LVTN NguyenPhungQuang

Thuyt minh n tt nghip
B mn o lng v iu khin
Li ni u
ng c khng ng b c nhiu u im so vi ng c mt chiu : h s cng sut, trng lng cao vn hnh tin cy, gi
thnh ch to v chi ph vn hnh thp. Tuy nhin do cu trc phi tuyn vi a thng thng s,nn iu khin ng c khng ng b l kh khn. Nhng nm gn dy, vi s pht trin mnh m ca k thut bn dn cng sut ln.Vi mch in t v l thuyt iu khin, nhiu phng php iu khin hiu qu c sut cho iu khin ng c khng ng b. Chnh v vy ng c khng ng b c s dng rng ri trong h thng truyn ng iu chnh tc ca cc my sn xut, thay th dn ng c mt chiu, v d nh cc thit b ca dy chuyn sn xut xi mng... iu khin in p ngun cumg cp cho ng c l mt phng php n gin v kinh t, nhng cht lng iu chnh tnh v ng khng cao. Phng php ny thch hp cho cac ph ti l qut gi, my bm . Phng php iu khin hiu qu l thay i tn s in p ngun cung cp cho ng c. Do tc ng c xp x bng tc ng b, nn ng c lm vic vi trt v tn hao cng sut trt trong mch rotor nh. Tuy nhin phng php ny phc tp v t tin. Trong h thng chuyn ng in iu khin tn s phng php iu khin theo t thng c th to cho ng c c tnh tnh v ng tt . Vi phng php iu khin gin tip , cc h thng iu khin in p/ tn s v dng in / tn s trt c s dng rng ri trong cng nghip. h thng iu khin in p tn s, sc in ng khe h ng c c iu chnh t l vi tn s. ng c c kh nng sinh mmen nh nhau mi tn s di nh mc. C kh nng iu khin hai vng, vng di tc c bn iu chnh t thng khng i, iu khin gi t s sc in ng khe h tn s l hng s, vng trn tc c bn in p c duy tr khng i, t thng ng c c dm theo tc m bo cng sut ng c khng i. Sinh vin thit k: Nguyn
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Phng php trn c th to ra c tnh tnh tt nhng khng p ng c cht lng iu chnh trong thi gian qu trnh qu . H thng iu chnh nh hng theo t trng cn gi l iu khin vector c th p ng c cc yu cu iu chnh trong ch tnh v ng. N cho php iu chmh ch mmen v t thng ng c bng iu chnh c hai thnh phn dng in stato tng ng. H thng iu chnh gm hai knh iu khin c lp: iu khin mmen v iu khin t thng rotor, knh iu khin mmen gm mt mch vng iu chnh tc v mch vng iu chnh thnh phn dng in sinh t thng. Nh vy h thng truyn ng in ng c khng ng b c th to ra c nhng c tnh tnh v ng cao so snh c vi ng c mt chiu . Sau qu trnh hc tp em c nhn ti tt nghip: Nghin cu v kho st h thng cn bng nh lng ca nh my xi mng Lu X. Trn c s thit k h thng truyn ng cho bng ti ca h thng cn bng nh lng ny . Qua mt thi gian tm hiu, nghin cu v c s gip ca thy gio hng dn thc s: L Xun Qu cng vi cc thy c gio trong b mn, bn b ng nghip, n nay n ca em hon thnh. Vi kh nng kin thc cn c hn, do bn n ny ca em chc s khng trnh khi nhng thiu st. Bi vy em rt mong c s gp , gip ca cc thy cc c v cc bn ng nghip. Ngy 6 thng 3 nm 2000
Nguyn c Quyn
Sinh vin thit k: Nguyn
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Phn I Tng quan v ng c khng ng b v cc phng php iu chnh tc

I.tng quan v ng c khng ng b 1- c im chung: ng c in khng ng b l ng c in xoay chiu hai dy qun m ch c mt dy (s cp) nhn in t li vi tn s f1 khng i cn dy th hai (th cp) c ni tt hay ngn mch trn in tr. Dng in trn dy th cp c sinh ra nh hiu ng cm ng in t vi tn s f2, tn s f2 ny l hm ca tc gc rotor m tc ny ph thuc vo mmen quay trn trc. Ph bin nht ngi ta s dng my in khng ng b dy qun stator ba pha i xng c cc tnh xen ch ly in t mt li in xoay chiu v dy qun rotor ba pha hoc dy qun nhiu pha c cc t xen ch. Nhng loi my in nh vy gi l ng c in khng ng b. Cc my in khng nh b kiu khc c gi l My in khng ng b c bit cc my khng ng b c th lm ng c hoc my pht in nhng ch yu dng lm ng c, cn dng lm my pht th rt hn hu, ng c khng ng b xoay chiu l thng dng nht. Trong phm vi ca n ny ta khng i su vo my pht m ch ch yu i su v ng c khng ng b (CKB). 2. Tm tt qu trnh pht trin ca CKB. Nguyn l lm vic dng ca ng c khng ng b da trn hin tngt tnh quay m F..Arag pht hin nm 1824 v sau c M.Faray gii thch vo nm 1931. Nhng trong nhng th nghim ca Arag a bng ng c chuyn ng nh nam chm quay ch khng phi t trng quay to nn bng thit b ng yn l stator nh cc my in hin ny. Mt thi gian di hin tng Arag khng c ng dng vo thc t. Ch n nm 1979 U.Bli ( ngi Anh ) mi pht hin ra 1 dng c trong vic dch chuyn trong khng gian ca t trng nh 1 thit b ng yn gm 4 nam chm in c t trn khong cch nh nhau so vi tr quay ca a bnh ng. to nn t trng quay ngi ta s dng mt thit b chuyn mch c bit cung cp cho nam chm in nhng xung in mt chiu c bin v chiu tng ng. Nh bc hc ngi Gfecrarix, nh bc hc ngi Nam t Ntecla sng v lm vic ch yu M pht hin ra hin tng t Sinh vin thit k: Nguyn
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trng quay nh nhn thc hin nay vo nm 1888 mt cch c lp nhau. H chng minh thnh cng hai cun dy t vung gc vi nhau v c cung cp cc dng in hnh sin lch nhau 900 s to ra c t trng quay, vector cm ng ca t trng ny t im giao nhau ca trc cc cun dy s quay u khng thay i bin . Nhng ng c hai pha ca Fecrarix c mch t h cn rotor l 1 hnh tr bng ng sinh ra cng sut tt c chng 3w ch gn ging vi ng c khng ng b hin nay. Mt khc xut pht t cc gi thit khng ng v vic cn thit phi vn hnh. ng c vi cng sut cc i Fecrarix kt lun sai lm rng hiu sut ca ng c khng ng b khng vt qu 50%. iu ny lm gim i mt cch r rt vic quan tm nhng cng trnh ca fecrarix v 1 thi gian s pht trin mang tnh k thut ca ng b km hm. Cc ng c khng ng b do Ntesla thit k c hng westing haox nhn ch to. Nhc im ca ng c fteslamaf sau ny n phi chu thua km ng c ba pha l vic dng dy qun tp trung stator v rto ca my. iu lm cho vic m my xu i v mmen ph thuc r rt vo v tr ban u ca rotor. Vic pht minh ra ng c khng ng b 3 pha v nhng c im v kt cu c tnh nguyn l vn c gi n ngy nay gn lin vi tn tui ca M.D.aliv abrovlxki sau khi tm hiu nhngkt lun ca fecrarix v tnh khng trin vng ca ng c khng ng b, ng khng ng vi quan im trn v bt tay vo nghin cu ng c khng ng b c cung cp t h 3 pha do ng ngh. Trong mt thi gian ngn nghin cu theo hng ny hon thnh mt pht minh rt quan trng l ng c khng ng b ba pha c : rotor kiu lng sc (nm 1889), stator c kiu dy qun di kiu hnh trng, cn loi rotor c dy qun ba pha c ly ra t cc vnh trt, bin tr m my c a vo rotor khi m my ng c (nm 1890). Vy nguyn l c pht hin nm 1824 v nm 1890. ng c khng ng b ngy cng c xy dng v pht trin v kt cu, cng sut, cng ngh v ngy cc cng pht huy c tc dng, cng hon thin hn cho ti ngy nay. ng c khng ng b c s dng rt rng ri trong thc t bi u im ni bt ca n l : cu to n gin hn so vi ng c in mt chiu th gi thnh h, vn hnh tin cy chc chn. Ngoi ra cc nh my in hin nay thng sn sut ra li in xoay chiu do vy khng cn trang b cc b bin i km theo. Tuy nhin hin nay vi s pht trin nh v bo ca k thut vi tnh c bit l k thut vi s l tn hiu (Digital Signal processer ). cho php gii quyt cc thut ton phc tp iu khin ng c xoay chiu ba pha, trong thi gian thc vi cht lng iu khin cao. iu dn n xu hng thay th trit truyn ng in mt chiu bi truyn ng in ba pha c mi u im ca truyn ng in mt chiu v t phi bo dng d ch to cng mi u im khc. Trang 4 Sinh vin thit k: Nguyn c Quyn
II. cc phng php iu chnh tc ng c khng ng b Bn cnh cc truyn ng v thu lc, dng kh nn, truyn ng in ngy cng c s dng rng ri v tr thnh khu khng th thiu c trong cc dy truyn sn sut, c bit l trong cc dy sn sut truyn t ng hin i. u c chuyn ng hc (cc dy truyn, tay my, ngi my ...). u c ng c in lm thit b trung gian chuyn ho in nng thnh c nng vi nhng c tnh cn thit. Vic iu chnh chnh xc dng c nng to nn cc chuyn ng phc tp ca dy truyn cng ngh nng cao cht lng v gim gi thnh sn phm l nhim v ca h thng truyn ng. Bi vy truyn ng in l mn khoa hc ng dng cc kiu thc mi nht, ca l thuyt iu khin cc cng ngh mi nht ca vi in t vi tnh, nhm gn cho ng c cc tnh nng cao p ng c cc i hi ngy cng cao ca t ng ho t ra cho cc h truyn ng. Khc vi ng c in mt chiu ng c khng ng b c cu to phn cm ng khng tch bit t thng ng cng nh mmen ng c sinh ra ph thuc vo nhiu tham s. Do vy h iu chnh t ng truyn ng in ng c khng ng b l h iu chnh nhiu tham s c tnh phi tuyn mnh. nh gi cht lng iu chnh tc , trn ca iu chnh tc , s ho hp gia c tnh iu chnh vi c tnh ti ... s dng tt ng c c iu chnh tc ta cn quan tm n cc ch tiu so vi cc thng s nh mc . iu chnh c tc ng c khng ng b ta xt n ng c tnh c ca ng c khng ng b cng cc thng s nh hng. ng c tnh c ca ng c khng ng b l quan h gia tc quay v mmen ng c : = f(M). S
1 Sdm Sth 0 Mdm Mth M
Hnh 1. c tnh c ca ng c khng ng b Sinh vin thit k: Nguyn
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Phng trnh c tnh c :
3.U 2 f1R 2 s M= R 1S[(R 1 + 2 ) 2 + X 2 ] nm S

Ta thy cc thng s nh hng n c tnh c ca CKB l : - nh hng in tr in khng stator ( ni thm in tr ph R1 v in khng X1 vo mch stator). - nh hng ca mch rotor ( ni thm in tr R2 v in khng X2 vo rotor vi ng c rotor dy qun). - nh hng ca s suy gim in p li cp cho ng c. - nh hng ca trt S. - nh hng ca tn s li in cp cho ng c. - Ngoi ra vic thay i s i cc s thay i tc ng c v nh hng n c tnh c(xy ra vi ng c nhiu tc ). Hin nay trong cng nghip thng s dng 4 h truyn ng iu chnh tc l : + iu chnh in p cp cho ng c. + iu chnh in tr rotor. + iu chnh cng sut trt. + iu chnh tn s ngun. Trong phm vi ca n ny ch yu ta i su nghin c phng php iu chnh tc ng c bng cch thay i tn s ngun bng cc b bin tn, cn cc phng php khc ch nu qua. 1. iu chnh tc ng c khng ng b bng cch thay i in p U1. Mmen quay ca ng c t l vi bnh phng in p stator (M = U2 ) do c th iu chnh c mmen v tc ng c khng ng b bng cch iu chnh gi tr in p stotor trong khi gi nguyn tn s. Khi thay i U1 gi nguyn c tnh ti MN = f(S) c tnh c bin i dn n lm thay i h s trt. Khi ti khng i ngha l khi M = const h s trt bin i coi nh t l nghch vi bnh phng in p S = 1/ U2 cch s trt khc nhau ng vi ccUm khc nhau. iu chnh in p khng ng b ta phi dng cc b bin i in p. Sinh vin thit k: Nguyn
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S nguyn l ca h iu chnh nh sau: Rf Uli

