power electronics ppts_0

7/26/2019 Power Electronics Ppts_0

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SINGLE PHASE HALF CONTROLLEDCONVERTERS


2/434

Sl.

No.

Module as per Session Planner Lecture

No.

PPT Slide No.

1 Introduction to Line commutated Inverter L-1 3-12

2 Principle of Phase Controlled Rectifier Operation L-2 13-19

3 Expression for the RMS Value of Output Voltage

of a Single Phase Half Wave Controlled RectifierWith Resistive Load

L-3 20-33

4 Performance Parameters

Of Phase Controlled RectifiersL-4 34-45

5 Single Phase Half Wave Controlled Rectifier

With An RL LoadL-5 46-61

6 Single Phase Half Wave Controlled Rectifier

With RL Load & Free Wheeling DiodeL-6 62-80

7 Single Phase Full Wave Controlled Rectifier

Using A Center Tapped TransformerL-7 81-113

8 Single Phase Full Wave Bridge Controlled

RectifierL-8 114-129

LECTURE PLAN


3/434

Lecture-1

Introduction to Line commutatedInverter


4/434

4

Type of input: Fixed voltage, fixed

frequency ac power supply.

Type of output: Variable dc output voltage

Type of commutation: Natural / AC line

commutation

Line

CommutatedConverter

+

-

DC Output

V0(dc)

AC

Input

Voltage


5/434

5

Different types of

Line Commutated Converters

AC to DC Converters (Phase controlled

rectifiers) AC to AC converters (AC voltage controllers)

AC to AC converters (Cyclo-converters) at

low output frequency.


6/434

6

Differences Between

Diode Rectifiers&

Phase Controlled Rectifiers


7/434

Cntd

7

The diode rectifiers are referred to asuncontrolled rectifiers .

The diode rectifiers give a fixed dc outputvoltage .

Each diode conducts for one half cycle.

Diode conduction angle = 1800or radians.

We can not control the dc output voltage orthe average dc load current in a dioderectifier circuit.


8/434

8

Single phase half wave diode rectifier gives an

Average dc output voltage

Single phase full wave diode rectifier gives an

2Average dc output voltage

m

O dc

mO dc

V

V

VV

Cntd


9/434

9

Applications of


DC motor control in steel mills, paper and

textile mills employing dc motor drives. AC fed traction system using dc traction

motor.

Electro-chemical and electro-metallurgicalprocesses.

Magnet power supplies.

Portable hand tool drives.


10/434

10

Classification of


Single Phase Controlled Rectifiers. Three Phase Controlled Rectifiers.


11/434

11

Different types of Single

Phase Controlled Rectifiers. Half wave controlled rectifiers.

Full wave controlled rectifiers.

Using a center tapped transformer.Full wave bridge circuit.

Semi converter.

Full converter.


12/434

12

Different Types of

Three Phase Controlled Rectifiers

Half wave controlled rectifiers.

Full wave controlled rectifiers.

Semi converter (half controlled

bridge converter).

Full converter (fully controlled

bridge converter).


13/434

Lecture-2

Principle of Phase Controlled Rectifier

Operation


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14

Principle of Phase Controlled

Rectifier Operation


15/434

15

Single Phase Half-Wave Thyristor

Converter with a Resistive Load


16/434

16

Supply Voltage

Output Voltage

Output (load)

Current


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17

Supply

Voltage

Thyristor Voltage


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18

Equations

sin i/p ac supply voltage

max. value of i/p ac supply voltage

RMS value of i/p ac supply voltage2

output voltage across the load

s m

m

mS

O L

v V t

VV

V

v v


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19

When the thyristor is triggered at

sin ; to

Load current; to

sinsin ; to

Where max. value of load current

O L m

O

O L

mO L m

mm

t

v v V t t

v

i i tR

V ti i I t t

RV

IR


20/434

20

To Derive an Expression for the

Average (DC)Output Voltage Across The

Load

Lecture-3


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21

2

0

1. ;2

sin

1sin .

2

1sin .

2

dc OO dc

O m

dc mO dc

mO dc

V V v d t

v V t for t to

V V V t d t

V V t d t


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22

sin .2

cos2

cos cos ; cos 1

2

1 cos ; 22

m

O dc

m

O dc

m

O dc

mm SO dc

VV t d t

VV t

VV

VV V V


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23

max

max

Maximum average (dc) o/pvoltage is obtained when 0

and the maximum dc output voltage

1 cos0 ; cos 0 12

mdmdc

mdmdc

VV V

VV V


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24

0

1 cos ; 22

The average dc output voltage can be varied

by varying the trigger angle from 0 to a

maximum of 180 radians

We can plot the control characteristic

v by using the eq

mm SO dc

O dc

VV V V

V s

uation for

O dcV

Cntd


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25

Control Characteristicof

Single Phase Half Wave PhaseControlled Rectifier

with

Resistive Load


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26

The average dc output voltage is given by the

expression

1 cos2

We can obtain the control characteristic by

plotting the expression for the dc output

voltage as a function of trigger angle

mO dc

VV

Cntd


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27


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28

Control CharacteristicVO(dc)

Trigger angle in degrees

0 60 120 180

Vdm

0.2 Vdm

0.6Vdm


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29

Normalizing the dc output

voltage with respect to , the

Normalized output voltage

1 cos2

11 cos

2

dm

m

dcn

mdm

dcn dcn

dm

V

V

VV

VV

VV V

V


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30

To Derive An Expression for the RMS Value

of Output Voltage of a Single Phase HalfWave Controlled Rectifier With Resistive

Load


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31

2

2

0

1

22 2

The RMS output voltage is given by

1.

2

Output voltage sin ; for to

1sin .

2

OO RMS

O m

mO RMS

V v d t

v V t t

V V t d t

Cntd


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32

2

1

22

1

2 2

12 2

1 cos 2By substituting sin , we get

2

1 cos 21.

2 2

1 cos 2 .4

cos 2 .4

mO RMS

m

O RMS

m

O RMS

tt

tV V d t

VV t d t

VV d t t d t

Cntd


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33

1

2

1

2

1

2

1

2

1 sin2

22

sin 2 sin 21;sin2 0

2 2

1 sin 2

2 2

sin2

22

m

O RMS

m

O RMS

m

O RMS

m

O RMS

V t

V t

VV

VV

VV

Cntd


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34

Performance Parameters

OfPhase Controlled Rectifiers

Lecture-4


35/434

35

Output dc power (avg. or dc o/p

power delivered to the load)

; . .,

Where

avg./ dc value of o/p voltage.

avg./dc value of o/p current

dc dc dcO dc O dc O dc

dcO dc

dcO dc

P V I i e P V I

V V

I I

Cntd

C


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36

Output ac power

Efficiency of Rectification (Rectification Ratio)

