Power Transfomer

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KKKF1063 Introduction to ElectricalEngineering

LECTURE 11


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2

WK TOPICS LECT

9 Polyphase System BB

10 Digital Logic Circuits MBI

11 Applications - Digital System MBI

12 Applications Power Supply,

Transformer, Electric Machines

BB

13 Measurement andInstrumentation

BB

14 Circuit Disturbance Noise,

EMI

MBI

15 Other Related TechnologiesIC, Optics, GPS, Acoustic

BB/MBI


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Badariah Bais KKKF163 Introduction to EESem II 2006/07

3

Reading Assignments

HambleyCh 10 p. 460 464Ch 15 p. 712 725

Ch 16 p. 735 777Ch 17 p. 787 824RizzoniCh 9 p. 513 524

Ch 18 p. 936 940Ch 19 p. 971 1027Ch 20 (Selected refer to lecture notes)


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Badariah BaisKKKF163 Introduction to EE

Sem II 2006/07 4

Contents

1. Power Supply2. Transformer3. Electric Machines


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Sem II 2006/07 5

Introduction

Power supply is a group of circuits thatconvert ac energy to dc energy.

Two types:

linear power supply: provides constantcurrent path between its input and itsload.

- switching power supply: providesintermittent current path between itsinput and its output.


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6

Linear Power Supply

Basic components:1. Rectifier diode circuit that converts the ac to what is

called a pulsating dc.2. Filter circuit that reduces the variations in the output

of the rectifier.3. Voltage regulator maintain a constant power supply

output voltage.


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Sem II 2006/07 7

Rectifier

2 types:

Half-wave rectifier

Full-wave rectifier


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8

Half-wave Rectifier


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9

Full-wave Rectifier


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Sem II 2006/07 10

Full-wave Rectifier

Also known as a full-wave bridge rectifier


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Badariah Bais KKKF163 Introduction to EESem II 2006/07 11

Full-wave Rectifier - operation


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Filter

Vr=ripple voltage


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Basic Capacitive Filter

RC

Time constant


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14

Filter output

Note:Ripple is minimized for higher values of RL or CF


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15

Surge Current

Surge current can be resolved by incorporating a seriescurrent limiting resistor, Rsurge.

Current limiting resistor are usually a low resistance, highwattage component.

The disadvantage reduces the output voltage from thecircuit.


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16

Filter Output VoltageVpk: peak rectifier output

voltageVdc : average (or dc) value

Vr : ripple

Example:

Assuming the linefrequency is 60 Hz, thetime between chargingpeak for half-wave rectifier,

t= 1/f= 1/60 = 16.7 ms

For full-wave rectifier, thefrequency is 2x60 = 120Hz. Therefore,

t= 1/120 = 8.33 ms


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RippleVoltage

Given by the equation:

C

tIV Lr

where IL= the dc load current

t = the time between charging peaks

C = the capacitance (in Farads)


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The full-wave rectifier has approximately half the ripple outputproduced by the half-wave rectifier. This is due to the shortened timeperiod between capacitor charging pulses. Therefore, full-wave rectifierare typically used in power supplies.


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Zener Voltage Regulator


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Zener Reduction of Ripple Voltage

ppppppr

SLZ

LZ

outr mVVVVRRZ

RZV 129)5.1(

518.4

8.4)5.1(

511205

1205

)(

)()(


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Putting It All Together


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Transformer

Made up of inductors.

Not electrically connected.

An ac voltage applied to the primary induces an ac voltage in the secondary.


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24

Types of Transformer

Step-up transformer

- provides a secondary voltage that isgreater thanthe primary voltage.

Step-down transformer

- provides a secondary voltage that isless thanthe primary voltage.

Isolation transformer

- provides a secondary voltage that is equaltothe primary voltage.

- to isolate the power supply electrically fromthe power line, which serves as a

protection.


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Transformer secondary voltage

The turns ratio of a transformer is equal to the voltage ratio of the component:

)(

)(

1

2

1

2

tv

tv

N

N

or )()(1

1

2

2 tvN

N

tv

For example:acac VVtv

N

Ntv 30)120(

4

1)()(

1

1

2

2


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Transformer secondary current

Assuming the transformer is 100% efficient, then

or

12PP

)()()()(1122titvtitv

)()(

)()( 1

2

1

2 titv

tvti )()( 1

2

12 ti

N

Nti


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ExampleConsider the source, transformer, and load shown in the circuit below. Determinethe rms values of the currents and voltages (a) with the switch open and (b) withthe switch closed.

VrmsV 110)(1

Solution

Voltage applied to the primary,

VrmsVN

NrmsV 22)110(

5

1)()(

11

2

2

(a) With the switch open, the secondary current is zero. Hence, the primarycurrent is also zero.

(b) With the switch closed, AR

rmsVrmsI

L

2.210

22)()( 22

ArmsIN

NrmsI 44.0)2.2(

5

1)()( 2

1

21


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Transformer Rating

The rating of a transformer is statedas Volt Ampere (VA) that it cantransform without overheating.

The transformer rating can becalculated as either V1I1 or V2I2where I2 is the full load secondary

current.


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Impedance Transformation

LZ

2

2

I

V

LZ

N

N

N

N

11

2

12

1

I

V

LL ZN

NZ

2

2

1

1

1'

I

V

The phasor current and voltage in the secondary are related to the loadimpedance by

Then, The impedance seen by the source,


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ExampleConsider the circuit shown below. Find the phasor currents and voltages. Also, findthe power delivered to the load.

