2010. Konverter Thyristor. Pekik Argo Dahono. Konverter Satu-Fasa Setengah-Gelombang. v o. sin. π 2π. Pekik A. Dahono : Konverter thyristor 2

3/12/2010

Konverter Thyristor Pekik Argo Dahono

Konverter Satu-Fasa Setengah-Gelombang i s

T

io

+

vs

vo

R

(a) Skema

vs

is 0

π

2π

ωt

π

2π

ωt

vo

Tegangan keluaran rata - rata : 1 π Vo = 2Vs sin (ωt )d (ωt ) 2π ∫α 2Vs (1 + cosα ) = 2π

io 0

α

(b) Bentuk gelombang.

Pekik A. Dahono : Konverter thyristor

2

1

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Gate Signal Generation AC line

Saw - tooth Generator

vsw

vsyn

Comparator and logic

Gate signal

vc

vsyn

0

vsw

vc

Gate signal


3

Linearizing the Phase-Control Characteristic

sin ωt

∫

+ −

+

FF +

1


4

2

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Konverter Satu-Fasa Setengah-Gelombang vs

io

is

is +

vL

T vd

vs

vo

R

0

2π

π

ωt

vd

•Arus masukan dan keluaran mempunyai bentuk yang sama dan diskontinyu •Harmonisa arus masukan mengandung komponen dc dan semua orde harmonisa •Nilai rata-rata tegangan keluaran tidak hanya ditentukan oleh sudut penyalaan tetapi juga parameter beban.

io

0

ωt

2π

π

α


5

Konverter Satu-Fasa Setengah-Gelombang io

is +

vs

vL

T FD

vd

vs

R

vo

is 0

Nilai rms arus masukan lebih kecil dibanding arus keluaran Arus masukan mengandung semua orde harmonisa termasuk komponen dc Nilai rata-rata tegangan keluaran hanya ditentukan oleh sudut penyalaan

2π

π

ωt

vd

io

0

α

π


2π

ωt

6

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Konverter Satu-Fasa Setengah Gelombang io

is +

vL

T

vs

vd

FD

R

vs

vo

is 0

2π

π

ωt

vd

io

0

2π

π

α

ωt


7

Konverter Satu-Fasa Setengah-Gelombang is +

vs

vs Ls

is

T FD

vd

Io

0

2π

π

ωt

vd

Induktansi sumber menyebabkan nilai rata-rata tegangan keluaran berubah sebagai fungsi arus beban

io

0

α

µ

2π

π

Vd =


2V s 2π

ωt

(1 + cos α ) − fLs I o

8

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Konverter Satu-Fasa Setengah-Gelombang io

is +

vs

vL

T vd

vs

Vo

Vo is

0

π

2π

ωt

vd

Nilai rata-rata tegangan keluaran dipengaruhi oleh emf beban

Vo

io 0 α

π

2π

ωt


9

Konverter Satu-Fasa Jembatan


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Konverter Satu-Fasa Jembatan Nilai rata - rata tegangan keluaran : Vo = =

1 π

π ∫α

2Vs

π

2Vs sin (ωt )d (ωt )

(1 + cos α )

Arus sumber hanya mengandung harmonisa orde ganjil Tegangan dan arus beban mempunyai jumlah pulsa dua dan hanya mengandung harmonisa orde genap


11

Konverter Thyristor Satu-Fasa io

T1

Ld

T3

Tegangan keluaran rata-rata:

is +

vo

vs

Vo =

R

2 2

π

V s cos α

Arus thyristor rata-rata: T2

T4

IT = I o / 2 Arus rms sumber :

vs

Is = Io

is

ωt

PF sumber :

PF = vo

α io 0

π

T1& T 4

2π

ωt

Vo I o 2 2 = cos α Vs I s π

Tegangan keluaran bisa diatur dari minus maksimum sampai plus maksimum tetapi arus selalu positip.

T 2&T3


12

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Konverter Thyristor Satu-Fasa vs is

T1

T3

ωt

is +

vs

vo

Io

α

vo

Io

T2

T4

0

ωt

π 2π


13

Output Voltage


14

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3/12/2010

Konverter Thyristor Satu-Fasa is

vs

vs

is

ωt

ωt

vo

vo Io

Io

0

π

0

ωt

2π

π

2π

ωt

vT 1

vT 1

ωt

ωt

vs

α=

π 4

α =

is

vs

π 2

is

ωt

ωt

Io 0

π

Io

ωt

2π

vo

0

π

2π ωt

vo

vT 1 ωt 3π α= 4

vT1


15

α =π

Analisis Daya dan Harmonisa Daya Aktif : P = Vs I s1 cos α Q = Vs I s1 sin α

P

Analisis Harmonisa Arus Masukan : Q

∞

Ik = = I1 =

∑

I k sin (kωt ) k = 2n −1 4 π /2 I o sin (kωt )d (ωt ) 0

is = 2

α

0

2π ∫

π

π

2

4Io [1 − cos(kπ / 2)] 2kπ 2 2

π

Io

I k = I1 / k Pekik A. Dahono : Konverter thyristor

16

8

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Pengaruh Induktansi Sumber T1