Efficiency ; % Efficiency 100

The o/p voltage consists of two components

The dc component

The ac

O ac O RMS O RMS

O dc O dc

O ac O ac

O dc

P V I

P PP P

V

/ripple component

ac r rmsV V

Cntd


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37

2 2

2 2

The total RMS value of output voltage is given by

Form Factor (FF) which is a measure of the

shape of the output voltage is given by

RMS output l

O RMS O dc r rms

ac r rms O RMS O dc

O RMS

O dc

V V V

V V V V

VFF

V

oad voltage

DC load output load voltage

Cntd


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38

22 2

2

The Ripple Factor (RF) w.r.t. o/p voltage w/f

1

1

r rms acv

dcO dc

O RMS O dc O RMS

v

O dc O dc

v

V Vr RF

V V

V V Vr

V V

r FF

Cntd


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39

2 2

max min

max min

Current Ripple Factor

Where

peak to peak ac ripple output voltage

peak to peak ac ripple load current

r rms ac

idcO dc

acr rms O RMS O dc

r pp

r pp O O

r pp

r pp O O

I Ir

I I

I I I I

V

V V V

I

I I I

Cntd


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40

Transformer Utilization Factor (TUF)

Where

RMS supply (secondary) voltage

RMS supply (secondary) current

O dc

S S

S

S

PTUF

V I

V

I

Cntd


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41

Cntd


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42

1

Where

Supply voltage at the transformer secondary side

i/p supply current

(transformer secondary winding current)

Fundamental component of the i/p supply current

Peak value of the input s

S

S

S

P

v

i

i

I

upply current

Phase angle difference between (sine wavecomponents) the fundamental components of i/p

supply current & the input supply voltage.

Cntd


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43

1

Displacement angle (phase angle)

For an RL load

Displacement angle = Load impedance angle

tan for an RL load

Displacement Factor (DF) orFundamental Power Factor

L

R

DF Cos

Cntd


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44

11

2 22 2 21

2

1 1

1

Harmonic Factor (HF) or

Total Harmonic Distortion Factor ; THD

1

Where

RMS value of input supply current.

RMS value of fundamental component of

the i

S S S

S S

S

S

I I IHF

I I

I

I

/p supply current.

Cntd

Cntd


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45

1 1

Input Power Factor (PF)

cos cos

The Crest Factor (CF)

Peak input supply c

For an Ide

urrent

RMS input supply current

1; 100% ;

al Controlled Rectifier

0 ; 1;

S S S

S S S

S peak

S

ac r rms

V I IPFV I I

ICF

I

FF V V TUF

R

0 ; 0; 1v

F r HF THD PF DPF

Cntd


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46

Single Phase Half Wave ControlledRectifier With An RL Load

Lecture -5


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47

Cntd


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48

Input Supply Voltage (Vs)

&Thyristor (Output) Current

Waveforms


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49

Cntd

O t t (L d)


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50

Output (Load)

Voltage Waveform


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51

1

To Derive An Expression For

The Output(Load) Current, During to

When Thyristor Conducts

t

T

Cntd


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52

1

1

Assuming is triggered ,

we can write the equation,

sin ;

General expression for the output current,

sin

OO m

t

mO

T t

diL Ri V t tdt

Vi t A e

Z

Cntd

Cntd


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53

22

1

1

2 maximum supply voltage.

=Load impedance.

tan Load impedance angle.

Load circuit time constant.

general expression for the output load current

sin

m S

Rt

m LO

V V

Z R L

L

RL

R

Vi t A e

Z

Cntd

Cntd


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54

1

1

1

1

1

Constant is calculated from

initial condition 0 at ; t=

0 sin

sin

We get the value of constant as

sin

O

Rt

m L

O

Rt

mL

R

mL

A

i t

V

i A eZ

VA e

Z

A

VA e

Z

Cntd

Cntd


55/434

55

1Substituting the value of constant in the

general expression for

sin sin

we obtain the final expression for the

inductive load current

sin sin

O

Rt

m mLO

Rt

m LO

A

i

V Vi t e

Z Z

Vi t eZ

;

Where t

Cntd


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56

Extinction angle can be calculated by using

the condition that 0

sin sin 0

sin sin

can be calculated by solving the above eqn.

O

Rtm L

O

R

L

i at t

Vi t eZ

e

Cntd


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57

To Derive An ExpressionFor

Average (DC) Load Voltage of aSingle Half Wave Controlled

Rectifier with

RL Load


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58

2

0

2

0

1

.2

1. . .

20 for 0 to & for to 2

1

. ;2

sin for to

L OO dc

L O O OO dc

O

L OO dc

O m

V V v d t

V V v d t v d t v d t

v t t

V V v d t

v V t t


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59

1sin .

2

cos

2

cos cos2

cos cos2

L mO dc

mLO dc

mLO dc

mLO dc

V V V t d t

VV V t

VV V

VV V

Effect of Load


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60

During the period to the

instantaneous o/p voltage is negative

reduces the average or the dc output

vo

and

this

when compared to a purely

resist

ltage

ive load.

t

Effect of Load

Inductance on the Output


61/434

61

Average DC Load Current

cos cos2O dc

mO dc L Avg

L L

V VI I

R R


62/434

Lecture-6

Single Phase Half Wave Controlled RectifierWith RL Load & Free Wheeling Diode


63/434

63

V0

i0T

R

L

Vs ~+

+

FWD


64/434

64

0

0

0

0

vS

iG

vO

t

t

t

t

Supply voltage

Load current

Load voltage

t=

2

Gate pulses

iO


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65

The followi

The average

ng points a

output voltage

1 cos which is the same as that2

of a purely resistive load.

For low value of inductance, the load currenttends to become dis

re to be noted

cont

mdc

VV

inuous.


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66

During the period to

the load current is carried by the SCR.

During the period to load current is

carried by the free wheeling diode.

The value of depends on the value of

R and L and the forwa

rd resistanceof the FWD.


67/434


68/434

68

Single Phase Half Wave

Controlled Rectifier With

A

General Load


69/434

69

R

vS~+

L

E+

vO

iO


70/434

70

1sin

For trigger angle ,the Thyristor conducts from to

For trigger angle ,

the Thyristor conducts from to

m

EV

t

t


71/434

71

0

0

iO

t

t

Load current

E

vO

Load voltage

Vm

Im


72/434

72

Equations

sin Input supply voltage.

sin o/p load voltagefor to .

for 0 to &for to 2 .

S m

O m

O

v V t

v V tt

v E tt


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73

Expression for the Load Current

When the thyristor is triggered at a delay angle of

, the eqn. for the circuit can b

sin +E

e written as

The general expression for the output load

current can be writte

;

n

Om O

diV t i R L t dt

s

as

int

mO

V Ei t Ae

Z R


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74

22

1

Where

= Load Impedance.