Bin i in p xoay chiu
Ub
Rf Rf
Nu coi in p xoay chiu l ngun p l tng th theo biu thc mmen ti hn ta c quan h sau:
M thu U b = M th U dmm
2
(1)
Nu tc ca ng c l khng i : = const th MN = MU /Mgh cng thc (1) ng vi mi gi tr in p mmen trong : - Um l in p nh mc ca ng c. - Ub l u ra ca in p xoay chiu. - Mth l mmen ti hn khi in p l nh mc. - Mu l mmen ti hn khi in p l in p iu chnh. - Mgh l mmen ng Um v in tr ph Rf. V gi tr ca trt ti hn Sth ca c tnh c t nhin l nh nn khng p dng cho ng c khng ng b rotor lng sc. Khi thc hin cho ng c rotor dy qun ta cn ni thm Rf m rng gii hn iu chnh tc v mmen. Lc cng c tnh c gim i trong tc khng ti l tng ca mi c tnh c u nh nhau v bng tc t trng quay. Ta c ng c tnh c nh sau: 1 Sth Sthgh Mc() 0 Sinh vin thit k: Nguyn M Mth2 Mth1 Mth Trang 7 Ub2 Ub1 Um ,Rf = 0 Um ,Rf 0
c Quyn
Nhng phng php ny c nhc im l gim kh nng qu ti ca ng c di iu chnh tc hp (t S = 0 Sm), tng tn hao rotor (P2 = Pt .S = S .M . 1 , 1 bng tc t trng quay) nn phng php ny ch yu c p dng cho nhng ng c cng sut va v nh c h s trt ti hn Sth ln. 2- Phng php iu chnh tc bng cch thay i in tr mch dy qun rotor. Khi a thm in tr ph Rf vo mch dy qun rotor th : Mth = const ; 1 = const v Sth =
Rr + Rf . R 12 + X 2 nm
Trong : - Sth l h s trt ti hn. - Rv l in tr rotor. - Rt l in tr a vo thm. - Xnm l in khng mch stator. T biu thc trn ta thy: khi Rf cng ln th S th cng ln m mmen ti hn khng i do ng c tnh c mm hn v c dng nh hnh v:
Mth
Um ,Rf = 0
Um ,Rf 0
Sth
Stth
Khi iu chnh tc bng phng php ny tn hao trong mch dy qun rotor (S. Pt) c phn phi gia bn thn ng c v bin tr iu chnh t l vi cc h s trt ln khi R2 = Rf v ch yu c pht ra trn bin tr. Sinh vin thit k: Nguyn
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Nu gi cho dng rotor khng i th mmen khng i v khng ph thuc vo tc ng c. V vy ta c th p dng phng php ny cho nhng truyn ng c mmen ti khng i. 3. iu chnh cng sut trt lm thay i tc ng c . Trong trng hp iu chnh tc ng c khng ng b bng cch lm mm c tnh v gi nguyn tc khng ti l tng th cng sut trt Ps = S.Pdt c tiu tn trn mch rotor. Cc h truyn ng cng sut ln tn hao ny l ng k v th va iu chnh c tc truyn ng va tn dng c cng sut trt ngi ta s dng cc s cng sut trt gi l s ni tng . C nhiu phng php xy dng h ni tng. Hin nay ngi ta thng dng h ni tng thyristor . Vi s ni tng dng thyristor th sc in ng rotor c qua b chnh lu thnh mt chiu qua cun khng cp cho nghch lu ph thuc. in p v tn s xoay chiu ca nghch lu l khng i do c xc nh bi tn s v in p ca li in. Gc iu chnh ca nghch lu thay i trong khong 900 1400. ln ca dng rotor hon ton ph thuc vo mmen ti m khng ph thuc vo gc iu khin ca nghch lu. ch khng ti l tng khi dng rotor bng 0 th tc khng ti tng s c tnh : S0 = Ur0 / Um Trong : Ur0 l sut in ng chnh lu khng ti pha rotor. Um l bin in p pha stator. Nn nu ta iu chnh c sc in ng chnh lu mch pha rotor th c th iu chnh c trt khng ti l tng tc l iu chnh c tc khng ti l tng m gi tr ca sc in ng chnh lu c iu chnh bi gc m ca nghch lu ph thuc. c tnh c ca h ni tng dng thiristor ng vi cc gc m khc nhau nh hnh v :
1 =900 2 =1000 n =1400
c tnh c t nhin
Mtht Sinh vin thit k: Nguyn
Mth
M Trang 9
c Quyn
Do in cm cun lc trong mch mt chiu c gi tr l hu hn nn dng mt chiu c th b gin nn c tnh on ny c dc ln. Mt khc do st p gy nn bi in tr stator. in tr mch mt chiu, in tr in khng ca my bin p cung cp cho mch nghch lu, st p do chuyn mch ca nghch lu v chnh lu cho nn cc c tnh c iu chnh u c cng v mmen ti hn nh hn c tnh c t nhin. Nguyn l iu cng sut trt thng p dng cho cc truyn ng cng sut ln v khi tit kim nng lng c ngha rt ln. Phm vi iu chnh tc khng ln lm v mmen ng c b gim khi tc thp . Mt khc cc h thng ln khi khi ng ng c thng dng in tr ph kiu cht lng sau chuyn sang ch iu chnh cng sut trt do nn p dng cho cc h thng c s ln khi ng v o chiu t. 4- iu chnh tc ng c khng ng b bng phng php thay i tn s. Cc h thng iu chnh tc ng c xoay chiu c yu cu cao v di iu chnh v tnh cht ng hc, ch c th thc hin c vi b bin tn. Cc h thng ny s dng ng c khng ng b rotor lng sc, kt cu n gin, vng chc gi thnh r, c th hot ng trong mi mi trng, k c mi trng c hot tnh cao nh nc .
100%
Gi thnh truyn ng 1 chiu Gi thnh phn iu khin
100%
Gi thnh truyn ng tn s
Gi thnh phn iu khin
50%
Gi thnh ng c
30%
Gi thnh ng c
P(kw) t(nm)
Biu so snh kinh t trn cng vi s pht trin ca v s tin b ca khoa hc k thut, ngy nay cho ta thy truyn ng tn s dn thay th trit truyn ng in mt chiu. Yu cu chnh i vi c tnh truyn ng tn s l : m bo Trang 10 Sinh vin thit k: Nguyn c Quyn
cng ca c tnh c v kh nng qu ti trong ton b trong to b di iu chnh tn s v ph ti. Ngoi ra cn c vi yu cu v iu chnh ti u trong ch tnh, th d nh yu cu m bo c mmen max ng vi dng in cho hoc m bo c mmen ln khi tc bng khng... Tu theo yu cu v nn kinh t ta c th xc nh c cc cu trc c bn ca h bin tn, ta c thn chia thnh cc loai b bin i l : bin tn trc tip v bin tn gin tip. 4.1 - Xt mt s lut bin i tn s. 4.1.1- Bin tn s-in p. Khi iu chnh tn s th tr khng v t thng dng in,...ca ng c thay i, m bo mt s ch tiu iu chnh m khng lm cho ng c b qu dng th phi iu chnh c in p. Vi h bin tn ngun p thng c yu cu gi cho kh nng qu ti v mmen l khng i trong sut di iu chnh tc . Mmen cc i m ng c sinh ra chnh l mmen ti hn, kh nng qu ti v mmen c quy nh bi h s qu ti mmen. Cc kho bn dn thng s dng cc van l tranzitor, cn cng sut ln thng s dng cc van thiristor. Vic ng ngt cc van thng c thc hin bng cc mch in ngt c bit nh t in hay thiristor ph.thi gian gn y c s dng cc van thiristor c bit l cc van kho c bng cc xung iu khin (GTO). Gi tr in p ca ng c c iu chnh hoc bi iu chnh bin in p mt chiu bng chnh lu iu khin hoc bi b bm xung p. in p cng c th iu chnh bng iu chnh thi gian ng ca cc van hoc iu bng ch rng cc xung p bng chnh nghch lu. Phng php iu chnh iu chnh t thng l phng php iu chnh c s dng rng ri nht trong cc truyn ng, nht l cc truyn ng cng sut nh. Do c u im ni bt l va iu chnh c in p, va lm sin ha in p t vo ng c. Vi s lng cc xung c th lm trit tiu cc sng hi bc cao. Ch nh mc l ch lm vic ti u ca ng c khng ng b, trong ch ny t thng l nh mc v mch t l ti a v cng sut. Lut iu chnh in p tn s l lut gn ng gi t thng khng i trong ton di iu chnh. Tuy nhin t thng trn ng c, trn mi c tnh cn ph thuc rt nhiu vo trt tc l mmn ti trn trc ng c. v th trong cc h iu chnh yu cu cht lng cao ta phi tm cch b t thng .
c Quyn
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Ta c mi quan h gia dng in stator v t thng rotor qua cng thc sau: L Tr = r I z = r 1 + (T.2 ) 2 ; Lm Rr Is : dng stator r : t thng rotor s : vn tc gc stator Lm : h cm gia stator v rotor Tr : hng s thi gian rotor Mun gi cho t thng khng i th dng in stator phi c u chnh theo trt l quy lut iu chnh t thng. Nguyn l iu chnh t thng ng c c gin tip thng qua in p ca ng c. NMC NL L K
Id
Ud /
D
1
D3
D5
S1
S3
S5
UAN
a
Ia
b c
ua
Ud
Ud /
D6 D4 D2 S6
S2
Uk
S nguyn l bin i tn s-in
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4.1.2- bin i tn s-dng in. Bin tn ngun dng c u im l tng c cng sut n v my, mch ng lc n gin m vn c th thc hin hm ti sinh ng c. Ngun in mt chiu cp cho nghch lu phi l ngun dng, tc l dng in khng ph thuc vo ti m ch ph thuc vo tn hiu iu khin. to ngun dng mt chiu thng dng chnh lu iu khin hoc bm xung p mt chiu c b iu chnh dng in c cu trc t l tch phn ( PI ) . Mch lc l in khng tuyn tnh c tr s in cm ln. Do c ngun dng 1 chiu nn vic chuyn mch cc van bn dn c th thc hin bng cc t trn van chuyn mch. h thng bin dng ngung dng th mmen cc i hay mmen ti hn khng ph thuc vo tn s m ch ph thuc vo bnh phng dng in stator khi bin tn ngun dng vic vi ti l ng c xoay chiu th in p ra ca ti xut hin cc xung nhn ti cc thi im chuyn mch gia cc pha. Trong thc t k thut s dng khng hon ton nh vy cn c cc mch kho cng bc van bn dn, bo m chuyn mch dng in gia cc pha mt cch chc chn trong phm vi iu chnh tn s v dng in rng . Tng t nh bin tn ngun p bin tn ngun dng cng c yu cu gi cho kh nng qu ti v mmen khng i trong sut di iu chnh. Ta c th dng cng thc dng stator: I z = r 1 + (T.2 ) 2 Lm Mt khc mi c tnh c ng vi mt gi tr c nh ca dng in stator, do ng c lm vic im ti hn ca c tnh, th kh nng sinh mmen l max. S nguyn l ca b bin i tn s-dng in c trnh by nh hnh v sau:
CL L T1 NL T3 T5 K
a K CL RI id
D1
D3
D5
ia b c ib
D6 D2
D4
T4
T6
T2
id
-id
c Quyn
Trang 13
Vi cu trc v thng s thch hp ca cc b iu chnh th phng php iu chnh tn s ngun p, h thng bin tn ngun dng cng c th iu chnh t thng bng cch iu chnh vector dng in. iu khc bit l trong h thng bin tn th dng in l lin tc v chuyn mch ca cc van l ph thuc ln nhau. 4.2.- Cc b bin i tn s. 4.2.1- B bin i tn s trc tip. loi bin tn trc tip in p xoay chiu u1( vi tn s f1 ) ch cn qua mt van chng lu l chuyn ngay ra ti vi tn s khc. V vy bin tn ny c hiu sut bin i nng lng cao nhng s lng van ln s mch van phc tp. Vic thay i tn s ra l phc tp v ph thuc vo tn s f1. Loi bin tn ny lm vic ch yu trong phm vi tn s ra f2 f1 . Tuy nhin trong l thuyt hay nguyn tc c th thit lp bin tn vi tn s f2 c th ln hn f1 nhng phc tp rt cao. Trong bin tn trc tip th ng cong in p ra l ng ghp ni cc on hnh sin ca in p ngun bng cch ni ti vo cc pha ca ngun mt cch lun phin nh cc van bn dn. Cc van bn daaxntrng bin tn c chuyn mch mt t nhin. Bin tn trc tip c hiu sut rt cao v ch c mt ln bin i nng lng v cho php thc hin hm ti sinh m khng cn cc mch ph cng c th thc hin iu chnh ip p v tn s u ra ca bin tn vi dng sng gn sin. Tuy nhin cng c nhiu nhc im nh : h s cng sut thp, di iu chnh tn s b gii hn trn ca ngung cung cp v iu khin chuyn mch ca cc van. Trong bin tn chuyn mch t nhin mi na sng dn ca cc van thyristor ,v vy tn s c bn ca dng ti c ly cc gi tr gin on. V nguyn tc cng c th iu chnh trn cc tn s in p ra, tuy vy khi thng sut hin s mt i xng gia na sng dng v na sng m ca chu k dng v p. cc tn s khng ti l c s ca tn s ngun c th xut hin bin iu tn s thp. Vic iu chnh in p ra ca bin tn trc tip c thc hin bng cch thay i gc pha xung, trong ch chnh lu v nghch lu ph thuc. th in p u ra c cha ph bn cc sng hi bc l, bin cc sng hi t l nghch vi bin ca chng. Nu ti ba pha i xng v khng dy trung tnh th in p ra khng cha thnh phn sng hi bc hai v bc ba. Di y l s ba pha ca bin tn trc tip 6 nh xung v 3 nh xung. Sinh vin thit k: Nguyn
c Quyn
Trang 14
Ngun ba pha A B C pha 1 pha 1
pha 2 A B C
pha 2
pha 2 pha 3
N S ch s 3 S ch s 6
4.2.2- B bin i tn s gin tip. Trong bin tn loi ny in p xoay chiu u tin c bin thnh in p mt chiu nh chnh lu sau qua b lc v cui cng thnh in p xoay chiu nh nghch lu. Bin tn loi ny cho php bin i v tn s f2 khng ph thuc vo tn s trong mt di rng c trn v di f1 ch ph thuc vo mch iu khin. Sinh vin thit k: Nguyn
c Quyn
Trang 15
Tn s v hnh dng in p xoay chiu u ra cung cp cho mch ti do nghch lu quyt nh. Cn c vo tnh cht ca mch 1 chiu cung cp cho nghch lu m ta c 2 loi nghch lu: nghch lu p v nghch lu dng. Nghch lu p: l b nghch lu nh hnh in p trn ti l nhng xung vung, cn hnh dng dng in v gc pha ca ti ph thuc vo thng s ca ti. Ngun cung cp cho nghch lu p l ngung in p 1 chiu, n nh nh mt b chnh lu c cc t song song vi dung lng ln u ra, s nguyn l ca nghch lu p nh sau: L
CL
C=
NL
T c = trong s thc hin vic tr nng lng phn khng ca ti v ngun in 1 chiu khi ti khng phi l ti thun tr v duy tr in p cung cp cho nghch lu l n nh. in p ra ca nghch lu p khng c dng hnh sin m a s l dng ch nht cc van bn dn chuyn dng tranzitor ngi ta tnh dng tristor. Nghch lu p 3 pha ngi ta thng dng s cu, qu trnh in t trong nghch lu ph thuc vo nhiu yu t khc nhau nh c tnh ti, cch u ti, kiu u bin p ra ngun cung cp v nguyn tc iu khin. Nghch lu dng : l l b nghch lu nh hnh dng in trn ti l nhng xung vung cn hnh dng in p v gc pha ca ti th do thng s ca ti quyt nh. Trong cng nghip ngi ta thng s dng chnh lu c mc cun khng c gi tr d ln u ra. L=
CL
NL
Dng in ti c dng nht hnh sin v do tnh t ng ngt t quyt nh. Ngy nay vi s pht trin ca khoa hc k thut v ng dng k thut s, vi x l dng van tranzitor pht huy c nhng u im ni bt ca loi bin tn ny v a s cc bin tn l c khu trung gian l 1 chiu ngi ta trnh dng thyritor. Di y l s nguyn l b bin i tn s gin tip ngun p ba pha dng cc van dn l tranzitor. Trang 16 Sinh vin thit k: Nguyn c Quyn
T1
D1
T3
D3
T5
D5 ZA
C T6 D6 T4 D4 T2 D2
ZB ZC
III. kt lun
Nh vy : trong truyn ng in vi nhng ch tiu kinh t v k thut c u im ni bt ca ng c in khng ng b xoay chiu ba pha, ngy nay c s dng rng ri v dn thay th ng c in mt chiu trong lnh vc iu chnh truyn ng in xoay chiu. Bi v nhc im chnh ca ng c in khng ng b xoay chiu ba pha l iu chnh tc v khng ch qu trnh qu kh khn. Nhng ngy nay vi s pht trin ca khoa hc k thut th gii quyt c cc thut ton phc tp, iu chnh tc khng cn l kh khn. Trong cc phng php iu chnh tc ng c xoay chiu c yu cu cao v di iu chnh v tnh cht ng hc ch c th thc hin vi phng php iu chnh tn s bng thit b bin tn. Cc h thng ny s dng ng c khng ng b rotor lng sc c kt cu n gin v cc u im khc. Trong cc lut, cc phng php iu chnh tn s v y ng c in khng ng b xoay chiu ba pha nn khng tn ti cc quan h minh bch gia dng xoay chiu v t thng, gia dng xoay chiu v mmen nh ng c mt chiu. M y tn ti mt cu trc mch v cc i lng ba pha phc tp, th phng php iu chnh tn s da trn phng php iu chnh tc ng c bng phng php thay i tn s da trn phng thc thay i gc pha ca t thng rotor l ti u nht, bi n c cc u im ca phng php iu chnh tn s khc khng c c. Cng vi cc phng php iu chnh tc ng c khng ng b xoay chiu ba pha, phng php iu chnh theo hng ca t thng rotor s l phng php ch cht ca truyn ng in hin i trong nhng nm ti. Vi ti cho trong phm vi n ny ta s gii thiu ng dng ca phng php bin tn vector trong h thng cn bng nh lng ca nh my xi mng Lu X. Trong thc t phng php bin tn vector c ng dng rt rng ri trong nhiu lnh vc khc nhau. Mun vy trc ht ta i xy dng h thng iu khin bin tn to ra tn s mong mun t ngun in p mt chiu da trn c s ca phng php iu ch vector khng gian. Sinh vin thit k: Nguyn
c Quyn
Trang 17
Phn II M t cc i lng ba pha ca ng c khng ng b rotor lng sc

I. Xy dng vector khng gian ng c khng ng b rotor lng sc c ba cun dy stator vi dng in ba pha b tr trng khng gian tng qut nh sau.
Pha U
Pha V
Pha W
isu
isv
isw
Rotor
Stator
Hnh 2.1. M hnh n gin ca CKBBP rotor lng sc. Ta khng cn quan tm n vic ng c c cu to theo hnh sao hay hnh tam gic, y ba dng in isu, isv, isw l ba dng chy t li qua u ni vo ng c. Khi ng c chy bng bin tn, l ba dng u ra ca bin tn. Ba dng in tho mn phng trnh sau: (2.1) isu+ isv + isw = 0 Trong tng dng in tho mn cc cng thc sau: (2.2a) isu(t) = is.cos(st) isv(t) = is.cos(st + 1200) (2.2b) isw(t) = is.cos(st +2400) (2.2c) Sinh vin thit k: Nguyn
c Quyn
Trang 18
Mt cnh l tng th ba cun dy ca ng c xoay chiu ba pha t lch nhau 1 gc 1200 trn mt ct ngang nu trn mt phng ta thit lp mt h trc to phc vi trc i qua cun dy u ca ng c ta c th dng vector khng gian sau y: is(t) =
2 [ isu(t) +isv(t) ej120 + isw(t). ej240 ] = is.ej 3
(2.3)
Theo cng thc (2-3) vector is(t) l vector c mdul khng i quay ttrn mt phng phc (c hc) vi tc s v to vi trc thc (i qua trc cun dy pha U) mt gc = s(t) (ts l tn s mch stator). Vic xy dng vector khng gian is(t) c m t nh sau:
j 120
v w
0
u 1
is(t) Re
2 isu(t) 3 2 isv(t). e 3
j 120
0 0 2 isw(t). e j j 240 3
j 240
Hnh 2-2. Thit lp vector khng gian t cc i lng ba pha
Qua hnh (2-2) ta thy rng cc dng in ba pha chnh l hnh chiu ca vector mi thu c trn cun dy pha . i vi cc i lng khc ca ng c nh in p, dng rotor, t thng stator, t thng rotor u c th xy dng cc vector tng ng i vi dng in k trn. Ta t tn cho trc thc ca mt mt phng l cn trc o l , ta hy quan st hnh chiu ca cc vector dng trn xung hai trc , hai hnh chiu c t tn l is v is, trn hnh v sau y. Sinh vin thit k: Nguyn
c Quyn
Trang 19
Cun dy pha V i s 1200 isv isw is Cun dy pha U
1200
1200 isu = is
Cun dy pha W
Hnh2-3. Biu din dng stator di dng vector khng gian vi cc phn t is v is thuc h to stator c nh Hai dng k trn l hai dng hnh sin, ta c th hnh dung ra mt ng c in tng ng vi cun dy c nh , thay th cho ba cun dy U, V,W. Trn c s cng thc (2.1) km theo iu kin im trung tnh ca ba cun dy stator khng ni t, ta ch cn o hai trong ba dng stator l ta c y thng tin v vector is(t) vi cc thnh phn trong cng thc (2.4). Cng thc (2.4) ch ng khi trc cun dy pha U c chn lm trc quy chiu chun nh trong hnh v (2.3), iu ny c ngha trong ton b qu trnh xy dng h thng iu khin/ iu chnh sau ny: (2.4.a) is = isu is=
1 3
(isu + isv)
(2.4.b)
Tng t vi cc vector stator cc vector in p stator, dng rotor, t thng stator, t thng rotor u c biu din bng cc phn t thuc h to c nh(TST). (2-5a) is = is + jis us= us + jus (2-5b) ir = ir + ir (2-5c) r = r + r (2-5d) s = s + s (2-5e) Trang 20 Sinh vin thit k: Nguyn c Quyn
II. chuyn h to cho vector khng gian. Mc ch y l a ra cch quan st cc i lng vector trn h to statar v quan st trn h to quay ng b vi t thng stator . Xt h to tng qut xy hnh dung c mt h to khc l x*y* c chung im gc v lch i mt gc l V* so vi h to xy. Quan st 1 gc V bt k ta thu c : Trn h xy : Vxy = x + j y (2-6) * * * * * Trn h x y : v =x +jy (2-7) jy v y x y* x x Hnh 2-4. Chuyn h to cho vector khng T hnh (2-4) ta c x* = xcosv* + ysinv* (2-8.a) * * * y = -xsinv + ycosv (2-8.b) thay (2-8.a) v (2-8.b) vo (2-7) ta c: (2-9.a) v* = (xcosv* + ysinv*) + j(-xsinv*+ ycosv* ) * * xy -jv* = (x+jy)(cosv - jsinv ) = v . e (2-9.b) Mt cch tng qut ta thu c t (2-9.a) v (2-9.b) cng thc chuyn h sau y. (2-10) Vxy = V* . ejv* V* = Vxy e-jv* * * Hai h to xy v x y c coi l c nh hay gc lch pha v* c coi l khng i. Tuy nhin trong thc t v* c th l 1 gc bin thin vi tc gc dv * * = trng hp h to x*y* l h to qay trn vi tc dt * gc xung quanh im gc ca h to xy.
*
jy
x*
* =
dv * dt
c Quyn
Trang 21
dv dt Trong v l gc to bi trc rotor v trc chun (trc i qua cun dy pha U ). Ta m t vector dng stator is v cc vector t thng rtor r vi mdul dv v gc pha ngu nhin. Vector t thng vi tc : s = 2fs = dt Trong : fs l tn s mch in stator Ta thy hnh v di y vi trng hp CXCBP l ng c ng b th trc ca t thng rotor cng chnh l trc rotor, d ng c l loi kch thch ngoi hay vnh cu. Trong trng hp y ta c = s nu CXCBP l CKB th s chnh lch gia v s s to nn dng in vi tn s fs dng in cng c th biu din di dng vector ir quay vi tc gc =2f .
By gi gi s ta quan st CXCBP ang quay vi tc = j Cun dy pha V is i s isd Trc t thng rtor Trc rtor
r
jq
isq is
Cun dy pha U
Cun dy pha W
Hnh 2-5. Biu din vector khng gian trn h to t thng rotor cn gi l h to dq Sinh vin thit k: Nguyn
c Quyn
Trang 22
hnh (2-5) nu ta xy dng mt h to mi c trc thc c hng trng vi hng ca vc t r v gc to trng vi gc to , v t tn cho cc trc mi l dq. Ta thy rng h trc to mi ny quay xung quanh im gc vi tc gc s v vector is c cc phn t mi s l isd v isq. r nhn thy vector ang c quan st h to no th ta quy c thm 2 ch s f v s v h to t thng rotor th d iss : vector dng quan st trn h to , isf l vector dng quan st trn h to dq. T ta c : iss = is + jis (2.11.a) f is = isd + jisq (2.11.b) Nu bit gc is th ta c th r dng tnh c isf bng cng thc (2.12) isf = iss .e-js Hoc mt cch chi tit hn ta c: (2.13.a) isd = issins +iscoss (2.13.b) isq = iscoss +issins s Trong is cng nh cc phn t is v is c tnh bng cng thc (2.4.a) v (2.4.b) trn c s cc dng o c isu ,isv . Ton b qu trnh din gii trn c tng kt trn hnh sau.
Khin bin tn isd isq Pt(2.10) e-js i s i s Pt(2.4) 3 2 M 3 3 u
= Bin tn v w
CXBP
Hnh 2.6.thu thp cc gi tr dng stator trn h to t thng thc ca vector ng rotor Sinh vin thit k: Nguyn
c Quyn
Trang 23
Mt u im r nhn thy ca h to mi ch do cc vector is v r cng nh h to dq quay ng b vi nhau vi tc s quanh im gc, cc phn t ca vector (v d isd, isq) l cc i lng mt chiu. Trong ch vn hnh xc lp, cc phn t thm ch c th l khng i. Trong qu trnh qu chng c th bin thin theo thut ton c nh trc. Mt khc trn c s ca hnh (2.6) ta c th nhn thy ngay c kh khn ca vic xc nh isd , isq v gc s . trong trng hp ng c khng ng b gc s c xc nh r dng bng thit b o tc vng quay(my pht xung km vch 0, resolver). Trong trng hp CKB, gc
s c to nn bi tc gc s = + r , trong ch c c th o
c. Ngc li r = 2fr vi fr l tn s ca mch rotor ta cha bit. Vy phng php m t trn h to dq i hi phi xy dng c phng php tnh r mt cch chnh xc, l c s ca h thng iu khin/ iu chnh ta theo t thng rotor(vit tt : T4R). Mt cch tng t nh i vi vector dng stator, ta c th biu din tt c cc vector cn li trn h to dq. usf = usd + jusq isf = isd + jsq (2.14.a) (2.14.b) (2.14.c) (2.14.d)
rf = rd + jrq rf = sd +jsq
Ta r nhn thy trong phng trnh (2.14.c) c rd = 0 do trc q ng vung gc vi bn thn vector r . Tuy nhin trn thc t rt kh khn tnh tuyt i chnh xc gc s , do vy ta vn gi rq m bo tnh khch quan trong khi quan st. III. Khi qut u th ca vic m t ng c xoay chiu ba pha trn h to t thng rotor Vi CXCBP khng tn ti cc quan h minh bch ( gia dng v t thng , gia dng v mmen) nh ng c mt chiu m l cu trc mch v cc i lng phc tp. Phng php m t trn h dq cho php m t dn ti cc tng quan ging nh ng c mt chiu nhm t c cc tnh nng iu chnh/ iu khin ging nh ng c mt chiu. Sinh vin thit k: Nguyn
c Quyn
Trang 24
Sau khi xy dng vector khng gian cho cc i lng dng p t thng ng c v chuyn cc vector sang quan st trn h to dq ta thu c cc quan h n gin gia mmen quay, t thng v cc phn t ca vector dng stator.
r =
mM = Vi :
L M .i sd 1 + PTr
3 LM Pc . rs . isq 2 Lt
(2.16.a) (2.16.b)
rd - phn t d ca vector t thng rotor (cng chnh l mdul ca

