Dr. Asrul Izam AzmiFaculty of Electrical Engineering

Universiti Teknologi Malaysia

Mechanical and Electrical Systems SKAA 2032

Transformers

Transformers

• A transformer is a static machine which change voltage: step up or down.

• There is no electromechanical energy conversion.• The transfer of energy takes place through the

magnetic field and all currents and voltages are AC

• Transformer can be categorized as: The ideal transformers Practical transformers Special transformers Three phase transformers

Transformers

Transformers

Transformers

Applications of the transformer

•A typical power system consists of generation, transmission and distribution•Power from plant/station is generated around 11-13-20-30kV (depending upon manufacturer and demand)

•This voltage is carried out at a distance to reach for utilization through transmission line system by step up transformer at different voltage levels depending upon distance and losses•The distribution is made through step down transformer according to the consumer demand.

Functions of transformer

• Raise or lower voltage or current in AC circuit• Isolate circuit from each other• Increase or decrease the apparent value of a

capacitor, inductor or resistor• Enable transmission of electrical energy over great

distances• Distribute safely in homes and factories

Transformer is one of the most useful electrical devices ever invented

Principles of Transformer

• A transformer consists of two electric circuits called primary and secondary

• A magnetic circuit provides the link between primary and secondary

Principles of Transformer

The basic idea:• AC voltage is applied to the primary circuit, AC

current Ip flows into the primary circuit.• Ip sets up a time-varying magnetic flux Ф in the core• An AC voltage is induced to the secondary circuit, Vs

according to the Faraday’s law

Core Types of Transformer

• The magnetic (iron) core is made of thin laminated steel sheet. The reason of using laminated steel is to minimize the eddy current loss by reducing thickness

• There are two common cross section of core which include square or (rectangular) for small transformers and circular (stepped) for the large and 3 phase transformers.

Configuration of Single phase

PrimaryWinding

SecondaryWinding

Multi-layerLaminatedIron Core

X1X2H1 H2

WindingTerminals

Three Phase Transformer

The three phase transformer iron core has three legsA phase winding is placed in each leg

Construction of Transformer

• Core (U/I) Type: Is constructed from a stack of U and I shaped laminations. In a core-type transformer, the primary and secondary windings are wound on two different legs of the core

• Shell Type: Is constructed from a stack E and I shaped laminations. In a shell-type transformer, the primary and secondary windings are wound on the same leg of the core, as concentric windings, one on top of the other.

Construction of Transformer

• Core and Shell Type Transformer

Core Type

Shell Type

Construction of a Small Transformer

Iron core

Terminals

Secondarywinding

Insulation

Transformer with Cooling System

Oil tank

Coolingradiators

High voltagebushing

Low voltagebushing

Winding

Iron corebehind the steel

bar

Radiator

Steeltank

Insulation

Bushing

The Ideal Transformer

For an ideal transformer, we assume:• No losses • Flux produced by the primary is completely linked by

the secondary and vice versa• Core is infinitely permeable• No leakage flux of any kind

The Ideal Transformer

• Primary and secondary posses N1 and N2 turns respectively

• Primary is connected to a sinusoidal source Eg • Magnetizing current Im creates a flux of Φm

The Ideal Transformer

• The flux is completed linked by the primary and secondary windings – mutual flux

• Flux varies sinusoidally and reaches a peak value of Φm

The Ideal Transformer

• A transformer with more turns in its primary than its secondary coil will reduce voltage and is called a step-down transformer

• One with more turns in the secondary than the primary is called a step-up transformer

The Ideal Transformer

• The sinusoidal current Im produces a sinusoidal mmf (magnetomotive force )= NIm which in turn creates a sinusoidal flux. The flux induces an effective voltage E across the terminals of the coil

Induces Voltages:• The effective induced emf in primary winding is

• Where N1 is the number of winding turns in primary winding, Фm the maximum (peak) flux and f is the frequency of the supply voltage

mfN.E 11 444

The Ideal Transformer

• This equation shows that for a given frequency and a given number of turns, Фm varies in proportion to the applied voltage, Eg

• This means that if Eg is kept constant, the peak flux must remain constant

• Similarly, the effective induced emf in secondary winding:

mfN.E 22 444

The Ideal Transformer - at No Load

aNN

EE

2

1

2

1

= Voltage induced in the secondary [V] 1E

2E

1N

2N

a = Turn ratio= Voltage induced in the primary [V]

= Number of turns on the primary= Number of turns on the primary

Example problem

A transformer has 500 primary turns and 3000 secondary turns. If the primary voltage is 240V, determine the secondary voltage, assuming an ideal transformerN1=500, N2=3000, V1=240VFor ideal transformer, V1=E1 and V2=E2

2

1

2

1

2

1

NN

VV

EE

3000500240

2

V

VV 14402

Ideal Transformer Under Load: Current Ratio

Let us connect a load Z across the secondary of the ideal transformer. A secondary current I2 will immediately flow.1) In an ideal transformer the primary and secondary windings are linked by the mutual flux, Фm, consequently voltage ratio will be the same as at no load

2) If the supply voltage Eg is kept fixed, then the primary induced voltage E1 remain fixed. Consequently, Фm also remains fixed. It follows that E2 also remain fixed

We conclude that E2 remains fixed whether a load is connected or not

Ideal Transformer Under Load: Current Ratio

• Let us now examine the mmf created by the primary and secondary windings. First, current I2 produces a secondary mmf N2I2. If it acted alone, this mmf would produce a profound change in the Фm. But we just saw that Фm does not change under load.