The general expression for the o/p current can

be written as sinR

tm L

O

V Ei t Ae

Z

Z R

L

R

L

R

R

L


75/434


76/434

76

We obtain the value of constant 'A' as

Substituting the value of the constant 'A' in theexpression for the load current; we get the

complete expression for the output load c

si

ur

nR

m LVE

A eR Z

sin

rent

in

s

as

R tm m L

O

V VE Ei t e

Z R R Z


77/434

77

To Derive

An

Expression For The AverageOr

DC Load Voltage


78/434

78

2

0

2

0

2

0

sin Output load voltage

1

.2

1.

for 0 to & for to 2

for

. .

to

2

1. sin .

2

O

OO dc

O O OO d

m

m

c

O dc

O

V v d t

V v d t v d t v d t

V E d t V t E d t

v V t

v E t t

t


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79

2

0

1 cos2

10 cos cos 2

2

cos cos 22 2

2cos cos2 2

mO dc

mO dc

m

O dc

m

O dc

V E t V t E t

V E V E

V EV

VV E


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80

2

2

0

Conduction angle of thyristor

RMS Output Voltage can be calculatedby using the expres

1 .

sion

2 OO RMS

V v d t

L t 7


81/434

Lecture-7

Single Phase Full Wave Controlled

Rectifier Using A Center Tapped

Transformer


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82

ACSupply

O

A

B

T1

T2

R L

vO

+


83/434

83

DiscontinuousLoad Current Operation

without FWDfor


84/434

84

vO Vm

0

( ) ( )

iO

t

t0


85/434

85

1


The Output(Load) Current, During to


t

T


86/434

86

1

1



sin ;


sin

OO m

t

mO

T t

diL Ri V t tdt

Vi t A e

Z


87/434

87

22

1

1


=Load impedance.




sin

m S

Rt

m LO

V V

Z R L

L

RL

R

Vi t A e

Z

Constant is calculated fromA


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88

1

1

1

1

1

Constant is calculated from


0 sin

sin


sin

O

Rt

m LO

Rt

mL

R

mL

A

i t

Vi A e

Z

VA e

Z

AV

A eZ


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89



sin sin



sin sin

O

Rt

m mLO

Rt

m LO

A

i

V Vi t e

Z Z

Vi t eZ

;

Where t


90/434

90

Extinction angle can be calculated by using

the condition that 0

sin sin 0

sin sin

can be calculated by solving the above eqn.

O

R

tm LO

R

L

i at t

Vi t eZ

e


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91

To Derive An Expression For The

DC Output Voltage OfA Single Phase Full Wave

Controlled Rectifier With RL Load

(Without FWD)


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92

vO Vm

0

( ) ( )

iO

t

t0

1


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93

1.

1sin .

cos

cos cos

dc OO dc

t

dc mO dc

mdcO dc

mdcO dc

V V v d t

V V V t d t

VV V t

VV V


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94

When the load inductance is negligible i.e., 0

Extinction angle radians

Hence the average or dc output voltage for R load

cos cos

cos 1

1 cos ; for R load, when

mO dc

m

O dc

m

O dc

L

VV

VV

VV


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95

2 2

To calculate the RMS output voltage we use

the expression

1sin .mO RMSV V t d t


96/434

96

Discontinuous Load Current

Operation with FWD

V


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97

vO Vm

0

( ) ( )

iO

t

t0


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98

2

2

1

1

Thyristor is trigger

Thyristor is triggered at ;conducts from to

FWD conducts from to &0 during discontinuous loa

ed at ;conducts from t

d current.

o 2

O

T tT

T t

t

T t

tv


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99

To Derive an Expression

For The

DC Output Voltage For

ASingle Phase Full Wave

Controlled RectifierWith RL Load & FWD

1


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100

0

1.

1sin .

cos

cos cos ; cos 1

1 cos

dc OO dc

t

dc mO dc

mdcO dc

m

dcO dc

mdcO dc

V V v d t

V V V t d t

VV V t

VV V

VV V


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101

The load current is discontinuous for low

values of load inductance and for large

values of trigger angles.

For large values of load inductance the load

current flows continuously without falling to

zero.

Generally the load current is continuous for

large load inductance and for low trigger

angles.


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102

Continuous Load Current

Operation(Without FWD)


103/434

103

vO Vm

0

( )

iO

t

t0

( )


104/434

104

To DeriveAn Expression For

Average / DC Output Voltage

OfSingle Phase Full Wave Controlled

Rectifier For Continuous Current

Operation without FWD


105/434

105

vO Vm

0

( )

iO

t

t0

( )


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106

1

.

1 sin .

cos

dc OO dc

t

dc mO dc

mdcO dc

V V v d t

V V V t d t

VV V t

V V


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107

cos cos ;

cos cos

cos cos

2cos

dcO dc

m

mdcO dc

mdcO dc

V V

V

VV V

VV V


108/434

108

By plotting VO(dc)versus ,

we obtain the control characteristic of a

single phase full wave controlled rectifier

with RL load for continuous load currentoperation without FWD


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109

cosdc dmV V

V V V


110/434

110

VO(dc)


030 60 90

Vdm

0.2 Vdm

0.6Vdm

-0.6 Vdm

-0.2Vdm

-Vdm

120 150 180

cosdc dmV V


111/434

111

00

By varying the trigger angle we can vary theoutput dc voltage across the load. Hence we can

control the dc output power flow to the load.

For trigger a . .,ngle , 0 to 90

cos is positive

0 90 ;

i e

and hence is positive

Converter

& are positive ; is positive

Controlled Rectifoperates as a

Power flow is from the

ie

ac source to the d.

r.

loa

dc dc dc dc d

d

c

cV

V I P V I

0 0F t i l 90 t 180


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112

0

0

0

0

. ., 90 180 ,

is negative; is positive

For trigger angle , 90

;

is negative.

Line

to 180

cos is negative and hence

In this case the conve

Co

rte

mmutated In

r operates

s vea a

dc dc

dc dc dc

i e

V I

P V I

Power flows from the load ckt. to the i/p ac source.The inductive load energy is fed back to the

i/p sou

rter.

rce.

Drawbacks Of Full Wave


113/434

113


Controlled RectifierWith Centre Tapped Transformer We require a centre tapped transformer

which is quite heavier and bulky. Cost of the transformer is higher for the

required dc output voltage & output power.

Hence full wave bridge converters arepreferred.


114/434

Single Phase


115/434

115

Single Phase

Full Wave Bridge Controlled

Rectifier2 types of FW Bridge Controlled Rectifiers are

Half Controlled Bridge Converter

(Semi-Converter)

Fully Controlled Bridge Converter

(Full Converter)

The bridge full wave controlled rectifier does

not require a centre tapped transformer


116/434

116

Single Phase

Full Wave Half ControlledBridge Converter

(Single Phase Semi Converter)


117/434

117

Trigger Pattern of Thyristors


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118


1

2

0

1 2

, 2 ,...

, 3 ,...

& 180

Thyristor T is triggered at

t at t


t at t

The time delay between the gating

signals of T T radians or


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119

Waveforms of

single phase semi-converter

with general load & FWD

for > 900


120/434

120

Single Quadrant

Operation


121/434

121


122/434

122


123/434

123

1 1

2 2

Thyristor & conductfrom

Thyristor & conductfrom 2

FWD conducts during0 to , ,...

