Prepared By:NABILAH BINTI MAZALAN

OBJECTIVE

• To understand mutual inductance

• To understand the construction and the operation ofa transformer

• To understand how transformer increases anddecreases voltagedecreases voltage

• To understand the effect of a resistive load across thesecondary winding

• To understand the non-ideal transformer

• To understand the several types of transformer

MUTUAL INDUCTANCE

• When coils (inductors) are connected together, aproperty called mutual inductance (M) must beconsidered. Mutual inductance is the magnetic fieldinteraction or flux linkage between coils.

• The amount of flux linkage is called the coefficient ofcoupling (k)

• The amount of mutual inductance between coils isfound by using the formula

M = k x L1 x L2

• MAGNETIC COUPLING

– When two loops with or without contacts between themaffect each other through the magnetic field generated byone of them

• ELECTRICAL ISOLATION– The transformer which are particularly designed to provide electrical

MUTUAL INDUCTANCE

– The transformer which are particularly designed to provide electricalisolation between primary and secondary circuits, without change involtage and current level are called isolation transformer

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THE BASIC TRANSFORMER

• Source voltage is applied to the primarywinding

• The load is connected to the secondarywindingwinding

• The core provided a physical structure forplacement of windings and a magnetic path sothat the magnetic flux lines are concentratedclose to the coils

• Typical core materials are : air, ferrite and iron

THE BASIC TRANSFORMER

When an AC voltage is applied in a coil, theelectromagnetic field expands, collapses andreverses as the current increases, decreases andreverses.

When two coils are placed very close to eachother, the changing magnetic field produced by

When two coils are placed very close to eachother, the changing magnetic field produced bythe first coil will cause an induced voltage in thesecond coil because of the mutual inductancebetween the two coil.

HOW???

THE BASIC TRANSFORMER When two coils are placed very close to each

other, the changing magnetic flux line producedby the first coil will cut through the second coil.The two coils are said to be magneticallylinked or coupled. As a result, a voltage isinduced.induced.

THE BASIC TRANSFORMER

• Because there is no electrical connectionbetween the two coils (only magnetic link), sothey are electrically isolated.

• The amount of voltage induced in the second• The amount of voltage induced in the secondcoil depends on the mutual inductance, LM

which is the inductance between the two coils.LM

k

1 2L1 L221LLkLM

THE BASIC TRANSFORMER

• The coefficient of coupling is a measure of how well

LM

k

1 2L1 L221LLkLM

• The coefficient of coupling is a measure of how wellthe coils are linked; it is a number between 0 and 1.

• The coefficient of coupling depends on factors suchas the orientation of the coils to each other, theirproximity, and if they are on a common core.

THE BASIC TRANSFORMER

• The basic transformer is formed from two coilsthat are usually wound on a common core toprovide a path for the magnetic field lines.Schematic symbols indicate the type of core.

Air core Ferrite core Iron core

Small powertransformer

TRANSFORMER BASICSThe parts of an ideal transformer

BASIC PRINCIPLES• The transformer is based on two principles:

– first, that an electric current can produce a magnetic field(electromagnetism), and

– second that a changing magnetic field within a coil of wire inducesa voltage across the ends of the coil (electromagnetic induction).

• Changing the current in the primary coil changes themagnetic flux that is developed.magnetic flux that is developed.

• The changing magnetic flux induces a voltage in thesecondary coil.

• Current passing through the primary coil creates amagnetic field.

• The primary and secondary coils are wrapped around acore of very high magnetic permeability, such as iron, sothat most of the magnetic flux passes through both theprimary and secondary coils.

Transformer action: based on the laws of electromagneticinduction.

There is no electrical connection between the primary andsecondary. The AC power is transferred from primary tosecondary through magnetic flux.

If an ac voltage is applied to the primary coil, magnetic flux will

OPERATING PRINCIPLES

If an ac voltage is applied to the primary coil, magnetic flux willbe created.

When the magnitude of the applied flux changed, then thegenerated flux changed.

This changing flux will link the primary and secondary coil andinduce a voltage across the secondary windings.

There is no change in frequency (output power has the samefrequency as the input power).

TRANSFORMER SYMBOLS ANDTESTING

(a) (b)

The easiest way to test a transformer is to test the primaryand secondary coils for continuity.

Since the primary and secondary of a transformer arecoils, both of them should have continuity between eachend of the coil.

THE FUNCTION

Function of a transformer: To transfer current, voltage and power from

one series of windings (coils) to another.

TRANSFORMER CONSTRUCTION

The transformer consists of core, surrounded by atleast two series of coils.

The core is used as to aid in linking the flux fromthe primary coil to the secondary coil.

From the principle of mutual induction, when twocoils are inductively coupled and if the current incoils are inductively coupled and if the current inone coil is change uniformly, an emf(electromagnetic force) is induced in the other coil.