vector). isd , isq : phn t d v q ca vector dng stator. MM Pc Tr P : mmen quay ca ng c. : s i cc ca ng c. : hng s thi gian ca rtor. : ton t laplace. Lr ,Lm : in cm rtor, h cm gia stator v rtor.
Phng trnh (2.16.a) cho ta thy l t thng rotor c th tng gim gin tip thng qua tng gim isd, quan h trong (2.16.a) l quan h tr bc nht vi hng s thi gian Tr . Nu thnh cng trong vic p t nhanh v chnh xc dng isd c th coi isd l i lng iu khin ca t thng rotor. Dng isd c gi l dng kch t v gi vai tr tng t nh ik (dng kch t) trong cng thc tnh t thng M = k.ik ca my in mt chiu. Nu bng dng isd thnh cng trong vic iu chnh n nh rd ti mi thi im cng tc ca ng c, ng thi thnh cng trong vic p t nhanh chnh xc dng isd theo phng trnh (2.16.b) s c th coi isd l i lng iu khin mmen ng c. Do isd c gi l dng to mmen quay, tng t iM ( dng phn ng ) trong cng thc tnh mmen quay mM = k. M .iM ca ng c mt chiu. Cc c tnh trn cho php ta xy dng cu trc h thng nh hnh v sau: Ton b qu trnh din gii trn c tng kt trn hnh sau.
c Quyn
Trang 25
rdisd usd
*
iu chnh dng iu chnh v tr MMisq
usq
Khin bin tn
= 3
u v Bin tn
iu chnh tc
M 3 o tc quay CXBP
Hnh 2.8.cu trc n gin h thng TXCBP trn c s ca phng php iu chnh ta t thng rotor. Trn c s cc tnh nng l tng n nay ta vn lun gi thit cho b iu chnh dng (vit tt CD) ta thu c mt cu trc h thng iu chnh hon ton ging nh h thng truyn ng in mt chiu. im khc bit l b CD. Tuy vy y l vn ginh cho phn sau ca n ny s cp n. Vi cch quan st mi ta khng cn quan tm n tng pha ring l na m quan tm n ton b vector: Ti tng im lm vic ca ng c vector is phi cp hai thnh phn thch hp isd iu khin t thng rotor , isq sn sinh mmen quay nh b iu chnh tc quay i hi. Vy vic chuyn cc vector quan st ln h to dq h to quay trn vi tc gc s = 2fs (fs l tn s mch stator) quanh im gc c trc thc i qua trc t thng rotor a ti cc quan h t l gia mmen quay, t thng v thnh phn ca vector dng stator, cho php xy dng h thng iu chnh truyn ng in tng t nh trng hp s dng ng c mt chiu. Qua cu trc h thng trn hnh (2-8)ta thy khu then cht m bo cho truyn ng xoay chiu ba pha t c cc tnh nng nh ng c in mt chiu l khu iu chnh dng. Khu bin tn c nhim v tnh chuyn cc phn t vector in p m b diu chnh dng i hi thnh thi gian ng ngt cc van bn dn ca bin tn. Phng php iu ch vector khng gian v l ni dung chnh ca n ny. Trang 26 Sinh vin thit k: Nguyn c Quyn
Phn III M hnh lin tc ca ng c khng ng b ba pha rotor lng sc

xy dng thit k b iu chnh cn phi c m hnh m t chnh xc n mc ti a i tng iu chnh. M m t cn phi th hin r cc c tnh v thi gian ca i tng iu chnh tuy nhin khng nhm mc ch m t chnh xc v mt ton hc i tng iu chnh m ch phc v cho vic xy dng thut ton iu chnh. V phng din ng c khng ng b c m t bi mt h phng trnh vi phn bc cao, v cu trc phn b cc cun dy phc tp v mt khng gian, v cc mch t mc vng ta phi chp nhn iu kin sau trong khi m hnh ho ng c. - cc cun dy stator c b tr i xng v mt khng gian. - Cc tn hao st t, v s bao ho t c th b qua. - Dng t ho v t trng c phn b hnh sin trn b mt khe t. - Cc gi tr in tr v in cm tm c coi l khng i. Trc chun ca mi quan st c quy c l trc i qua tm cun dy pha U. Ta s s dng cc m hnh trong khng gian trng thi r m t ng c. I. H phng trnh c bn ca ng c. Quan st cc i lng ba pha cch qua st ba pha kinh in ch khng phi l quan st di cc i lng vector ta s thu c ba phng trnh in p ca ba cun dy stator nh sau:
dsu ( t ) u SU ( t ) = R s .i su ( t ) + dt dsv ( t ) u sv ( t ) = R s .i sv ( t ) + dt dsw ( t ) u (t) = R i (t) + s sw sw dt
31.a 31.b 31.c
Vi cc su , sv , sw : t thng stator ca ba cun dy s, u,w. Rs : in tr cun pha stator. p dng cng thc is(t) = isej cho in p ta thu c : Sinh vin thit k: Nguyn
c Quyn
Trang 27
2 usv(t).e j120 +uswe j240 us = usu(t) + 3
(3.2)
Thay cc in p pha trong(3.1) v (3.2) ta thu c phng trnh in p stator di dng vector nh sau: us = Rs.is + Hay: us = uss = Rss.iss + ds dt dss dt (3.3) (3.4)
Tng t vi cun dy stator ta thu c phng trnh in p rotor nh sau : 0=Ri +

r r r
dss dt
(3.5)
vi Rr : in tr ca rotor d tnh quy i v stator. 0 : vector 0 (vector c mdul = 0 ). Cun ng c c cc in cm sau y: Lm : h cm rotor v stator. LS : in cm tiu tn pha cun dy stator. Lr : in cm tiu tn pha cun dy rotor ( quy i ). T cc in cm ta c thm cc thng s sau: : in cm stator. L = Lm + LS Lr = Lm + Lr : in cm rotor. : hng s thi gian stator. Ts = Ls/Rs Tr = Lr/Rs : hng s thi gian rotor. = 1-Lm2/ (LsLr) : h s tiu tn tng. T cc nh ngha trn ta c phng trnh t thng rotor v stator nh sau: s = isLs + irLm (3.6a) r = isLm + irLr (3.6b) Phng trnh mmen: 3 mm = Pc (s.is) = -3/2.Pc (r.ir) (3.7) 2 Sinh vin thit k: Nguyn
c Quyn
Trang 28
Phng trnh chuyn ng: jd mm = mT + Pc.dt Vi : mT l mmen ti, mmen cn. J l mmen qun tnh c.
(3.8)
l tc ca gc rotor.
By gi ta hnh dung mt h to vung gc quay trn xung quanh gc to chung vi tc k bt k v tm cch chuyn cc phng trnh va tm c sang h to k . V c bn ta c th chn c 1 trong 4 h to sau y c li v phng din vt l t dn n c li cho vic thit k h thng iu chnh/ iu khin. Bn h to l: 1. H to c nh stator (h ). 2. H to ta theo t thng rotor ( h dq). 3. H to ta theo t thng stator. 4. H to c nh rotor. Tuy nhin trong thc t hai h sau cng khng mang li ngha thc t g. do ta xt cc phng trnh in p rotor v stator trn hai h to
v dq.
I.1- Phng trnh in p stator. p dng cng thc chuyn h to (2.10) v sau mt s php bin i ta thu c phng trnh tng qut cho in p stator l :
USK = RS
isk+
dsk dt
JKSK
(3.9)
Vi h to : trng hp ny sy ra khi k = 0 phng trnh in p stator gia nguyn gi tr can n nh ban u v y l h duy nht ng c nh trong bn h k trn. Vi h to dq : trng hp ny sy ra khi k =s thay ch s k = f trong phng trnh (3.9) ta c:
USF = RS
isf+
dsf dt
JFSF
(3.10) Trang 29
c Quyn
I.2- Phng trnh in p rotor. Tng t nh in p stator ta p dng cng thc chuyn h to cho (2.10) v sau mt s php bin i ta c: 0=R i +
k R r
drk dt
JKRK
(3.11)
Vi h to : v h to nm c nh trn stator chuyn ng tng i so vi rotor bi tc gc . vy thu c phng trnh in p rotor trn h ta thay k = - vo phng trnh (3.11) ta c: 0=R i +
s R r
drs dt
JRS
(3.12)
Vi h to dq : Ta xt trng hp k =s - = r h to chuyn trng ng vt trc rotor bi tc gc r = 2f1 (f1 l tn s mch rotor chnh l h c trc trng vi trc t thng rotor h dq. Thay vo (3.11) ta c: 0 = RR irf+ drf dt
RRF
(3.13)
II. M hnh trng thi ca ng c trn h to stator. Vi cc vector thu c trong h to nh trong h phng trnh (2.5) ta c: Us (t) = Rsis +
s
dss dt drs dt JRS
(3.14a)
0= Ri
s r r
(3.14b) (3.14c) (3.14d) (3.15a) (3.15b) Trang 30
ss
= issLs + irsLm
rs = issLm + irsLr T phng trnh (3.14c) v (3.14d) ta rt ra c : isr = 1/ Lr ( rs -iss Lm ) ss = is Ls + Lm / Lr(rs - iss Lm)
c Quyn
Thay isr v ss vo 2 phng trnh u trong h s dng cc tham s , Ts, Tr sau vi php bin i ta thu c h sau: us = Rsis +Ls
s s
diss dt
s r
+ Lm
drs dt drs dt
(3.16a)
0 = - i Lm/ Tr+ (1/ Tr - j) +

s s
(3.16b)
Ta nh ngha thm hai i lng mi sau y: r = r/ Lm v r = r/ Lm Thay cc i lng mi nh ngha vo h phng trnh (3.16) v chuyn sang vit di dng cc phn t vector sau vi php bin i ta thu c h phng trnh: dis dt dis dt dr dt dr dt 1 1- 1- 1- 1
=-
Ts
1
Tr
1-
i s
Tr
r r
Ls
1
U s
(3.19a)
=1 =
Ts
is
Tr
1 Tr
i s
1- r
1-
Tr
Ls
Us
(3.19b) (3.19c)
Tr 1
r
1 Tr
Tr
i s
(3.19d)
Rt irs t phng trnh (3.6b) thay vo phng trnh (3.7) sau khi ta bin i th ta c c phng trnh mmen quay nh sau : mM = (3/ 2 ) . ( Lm2/ Lr)Pc(r iS - r is ) (3.18) H phng trnh (3.17) l m hnh m t y phn h thng in ca ng c khng ng b. Kt hp vi phng trnh (3.18) s m t y h thng c - in ca ng c khng ng b. H phng trnh (3.17) cn c vit di dng : dxs dt = As xs + Bs uss (3.19)
c Quyn
Trang 31
Phng trnh trn c gi l phng trnh trng thi ca ng c khng ng b. vi: uss : Vector i lng u vo (ng thi cng l vector in p stator) xs : Vector trng thi vi cc phn t l s thc. As : Ma trn h thng. B : Ma trn u vo. Cc vector mi nh ngha c dng nh sau :
xsT = [ is , is , r , r ] ussT = [us , us ]
Cc ma trn As , Bs c th vit di dng ma trn con vi cc cng thc c th sau : A11s A12s s A = A21s A22s (3.20) 1 1- 0 Ts Tr A11s = 1 1- 0 Ts Tr 1- ; 1 Tr 0 ; 1 Tr ; B2=
s
1 A22 =
s
Tr
1-
A12 =
Tr
1-

1-
1 A21 =
s
1 B1 =
s
Tr
Tr 0 B1s Bs2
Tr
0 0 0 0
0 1
Tr
(3.20) 0
B =
c Quyn
Trang 32
Cc m hnh trng thi c minh ho bng cc s sau: dxs (t) dt
Uss(t) Bs
xs (t)
diss (t) dt
As
Hnh 3.1. M hnh tng quan ca ng c khng ng b trong khng gian trng thi
As11 usr(t) iss (t)
s 1
As21
Na m hnh trn Na m hnh di As12 dsr (t) dt
As22
sr(t)
Hnh 3.2. M hnh trng thi ca ng c khng ng b minh ho bi cc ma trn con Trang 33
c Quyn
Hnh 3.2 cho ta nhn r hn tng quan bn trong ca ng c khng ng. Na m hnh trn l m hnh iu chnh dng vi 2 i lng u vo: in p stator trong vector t thng rotor c th coi l i lng ngu nhin c mdul bin thin chm. Na di c i u vo l dng stator cn u ra khng th o c l t thng rotor. M hnh cho thy r l i lng t thng rotor l i lng ra khng th o c song ( theo na m hnh di ) c th c lng c nu bit dng stator.
1- M hnh trng thi ca ng c khng ng b trn h trc to
dq. Ta tp hp cc phng trnh (3.10); (3.12); (3.6a); (3.6b) thnh mt h phng trnh m t ng c khng ng b trn h dq.
Us = R i +
f s s
dss dt drs dt irfLm
JRF
(3.22a)
0 = Rrirf
JR RF
(3.22b) (3.22c) (3.22d)
sf ss
= isfLs +
= isfLm + irfLr
T hai phng trnh cui ca h ta rt ra c: isf = 1/ Lr ( rf -isf Lm ) (3.23a) (3.23b)
sf = is Ls + Lm / Lr(rf -isf Lm)
Thay isf v sf vo 2 phng trnh u trong h ta thu c h phng trnh mi vi cc vector mi c nh ngha l:
rd = rd/ Lm
v rq = rq/ Lm
c Quyn
Trang 34
Sau khi ta bin i th ta thu c h mi l:
1 di sd 1 1 1 1 =-[ + isq+ rq+ rq + rq U (3.25a) L s sd dt Tr Tr Tr

di sq dt d rq dt
=-[ = =
1 1 1 1 1 1 + isq rq + rq + Usq (3.25b) Tr Tr Tr L s L s

(3.25c) (3.25d)
1 1 isq rq + ( s - ) rd Tr Tr
d rq dt
1 1 isq - ( s - ) rd rq Tr Tr
Tng t nh trong h to stator (), qua mt s php bin i ta thu c phng trnh tnh mmen trn c s dng stator v t thng rotor nh sau : mM = (3/ 2 ) . ( Lm2/ Lr)Pc(rd iSq ) (3.26)
H phng trnh (3.25) v kt hp vi phng trnh (3.26) ta thu c m hnh m t y ca ng c khng ng b trn h to dq(trong trng hp ng c c nui bi bin tn ngun p). H phng trnh (3.25) c th vit li c di dng M hnh trng thi phi tuyn yu nh sau: dxf dt vi : usf : Vector i lng u vo (ng thi cng l vector in p stator) vi cc phn t l s thc. xsf : Vector trng thi vi cc phn t l s thc. Af : Ma trn h thng. Bf : Ma trn u vo. N : Ma trn php phi tuyn Sinh vin thit k: Nguyn = Af xf + Bf ufs +N xf s (3.27)
c Quyn
Trang 35
s : i lng vo th 3 ngoi 2 phn t ca usf (= 2f )

Cc vector mi nh ngha c dng nh sau :
xfT = [ isd , isq , rd , rq ]

usfT = [usd , usq ] Cc ma trn As , Bs c th vit di dng ma trn con vi cc cng thc c th sau : 1 1- 1- 1-
Ts
0 Af = 1 Tr 0 1
Tr
1
0 1-
Tr
1-

1-
Ts
Tr
Tr
(3.28) - 1 Tr
0 1 Tr 0 0 1 -1 N= 0 (3.29) 0
Tr
1 0 0 0
0 0 0 -1
0 0 1 0
Tr
Bf = 0 0 0
(3.30)
Tr
0 0
c Quyn
Trang 36
M hnh trng thi c minh ho bng s sau:
Phn phi tuyn
N dxf (t) dt xf (t)
s
Bf usf(t) Af
Hnh 3.3. M hnh tng quan ca ng c khng ng b trong khng gian trng thi trn h to dq.
IV. Cu trc c bn ca 1 h truyn ng dng ng c khng ng b iu khin ta t thng rotor Trn c s ca ng c xy dng ta i xy dng cu trc c bn ng c da trn nguyn l ta theo t thng rotor ( T4R ). xy dng s cu trc ta xut pht t cc quan sau: 1- ng c c nui bng bin tn ngun p. iu c ngha : L i lng iu khin phi l ngun p, in p s thng qua khu iu ch vector khng gian ( cVTKG ) v bin tn t ln stator ng c. Khu iu ch vector khng gian c coi l khu truyn t trung thnh v pha, v bin v lc ny ta tm thi cha quan tm n. Nu in p c cho trc di dng usd v usq ta cn s dng khu chuyn h to in p (CTu) tnh chuyn sang us v us trc khi a ti khu iu ch vector khng gian. Nu l dng us v us th khng cn phi chuyn h to in p. Sinh vin thit k: Nguyn
c Quyn
Trang 37
2- Ta s dng mt khu iu chnh no nhm p t nhanh 2 dng isd v isq thc s bin chng thnh 2 i lng iu khin t thng rotor v mmen. l c vic ny ta s dng ring r 2 b iu chnh dng ring r Cid, Ciq kiu PI gii quyt vn t ra. Nhng ta cha c bit nu nh cc i lng u ra Cid v Ciq l yd v yq th kch c v n v nh cc i lng u vo, nh dng in v ghp chng vo chuyn to in p nh th no? v u vo ngoi gi tr phn hi cn mt gi tr cn nhn bit isd* v isq* c cung cp t u. gii quyt vn trn th ta phi : - Dng mt mng tnh cc gi tr usd v usq t yd v yq. Mng gi l mng tnh p (MTu). - Dng mt mng tnh dng(MTi) tnh isd* v isq* t t thng rotor rd* v mmen quay mM* . Xut pht t quan im xy dng h thng cu trc nh vy ta c th ghp cc khu l t thnh mt khi hon chnh nh sau:
rd*
isd* MT i isq* mM*
isd isq
CTu yd u MT sd usq e-js u
CVTKG us us 3 u is isu
3
yq Ciq isv v w
e K
-js
i s
CTi
* s
CXBP
M 3
khu tch phn
my o tc quay
Hnh 3.4. Cu trc kinh in ca mt h thng truyn dng dng CKB ngun nui bi bin tn ngun p iuchnh theo T4R.
c Quyn
Trang 38
Trong thc t cu trc kinh din trn ch lm vic tt trong ch tnh, ch xc lp, ch ng nh qu trnh qu cn bc l nhiu nhc im. Hin nay ngi ta s dng cu trc khc vi : Hai i lng isd , isq l 2 i lng khng c lp m l ph thuc ln nhau thng qua phn phi tuyn. V vy khu iu chnh dng phi c thit k di gic coi CKB l i tng iu chnh a thng s- s lng iu chnh ln hn 1, lc khu iu chnh dng s l khu iu chnh a thng s. Khu iu chnh a thng s lm nhim v p cho tng thnh phn dng cc c tnh truyn t cho trc - thng qua 2 cm iu chnh nhnh dcng thi c nhim v cch ly thng qua 2 cm iu chnh ni ti nhnh ngang- hai i lng khi tc ng ni ti ln nhau. Ngoi vic th hin khu iu chnh dng cu trc ny cn s dng trong h thng ba khu l : khu m hnh t thng(MHTT), khu iu chnh dng t thng(CTT) v khu dn t thng (DTT) S cu trc nh hnh v sau: DTT CTT
rd*
isd* C D
CTu usd usq e-js
CVTKG u s u s 3 u v w =
rd s
isd MH TT i s e
-js
3
isu isv
2
isq
i s
CTi
isq*
CXBP my o tc quay
M 3
CTQ Hnh 3.5. Cu trc i ca mt h thng truyn ng dng CKB nui bi bin tn ngun p iu chnh ta theo t thng rotor (T4R).
c Quyn
Trang 39
Phn IV M hnh gin on ca ng c xoay chiu ba pha

