Phase-Controlled AC-DC Converterspemclab.cn.nctu.edu.tw/W3news/開授課程-old/電力... · 2005-04-19 · Phase-Controlled AC-DC Converters NCTU 2005 Power Electronics Course Notes

Phase-Controlled AC-DC Converters

NCTU 2005 Power Electronics Course Notes 1

page 1

2005年4月19日

鄒應嶼教授

Filename: \PEMC-03：投影片\A01 投影片：電力電子 (研究所)\PE-07.整流器.相位控制.ppt


國立交通大學電機與控制工程研究所

POWERLABNCTU

電力電子晶片設計與DSP控制實驗室Power Electronics IC Design & DSP Control Lab.

台灣新竹交通大學 • 電機與控制工程研究所

國立交通大學電力電子晶片設計與DSP控制實驗室Power Electronics IC Design & DSP Control Lab., NCTU, Taiwan

http://powerlab.cn.nctu.edu.tw/

page 2


Introduction

Thyristor Circuits and Their Control

Single-Phase Converters

Three-Phase Converters

Other Three-Phase Converters



page 3

Line-Frequency Controlled Converter

+

−

Vd

Id

1- or 3-phase (50/60 Hz) Id

Vd

0

Rectification

Inversion

Limited by circuit configuration and the input voltage

Limited by the components′current rating

(a) (b)

page 4

Single-Phase Thyristor Phase-Controlled Converter

Circuit Type Typical hp Ripple frequencyQuadrant operation

Half-wave Below 1/2 hp fs

Semi-converter

Up to 20 hp (100

hp in traction

systems)

2fs

Full-converter

Up to 20 hp (100

hp in traction

systems)

2fs

2fsUp to 20 hpDual-

converter

Ia

EaFrequencyfs

~

~

~

~

+

−

Ia

Ea

+

−

Ea

+

−

Ea

+

−

Ia

~

Ia

Ia

Ia

Ia

Ea

Ea

Ea

Ea



page 5

Three-Phase Thyristor Phase-Controlled Converter

Half-wave 3fs

Semi-converter 3fs

Full-converter 6fs

6fsDual-

converter

Circuit Type Typical hp Ripple frequencyQuadrant operation

10-50

15-150

100-150

200-2000

Ea

+

−

Ia

Ea

+

−

Ea+

−

Ea+

−

Supply frequencyfs

Ia

Ia

Ia

ABC

ABC

ABC

ABC

Ia

Ea

Ia

Ia

Ia

Ea

Ea

Ea

page 6

Basic Thyristor Converters

~

+

−

vd

~

id

RiG+

−vs

R

ivL+ −L

vR=Ri

+

−

+

−

vdiG

+

−vs

0 0

0

00

0

ω t

ω t

ω t

ω t

α π 2π

vd

ii , vd

vsiG

Area A1

Area A2

2π

vR , vd

vd vR

vsvdvd

π vd

vL

θ1

iG

( ) RsL vvdtdiLtv −==

( ) ( ) ζζω

ωω

αdv

Lti

t

L∫=1



page 7

Basic Thyristor Converters (continued)

~

i

+

−vs

v thy+

+

−

vd

−vL

+ −L

+

−

EdiG

0 ω t

vd

iθ1

Area A2

0

vd

Area A1

Ed i2π

vs

vdθ2 θ3 θ4

0v thy

ω t

0 ω tiG

( ) dsL EvdtdiLtv −==

( ) ( )[ ]∫ −=t

ds dEvLti

ω

θζζ

ωω

2

1

Ed

page 8

Thyristor Gate Triggering

t

t

t

0

0

0

vcontrol

stVωα

stvωα

v synchronization

Gate-triggersignal

Gate-trigger signal

vcontrol

Comparatorand Logic

v synchronization

Saw-tooth GeneratorAC line

voltage

t

vst

t

vst

vcontrol

+−+

−

stVv

ˆ180control°=°α



page 9

Practical Thyristor Converters

~+

−vs

Ls

id

+

−

vd load

~

id

+

−

vd

Ls

ia~

~

a+−

n load

Ldvan

(a) single-phase thyristor converter (b) three-phase thyristor converter

Line inductanceLoad inductance

Line InductanceLoad InductanceLoad Bias VoltageContinuous/Discontinuous Thyristor Current

page 10

Single-Phase Converter

Single-phase thyristor converter with Ls = 0 and a constant dc current.

