ee8407 topic 3 power converter systems - ryerson …bwu/courses_teaching/8407/xslides_topic... · 1...

1

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

EE8407 Topic 3Power Converter Systems

Graduate Course EE8407

Ryerson Campus

Bin Wu PhD, PEng

ProfessorELCE DepartmentRyerson University

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


EE8407 Topic 3

di

Av

Bv

Cv

AisL1N

ai~

2N

3N

ai

°= 30δ

°= 0δ

aa iiiA ~′+′=

dCdV

lkL

∞

lkL

a~

b~

c~

Topic 3

Multi-pulse Diode Rectifiers

2


EE8407 Topic 3

Lecture Topics

• Six-pulse Diode Rectifier (Building Block)

• Series-type 12-, 18- and 24-pulse rectifiers

• Separate-type 12-, and 18-pulse rectifiers



EE8407 Topic 3

Why Use Multipulse Diode Rectifiers?

• To reduce line current THD;

• To improve input power factor; and

• To avoid semiconductor devices in series.


3


EE8407 Topic 3

• Resistive Load

)3/4sin(2

)3/2sin(2

)sin(2

πω

πω

ω

−=

−=

=

tVv

tVv

tVv

PHc

PHb

PHa

)6/sin(2 πω +=−= tVvvv LLbaab

Supply Voltages:

ai

bi

ci

1D 3D 5D

4D 6D 2D

dv

av

bv

cv

di

dR

a

b

c

Six-pulse Diode Rectifier


EE8407 Topic 3

• Waveforms

dvabv acv

1Abcv cav cbv

dv

bav

doV

I II

π

bv cv avv

0

0

0

ai

61, DD 21, DDON

π2tω

tω

tωπ2π

abv

6π

2π

av

ON

( ) )(6/sin23/

13/

area 2/6/ tdtVV LLdo ωπω

ππππ +== ∫1A

LLLL VV 35.123≈=

π


4


EE8407 Topic 3

• Capacitive Load

dCdV

(a)

dV

(b)

R∞

ai

bi

ci

1D 3D 5D

4D 6D 2D

di

a

b

c

sLav

bv

cv

sL

sL

Assumption:constant=⇒∞= dd VC



EE8407 Topic 3

• Waveforms

tω

tω

tω

tω

tω

abv acv bcv bav cav cbv abv acv

π π2ai

bi

di

61,DDπ π2

ci

ON21, DD

ON

v

0

0

0

0

0

dV


5


EE8407 Topic 3

• Discontinuous Current Operation

π3/π

1θ 2θ 3θ

di

v abv acv

tω

tω

dV

3/π

4θ

0

0

pI

5θ

⎟⎟⎠

⎞⎜⎜⎝

⎛= −

LL

d

VV2

sin 11θ

12 θπθ −=



EE8407 Topic 3


π3/π

1θ 2θ 3θ

di

v abv acv

tω

tω

dV

3/π

4θ

0

0

pI

5θ

Differential Equation: dabd

s VvdtidL −=2 31 θωθ <≤ t

( ) ( )( ) ( )

( ) ( )( )θθθθω

ωωω

θθ

θ

−+−=

−= ∫

11 coscos22

1

sin22

11

dLLs

dLLs

d

VVL

tdVtVL

iSolution:

dV

ai

bi

ci

1D 3D 5D

4D 6D 2D

di

sLav

bv

cv

sL

sL


6


EE8407 Topic 3


π3/π

1θ 2θ 3θ

di

v abv acv

tω

tω

dV

3/π

4θ

0

0

pI

5θ

Peak dc current:

Average dc current:

( ) ( )( )2121 coscos22

1 θθθθω

−+−= dLLs

p VVL

I

Voltage – theta relation:31

31 coscos2 θθ

θθ−

−=

LL

d

VV

( ) ( )θθπ

θ

θdiI dd ∫= 3

13/1



EE8407 Topic 3

• Continuous Current Operation

di

ici

D2 , D3D1 , D2

D1D2D3

I IIIII

On-stateDiode

π π2

pIbi

tω

tω

ai

pI

aibi ciγ

cθdV

ai

bi

ci

1D 3D 5D

4D 6D 2D

di

sLav

bv

cv

sL

sL

Note: - With the increase of the load current, the rectifier will

enter into continuous current operation. - During commutation interval, three diodes are on.