Chng ta bit rng vi x l cng nh my tnh l cc phn t khng lin tc m gin on. C ngha l cc thit b iu chnh/ iu khin trn c s vi x l ch c kh nng x l cc gi tr thc/ gi tr cn thu c cc thi im gin on tng ng. Ngay t nhng ngay u ca k thut iu khin trn c s vi x lu / my tnh tn ti v vn cn tn ti: Xu hng coi c i tng iu chnh/ iu khin ln thit b iu khin/ iu chnh l cc khu gn nh lin tc. iu cho php ta tn dng cc kinh nghim, cc phng php thu c trn c s ca cc phng php trn cc h thng tng t. Tuy nhin cc phng php trn tr ln lc hu, phng thc hp l l ta phi xy dng cc thut ton iu khin/ iu chnh ti cc thi im gin on cch u i tng c quan st. Vn l ta phi i tm cc phng php thu thp cc m hnh gin on cn thita k t cc m hnh lin tc xy dng. Mc ch ca vic xy dng cc m hnh gin on l thit k cc khu iu chnh. iu cho php ta n gin ho m hnh mt chng mc no trnhs vt kh nng ca vi x l / vi tnh. Cc sai s ca m hnh s c trit tiu bng phng php b hiu chnh. I .M hnh gin on ca ng c khng ng b trn h to stator. T m hnh lin tc ca ng c khng ng b ta c phng trnh trng thi nh sau: dxs dt vi : uss : vector i lng u vo (ng thi cng l vector in p stator) vi cc phn t l s thc. s x : vector trng thi vi cc phn t l s thc. As : ma trn h thng. Bf : ma trn u vo. Tch phn phng trnh trn trong phm vi gia hai thi im trch mu ta thu c m hnh gin on sau y ca ng c khng ng b ba pha. Sinh vin thit k: Nguyn = As xs + Bs uss (4.1)
c Quyn
Trang 40
xs (k+1) = s(,T) xs(k) +Hs (,T )uss(k)

Trong :
(4.2) (4.3a)
s(,T) = exp[As()T] =
( k +1)T
s (A ) T / !
=0
H (,T) = vi
kT
s(,)d Bs =

=1
(As) -1 s TB ! (4.3b)
k = 0,1,2,... T : l chu k trch mu. D nhn thy vector u vo uss s do vi tnh cung cp cc thi im cch u, v th cc phn t ca vector s c dng bc thang. Ma trn s c gi l ma trn trng thi, s ph thuc vo chu k trch mu T v tc gc ca ng c ging nh ma trn u vo Hs. Hai ma trn c dng hm s m c s e, bi vy c th khai trin thnh chui, vn l chui s ly n s hng th my, sai s s x l nh th no. trn thc t cho thy i vi h c tn s trch mu nh hn 400 s th ch cn s dng m hnh gn ng bc nht cho s v Hs l . H c chu k trch mu cng nh th h cng n nh, do ta phi tm tn s trch mu ph hp, p ng nhu cu v tn s cng tc n nhng thi ph hp vi kh nng tnh ton ca vi x l chn. T 1 1 1- Tr T 0 1 1 Ts 0 1- Tr (1-)T 1
Ts
Tr
1 T
(1-)T
Tr
(4.4a ) -T T 1 Tr
s =
1 Tr 0 0 T Tr 1 Tr
c Quyn
Trang 41
Ts
Hs= 0 0 0
0 T
Ts
0 0
(4.4b)
Hai ma trn (4.4a) v (4.4b) cng c biu din di dng ma trn con nh sau:
s11 s (,T) = 21
s
s12 22
s
Hs1 (4.4c) Hs(,T)= Hs2 (4.4d)
Vit li cc m hnh gin on di dng ma trn con cho php d dng nh gi cc qu trnh ct l xy ra trong ng c. so snh (4.3b) v (4.4b) cho thy H2s l mt ma trn rng vi cc phn t c gi tr = 0, do ta th vit li phng trnh (4.2) nh sau:
iss (k+1) = s11iss(k) + s12sr(k) + H1suss(k) sr (k+1) = s21is(k) + s22sr(k)
(4.5a) (4.5b)
H phnh trnh (4.5a.b) thu c, c th hin trn hnh (4-1) tng ng vi m hnh trng thi lin tc trn h .
c Quyn
Trang 42
s11
uss (k) H
s 1
iss (k+1) Z-1I iss (k)
s21
Na m hnh trn Na m hnh di
s12 rs (k+1)
Z-1I
sr(k)
s22
Hnh 4.1. M hnh gin on ca ng c khng ng b trn h trc (stator) M hnh trn l mt cng c rt quan trng dng : -Thit k khu iu chnh dng do c m hnh dng na trn -c lng t thng rs ca rtor bng m hnh t thng na di minh ho r rng kt qu va thu c . hnh 4.1 c chia thnh hai hnh (4.2a, 4.2b ) m hnh dng ca CKB ( hnh3.2a) c 2 tn hiu vo trong tn hiu rs c coi l i lng bin thin chm v s c b san bng ngay u vo ca khu CD sau ny.
c Quyn
Trang 43
sr(k) s12
uss(k) iss(k+1) H1
s
Z I
-1
iss(k)
Hnh 4.2a
s11
iss(k)
H1
sr(k+1)
Z I
-1
sr(k)
Hnh 4.2b
s22
Hnh 4.2: a, M hnh dng ca ng c khng ng b trn h to ; b, M hnh t thng ca ng c khng ng b trn h to Nh ta bit r l i lng khng th o c bng phng thc o thng thng. vi m hnh 4.2b ta c th gin tip thng qua dng stato is v tc gc c hc tnh ton (c lng ) r. v vy m hnh cn gi l m hnh i. (sau ny cn c m hnh i trn h trc to dq ) l chng lm vic chnh xc mi di tn s cng tc ca CKB, k c khi ng yn. v l do chng c s dng xuyn sut trong n ny . i vi ng c c cng sut ln ( di MW ), do cc tham s ca C (in tr stator ch di m ) rt b, khng khc cc thng s ca cp dn l bao. nng cao d chnh xc, loi tr nh hng ca tham s ngoi lai (ca cp dn ) v ca nhiu, ta c th dng khu quan st (observer ) c lng r . Cht lng tnh ton r ca cc khu quan st thng cao hn nhiu m hnh i . Sinh vin thit k: Nguyn
c Quyn
Trang 44
II M hnh gin on ca CKB trn h to t thng rotor Vic xy dng m hnh gin on ca (CKB ) h phi tuyn yu c th c thc hin mt cch n gin nu tho mn iu kin. Cc i lng u vo c coi nh l khng i trong phm vi mt chu k ly mu.ta ta bit rng iu kin trn ho ton tho mn vi cc h truyn ng hin i vi chu k ly mu nhhn 400 s. T m hnh lin tc ca ng c khng ng b ta thu c phng trnh m t trng thi phi tuyn ca ng c khng ng b nh sau: dxf dt vi : usf : vector i lng u vo (ng thi cng l vector in p stator) vi cc phn t l s thc. f x : vector trng thi vi cc phn t l s thc. Af : ma trn h thng. Bf : ma trn u vo. N : Ma trn phi tuyn. s : Di lng th ba ngoi hai phn t ca uf ( =2fs ) Tch phn phng trnh trn trong phm vi gia hai thi im trch mu ta thu c m hnh gin on sau y ca ng c khng ng b ba pha. = Af xf + Bf uf + NsXf
(4.7)
xf (k+1) = f(,s,T) xf(k) +Hf (,s,T )usf(k)

Trong :
(4.8) (4.9a)
(4.9b)
s (,s,T) = exp[ (Af +Ns)T]

( k +1)T
Hs (,s,T) = vi
kT
exp[ Af +Ns)]d BF
k = 0,1,2,... T : l chu k trch mu. M hnh (4.8)vi cc ma trn cng thc (4.9.a.b) chnh l mt m hnh tuyn tnh.Trong i lng u vo s tr thnh tham s ca h vi tnh cht bng trong phm vi mt chu k ly mu T vi tnh cht hm trong ton b di tn s cng tc ca ng c. Tng t cc ma trn (4.3.a.b) ca m hnh (4.2) ti y ta thu c cc cng thc gn ng bc nht thng qua gii tch hai hm s m c s e. ng thi biu din cc ma trn gn ng di dng cc ma trn con nh sau: Trang 45 Sinh vin thit k: Nguyn c Quyn
T 1
1- Tr T 1
Ts
-sT
sT
(1-)T
T
1- Tr 1 T
1 T
1 Ts
(1-)T
T
(4.10a) (s - )T T 1 Tr
s =
1 Tr 0 0 T Tr T 1 Tr
-(s-)T
Ls
Hs= 0 0 0
0 T
Ls
0 0
(4.10b)
Hai ma trn (4.10a) v (4.10b) cng c biu din di dng ma trn con nh sau:
f11 s (,s,T) = f21
f12
(4.11a)
Hf1 Hs(,s,T)= Hf2
(4.11b)
f22
Tng t nh h to m hnh (3.8) c th biu din chi tit hn di dng cc ma trn con nh sau:
ifs (k+1) = f11isf(k) + f12fr(k) + H1fusf(k) (4.12a) fr (k+1) = f21isf(k) + f22fr(k)

(4.12b)
Trang 46
c Quyn
H phnh trnh (4.5a.b) thu dc, c th hi trn hnh (4-1) tng ng vi m hnh trng thi lin tc trn h .
f11
ufs (k) H
f 1
isf (k+1) Z I
-1
ifs (k)
f21
Na m hnh trn Na m hnh di
f12 rf (k+1)
Z-1I
fr(k)
f22
Hnh 4.1. M hnh gin on ca ng c khng ng b trn h trc dq(to t thng)
fr(k) f12
ufs(k) isf(k+1) H
f 1
Z I
-1
ifs(k)
Hnh 4.3a
f11
c Quyn
Trang 47
ifs(k)
Hf1
fr(k+1)
Z-1I
fr(k)
Hnh 4.2b
f22
Hnh 4.2 a, M hnh dng ca ng c khng ng b trn h to dq. b, M hnh t thng ca ng c khng ng b trn h to dq
Qua cc hnh v trn ta thy s ging nhau v mt hnh thc gia cc m hnh: trn h to v dq. s gng nhau sau ny chnh l c s cho php xy dng khu iu ch dng(CD) d trn h to no theo mt phng php thng nht, trc khi chi tit ho cho tng trng hp p dng c th.tuy nhin cn lu cc im khc nhau c bn sau: 1. Cc thnh phn ca vector u vo(in p stator), vector trng thi (ngun dng stator) trn h to dq l cc lng 1 chiu v ch thay i khi ng c lm vic qu trnh qu (ch ng). Ngc li trn h l cc i lng hnh sin, chnh iu ny dn n s khc nhau c bn v cht lng truyn ng gia hai phng n iu chnh. 2. Quan st cc ma trn qu trng thi, ta s thy f c cha trong s cn trong s th khng. Ta d dng nhn thy rng y l yu t qua li gia dng isd v isq v y chnh l nguyn nhn lm cho cu trc kinh in ch lm vic tt ch tnh. Trong cu trc kinh in khng c kh nng kh trit tc ng qua li , dn n nh hng ln nhau gia hai qu trnh t ho v sn sinh mmen quay. III M h nh tng qut cho c hai loi ng c ng b v khng ng b. Trc ht ta quy c mt cch k hiu thng nht cho tt c cc phn t h thng, nhm mc ch x l, l thuyt chung cho tt c mi i tng, trc ht p dng voc th. 1. Cc ma trn qu trnh qu trng thi c k hiu chung l . 2. Cc ma trn u vo l H . 3. Cc ma trn (hoc vector) nhiu (c trng cho nh hng ca t thng rotor vo m hnh dng)l h . Sinh vin thit k: Nguyn
c Quyn
Trang 48
vi cc quy c mi ta thu c m hnh dng tng qut cho ng c xoay chiu ba pha sau y: Tch phn phng trnh trn trong phm vi gia hai thi im trch mu ta thu c m hnh gin on sau y ca ng c khng ng b ba pha.
is (k+1) = is(k)+ H us(k) + h(k)

m hnh (4.13) c th c vit tren mt phng Z nh sau: Zis(z) = is(z) +H us(z) + h(z)
(4.13)
(4.14)
vi phng trnh c trng sau y: det[zI - ] = 0

Trong I l ma trn n v. M hnh thng nht cho ng c xoay chiu ba pha c th hin trong hnh (4-5). thng qua phng trnh (4.14), ta s dn lm quen vi phng thc m t trn mt phng Z, mt cng c m t i tng ti cc thi im gin on cch u,phc v mc ch xy dng thut ton chy trn vi x l / vi tnh.
(p)
h ufs(k) isf(k+1) Hf Z I
-1
ifs(k)
f
Hnh 4.5. M hnh tng qut ca ng c xoay chiu ba pha trn h to bt k
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Phn V iu khin bin tn trn c s phng php iu ch vector khng gian

I. S nguyn l ca CXCbP nui bi bin tn ngun p Trong gii hn ca n ny ta ch yu nghin cu iu khin bin tn trn c s phng php iu ch vector khng gian. Trn hnh (5-1) cho thy s nguyn l ca CXCBP nui bi bin tn dng van bn dn. Thng thng, cc i van c x l/ vi tnh iu khin sao cho in p xoay chiu ba pha vi bin cho trc, vi tn s cng nh l gc pha cho trc, c t ln ba cc ca ng c ng theo yu cu. Bin tn c nui bi in p 1 chiu UMC . i tng bin tn c cp n y hot ng theo kiu ct xung vi tn s ct cao, phn bit vi hai loi hot ng theo nhp dnh cho cc h thng vi cng sut rt ln, ni m thi gian tr ng ngt ca van bn dn rt ln. Cc van bn dn y c dng ch yu l transistor (IGBT, MOSFET). y ta c th p dng mt cch hn ch i vi bin tn dng thyristor.
UMC
u2 v2 w2 CXCBP
Chm xung kch do vi x l gi n Hnh 5-1. S nguyn l ca CXCBP nui bi bin tn ngun p Sinh vin thit k: Nguyn
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Mi mt pha ca ng c ta c th nhn c trong 2 trng thi : 1 (ni vi cc + ca UMC) hoc 0 (ni vi cc - ca UMC). do 3 pha (3cp van bn dn) nn s tn ti 23 = 8 kh nng ni cc pha ca ng c vi UMC nh bng 5.1 sau:
s th t cun dy pha
1 1 0 0
2 1 1 0
3 0 1 0
4 0 1 1
5 0 0 1
6 1 0 1
7 1 1 1
Pha U Pha V Pha W
0 0 0
Ta quay tr li vi cch b tr hnh hc ca ba cun dy pha trn mt phng, ta thy rng t hp th 4 tng ng vi trng hp ta p t ln ba cun pha vector in p Us vi mdul 2UMC nh trong hnh 5-2b. tm in p thc s ri trn tng pha, ta ch vic tm hnh chiu ca vector Us ln trc ca cun dy.
cunn dy pha V USV UMC US USU cunn dy pha W Hnh a USU = -2UMC / 3 USU = USW= UMC / 3 US = 2UMC / 3 Hnh b cunn dy pha U USW
Hnh 5-2.a) s ni 3 cun dy pha theo kh nng th 4 ca bng 5.1 b) vector khng gian ng vi kh nng th 4 ca bng 5.1
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Tng t ta c th xy dng c vector in p tng ng cho cc trng hp cn li (hnh 5.3). cc vector chun c nh s U0 , U1 , ... , U7 nh s th t ca bng. y cn lu n 2 trng hp c bit : U0 c 3 cun dy pha ni vi cc - c 3 cun dy pha ni vi cc + U7 ca UMC . Hai vector c mdul bng khng v gi mt ngha rt quan trng sau ny. cun dy pha V
S2 U2
U3
S3 U3 Q0 U7 S4 U4 Q2 Q3 Q1 Q4
S1 U1 Cun dy pha U
S6
U5 cun dy pha W
S5
U6
Hnh 5.3. Tm vector do ba cp van bn dn ca bin tn to nn. Q1... Q2 : cc gc phn t . S1 ... S6 : cc gc phn su. Hnh5.3. Cho ta thy r rng v tr ca tng vector chun trong h to . Ta cn ghi nh rng, mdul ca tng vector lun c gi tr : 2UMC / 3 . Ngoi quy c thng thng v cc gc phn t Q1...Q4 (Q: quadrant) phn chia bi hai trc ca h to , cc vector chia ton b khng gian thnh cc gc phn su S1... S6 (S : sector). Ch bng tm vector chun ca hnh 5.3. Ta phi to nn in p stator vi bin , gc pha bt k m khu iu ch dng sau ny yu cu. Sinh vin thit k: Nguyn
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II . Nguyn l ca phng php iu ch vector khng gian Gi s ta phi thc hin vector uS bt k nh hnh v 5.4 sau vector c th nm gc phn su bt k no , y uS nm S1 . uS c th tch thnh tng hai vector con uP , ut ta theo hng ca hai vector chun u1 u2 . Cc index vit thp bn phi c ngha nh sau: p : l vector bn phi. t : l vector bn tri. Ta bit rng in p s phi c tnh i thnh thi gian ng ngt van trong phm vi mt chu k ct xung no . Gi thit ton b chu k l chu k c ch, c php dng thc hin vector khi mdul ti a cng khng th vt qua 2UMC / 3 . Do vyta c : u2 ut
us
uP u1 (5.2)
USmax = U1 = ... = U6 = 2/ 3 UMC
Nu thi gian ti a l T ta c nhn xt sau: 1.uS l tng ca 2 vector bin uP , ut : uS = uP + ut 2.Hai vector bin c th c thc hin bng cch thc hin u1 (cho uP) v u2 (cho ut) trong hai khong thi gian sau.
TP =
us T u s max
; Tt =
ut T u s max
(5.3)
tnh c TP ,Tt ta phi bit c mdul ca cc vector bin phi uP v bin tri ut . Xut pht im tnh mdul v gc pha ca vector do khu iu ch dng i hi. m y cn hai vn ta cha cp n l: 1.Thi gian thc hin cc vector bin phi TP v bin tri Tt . Vy th trong khong thi gian cn li T- (TP + Tt) bin tn lm g ? 2.Ta cha cp n vector no thc hin trc vector no thc hin sau. Ta c th tr li cho cu hi 1 : Trong khong thi gian - (TP + Tt) cn li bin tn thc hin mt trong 2 vector c mdul bng khng u0 hoc u7 . Sinh vin thit k: Nguyn
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Trn thc t ta thc hin cng vector sau:
u S = u p + u t + u 0 (u 7 ) uS = T T TP u1 + t u 2 + T T (TP + Tt ) u0 T
(5.4)
Trnh t thc hin 3 vector u1 , u2 hoc u7 c a ra trong bng sau: u1 u2 u7 u0 u v w 0 0 0 Bng5.2 y mi cp van ch chuyn mch mt ln. Nu nh trng thi cui cng l u0 , trnh t thc hin s l: u1 u2 u 7 Nu trng thi cui cng l u7 , trnh t s l: u2 u1 u 0 Bng phng php thc hin in p nh vy s gy tn hao ng ngt cc van ca bin tn mc t nht. Nu ta v ghp tng trng 2 chu k ni tip nhau thuc gc phn su th nht S1 trong hnh 5.5 ta c biu xung quen thuc ca phng php iu ch b rng xung thc hin bng k thut tng t ( analog ). 000 100 110 111 u v w u1 TP u2 Tt Tx / 2 u7 T7 u2 Tt u1 TP Tx / 2 Trang 54 u0 T0 110 100 000 100 Hnh 5.5. Biu xung ca vector in p thuc gc phn su th nht S1. Sinh vin thit k: Nguyn 1 0 0 1 1 0 1 1 1
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Trong tt c cc gc phn su cn li S1 ... S6 , cch thc hin l ging nh S1. Ta c th khi qut cc vector .
u u3 uS u2 v w ut uP T1 TP T7 TP T1 To u3 u2 u7 u2 u 3 u0 Tx Hnh 5.5a Biu xung kch thch thuc S2
u3
u v
uS
uP
w TP Tt T7 Tt TP To u4 u3 u0 Tx
u4
ut
u3 u4 u7
Hnh 5.5b Biu xung kch thch thuc S3
uP u4 uS ut u u5
w v w Tt TP T7 u5 u4 u7 Tx TP Tt To
u4 u5 u 0
Hnh 5.5c Biu xung kch thch thuc S4 Sinh vin thit k: Nguyn
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uP
ut
u v w
uS
u6
TP Tt T7 Tt TP To u5 u6 u7 u6 u5 u0 Tx
Hnh 5.5d Biu xung kicks thch thuc S5 ut
u1
u v
uP
uS
w Tt TP T7 TP Tt To u1 u6 u7 u 6 u1 u0 Tx
u6
Hnh 5.5e Biu xn kch thch thuc S6
Vic tch vector in p us thnh hai vector bin c hng nh hai vector chun k bn vi us, dng nh cho php thc hin us mi gc ca khng gian. iu ny thc ra khng ng, cn nhiu hn ch khng cho php tn dng 100% kh nng ti a. Ngoi ra ta ch c th tnh ton vi chnh xc hu hn( phn gii l 16 Bit hoc 32 Bit), hoc nu o tc quay bng my o gc tuyt i(RESOLVER) th cng s vp phi hn ch v phn gii. Nhim v ca ngi k s thit k lafphair nhn bit y cc kh nng hn ch a ra cc gii php k thut ti u nht trong phm vi kinh t cho php. Hnh 5.6 cho ta thy dng ca dng p to theo phng php CVTKG vi tn s ct xung fx = 1/ Tx. Sinh vin thit k: Nguyn
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III . Cch tnh v thc hin thi gian ng ngt van bn dn ca bin tn Xt cho cng khi thc hin nguyn l CVTKG ta phi tra li hai cu hi khi iu khin cc van bn dn : cc van cn phi c chuyn
mch nh th no v trng thi tn ti trong bao lu? nu

nh mdul v gc pha ca in p cho trc. Bng cch thng tin v gc pha cng nh v tr (gc phn t, gc phn su) ca vector in p ta c th tr li cu hi u tin vi s gip ca phn 5.1. Cc cng thc (5.2) v (5.3) cho ta thy vic tnh ton thi gian hon ton ph thuc vo cc thng tin v mdul ca cc vector uP, ut . Vector in p stator uS thng c cho bit trc di mt trong hai dng sau: -Hai thnh phn mt chiu usd v usq trn h to T4R gc pha gm c S ca h to (hnh 2.5) cng vi gc ring ca uS ( so vi trc d) theo cng thc sau:
u = S + arctg
uSd uSq
(5.4)
-Thnh phn uS , uS . dng ny thng tin v gc pha tn tai khng tng minh m n trong uS , uS. V l do cng tn ti 2 phng php tnh mdul ca UP v Ut nh sau: III.1- Phng php 1: Xt vector uS bt k thuc gc phn su th nht nh trong (hnh 5.5a,b). Trn c s cng thc (5.3) ta tnh c Su do tnh c . Mdul ca cc vector bin phi, bin tri s c nh sau: 2 uP = us sin(600-)
3
2 ; uP = = t
us sin()
3
(5.5a,b)
Mdul ca uS trong (5.5a,b) c tnh nh sau: uP = Sinh vin thit k: Nguyn