Id

+

−

vd

T1

T2

~ Id

T3

T4

is

+

−vs

~

is

+

−vs

vd

+

−

T1

T3

T4

T2(a) (b)



page 11

Waveforms of the Single-Phase Converters

is

is

vs

0

1.2 3.4

ωt=0ωt

(a) α = 0

vd

vd

3.41.23.4

ω t0

ωt=0

0

α α

α

Aα

ω tπ

vsvs

is

(b) α = finite

ααπ

ωωπ

απ

ααcos9.0cos22)(sin21 sssd VVtdtVV === ∫

+

page 12

Normalized Output DC Voltage

Vdo is the average voltage with α=0 and Ls=0

Normalized Vd as a function of α.

0 α

do

d

VV

Rectifier mode

Inverter mode

1.0

-1.0

0° 90° 180°

sssdo VVtdtVV 9.022)(sin21

0=== ∫ πωωπ

π

)cos1(9.0 αα −=−=∆ sddodo VVVV



page 13

Average Output Power

∫ ∫==T T

dd dtivTdttp

TP

0 0

1)(1

αcos9.0)1(0 ds

T

dddd IVVIdtvTIP === ∫

+

−

vd

T1

T2

~ Id

T3

T4

is

+

−vs

page 14

Harmonics and THD of Line Current is

φ1= α

vs

ωt0

isis1

12 sI

1.0

h

1s

sh

II

1 3 7 9 11 13 15 17 19 21 235

31

51

71

91

L+−+−+−= )](5sin[2)](3sin[2)(2)( 531 αωαωαωω tItItIti ssss

dds IIi 9.022

1 == π

hII shsh =

ds II =

%43.48

100THD%1

21

2

=

−×=

s

ss

III



page 15

Power, Power Factor, and Reactive Volt-Amperes

S

α135° 180°90°45°0

S1Q1P

αφ coscosDPF 1 ==

αcos9.0DPFPF 1 ==s

s

II

1cosφssIVP =

αcos9.0 ss IVP =

αφ sin9.0sin 111 dsss IVIVQ ==

2/121

2111 )( QPIVS s +==

φ1= α

vs

0

isis1

12 sI

page 16

Effect of Ls

Single-phase thyristor converter with a finite Ls and a constant dc current.

+

−

vd~ Idis

+

−vs

vLs+ −

What is the effect of the line inductance?

Ls



page 17

Waveforms of the Single-Phase Converters with Line Inductance

vs

ωt0

isis1

(a)

(b)

u21

1 +=αθ

vd

ωt0

Au

vsuα

dtdiLvv sssLs ==During the commutation interval:

∫∫ −+

== dd

I

I dssss

uILdiLtdtV ωωωω

α

α2)()(sin2

∫+

=u

su tdtVAα

αωω )(sin2

page 18

Voltage Drop by the Line Inductance

dssu ILuVA ωαα 2)]cos([cos2 =+−=

s

ds

VILu

22cos)cos( ωαα −=+

πω

πdsu

duILAV 2==∆

dssd ILVV ωπα 2cos9.0 −=



page 19

Example 6.1: Commutation Interval

φ1= α

vs

ωt0

isis1

12 sI

In the converter circuit as the following, Ls is 5% with the rated voltage of 230 V at 60 Hz and the rated volt-amperes of 5 kVA. Calculate the commutation angle u and Vd/Vdowith the rate input voltage, power of 3 kW, and α = 30°

Example 6.1

page 20

Solution of Example 6.1

A74.21230

5000rated ==I

Ω== 58.10rated

ratedbase I

VZ mH4.15.0 base ==ωZLs

kW32cos9.0 2 =−== dsdsddd ILIVIVP ωπα

06.892853.5332 =+− dd II

A3.17=dI

°= 9.5u

The rated current is

The base impedance is

The average power is

s

ds

VILu


dssd ILVV ωπα 2cos9.0 −= V5.173=dV



page 21

Input Line Current is

)21cos(DPF u+≅ α

ddss IVIV =DPF1 )2/cos()/2(cos9.0 2

1 uVILIVI

s

dsdss +

−≅α

ωπα

vs

ωt0

isis1

u21

1 +=αθ

dssd ILVV ωπα 2cos9.0 −=

page 22

Practical Thyristor Converters

~ is+

−vs

+

−

vd

Ed

Ls

Ld

+−

rd = 0

Line InductanceLoad InductanceLoad ResistanceLoad Back emf



page 23

Waveforms

vd

ωt0

vs

uα

id

vdEd

min,)/2(cos9.0 dssd ILVV ωπα −≅

dd

dddd EdtdiLirv ++=

∫ ∫∫ ++=T TI

I ddd

dd

T

dd

dEdi

TLdti

Trdtv

T 0)(

)0(0

1dddd EIrV +=

where Id,min is the minimum value of id that occurs at ωt=α.

