large resistance.

• Leakage flux should be zero. − Leakage fluxes play no part in transformer action, but they require

an mmf to establish them and hence a power loss. − Also cause voltage drop in transformer and thus adversely affect

transformer voltage regulation. − In practice, leakage reactances are very small impedances.

ci cr

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Transformers 31

Transformer Parameters• No mmf should be required to establish the working flux.

− In practice, this is not possible but by using good quality steel for cores, magnetising reactance can be made a very large impedance.

• In summary

cm rrrXXX <<′<<′ 2121 and and and

Transformers 32

Analysis of Transformers

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Transformers 33

Transformer Phasor Diagramn• Being able to sketch phasor diagram of an electrical machine is

often helpful in analysing systems.• The phasor diagram for transformer, with secondary terminal

voltage as reference, is

Transformer Phasor Diagram

Transformers 34

Approximate Equivalent Circuit• Due to relative magnitudes of parameters, voltage drop across

primary leakage impedance is small compared to applied voltage.• Voltage across magnetising branch is almost equal to source.• If magnetising branch of equivalent circuit is moved to input

terminals, there is no introduction of significant errors in analysis.• Results in simplified circuit analysis. • It is also impossible to split primary and secondary leakage

inductances from simple tests on a transformer.

Approximate Equivalent Circuit

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Transformers 35

Transformer Voltage Regulation• A measure of the variation in secondary voltage magnitude when

load is varied from zero to rated value at a constant power factor.• Usually represented as a percentage of rated load voltage

( ) ( )

( )%100

2

22 ×−

= −

rated

ratedloadno

v

vvRegulationVoltage

Voltage Regulation Curves

Transformers 36

Transformer Efficiency• Defined as the ratio of transformer output power to input power

• Transformer output power to a load with phase angle θ

• Input power = sum of output power plus transformer power losses:

• Losses comprise resistance losses in primary and secondary windings + core losses (hysteresis and eddy current).

• If the approximate equivalent circuit is considered

• And efficiency will be

1

2

PP

=η

θθ coscos 22222 IVIVP ′′==

lPPP += 21

( ) totalcc

l rIPrrIr

VP 2221

22

21 ′+=′+′+=

totalc rIPIV

IV2

222

22

cos

cos′++′′

′′=

θ

θη

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Transformers 37

Transformer Efficiency• A typical curve of efficiency versus load is

• Efficiency reaches a maximum value somewhere between half-load and full-load.

• Where maximum efficiency occurs is at discretion of the designer but is usually dictated by transformer application.

Transformers 38

Transformer Maximum Efficiency• Transformer maximum efficiency occurs at a load current when

i.e. when

• Simplifying

• Maximum efficiency occurs when the load is such that iron losses of transformer are equal to copper losses in the windings.

• This is true for most electromechanical energy conversion devices

02

=′Id

dη

( ) ( )( )0

2coscoscoscos2

2

222222

222 =′+′′′−′′++′′

PrIVIVVrIPIV totaltotalc θθθθ

022 =′− totalc rIP

totalc rIPIV

IV2

222

22

cos

cos′++′′

′′=

θ

θη

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Transformers 39

Transformer Parameter Evaluation

Transformers 40

Transformer Parameter Evaluation• To use transformer equivalent circuit to predict transformer

performance, we need to know parameters of the transformer.• Simple tests can be performed for parameter evaluation. • Identification of individual leakage impedance is not possible, only

their sum can be evaluated from simple test data.• If necessary to use exact equivalent model of transformer, it is usual

to assume that primary and referred secondary leakage impedances are equal.

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Transformers 41

Open-Circuit Test• In open circuit test, rated voltage is applied to primary winding of

transformer with secondary on open-circuit. • Primary voltage, primary current and the input power is measured.• The model for the transformer on open-circuit is• The core loss is

• The input apparent power is

• The input reactive power is

• The magnetising reactance is

1

21

PVrc =

1

21

QVX m =

111 IVS =

21

211 PSQ −=

Transformers 42

Short-Circuit Test• For short circuit test, rated current is applied to primary winding with

secondary on short-circuit.• Input power, input current and input voltage measured.• This test must be done at reduced primary voltage.• Equivalent circuit for transformer under short circuit conditions is

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Transformers 43

Short-Circuit Test• Magnetising branch current is very small compared to input current:

• Therefore, 12 ii ≈′

22

1

12

1

1

totaltotal

total

rZX

IVZand

IPr

−=

==

Transformers 44

Example

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Transformers 45

Example

Transformers 46

Example

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Transformers 47

Example

Transformers 48

Example

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Transformers 49

Per-Unit Values

Transformers 50

Per-Unit Values• A per-unit system is the expression of system quantities as fractions

of a defined base unit quantity:

• Different types of quantities are labeled with same symbol (pu); − it should be clear from context whether quantity is a voltage,

current, etc.• Calculations are simplified because quantities expressed as per-unit

are the same regardless of the voltage level.• A per-unit system provides units for power, voltage, current,

impedance, and admittance − admittance (Y) = inverse of impedance (Z); Unit: siemens (S).

• Only two base-values are independent, usually apparent power (Sbase) and voltage (Vbase). Then,

quantity valuebasequantity actualpuin Quantity

−=

basebasebasebasebasebase

basebasebasebasebasebasebase

IVZZXR

SSQPVSI

====

====

,

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Transformers 51

Per-Unit Values

Transformers 52

Per-Unit Values

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Transformers 53

Per-Unit Values

Transformers 54

Per-Unit Values

4873 WSbase

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Transformers 55

Per-Unit Values

Transformers 56

Per-Unit with Multiple Base Values• For per-unit impendence of a transformer to have same value when

refereed to high- or lower-voltage side, rated (or nominal) voltages of respective sides of transformer are chosen as base voltages.

• Thus, for a transformer, values of Vbase are different on the two sides and are in same ratio as turns of the transforms.

• Advantage of such a choice of base quantities is that equivalent circuit in per-unit quantities will be same whether referred to high- or low-voltage side.

• Next example illustrates this with numerical values.

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Transformers 57

Per-Unit with Multiple Base Values

220

Transformers 58

Per-Unit with Multiple Base Values

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Transformers 59

Per-Unit with Multiple Base Values

Transformers 60

Per-Unit with Multiple Base Values

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Transformers 61

Per-Unit with Multiple Base Values

Transformers 62

Summary• Mutually Coupled Coils

− Mutual Flux and Mutual Inductance− Coupling Coefficient

• Transformer Construction and Ideal Transformer• The Practical Transformer Model• Transformer Parameter Evaluation• Per-Unit Values