• We conclude that flux Фm can only remain fixed if the primary develops a mmf which exactly counterbalances N2I2 at every instant. Thus, a primary current I1 must flow so that

2211 ININ

Ideal Transformer Under Load: Current Ratio

To obtain the required instant-to-instant bucking effect, current I1 and I2 must increase and decrease at the same time

(a)Ideal transformer under load (b) Phasor relationships under load

Ideal Transformer Under Load: Current Ratio

INSPIRING CREATIVE AND INNOVATIVE MINDS

aNN

II

2

1

1

2

1I = Primary current [A]

2I = Secondary current [A] 1N = Number of turns on the primary2N = Number of turns on the secondary

I1 I2

Primary Secondary

Single phase transformer

Transformer Impedance

INSPIRING CREATIVE AND INNOVATIVE MINDS

Primary Impedance: Primary impedance in terms of secondary impedance : Primary Voltage : Primary Current :

P

PL I

V'Z

LL Za'Z 2

SP aVV

aII S

P

Example problem

An ideal transformer has a turns ratio of 8:1 and the primary current is 3A when it is supplied at 240V. Calculate the secondary voltage and current.

2

1

2

1

2

1

NN

VV

EE

1

212 NNVV

VV 30812402

Example problem

1

2

2

1

II

NN

ANNII 24

183

2

112

Example problem

A transformer coil possesses 4000 turns and links an ac flux having a peak value of 2 mWb. If the frequency is 60 Hz, calculate the effective value of the induced voltage E.

mfNE 44.4)002.0)(4000)(60(44.4

V2131

Example problem

A coil having 90 turns is connected to a 120V, 60 Hz source. If the effective value of the magnetizing current is 4 A, calculate the following:a. The peak value of fluxb. The peak value of the mmfc. The inductive reactance of the coild. The inductance of the coil.

Real Transformer

Leakage Flux: Not all of the flux produced by the primary current links the winding, but there is leakage of some flux into air surrounding the primary. Similarly, not all of the flux produced by the secondary current (load current) links the secondary, rather there is loss offlux due to leakage. These effects are modeled as leakage reactance in the equivalent circuit representation.

Voltage Regulation

• Voltage Regulation is defined as the change in the magnitude of the secondary voltage as the load current changes from the no-load to full load

• The primary side voltage is always adjusted to meet the load changes; hence V’s and Vs are kept constant

100VV

100aVV

100%

's

'p

sp

s

s

NL

FLNL

V

aVVVVVR

Efficiency of Transformer

• As always, efficiency is defined as power output to power input ratio

• Pcopper represents the copper losses in primary and secondary windings. There are no rotational losses.

coppercoreoutin

inout

PPPP%PP

100

Example Problem

A not-quite-ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 120 V, 60 Hz source. The coupling between the primary and the secondary is perfect but the magnetizing current is 4 A. calculate:a. The effective voltage across the secondary terminalsb. The peak voltage across the secondary terminals.c. The instantaneous voltage across the secondary when

the instantaneous voltage across the primary is 37 V.Ans: 3000V, 4242 V, 925 V.

Example Problem

An ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 200 V, 50 Hz source. The load across the secondary draws a current of 2 A at a power factor of 80 per cent lagging. Calculate :• The effective value of the primary current• The instantaneous current in the primary when the

instantaneous current in the secondary is 100 mA.• The peak flux linked by the secondary winding.

Ans: 50 A, 2.5 A, 10 mWb

Example Problem

A 125 kVA transformer has 500 turns on the primary and 80 turns on the secondary. The primary and secondary resistances are 0.5 Ω and 0.025 Ω respectively, and the corresponding leakage reactances are 2.5 Ω and 0.025 Ω respectively. The supply voltage is 2.2 kV. Calculate the voltage regulation and the secondary terminal voltage for full load having a power factor of 0.85 lagging

Example Problem

The primary and secondary windings of a 400 kVA transformer have resistances of 0.3 Ω and 0.0015 Ω respectively. The primary and secondary voltages are 15 kV and 0.4 kV respectively. If the core loss is 2.5 kW and the power factor of the load is 0.80, calculate the efficiency of the transformer on full load.

Summary

At the end of this topic, you should know:• The functions of a transformer in electrical

distribution system• The working principles of a transformer• The voltage and current ratios• The difference between the ideal and real

transformer