2 2 U Sd + U q
(5.6)
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III.2 Phng php 2: Cc vector bin phi, bin tri c tnh trc tip t uS , uS theo cng thc sau: uP = uPS =u 2 uS
3
2 uS ; uP == t
3
(5.6a,b)
q u2
usq u s ut uS u1 us d
uP usd
Hnh 5.6 Cc kh nng cho bit trc v veector in p stator us iu cn phi lu trong hai phng php k trn : phng php 1 vi cng thc(5.5a,b) v (5.6) c hiu lc trong ton b khng gian vector. Ngc li cng thc (5.7a,b) ch c gi tr i vi S1 theo phng php hai, ta s phi tu theo gc phn t, gc phn su c th m p dng cc cng thc thuc bng 5.3. Cc hia phng php u c th s dng mt cch thun li nh nhau, vic la chn phng php no l ph thuc ch yu vo phn cng v h to (dq hay ) ca cu trc iu khin / iu chnh.
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uP
ut 2 uS
3
S1
Q1Q1
uS -
uSS u
3
Q1 S2 Q2
uS +
uS
3
- uS +
uS
3
- uS +
uS
3
uS +
uS
3
2 uS S3 Q2
3
uS -
uS
3
S1 4
Q13 Q
uSS - u
uSS u
33
2 uS
3
Q3 S5 Q4
uS +
uS
3
- uS +
uS
3
- uS +
uS
3
uS +
uS
3
S6
2 uS Q4
3
uS -
uS
3
Bng 5.3: Mdul ca vector bin tri v bin phi tnh bng cc thnh phn in p us ,us . Vic p dng phng php hai tng chng phc tp hn do phi dng nhiu cng thc khc nhau trong bng (5.3). Nhng quan st k ta Trang 59 Sinh vin thit k: Nguyn c Quyn
thy tt c quy t v 3 cng thc sau:
a = uS +
2 u S 3
; b = uS -
2 u S 3
; c=
2 u S 3
(5.6a,b, c)
Trong c 3 cng thc u khng cha php chia nh php tnh lng gic, v th p dng s c li nhiu v thi gian tnh. Vn tn ti l cn phi bit us nm gc phn t, gc phn su no ca khng gian vector la chn ng cng thc y l vic khng kh khn ta ch vic p dng suy ngh sau: 1- Bng vic xt du ca us v us ta c th r dng nhn bit vector us nm gc phn t th my. 2- Biu thc tnh b trong (5.8,b) s i du mi khi vector us i qua ranh gii gia hai gc phn su bt k. Sau khi bit gc phn t ( bc 1) bng vic xt du bin i ta c th bit c gc phn su c th thuc gc phn t .
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phn VI Cc vn ta theo t thng rotor, o c gi tr thc v xc nh gi tr cn

I. Cc vn ta theo t thng rotor. Ta bit c s ca vic thc hin cc h thng truyn ng in XCBP hin i l phng php ta theo t thng rotor( vit tt l:T4R).Thc cht ca k thut T4R l k thut iu khin/ iu chnh CXCBP, xy dng trn c s phng php m t cc i lng ca CXCBP trn h to quay trn c trc thc nm dc theo trc ca vector t thng (iu khin iu chnh ta theo t thng rotor) . lm c iu Ta cn phi bit hng( bit gc pha, cn gi l ta ) ca vector t thng rotor. i vi ng c ng b, gc pha ca t thng rotor q-do h thng kch t gn c nh trn rotor- chnh l gc c hc ca rotor do c th o c r rng bng cch tch phn hiu ra ca my o tc quay vi t gi tr ban u no . Nhng ng c khng ng b . c th bit hng ta cn phi bit mdul ca vector t thng l mt i lng khng th o c m phi dng phng php c lng. Vic tm gii php c lng chnh xc mdul r ca CKB v ang l n lc ca nhiu cng trnh nghin cu vi s gip ca khoa hc k thut. Di y ta khi qut cc gii php tn ti trong nghin cu v nhn mnh cc gii php c th s dng thc tin trong cng nghip.
* Cc phng php c lng t thng rotor

Xut pht im ca phng php c lng l cc i lng o c c chia lm 3 nhm chnh nh sau: + Phng php dng m hnh ng c thu c t cc phng trnh m t m ta xy dng phn III v IV. Do cc m hnh nh trn khng trnh khi cc sai s (sai s ban u ca m hnh, sai s do nng hoc do bin ng t ho ca ng c). Chng c b sung thm cc khu hiu ca ng c : dn stator, p stator, tc quay rotor. Cc phng php chnh sai s trn c s so snh cc i lng u vo ca m hnh vi cc i lng o c ca h thng. M hnh m rng c gi l khu quan st (observer stator estimator) . Sinh vin thit k: Nguyn
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+ Do vic o c cc gi tr i lng u vo (v d nh dng stator) khng trnh khi tp m o lng. Do vic m t i tng ng c c sai s hu hn, khu quan st trn c ni rng lc ny c gi l khu lc kalman(kalman filter). Ni chung c lng t thng rotor l mt phm vi nghin cu rng v a dng. V n ny khng th m cng khng t ra mc ch gii thiu tt c cc gii php nghin cu v ang tn ti. II. Tnh dng kch t v t thng rotor: Trong phm vi n ta xt t thng rotor rd khng i hi nn dng kch t IsdN s c vo t bn phm v tnh theo cng thc sau: (6.1) PN =3Upha.I phacos Dng danh nh : ISN = 2 IN=
2 I 2 + I sqN dN
(6.2)
in khng phc 1 pha ca ng c U ZN = pha I pha in tr phc1 pha ca ng c: Rr = S.ZN Trong :
(6.3)
(6.4) (6.5) (6.6) (6.7)
S = r / s
PN = 3(Rr/S)I2sqN thay (50 vo (6.4) I2sqN = PN/ 3.ZN Thay (6.1) vo (6.6) I2sqN = 3Upha.Ipha.cos / 3ZN
3ZN IsdNc th tnh gn ng theo cng thc sau(vit sau) Mt khc theo cng thc ta c: V rd v IsdN khng i nn rd =LmIsdn
III. Cch tnh gc pha ca t thng rotor: III.1. Tnh gc pha trn h trc to . s ca vecto t thng r c tnh bng php tch phn tc gc s ca vecto s=s(o) + s dt. Trong s(0) l v tr ban u ca r v c gi tr bng 0. Cn tc sgm hai thnh phn: Sinh vin thit k: Nguyn
Thay (6.7) vo (6.2) IsdN =
I2 N
3U pha I pha cos
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s = + R = 2Pc nM /60
Pc : S i cc ca ng c NM: Tc quay c ng c T phng trnh(4.2)ta c:
r(k) = isq(k)/Tr,rd(k)
Nh vy ta thy vector t thng r quay trn vi tc s trong trng hp + Khng ti: Isq=0 Vy r=r quay vi tc bng tc gc ca rotor. + c ti:Isq0.Vy r0 r vt trc rotor mt gc ng vi r III.2. Hiu chnh gc pha s c dng vo vic chuyn h to cho dng v p: Gc pha s c dng vo vic chuyn h to cho dng Is v p Us , gc pha s khng th dng c mt cch ging nhau cho c 2 i lng dng v p. Ta bit gc s(k) c tnh t Is(k), nh vy u nhp(k) ta cha th c gc mi s(k) m ch tnh c gc c s(k-1), chuyn i h to cho dng Is(k)mi o c.Trong khong thi gian nhp th (k-1) tri qua vecto r quay thm 1 gc vi gi tr Ts(k-1) .Vy gc thc s c dng ti thi im K s l sI:
sI = s(k+1) +Ts(k-1)
+ Trn c s dng Is(k) v tc gc rotor (k) o u nhp(k), thi im qu gia (k-1) v (k), khu CD tnh c vector in p Us(k). Khu CVTKG thc hin Us(k) v Us(k) ch bt u c tc dng ln h thng k t nhp th (k+1), thi im qu gia (k) v (k+1) .n thi im vc t r quay tip c mt gc Ts(k).Vy gc thc s c dng l: SU(k)=s(k)+Ts(k) .
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Phn vII Thit k b iu chnh dng Stator c p ng tc thi

I. Cc vn c bn. iu chnh dng stator gi mt vai tr quan trng trong h thng TXCBP kiu T4R. khi thit k cc h thng iu khin/ iu chnh, u xut pht t gi thit khu CD l l tng- l tng ch p t nhanh khng chm ch dng stator cho ng c. Nhng ngay c vic thay th mch vng iu khin ( khu CD) bng khu tr bc nht c trng bng hng s thi gian l mt vic nh, s kh khn ny cng ln khi h thng c: *L mt h thng n hi d dao ng c kh nng tt dn km. * c iu chnh vi hi tip tc quay pha ng khng c hi tip tc quay pha ph ti, tc quay pha ph ti c xc nh bi mt khu quan st. *C cha cc khu phi tuyn nh ma st Coulomb hay d lc gia cc chi tit my. Vic c th thay bng mch vng bng mt khu t l P(Proporontinal) s gp phn n gin ho khi thit k h thng c phc tp trn. Thng thng tc p t nhanh nht thu c khi khu CD c p ng tc thi -c ngha gi tr thc theo kp gi tr cn sau 2 chu k T . Ngoi i hi v tc p t nhanh khu CD cn c kh nng cch ly tc ng qua li gia dng kch t iSd v dng to mmen quay iSq . y l yu cu m TCMC khng h c bi do cu trc c n cch ly dng kch t ik v dng to mmen quay iM . y l yu cu m cu trc kivnh in (hnh 3.10) khng p ng c trn vn. Trong phn ny ta s lm quen v khu CD c thit k di gic coi ng l mt i tng a thng s. Ngoi ra khu iu chnh dng phi tho mn yu cu phi xt tt c cc iu kin bin ca h thng khi t ra gi thit ban u cho khu cng nh t gi tr phn hi gi tr thc ca i lng b iu chnh . gii php kinh in khu iu chnh PI khng gii quyt c yu cu ny cc iu kin bin hay gp l : *Tr ro thi gian tnh ton ca vi x l gy ln thng thng thi gian tr l mt chu k T . *K thut o dng stator khi dng gii php bin i tng t / s ADC khng gy kh khn th gii php dng bin i in p/ tn s th ta phi cn nhc k hn . Sinh vin thit k: Nguyn
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*Gii hn trn in p mt chiu nui bi bin tn .bin tn ch c kh nng cung cp p trong mt phm vi hn ch ph thuc vo ln p 1 chiu u vo . ii. khi qut v cc phng php CD c s dng . V c bn, cc gi php tn ti c th c chia lm hai nhm : iu chnh dng phi tuyn v iu chnh dng tuyn tnh. cc gii php tu theo iu kin c th - c p dng trong thc tin. Tuy vy phn ny ta khng i su vo phn tch cc gii php m ch im qua . II.1.iu chnh dng phi tuyn . Cc gii php thuc nhm ny c th chia thnh hai nhm nh vi c tnh hai im hoc c tnh ba im . Mt gii php c bit trong nhm l iu chnh dng kiu d bo trc, nu vector dng di xa mt phm vi sai s nht nh no (c th dng ng trn) khu CD x c phn ng tr li ngay bng mt mu xung ti u c bo trc . II.2.iu chnh dng phi tuyn . Cng nh phng n trn y l mt gii php c ng dng rng ri trong thc tin bi c im n gin d thc hin . Tuy nhin hai gii php trn cha c s ho ton b m phi dng bng k thut tng t . III. Thit k b iu chnh dng c p ng vi tc tc thi thit k b iu chnh ta thc hin hi tip m dng stator o c iS (k), khp kn mch vng iu chnh v so snh gi tr is(k) vi gi tr cn iS*(k) sau , sai s ca php so snh c a ti u vo ca khu iu chnh RI no , vi vector u ra l y(k) . Ta s thu c cu trc trn vn u tin ca CXCBP nui bng bin tn ngun p c iu chnh dng nh sau: (k) h xW RI y(k) H-1 uS(k) Z-1I H iS(k+1) Z-1I iS(k)
Pha iu chnh Pha ng c
Hnh 7.3 S khi khu iu chnh vector dng is cho CXCBP. Trang 65 Sinh vin thit k: Nguyn c Quyn
T hnh (7.3) ta vit c phng trnh iu chnh tng qut nh sau: y(Z) = RI[ iS*(Z) - iS(Z)] (7.1) T m hnh ta thu c phng trnh (7.2) Is(k+1)=i+s(k) +y(k-1). Trn khng gian phng trnh ny c dng :
[ZI ] .is(Z) =Z-1.y(Z)
(7.3)
Nu thay phng trnh trn vo (7.1) ta c hm truyn t cho CCBP c iu chnh dng nh sau: iS(Z) = Z-1[ZI - +Z-1RI]-1 RI iS*(Z) (7.4)
Mc tiu t ra cho hm (7.16) l (7.7) ta thy (7.16) ging (7.7) nu nh RI c dng : I -Z-1 (7.5) RI(Z) = 1-Z-2 p dng y(Z) = RIxw(Z) vo (7.5) ta chuyn phng trnh ca ma trn iu chnh sang dng c th vit phn mm nh sau: y(k) = xw(k) - xw(k-1) + y(k-2) Vi xw l sai lch iu chnh dng. xW (z) = i*S (z)-iS (z) = xwd+ixwq = xw+xw xwd = i*S d-iS d ; xwq = i*S q-iS q ; xw = i*S d-iS ; xw =i*S -iS ; Thay y(Z) (7.6) vo phng trnh: us(K) = H-1 ( y ( K 1) h ( K ) ) (7.8) (7.7) (7.6)
Ta s thu c i lng iu khin in p stator uS l i lng s c a ti ng c nh khu iu ch vector khng gian us(K+1) = H-1
[X
w( K )
. X w( K 1) + y ( K 2 ) h( K 1)
(7.9)
- Sau y ta p dng (7.9) cho tng trng hp c th. III.1 -p dng cho trng hp CKBBP rotor lng sc trn h to dq : Trc khi a vo p dng c th, taln lt thay cc k hiu tng qut bng cc k hiu c quy c c th mc 3.2. Thay bi f11 , Sinh vin thit k: Nguyn
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H bi H1f , h bi f12 . T phng trnh (3.3a, b) ta c.
T 1 1 T 1 + T s Tr 11 12 f ; 11= = 12 11 1 1 T T 1 + Ts Tr
1 1 T T 13 14 Tr f12 = ; = 1 T 1 14 13 T Tr
T h 0 l Hf11 = 11 T = s 0 l 0 s 0 T l s
(7.10a, b, c)
Vi cc phn t ma trn cng thc (7.10) ta thu c hai phng trnh in p sau: usd (k+1) = h-111[ xwd (k) - 11xwd (k-1)- 12xwd (-1) + yd(k-2)- 13rd (k+1) ] usq (k+1) = h-111[ xwq (k) + 12xwq (k-1) - 11xwq (-1) + y q (k-2) + 14rq (k+1) ] + Bc 1: Tnh vector y(k) . yd(k) = xwd(k) -11xwd(k-1) -12xwq(k-1) +yd(k -2) yq(k) = xwq(k) -11xwq(k-1) -12xwq(k-1) +yq(k -2) + Bc 2: Tnh vector us(k+1) . usd(k+1) = h11-1[ yd(k) - 13rd (k+1) ] usq(k+1) = h11-1[ yq(k) + 14rq (k+1) ] h to : Trc khi a vo p dng c th, ta ln lt thay cc k hiu tng qut bng cc k hiu c quy c c th mc 3.2. Thay bi s11 , Sinh vin thit k: Nguyn (7.13) (7.12a, b) (7.11,a) (7.11b)
III.2. p dng cho trng hp CKBBP rotor lng sc trn
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H bi H1s , h bi s12 . T phng trnh (3.3a, b) ta c.
T 1 1 0 1 + Tr 11 0 Ts s ; 11= = T 1 1 0 11 0 1 + Ts Tr 1 1 T T T r s12= 13 14 = ; 14 13 1 T T 1 Tr T 0 h 0 l (7.14a, b, c) Hf11= 11 T = s T 0 l 0 s l s Vi cc phn t ma trn cng thc (7.14) ta thu c hai phng trnh in p sau: us (k+1) = h-111[ xw (k) - 11xw (k-1)- 12xw (-1) + y(k-2)- 13r (k+1) ] (7.15a) (7.15b) us (k+1) = h-111[ xw (k) + 12xw(k-1) - 11xw(-1) + y (k-2) + 14 r (k+1) ]
tin cho lp trnh ta tnh cng thc (7.15a, b) theo hai bc sau: + Bc 1: Tnh vector y(k) . (7.16a) y(k) = xw(k) -11xw(k-1) -12xw(k-1) +y(k -2) y(k) = xw(k) -11xw(k-1) -12xw(k-1) +y(k -2) (7.16b) + Bc 2: Tnh vector us(k+1) . us(k+1) = h11-1[ y(k) - 13r (k+1) ] us(k+1) = h11-1[ y (k) + 14r (k+1) ]
(7.17a) (7.17b)
Bng l thuyt v thc nghim c kt qu v bng kt qu thnh cng trn thng mi ch ra rng: Khu CD vi p ng tc thi tho mn c rt tt i hi v tnh ng ca cc mch vng iu chnh vng ngoi (CTQ, CVT, CTT) i vi mch vng trong cng. Sinh vin thit k: Nguyn
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IV. Kt lun. n y v c bn n ca ta hon thnh: gii quyt c khu then cht ca n, khu then cht thit k thut ton iu chnh dng trong h TXCBP theo nguynl iu chnh ta t theo t thng rotor T4R. V c bn khu iu chnh dng c chia lm 2 h chnh. + Mt h hot ng trn c s x l vector sai lch iu chnh (mc 7.1) v (mc 7.2) . + Mt h hot ng trn c s hi tip vector trng thi. Kinh nghim thc t cho thy: Khu iu chnh dng hot ng trn cs iu chnh vector sai lch iu chnh hot ng vi tin cy rt cao, ngay c khi cc thng s iu chnh km chnh xc (v d nh: tham s ch c tnh t nhn my ch khng c xc nh thng qua o lng). u im c th cho khu tr thnh gii php chun cho loi bin tn van nng phi hot ng tt iu kin ng c t quen bit . Ngc li khu iu chnh trng thi (khng phn tch trong n) kh nhy vi sai lch tham s iu chnh. Nu c cung tham s mt cch chnh xc khu cho php t c mt vi ch tiu cht lng cao hn khu trn v v th c th s dng c li trong h thng cht lng cao ni m ph tn ph khng ng k. Trong h TXCBP c cc khu iu chnh vng trong v vng ngoi khu iu chnh vng trong ta s dng song. Sau y ta xy dng cc b iu chnh vng ngoi cng cc gii php i theo da trn c s iu chnh vng trong - iu chnh dng (CD).
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Phn VIII Cc b iu chnh vng ngoi

