EE631 – Spring 2005 1

ECE631/EE631QLecture 9 – The Rectifier

S.D.Sudhoff

Purdue University

EE631 – Spring 2005 2

Restrictions on Firing And Commutation

• Let’s think about the a-phase current

EE631 – Spring 2005 3

Restrictions on Firing and Commutation Angle

• Thus, we must have– (1) >0– (2) +u<

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Modification for Mode II Operation

• We can use our work for Mode 2 as well as Mode 1 if we make some additional modifications…

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Modification for Mode II Operation

• Solution approach

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Modification for Mode II Operation

EE631 – Spring 2005 7

Modification for Mode II Operation• Thus, the effective firing

angle must obey

E

il dgceff

6

2)3/cos(

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Numerical Example

• Consider a system with a 560 V l-l rms voltage source, 20 mH commutating inductance, and operating at 60 Hz

• Let’s look at the output characteristics as dc link current varied from 0 to 100 A

EE631 – Spring 2005 9

Operating Mode

0 10 20 30 40 50 60 70 80 90 1000

20

40

60

80

100

120

140

160

180

DC Current, A

Firi

ng A

ngle

, D

egre

es

EE631 – Spring 2005 10

Output Voltage

0

20

40

60

80

100

020

4060

80100

120140

160180

-1500

-1000

-500

0

500

1000

1500

DC Current, A

Rectifer Output Characteristics

Firing Angle, Degrees

Out

put

Vol

tage

, V

EE631 – Spring 2005 11

Commutation Angle

0

10

20

30

40

50

60

70

80

90

100

020

4060

80100

120140

160180

-10

0

10

20

30

40

50

60

DC Current, A

Firing Angle, Degrees

Com

mut

atio

n A

ngle

, D

egre

es

EE631 – Spring 2005 12

Effective Firing Angle

010 20

3040

5060

7080

90 100

0

50

100

150

200

0

50

100

150

200

DC Current, A

Firing Angle, Degrees

Eff

ectiv

e F

iring

Ang

le,

Deg

rees

EE631 – Spring 2005 13

Calculation of AC Currents

• Let’s consider the current in the generation source reference frame, i.e. the reference frame wherein

)3/2cos(

)3/2cos(

)cos(

2

g

g

g

abcg Ev

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General Approach

• The average q- and d-axis current may be expressed

2

3

3

3( )g g

qg qg g gi i d

2

3

3

3( )g gg gdg dgi i d

EE631 – Spring 2005 15

Commutation and Conduction Components

• Breaking the integrals up we have

• where

, ,gg g

qg qg com qg condi i i

, ,g g gdg dg com dg condi i i

3

3

, ,3

( )u

g gqg com qg com g gi i d

2

3

3

, ,3

( )u

g gg gqg cond qg condi i d

3

3, ,

3( )

ug gg gdg com dg comi i d

2

3

3

, ,3

( )u

g gg gdg cond dg condi i d

EE631 – Spring 2005 16

Calculation of Commutation Component

• We start with

d

agd

ag

abcg

i

ii

i

i

1 3( ) 2 cos( ) cos

2 3ag g d gc e

i i El

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Calculation of Commutation Component

EE631 – Spring 2005 18

Calculation of Commutation Component

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Calculation of Commutation Component

• Finally, we arrive at

(11.5-53)

)22cos()2cos(23

4

1

)cos()cos(cos23

)6/5sin()6/5sin(32

,

ul

E

ul

E

uii

gc

gc

dgcomqg

(11.5-54)

ul

Eu

l

E

ul

E

uii

gcgc

gc

dgcomdg

2

23)22sin()2sin(

23

4

1

)sin()sin(cos23

)6/5cos()6/5cos(32

,

EE631 – Spring 2005 20

Calculation of Conduction Component• During the conduction interval

d

dabcg

i

ii

0

EE631 – Spring 2005 21

Calculation of Conduction Component

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Calculation of Conduction Component

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Calculation of Conduction Component

• Thus we arrive at

,2 3 7 5

sin sin6 6

gdqg condi i u

,2 3 7 5

cos cos6 6

gddg condi i u

(11.5-55)

(11.5-56)