If a closed path is provided at the secondary circuit,this induced emf at the secondary drives a current.

Dots are used in diagrams of transformers toindicate the current polarities for the windings.

Dotted terminals have the same polarity.

DOT NOTATION

(a) Voltage are in phase (b) Voltage are out of phase

DIRECTION OF WINDINGS

DIRECTION OF WINDINGS• Typical circuits illustrating polarity for voltages and

direction of currents of an ideal transformer

INDUCTION LAW• The voltage induced across the secondary coil may

be calculated from Faraday's law of induction, whichstates that:

• where Vs is the instantaneous voltage, Ns is the• where Vs is the instantaneous voltage, Ns is thenumber of turns in the secondary coil and Φ is themagnetic flux through one turn of the coil.

• Since the same magnetic flux passes through boththe primary and secondary coils in an idealtransformer, the instantaneous voltage across theprimary winding equals

INDUCTION LAW

• Taking the ratio of the two equations for Vs and Vpgives the basic equation for stepping up or steppingdown the voltage

IDEAL POWER EQUATION

• If the secondary coil is attached to a load that allowscurrent to flow, electrical power is transmitted from theprimary circuit to the secondary circuit.

• Ideally, the transformer is perfectly efficient; all theincoming energy is transformed from the primary circuitincoming energy is transformed from the primary circuitto the magnetic field and into the secondary circuit. Ifthis condition is met, the incoming electric power mustequal the outgoing power:-

IDEAL TRANSFORMER EQUATION

giving the ideal transformer equation

Transformers normally have high efficiency, so this formulais a reasonable approximation.

For an ideal transformer, the PPrimary is equal to the PSecondary, orthe power input is equal to the power output.

Pout = Pin

VsIs = VpIp

Is/Ip = Vp/Vs

IDEAL TRANSFORMER EQUATION

Is/Ip = Vp/Vs

Efficiency of transformer,

A useful parameter for ideal transformers is theturns ratio defined as

Turns ratio

sec

pri

Nn

N

TURN RATIO

priN

Nsec = number of secondary windings

Npri = number of secondary windings

Most transformers are not marked with turns ratio,however it is a useful parameter for understanding

transformer operation.

STEP-UP TRANSFORMER

• A transformer that has more turns on the secondarythan the primary side of transformer will increase theinput voltage.– This is called a step-up transformer.

– Nsec > Npri– Nsec > Npri

– ŋ > 1

– Symbol

Npri Nsec

STEP-DOWN TRANSFORMER

• A transformer that has more turns on the primarythan the secondary side of the transformer willdecrease the input voltage.– This is called a step-down transformer.

– Nsec < Npri– Nsec < Npri

– ŋ < 1

– Symbol

Npri Nsec

TRANSFORMER AS ANISOLATION DEVICE

DC ISOLATION

REFLECTED LOAD

REFLECTED LOAD

andpri secV V

R R

• A transformer changes both the voltage and current onthe primary side to different values on the secondaryside. This makes a load resistance appear to have adifferent value on the primary side.

andpri secpri L

pri sec

V VR R

I I

2

1 1 1=pri pri sec

L sec pri

R V I

R V I n n n

REFLECTED LOAD

IMPEDANCE MATCHING

MATCHING TRANSFORMER

TRANSFORMER POWERRATING & EFFICIENCY

NON IDEAL TRANSFORMER

• An ideal transformer has no power loss; all power applied tothe primary is all delivered to the load. Actual transformersdepart from this ideal model. Some loss mechanisms are:

• Winding resistance• Hysteresis loss• Core losses

Flux leakageCore losses

• Flux leakage• Winding capacitance

• The ideal transformer does not dissipate power. Powerdelivered from the source is passed on to the load by thetransformer.• The efficiency of a transformer is the ratio of power deliveredto the load (Pout) to the power delivered to the primary (Pin).

EMF EQUATION OF ATRANSFORMER• The rms value of induced emf in primary winding is

given by

p p

• Where B = Ø / A

EMF EQUATION OF ATRANSFORMER

• Similarly, RMS value of induced emf insecondary winding is

s s

POWER LOSSES

• A practical transformer differs from the idealtransformer in many respects.

• The practical transformer has:-• iron or core losses• copper losses.• copper losses.

• Iron or core losses - eddy current andhysterisis losses

• copper losses - in the resistance of thewindings

POWER LOSSES

Iron or core losses• The magnitude of these losses is quite small.This losses due to eddy current and hysteresisloss in it.

Copper losses•Is the energy loss in the windings when thetransformer is loaded.

Total copper loss, Pc = Ip2Rp + Is

2Rs

IRON OR CORE LOSSES

Eddy Current

• The magnetic core of a transformer consists ofmany laminations of a high-grade silicon steel.

• When the alternating flux cuts the steel core, anemf is induced in each lamination, causing aemf is induced in each lamination, causing acurrent (eddy current) to flow in the closedelectrical.

