chương5_nghịchlưuffff

7/24/2019 chng5_nghchluffff

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.

B mn T ng ha,Khoain, HBK H ni

H ni, 9 - 2010


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Khi nim vnghch lu c lp ,

NLL ngun dngNLNA mt pha, phng php iu chPWMNLNA ba pha, PWM, SVM.

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Chng 5

Nghch lu c lp V.1 Nhng vn chung

V.1.1 Nghch lu c lp l g?

V.1.2 Phn loi v ng dng

V.1.3 Khi nim vngun p, ngun dng

V.2 Nghch lu c lp ngun dng song song V.2.1 Nghch lu c lp ngun dng song song mt pha

. . g c u c p ngu n ng song song a p a

V.3 Nghch lu c lp ngun p V.3.1 Nhng vn chnh vnghch lu ngun p

V.3.2 VSI smt pha na cu (Half Bridge)

V.3.3 VSI scu mt pha (H Full Bridge)

V.3.4 Phng php iu chrng xung (PWM)

V.3.5 iu chPWM dng iu khin s

V.3.6 Nhn xt chung vPWM.

V.3.7 Tnh ton sNLNA PWM.

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Chng 5

Nghch lu c lp V.3.8 M hnh m phng NLNA PWM

V.4 VSI ba pha

V.4.1 VSI ba pha su xung

V.4.2 VSI ba pha PWM

V.4.3 iu chPWM vi thnh phn thtkhng ZSS-PWM

V.4.4 Cc thng scbn ca PWM

. .

V.5.1 Khi nim vvector khng gian

V.5.2 Cbn vSVM

V.5.3 Phng php iu chvi vi to = t7 SVPWM.

V.5.4 Qu iu ch.

V.5.5 Nhn xt chung vSVM.

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V.1 Nhng vn chung

V.1.1 Nghch lu c lp l g NLL: bbin i DC/AC, tn sv in p ra thay i c.

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N ,

DC/AC


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V.1 Nhng vn chung

V.1.1 Nghch lu c lp l g? Ti sao li cn n BBDC/AC?

Chc ngun l DC: v d, khi ngun duy nht ta c l tacquy.

Khi phti AC yu cu ngun cp c cc thng snhin p, tn sthay itrong di rng, khc xa cc thng sca ngun in p li.

Khi c yu cu viu chnh ctn sln in p xoay chiu, v dtrong cc htruyn ng ng ckhng ng bhoc ng cng b.

Khi trong cc b bin i cng sut yu cu c tn scao (Tn scao slm

cho cc phn tin tnhMBA, cc phn tphn khng nhtin, incm c gi trnh).

Mt sngun pht scp c u ra l mt chiu hay c chuyn vdng mtchiu tch trtrong acquy: pin mt tri (Photocell), pin nhin liu (Fuel cell),in sc gi (Wind Turbine Generator),

Mt sdng nng lng tch ly di dng acquy (Battery Energy StorageSystem BESS).

u cui ca hthng truyn ti in mt chiu HVDC.

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V.1 Nhng vn chung

V.1.2 Phn loi v ng dng Phn loi:

Da theo c tnh ca ngun mt chiu u vo: Nghch lu ngun dng: Current Source Inverter CSI,

Nghch lu ngun p: Voltage Source Inverter VSI, Nghch lu ngun Z, ZSI, trung gian gia CSI v VSI.

Da theo cc c im ca phng php iu chnh in p v tn su ra,phbin l nghch lu PWM.

Da theo c im ca mch ti: mt lp cc nghch lu lm vic vi ti lmch vng cng hng LC, gi l nghch lu cng hng.

ng dng: rt rng ri, Trong lnh vc truyn ng xoay chiu. Cng vi chnh lu to nn cc bbin

tn.

Trong lnh vc xe chy in (Electric Vehicle EV), hin nay pht trinthnh mt xu hng xe mi cho tng lai gn.

Thm nhp vo hthng iu khin trong hthng in (FACTS v D-FACTS).

Cc hthng cp ngun AC-DC-AC-DC thay cho cc hAC-DC thng thng.

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V.1 Nhng vn chung

V.1.3 Khi nim vngun p, ngun dng Ngun dng

Ngun in c dng in ra khngi, khng phthuc vo ti v

tnh cht ca ti. To ra bng mc ni tip ngun

DC vi in cm ln,

Hon ton c thngn mch,

Ngun p

Ngun in c in p ra khngi, khng phthuc vo ti v

tnh cht ca ti. To ra bng mc song song u ra

ngun DC vi tin ln,

Hon ton c thhmch, khngkhng c hmch. c ng n mch.