dddd EIrV +=

page 24

Example 6.2: PSPICE Simulation

The signal-phase thyristor converter as the following is supplied by a 240 V, 60 Hz source.

Assume Ls = 1.4 mH and the delay angle α = 45°. The load can be represented by Ld = 9 mH and Ed = 145 V. Using Pspice, obtain the vd and is waveforms and calculate Is1, Is, DPF, PF, and %THD.

Example 6.2

~is+

−vs

+

−

vd

Ed

Ls

Ld

+−

rd = 0



page 25


φ1= 54.84°

vs

is

α=45°

ωt0

is1

Is1 = 59.68 AIs = 60.1 ADPF = 0.576PF = 0.572

THD = 12.3 %

Solution

page 26

PSPICE Input File for Example

* Signal-Phase, Thyristor-Bridge Rectifier.PARAM PERIOD = {1/60}, ALFA = 45.0, PULSE_WIDTH=0.5ms.PARAM HALF_PERIOD ={1/120}*LS1 1 2 0.1mH IC = 10ALS2 2 3 1.3mH IC = 10ALD 4 5 9mHVD 5 6 145V*XTHY1 3 4 SCR PARAMS: TDLY= 0 ICGATE=2VXTHY3 0 4 SCR PARAMS: TDLY= {HALF_PERIOD} ICGATE=0VXTHY2 6 0 SCR PARAMS: TDLY= 0 ICGATE=2VXTHY4 6 3 SCR PARAMS: TDLY= {HALF_PERIOD} ICGATE=0V*VS 1 0 SIN(0 340V 60 0 0 {ALFA})*.TRAN 50us 100ms 0 50us UIC.PROBE.FOUR 60.0 v(1) i(ls1) i(ld)

.SUBCK SCR 101 103 PARAMS: TDLY = 1ms ICGATE = 0V*Power Electronics: Simulation, Analysis Education……by N. Mohan.SW 101 102 53 0 SWITCHVSENSE 102 103 0VRSNUB 101 104 200CSNUB 104 103 1uF*



page 27

PSPICE Input File for Example (continued)

VGATE 51 0 PULSE(0 1V {TDLY} 0 0 {PULSE_WIDTH} {PERIOD})RGATE 51 0 1MEGEGATE 52 0 TABLE {I(VESNSE) + V(51)} = (0.0, 0.0) (0.1, 1.0) (1.0, 1.0)RSER 52 53 1CSER 53 0 1uF IC={ICGATE}*.MODEL SWITCH VSWITCH (RON=0.01).ENDS

.END

PSPICE input circuit for example 6.2

~ is+

−vs

+

−

vd

XTHY1Ls1

VD

3

24

Ld

1

0

23

4 5

6

+

−vm

Ls1

id

page 28

Discontinuous-Current Conduction

Waveforms in a discontinuous-current-conduction mode.

α

ωt

0id

vdvs

Ed

Period of discontinuity

At light loads with low values of Id , the id waveform becomes discontinuous!



page 29

Vd versus Id in the Single-Phase Thyristor Converter

0

Vd

Id0 10 20 30 40 50 60 70 80

50

100

150

200

250

300

350

α = 45°α = 30°α = 0°

α = 60°

page 30

Inverter Mode of Operation

~ is+

−vs

+

−

vd Id

Ls 1 3

4 2

ω t

0 ωtωt

0

0

vsis

vd3, 4

1, 23, 4

α 3, 4α 1, 2(iG) 1, 2 (iG) 3, 4

(a) Inverter, assuming a constant dc current (b) Waveforms

°



page 31

Waveform Analysis

~is+

−vs

+

−

vd Ed

Ls 1 3

4 2

+

−

id

Ld(vthy)3−

+ Id

α 3

Vd

α 2

α 1

Id=Id10

Ed=Ed1

(a) Thyristor inverter with a DC voltage source (b) Vd versus Id

dsdodd ILVVE ωπα 2cos −==

Average voltage across the output:

page 32

Voltage Across a Thyristor in the Inversion Mode

0ω t ω t0(iG) 1.2(iG) 3.4

0 ωtis

vs

α 3.4

(vthy)3

u γ

)(180 u+−°= αγ

The extinction time interval tγ = γ/ω should be greater than the thyristor turn-off time tq . Otherwise, the thyristor will prematurely begin to conduct, resulting in the failure of current to commutate from one thyristor pair to the other.