T3

is

vs

is

+

Ls

vs

vo

Io ωt

T2

T4

vo

α µ Io

π

0

2π

Vo =

2 2

π

ωt

Vs cos α − 4 fL s I o


17

Sudut Pemadaman is

vs

vs

is

ωt

ωt

vo

α µ

α

µ

Io

Io

vo

0

π

2π

ωt

vT 1

π

0

2π

ωt

vT 1

γ Vo =

2 2

π

Vo =

Vs cos α − 4 fLs I o


2 2

π

V s cos α − 4 fLs I o

18

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Konduksi Tak Kontinyu vs io

is T1

Ld

T3

ωt

is +

vo

vs

Vo

vo

α Vo

T2

io

T4 0

π

2π


ωt

19

Output voltage characteristic


20

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Penyearah Satu-Fasa Setengah-Terkendali io

Ld

D3

T1

is +

vs

vo

T2

R

D4


21

Penyearah Satu-Fasa Setengah-Terkendali

Tegangan keluaran rata-rata: Vo =

2

π

Vs (1 + cosα )

Arus thyristor rata-rata: IT = I o

π −α 2π

Arus rms sumber : 1/ 2

π −α  I s = Io   π 

PF sumber :

PF =

Vo I o 2 1 + cosα = Vs I s π  π − α 1/ 2    π 


22

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Penyearah Satu-Fasa Setengah-Terkendali


23

Konverter Thyristor Tiga-Fasa Setengah-Gelombang vun

0

u

iw

T3

iu

T1

π

vwn

2π

ωt

io

T2

R

AC source

vvn

Load

n

vo

iv T w

v

S


24

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3/12/2010

Persamaan Tegangan Nilai rata - rata tegangan : Vo =

3 2π

5π +α 6

∫π6 +α

2Vln s sin (ωt )d (ωt )

3 2 Vlls cos α 2π 6 Vo = Vlls 1 + cos π6 + α 2π

0 ≤α ≤

Vo =

[

(

π

)]

π 6 ≤α ≤

6

5π 6


25

Konverter Thyristor Tiga-Fasa Setengah-Gelombang

u

iw

T3

iu

T1

io

α

vun

vvn

vwn

T2

R

AC source

Load

n

vo

iv

0

T

w

π

ωt

2π

v

S

iu

Jumlah pulsa tegangan keluaran sama dengan tiga Arus sekunder trafo mengandung komponen dc.

ωt iv

iw

Nilai rata - rata tegangan Vo =

3 2 Vlls cosα 2π

iR


26

13

3/12/2010

Konverter Thyristor Tiga-Fasa Setengah-Gelombang

α

vun

α

vwn

vvn

vun

0

π

ωt

2π

vwn

vvn

0

π

T1

ωt

2π

T3

T2

T3

T1

T2

vT 1

ωt

vT 1

ωt


27

Konverter Thyristor Tiga-Fasa io T3

T1

iu

n

α

0

iv

R

v

iw

w T4

vwn

T5

u

T2

vvn

vun

π

ωt

vo vd

ωt

T6

Jumlah pulsa tegangan keluaran sama dengan enam. Harmonisa arus masukan adalah 5,7, 11, 13,…

2π

iu


28

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3/12/2010

Konverter Thyristor Tiga-Fasa Beban Resistif

α


29

Konverter Thyristor Tiga-Fasa Beban Resistif Tegangan rata - rata : Vo = Vo = Vo = =

π

3

+α

(

2 π ∫π6 +α

3 2

π 3 2

π

)

2Vll sin ωt − π6 d (ωt )

Vll cos α π

Vll ∫π

3

+α

0 ≤α ≤

π 3

sin (ωt )d (ωt )

 π  Vll 1 + cos + α  π 3  

3 2


π 3

≤α ≤

2π 3

30

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Konverter Tiga-Fasa Jembatan Beban Induktif vun

vun

iu iu

ωt

2π

0

2π

0

ωt

π

π

vun

vun

iu

iu ωt

2π

0

2π

0

π

ωt

π


31

Analisis Nilai rata - rata tegangan output : Vo =

3 2

π

Vllscosα

Faktor daya : PF =

Arus masukan :