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C ,

.


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Phi hp ngun vi ti: ngun p,ngun dng. Khng thni song song hai

ngun p vi nhau v dng sanbng in p srt ln.

Khng thni ni tip hai ngundng vi nhau v gy t bin

V.1 Nhng vn chung

V.1.3 Khi nim vngun p, ngun dng

.

Khi nim vngun p, ngundng cng p dng cho ti: Song song vi t- ngun p;

Ni tip vi cun cm ngun

dng. BBl khu khng qun tnh:

Nu u vo l ngun p th u ral ngun dng v ngc li.

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N

N


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V.2 Nghch lu ngun dng

V.2.1 Nghch lu ngun dng song song mt pha Sdng thyristor V1, , V4.

Ngun u vo c in cm L gitrln, to nn ngun dng.

TC song song vi ti, to khnng chuyn mch.

(V1, V2) v (V3, V4) mtrongmi na chu k .

thdng dng in, in p

D NL

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,

=>, (

T


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V.2.1 Nghch lu ngun dng song song mt pha Phn tch sbng phng php

gn ng sng hi bc nht: Chxt n thnh phn sng hi

bc nht ca dng in v in p. C thbiu din cc i lng

bng biu vector.

iu kin shot ng c

thvector

t

ng t p mang t n ung,

vt trc in p. Gc vt trcny chnh l gc kha ca van.

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C C

, (

)

( )C L CC L C LR R C t

I I UI I Q Qtg

I I U P

= = =

C t tQ Q Ptg= +


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V.2.2 Nghch lu ngun dng song song ba pha V.2.2 Nghch lu ngun dng song

song mt pha, c it cch ly. itc tc dng cch ly mch chuyn

mch khi mch ti. Phng n tng tcng c NL

ba pha.

NLND ba pha

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60

120

180

240

300

360


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V.3 Nghch lu ngun p mt pha

V.3.1 Nhng vn chung vNLNA Nhc im ca NLND:

in p ra phthuc vo ti, vvy rt kh ph hp vi cc ph

ti thng thng. Thit binthng c sn xut cho cc cpin p tiu chun nn khng thhot ng khi in p bin ngmnh.

NLNA xy dng chyu trnMOSFET v IGBT, mch lcc chto chun, to thnh ccmodul, dsdng.

NLND chc thit kcho mtphti cth, c thc cng sutln hoc rt ln.

NLNA c thc chto dngcho mt lp rng ri cc phti. NLNA m bo in p ra c dng

khng i, p ng cho cc phtisn xut hng lot.

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Van V1, V2 ON/OFF ngc nhau,

D1, D2 it ngc, dn dng tdovtDC,

in p trn ti:VOC = +/- VDC.

M hnh ti Ls, Rs, Es (Es c thl

S


V.3.2 SNL na cu (Half bridge)

trng hp: ng c, ngun dngAC iu khin c, chnh lutch cc. S..Es thhin chnh lphti, ni in nng bin i

thnh dng nng lng khc. C thiu khin dng Io theo

hnh dng bt k.

Gii hn: VOC cht-VDCn+ VDC

dIo/dt


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V.3.3 Nghch lu ngun p cu mt pha (H Full Bridge)

thdng dng in, in p.

V1, V2, V3, V4 van /k hon ton,nhBJT, MOSFET, IGBT.

D1, , D4 cc it ngc.

TC u vo c gi trln.

iu khin: 0 T/2 m(V1, V2),

T/2 T m(V3, V4).

in p trn ti c dng +/-E.

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V.3 Nghch lu ngun p 1 pha

V.3.4 iu chPWM Vn t ra i vi NLNA:

1. Lm thno c thiuchnh c in p cng nhtn

sca in p ra? 2. Dng in p ra dng xung ch

nht, nu phn tch ra chuiFourier cha nhiu thnh phn

thdng dng in, in p.

sng hi bc cao.

Lm thno gim c sng

hi bc cao? Dng mch lc. Tuy nhin tc

dng ca lc phthuc ti.

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( )

1

sin 2 14( )

2 1k

k tEu t

k

=

=


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iu chPWM: iu khin mc thp nht.