i. M u Cc b iu chnh vng ngoi l cc khu iu chnh cp trn m cc i lng u ra ca chng l cc i lng u vo ca khu CD CXCBP km theo bin tn v in tr iu khin l mt khu thng nht nhm phc v vic ch ng mt i tng c hc nht nh. H thng c hc t khi c dng ghp ni cng n gin, m phn nhiu km theo r lc, ma st trt, cc chi tit c tnh n hi v... V vy ta phi gii quyt 2 cm vn ln: - Ch ng CXCBP ( xy dng cng c truyn ng) - Ch ng h thng c hc ( s dng cng c truyn ng) Vn ch ng CXCBP bao gm cc khu d c gii quyt t u n n nay v c khu CTT ca mc ny. Phn mm iu khin iu chnh CXCBP phi cung cp c cc ch tiu ch ng i tng c hc phc tp. V th nn phn ny mt mt tip tc a ra cc gii php iu chnh tuyn tnh mt khc a ra cch quan st mi nhm to cho vic thit k h thng c d dng hn. Ni cch khc v tnh a dng ca c tnh c, khng th da ra mt gii php tng qut c tnh vn nng cho cc khu iu chnh gc (CG), iu chnh gc quay(CGQ). Ii. B iu chnh t thng rotor. Ta bit khu iu chnh tuyn tnh ch c ngha i vi trng hp CKB. Ta cng bit khu CD cho php p t dng t ho isd vi thi gian p ng l 2T hoc 3T hoc 4T. V vy i tng iu khin t thng rotor km theo mch vng trong iu chnh dng c th c th hin tng qut nh trong hnh (8.1a). Nu khu iu chnh tuyn tnh lm vic vi thi gian vi thi gian trch mu T 10T ngha l vi tn s trch mu b t nht 10 ln so vi trch mu ca khu iu chnh dng ta c th b qua thi gian tr ca CD v s dng quan h h lin tc gia isd v rd trong m a) hnh i tng hnh (81b) i*sd(k) isd(k) Z-k k = 2, 3, 4 1T Tr T Tr
rd(k+1)
Z-1
rd(k)
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b)
*
rd
CTT
i*sd(k)
i tng iu chnh Hnh 8.1a hoc 8.1b
c) i*sd(p) e-p(kT) k = 2, 3, 4
isd(p) 1 1+pTr
rd(p)
Hnh 8.1a) M hnh gin on ca i tng iu chnh t thng. b) M hnh lin tc ca i tng iu chnh t thng. c)Cu truc iu chnh mch vng t thng . V nguyn tc ta c th xy dng khu iu chnh tuyn tnh hot ng trn khng gian trng thi xut pht im thit k m hnh (8.1a).
T ' T * (8.1) Z 1 rd ( Z) = i d ( Z) T Tr r Tuy nhin i tng l mt khu tr bc nht n gin (phng trnh(2.16a)), ng thi khu iu chnh dng l nhanh c th b qua thi gian tr p ng ta ch cn s dng mt khu vi tnh nng PI l . Nu s dng khu PI th khu s c cng thc tng qut sau. 1 D Z 1 R(Z) = (8.2) 1 Z 1 Khu iu chnh nhn (8.2) c th c xc nh d dng theo tuu chun mdul ti u, l phng php xc nh ti u kinh in c cpm rt nhiu ti liu khc nhau. V d nu i tng c hm truyn nh (2.16a) ta thu c cc cng thc sau cho 2 tham s v , D ca R(Z). * H s khuch i v : v = 1/ [3(1-e-T / Tr)] (8.3) -T / Tr * Thnh phn tch phn: D = e (8.4)
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Trong T : L chu k iu chnh ca CTT Tr : L hng s thi gian rotor Vi sai lch iu chnh c nh ngha nh sau X(k) = *rd (k) d(k) (8.5) ng thi vi i*Sd lad i lng u ra ca R , ta thu c cng thc tnh cho thut ton CTT c th dng trc tip vit phn mm nh sau: (8.6) i*sd(k) = v[x(k) -Dx(k-1)]+ i*sd(k-1) Trong cc cng thc trn gi tr thc rd l gi tr khng o trc tip c m chnh tnh c nh m hnh hoc khu quan st. Nh mi khu iu khin y ta phi gii quyt vn gii hn i lng u ra ca khu CTT c ngha l gii hn i*sd.. L do rt n gin bin tn khng th cung cp dng ln tu , dng ra b kh nng dn ti a ca van hn ch. Mt mt ta c kh nng p dng cch ngng khu tch phn kinh in mt khc ta c th da theo gii php x l c sm dng cho CD. Theo phng php : Nu nhp th (k) i lng u ra i*sd i vo gii hn phng trnh (8.6) c dng : (8.7) i* sdr (k) = v[xc(k) - Dx(k-1)]+i*sd(k-1) Cc ch s ph r, c c nh ngha cng thc(7.44). tr (8.6 )v (8.7) cho nhau ta thu c cng thc sa sai lch iu chnh. (8.8) xc(k)= x(k)-1/ v [i*sd(k) -i* sdr(k)] Cng thc 8.8 ch c hiu lc khi tn ti chnh lch gia i*sdv i*sdr lc gi tr do (8.8) cung cp s c (8.6) s dng nhp iu k tip. Phng pohp iu chnh tn cn c p dng cho khu CTQ nu nh khu tim n gi tr tch phn. iii. Cc b iu chnh tc quay, iu chnh v tr. Mc ch ca phng php T4R l thng qua vic chuyn to quan st t h to gn lin 3 pha a_b_c c nh vi stator sang h to 2 pha x_y quay ng b vi vector t thng rotor r c trc thc ng hng vi vector t thng is. Ta c th phn tch is thnh 2 thnh phn, thnh phn kch t isdv thnh phn to mmel quay isq vi s dp ca khu iu chnh dng ta cn c kh nng p t hai thnh phn c pp nhau v gn nh khng th. Bng kt qu o CKB lc ny c tnh nngiu chnh gn ging nh dng phn ng . i vi cc khu iu khin /iu chnh truyn t c hc vic thit k h thng lc ny tr ln n gin rt nhiu . Ta xt trng hp CKB vi c cu ghp cng gia ng c v ph ti c th bao gm c CTQ ,CG (CVC) hoc ch dng mt trong hai khu. i vi cc h c c cu ghp cng nh trn CTQ ch cn c c tnh PID hoc PI l . Khu CG thng l khu (P) .h thng k c CKB v CB, CD c th c m t y bng cu trc sau . Sinh vin thit k: Nguyn
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3pcL2
r
2L
CG&CV T
*
CT Q
mT
Tr isq m +1
pc pi
Hnh(8.2) Cu trc h TKB vi phn c ghp cng v dng khu CD vi thi gian p ng nh c th b qua . Vi gi thit rd l khg i ta c d dng thu c t hnh (8.2). hm truyn t sau y trn khng gian Laplace.
(P)/ *r (k) = 2rd
3 1 1 1 . = p2c (8.9) 2 R r p i pTM
Trong hng s thi gian c TM c tnh theo cng thc . TM = 2jRr/ 32rdpc2 (8.10)
Hm truyn t trn c dng hon ton tng t nh h CMC vic tht k CTQ cng nh khu CG khng c g khc bit so vi CMC. Tng t ta r dng xy dng hm truyn t cho h dng CB. Ngay c h thng c c dng phc tp dng mm, d dao ng cng khng c g khc bit gia TMC v TXCBP . n y vic kho xt thit k h truyn ng in xoay chiu ba pha iu chnh tn s CKB dng iu khin vector -nhim v c bn ca n v c bn hon thnh.
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Phn IX V d thc tin-lp trnh tnh thi gian ng ngt van

* V d s dng vi x l tn hiu TMS 320C25 v vi iu chnh SAB 167 Bn cht ca CTKG l khu ghp ni cng ngh gia thit b iu khin/ iu chnh gin on vi i tng iu khin/ iu chnh gin on. (vi tnh/ vi x l vi bin tn/ ng c). Trc khi i vo v d chi tit ta quy c mt s iu sau: - trnh i li li so vi tc pht trin ca khoa hc k thut v d trong mc ny nhm phc v ch yu i tng van: IGBT,MOSFET ca gia thp k 90 l cc van c c ch bo v t hp ngay trn phin dn. Trong cng thc tnh do khng cn quan tm n b tD . Mt khc sai s gy nn do b qua b tD l nh khng phi tnh n chng m khng h nh hng n cht lng h thng.Cc thnh phn in p c chun ho bi gi tr ti a 2uMC / 3 trong cng thc 5.1 nh sau: usN = us /(2uMC/3) (9.1a, b) usN = us /(2uMC/3) ; Vic chun ho nhm chuyn cc phn t in p thnh cc i lng tnh ton khng n v vi gi tr nm trong khong t 0...1 . Lc ny cc i lng khng mang ngha in p ban u na chng ch c trng cho m ca bin tn. Do c th din t c bng li nh sau: Khi in p mch mt chu uMC thp bin tn s phi m nhiu hn khi uMC ln t c cng mt in p nh nhau. Di y l biu tnh tng qut ca thut ton CVTKG. Nhp s liu uS v uS Tnh a,b theo cc cng thc (5.8a,b,c) USP < 0 ? Sai ng uS < 0 ? uS < 0 ? Sai Q1 ng Q2 ng Q3 Sai Q4 Sai ng ng Sai ng Sai ng Sai S1 S1/ Q1 S2/ Q2 S3 S5/ Q3 S4 S5/ Q4 S6 Tnh thi gian ng ngt van theo cng thc Chun b sn phu hp vi phn cng Xut s liu v thi gian ng ngt van Biu tnh tng qut ca thut ton CVTKG. Sinh vin thit k: Nguyn
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*V d s dng h vi x l kp dng vi iu chnh SAB C167v vi x l tn hiu TMS 320C25.

SAB C167 l vi iu chnh ra i sau do c ci tin rt nhiu nhm mc ch phc v tt hn cc nhim v iu khin, iu chnh trong cng nghip cu trc c bn c gii thiu trong hnh (5.6a). Trong h nhim v ca hai vi x l c phn chia nh sau: -Vi x l tn hiu TMS 320C25 thc hin cc bi ton iu khin vi thi gian thc, phc v mc ch ch ng ng c. -Vi iu chnh SAB C167, bn cnh nhim v dng h thng thanh ghi iu ch xut thi gian ng ngt van bin tn ng c do TMS 320C25 tnh, thc hin ton b cc bi ton cp trn v khu i thoi vi th gii bn ngoi.
Mch iu khin RAM
My tnh
TMS 320C25 Tng m
Khu ghp ni
SAB C167
= 3
CXCBP
My o tc quay Hnh 5.6a Cu trc c bn ca h truyn ng dng vi x l kp SAB C167 v TMS 320C25. Sinh vin thit k: Nguyn
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C167 c 4 thanh ghi ng h PT0...PT3 ch iu ch i xng, bn thanh ghi v bn cht l 4 b m tin li. Khi m tin, sau li v ti 0, PT0... PT3 s t ng nhn gi tr ti a cho chu k mi t 4 thanh ghi chu k PP0...PP3. iu ch in p ta ch cn 3 trong 4 ca mi loi thanh ghi k trn hnh (5.6b). Ta thy rng t c chu k m tin, m li nh nhau, ba thanh ghi PP0, PP1 v PP2 s phi dng thi nhn ba gi tr nh nhau. Hn th na nu l chu k ct xung Tx khng i nh khi iu ch khng ng b, ta s ch phi ghi mt ln duy nht gi tr Tx vo PP0, PP1 v PP2.
uMC
DDCXCBP
16
PW0 PP0 PW0
16 16 Comp POUT0
PP0
16
16 16 Comp POUT0
PP0
16
PP0 PW0
16 16 Comp POUT0
PP0
16 16
PP0 Bng thi gian

16
Xung nhp BUS s liu
Hnh 5.6b Cu trc h thng thanh ghi siu ch ca SAB C167 dng thc hin thut ton CVTKG Sinh vin thit k: Nguyn
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Vi SAB C167 y ba thanh ghi PT0, PT1 v PT2 gi vai tr ca mt mnh thanh ghi T0, 3 thanh ghi PP0, PP1 v PP2 ca TOREL y nu nn tnh linh hot rt cao ca SAB C167, do 4 thanh ghi hot ng c lp nhau, ta c th thc hin dc ti 4 ch iu ch khc nhau, phc v cho 4 i tng cng mt lc, mt nhu cu thc t khng phi him trong iu khin cng nghip (TOREL: thanh ghi np chu k cho T0). Ngoi cc thanh ghi k trn, ta cn 4 thanh ghi PW0... PW3 dng vo mc ch to b rng xung. v d ny ta ch dng ba thanh PW0...PW2 v sp xp chng ng vi cc nhnh van bin tn hnh (5.6,b).Trong qu trnh PT0... PT2 m, gi tr ca cc cp thanh ghi PT0/ PW0, PT1 /PW1, PT2/ PW2 lun c so snh vi nhau, mi khi gi tr ca tng cp bng nhau, s kin s o trng thi ca ca ra tng ng POUT0, POUT1,POUT2. (k-1) (k) (k+1)
PWM-Timer u v w
Tu ngt Tvngt
Twngt Tx Tx
Hnh 5.6c nh ngha thi gian ng ngt van cho cu trc hnh 5.6b Qu trnh m t bng li trn c th hin trong hnh (5.6,c), trong vai tr ca PT0, PT1,PT2 c th hin c tnh cht i din qua PWWM-TIMER do kh nng m tin li ca PWM- TIMER, i vi mt chu k xung Tx ta ch cn np mt ln duy nht thi gian Tu ngt (vo PW0), Tv ngt ( vo PW1), Tw ngt (vo PW2). Cc s kin bng nhau ca tng cp : khi m tin s a trng thi nhn van v 0( ni pha vi cc -ca uMc , khi m li v 1( ni pha vi cc + ca uMC ). Thi gian theo nh ngha nh hnh (5.6c) c tnh theo cc cng Sinh vin thit k: Nguyn
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thc sau: ri mt ta b qua cc ch s ngt vit li pha di bn phi.
Gc phn su th 1(s1):
Tu = Tx T T (1+a) ; Tv = x (1-a+2c) ; Tw= x (1-a) 2 2 2 Tx (1+a+2c) 2 ; Tx (1-a+2c) 2 (5.10a,b,c)
Gc phn su th 2(s2):
S2/Q1: Tu= Tv = ; S2/Q2: Tu= Tx (1-c) 2 (5.11a,b)
Tx (1+c) 2
Tw=
(5.11c,d)
Gc phn su th 3(s3):
Tu = Tx (1-a) 2 ; Tv = Tx (1+a) ; 2 Tw= Tx (1+a-2c) (5.12a,b,c) 2
Gc phn su th 4(s4):
Tu = Tx (1-a) 2 ; Tv = Tx T (1+a-2c) ; Tw= x (1+a) (5.13a,b,c) 2 2
Gc phn su th 5(s5):
S5/Q3: Tu= Tx (1-a+c) 2 T Tv = x (1-c) 2 ; ; S5/Q4: Tu= Tw= Tx (1+2a+c) 2 (5.14a,b) (5.14c,d)
Tx (1+c 2 )
Gc phn su th 6(s6):
Tu = Tx T (1+a) ; Tv = x (1-a) ; 2 2 Tw= Tx (1-a+2c) (5.12a,b,c) 2
Hnh (5.6a) gip ta c cm gic c th v tng quan gia in p pha usu v thi gian Tu trong phng n hnh (5.6a) v (5.6b) quan h v gc lch pha gia 3 thi gian Tu , Tv, Tw c th hin trong hnh (5.6b). Ta c thm nhn xt :Hai hnh trn thu c khi in p stator gn nh t bin ny, CVTKG s dc coi l mt khu truyn t 1:1 trung thnh v pha Sinh vin thit k: Nguyn
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v mdul, do khng c th hin trn s truyn t ca h thng.
ti a. Ta thy r iu thng qua cc thi im Tu , Tv, Tw gn bng

0. n y ta l quen vi phng php CVTKG , mt phng php mnh trong qu trnh s ho h thng truyn ng in hin i dng CXCBP. gin ho khi thit k khu iu chnh dng(CD) sau
* Tnh ton gn ng cc tham s t nhn ng c:

Trng hp s dng ng c khng ng b c cc s liu sau: - cng sut danh nh: PN = 12 KW - in p pha danh nh: UN = 220 V - Dng in pha danh nh: IN = 21,3 A - Tn s danh nh: fN = 50 Hz - Tc quay danh nh: nN = 2880 v/p - H s cng sut: cos = 0,85 + Tnh dng kch t danh nh IsdN: ISqN= 2 IN 1 cos = 2.21,3. 1 0,85 = 11,67 (A) + Tnh dng danh nh to mmen quay ISqN :
2 2 ISqN = 2 I N I SqN = 2.(21,3) 2 (11,67) 2 = 22,77
(A)
+ Tnh tc gc cng nh tn s danh nh rN ca mch rotor: Pn rN =2 f N c N 60 Trong : Pc l s i cc, n khng tn ti tng minh trn nhn my, trn c s fN v tc khng ti l tng ta c: n0 3000 2880 nN = 1 rN=2 fN Pc= = = 12,6 =2.3,14 50 60.50 60 60 60fN + Tnh hng s thi gian roto Tr ch danh nh: I 22,77 = Tr sqN = 0,19 (s) rN I dN 12,56.11,67 + Tnh khng phc tiu tn ton phn X ch danh nh: tnh gn ng X = sLs ta c th b qua (Rs) l do:Ti thi Sinh vin thit k: Nguyn
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im cng tc danh nh vi tn s fs= fr, in p ri trn X ln hn nhiu ln so vi in p ri trn(Rs). - in p stator danh nh USN= 2 N = 2 = 179,6 (v) 3 3 - Quan h gia v IsdN,IsqN:
I sqN 22,77 = = 0,75 sin = I sN 37 cos = I SdN = 11,67 = 0,31 I sN 37
U 220
Quan h gia 3 gc , v - Tnh sin=sin[ -(900-)]=sinsin- cossin
sin = sin
I qN I SN
cos
I sdN I SN
vi cos = 0,85 sin = 0,52 sin = 0,52.0,75 - 0,85.0,31= 0,12 Tnh X t hnh v:
X sin U SN I sin cos sdN I sqN I sqn U SN 11,67 .220 = 0,52 0,82 = 1,16( s) I 27,77 21,3 SN
+ Tnh in khng phc Xh in khng phc Xh=sLm l in khng m sc t ng estdri trn . tnh c Xhta xt trng hp ng c quay tc danh nh nN,tn s fN v khng ti tc l isq=0,s=0.in tr thay th Rs tin ti v cng.Mt khc do fS=fN ta b quaR(s) cng ging nh tm X.T hnh v trn ta thu dc kt qu gn ng sau y :
Xh U BN X = I sdN 2 .U N 3I sdN X =
2. 220 3.11,76
1,68 = 13,7()
+ Tnh in tr stator (Rs) Chp nhn gn ung sc t ng danh nh:

estd = X h I sdN =
rN sN
Rr
I sqN
T ta s thu c gi tr gn ng cho Rs
R s Rr
rN I sdN 12,56.11,67 .X h .13,7 = 0,2() 2f N I sqN 2.3,14.50.22,77
+ Tnh h s tiu tn tng v hng s thi gian stator Ts: - Trn c s Xh v X ta thu c kt qu gn ng Sinh vin thit k: Nguyn
c Quyn
Trang 80
X 1,68 = 0,12 = X h 13,7

Xh 13,7 = = 0,04( H ) 2f N 2.3,14.50 Xh 13,7 = = 0,21( s ) 2f N Rs 2.3,14.50.0,2
in cm stator Ls:
Ls =
Hng s thi gian stator Ts:

Ts =
n y ta c y s liu gn ng c th tnh cc tham s h thng. tin cy cao ca phng php tnh t nhn my k trn c chng nghim qua thc t.
*Lp trnh tnh thi gian ng ngt ca van - T biu tnh tng qut ca thut ton CVTKG. Lp trnh bng ngn ng lp trnh Turbo C *
# include<stdio.h > # include<conio.h > # include<math.h > # include<dos.h> # include<graphics.h> int xo,yo, void_ ve ( float thietas, unsigned dai, unsigned day, unsigned mau) { unsigned x,y ; x =( x0Int) +(int)(dai* cos(thieta_s) +0.5); y = y0-(int)(dai* sin(thieta_s) +0.5); setcolor(mau) // setlinestyle(0,0,day); /* x1 = x0 +(int)(dai* cos(thieta_s)/2 +0.5); y1= y0-(int)(dai* sin(thieta_s)/2 +0.5); x2 = x0 +(int)(dai* cos(thieta_s)/2 +0.5); y2 = y0-(int)(dai* sin(thieta_s)/2 +0.5); setfill style(11,13); bar (x0,y0,x,y);*/ setlinestyle(0,0,day); line(x0,y0,x,y); } void vectoado(unsigned dai,unsigned day,unsigned mau) { Sinh vin thit k: Nguyn
c Quyn
Trang 81
Unsigned x1,y1,x2,y2,x3,y3,x4,y4; x1=x0-dai x2=x0+dai y1=y0-dai y 1=y0+dai x3=x0+(dai+30)*cos(45*3.1416/180); y3=y0-(dai+30)*sin(45*3.1416/180); x4=x0-(dai+30)*cos(45*3.1416/180); y4=y0+(dai+30)*sin(45*3.1416/180); setcolor(mau); setlinestyle(0,0,day); line(x0,y0,x0,y1); line(x0,y0,x0,y2); line(x0,y0,x1,y0); ine(x0,y0,x2,y0); setcolor(mau+5); line(x0,y0,x3,y3); line(x0,y0,x4,y4); } main() { float xwdc,xwd,h11,usd,usdr,xwqc,xwq,usq,usq,usqr,isdsao,sqsao,yd,yq, Fi11,Fi12,Fi13,Fi14,T_ UN,IN,PN,fN,nN,ndb,s,cosfi,Isdn,Isqn,Omega_rN,Tr,Xbeta,Xh, Rs, beta,Ts,beta,Ts,Sinfi,Pc,thieta_s,thieta_si,thieta_su,fi,psIrd,isU,i sv, sanpha,ispeta,PSIphay, omegas,omegar,isd,isq,dorad,T; char ch; int mh=0,mode=0; initgraph (& mh,& mode." "); x0=getmaxx0()/2; y0=getmaxy()/2; dorad = 180/3.1416; UN=220; IN=21.3; cosfi=0.9; S=0.02; nN=12; fN=50; nN=2880 ;fi=cosfi*dorad; Sinh vin thit k: Nguyn
c Quyn
Trang 82
pc=1;ndb=3000; T=100e-6; /*printf("\n Nhap tham so tu nhan dong co "); printf("\n Nhap dien ap'pha/pha' danh dinh UN= "); scanf("%f",&Un); printf("\n Nhap dong pha danh dinh IN= "); scanf("%f",&In); printf("\n Nhap cong suat danh dinh PN= "); scanf("%f",&Pn); printf("\n Nhap tan so danh dinh fN= "); scanf("%f",&fn); printf("\n Nhap toc do quay danh dinh nN= "); scanf("%f",&nn); printf("\n Nhap he so cong suat danh dinh cosfi= "); scanf("%f",&cosfi);
/*tinh tong kich tu danh dinh*/

IsdN=sqrt(2)*IN*sqt(1 -cosfi);
/*tinh dong danh din tao momen quay*/

IsqN=sqrt(2*IN*in-Isdn*Isdn);
/* Tinh toc do goc va tan so danh dinh cua RoTor*/

Pc=60*fN/nbd; omega=2*3.1416*pc*(nN/60) omega_rN=2*3.1416*(fn-(Pc*nN/60));
/*Tinh hang so thoi gian Rotor o che do danh dinh*/

Tr=Isqn/(omega_rN*IsdN); /*tinh khac phuc tieu tan toan phan o che do danh dinh*/ sinfi=sqrt(1 - cosfi); xbeta=fabs((sinfi-cosfi)*(Isdn/IsqN)*UN/(sqrt(3)*IN))
/*Tinh dien khang phuc*/

xh=(sqrt(2)*UN/(sqr(3)*isqN))-xbeta;
/* tinh dien tro rotor va stotar*/

Rs=(omega_rN*IsdN*xh)/(2.*3.1416*fn*IsqN); Rr=Rs; Sinh vin thit k: Nguyn
c Quyn
Trang 83
/* tinh hieu so tieu tan tong va hang so thoi gian stator*/

beta=xbeta/xh; Ls=xh/(2*3.1416*fn); Ts=xh/(2*3.1416*fn*rs);
/*Dieu chinh dong*/

h11=T/(beta*Ls); // Fi11=1-(t/beta)*(1-Ts-(1-beta))/Tr; // Fi12=omega_s*t; // Fi13=fabs((1-beta)/beta*(t/Tr)); // F14=fabs((1-beta)*omega*T); // printf("\n nhap dong dien can isd_sao= "); // scanf("%f"& isd_sao); // Printf("\n nhap dong dien can isd_sao="); // scanf("%f"&isq_sao); xwdc=0;xwqc=0;id=0; usqr=0;usdr=0; id_sao=5;iq_sao=5;
/*uoc luong tu thong va tinh goc*/

isu=15; isu=15; isanpha=ius; ispeta=(isu+isv82)*1/sqrt(3); thieta_s=0; omega=0; pstr_phay=0; isq=0; isd=0 PSIrd=0; f1:
/* chuong trinh uoc luong tu thong va rotor va tinh goc*/

vetoado(150,2,10); thieta_s=(thieta_s+T*omega); PSIrd=PSId+thieta_s; gotoxy(3,6); printf("thieta_s=%f\n",PSIrd); ve(PSIrd,120,4,14); delay(20); ve(Psird,120,4,1); Sinh vin thit k: Nguyn
c Quyn
Trang 84
psird_phay=fabs(isd)*(t/Tr)+Psi+phay*(1-t/tr); gotoxy(26,1); printf("Psird_phay=%f\n",psird_phay);
/* chuyen toa do cho dong dien*/

isq=ispeta*cos(thieta_s*dorad)+isanpha*sin(thieta_s*dorad); gotoxy(2,1); printf("isq=%f\n",isq); isd=ispeta*sin(thieta_s*dorad)+isanpha*cos(thieta_s*dorad); gotoxy(2,2); printf("isq=%f\n",isd); omega_r=fabs(isq/(Tr*PSIrd_phay)); omega_s=omega+omega_r; gotoxy(50,1); printf("omega=% f\n",omega); thieta_s=omega*Y; //gotoxy(2,12)//; //printf("thieta_s=%f\n",thieta_s);
/*hieu chinh goc pha*/

thieta_s=thieta_s+t*omega_s; gotoxy(2,4); printf("thieta_s=%f\n",thieta_su);
/* chuong trinh dieu chinhr dong */

//xwd=0;usd=0;usdr=0;isd_sao=5; yd=0;xwq=0;usq=0;usqr=0;isq_sao=5; usdr=usd-Usdr;Usqr=Usq-Usqr; Usapha=Usdn*cos(thieta_sU*dorad)sqr*sin(thieta_sU*thieta_s*dorad); us_beta=Usqr*sin(thieta_s+dorad)+Usqr*cos(thieta_su+rorad);
/* hieu chinh sai so dieu chinh tren truc d$q*/
xwdc=xwd-h11*(Usd-Usdr); xwqc=xwq-h11*(Usd-Usdr); gotoxy(2,11); printf("xwdc=%f\n"xwdc) gotoxy(2,12); printf("xwqc=%f\n"xwqc); Fi11=fabs((1-(t/beta))*((1/Ts)+((1-beta)/Tr)))*xwdc; Fi12=fabs(omega*T)*Xwqc; Sinh vin thit k: Nguyn
c Quyn
Trang 85
/*Tinh sai so dieu chinh moi*/
Xwd= Is_sao-isd; Xwq=is_sao-isq;

/* tinh gia tri Yq(k+1) $usq(k+1)*/
Ud=xwd-fi121-Fi12; yq=xwq+Fi12-Fi11;
/* tinh dien ap Usd(k+1) $ Usq(k+1); */
tinh dien ap Usq(k+1) $ Usd(k+1); Usd=(1/h11)*(yd-Fi13*PSIrd_phay; Usq=(1/h11)*(yd-Fi14*PSIrd_phay; gotoxy(2,8); printf(usdd=%F|N",usd); gotxy(2,9) printf("Usq=%f\n",usq);
/* chuong trinh dieu che vector khoang gian*/