page 33

Inverter Start-up

0 ω tω t 0(iG) 1.2 (iG) 3.4

0 ω tid

vs vsid

vdvd

Ed

The soft start of a phase-controlled thyristor converter operating in inverter mode can be achieved by setting a large delay angle close to 180°.

α

page 34

Three-Phase Converters

Three-phase thyristor converter with Ls = 0 and a constant dc current.

~+

−

vd

ia

~

~

a

b

c

T1 T3 T5

T4 T6 T2

nib

ic

P

Id

N

~

~

~

n

ia

ib

ic

a

b

c

Id

+

−

vd

P

N

T1T3T5

T4T6T2

(a) (b)



page 35

Waveforms

0 ω t

iavcn

vPnvan vbn Vdo

vcn

ω t=0

ωt=0

iavNnvPn van vbn vcn

0

0

ω t

ω t

6 2

1 3

4

5

6

α

Aα

vNnvcn

ia

ia iavcbvcavbavbavacvabvcb vd

van vbnvcn

vdα

ω t=00ω t

vcn

α

(a)

(b)

(c)

(d)

Aα

NnPnd vvv −=

page 36

Idealized Circuit with Ls=0 and id (t)=Id

LLLLdo VVV 35.123 ==

π

3/πα

αAVV dod −=

tVv LLac ωsin2=

)cos1(2)(sin20

αωωα

α −== ∫ LLLL VtdtVA

αααπα

coscos35.1cos23 doLLLLd VVVV ===

αcos35.1 dLLdd IVIVP ==

Vdo is the average voltage with α=0 and Ls=0

Average output voltage:

Note: Vdo is independent of the current magnitude Id so long as id flows continuously and Ls=0.

Average output power:



page 37

DC-Side Voltage as a Function of α where Vdα = A/(π/3)

0 (a) ω =0

0

0

0

ω t

ω t

ω t

ω t

α

α

α

α

(b) ω =30°

(c) ω=60°

(d) ω =90°

A

A

απα

cos23 LLd VV =

At α= 90°, Vdα 0

page 38

DC-side Voltage as a Function of α where Vdα = A/(π/3)

0 ωt

0 ωt

0 ωt

α

(e) ω=120°

(f) ω=150°

(g) ω=180°

α



page 39

Input Line Current

L)](19sin[2)](17sin[2

)](13sin[2)](11sin[2

)](7sin[2)](5sin[2)sin(2)(

1917

1311

751

αωαω

αωαω

αωαωαωω

−−−−

−+−+

−−−−−=

tItI

tItI

tItItIti

ss

ss

sssa

),2,1(16 K=±= nnh

0 ωt

vcnvbnvanvcnia

ia1

φ1=αα

ωt=0

The input line current ia, ib, and ic have rectangular waveforms with an amplitude Id.

where only the nontriplen odd harmonics h are present and

page 40

Harmonics of Line Current

955.031 ==πs

s

II

%08.31TDH =

dds III 816.032 ==

ddd

ds

III

DII

78.0623

214

2sin

214

1

===

=

ππ

ππ

16 where1 ±== nhh

II ssh

1.0

h

1s

sh

II

1 7 11 13 17 19 235

131

51

71

111

25



page 41

Power, Power Factor, and Reactive Volt-Amperes

αφ coscosDPF 1 ==

απ

cos3PF =

0 ωt

vcnvbnvanvcnia

ia1

φ1=α

αωt=0

page 42

Line Current as a Function of α

φ

(a) α = 0

(b) α = 30°

(c) α = 60°

VanIa1

VanIa1

Ia1φ

Van

(d) α = 90°φ

Van

Ia1

(e) α = 120°

Van

Ia1φ

vania1ia

α

ωt

ωt

ωt

ωt

ωt

ωt

Id

α

α

α

α φ

Van

Ia1

(f) α = 150°



page 43

Effect of Ls

Three-phase converter with Ls and a constant dc current.

