power converter systems graduate course ee8407

1 Textbook: Bin Wu, ‘High-Power Converters and AC Drives’, Wiley - IEEE Press, 2006

EE8407 Topic 4

Power Converter Systems

Graduate Course EE8407

Ryerson Campus

Bin Wu PhD, PEng

ProfessorELCE DepartmentRyerson University

Contact Info Office: ENG328Tel: (416) 979-5000 ext: 6484Email: [email protected] http://www.ee.ryerson.ca/~bwu/


EE8407 Topic 4

AisL

Z ai

20

0

Z20

ai

ai~

lkL

lkL

lkL

dI

dvLOAD

dL

Topic 4

Multi-pulse SCR Rectifiers


EE8407 Topic 4

Lecture Topics

• Six-pulse SCR Rectifier (Building Block)

• 12-, 18- and 24-pulse SCR Rectifiers

• THD and PF



EE8407 Topic 4

Why Use Multi-pulse SCR Rectifiers?

• To reduce line THD

• To improve input power factor

• To avoid semiconductor devices in series.



EE8407 Topic 4

ai

bi

ci

1S 3S 5S

4S 6S 2S

dvGND

av

bv

cv

(a) (b)

dI

1g

dI

3g 5g

4g 6g 2g

a

b

c

P

N

dL

LOAD

• Converter Configuration

• Assumption: 1) Ideal SCRs - no power loss

2) dc current ripple free

• Use constant current source Id as a load

Ld

Six-pulse SCR Rectifier


EE8407 Topic 4

• Waveforms

• Average dc voltage

)(3/

13/

area

2/

6/tdv

V

ab

d

1A

cos35.1

cos23

LL

LL

V

V

av bv cv

6

6

2

6

5

6

7

6

9

6

11

0

2gi

3gi

4gi

5gi

6gi

1gi

v

2 3t

t

t

t

Pvav bv

Nv

cvav bv

v

0

dv

0

abv acv bcv bav cav cbv abv acv

dV

dI

1A

ai1

1ai

0

bi

0

ci

t

t

t

dI

I II

1

2 16 , SSON

21 , SSON



EE8407 Topic 4

• Fourier Analysis

....19sin19

117sin

17

113sin

13

1

11sin11

17sin

7

15sin

5

1sin

32

111

1111

ttt

ttttIi da

• RMS current

dd

ddaa

II

tdItdItdiI

816.03

2

)()(2

1)(

2

12/1

611

67

265

6

22/1

2

0

2

• THD

311.078.0

)78.0()816.0( 22

1

2

1

2

d

dd

a

aa

I

II

I

IITHD



EE8407 Topic 4

• Power Factor (PF)

• Displacement Power Factor

coscos 1

121

DPF

• Overall Power Factor

cos955.0

1

cos2

1

THD

PF



EE8407 Topic 4

• Waveforms at Various Firing Angles Pv

90

av bv cv

2Nv

2

dv

t 2

Pv

Nv

45

avbv cv

Pv

135av bv cv

Nv

0

2

abv acv bcv cav cbvbavdv

dV

2

abv acv bcv cav cbvbav


2

NPd vvv

dV

dv

dV

2


dv

Pv

180

Nv

2av bvcv

t

t

t

0

0

0

NPd vvv

v

v45(a) 90(b)

135(c) 180(d)



EE8407 Topic 4

• Effect of Line Inductance Ls

ai

bi

ci

1S 3S 5S

4S 6S 2S

GND

sL

sL

sL

av

bv

cv

sLv

dI dL

dvLOAD

P

N

a

c

b



EE8407 Topic 4

• Waveform During SCR Commutation

t

ai bici

1ai

dI

dI

62

1

2

av

cvav

Pv

6/5 6/9

t

Pv

0

0

61

2/)( ca vv

A

i



EE8407 Topic 4

• DC Voltage Reduction (caused by commutation)

• S1 and S5 on (during commutation)

dt

diLv

dt

diLvv c

sca

saP

from which

dt

di

dt

diLvvv casca

P 22

• Since ia + ic = Id, and id is constant

0dt

di

dt

di ca

from which

2ca

P

vvv



EE8407 Topic 4

• DC Voltage Reduction

6

6

)( tdvvA Pa

dtdiLvv aspa /• Since

dsa

I

s ILdiLAd 0

• Shaded Area Aγ

• Average dc voltage loss

ds I

LAV

3

3/

• Average dc output voltage

ds

LLd IL

VV 3

cos35.1



EE8407 Topic 4

• Commutation Interval

• Aγ can be rearranged as

)(coscos2

)(2

6

6

LLac Vtd

vA

)()6

sin(2 tdtVv LLac where

• Substitute

)(coscos2

LLds

VIL

from which

d

LL

s IV

L2coscos 1


ds IL into the above equation


EE8407 Topic 4

• Line Current Waveforms

ai bi

1ai

ai bi

2 23 20

2.00

1.00

0.00

-1.00

-2.00

2.00

1.00

0.00

-1.00

-2.00

(b) = 30°, Ia1 = 1 pu

(a) = 0°, Ia1 = 1 pu

1.41pu

Ls = 0.05pu

Harmonics n 5 7 11 13 17 19 23 25 THD

%

1/ aan II (%)

0 18.6 12.4 6.32 4.58 2.40 1.73 1.02 0.87 23.9

1/ aan II (%)

30 19.7 14.1 8.58 7.27 5.16 4.62 3.43 3.16 28.3



EE8407 Topic 4

• Line Current THD

0.2 0.4 0.6 0.80

THD (%)