S


V.3.4 iu chPWM cho NLNA

c(t) rng ca, gi l sng mang;cPKbin rng ca;

m(t) tn hiu chun mong mun,gi l sng iu ch;

Ts chu kiu ch, cn gi l chuktrch mu.

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PK

s s

m cdT T

=


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Trong mi chu kng ct in pu ra c gi trtrung bnh, gi ltrung bnh trt:

Gi trtrung bnh ca in p ura nghch lu PWM:

th


V.3.4 iu chPWM cho NLNA

( ) ( )1

st T

s t

v t v d T

+

=

Tsmch in tng ngc ththy quan hhm truyn t

gia in p ra nghch lu v dngu ra l mch lc tn thp bcnht.

Trong mi chu kTsin p raVOC sphn ng lp tc vi tnhiu mong mun ngay trong chu

kiu ch. Nu hng sthi gian Ls/Rs >> Ts

dng in sun theo dng ca tnhiu m(t).

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( ) ( ) ( )( )

( )( )

1

2 1

OCDC s DC s

s

DC

V t V T d t V T d t T

V d t

=

=


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Biu khin sPWM,thng c trong cc viiu khin hin i:

thdng sng:


V.3.5 iu chPWM dng iu khin scho NLNA

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Uniformly sampled with single update mode (Khc analog naturally

sampled PWM). Chtrch mu u (Khc vi trch mu tc thi). 1. Trailing edge modulation, (Hnh b). Biu chsn sau.

2. Leading edge modulation, (Hnh c). Biu chsn trc 3. Triangular carrier modulation, (Hnh d). Biu chsng mang i xng.



T

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Uniformly double update. Trch mu hai ln, nguyn l thc hin:



M hnh:

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T


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Cc chsnh gi hiu nng ca PWM

1. Hsiu ch, tsgia bin sng iu chm(t) so vi bin sngrng ca:

2. Hsmo tng: THD chnh l tsgia tng gi trhiu dng ca cc thnh phn sng hi bc


V.3.6 Cc chsnh gi PWM

; 0 1rm

cm

U

U=

22 2

2,3,... 1

2

1 1

k

k o

UU U

THDU U

= = =

.

3. Hstn s: kf = fs/f1 , tsgia tn sca sng mang so vi tn ssng ra mong mun. Thng thng c hsmo tng THD trong phm vi cho php cn c kf

20. Vi cng sut lnfs c2 4 kHz, trong khi di cng sut nhhn

thng phi chnfs t10 - 20 kHz. iu ny cng l v m bo p mch dng ra trong phm vi cho php th

vi dng cng nhin cmLs cng phi ln. Tuy nhin nuLs ln th st p tn scbn cng ln. tha hip, do phi chnfs ln.

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Vic tnh ton thng da trn cc sliu ban u: Gi trin p hnh sin ra mong mun Uo (V) v tn ssng cbnf1 (Hz).

Cng sut hoc dng u ra mong mun Po (W), Io (A), hscng sut ca ti

cos. Thng thng hscng sut c0,8. V dtnh ton:

Cc bc v cc thng scn tnh ton:


V.3.7 Tnh ton cc thng sca sNLNA PWM

1220 2 311( ); 50 ; 1 ;cos 0,8om oU V f Hz P kW = = = = =

. DC .

Vi PWM trong di lm vic tuyn tnh, 1, gi trbin in p u ra ctht ln nht l UDC, khi tn sng ct fs coi l v cng ln. dphngin p mt chiu thay i trong phm vi +/-10% cn chn max = 0,9.

Vy: UDC = Uom/0,9 = 311/0,9 = 346 V.

Trong mch thng c mch lc LC to in p ra hnh sin. Dphng st ptrn cun cm lc Ls c10% in p ra nn phi chn UDC = 1,1.346 = 380 V.

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2. Tnh ton bin ddng u ra yu cu:Iom (A). Cng sut ton phn ca ti So = Po/ cos= 1000/0,8 = 1250 (VA);

Dng ti yu cu:Io = So/Uo = 1250/220 = 5,68 (A).

Bin ca dng tiIom = Io.sqrt(2) = 5,68*1,4142 = 8 (A).

3. Chn tn sng ct:fs (Hz), Vi cng sut nhchn tn sng ctfs = 20 kHz, Ts = 0,5.10-4 (s).



. V, D Dng trung bnh qua van:

IV= 2,29 A.

Dng trung bnh qua it:

ID = 0,26 A.