~+

−

vd

Ls

~

~

a

b

c

T1 T3 T5

T4 T6 T2

n

N

P

Id

1

3/05.0s

LLs I

VL ≥ω

The ac-side inductance can not be ignored in practical thyristor converters. In fact, the German VDE standards require that this inductance must be a minimum of 5%.

page 44

Commutation in the presence of Ls

~Ls

~

a

b

c

n

~

id=Id

ia T1

T6

T5

idicLs

Ls

vLs P

N

ω t

ω t

0

0

vcn van Au

vPn

vbn5 1

(ωt=0)α

6u

2

3

2

vcn5

4

Id Id

α

ic=i5 ia=i1

(α+u)

(a)

(b)

(c)

∫+

=u

Lu tdvA sα

αω )(

ds

I

asu ILdiLAd

ωω ∫ == 0



page 45

Line current in the presence of Ls

0 ω t

α

vanvcn

ia

u

vcn

ia1

21u+= αφ

sLanPnvvv −=

dtdiLv asLs =

During the commutation interval α < ωt < α + u

where

∫+

=u

Lu tdvA sα

αω )( ds

I

asu ILdiLAd

ωω ∫ == 0

ds

LLu

LLd ILVAVV

πωα

ππα

π3cos23

3/cos23 −=−=

page 46

Calculation of Commutation Transient

dtdiLvv asanPn −=

dtdiLvv csanPn −=

)(22

)ncommutatio during(dtdi

dtdiLvvv cascnanPn +−

+=

dtdi

dtdi ca −=

)(21

cnanPn vvv +=

cad iiI +=



page 47

Calculation of Commutation Interval

222accnana

svvv

dtdiL =−=

s

LLa

LtV

tddi

ωω

ω 2sin2

)(=

∫∫+

=uI

s

LLa tdtL

Vdidα

αωω

ω)(sin

22

0

dLL

s IVLu


Solving for the commutation interval u

page 48

Line Current in the Presence of Ls

)21cos(DPF u+≅ α

)]cos([cos21DPF u++≅ αα

0 ω t

α

vanvcn

ia

u

vcn

ia1

21u+= αφ

The ac-side inductance reduces the magnitudes of the harmonic currents.



page 49

Normalized harmonic current in the presence of Ls

u20°10°0

10

12

14 h = 7

α = 0°

5°15°30°- 90°

(%)1

7

II

u20°10°0

u20°10°0

u20°10°0

16

18

20

(%)1

5

II

α = 0°5°15°30°- 90°

h = 5

10

4

5

6

7

8

9

(%)1

11

II

h = 11

α=0°

5°15°30°- 90°

4

6

8

(%)1

13

II

h = 13

2

α = 0°

5°15°60°

(a) (b)

(c) (d)

page 50

Typical and Idealized Harmonics

Typical

Idealized

h 5 7 11 13 17 19 23 25

Ih / I1

Ih / I1

0.17

0.12

0.10

0.14

0.04

0.09

0.03

0.07

0.02

0.06

0.01

0.05

0.01

0.04

0.01

0.04



page 51

Example 6.3: Simulation of Practical Thyristor Converter

A three-phase thyristor converter is supplied by a 480 V (line-to-line) 60 Hz source. Its internal

inductance Ls1 = 0.2 mH. The converter has a series inductance Ls2 = 1.0 mH. The load is

represented as the following, with Ld = 5 mH, rd = 0, and Ed =600 V and the delay angle α = 20°. Using Pspice simulation, obtain vs, is and vd waveforms and calculate Is1, Is, DPF, PF, and

%THD.

Example 6.3

A practical thyristor converter.