7


EE8407 Topic 3

• Definition of Total Harmonic Distortion (THD)

Phase voltage (pure sine):

Line current (distorted):

RMS line current:

Line current THD:

( )nnann

a tIi φω −= ∑∞

=

)(sin2,...3,2,1

( )2/1

,...3,2,1

22/1

2

0

2 )(21

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟

⎠⎞

⎜⎝⎛= ∑∫

∞

=nanaa ItdiI ω

ππ

1

21

2

a

aa

III

THD−

=

tVv aa 1sin2 ω=



EE8407 Topic 3

• Definition of Power Factor (PF)

Per-phase average (real) power:

Per-phase apparent power:

Total power factor (PF):

Distortion factor (DF) :

Displacement power factor (DPF) :

( )tdivP aa ωπ

π×= ∫

2

021

11 cosφaa IV=

aa IVS =

DPFDFII

IVIV

SPPF

a

a

aa

aa ×==== 1111 coscos φφ

aa IIDF /1=

1cosφ=DPF

PF = f (THD) :21 THD

DPFPF+

=


8


EE8407 Topic 3

• Per Unit System

Rated power, rated lint-to-line voltage:

Base voltage and frequency:

Base current and impedance:

Base inductance and capacitance:

ExampleRectifier ratings: 4160V, 60Hz, 2MVA.Base current = 277.6A, Base inductance = 22.9mH.Line inductance = 2.29mH = 0.1puLine current = 138.8A = 0.5pu

RR VS ,

.and3 B

BB

B

RB I

VZVSI ==

3R

BVV = 12 fB πω =and

BBB

B

BB Z

CZLωω

1and ==



EE8407 Topic 3

ai

1ai

π0 π2 π3 π4

ai

1ai

Harmonics n

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

1/ aan II (%)

pu2.01 =aI 63.4 38.7 8.99 8.64 4.22 3.61 2.48 2.02 75.7

1/ aan II (%)

pu11 =aI 30.4 8.79 6.31 3.40 2.30 1.89 1.04 1.03 32.7

• Typical Waveforms / Harmonic Content


9


EE8407 Topic 3

A

B

C

1.0 PF

0.6

0.8

0.7

0.9

A

B

C

(b) PF

100 THD(%)

20

60

40

80

(a) THD

A: Ls = 0.05B: Ls = 0.10C: Ls = 0.15

A: Ls = 0.05B: Ls = 0.10C: Ls = 0.15

0.2 0.4 0.6 0.8 1aI0 (pu) 0.2 0.4 0.6 0.8 1aI0 (pu)

• THD and PF



EE8407 Topic 3

di

Av

Bv

Cv

AisL1N

ai~

2N

3N

ai

°= 30δ

°= 0δ

aa iii A ~′+′=

dCdV

lkL

∞

lkL

a~

b~

c~

• Rectifier Topology

• Series type:Two six-pulse rectifiersare in series at the output.

• Phase shifting transformer: °=∠−∠= 30~~ ABba VVδ

• Secondary line-to-line voltage: 2/~~ ABbaab VVV ==

• Turns ratio: 22

1 =NN .

32

3

1 =NNand

12-pulse Series-type Diode Rectifier

10


EE8407 Topic 3

• Simplified Block Diagram

sL

ai

°= 0δ

°= 30δ

di

Av

Bv

Cv

AisL1N

ai~

2N

3N

ai

°= 30δ

°= 0δ

ai~

aa iiiA ~′+′=

aa iii A ~′+′=

dCdV

lkL

lkL

lkL

∞

lkL

A

B

C

a

b

c

Ai

(a) 12-pulse diode rectifier

(b) Simplified diagram

a~

b~

c~

di

dC

dV∞



EE8407 Topic 3

di)( pu111 =aa Iiai

2/aa ii =′

ai~′

aaA iii ~′+′=

π π2 π3 π4

)( pu111 =AA Ii

• Waveforms and FFT

• No 5th or 7th harmonics in the line current.• Primary line current THD: 8.38%

0

0.20.40.6

0.81/ aan II ′′

n = 1

n = 5 711 13 17 19

00.20.40.6

0.8

711 13 17 19

00.20.40.6

0.8

11 13

0

(a)

(b)

(c)

1~~ / ana II ′′

1/ AnA II

n = 1

n = 5

n = 1

250 500 750 1000 f (Hz)

THD = 24.1%

11 // aanaan IIII =′′

THD = 24.1%

1~~1~~ // anaana IIII =′′

THD = 8.38%


11


EE8407 Topic 3

5

10

15

20

25 THD(%)

(a) THD

A: Ls = 0.00B: Ls = 0.05C: Ls = 0.10(Llk = 0.05)

A

B

C

A: Ls = 0.00B: Ls = 0.05C: Ls = 0.10(Llk = 0.05)

1.00 PF

0.92

0.96

0.94

0.98

(b) PF

A

B

C

0.2 0.4 0.6 0.8 1AI0 (pu) 0.2 0.4 0.6 0.8 1AI0 (pu)

Comparison with six-pulse rectifier:• THD is reduced; and• PF is improved.