5. Xc nh dng nh ln nht qua van v it.

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( )1 1 cossin2 2

V om omI I d I

+= =

( )0

1 1 cossin

2 2D om om

I I d I

= =


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5. Xc nh dng nh ln nht qua van v it. Dng ti thhin chnh l gi trdng trung bnh u ra nghch lu trong mi

chu kct mu. V vy chcn xc nh p mch ln nht ca dngIo(t).

Bqua nh hng ca Rsi vi p mch dng ti, ta c:

Trong NLNA PWM . Dng in c p mch ln nht khi hslp y xung (Duty ratio) l d = 0,5. Do :



( ) ( )os odi t

L u tdt

,max 2

o DCU U =

6. Xc nh gi trin cmLs. Ly st p ti tn scbn bng 10%Uo.(i vi cng sut nh).

ULs = Io.XLs = 0,1.Uo = 0,1.220 = 22(V)XLs = 22/5,68 = 3,8732()Ls = 12 (mH);

p mch dng ti bng: Io,max = 380.0,5.10-4/(2.12.10-3)= 0,79 A.

So vi bin dng in th p mch bng IL 100% = 0,79/8 = 20 %. yc thcoi l gi trchp nhn c.

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,max

,max

/ 24

os

o DC s ss

I U T LL


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7. Tnh ton tC ca mch lc LC. Trong NL PWM in p ra chyu l sng cbn. Cc thnh phn sng hi bc

cao xut hin chung quang tn sng ctfs, cthl h.fs +/- l.f1, trong h

= 1, 2, ., l = 1, 2, Nhng tn ssng hi thp nht lfs f1, fs -2.f1, Tuynhin dofs >> f1 nn cc sng hi ny chyu tp trung quanhfs, ngha l rtxa so vif1. iu ny lm n gin vic tnh ton mch lc LC u ra nghchlu rt nhiu.



1= =

Khng cn n iu kin trnh cng hng cc sng hi c thc trnsng in p ra.

Chn CL = 0,1s CL = 12,5664.103 (rad/s) . Vy:

C thchn trstC ln hn, v d1F.

m bo tn sct CL gi tr tphi chn ln hn b vo cng sut phnkhng ca ti.

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LC s sLC

( )

( )22 3 31 1 1 1

0,53

12.10 12,5664.10CL

C F

L

= = =


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8. B cng sut phn khng ca ti:

Nu b bng tC th phi c QC= QL;

So vi gi trtC tnh mc (7) thy rng c thchn tC=50F l ph hp.



( )2 2

75049,35

2. .50.220

C

C

QC F

U = = =

2 2 2 21250 1000 750( )Var

L o oQ S P= = =

22C

C C

C

UQ CU

X= =

9. C n k m tra l u k n t n s c nXC>> XL: Nu khng sto nn phn p giaXCvXL, khng tht c in p 220 V

u ra.

Thc slXC>> XL .

10. Kim tra li sliu tnh ton ca sbng m hnh m phng. y l phng php rt hiu qukim chng cc tnh ton tmc (1) n (9)

trn y.

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( )

3

6

2. .50.12.10 3,768 ;

1 / 2. .50.50.10 63, 7

L

C

X

X

= =

= =


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11. Tnh ton tC ca mch mt chiu. TC trong mch mt chiu dng vai tr l tlc ca mch chnh lu pha trc,

va ng vai tr tip nhn cng sut phn khng tmch nghch lu do cc it

ngc a v. Vy gi trca tl gi trno cn ln hn. Trng hp nng nnht l dng ti gi trbin , hsd = 0,5 (tng ng

khi ti thun cm, in p iu chqua khng), khi :



xt

U I

= / 2;t T I I = =

Thng chn UC= 0,050,1UDC. C thtnh c:

TC tnh c c gi trkh nh, chng tu vit ca PWM. Trong trng hpny tmt chiu C sc xc nh chyu tiu kin san bng in p ura chnh lu.

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C ,

( )638

10,53.10 102 2.20.10 .0, 05.380

C

s C

IC F

f U

= = =


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V.3.8 M phng sNLNA PWM M hnh

Trn MATLAB

S 1,

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S 2,C


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V.3.8 M phng sNLNA PWM Kt qum hnh 1, sna

cu (m=0,8; UDC=200V)

Tn siu chchn thp 1

kHz minh ha r hn p mch ca dng ti.

Dng p mch ln nht thiim in p iu chm(t) qua

thdng, p ra NL.

