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Controlled Semi-

ControlledDual

Io

Vdc

-V

-Io

Chapter 3 Chapter 10

RectifierMode

Inverter

Mode

Rectifier

Mode

InverterMode

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02

1)t(dtsinVm

0

1

)t(dtsinVV mdc

3

0

3

32

2/

mdc)t(dtcosVV

6

03

62

2/

mdc )t(dtcosVV

)t(dtsinVm2

1

)t(dtsinVV mdc

1

).(

/

/mdc )t(dtsinVV

1910

65

62

3

2

66

33

/

/mdc )t(d)tsin(VV

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rmsdc

max

dc

V.V

V

V

450

rmsdc

maxdc

V.V

VV

90

2

)LL(rmsdc

)LLmax(dc

V.V

VV

1712

33

cosV.V

cosV

V

rmsdc

max

dc

450

cosV.V

cosV

V

rmsdc

maxdc

90

2

cosV.V

cosV

V

)LL(rmsdc

)LLmax(dc

1712

33

)LL(rmsdc

)LLmax(dc

V.V

VV

351

3

cosV.V

cosV

V

)LL(rmsdc

)LLmax(dc

351

3

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To Derive an Expression for the

Average Output Voltage of a

3-Phase Half Wave Converter withRL Load for Continuous Load Current

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Vector Diagram of 3 Phase Supply

Voltages

VAN

VCN

VBN

1200

1200

1200 RN AN

YN BN

BN CN

v v

v v

v v

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0

0

0

sin ;

Max. Phase Voltage

2sin

3

sin 120

2sin

3sin 120

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

VAN

VCN

VBN

1200

1200

1200

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8/30120o conduction

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5

6

6

5

6

6

3

sin .2

3cos

2

3 5cos cos

2 6 6

m

dc

m

dc

m

dc

VV t d t

VV t

VV

120o conduction

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0 0

0

Note from the trigonometric relationship

cos cos .cos sin .sin

5 5cos cos sin sin

6 63

2

co

cos 150 cos sin 150 sin3

2 cos 30

s .cos sin sin6 6

.cos

m

dc

m

dc

A

VV

B A B A B

VV

0

sin 30 sin

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0 0

0 0 0 0

0 0

0 0

0

0

0

0

0 0

Note: cos 1

cos 180 30 cos sin 180 30 sin3

2 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

m

dc

m

dc

V

V

VV

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03 2cos 30 cos2

3 32 cos

2 2

3 3 33 cos cos

2 2

3

cos2

Where 3 Max. line to line supply voltage

m

dc

m

dc

m m

dc

Lm

dc

Lm m

VV

VV

V VV

VV

V V

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max

The maximum average or dc output voltage is

obtained at a delay angle 0 and is given by

3 3

2

Where is the peak phase voltage.And the normalized average output voltage is

m

dmdc

m

d

dcn n

VV V

V

VV V

cosc

dmV

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15 2

62 2

6

1

2

The rms value of output voltage is found byusing the equation

3sin .

2

and we obtain

1 33 cos 2

6 8

mO RMS

mO RMS

V V t d t

V V

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Three Phase Full Converter

3 Phase Fully Controlled Full Wave Bridge Converter.

Known as a 6-pulse converter. Used in industrial applications up to 120kW output

power.

Two quadrant operation is possible.

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0

0

0

We deifine three line neutral voltages(3 phase voltages) as follows

sin ; Max. Phase Voltage

2sin sin 120

3

2sin sin 1203

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

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2

6

mL

max

33 sin .6

3 3 3cos cos

Where V 3 Max. line-to-line supply vo

The maximum average dc output voltage isobtained for a delay angle

ltage

3 3

0,

3

dc m

m mL

dc

m

m m

dmdc

V V t d t

V V

V

V

V VV V

L

Th li d d l i

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1

222

6

The normalized average dc output voltage is

cos

The rms value of the output voltage is found from

6.

2

dc

dcn n

dm

OO rms

VV V

V

V v d t

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=-/m

=-/m+

=0

=/m+

=/m

cosm/

)m/sin(VV mdc

m/

m/

dcosVV mm

dc 2

1

cosm

sinVm

V rmsdc2

E l Si l

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m/

m/

dcosVV m

m

dc 2

1

cos

m

sinVm

V rmsdc2

2

2

1/

/mdc )(dcosVV

Example: Single

Phase Full Wave

controlled rectifier

(called2-Phase or

2-pulse rectifier)

m=2

cossinVV rmsdc2

22

cosV.V rmsdc 90

900

conduction2-ripples or

pulses

Each cycle Draw

?cosV.V rmsdc 090

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E ample 3 Phase

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m/

m/

dcosVV m

m

dc 2

1

cosm

sinVm

V rmsdc2

6

662

1/

/

mdc )(dcosV/V

Example: 3-Phase

Half Wave controlled

rectifier

(called3-Phase or

6-pulse rectifier)

m=6

cossinVV rmsdc6

62

cosV.V rmsdc 351

600

conduction

6-ripples or

pulses

Each cycle

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Three Phase Dual Converters

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3 Ph H lf C t ll d B id C t (S i C t ) ith

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3 Phase Half Controlled Bridge Converter (Semi Converter) with

Highly Inductive Load & Continuous Ripple free Load Current

3 Phase semiconverters are used in Industrial dc drive

applications up to 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.

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Practice Problem

For controlled RL rectifier, the source is120Vrms at 60Hz, R=20, L=0.04H, delay

angle is 45o and extinction angle is 217o.

Determine

i. An expression for i(t)

ii. Average current and voltage

iii. Power absorbed by load

iv. Power factor

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