• THD and PF



EE8407 Topic 3

∞AisL

Z°−= 20δ

°= 0δ

Z°= 20δ

dC

ai

ai~

ai

di

aaa iiiiA ′+′+′= ~

lkL

lkL

lkL

dV

Phase-Shifting (Zigzag) Transformer

• Converter Topology


12


EE8407 Topic 3

• Simulated Waveforms

0

ai′ ai′3/~~ aa ii =′

aaaA iiii ′+′+′= ~

π π2 π3 π4

• No 5th, 7th, 11th, or 13th harmonics in the line current.• Lowest harmonic: 17th

• Line current THD: 3.06%



EE8407 Topic 3

• Measured Waveforms

ai

ai~

Ai

5ms/divpu/div,2

200Hz/divpu/div,5/2

ai


13


EE8407 Topic 3

• THD and PF

2

4

6

8

10 THD(%)

(a) THD

A: Ls = 0.00B: Ls = 0.05C: Ls = 0.10(Llk = 0.05)

A

B

C

A: Ls = 0.00B: Ls = 0.05C: Ls = 0.10(Llk = 0.05)

1.00 PF

0.92

0.96

0.94

0.98

(b) PF

AB

C

0.2 0.4 0.6 0.8 1AI0 (pu)0.2 0.4 0.6 0.8 1AI0 (pu)

Comparison with 12-pulse:Improved THD



EE8407 Topic 3

• Converter Topology

AisL

Z°−= 15δ

°= 0δ

Z°= 15δ

°= 30δ

aaaa iiiiiA)′+′+′+′= ~

ai

ai~

ai

ai )

lkL

lkL

lkL

lkL

∞dC

di

dV

Phase-Shifting (Zigzag) Transformer


14


EE8407 Topic 3

• Typical Waveforms

0.30

-0.30

-0.60

0.60

0.30

-0.30

-0.601.50

0.75

-0.75

-1.50 0

(a)

(b)

(c)

aaaaA iiiii )′+′+′+′= ~

ai′ ai )′

4/~~ aa ii =′

ai′

π π2 π3 π4

π π3

THD = 24.0%

THD = 24.0%

THD = 1.49%

0

0

0

(pu)0.60

• Lowest line current harmonic: 23th

• THD: 1.49%



EE8407 Topic 3

• THD and PF

0

2

4

6

8 THD(%)

(a) THD

AB

C

1.00 PF

0.92

0.96

0.94

0.98

(b) PF

A

B

C

A: Ls = 0.00B: Ls = 0.05C: Ls = 0.10(Llk = 0.05)

A: Ls = 0.00B: Ls = 0.05C: Ls = 0.10(Llk = 0.05)

0.2 0.4 0.6 0.8 1AI0 (pu) 0.2 0.4 0.6 0.8 1AI0 (pu)


15


EE8407 Topic 3


dC

sL

ai

°= 0δ

°= 30δ

dC

diLOAD

LOAD

ai~

aa iiiA ~′+′= lkL

lkLd

i~

dV

dV ~

Ai

• Separate TypeEach six-pulse rectifier feeds a separate dc load.

12-pulse Separate-type Diode Rectifier


EE8407 Topic 3

°= 0δ

°= 0δ

°= 0δ

M

°= 30δ

°= 30δ

°= 30δ

• Application Example

Multilevel Cascade H-bridge Inverter Fed Drive

Phase ShiftingTransformer


16


EE8407 Topic 3

• Typical Waveforms

• Comparison with series-type:- DC current ripple: higher- Line current THD: close

π π2 π3 π4

aidi di~

ai~′

0

π π2 π3

aaA iii ~′+′=

2/aa ii =′

1/ aan II ′′

1/ AnA II

11~~ // aanana IIII ′′=′′



EE8407 Topic 3

• Measured Waveforms

ai

ai~

Ai

5ms/divpu/div,2

200Hz/divpu/div,10/2


17


EE8407 Topic 3

• Line Current THD and Input PF

0.2 0.4 0.6 0.80 IA1 (pu)

THD(%)

0.92

0.94

0.96

0.98

1.00

0.2 0.4 0.6 0.805

10

15

20

25

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

( Llk= 0.05pu )

PFContinuous

Current Mode

A

B

C

A

B

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

( Llk= 0.05pu )

IA1 (pu)



EE8407 Topic 3


AisL

Z°−= 20δ

°= 0δ

Z°= 20δ

dCai

dC

LOAD

LOAD

ai~

dCLOAD

ai

di

aaa iiiiA ′+′+′= ~

lkL

lkL

lkL

di~

di

dV

dV ~

dV


18


EE8407 Topic 3

• Simulated Waveforms

3/~~ aa ii =′

ai′ai′

0 π π2 π3 π4

aaaA iiii ′+′+′= ~



EE8407 Topic 3



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EE8407 Topic 3




EE8407 Topic 3

ee8407 topic 3 power converter systems - ryerson …bwu/courses_teaching/8407/xslides_topic... · 1...

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