0 (khi d=0,5). Nu lc by gidng t gi trbin (ti gnthun cm) th chu kiu chny xc nh dng nh lnnht (Trng hp xu nht).

y l cstnh ton dngnh qua van v it mc (5),phn V.3.7.

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V.3.8 M phng sNLNA PWM Kt qum hnh 2, scu

mt pha. Tham stnh ton theophn 3.7.

Tn siu ch20 kHz. Mch lc LC tnh ton theo:

1. Cun cm L m bo pm ch dn ti tron h m vi

thdng, p u ra.

20%. Tn sct ca mch lc bng

1/10 tn sfs.

Tlc C tnh theo tn sct

ca mch lc v hiu chnh b cng sut phn khng cati.

L = 12 mH, C = 50 uF.

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VSI cu ba pha c thcoi gm banhnh van na cu (V1, V4), (V3,V6), (V5, V2). Cc van trn cng

nhnh cu khng bao gic mcng nhau.

Ti pha xoay chiu ni gia ccim ra ca na cu nn khng

VSI cu ba pha

V.4 Nghch lu ngun p cu ba pha

V.4.1 Scu ba pha

c n n i m gia pha mtchiu nhsna cu thngthng.

sdng cc kt quvPWMca sna cu cho scu

ba pha ta vn sdng mch intng ng cu ba pha nhbana cu, vi im gia pha DC.

Cu ba pha = 3 na cu.

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1

2 DC

U

1

2 DC

U


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Dng in p ra 6 xung ca VSI cu ba pha.

uAn, uBn, uCn l ba in p ra ca snacu (+/-UDC/2), lch pha nhau 120.

uZn=1/3.(uAn+ uBn+ uCn ); uZn c dng xungchnht, tn s3f, bin +/-1/6UDC.

uA=uAn-uZn ; uB=uBn-uZn ; uC=uCn-uZn;

V.4.1 Scu ba pha

Phng php iu khin cbn 2

DCU

2

DCU

2

DCU

2

DCU

2

DCU

2

DCU

2

2 DCU

uAB=uAn-uBn ; uBC=uBn-uCn ; uCA=uCn-uAn.

Sng hi cbn in p pha u ra:

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3

DCU

3

6

DCU

(1)

6

/3 2 /3

0 /3 2 /3

1sin

2 1 2 1

sin sin sin3 3 3

2

s

DC

DC

U u d

U d d d

U

=

= + +

=


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SPWM (sinusoidal PWM) chocu ba pha c thc hin cho basna cu: vi ba sin chun,cng mt hthng in p rngca (Carrier based PWM).

Hsiu ch: m = mref/ms , binsng sin chun trn bin rn ca. Tron di iu ch

V.4.2 iu chPWM cho nghch lu cu ba pha

Siu khin SPWM

tuyn tnh in p ra hnh sin,yu cu 0 m 1. Cc tiu chun nh gi: M = U1m/U1m,6s bin sng hi

bc nht so vi sng bc nht ca

dng in p ra 6 xung. 0 M 0,785.

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S CBPM


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Mu xung iu khin trong PWMvi rng ca i xng:

Mu xung cho thy dng ti u

vchuyn mch, mi ln chcmt pha phi ng ct.

Trng thi van cho ra in pbn 0 n vi vector khn

V.4.2 iu chPWM cho nghch lu cu ba pha

Siu khin SPWM

trong SVM) phn bi xng hai u v gia chu kTs.

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Vi iu chin p ra hnh sin theo mch in tng ng vi snacu in p ra trn mi pha u ra chthay i gia +/- UDC/2, l bin ln nht ca in p ra. Chnh v vy theo SPWM hsiu chln nht

chlMmax= (UDC/2)/ (2/.UDC)=/4=0,785 (m=1). Thc ra vi scu khng cn im gia ca mch DC v in p ra l

+UDCvUDC. iu ny ngha l bin in p sng sin cbn iu chra nghch lu c thln hn, t nht l n 2/.UDCnhdng in p ra 6

V.4.3 iu chPWM vi thnh phn thtkhng

Khi nim vZSS-PWM

xung.

Phng php iu chc thnh phn tht0 (Zero Sequence Signal PWM ZSS PWM) da trn csl trong hthng ba pha cn bng thnh phnthtkhng c trkhng v cng ln. iu ny ngha l nu trong dngsng chun mong mun c thnh phn sng hi bc 3 th thnh phn nykhng thxut hin dng sng in p ra. Thnh phn sng hi bc 3 trnmi pha thhin trn thca im trung tnh ti, uZn . Nu uZn c sng hibc 3 th in p ra cng khng bnh hng g.