T Dt to

T Dt to

t to


124/434

124

Load Voltage & Load CurrentWaveform of Single Phase Semi

Converter for< 900

& Continuous load current

operation

v Vm


125/434

125

vO Vm

0

iO

t

( )

t0

( )



126/434

126


For TheDC Output Voltage of

A

Single Phase Semi-Converter

With R,L, & E Load & FWD

For Continuous, Ripple FreeLoad Current Operation

1

V V d t


127/434

127

0

.

1sin .

cos

cos cos ; cos 1

1 cos

dc OO dc

t

dc mO dc

mdcO dc

mdcO dc

mdcO dc

V V v d t

V V V t d t

VV V t

VV V

VV V

can be varied from a maxV


128/434

128

max

can be varied from a max.

2value of 0 by varying from 0 to .

For 0, The max. dc o/p voltage obtained is

Normalized dc o/p voltage is

2

11 cos2

dc

m

m

dc

mdn

m

dmdc

dcn n

V

Vto

V

V

V V

VVV

V

V

1 cos2

RMS O/P Voltage VO(RMS)


129/434

129

RMS O/P Voltage VO(RMS)

1

22 2

1

2 2

1

2

2sin .

2

1 cos 2 .2

1 sin 2

22

mO RMS

m

O RMS

m

O RMS

V V t d t

VV t d t

VV


130/434

130

Single Phase Full Wave

Controlled Rectifier

LECTURE PLANSl.No Module as per Session Planner Lecture No. PPT Slide No.


131/434

131

1 Single Phase Full Wave Controlled Rectifier

Using A Center Tapped Transformer

L 1 2-12

2 Derivation for Expression For The DC OutputVoltage Of A Single Phase Full Wave

Controlled Rectifier With RL Load

L2 13-20

3 Derivation for Expression For The DC Output

Voltage For A Single Phase Full Wave

Controlled Rectifier With RL Load & FWD

L3 21-34

4 Single Phase Full Wave Bridge ControlledRectifier

L 4 35-44

5 Load Voltage & Load Current Waveform of

Single Phase Semi Converter for < 900&

Continuous load current operation

L 5 45-51

6 Single Phase Full Converter L 6 52-58

7 Derivation for Expression For The Average DC

Output Voltage of a Single Phase Full

Converter assuming Continuous & Constant

Load Current

L 7 59-68

8 Two Quadrant Operation

of a Single Phase Full Converter

L 8 69-78

Lecture-1


132/434

132

Single Phase Full Wave Controlled

Rectifier Using A Center TappedTransformer

Single Phase Midpoint type

Fully controlled Rectifier


133/434

133133

ACSupply

O

A

B

T1

T2

R L

vO

+

Fully controlled Rectifier


134/434

134134

DiscontinuousLoad Current Operation

without FWDfor

v Vm


135/434

135135

vO Vm

0

( ) ( )

iO

t

t0


136/434

136136

1


The Output

(Load) Current, During to


t

T


137/434

137137

1

1



sin ;


sin

OO m

tm

O

T t

diL Ri V t tdt

Vi t A e

Z

2 maximum supply voltageV V


138/434

138138

22

1

1


=Load impedance.




sin

m S

Rt

m LO

V V

Z R L

L

R

L

R

Vi t A e

Z

1Constant is calculated fromA


139/434

139139

1

1

1

1


0 sin

sin


sin

O

Rt

m LO

Rt

mL

R

mL

i t

Vi A e

Z

VA e

Z

A

VA e

Z

Substituting the value of constant in theA


140/434

140140



sin sin



sin sin

O

Rt

m mLO

Rt

m LO

A

iV V

i t eZ Z

V

i t eZ

;

Where t


141/434

141141

Extinction angle can be calculated by usingthe condition that 0

sin sin 0

sin sincan be calculated by solving the above eqn.

O

Rt

m LO

R

L

i at t

Vi t eZ

e

Lecture-2


142/434

142142

To Derive An Expression For The DC

Output Voltage OfA Single Phase Full Wave Controlled

Rectifier With RL Load

(Without FWD)

Lecture 2

vO Vm


143/434

143143

vO m

0

( ) ( )

iO

t

t0

1

V V v d t


144/434

144144

.

1sin .

cos

cos cos

dc OO dc

t

dc mO dc

mdcO dc

mdcO dc

V V v d t

V V V t d t

VV V t

VV V


145/434

145145

When the load inductance is negligible i.e., 0

Extinction angle radians

Hence the average or dc output voltage for R load

cos cos

cos 1

1 cos ; for R load, when

mO dc

m

O dc

m

O dc

L

VV

VV

VV


146/434

146146

2 2

To calculate the RMS output voltage we usethe expression

1 sin .mO RMSV V t d t


147/434

147147

Discontinuous Load Current

Operation with FWD

vO Vm


148/434

148148

0

( ) ( )

iO

t

t0


149/434

149149

2

2

1

1

Thyristor is trigger

Thyristor is triggered at ;conducts from to

FWD conducts from to &0 during discontinuous loa

ed at ;conducts from t

d current.

o 2

O

T tT

T t

t

T t

tv

Lecture-3


150/434

150150

To Derive an Expression For The DC

Output Voltage For A Single Phase FullWave Controlled Rectifier With

RL Load & FWD

Lecture 3

1

.dc OO dcV V v d t


151/434

151151

0

1sin .

cos

cos cos ; cos 1

1 cos

dc OO dc

t

dc mO dc

mdcO dc

mdcO dc

mdcO dc

V V V t d t

VV V t

VV V

VV V


152/434

152152

The load current is discontinuous for low

values of load inductance and for large

values of trigger angles.

For large values of load inductance the loadcurrent flows continuously without falling to

zero.

Generally the load current is continuous forlarge load inductance and for low trigger

angles.


153/434

153153

Continuous Load Current

Operation(Without FWD)

vO Vm


154/434

154154

O

0

( )

iO

t

t0

( )


155/434

155155

To DeriveAn Expression For

Average / DC Output Voltage

OfSingle Phase Full Wave Controlled

Rectifier For Continuous Current

Operation without FWD

vO Vm


156/434

156156

O

0

( )

iO

t

t0

( )

1


157/434

157157

1

.

1sin .

cos

dc OO dc

t

dc mO dc

mdcO dc

V V v d t

V V V t d t

VV V t

dcO dcV V


158/434

158158

cos cos ;

cos cos

cos cos

2cos

m

mdcO dc

mdcO dc

V

VV V

VV V


159/434

159159

By plotting VO(dc)versus ,we obtain the control characteristic of a

single phase full wave controlled rectifier

with RL load for continuous load currentoperation without FWD


160/434

160160

cosdc dmV V

VO(dc)

V

cosdc dmV V


161/434

161161


030 60 90

Vdm

0.2 Vdm

0.6Vdm

-0.6 Vdm

-0.2Vdm

-Vdm

120 150 180

By varying the trigger angle we can vary the


162/434

162162

00

output dc voltage across the load. Hence we cancontrol the dc output power flow to the load.