• Due to this eddy current and resistance in eachlamination, a certain amount of power will beabsorbed, producing heat in each lamination andalso in the core.

IRON OR CORE LOSSESHysteresis

• The alternating flux causes changes in the alignmentof the molecules in the magnetic cores.

• The change is energy consuming and heat isproduced within the core.

• The energy loss is referred to as hysteresis loss, the• The energy loss is referred to as hysteresis loss, thedegree of loss being dependent on the materialused .

Total losses in transformer

• Is the summation of core/iron losses and copperlosses

POWER LOSSES IN A NON-IDEAL TRANSFORMER

Total transformer losses,

PT = Ip2Rp + Is

2Rs +core losses

TYPE OF TRANSFORMER

• Several types of transformer:

– Center tapped transformer

– Multiple winding transformer

– Auto transformer– Auto transformer

CENTER TAPPED TRANSFORMER

MULTIPLE WINDINGTRANSFORMER

AUTOTRANSFORMER

TRANSFORMER APPLICATION

– Step-up or step-down voltage and current.

– Isolation device

» electrical isolation. Electrical isolation can beused to improve the safety of electricalused to improve the safety of electricalequipment.

» DC isolation. Coupling transformers can beused to block DC signals from reaching thesecondary circuit.

– Impedance matching for max power transfer

SPECIAL TRANSFORMERAPPLICATIONS• Autotransformers

– Use in low-voltage applications since has only one winding.

• Induction Circuit Breaker– Use to shut off the current in a circuit, thus prevent fires

cause by a short in an electrical device.cause by a short in an electrical device.

• Lighting Ballast– Use to start the lamp by producing the necessary voltage

and limits current thro the lamp.

• Coupling Transformer– Isolates each section of the amplifier – so individual

amplifier characteristics will not interfere with others

EXAMPLE 1

A transformer has 800 turns on the primary anda turns ratio of 0.25. How many turns are on thesecondary?

Solution

n = Ns / Np

NS = Np x n = 800 x 0.25 = 200 turns

EXAMPLE 2

A doorbell requires 0.4 A at 6V. It isconnected via a transformer whoseprimary coil contains 2000 turns to a 120VAC line. Calculate:-AC line. Calculate:-

(a) NS

(b) IP

SOLUTION 2

a)

b)

EXAMPLE 3

The voltage ratio of a single-phase, 50Hztransformer is 5000/500V at no load. Calculatethe number of turns in each winding if themaximum value of the flux in core is 7.82mWb.maximum value of the flux in core is 7.82mWb.

SOLUTION 3

Ep=Vp=5000V, Es=Vs=500V, Øm=7.82mWb

Ep = 4.44 f Øm Np

Np = Ep / 4.44 f Øm = 5000 / (4.44 x 50 x 7.82m)

= 2880 turns= 2880 turns

EXAMPLE 4

A single transformer is connected to a 660Vsupply. The voltage/turn of the transformer is1.1V. The secondary voltage of the transformeris found to be 440V. Determine the following:-is found to be 440V. Determine the following:-

a)Primary and secondary turns

b)Cross-section of the core if the maximum fluxdensity is 1.4T

SOLUTION 4

Voltage/turn=1.1V, Vp=660V, Vs=440V, Bmax=1.4T

i) Np = Vp / 1.1 = 660 / 1.1 = 600 turns

Ns = Vs / 1.1 = 440 / 1.1 = 400 turns

ii) Ep = 4.44 f Øm Npii) Ep = 4.44 f Øm Np

Øm = Bmax A

A = Vp / 4.44 f Bmax Np

= 660 / (4.44 x 50 x 1.4 x 600)

= 35.39 x 10-4 m2 = 35.39cm2

EXAMPLE 5

a) Illustrate the parts of an ideal transformer.(5m)b) Identify the importance of the core material. (3m)c) A 50 kVA, single-phase transformer has 500 turns on the primary

and 100 turns on the secondary. The primary is connected to 2500V,50Hz supply. Calculate the following :- (9m)

i) The secondary voltage on open circuitii) The current flowing through the windings on full loadii) The current flowing through the windings on full loadiii) The maximum value of flux

d) A single phase ideal transformer contains 3000 circumferences atprimary coil and 800 circumferences at secondary coil. Primary coilis connected to 240VAC, 50Hz supply. Calculate:- (8m)

i) Secondary voltageii) Current on secondary coil if the current on primary coil is 5A.iii) Transformer power on primary and secondary coils

SOLUTION 5(a)

a)

SOLUTION 5(b) & (c)

b) The importance of the core material is to reducelosses of eddy current. The losses in transformer areconstant but not affected by load changes

c)c)

SOLUTION 5(c)

c)

SOLUTION 5(d)

d)

SOLUTION 5(d)

d)

THE ENDTHE END

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