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Nu thm vo thnh phn sng hi bc 3 trn dng in p sng sin chun,c thmrng c di thay i ca bin sng hi bc nht in p ram khng nh hng g n di iu chtuyn tnh ca VSI ba pha.

Sng bc 3 thm vo c thc dng sin, tam gic, hoc chnht. Bin sng bc 3 hnh sin bng bin sng ra mong mun cbn

tng ng vi hssng hi dng in ra nhnht.


Khi nim vZSS-PWM

ra n ln nht n . Hsiu chmmax mrng n1,154, tc l tng thm c 15,4%.

Hsmmax mrng c n gi trno m dng sng iu chthu cmref cn nhhn hoc bng 1, ngha l vn trong vng tuyn tnh i vi tn

hiu rng ca.

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max / 2 3 0, 907M = =


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Minh ha phng phpto tn hiu iu khintrong iu chvi thnh

phn tht0. Hai dngtn hiu sng bc ba cdng:

- Sng bc 3 hnh sin

thdng tn hiu iu chZZS PWM.


Dng tn hiu cho trong ZSS-PWM

(bin hoc 1/6 binsng cbn).

- Sng bc 3 hnh tamgic. Tng ng viiu chvector khnggian SVPWM.

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Dng tn hiu cho trong ZSS-PWM thdng tn hiu iu chZZS PWM.

C ththy cc tn hiu iu chsin mong mun c dng mo ln sng hibc ba

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Dng tn hiu cho trong ZSS-PWM thdng xung ca biu chZZS PWM.

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Thng s K hiu nh ngha Gii thch

1. H siu ch, sdnghai loi hsiu ch:

- Binsng ra bc nhtso vi dng p ra 6 xung.

M i vi SPWM inp ra hnh sin

- T sbin sng sin m Trong diiu ch

V.4.4 Cc thng scbn ca PWM cu ba pha

( )

1

1 ,6

1

2 /

m

m s

m

DC

UM

U

U

U

=

=

U

0 0, 785M

( )/ 4 0,785 =

iu chso vi bin sng rng ca.

tuyn tnh SPWM

2. Diiu chnh tuyntnh ln nht

Mmaxmmax

0 0,907

0 1,154

Ph thuc dng tnhiu iu chchoZSS-PWM

3. Quiu ch M > Mmaxm > mmax

Diiu chphi tuyn(in p ra mo dng)

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,

mc

m

U

=

0 1m


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Thng s K hiu nh ngha Gii thch

4. Tsgia tn siuchso vi tn scbn

mf mf= fs/f1 mfl snguyn l ttnht, mf>20.

5. Tn sng ct fs fs=1/Ts Ts l chu kiu ch

6. H smo phi tuyn THD THD%=Ih/Is1*100

Dng cho dng inv in p.

V.4.4 Cc thng scbn ca PWM cu ba pha

7. H smo dng in d Ih/Ih,6s Khng phthuc trkhng ti.

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V 5 Ph h i h t kh i SVM


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Mt hthng in p, dng inba pha bt kX = (XA, XB, XC), nutha mn ,Qua php bin i Clark trthnhmt vector:

Biu din di dng ma trn:

V.5 Phng php iu chvector khng gian - SVMV.5.1 Khi nim vvector khng gian Space vector

0a b cX X X+ + =

( )22

3 A B Cu au a u= + +u

[ ]

[ ]1

1 11

2 2 2

3 3 302 2

.

T

A B C

T

A B C

uu u u

u

T u u u

=

=

Trong : Biu din trn trc ta vector

trthnh:

N u:

Vector trthnh vetor quay:

10/22/2010 43

3 1 3

2 2

j

a e j= = +

( )

( )

12

31

3

A B C

B C

u u u u

u u u

=

=

( )cos2

cos -3

2cos

3

m

A

m

B

m

C

u U t

u U t

u U t

=

=

= +

( )j tmU e

=u

u

u



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Tng tvector in pvector dng in c thl:

Vi l gc pha gia dng in viin p.

Vector khng gian tng qut: trongh th n i n vector c bi u

di ca vetor chnh l bin ca cc thnh phn tng ng.