/*Lap giatri tuyet doi cho Us_anpha va us_anpha*/
Us_beta=fabs(U s_beta); Us_anpha=fabs(U sanpha); tinh gia tri a,b,c:cong thuc(58)*/ a=Us_anpha+Us_beta/sqrt(3); b=Us_anpha-Us_beta/sqrt(3); c=a*t;b=b*t;C=c*t; if (us_beta>0) { if (us_anpha0) { if (b0) { Tu=(1+a)*t/2; Tr=(1-a+2*c)*T/2; Tw=(1-a)*T/2; } else { Tv=(1+c)*t/2; Tw=(1-a)*t/2; if(Us_anpha0) Tu=(1+2*b+c)*t/2; else Tu=(1-2*a+c)*t/2; Sinh vin thit k: Nguyn
c Quyn
Trang 86
} } else { if(b0) { Tu=(1-a)*t/2; Tv=(1+a)*t/2; tw=(1+a2*c)*t/2; } else { Tv=(1+c)*t/2; Tw=(1-a)*t/2; if (Us_anpha0) Tu=(1+2*b+c)*t/2; else Tu=(1-2*a+c)*t/2; } } } else { if (us_anpha0) { if (b0) Tu=1+a)*t/2 Tv=(1-a)*t/2; Tw=(1-a+2*c)*t/2; } else { Tv=(1-c)*t/2; Tw=(1+c)*t/2; Tu=(1+2*b+c)*t/2; else Tu=(1-2*a+c)*t/2; } } else { Sinh vin thit k: Nguyn if (us_anpha0)
```
c Quyn
Trang 87
if (b0); { Tu=(1-a)*t/2; Tw=(1+a)*T/2; } else { Tv=(1+c)*t/2;; Tw=(1-a)*t/2; Tu=(1+2*b+c)*t/2; else Tu=(1-2*a+c)*t/2; } } gotoxy(2,14); printf("Tu=%f\n",tu); gotoxy(2,15); printf("Tv=%f\n",tv); gotoxy("Tw=%f\n",Tw); if(kbhit()) gotof1; if(getch()!=27) gotof1; closegraph(); }
Tv=(1+a-2*c)*t/2;
if
(us_anpa0)
c Quyn
Trang 88
Phn X: Gii thiu cng ngh h thng truyn ng cn bng nh lng nh my xi mng lu x
H thng cn bng nh lng l h thng rt quan trng trong giy truyn sn xut xi mng ca nh my xi mng lu x .H thng nm ngay u dy truyn sn xut . Nguyn l hot ng ca h thng cn bng nh lng : Nguyn liu sau khi sy c a vo cc xil. H thng cn bng ny gm 6 khu .mi cn lm nhim v cn mt loi nguyn liu (than ,t , 1, 2 ,ct ,ph gia ).nguyn liu c chy xung bng ti t xil cp liu qua h thng van ngn ko do vylng liu ch mang tnh cht tng i ,khng phi lc no cng u nhau .khi dng liu chy xung bng ti, trn bng ti c mt on bn cn, vic ly tn hiu trn on bng ti nh mt u o khi lng khi dng liu xung t th bng ti phi chy nhanh ln v ngc li m bo lu lng ng u khng thay i .vic iu chnh tc ca bng ti ph thuc vo tc ca ng c truyn ng cho bng ti . Tc ng c c iu chnh nh vic thay i tn s ca bin tn .Do vy mi cn bng c iu chnh vi tc khc nhau ,ph thuc vo t l phi liu .Tt c nguyn liu t cc cn c a vo my nghin nh h thng bng ti ,gu ti v c nghin ra thnh bt cung cp cho cc cng on tip theo ca nh my. m bo cc yu cu k thut th vic iu chnh cc cn phi m bo chnh xc < ,=1%.vic ny c thc hin a vo vic ly tn hiu t u o tn hiu ,qua card chuyn i ,sau dc a vo my tnh. My tnh a ra tn hiu Uk sang b bin tn, bin tn s a ra tn s ph hp iu chnh tc ng c truyn ng cho bng ti.H thng gm 6 cn tng t nhau khi mt cn c s c th h thng x t bo ng nh phn hi m tc , tn hiu c a vo my tnh v my tnh s t x l s c ny, lc ny trn
mn hnh my tnh s bo ng v t ng thay i tc ca cn a cn v v tr cn bng mi. Sau 30 giy nu s c khng c khc phc th my tnh s pht lnh dng ton b. Sau y ta s nghin cu s khi c th ca mt cn bng trong h thng trn hnh 1.1
c Quyn
Trang 89
u o k.lng
1
RS 232/485
220
K.i
2
Convert A/D
V
B nhn Q= mv Lng h/c Q
7
Bin Tn
Convert A/D
5 3
My tnh
6
Q= m
10 8 9
Convert A/D
M
LED
LED
(m/ph ) (kg/m ) Hnh1-1. S khi h thng cn bng nh lng
11
I. Gii thiu tng khi trong s : *Khi 1: u o khi lng Trong thc t ngi ta thng dng chuyn i in tr o cc di chuyn thng , cc di chuyn gc ca cc i tng o, ngoi ra cn dng trong cc dng c do lc, o p sut, o gia tc hoc cc chuyn i trong mch cu... Chuyn i in tr l loi chuyn i bin i lng khng in cn o thnh s thay i in tr ca n. Chuyn i in tr: L mt bin tr gm mt li bng vt liu cch in nh Gm S ,Baklit hoc bng Nhm, ng c ph lp cch in vi hnh dng khc nhau. Trn li qun dy in tr bng Maganis,niken,crm,vofram. Cc dy in tr c trng Emay cch in c th qun st nhau. Trn li v dy qun c con trt,con trt c ch to bng hp kim Platin-Iridi c n hi v tip xc tt , lc t gia con trt v li rt nh. Di tc dng ca i lng vo, con trt di chuyn. Quan h gia i lng vo v i lng ra biu din di dng : R=f(XV). Chuyn i in tr lc cng: Khi dy dn chu bin dng c kh th in tr ca n thay i , hin tng gi l hiu ng Tenro. Chuyn i in tr lc cng c ba loi: Chuyn i in tr lc cng dy mnh, l mng , mng mng . Ph bin nht l loi chuyn i in tr lc cng dy mnh. Sinh vin thit k: Nguyn
c Quyn
Trang 90
Khi d bin dng l / l chuyn i c dn ln i tng o, lc i tng o b bin dng chuyn i bin dng theo v chuyn i ca bin dng Tenro thay i mt lngR/R .
R l = f( ) R l
Thng thng chuyn i in tr lc cng c dng vi mch cu mt chiu, xoay chiu v mch phn p. Trong h thng ta dng chuyn i in tr lc cng dng mch cu mt chiu. p RT
R2 912v R3 R4 Ura (mv)
Hnh I-2. B chuyn i in tr.

Khi mch cu ch c mt nhnh hot ng( tc l ch mt chuyn i hot ng) vn b nhit phi c thc hin, do vy ngi ta dng mt chuyn i cng loi dn ln chi tit khng lm vic cng vt liu vi i tng o v t trong cng mt nhit . Khi cu khng lm vic trng thi cn bng: RT/R2=R3/R4=K. Khi i tng o lm vic Rt thay i thnh R.RT cu mt cn bng v din p ra l: (l+R).RT.R3-R2R4 Ur=U. [(l+R )RT+R4](R2+R3) Tn hiu ra t b chuyn i bin tr l rt nh do ta cn phi khuch i tn hiu ln. C rt nhiu b khuch i. Gia b khuch i thut ton v b khuch i thng thng v c bn khng c g khc nhau. C hai loi ny dng khuch i in p , dng in hoc cng sut. Trong khi tnh cht ca b khuch i thng thng ph thuc vo kt cu bn trong ca mch th tc dng ca b khuch i thut ton c th thay Trang 91 Sinh vin thit k: Nguyn c Quyn
i c v ch ph thuc vo cc linh kin mc mch ngoi. thc hin iu , b khuch i thut ton phi c h s khuch i rt ln, tr khng vo rt ln v tr khng ra rt nh. B khuch i thut ton c biu din nh hnh I-3: RN R1 a UV R1 b UV + Ur Ud
UR
+ RN
URmax Ur
-vi mv
Uv
+vi mv
URmin Hnh I.4.c tnh ca b khuch i thut Hnh I.3. a) s khuch i o, b) thun ton. Trong h thng c s dng mt s s khuch i, v d nh s khuch i o, s bin i in p - in p, s khuch i thun. *Khi 3,5,8: b chuyn i A/D Cc i lng iu khin thng l tn hiu tng t, trong h iu khin s cn phi bin i i lng thnh tn hiu s. Nh vy cn phi bin i tng t-s ( A/D). c ba phng php thc hin khc nhau v nguyn tc: 1. phng php song song: in p vo ng thi so snh vi in p chun v xc nh chnh xc xem n ang nm gia hai mc no, kt qu ta c mt bc ca tn hiu xp x, phng php ny c gi thnh cao bi v mi s ta cn phi c mt b so snh, phng php ny c u im l nhanh. 2. Phng php trng s : Vic so snh din ra cho tng bt ca s nh phn, cch so snh nh sau: u tin ta xc nh xem in p vo c vt in p chun ca bt ra hay khng. Nu n vt th kt qu c gi tr 1 v ly in p vo tr i in p chun, phn d em so vi cc bt tr ln cn. R rng l c bao nhiu bt trong mt s nh phn th cn ly bao nhiu bc so snh v ly bao nhiu in p chun. 3. Phng php s :Phng php n gin nht l phng php s, trng hp ny ta k n s lng cc tng s in p chun ca cc bt tr dng din t cc in p vo. Nu s lng cc i m t bng n th do cng cn ti a n bc d nhn c kt qu. Phng php ny n gin, r tin nhng chm. Hin nay c rt nhiu b bin i A/D nhng c mt vi nguyn tc Sinh vin thit k: Nguyn
c Quyn
Trang 92
c bn xy dng mch: Mch bin i A/D theo nguyn tc b, theo nguyn tc servo, b bin i A/D song song so snh,b bin i A/D vi hai ln tch phn, b bin i A/ theo bin tn s. V d b bin i A/D theo nguyn tc b: UR S Aj 1 R Q T2 Q T1 T1 2j R1 + U0
R2
Hnh I-5. B bin i A/D theo nguyn tc b. Mch bin i A/D ny mch m nh phn, mch so snh v b bin i A/D. Ti thi im u b dm t trang thi 0 bi xung chnh lu, nh vy u ra cng c tn hiu 0. Mch so snh thit lp gi tr 1 tn hiu nhp H qua cng AND c a vo mch m. Mch dm lm vic cho ra tn hiu s t Q0...Qm-1 , ng thi qua b bin i D/A s c in p U0 cho n khi in p U0/UV th b so snh lt gi tr u ra ca n c gi tr cng AND s kho b m s dng. Trn u ra b m Q0...Qm-1 dng s t l vi in p vo UV s ny c vit vo b ghi. Tip theo b m c xo v khng v chun b cho chu k bin i tip theo. Sau mi chu k ni dung b ghi s ghi s liu mi ca b m. B bin i A/D song song vi mch so snh trn hng I-6: B bin i A/D vi tc nhanh nht. in p vo UV c so snh vi tng gi tr URi tng ng vi tng bt ca s u ra. Nh vy thi gian bin i ch bng thi gian tc ng ca mch so snh. Tuy vy cn s lng ln b so snh v mch in tr chnh xc. Ux + So snh A/D Hnh I-6 b bin i A/D song song vi mch so snh Sinh vin thit k: Nguyn iu khin Ghi N
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Trang 93
Khi 4,9:
Mn hnh hin th s. Khi 4 hin th s kg/m. Khi 9: Hin th s m/ph.
Khi 6: my tnh
Card chuyn i A/D 8 knh 8 bt , trong h thng ny s dng 6 knh. Phn mm ng dng : - tc bng ti m/ph. - Tnh ton khi lng vt liu kg/ m ( bng ti ). - B nhn tnh ra c khi lng vt liu l kg/ph. Phn mm c thng s ca Dinverter (do hng bin tn cung cp). Phn mm ng dng trong my tnh do ngi lp ra chng trnh thc hin chc nng thu thp cc thng tin da n v x l cc thng tin o m khi h thng c s c, a ra tn hiu iu khin.
Khi 7: B chuyn i RS232/485.

. o lng v ghp ni vi my tnh qua cng RS- 232
Card ghp ni u o
COM
H1. S khi n gin ca mt h o lng v ghp ni vi my tnh qua cng RS-232 S khi n gin ca mt h o lng v ghp ni vi my tnh qua cng RS- 232 nh hnh 1. S liu t u o l analog s c a ti card ghp ni card ghp ni ny s c b bin i A/D v sau n s c bin i thnh tn hiu chun ca RS-232. Tc l mc +3V n +12V tng ng vi mc 0 v -3V n -12V tng ng vi mc 1. a d liu vo PC qua cng COM th ta phi c phn mm nht qun gia card ghp ni v cng COM v mt tc , cng mt format thng tin SDU.
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Trang 94
1. Vi nt v cng ni tip RS- 232

Cng ni tip RS- 232 l loi giao din ph bin, rng ri nht. Ngi dng my PC cn gi cc cng ny l COM1 v COM2. Vi COM2 t do cho cc ng dng khc. Ging nh cng my in v cng thng qua card ghp ni qua Slot m rng th cng ni tip RS- 232 cng c s dng mt cch ht sc thun tin cho cc mc ch o lng v iu khin cng nh khi c mt ng dy lin lc hay mt knh thng tin th n c th c rt nhiu ng dng nh truyn vn bn qua modem hay fax. Vic truyn d liu qua cng RS- 232 c tin hnh theo cch ni tip, ngha l cc bt d liu c gi i ni tip nhau trn mt ng dn. Trc ht, loi truyn ny c kh nng dng cho nhng khong cch ln hn, bi v cc kh nng gy nhiu l nh hn khi dng cng song song. H thng tin trao i ni tip c thc hin theo cc phng thc :
a cng: s liu ch c gi theo mt hng Bn song cng: s liu c gi theo hai hng nhng ti mi thi im th ch c th gi i theo mt hng Song cng: s liu c gi theo hai hng thng tin ni tip c phn lm hai loi: + Thng tin ng b + Thng tin khng ng b
Trong thng tin khng ng b (di b), mt t (hay mt cm t th c truyn hay thu theo tng khung mt v khung gi l SDU (serial data unit). Mt khung bao gm:
Mt bt start 5,6,7,8 bt d liu 1 bt pht hin li Bt stop (1,1.5 hoc 2)
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Trang 95
y bt pht hin li l dng bt chn-l
2. Chun ghp ni RS- 232

Cng ni tip RS-232 khng phi l mt h thng bus, n cho php d dng to ra lin kt im-im gia hai my cn trao i thng tin vi nhau. Mt thnh vin th 3 khng th tham gia vo cuc trao i thng tin ny.
H . 2 Sp xp chn ca cng ni tip my tnh PC
Chn(loi Chn(loi 9 chn) 1 2 3 4 5 6 7 8 9 25 chn) 8 3 2 20 7 6 5 4 22
Chc nng
DCD(Data carrier delect) RxD(Receiver data) TxD(Transmit data) DTR(Data terminal ready) GND(Ground) DSR(Data set ready) RTS(Request to send) CTS(clear to send) RI(Ring Indicator)
Li vo Li vo Li ra Li ra ni t Li vo Li ra Li vo Li vo
T hnh 2 ta thy cm RS-232 c tng cng 8 ng dn v mt ng t. Trn thc t c hai loi phch cm 9 v 25 chn, c hai loi ny Sinh vin thit k: Nguyn
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u chung c im lm vic. Vic truyn d liu xy ra trn 2 ng dn RxD v TxD. Qua chn TxD, my tnh gi d liu ca n n my kia. Trong khi th my tnh nhn cc d liu ti chn ni RxD. Cc tn hiu khc ng vai tr nh l nhng tn hiu h tr khi trao i thng tin v v th khng phi trong mi ng dng u dng c. Vic truyn d liu : Mc tn hiu trn chn ra RxD tu thuc vo ng dn RxD v thng thng nm trong khong -12V n +12V. Cc bt d liu c gi o ngc li. Mc in p i vi mc high nm gia -3V -12V v mc Low nm gia +3V +12V. Trn hnh 3 m t mt dng d liu in hnh ca mt byte d liu trn cng ni tip RS- 232. Khi trng thi tnh trn ng dn c in p -12V. Mt bt khi ng (bt start ) s m u vic truyn d liu. Tip l cc bt d liu ring l s n, trong nhng bt gi tr thp s c gi trc tin. Con s ca cc bt d liu thay i gia 5v 8 bt. cui ca dng d liu cn mt bt dng (stop bt), t tr li trng thi li ra (-12V) Bng tc baud ta thit lp tc truyn d liu v cc gi tr thng thng l 300; 600; 1200; 2400; 4800; 9600 v19200 baud. V y k hiu baud tng ng vi s bt c truyn trong mt giy. T ta thy rng c mt bt start v bt stop c gi cng bt d liu. Nh vy ta c th bit c tc cc i ca d liu c truyn nh vy vi mi byte c 10 bt c gi. Vi tc 9600 baud cho php truyn nhiu nht l 960 byte mi giy. Qua cch tnh ny ta thy mt nhc im khng nh ca cng truyn ni tip l tc truyn d liu b hn ch. Cn vi vn na l khun mu (Format) truyn d liu cn phi c thit lp nh nhau c hai bn (bn gi v bn nhn). Cc thng s truyn c th c thit lp trn cc my PC bng cc cu lnh trn DOS. Ngay c trn Window cng c chng trng ring s dng n. Sinh vin thit k: Nguyn
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Khi cc thng s truyn d liu nh: tc baud, s bt d liu, s bt dng, bt chn l (parity) c th c thit lp mt cch rt n gin. S trao i ca ng dn tn hiu y cc ng dn tn hiu ring bit cng Start bt 12V D0 -12V D1 D2 D3 D4 D5 D6 D7 HIGH Stopbt LOW
104 s 1,04 ms H.3 dng d liu trn cng RS- 232 vi tc 9600 baud
3. Vi nt v giao din 485.

Giao din 485 l kt qu ca vic m rng tiu chun RS-232. Vi cng di ng truyn v tc truyn cho php giao din ny cho php nhiu hn hai thnh vin tham gia ( c th n 32 thnh vin). C th xem giao din RS -485 nh l mt bus, bi v s lng thnh vin tham gia truyn thng trn giao din RS -485 khng b hn ch bi con s 2(nh giao din RS-323) do c tnh ba trng thi (tristate) ca tng thnh vin ring l. Mc logic 1 c n nh tng t nh giao din 422 -ngha l nm trong vng in p -1,5v n -6v, cn mc logic 0 th trong vng in p t +1,5 n +6v. B m ng dn ca giao din RS-485 to ra mt in p vi sai bng 5v trn hai dy dn truyn d liu, ging nh trong trng hp RS-422. Giao din 485 c dng nhiu trong cc h thng o lng, trong nhiu u o cn c thu thp v x l s liu ng thi. Mt trong cc Sinh vin thit k: Nguyn
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th d l h thng cn bng nh lng nh my xi mng lu x, h thng theo di lng hng c xut nhp cc kho xng du...cc s liu o lng c truyn cng vi m s tng ng qua cng RS-485 trn mt bus. B x l trung tm thu thp tt c cc s liu qua cng ni tip ri x l trn my vi tnh.
Khi 10: Bin tn . H thng s dng 6 bin tn iu khin tc

ca ng c - khng ng b xoay chiu 3 pha. Bin tn ca anh thuc hng Control Technic k hiu Din 12201508 vi tn s vo 50 60Hz ,tn s u ra 0 960Hz .in p u vo 220v 240v ,dng du vo 28,5A,dng u ra 7A.Bin tn ny s dng in p u vo xoay chiu 220v ,din p u ra 3 pha Trong b nin tn cn c b x l .u ra b bin tn c tn s c th thay i t 0,1,2...,960Hz thay i vi mt lng rt nh gn nh thay i trn .Trong h thng ny ngi ta ch s dng vi s thay i t 050Hz .
Khi 11: Trong h thng ny s dng 6 ng c khng ng b

xoay chiu 3 pha v loi ng c ny c cu to n gin, kch thc v trng lng nh, gi thnh h vi nguyn l lm vc n gin, tin cy cao tuy nhin Ik ln Mk nh, vic iu chnh tc ng c l kh khn. c tnh c ca ng c khng ng b c dng nh sau: Hnh I-7. c tnh c ca ng c khng ng b. Phng trnh c tnh c : 3.Uf2R2/ M=
0S [( r1+ R2 // S ) 2 + Xnm 2]
Trong : r1 , r2 : L in tr dy qun 1 pha mch stato, rto quy i. x1 , x2 :L in khng ca mt pha dy qun mch stato, mch rto quy i v mch stato. Ur : L in p pha t len mt pha dy qun mch stato ng c. Sinh vin thit k: Nguyn
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Rr/ :in tr ph a vo mch rto ng c. S : trt tc . S=
0 - 0
0 : tc t trng quay. : tc trn trc ng c.

R2/ = r2/ + Rf/ Xnm/ = x1 + x2/ Do yu cu v k thut, do yu cu cng ngh ca nh my nn trong h thng truyn ng cn bng nh lng s dng ng c khng ng b xoay chiu 3 pha Roto dy qun truyn ng cho bng ti. ng c ny ca Trung Quc Y802-4-4 vi cng sut 0,75kw, n = 1450 v/pht, f = 50 Hz, U = 380 v. iu khin ng c khng b dng phng php tn s, khi ng c khng ng b khng nhn in t li chung m t mt b bin tn. B bin tn c ni vi li c tn s l f = 50 Hz , in p xoay chiu 220v. u ra ca n khng nhng c tn s f1 bin thin m c U1 bin thin. Bng cch bin i tn s ngun cung cp, tc ng c c iu chnh cao hn v c thp hn tr s c bn, tn s ngun cung cp khng vt qu 1,5 2 ln tn s nh mc. Gii hn trc ht l do bn vng ca bi dy rto quyt nh. Hn na khi tn s tng, tn tht cng sut trng li thp stato cng tng ln r dt. Phm vi iu chnh tc trong vng thp hn tr s c bn thng nh hn 10 15 ln. Gii hn di ca tn s b hn ch bi phc tp ca b ngun c tn s thp, kh nng quay khng u v mt s yu t khc na. do iu chnh tc ng c khng ng b bng phng php tn s c th thc hin c trong phm vi di (20-30) :1 . Nu s dng cc ng c c cu trc dc bit th phm vi iu chnh c m rnh rng thm v gii hn trn ca tc Sinh vin thit k: Nguyn
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ln ln , gii hn di ca tc c c th gim nh c bng cch a cc mch hi tip vo s iu khin . Trong qu trnh iu khin tn s, ta bin i in p sao cho = const , th m men cho php trn trc ng c khng ng b cng dc gi khng i. ng c M l ng c khng ng b xoay chiu 3 pha thc hin nhim v truyn chuyn ng cho bng ti .
II. Nguyn l ca h thng :

Trong s khi trn mi bng ti cn c lp t cc u cn in t o ti trng trn chiu di bng, tn hiu o c o qua b khuch i a vo my tnh. Cc tn hiu m, V c c vo my tnh theo cc ng khi lng( S dng Card chuyn i A/D 6 knh ), ng tc ( cng COM 1 t b chuyn i RS-232/RS-485) My tnh s tnh c nng sut thc ca cc cn Q=mV, so snh vi nng sut nh mc Q ca chng c nhp vo t bn phm sau khi c hiu chnh th s a ra tn hiu iu khin Uk iu khin cc ng c thng qua b bin tn . Mc ch l iu chnh tc ng c hp l cho cc bng ti cn sao cho sai s thc ca cn vi nng sut nh mc ca chng t c 1%. Khi h thng c s c nh mt cn ngng hot ng hoc do lng nguyn liu xung qu t tc bng ti nhanh ( ng c iu chnh tc bng ti c tc quay ln hn nh mc ) hoc lng nguyn liu xung qu nhiu lm cho bng ti dng ... th u o khi lng s a tn hiu vo trong my tnh. Lc ny trong my tnh c chng trnh c lp sn s x l ngay s c xy ra lm cho tt c h thng u dng lm vic.
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kt lun ti nghin cu iu khin CKB rotor lng sc theo phng php ta theo t thng rotor l mt ti rt phong ph v phc tp. Trong bn n ny em trnh by c mt s vn c bn ca ti. - Xy dng c m hnh ton hc ca b bin tn theo nguyn l iu ch rng xung. - Xy dng m hnh ton hc ca ng c khng ng b rotor lng sc hai dng lin tc v gin on. - V d thc tin v lp trnh tnh thi gian ng ngt van. Tuy vy cc kt qu nhn c da trn c s cc gi thit n gin ho. + Coi mch t ca ng c l i xng v khng bo ho. + B qua sng hi bc cao b bin tn. + Coi in p ngun mt chiu b bin tn l khng i. + Coi m men qun tnh l khng i. Ni dung ti cn nhiu kha cnh cha xt n. V vy m ti ny cn nhiu vn cn phi hon thin. Hng pht trin ca ti c th cp n cc vn sau: + Xt nh hng ca cc yu t m ta n gin ho. + Thit k cc thng iu khin ti u v iu khin thch nghi. + Lp trnh iu khin cho c h thng. c th hon thin ti ny cn phi u t nhiu thi gian v cng sc. V vy em rt mong c s gip tip tc ca thy c v s cng tc ca cc ng nghip. Em xin chn thnh cm n!
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Ti liu tham kho

1. iu khin t ng truyn ng in xoay chiu ba pha tc gi Nguyn Phng Quang. Nh xut bn gio dc. 2. iu khn t ng truyn ng in tc gi: Bi Quc Khnh, Nguyn Vn Lin, Phm Quc Hi, Dng Vn Nghi. Nh xut bn khoa hc v k thut. 3. Cc phng php iu chnh dng trong truyn ng in xoay chiu ba pha. Nguyn l v hn ch chng-Tuyn tp hi ngh chuyn ngnh iu khin cc h thng c hc . 4. Truyn ng in-Bi Quc Khnh, Nguyn Vn Lin, Nguyn Th Hin. Nh xut bn khoa hc v k thut. 5. My in - A.V.IVANOV SMOLENSKI. Bin dch ,Phan t Th ,V Gia Hanh. Nh xut bn khkt.
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Ph lc
Phn I: Tng quan v ng c khng ng b v cc phng php iu chnh tc . I.Tng quan v ng c khng ng b. 1.c im trung. 2.Tm tt qu trnh pht trin ca CKB. II.Cc phng php iu chnh tc CKB. 1.iu chnh tc CKB bng cch thay i in p u1. 2.Phng php iu chnh tc bng cch thay i in tr mch day qun rotor. 3.iu chnh cng sut trt lm tc ng c. 4.iu chnh tc CKB bng phng php thay i tn s. 4.1.Xt mt s lut bin i tn s. 4.1.1.B bin di tn s. 4.1.2.Bin i tn s Dng in. 4.2. Cc b bin i tn s. 4.2.1.B bin i tn s trc tip. 4.2.2b bin i tn s gin tip. III.Kt lun. Phn II.M t cc i lng ba pha ca CKB rotor lng sc. I.Xy dng vector khng gian. II.Chuyn h to cho vector khng gian. III.Khi qut u th ca vic m t CXCBP trn h to t thng rotor. Phn III.M hnh lin tc ca CKBBP rotor lng sc. I.H phng trnh c bn ca ng c. I.1.Phng trnh in p stator. I.2.Phng trnh in p rotor. Sinh vin thit k: Nguyn
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II.M hnh trng thi ca ng c trn h to stator III. M hnh trng thi ca ng c trn h trc to dq. IV.Cu trc c bn ca 1 h truyn ng dng CKB u khin t thng rotor Phn IV.M hnh gin on ca ng c xoay chiu ba pha. I.M hnh gin on ca ng c khng ng b trn h to t stator II.M hnh gin on ca ng c khng ng b trn h to t rotor. III.M hnh tng qut cho c hai loi n c ng b v khng ng b. Phn V.iu ch bin tn trn c phng php iu ch vector khng gian. I.S nguyn l ca ng c xoay chiu ba pha nui bi bin tn. II.Nguyn l ca phng php iu ch vector khng gian. III.Cch tnh v thc hin thi gian ng ngt van bn dn ca bin tn. Phn VI.Cc vn ta theo t thng rotor, o c gi tr thc v xc nh gi tr cn. I.Cc vn ta theo t thng. II.tnh dng kch t v t thng rotor. III.Cch tnh gc pha ca t thng rotor. III.1.Tnh gc pha trn h trc to . III.2.Hiu chnh gc pha s c dng vo vic chuyn h to cho dng v p. Phn VII.Thit k b iu chnh dng stator c p ng tc thi. VII.1.Cc vn c bn. VII.2.Khi qut v cc phng php CD c s dng VII.2.1.iu chnh dng tuyn tnh. VII.2.2.iu chnh dng phi tuyn. Sinh vin thit k: Nguyn
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VII.3.Thit k b iu chnh dng c p ng vi tc tc thi. VVII.4.Kt lun. Phn VIII.Cc b iu chnh vng ngoi. VIII.1.M u. VIII.2.B iu chnh t thng rotor. VIII.3.Cc b iu chnh tc quay, iu chnh v tr. Phn IX.V d thc tin-lp trnh tnh thi gian ng ngt van. *V d s dng vi x l tn hiu TMS 320C25 v vi iu chnh SAB 167. *V d s dng vi x l kp dng vi iu chnh Sab C167 v vi x l tn hiu TMS 320 C25. *Tnh ton gn ng cc thng s t nhn ng c *Lp trnh tnh ti gian ng ngt ca van T biu tnh tng qut ca thut ton CVTKG.(lp trnh bng ngn ng Turbo c ) Phn X.Gii thiu cng ngh h thng cn bng nh lng nh my xi mng Lu X X.1.Gii thiu cc khi trong s . X.2.Nguyn l ca h thng.
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Thuyt minh n tt nghip

Thuyt minh n tt nghip

Sinh vin thit k: Nguyn

Thuyt minh n tt nghip

Phn I Tng quan v ng c khng ng b v cc phng php iu chnh tc

Thuyt minh n tt nghip

Thuyt minh n tt nghip

1 Sdm Sth 0 Mdm Mth M

Hnh 1. c tnh c ca ng c khng ng b Sinh vin thit k: Nguyn

Thuyt minh n tt nghip

Phng trnh c tnh c :

3.U 2 f1R 2 s M= R 1S[(R 1 + 2 ) 2 + X 2 ] nm S

Thuyt minh n tt nghip

S nguyn l ca h iu chnh nh sau: Rf Uli

Thuyt minh n tt nghip

Thuyt minh n tt nghip

1 =900 2 =1000 n =1400

Mtht Sinh vin thit k: Nguyn

Thuyt minh n tt nghip

Gi thnh truyn ng 1 chiu Gi thnh phn iu khin

Gi thnh phn iu khin

Thuyt minh n tt nghip

Sinh vin thit k: Nguyn

Thuyt minh n tt nghip

S nguyn l bin i tn s-in

Sinh vin thit k: Nguyn

Thuyt minh n tt nghip

Sinh vin thit k: Nguyn

Thuyt minh n tt nghip

Thuyt minh n tt nghip

Ngun ba pha A B C pha 1 pha 1

Thuyt minh n tt nghip

Thuyt minh n tt nghip

Thuyt minh n tt nghip

Phn II M t cc i lng ba pha ca ng c khng ng b rotor lng sc

Thuyt minh n tt nghip

Hnh 2-2. Thit lp vector khng gian t cc i lng ba pha

Thuyt minh n tt nghip

Cun dy pha V i s 1200 isv isw is Cun dy pha U

Thuyt minh n tt nghip

Sinh vin thit k: Nguyn

Thuyt minh n tt nghip

Thuyt minh n tt nghip

Khin bin tn isd isq Pt(2.10) e-js i s i s Pt(2.4) 3 2 M 3 3 u

Thuyt minh n tt nghip

Thuyt minh n tt nghip

rd - phn t d ca vector t thng rotor (cng chnh l mdul ca

Sinh vin thit k: Nguyn

Thuyt minh n tt nghip

iu chnh dng iu chnh v tr MMisq

Thuyt minh n tt nghip

Phn III M hnh lin tc ca ng c khng ng b ba pha rotor lng sc

dsu ( t ) u SU ( t ) = R s .i su ( t ) + dt dsv ( t ) u sv ( t ) = R s .i sv ( t ) + dt dsw ( t ) u (t) = R i (t) + s sw sw dt

31.a 31.b 31.c

Thuyt minh n tt nghip

2 usv(t).e j120 +uswe j240 us = usu(t) + 3

Tng t vi cun dy stator ta thu c phng trnh in p rotor nh sau : 0=Ri +

Thuyt minh n tt nghip

Sinh vin thit k: Nguyn