For trigger a . .,ngle , 0 to 90

cos is positive

0 90 ;

i e

and hence is positive

Converter

& are positive ; is positive

Controlled Rectifoperates as aPower flow is from the

ieac source to the d.

r.loa

dc dc dc dc d

d

c

cV

V I P V I

0 0For trigger angle , 90 to 180


163/434

163163

0 0. ., 90 180 ,

is negative; is positive ;

is negative.

Line

cos is negative and hence

In this case the conve

Co

rte

mmutated In

r operates

s vea a

dc dc

dc dc dc

i e

V I

P V I

Power flows from the load ckt. to the i/p ac source.The inductive load energy is fed back to the

i/p sou

rter.

rce.

Lecture-4


164/434

164

Single PhaseFull Wave Bridge Controlled Rectifier



165/434

165165


With Centre Tapped Transformer We require a centre tapped transformer

which is quite heavier and bulky. Cost of the transformer is higher for the

required dc output voltage & output power.

Hence full wave bridge converters arepreferred.

Single Phase Full Wave Bridge



166/434

166166


2 types of FW Bridge Controlled Rectifiers are

Half Controlled Bridge Converter

(Semi-Converter)

Fully Controlled Bridge Converter

(Full Converter)

The bridge full wave controlled rectifier does

not require a centre tapped transformer


167/434

167167

Single Phase

Full Wave Half Controlled

Bridge Converter

(Single Phase Semi Converter)

Single Phase Full Wave Half Controlled

Bridge Converter


168/434

168168

g



169/434

169169

1

2

0

1 2

, 2 ,...

, 3 ,...

& 180


t at t


t at t

The time delay between the gating

signals of T T radians or


170/434

170170

Waveforms of

single phase semi-converter

with general load & FWD

for > 900


171/434


172/434

172172


173/434

173173

Th i & dT D


174/434

174174

1 1

2 2

Thyristor & conduct

from

Thyristor & conduct

from 2

FWD conducts during0 to , ,...

T D

t to

T D

t to

t to

Lecture-5


175/434

175175

Load Voltage & Load Current Waveform

of Single Phase Semi Converter for

< 900& Continuous load currentoperation

vO Vm


176/434

176176

0

iO

t

( )

t0

( )



177/434

177177

For TheDC Output Voltage of

A

Single Phase Semi-Converter

With R,L, & E Load & FWD

For Continuous, Ripple FreeLoad Current Operation

0

1.dc OO dcV V v d t


178/434

178178

0

1sin .

cos

cos cos ; cos 1

1 cos

t

dc mO dc

mdcO dc

mdcO dc

mdcO dc

V V V t d t

VV V t

VV V

VV V

can be varied from a max.

2

dcV

V


179/434

179179

max

2

value of 0 by varying from 0 to .

For 0, The max. dc o/p voltage obtained is

Normalized dc o/p voltage is

2

11 cos

2

m

m

dc

mdn

m

dmdc

dcn n

V

to

V

V

V V

VVV

V

V

1 cos2

RMS O/P Voltage VO(RMS)1


180/434

180180

1

22 2

1

2 2

1

2

2 sin .2

1 cos 2 .2

1 sin 2

22

mO RMS

m

O RMS

m

O RMS

V V t d t

VV t d t

VV

Lecture-6


181/434

181

Single Phase Full Converter

Single Phase Full Converter


182/434

182182


183/434

183183

Waveforms ofSingle Phase Full Converter

Assuming Continuous

(Constant Load Current)

&

Ripple Free Load Current


184/434


185/434

185185

iOConstant Load Currenti =IO a


186/434

186186

a

i

iT1

T2&

Ia

t

t

t

Iai

i

T3

T4&

Ia

Ia


187/434

The average dc output voltage

b d t i d b i th i


188/434

188188

2

0

can be determined by using the expression

1. ;

2

The o/p voltage waveform consists of two o/p

pulses during the input supply time period of

0 to 2 r

dc OO dcV V v d t

adians. Hence the Average or dco/p voltage can be calculated as


189/434

Maximum average dc output voltage is


190/434

190190

0

max

max

Maximum average dc output voltage is

calculated for a trigger angle 0

and is obtained as

2 2cos 0

2

m mdmdc

mdmdc

V VV V

VV V

The normalized average output voltage is given by


191/434

191191

max

The normalized average output voltage is given by

2 cos

cos2

O dc dcdcn n

dmdc

m

dcn nm

V VV V

V V

V

V VV


192/434

192192

By plotting VO(dc)versus ,

we obtain the control characteristic of

a single phase full wave fully

controlled bridge converter

(single phase full converter)

for constant & continuousload current operation.


193/434


194/434

194194

VO(dc)

Vdm

cosdc dmV V


195/434

195195


030 60 90

0.2 Vdm

0.6Vdm

-0.6 Vdm

-0.2Vdm

-Vdm

120 150 180


196/434

196196

During the period from t = to the inputvoltage vSand the input current iSare both

positive and the power flows from the

supply to the load. The converter is said to be operated in the

rectification mode

Controlled Rectifier Operationfor 0 < < 900


197/434

197197

During the period from t = to (+),the input voltage vSis negative and theinput current iSis positive and the outputpower becomes negative and there will be

reverse power flow from the load circuit tothe supply.

The converter is said to be operated in theinversion mode.

Line Commutated Inverter Operation

for 900 < < 1800

Lecture-8


198/434

198

Two Quadrant Operation

of a Single Phase Full Converter

Two Quadrant Operationof a Single Phase Full Converter


199/434

199199

0


200/434

200200

p g

2

2

0

The rms value of the output voltage

is calculated as

1

.2 OO RMSV v d t

The single phase full converter gives two


201/434

201201

The single phase full converter gives two

output voltage pulses during the input supply

time period and hence the single phase full

converter is referred to as a two pulse converter.

The rms output vo

2

ltage can be calculated as

2 .2

OO RMSV v d t

2 21 sin .mO RMSV V t d t


202/434

202202

22

2

2

sin .

1 cos 2.

2

cos 2 .2

m

O RMS

m

O RMS

m

O RMS

VV t d t

tVV d t

VV d t t d t

2


203/434

203203

2

2

2

sin222

sin 2 sin 2

2 2

sin 2 2 sin 2;

2 2

sin 2 2 sin 2

mO RMS

m

O RMS

m

O RMS

V tV t

V

V

VV

2

sin 2 sin 2mO RMS

VV


204/434

204204

2 2

2 2

02 2 2

2

Hence the rms output voltage is same as therms input supply voltage

O RMS

m m m

O RMS

mSO RMS

V V VV

VV V


205/434

iOConstant Load Currenti =IO a

I


206/434

206206

i

i

T1

T2&

Ia

t

t

t

Iai

i

T3

T4&

Ia

Ia

The rms thyristor current can be


207/434

207207

calculated as

2The average thyristor current can be

calculated as

2

O RMS

T RMS

O dc

T Avg

II

II


208/434

power Electronics

unit-5

208

THREE PHASE LINE

COMMUTATED CONVERTERS

LECTURE PLAN

Sl.