~

id+

−

vd

Ls Rd

~

~

a

b

c

n

+− Ed

Ld

page 52


0 ω t

iava

Edvd

SolutionThe waveforms are shown as the following, and the calculated results are as follows:

Is1 = 22.0 A,

Is = 22.94 A,

DPF = 0.928, PF = 0.89, and THD = 29.24 %



page 53

PSPICE Input File for Example

* Three-Phase, Thyristor-Bridge Rectifier.PARAM PERIOD = {1/60}, DEG120={1/(3*60)}.PARAM ALFA = 20.0, PULSE_WIDTH = 0.5ms*LS1A 11 12 0.2mH IC = 45ALS2A 12 13 1.0mH IC = 45ALS1B 21 22 0.2mH IC = -45ALS2B 22 23 1.0mH IC = 45ALS1C 31 32 0.2mHLS2C 32 33 1.0mH*LD 4 5 5mH IC=45AVD 5 6 600.0V*XTHY1 13 4 SCR PARAMS: TDLY= 0 ICGATE=2VXTHY3 23 4 SCR PARAMS: TDLY= {DEG120} ICGATE=0VXTHY5 33 4 SCR PARAMS: TDLY= {2*DEG120} ICGATE=0VXTHY2 6 33 SCR PARAMS: TDLY= {DEG120/2} ICGATE=0VXTHY4 6 13 SCR PARAMS: TDLY= {3*DEG120/2} ICGATE=0VXTHY6 6 23 SCR PARAMS: TDLY= {5*DEG120/2} ICGATE=2V*VS 1 0 SIN(0 340V 60 0 0 {ALFA})*VSA 11 0 SIN(0 391.9V 60 0 0 {30+ALFA})VSB 21 0 SIN(0 391.9V 60 0 0 {-90+ALFA})VSC 31 0 SIN(0 391.9V 60 0 0 {-210+ALFA})

page 54

PSPICE Input File for Example (continued)

.TRAN 50us 50ms 0s 50us UIC

.PROBE

.FOUR 60.0 v(11) i(ls1A) i(ld)

.SUBCK SCR 101 103 PARAMS: TDLY = 1ms ICGATE = 0V*Power Electronics: Simulation, Analysis Education……by N. Mohan.SW 101 102 53 0 SWITCHVSENSE 102 103 0VRSNUB 101 104 200CSNUB 104 103 1uF*VGATE 51 0 PULSE(0 1V {TDLY} 0 0 {PULSE_WIDTH} {PERIOD})RGATE 51 0 1MEGEGATE 52 0 TABLE {I(VESNSE) + V(51)} = (0.0, 0.0) (0.1, 1.0) (1.0, 1.0)RSER 52 53 1CSER 53 0 1uF IC={ICGATE}*.MODEL SWITCH VSWITCH (RON=0.01).ENDS

.END



page 55

Pspice Input Circuit for Example

is

+−

+

−

vd

XTHY1Ls1

id

312

4 5

60

5

22

4 6 2

Ld

+−

+− 32

13

23

32

Ls211

21

31

+−

page 56

Discontinues-Current Conduction

Waveforms in a discontinuous-current-conduction mode.

vdEd

id0 ω t

A large value of Ed will result in a smaller Id. In order to regulate, α will have to be increased at lower values of Id.



page 57

Inverter Mode of Operation

Inverter with a constant dc current.

~+

−

vd

Ls

~

~

a

b

c

n Idia

1 3 5

4 6 2

P

N

page 58

Waveforms

ω t0

van vbn

1

φ1

id

α3

2

5

4

vcn vcn

4 6

u

Vdvd

ω t0ia1

ia

(a)

(b)



page 59

Inverter Start-up

~+

−

vd

Ls

~

~

a

b

cn

ia

1 3 5

4 6 2

Ld

vL+ −

Ed−

+

id

Power

Vd

Id

α0α1α2α3=90°

α4α5

0Rectifier

Inverter

(a) Thyristor inverter with a dc voltage source (b) Vd versus Id

page 60

Voltage Across a Thyristor in the Inverter Mode

vbn

1 3

24

vcn

6ωt

(a)

(b)

ωt

vcnvan

3

5

vPnα

vcav5

α

γ =extinction angleu

x

LLV2



page 61

Line Notching and Distortion

Line matching in other equipment voltage (a) circuit.