3

π

cos α

∞

iu = 2

∑ Ik sin(kωt ) k = 2n -1

2 2 π /2 I1 = ∫ Io sin(ωt )d (ωt ) π /6

π

I1 =

6

π

Io

I k = I1 / k I k = 0 untuk k kelipatan 3. Iu = I o 2 / 3 Pekik A. Dahono : Konverter thyristor

32

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3/12/2010

Input current


33

Input current characteristic


34

17

3/12/2010

Pengaruh Induktansi Sumber

T1

iu

T5

T3 u

iv v

n

iw

Ls

w T2

T4

T6


35

Pengaruh Induktansi Sumber vun

iu

ωt

2π

0

π

vuv

vuw

Vo =

3 2

π

Vll cos α − 6 fL s I o


36

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3/12/2010

Half-Controlled Thyristor Converters

io

io T1 T1

iu

T3

T2

iu

L

T5

T3

D1

L

u

u iv

iv

R

v

R

v

vo

vo

iw w

iw

w D1

D2

T2

T4

T6

D2

T3

Penyearah A

Penyearah B


37

Half-Controlled Rectifier (A) α

vun

vvn

vwn

α

vvn

vwn

vun

0

π

2π

0

ωt

vd

2π

ωt

vd

ωt

iu

π

ωt

ωt

iu

ωt

iw iw

iR

iR


38

19

3/12/2010

Konverter Tiga-Fasa Jembatan Half-Controlled (B)

ωt

ωt

ωt

ωt

iu

iu

iw

iw

iR

iR


39

Application Considerations • Single-phase rectifiers generate input harmonics at the order of 2p±1, where p is the pulse number. • The displacement power factor is reduced when the output voltage is reduced. • Commutation generates voltage notches across the source. • Input harmonics can be reduced by increasing the pulse number. Pekik A. Dahono : Konverter thyristor

40

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Current Controller • A thyristor converter is usually operated as a current source • A thyristor converter cannot be controlled faster than the thyristor can respond • After a thyristor is turned on, the thyristor can only be turned off by the input line voltage. • By operating as a current source, the thyristor converter is inherently overcurrent protected. • A current source can paralleled easily with other current sources Pekik A. Dahono : Konverter thyristor

41

Current controller for PhaseControlled Rectifiers es

Ls L

vd

io

Load

Gate driver

Reference current

+

PID

cos−1

−

Actual current Pekik A. Dahono : Konverter thyristor

42

21

3/12/2010

Current-Controlled PhaseControlled Rectifiers Vo (s)

I L* ( s )

+ PID −

1 1 + sTd

3 2

π

+

Vll

−

1 sL

I L (s )


43

Current Control of Phase-Controlled Rectifiers


44

22

3/12/2010

Dual Thyristor Converter ia

ia*

+

−

Current controller

cos−1


45

Application Considerations • At present, thyristor converters are used only for large power applications. • The AC side always need reactive power under both rectifier and inverter operations. • The AC side current has high harmonic content. The harmonic order is pk±1 where p is pulse number and k is integer. The harmonic current can be reduced by increasing the pulse number. • Thyristor converter also generates voltage nothches due to the commutation. • It is recommended to use a special feeder (or it is better if using a dedicated transformer) to supply a thyristor converter. Pekik A. Dahono : Konverter thyristor

46

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High-Current Rectifiers


47



48

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49

High-Current Rectifiers (PWM)


50

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3/12/2010

DC Arc Furnace Applications


51

HVDC Applications • • • • •

Environmental advantages Economical advantages Asynchronous interconnections Power flow control Added benefits to the existing transmission system


52

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HVDC History • Hewitt’s mercury-vapour rectifier, 1901 • Experiments with thyratrons in US and mercury-arc valves in Europe in 1940s. • First commercial HVDC operation, Gotland, Sweden in 1954. • First solid-state semiconductor switches, 1970. • First microcontroller applications for HVDC in 1979. • Highest DC voltage operation (600 kVdc) for Itaipu, Brazil, in 1984. • First dc active power filter, 1994. • First capacitor commutated converter for Argentina-Brazil interconnection, 1998. • First Voltage Source Converter for HVDC in Gotland, 1999. Pekik A. Dahono : Konverter thyristor

53

HVDC Topology


54

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HVDC Systems


55

HVDC Operation


56

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HVDC Station


57

AC and DC Comparison


58

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HVDC Technologies


59

HVDC Applications


60

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HVDC System

+/- 500 kV, 2800 MW, Kii Chanel, Japan Pekik A. Dahono : Konverter thyristor

61

Thyristors for HVDC


62

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The End

32

2010. Konverter Thyristor. Pekik Argo Dahono. Konverter Satu-Fasa Setengah-Gelombang. v o. sin. π 2π. Pekik A. Dahono : Konverter thyristor 2

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