Nu trong in p c cc thnh

phn sng hi bc cao th vectorbiu din qua cc thnh phn nhchui phc Fourie nhsau:

V.5 Phng php iu chvector khng gian - SVMV.5.1 Khi nim vvector khng gian Space vector

( )j tmU e

=u

( )j tmI e

=i

jk t jk t

*

din bi ba thnh phn: Thnh phn thtthun, Thnh phn thtngc,

Thnh phn thtkhng.

Trong :

10/22/2010 44

p n zeru=u +u +u

( )

( )

( )

0

1

;

;

1.

3

jm

p

jm

n

A B C

U e

U e

u u u

+

+

=

=

= + +

p

n

zer

u

u

u

0 1k k= =

pk nk

0

1, 0,1,...,

T

jk te dt k

T

= = pku u

0

1, 1, 2,...,

Tjk t

e dt k T

+= = nku u

V 5 Phng php iu ch vector khng gian SVM


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1. State switch: trng thi ca van.Trong bbin i trng thi cphp ca van c xc nh trongcc iu kin: Khng lm ngn mch ngun p;

Khng lm hmch ngun dng.

2. State vector: vector trng thi.

3. Vector in p ra mong mun cthbiu din di dng hta cc:

Hoc ta thnh phn:

4. Tng hp vector mong mun tcc vector tr n thi. Tron mi

V.5 Phng php iu chvector khng gian - SVMV.5.2 Cbn vSVM

m j

ref ref U e

=u

,ref

u u = u

ng v m trng t c a van

xc nh c gi trca vectorkhng gian in p ra. Tnh cht: Vector trng thi c di v hng c

nh trn mt phng.

Cc vector trng thi chia mt phngthnh nhng phn u nhau, gi l ccsector.

gc iu ch viT

s l chukiu ch, vector mong munc tng hp thai vector trngthi:

Thng thng vector trng thi lhai vector bin ca sector.

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k s

T =

1 2

2

sT

t t= +r 1 2

u U U



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No Van dn uA uB uC

U0 V2, V4, V6 0 0 0 0

U1 V6, V1, V2 2/3UDC -1/3UDC -1/3UDC

U2 V1, V2, V3 1/3U 1/3U -2/3U

V.5 Phng php iu chvector khng gian - SVM

V.5.2 Bng cc vector chun ca SVMu

02

3

j

DCU e

32 j

U3 V2, V3, V4 -1/3UDC 2/3UDC -1/3UDC

U4 V3, V4, V5 -2/3UDC 1/3UDC 1/3UDC

U5 V4, V5, V6 -1/3UDC -1/3UDC 2/3UDC

U6 V5, V6, V1 1/3UDC -2/3UDC 1/3UDC

U7 V1, V3, V5 0 0 0 0

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3 DC

23

2

3

j

DCU e

2

3

j

DCU e

2

32

3

j

DCU e

32

3

j

DCU e



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Cc vector trng thi c biudin trn mt phng ta 0.

u mt cc vector l nh mt lc

gic u. Vector chia mt phng thnh 6 gc

bng nhau, gi l cc sector, nhstI, II n VI.

V.5 Phng php iu chvector khng gian SVMV.5.2 Biu din cc vector trng thi trn mt phng 0

Hai vector khng V0, V7 nm gc ta .

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Gisvector in p ra nm trongsector I. Biu din vector uo quahai vector bin:

Trong :

di cc vector:

Tnh c thi gian sdng ccvector bin:

Gi m=Uo/Ui, trong 0m 1, lhsiu ch, c thtnh cthi ian:

V.5 Phng php iu chvector khng gian SVMV.5.2 Tng hp vector in p ra

ou u u

p t= +

2

; .p 1 t 2u u u up

t

s s

t t

T T= =

2 2sin ; sin .

33 3

o op s t s

i i

U Ut T t T

U U

= =

di cc vector:

l gc pha ca vector in pu ra, tnh trong gc phn su:

Trong vng iu chtuyn tnh

tp+ttTs

Trong khong thi gian cn li pdng vector khng

to = Ts (tp+tt).

10/22/2010 48

33

2sin .

3

p

tu u

=

1 2

2

3

u ui

U E= = = u oU=

2 2

sin ; sin .33 3p s t st T q t T q

= =

; 0,1,2,3,4,53

uo

k k

= =



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Thi gian t1, t2 thhin l thi giansdng cc vector tch cc. Thigian cn li t0/2=Ts/2-(t1+t2) pdng vector 0, V0 hoc V7.