No

Module as per Session Planner Lecture

No.

PPT Slide

No.


209/434

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209

No.

1Introduction to Three phase converters L 1 3-10

2 3-Phase Half Wave Converter

(3-Pulse Converter)L2 11-26

3 3 Phase Half WaveControlled Rectifier

Output Voltage Waveforms For RL

Load atDifferent Trigger AnglesL3 27-39

4 Three Phase Semi-converters L 4 40-52

5 Wave forms of 3 Phase Semi-converter

for600 and Discussion

L 5 52-63

6 Three Phase Full Converter L 6 64-77

7 Single Phase Dual Converter L 7 78-99

8 Three Phase Dual Converters L 8 100-119

Lecture-1


210/434

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210

Introduction to

Three phase converters

1-phase Controlled Rectifiers


211/434

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211

Single phase half controlled bridgeconverters & fully controlled bridgeconverters are used extensively inindustrial applications up to about

15kW of output power.

The single phase controlled rectifiersprovide a maximum dc output of

The output ripple frequency is equal tothe twice the ac supply frequency.

max

2 mdc

VV

Contd


212/434

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unit-5

212

The single phase full wave controlledrectifiers provide two output pulses during

every input supply cycle and hence are

referred to as two pulse converters

3 Phase Controlled Rectifiers


213/434

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unit-5

213

Three phase converters are 3-

phase controlled rectifiers

which are used to convert acinput power supply into dc

output power across the load

Features of 3-phase controlledrectifiers


214/434

power Electronics

unit-5

214214

Operate from 3 phase ac supply voltage. They provide higher dc output voltage.

Higher dc output power.

Higher output voltage ripple frequency. Filtering requirements are simplified for

smoothing out load voltage and load

current.


215/434

power Electronics

unit-5

215215

Extensively used in high power variablespeed industrial dc drives.

Three single phase half-wave converters

can be connected together to form a threephase half-wave converter.

Classification of 3-phaseconverters


216/434

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216

3-phase half wave converter

3-phase semi converter

3-phase full converter 3- phase dual converter

Classification according tono of pulses in the output wave


217/434

power Electronics

unit-5

217

3- pulse converter

6-pulse converter 12- pulse converter

3-Phase

Lecture-2


218/434

power Electronics

unit-5

218218

Half Wave Converter(3-Pulse Converter)

withR-L Load

Continuous & Constant

Load Current Operation

Circuit Diagram of 3- pulse converter


219/434

power Electronics

unit-5

219219

Vector Diagram of

3 Phase Supply Voltages


220/434

power Electronics

unit-5

220220

VAN

VCN

VBN

1200

1200

1200 RN AN

YN BN

BN CN

v v

v v

v v

3 Phase Supply Voltage

Equations


221/434

power Electronics

unit-5

221221

q

We deifine three line to neutral voltages

(3 phase voltages) as follows


222/434


223/434

power Electronics

unit-5

223223

van vbn vcn van

Constant Load

Each thyristor conducts for 2/3 (1200)


224/434

power Electronics

unit-5

224224

io=I

a

Constant Load

Current

Ia

Ia


225/434

power Electronics

unit-5

225225

To Derive anExpression for the

Average Output Voltage of a

3-Phase Half Wave Converterwith RL Load

for Continuous Load Current

0

1 30T is triggered at t


226/434

power Electronics

unit-5

226226

1

0

2

0

3

0

6

5 150

6

7 270

6

2Each thytistor conducts for 120 or radians3

gg

T is triggered at t

T is triggered at t


227/434


228/434


229/434

0 0 0 0cos 180 30 cos sin 180 30 sin


230/434

power Electronics

unit-5

230230

0 0

0 0

0 0

0

0

0

0

0 0

Note: cos 1

cos 180 30 cos sin 180 30 sin32 cos 30 .cos sin 30 sin

cos 30 cos sin 30 sin3

2 cos 30 .cos sin 30 s

80 30 cos 30

sin 180 30 sin 30

in

mdc

m

dc

VV

VV


231/434


232/434

The rms value of output voltage is found by

using the equation


233/434

power Electronics

unit-5

233233

15 26

2 2

6

1

2

3sin .

2

and we obtain

1 33 cos 26 8

mO RMS

mO RMS

V V t d t

V V

3 Phase Half Wave

Lecture-3


234/434

power Electronics

unit-5

234234

Controlled Rectifier Output

Voltage Waveforms For RL

Loadat

Different Trigger Angles

Van

V0

=300

Vbn Vcn

=300


235/434

power Electronics

unit-5

235235

0

0

300

300

600

600

900

900

1200

1200

1500

1500

1800

1800

2100

2100

2400

2400

2700

2700

3000

3000

3300

3300

3600

3600

3900

3900

4200

4200

V0

Van

=600

Vbn Vcn

t

t

=600


236/434

power Electronics

unit-5

236236

030

060

090

0120

0150

0180

0210

0240

0270

0300

0330

0360

0390

0420

0

V0

Van

=900

Vbn Vcn

t

=900


237/434

power Electronics

unit-5

237237

3 Phase Half WaveControlled Rectifier With

R Loadand

RL Load with FWD

T1 T1


238/434

power Electronics

unit-5

238238

a a

b b

c c

R

V0L

R V0

+

T2

T3

n n

T2

T3

3 Phase Half Wave


239/434

power Electronics

unit-5

239239

Controlled Rectifier Output

Voltage Waveforms For R Load

or RL Load with FWDat

Different Trigger Angles

Vs

Van

=0

Vbn Vcn

=00


240/434

power Electronics

unit-5

240240

0

0

300

300

600

600

900

900

1200

1200

1500

1500

1800

1800

2100

2100

2400

2400

2700

2700

3000

3000

3300

3300

3600

3600

3900

3900

4200

4200V0

=150

t

Van Vbn Vcn

t

=150

0

=300

Van Vbn Vcn

=300


241/434

power Electronics

unit-5

241241

0

300

300

600

600

900

900

1200

1200

1500

1500

1800

1800

2100

2100

2400

2400

2700

2700

3000

3000

3300

3300

3600

3600

3900

3900

4200

4200

V0 t

V0

=600Van Vbn Vcn

t

=600


242/434

power Electronics

unit-5

242242

To Derive An Expression For The

Average Or Dc Output Voltage Of A

3 Phase Half Wave Converter WithResistive Load Or RL Load With FWD

01 306

T is triggered at t


243/434

power Electronics

unit-5

243243

0 0

1

0

2

0 0

2

0

30 180 ;

sin

5 150

6

150 300 ;

sin 120

O an m

O bn m

T conducts from to

v v V t

T is triggered at t

T conducts from to

v v V t


244/434

power Electronics

unit-5

244244

0

3

0 03

0

0

7 270

6

270 420 ;

sin 240

sin 120

O cn m

m

T is triggered at t

T conducts from to

v v V t

V t

0

0

180

30

3.