IA1 (pu)20

23

26

29

32

A

B

C

A: Ls= 0 B: Ls= 0.05pu C: Ls= 0.10pu

= 0°

0.2 0.4 0.6 0.80

THD (%)

IA1 (pu)20

23

26

29

32

A: = 20°

B: = 40°

C: = 60°

Ls = 0.05pu

A

B

C

(a) (b)



EE8407 Topic 4

• Power Factor

0.2 0.4 0.6 0.80

PF

A: = 0°

B: = 20°

C: = 40°

Ls= 0.05pu

IA1 (pu)0.2

0.4

0.6

0.8

1.0

AB

C



EE8407 Topic 4

• Topology

AisL

ai

0

30

ai~

aa iiiA ~ lkL

lkL

dL

dvLOAD

dI

• Main Benefits

- Low line current THD

- No SCRs in series

12-pulse SCR Rectifier


EE8407 Topic 4

• Application Example

• Used for current source inverter (CSI) fed drive

0

30

dL

M

fC



EE8407 Topic 4

• Waveforms of Idealized Rectifier (Ls = Llk = 0)

2t

t

ai

ai

ai~

Ai

dI

dI

2dI

dI3

1dI

32

1

dI2

1dI

32

1

dI32

1

t

t

t

6/ 6/50

0

0

0

0

ai~



EE8407 Topic 4

• Fourier Analysis

....19sin19

117sin

17

1

13sin13

111sin

11

17sin

7

15sin

5

1sin

32

tt

tttttIi da

• Secondary currents

....)30(19sin19

1)30(17sin

17

1)30(13sin

13

1

)30(11sin11

1)30(7sin

7

1)30(5sin

5

1)30sin(

32~

ttt

ttttIi da



EE8407 Topic 4

• Fourier Analysis (continued)

• Secondary currents referred to the primary side

....}19sin191

17sin171

13sin131

11sin111

7sin71

5sin51

{sin3

tt

tttttIi da

....}19sin19

117sin

17

1

13sin13

111sin

11

17sin

7

15sin

5

1{sin

3~

tt

tttttIi da

....}25sin25

1

23sin23

113sin

13

111sin

11

1{sin

32~

t

ttttIiii daaA

• Primary line current



EE8407 Topic 4


• Secondary side

%1.31

...

1

2/12

13

2

11

2

7

2

5

1

2

1

2

a

aaaa

a

aai I

IIII

I

IITHD

a

• Primary side

%3.15

...

1

2/12

25

2

23

2

13

2

11

1

2

1

2

A

AAAA

A

AAi I

IIII

I

IITHD

A



EE8407 Topic 4

• Line Current Waveform (Ls 0 )

(b)

(a)

1.50

0.75

0.00

-0.75

-1.50

1.50

0.75

0.00

-0.75

-1.50

1Ai

ai2

1ai

2

0 2 3 4

Ai

Harmonics n

5 7 11 13 17 19 23 25 THD (%)

(%)/ 1aan II 18.8 12.7 6.78 5.05 2.77 2.01 1.01 0.75 24.6

(%)/ 1AAn II 0 0 6.78 5.05 0 0 1.01 0.75 8.61



EE8407 Topic 4

• THD and PF

0.2 0.4 0.6 0.80

THD (%)

IA1 (pu)4

7

10

13

16

A

B

C

A: Ls=0 B: Ls= 0.05pu C: Ls= 0.10pu

Llk= 0.05pu = 0°

(a)0.2 0.4 0.6 0.80

PF

A: = 0°

B: = 20°

C: = 40°

Ls= 0 Llk= 0.05pu

IA1 (pu)0.2

0.4

0.6

0.8

1.0A

B

C



EE8407 Topic 4

• Rectifier Configuration

AisL

Z ai

20

0

Z20

ai

ai~

aaa iiiiA ~

lkL

lkL

lkL

dI

dvLOAD

dL



EE8407 Topic 4

0.60

0.30

0.00

-0.30

-0.60

0.60

0.30

0.00

-0.30

-0.60

1.50

0.75

0.00

-0.75

-1.500 2 3 4

ai~

32

Ai

(a)

(b)

(c)

1~ai

ai ai

Harmonics n

5 7 11 13 17 19 23 25 THD

(%)

(%)/ 1aan II 18.8 12.7 6.78 5.05 2.77 2.01 1.01 0.75 24.6

(%)/ 1AAn II 0 0 0 0 2.77 2.01 0 0 3.54

• Waveforms



EE8407 Topic 4


0.2 0.4 0.6 0.80

THD (%)

IA1(pu)2

4

6

8

10

A

B

C

A: Ls = 0 B: Ls = 0.05pu C: Ls = 0.10pu

Llk = 0.05pu = 0°

0.2 0.4 0.6 0.80

THD (%)

IA1 (pu)2

4

6

8

10

A: = 20°

B: = 40°

C: = 60°

Ls = 0.05pu Llk= 0.05pu

A

B

C



EE8407 Topic 4

AisL

Z 15

0

Z15

30

aaaa iiiiiA ~

ai

ai~

ai

ai

lkL

lkL

lkL

lkL

dV

dI dL

• Rectifier Configuration



EE8407 Topic 4


0.2 0.4 0.6 0.80

THD (%)

IA1 (pu)0

2

4

6

8

A

B

C

A: Ls = 0 B: Ls = 0.05pu C: Ls = 0.10pu

Llk = 0.05pu = 0°

0.2 0.4 0.6 0.80

THD (%)

IA1 (pu)0

2

4

6

8

A

B

C

A: = 20°

B: = 40°

C: = 60°

Ls = 0.05pu Llk = 0.05pu



EE8407 Topic 4

power converter systems graduate course ee8407

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