Cc cch sp xp v sdngvector khng l tdo v khng nhhng n gi trvector mong

1. Sine wave SVM, gi lSVPWM - SVM with Symmetrical

Placement of Zero Vectors.

t V0, V7 i xng quang nachu kiu chTs. V dtrongsector I dng cc vector:

V0 V1 V2 V7 V7 V2

V.5 Phng php iu chvector khng gian SVMV.5.2 Tng hp vector in p ra

mu n. Cch dng vector khng l

ty theo mc tiu mun t c: Gim thiu mo in p,

Gim n ti thiu sln chuynmch ca van, tc l gim tn tht

trn van. Khng phi lc no gimmo in p cng l mc tiu caonht, khi c thp dng gimtn tht.

.

2. Gim tn tht, gi lDiscontinuous pulse width

modulation - DPWM.

Trong mt chu kTs chdngvector khng mt ln (V0 hocV7), nhvy gim c hai lnchuyn mch.

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Cc gii hn ca SVM in p rahnh sin trn mi nhnh na cu.

1.

in p ra sin. Quo vector trn.Chiu chny tng ng viPWM trong vng tuyn tnh, in p rahnh sin, gi l SPWM.

thgii hn ca Sine waveSVM.

V.5 Phng php iu chvector khng gian SVMV.5.2 Cc gii hn ca SVM

U U

02

DCr

U u

.

Mt pha bgii hn bin ti UDC/2.in p ra bmo. Quo vector itheo ng lc gic, nt chm.

3.

Hai pha bgii hn bin ti UDC/2.in p bmo.

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2 3r

u

3

DCr

U u

V.5 Phng php iu ch vector khng gian SVM


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y l SVM tng ng viPWM c iu chthtkhng,vi U3fc dng tam gic cn.

thdng in p iu ch

V.5 Phng php iu chvector khng gian SVMV.5.3 Phng php SVPWM vi t

0

= t7

( )0 7 1 21

2 st t T t t = =

( )1

3 1 cos32 2

2

rms

tt UT

t U

=

10/22/2010 51

0 1

( )

( )

( )

1 2

1 2

1 2

2;

2

2;

2

2.

2

DCAn

s

DCBn

s

DCCn

s

UU t t

T

UU t t

T

UU t t

T

= +

= +

=

3cos ;

2 6

3sin ;

2 6

3cos .

2 6

An rm

Bn rm

Cn An rm

U U t

U U t

U U U t

=

=

= =

( )1

3Zn An Bn Cn

U U U U = + +

;;

.

A An zn

B Bn zn

C Cn zn

U U UU U U

U U U

=

=

=



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Cc gii hn ca SVPWM

Khi in p ratrn cc pha ti lun c dng sin

hon ton. Khi cc in p

ra uAn, uBn,uCn sbgii hn bi+/-UDC/2.

Dng in p bin iu uAn, uBn, uCn,uZn v in p trn cc pha ti uA, uB,uC vi UDC= 300 V, Urm = 173 V.

V.5 Phng php iu chvector khng gian SVMV.5.3 Cc gii hn ca SVPWM

( )1 / 3rm DC U U

( )1 / 3rm DC U U>

Vectkhng gian in p ra bgiihn trong hnh lc gic c nh lcc vectbin.

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V.5 Phng php iu chvector khng gian SVM


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Cc gii hn ca SVPWM

Vectin p ra chcn bhn chbi hnh lc gic c nh l ccvectbin chun. Vectkhng gian in p ra vi UDC=

300 V, Urm

= 200 V.


g p p g g

V.5.3 Cc gii hn ca SVPWM

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Php iu chm vectin p ravt qu gi l quiu ch.(Overmodulation).


g p p g g

V.5.4 Qu iu chSVPWM

( )1 / 3 dU

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SVM l phng php dng shon ton. Thut ton n gin, dng dng trn vi xl.

Mrng c phm vi iu chsovi PWM.

C thqu iu chm khngphi thay i nhiu trong thut

Scu trc thc hin SVM.

g p p g g

V.5.5 Nhn xt chung vSVM

ton.

L phng php c thmrngcho cc nghch lu phc tp hnnhs3 pha 4 dy, cc snghch lu a cp, ngay ccho cc

nghch lu mt pha.

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V.6 Nghch lu cng hng


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SVM l phng php dng shon ton. Thut ton n gin, dng dng trn vi xl.

Scu trc thc hin SVM.

g g g

V.6.1 Cc vn chung vNLCH

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