2dc O

V v d t


245/434

power Electronics

unit-5

245245

0

0

0

0

30

0 0

180

30

180

30

sin ; for 30 to 180

3 sin .2

3sin .2

O an m

dc m

mdc

v v V t t

V V t d t

VV t d t


246/434

Lecture-4


247/434

power Electronics

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247

Three Phase Semi-converters

Three Phase Semi-converters


248/434

power Electronics

unit-5

248248

3 Phase semi-converters are used in

Industrial dc drive applications upto 120kW

power output.

Single quadrant operation is possible.

Power factor decreases as the delay angle

increases.

Power factor is better than that of 3 phase

half wave converter.

3 Phase

H lf C t ll d B id C t


249/434

power Electronics

unit-5

249249

Half Controlled Bridge Converter(Semi Converter)

with Highly Inductive Load &

Continuous Ripple free Load

Current


250/434

power Electronics

unit-5

250250


251/434

power Electronics

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251251

Wave forms of 3 Phase

Semiconverter for

> 600


252/434

power Electronics

unit-5

252252


253/434

power Electronics

unit-5

253253

3 phase semiconverter output ripple frequency of

t t lt i 3 f


254/434

power Electronics

unit-5

254254

0 0

1

output voltage is 3

The delay angle can be varied from 0 to

During the period

30 210

7 , thyristor T is forward biased6 6

Sf

t

t

1If thyristor is triggered at ,6

& d t t th d th li t li lt

T t

T D


255/434

power Electronics

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255255

1 1& conduct together and the line to line voltage

appears across the load.

7At , becomes negative & FWD conducts.6

The load current contin

ac

ac m

T D

v

t v D

1 1

ues to flow through FWD ;

and are turned off.

mD

T D

1

2

If FWD is not used the would continue to

d t til th th i t i t i d t

mD T

T


256/434

power Electronics

unit-5

256256

2

1 2

conduct until the thyristor is triggered at5

, and Free wheeling action would6

be accomplished through & .

If the delay angle , e3

T

t

T D

ach thyristor conducts

2for and the FWD does not conduct.

3 mD


257/434

3 sin6

RB ac an cn mv v v v V t


258/434

power Electronics

unit-5

258258

53 sin

6

3 sin2

3 sin6

YR ba bn an m

BY cb cn bn m

RY ab an bn m

v v v v V t

v v v v V t

v v v v V t

Lecture-5


259/434

power Electronics

unit-5

259259

Wave forms of 3 Phase

Semiconverter for

600


260/434

power Electronics

unit-5

260260


261/434

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261261


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To derive an Expression for the


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To derive an Expression for theAverage Output Voltage of 3 Phase

Semi-converter for > / 3

and Discontinuous Output Voltage

For and discontinuous output voltage:3

the Average output voltage is found from


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76

6

76

6

the Average output voltage is found from

3

.2

33 sin

2 6

dc ac

dc m

V v d t

V V t d t

3 31 cos

2

3 1

mdc

mL

VV

VV


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max

3 1 cos2

3 Max. value of line-to-line supply voltage

The maximum average output voltage that occurs at

a delay angle of 0 is

3 3

mLdc

mL m

mdmdc

VV

V V

VV V

The normalized average output voltage is

0 5 1 cosdcVV


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17 2

62

6

0.5 1 cos

The rms output voltage is found from

3.

2

dcn

dm

acO rms

VVV

V v d t

17

263


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262 2

6

1

2

33 sin

2 6

3 sin 23

4 2

mO rms

mO rms

V V t d t

V V


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For and continuous output voltage3


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562

6 2

For , and continuous output voltage3

3

. .2

3 31 cos2

dc ab ac

mdc

V v d t v d t

VV

0.5 1 cos

RMS value of o/p voltage is calculated by using

dcn

dm

VV

V


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15 2

62 2 2

6 2

1

22

RMS value of o/p voltage is calculated by usingthe equation

3 . .2

3 23 3 cos

4 3

ab acO rms

mO rms

V v d t v d t

V V

Lecture -6


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271

Three Phase Full Converter

Three Phase Full Converter

3 Phase Fully Controlled Full Wave Bridge

Converter


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272272

Converter.

Known as a 6-pulse converter.

Used in industrial applications up to120kW output power.

Two quadrant operation is possible.


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The thyristors are triggered at an interval of

/ 3.

Th f f i l l i


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The frequency of output ripple voltage is

6fS.

T1is triggered at

t = (

/6 +

), T6isalready conducting when T1is turned ON.

During the interval (/6 + ) to (/2 + ),

T1and T6conduct together & the outputload voltage is equal to vab = (vanvbn)

T2is triggered at t = (/2 + ), T6turns

off naturally as it is reverse biased as soon

as T is triggered


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as T2is triggered.

During the interval (/2 + ) to (5/6 + ),

T1and T2conduct together & the output

load voltage vO= vac= (vanvcn)

Thyristors are numbered in the order in

which they are triggered.

The thyristor triggering sequence is 12,

23, 34, 45, 56, 61, 12, 23, 34,

We deifine three line neutral voltages

(3 phase voltages) as follows

sin ; Max Phase VoltageRN an m mv v V t V


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0

0

0

sin ; Max. Phase Voltage

2sin sin 120

3

2sin sin 120

3

sin 240

RN an m m

YN bn m m

BN cn m m

m

v v V t V

v v V t V t

v v V t V t

V t

V

is the peak phase voltage of a wye-connected source.

m

The corresponding line-to-line

supply voltages are


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3 sin6

3 sin2

3 sin2

RY ab an bn m

YB bc bn cn m

BR ca cn an m

v v v v V t

v v v v V t

v v v v V t


The Average Output Voltage


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The Average Output Voltage

Of

3-phase Full ConverterWith Highly Inductive Load

Assuming Continuous And

Constant Load Current

The output load voltage consists of 6

voltage pulses over a period of 2radians,

Hence the average output voltage is


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2

6

6. ;2

3 sin6

dc OO dc

O ab m

V V v d t

v v V t

Hence the average output voltage iscalculated as

2

6

33 sin .

6dc mV V t d t


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mL

max

3 3 3cos cos

Where V 3 Max. line-to-line supply voThe maximum average dc output voltage is

obtained for a delay angle

ltage

3 3

0,

3

m mLdc

m

m mdmdc

V VV

V

V VV V

L

The normalized average dc output voltage is

cosdcVV V


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1

22

2

6

cos

The rms value of the output voltage is found from

6.

2

dcn n

dm

OO rms

VV VV

V v d t

1

22

2

6

6.