~Ls1

~

~

a

b

c

n

VLL

equipment

a′

b′

c′

(vequip)LL

Ls2

C3 ia

T3T1 T5

T4 T6 T2

A

B

C

id

id=continuous

snubber

(a)

21 sss LLL +=

page 62

Line-to-Line Voltage Distortion

(b) phase voltage, (c) line-to-line voltage vAB.

vbn

vAB

ωt0

vcnvan

LLV2

van

An=2ωL3Id

LLdn VILA 2

2Area 3ω==

u

α

ω t = 0

αsin2 LLn VV =0 ωt

(b)

(c)



page 63

Calculation of Notch Width

radians][volt 2 area notch Deep dsn ILA ω=

angle)delay ( sin2deptharea notch Deep =≅ ααLLV

rad sin2

2depth notcharea notch width Notch

αω

LL

ds

VILu ≅=

21

1

ss

s

LLL+

=ρ

305.012 LLas

VIL ≥ω

The German VDE standards require that this inductance must be a minimum of 5%.

The notch depths and notch areas are a factor of α

page 64

Voltage Distortion

Line notching and distortion limits for 460V systems.

[ ]100

)(%THD Voltage

amental)phase(fund

21 ×

×= ∑

VLI sn ω

Special applications

Dedicated system

General system

Class Line Notch Depthρ(%)Line Notch Area

(V-µs)Voltage Total

Harmonic Distortion(%)

10 16,400 3

20 22,800 5

1036,50050



page 65

References

[1] N. Mohan, “Power electronics: computer simulation, analysis, and education using the education version of Pspice,” Minnesota Power Electronics Research and Education, P. O. Box 14503, Minneapolis, MN 55414.

[2] E. W. Kimbark, Direct Current Transmission, Wiley-Interscience, vol. 1, New York, 1971.

[3] B. M. Bird and K. G. King, An Introduction to Power Electronics , Wiley, New York, 1983.

[4] Institute of Electrical and Electronics Engineers, “IEEE guide for harmonic control and reactive compensation of state power converters,” ANSI/IEEE Standard 519-1981, New York.

[5] D. A. Jarc and R. G. Schieman, “Power line considerations for variable frequency drives,” IEEE Transactions on Industry Applications, vol. IAS, no. 5, pp. 1099-1105, September/October 1985.

[6] Institute of Electrical and Electronics Engineers, “IEEE standard practice and requirements for general purpose thyristor DC drives,” IEEE Standard 597-1983, New York.

[7] M. Grotzbach, W. Frankenberg, “Injected currents of controlled AC/DC converters for harmonic analysis in industrial power plants,” Proceedings of the IEEE International Conference on Harmonics in Power Systems, pp. 107-113, Atlanta, GA, September 1992.

[8] N. G. Hingorani, J. L. Hays and R. E. Crosbic, “Dynamic simulation of HVDC transmission systems on digital computers,” IEEE Conf. Rec., vol. 113, no. 5, pp. 793-802, May 1966.

page 66

Home Work

~Ls1

~

~

a

b

c

n

VLL

equipment

a′

b′

c′

(vequip)LL

Ls2

C3 ia

T3T1 T5

T4 T6 T2

A

B

C

id

id=continuous

snubber

21 sss LLL +=

In the following circuit, Ls1 corresponds to the leakage inductance of a 60 Hz transformer with the following ratings: three-phase kVA rating of 500 kVA, line-to-line voltage of 480 V, and an impedance of 6%.Assume Ls2 is due to as 200-ft-long cable, with a per-phase inductance of 0.1 µH/ft. The ac input voltage is 460 V line to line and the dc side of the rectifier is delivering 25 kVW at a voltage of 525 V.



page 67

1. Calculate the notch width in microseconds and the line notch depth σ in percentage at the point of common coupling. Also, calculate the area for a deep line notch at the point of common coupling in volt-microseconds and compare the answers with the recommended limits in the following Table.

Line notching and distortion limits for 460V systems.

Special applications

Dedicated system

General system

Class Line Notch Depthρ(%)Line Notch Area

(V-µs)Voltage Total

Harmonic Distortion(%)

10 16,400 3

20 22,800 5

1036,50050

page 68

2. Repeat Problem 1 if a 480 V 1:1 transformer is also used at the input to the rectifier, which has a leakage impedance of 3%. The three-phase rating of the transformer equals 40 kVA.

3. Calculate the THD in the voltage at the point of common coupling in problem 1 and 2.

4. Using the typical harmonics in the input current given in Table 6-1, obtain the THD in the voltage at the point of common coupling in Problem 1.

Documents

Phase-Controlled AC-DC Converterspemclab.cn.nctu.edu.tw/W3news/開授課程-old/電力... · 2005-04-19 · Phase-Controlled AC-DC Converters NCTU 2005 Power Electronics Course Notes