EE207 Electrical Power - Lecture 7

7/30/2019 EE207 Electrical Power - Lecture 7

1/18

EE207 Electrical Power

Lecture 7

Power Electronics

Rajparthiban Kumar EE207 Electrical Power 2

INTRODUCTION

Electronic system are used extensively in power

systems and controls.

Basic operations of diodes and thyristors that are

used to convert DC to AC and vice versa.

In power electronics , these devices acts as highspeed switches.


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Diodes and diode circuits

A diode is an electronic device possessing two

terminals anode (A) and cathode (K). It has no

moving parts, it acts like a high speed switch

whose contact open and close.

E

A K


Rules of operation

1)When no voltage is applied across the diode, it

acts as an open circuit.

E =0

Open circuit


3/18


2) If a forward voltage of 0.7V or more is applied

across the terminal, anode more slightly positive

compared to cathode, the terminal short-circuits. It

acts like a closed switch, and a current I flows

through it. This is calledforward bias.

E0

I

+ -


3) If an inverse voltage is applied to the diode,

anode is negative compared to cathode, the diodes

remains an open circuit. This is called reversed

bias.

E

- +


4/18


Main characteristics of diode

Peak inverse voltage, PIV. Maximum inverse voltage

before it breaks down.

Maximum average current, limit to the average current a

diode can carry. Depends on construction, size and

assembly.

Maximum temperature, power dissipated as heat, can never

exceed permissible limit, otherwise it will be destroyed.

Use heat sinks or cooling systems.


Application of diodes

Diodes have many application in electronic powercircuits. We will analyse the following circuits toillustrate the functions of diodes and also methodsused in power technology.

Battery charger with series resistor

Battery charger with series inductor

Single phase bridge rectifier Filters

Others


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Battery charger with series resistor

R = 1

60V

D

100V

120V,

60Hz

2 4

3

1


Waveforms

2

3

4

3

44

2

2

1

I

V

t

100

60


6/18


Circuit analysis:-

Diode is in forward bias only when potential at 2 is morethan 4.

4 is always more than 60V and 3 is always at 60V withrespect to reference.

Peak current is at 40A, when the switch is closed.

When the diode opens, 4 follows 3. Since no current flowsthrough the circuit.

Positive current flows into the battery will charge it up.


Resistor produces losses and are inefficient, Useinductor instead.

L=

3.3mH

60V

D

100V

120V,

60Hz

2 4

3

1


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2

3

4

3

44

2

2

1

I

V

t


Circuit analysis:-

Diode begins to conduct the same ways as the

previous circuit.

The voltage accumulates, current increases

gradually. Maximum current is 40A.

When the diode is open, the inductor discharges.


8/18


Single phase bridge rectifier

R

ia ib

iaib

G = Em

I

The circuit enables us to rectify both positive and

negative half cycles of an ac source, to supply DC

power to the load.


ia ib ia

Em

Ed=0.9 Em, rms


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Filters

To reduce the pulsation in

voltage and current, filters

are used to smooth out the

waveforms.

The purpose of this DC

filters are to produce

smooth power flow into a

load.

ia ib

ia

G = Em

I

ib

L

R


The filter must be able to absord the energy

whenever the dc voltage or current rises, and

release the energy when voltage or current falls, so

tends to maintain a constant power flow.

ia ib

Em

Ed=0.9 Em, rms


10/18


Capacitor can be placed parallel to the load, and

maintain a constant voltage.

RC

I

Em


The peak to peak ripple in percent is given by:-

We wish to build a 135 V, 20A dc supply using a single

phase bridge rectifier using an inductive filter. The peak to

peak current ripple is 10%. If 60Hz source is used,calculate:

The inductance of the inductor and peak to peak current

ripple

LfW

Pripple 5.5=


11/18


HLJLLIW

Jripplef

PW

WVIP

L

124.075.24)20(2

1

2

1

75.24)10(60

27005.55.5

2700)20(135

22====

=

=

=

===

21A19A tofromFluctuates

2201.0 AI peakpeak ==


Three phase 3pulse diode rectifier

Composed of 3

diodes connected in

series with secondary

windings of a three

phase transformer

Em is the line-to-

neutral peak voltage R L

D1

D2

D3

I1

I2

I3

Id

Em

1

2

3


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1 2 3

i1 i2 i3

Ed=0.827Em


Current flow one-third of the time in given winding.

Observe that the line current flow intermittently, and the

sudden jumps produce rapid fluctuations in the magnetic

field, which can cause noise.

For example if a 100KVA transformer is used, it can

deliver only 74kW without overheating.

To overcome the drawbacks, 6 pulse rectifier is developed.


13/18


6 pulse bridge rectifier

T

D1D2 D3

D4 D5 D6

L

R

Ia

Ib

Ic

I1 I2

I3

I4

I5

I6


0 60 120 180 240 300 360 420 480 540 600 660-1

-0.5

0

0.5

1

Ed=1.654Em

i1 i2 i3

i6 i4 i5

Ed=

1.35E

1.414 E

1.225 E


14/18


Current flows two-third of the time, and if a

transformer of rating 100kVA is used, it can

deliver 95kW.

L

rmsd

fW

Pripple

EE

17.0

35.1

=

=


Example

A 3 phase bridge rectifier has to supply power

360kW, 240 V dc load. If a 600V, 3 phase, 60

Hz feeder is available, calculate

a) secondary line voltage

VE

E d 17735.1

240

35.1===


15/18


b) DC current per diode

500A1500/3diodepercurrentdc

1500240

360

==

== AkW

Id


C) peak to peak output voltage

From graph the voltage fluctuates btw 1.225E and

1.414 E

VE

VE

VE

peaktopeak 33217250

250)177(414.1

217)177(225.1

max

min

==

==

==


16/18


Fundamental line current

IF

The line current have amplitude of Id, and are rectangularwaves.

The effective current composed of fundamental componentplus all the harmonics.

I=0.816 Id and since IF=0.78 Id IF = 0.955I

Power factor is 0.955, but no reactive power is absorbed because thecurrent and voltage is still in phase.


Harmonic content and THD

The rectangular current wave occurs very frequently in power

electronics.

The degree of distortion of an ac voltage or current is defined as the

ratio of rms value of all the harmonic component divided by the rms

value of the fundamental component.

Total harmonic distortion =

currentlinetheofvaluerms

combinedcomponentharmonicstheallofvaluerms

currentlinelfundamentatheofrms

222

=

=

=

+=

I

I

I

III

H

F

HF

F

H

I

ITHD =


17/18


Amplitudes of the harmonics components are

equal to the amplitude of the fundamental, IF

divided by the order of the harmonics.


Example

The 3 phase, 6 pulse rectifier furnishes a dc current

of 400A to the load. Estimate:-

The rms value of line current

AII d 326)400(816.0816.0 ===


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The rms value of fundamental

The rms value of the 7th and 11th harmonic combined

AIIF 311)326(955.0955.0 ===

AI

III

AI

I

AI

I

HH

HH

FH

FH

HH

52

2811

311

11

447

311

7

117

222

117

11

7

117

=

+=

===

===

+

+

EE207 Electrical Power - Lecture 7

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