2

abO rmsV v d t


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6

1

22

2 2

6

1

2

3 3 sin .2 6

1 3 33 cos 22 4

mO rms

mO rms

V V t d t

V V

Lecture-7


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Single Phase Dual Converter

Single Phase Dual Converter


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Th d l f 1 i


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1 1

2 2

The average dc output voltage of converter 1 is

2cos

The average dc output voltage of converter 2 is

2

cos

m

dc

m

dc

VV

V

V

In the dual converter operation one

converter is operated as a controlled rectifier


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0

0

1

converter is operated as a controlled rectifier

with 90 & the second converter is

operated as a line commutated inverter

in the inversion mode with 90

dcV V

2dc

1 2 2

1 2

2 2 2cos cos cos

cos cos

m m mV V V


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2 1 1

2 1

1 2

2 1

or

cos cos cos

or

radians

Which gives

To Obtain an Expression


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To Obtain an Expression

for the

Instantaneous CirculatingCurrent

vO1 = Instantaneous o/p voltage of converter 1.

vO2= Instantaneous o/p voltage of converter 2.

The circulating current ir can be determined byi t ti th i t t lt diff


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g yintegrating the instantaneous voltage difference

(which is the voltage drop across the circulating

current reactor Lr), starting from t = (2- 1). As the two average output voltages during the

interval t = (+1)to (2- 1)are equal and

opposite their contribution to the instantaneouscirculating current iris zero.

1

1 2

2

1. ;

t

r r r O O

r

i v d t v v v

L


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1

2

1 2

1 2

2

1 1

As the o/p voltage is negative

1. ;

sin for 2 to

O

r O O

t

r O O

r

O m

v

v v v

i v v d t L

v V t t

t t


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1 12 2

1

sin . sin .

2cos cos

The instantaneous value of the circulating currentdepends on the delay angle.

t t

mr

r

mr

r

Vi t d t t d t

L

Vi t

L

1For trigger angle (delay angle) 0,

the magnitude of circulating current becomes min.

when , 0,2,4,.... & magnitude becomest n n


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g

max. when , 1,3,5,....

If the peak load current is , one ofp

t n n

I

the

converters that controls the power flow

may carry a peak current of

4 ,mp

r

VIL

where


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max

max

where

,

&

4

max. circulating current

mp L

L

m

r

r

VI I

R

V

i L

Different Modes Of Operation of

Dual converter


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Non-circulating current (circulating current

free) mode of operation.

Circulating current mode of operation.


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CirculatingCurrent Mode Of Operation


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In this mode, both the converters are

switched ON and operated at the same

time. The trigger angles 1and 2are adjusted

such that (1+ 2) = 1800; 2= (180

0-

1).

When 0


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p 2

In this case Vdcand Idc, both are positive.

When 900 900

Conv. 2

Rectifyin

g

2 < 900

Rectifyin

g

1 < 900

Conv. 1

Inverting1 > 90

0

Contd

There are two different modes of

operation.


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Circulating current free

(non circulating) mode of operation

Circulating current mode of operation

Non CirculatingCurrent Mode Of Operation

In this mode of operation only oneconverter is switched on at a time


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When the converter 1 is switched on,

For 1< 900the converter 1 operates in

the Rectification mode

Vdcis positive, Idcis positive and hence theaverage load power Pdcis positive.

Power flows from ac source to the load

When the converter 1 is on,For 1 > 90

0 the converter 1 operates in


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For 1> 90 the converter 1 operates in

the Inversion mode

Vdcis negative, Idcis positive and theaverage load power Pdcis negative.

Power flows from load circuit to ac source.

When the converter 2 is switched on,For 2< 90

0the converter 2 operates in


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2 pthe Rectification mode

Vdc

is negative, Idc

is negative and theaverage load power Pdcis positive.

The output load voltage & load currentreverse when converter 2 is on.

Power flows from ac source to the load

When the converter 2 is switched on,For 2> 90

0the converter 2 operates in


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2 pthe Inversion mode

Vdc

is positive, Idc

is negative and theaverage load power Pdcis negative.

Power flows from load to the ac source.

Energy is supplied from the load circuit to

the ac supply.

Circulating CurrentMode Of Operation Both the converters are switched on at the

same time. One converter operates in the rectification


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One converter operates in the rectificationmode while the other operates in the

inversion mode. Trigger angles 1& 2are adjusted such

that (1+ 2) = 1800

When 1< 900, converter 1 operates as a

controlled rectifier. 2 is made greaterth 900 d t 2 t


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than 900 and converter 2 operates as an

Inverter.

Vdcis positive & Idcis positive and Pdcispositive.

When 2< 900, converter 2 operates as a

controlled rectifier. 1 is made greaterthan 900 and con erter 1 operates as an


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than 900 and converter 1 operates as an

Inverter.

Vdcis negative & Idcis negative and Pdcispositive.

A C Voltage Controller


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Power Electronics Unit-6 327

A.C Voltage Controller

And

Cyclo-Converter

LECTURE PLANSl. No Module as per Session Planner Lecture No. PPT Slide No.

1 Introduction to AC voltage controllers L-1 3-12

2 Expression For The RMS Value OfOutput Voltage, For ON-OFF Control

Method

L-2 13-27


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3 Principle Of AC Phase Control And

Operation of single Phase half-Wave

Ac Voltage controller with R-Load

L-3 28-43

4 Single Phase Full Wave Ac VoltageController (Bidirectional Controller)

With R-Load

L-4 44-59

5 TRIAC and Its Modes of Operation L-5 60-73

6 Single phase full wave ac voltage

controller (Bi-directional Controller) usingTRIAC L-6 74-91

7 Cycloconverter Midpoint Type L-7 92-101

8 1-to 1-Bridge type Cyclo-converter

with R and R-L loadL-8 101-109

Lecture-1


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Ac Voltage Controllers

Ac Voltage controller circuits(RMS voltage controllers)

An ac voltage controller is a type of thyristor

power converter which is used to convert a

fi d l fi d f i l


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fixed voltage, fixed frequency ac input supply

to obtain a variable voltage ac output

Applications Of Ac Voltage

Controllers

Lighting / Illumination control in ac powercircuits.


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Induction heating.

Industrial heating & Domestic heating.

Transformer tap changing (on load

transformer tap changing).

Speed control of induction motors C

magnet controls.

Type Of Ac Voltage Controllers

Single phase half wave ac voltage controller(Uni-directional controller)


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(Uni-directional controller).

Single phase full wave ac voltage controller

(Bi-directional controller). Three phase half wave ac voltage controller

(Uni-directional controller).

Three phase full wave ac voltage controller(Bi-directional Controller)

A.C voltage control technique


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Principle of ON-OFF ControlTechnique


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Vs

Vo

i

wt

n m


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io

ig1

ig2

wt

wt

wt

Gate pulse of T1

Gate pulse of T2


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Lecture-2

Expression For The RMS Value Of


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Expression For The RMS Value Of

Output Voltage, For ON-OFF

Control Method


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Power Electronics U

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