Solution

Impedance at the secondary, )2010( jZL

Impedance reflected at the primary,

)20001000()2010(

1

1022

2

1' jjZ

N

NZ LL

Total impedance )20002000(200010001000'

1jjZRZ LS

452828SZ


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Example

Primary current and voltage:

452828SZ

A

S

S

453536.0

452828

010001

Z

VI

)20001000(453536.0

'

11 jL ZIV

V 43.186.790)43.632236(453536.0

Secondary current and voltage: AN

N

45536.3)453536.0(

1

101

2

1

2II

VN

N

43.1806.79)43.186.790(10

11

1

22 VV

)20001000(' jZL

Power delivered to the load: WRrmsIP LL 51.62)10(2

536.3)(

2

2

2


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Electric Machines

Motor and Generator


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Sem II 2006/07 33

Electric Machines

Machines that convert mechanicalenergy to and from electric energy.

- Motor convert electrical energyinto rotational mechanical energy

- Generator convert mechanicalenergy into electrical energy


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Electric Motor Basic construction

Basic components:

1. Stator stationary part

2. Rotor rotating part

3. Shaft coupled the machine to the mechanicalload.


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Electric Motor

Rotor is rotating inside the

stator and separated bymeans of an air gap.

The rotor and stator eachconsists of a magneticcore, some electrical

insulation and the windingsnecessary to establish amagnetic flux.

The windings carry theelectric currents thatgenerate the magneticfields and flow to theelectrical loads.


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Sem II 2006/07 36

Motor can be divided into

AC Motors

powered by AC sources which caneither be single phase or three phase.

- most common type is induction motorand synchronous motor.

DC Motors

powered by DC sources.

Electric Motor


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In most types of motor, a given windings canbe classified as field winding or as armaturewinding.

Field winding - to set up the magnetic fieldrequired to produce torque.

Armature windings - carry currents that varywith mechanical load. When the machine is

used as a generator, the output is taken fromthe armature windings.

Electric Motor


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Electric Motor - Basic classification


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Losses, Power Ratings, and Efficiency

The electrical input powerPin, in watts, supplied by the three-phase source isgiven by:

cos3 rmsrmsin IVP

where Vrms is the rms value of line-to-line voltage, Irms is the rms value of linecurrent and cos is the power factor.

The mechanical output powerPout, in watts, is: moutout TP

where Tout is the output torque in newton-meters, and m is the angular speed ofthe load in radians per second.

The rotational speed may be given in revolutions per minute denoted by nmorby radian per second denoted by m . These quantities are related by:

60

2 mm n


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Losses, Power Ratings, and Efficiency

The mechanical output power for a given electric motor is stated in horsepower(hp). To convert from watts to horsepower, we have

746

watts

horsepower

PP

The power rating of a motor is the output power that the motor can safely

produce on a continuous basis. Most motor can supply output power varyingfrom zero to several times their rated power, depending on the mechanical load.

The power efficiency of a motor is given by:

%100in

out

PP


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DC Machines


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DC Machines - Construction


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DC Motor

Can be divided into 2 types:

(a) Wound type

shunt

series

compound

(b) Permanent magnet type


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Characteristics of DC Electrical Motors


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DC motor

Shunt wound motor- field is connected in parallel with the armature

- has good speed regulation

Series wound motor

- field winding in series with the armature- very high starting torque and poor speed regulation.

Compound wound motor

- field winding has both series and shunt components

- offers better starting torque than the shunt motor butworse speed regulation


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DC motor

Permanent magnet

- field windings are replaced by permanentmagnets

- adequate starting torque- speed regulation somewhat worse than that ofthe compound wound motor


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KKKF163 Introduction to EE

Sem II 2006/07 50

Speed Control of DC Motors

Vary the voltage supplied to thearmature circuit while holding thefield constant.

Vary the field current while holdingthe armature supply voltageconstant.

Insert resistance in series with thearmature circuit.


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Sem II 2006/07 53

Induction motor

Widely used because of its relativesimplicity in construction

Does not require external electricalconnection to the rotor, thus slipsrings and brushes are not required

Operates at a lower speed than the

synchronous speed


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Synchronous Speed

For a P-pole machine, the angular velocity of thefield is given by:

2/Ps

This is also known as synchronous angular velocity

The synchronous speed (in rpm) is given by:

P

fns120

S i l C I d ti M t


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Squirrel-Cage Induction Motor

cross-section conductors inrotor

photograph

Views of Smokin Buckeye motor: rotor, stator and cross-section of stator


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Selection of Induction Motors

Some of the most important considerations inselecting an induction motor are:

1. Efficiency2. Starting torque3. Pull-out torque4. Power factor

5. Starting current


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Synchronous Motor

Generation of electrical energy by utilitycompanies is done almost exclusively withsynchronous machines.

Assuming a constant frequency source,the speed of a synchronous motor doesnot vary with load.

The stator windings of a synchronous

machine are basically the same as thoseof an induction machine


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Synchronous Motor

The synchronous motor can act as asource of reactive power

Proper use of synchronous motorscan lower energy costs of anindustrial plant by increasing thepower factor


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Basic Single-Phase Induction Motor

The pulsating fluxproduced by themain winding can be

resolved into twocounter-rotatingcomponents


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Auxiliary Windings

Two windings that are 90 apartphysically and carrying currents 90apart in phase produce a rotating

magnetic field.

Single-phase induction motorscontain an auxiliary winding

displaced by 90 electrical degreesfrom the main winding.


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Auxiliary Windings


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Stepper Motor

Are used for accurate, repeatablepositioning applications such as read/writeheads of a computer floppy drive or for

moving the head in an ink-jet printer By controlling the rate at which pulses are

applied to the windings of the steppermotor, speed can be varied continuously

from a standing stop to a maximum thatdepends on the motor and the load.


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Stepper Motor

Various types:

1. Variable-reluctance stepper motor

2. Permanent-magnet stepper motor

3. Hybrid stepper motor

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