Do an Dien Co Thay Dinh

MC LC

Nguyn Duy nh TH3 K49

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LI NI UGia cng mi l mt trong nm nguyn cng gia cng c bn gm: tin, phay, khoan doa, bo, mi, Nguyn cng mi lm tng bng, gim nhm b mt chi tit gia cng. mi c th gt mt lp kim loi rt mng nn khi mi c th t n chnh xc rt cao. Nguyn cng mi c thc hin trn my mi. C nhiu loi my mi, trong , c hai loi chnh l my mi trn v my mi phng. Ngoi ra, cn c cc my khc nh my mi v tm, my mi rnh, my mi ct, my mi rng, v.v My mi chim khong 25% trong s cc loi my ct kim loi. Hin nay, mi khng nhng dng cc nguyn cng gia cng tinh m cn dng cc nguyn cng gia cng th khi cn c nng sut v hiu qu kinh t cao. n ny s i su tm hiu cng ngh my mi trn v thit k h truyn ng quay trn chi tit cho my mi trn. n gm 5 phn chnh: Phn 1: Cng ngh gia cng ct gt kim loi v cng ngh mi. Phn 2: Tnh chn phng n truyn ng. Phn 3: Xy dng h thng iu khin cho h truyn ng Phn 4: M phng h thng Phn 5: Thc hin b iu khin bng mch in t. Do thi gian c hn v hn ch v kin thc cng nh thc nghim nn n khng trnh khi cn nhiu thiu st. Em knh mong nhn c s ch bo, gp ca cc thy c v cc bn mang n hon thin v st thc t hn. Em xin chn thnh cm n thy Nguyn Minh H v cc thy c trong b mn T ng Ha X Nghip Cng Nghip Khoa in i hc Bch Khoa H Ni n cn ch bo v hng dn em hon thnh n.


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CHNG 1:

CNG NGH GIA CNG CT GT KIM LOI V CNG NGH MI

1. Gii thiu cng ngh gia cng ct gt kim loi v cng ngh mi 1.1. Gii thiu cng ngh gia cng ct gt kim loi Gia cng ct gt kim loi l cng ngh gia cng bng cch ct bt cc lp kim loi tha, sau khi gia cng c hnh dng gn ng yu cu (gia cng th) hoc tha mn hon ton yu cu t hng vi chnh xc nht nh v kch thc v bng cn thit ca b mt gia cng (gia cng tinh). Cc my thc hin vic gia cng ct gt kim loi gi chung l my ct kim loi. Trn quan im cng ngh, gia cng ct gt kim loi c my loi c bn l tin, phay, khoan doa, bo, mi. Nm cng ngh c phn bit nhau bi 2 chuyn ng: chuyn ng chnh v chuyn ng n dao. Trong : Chuyn ng chnh l chuyn ng to ra lc ct. Chuyn ng n dao l chuyn ng ct to ra phoi.

Hnh 1.1: Cc dng gia cng trn my ct kim loi a) Tin. b) Khoan. c) Phay. d) Mi. e) Bo Nguyn Duy nh TH3 K49 -3-

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Gia cng trn my tin (a): chi tit quay trn, dao chuyn ng tnh tin. Gia cng trn my khoan (b): chi tit ng yn, dao chuyn ng quay trn. Gia cng trn my phay (c): chi tit chuyn ng tnh tin, dao chuyn ng quay trn. Gia cng mi (d): chi tit chuyn ng quay trn, dao chuyn ng quay trn. Gia cng bo (e): chi tit chuyn ng tnh tin, dao chuyn ng tnh tin.

Sau y, s ch i su gii thiu cng ngh mi trn. 1.2. Gii thiu cng ngh gia cng mi trn:

Hnh 1.2: Hnh dng chung ca my mi. Mi trn c hai loi: mi trn trong v mi trn ngoi. Trn my mi trn, chuyn ng chnh l chuyn ng quay ca mi; chuyn ng n dao l chuyn ng tnh tin ca dc trc (n dao dc trc) hoc di chuyn tnh tin theo hng ngang (n dao ngang trc) hoc chuyn ng quay ca chi tit (n dao vng). Chuyn ng ph l di chuyn nhanh hoc chi tit.

Hnh 1.3: Mi trn ngoi Nguyn Duy nh TH3 K49

Hnh 1.4: Mi trn trong

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C 2 phng php mi c bn: mi chy dao dc v mi chy dao ngang: Mi chy dao dc: chuyn ng chy - Mi chy dao ngang: l phng php c lng chy dao ngang, dng mi trn cc chi tit c chiu di ngn hn chiu rng ca dao hng knh ch c thc hin cui hnh trnh sang tri hoc sang phi sau mt hnh trnh kp ca bn my mang chi tit

Hnh 1.5: Mi chy dao dc

Hnh 1.6: Mi chy dao ngang mi.

C hai ch mi: mi th v mi tinh. Khi mi th, lm th no trong thi gian ngn nht gt c nhiu lp kim loi tha cn yu cu i vi cht lng mt ngoi v chnh xc gia cng th tng i thp. Khi mi tinh, chiu su ct rt nh; khi mi ht lp kim loi tha trn vt gia cng, khng c dng bc tin ngang m c tip tc mi cho n khi khng cn pht ra tia la mi thi. Mi lm tng bng, chnh xc ( bng t cp 8 10, chnh xc t n cp 2). Mi cng c th ct c cc loi thp ti cng cc loi thp dng c m cc gia cng khc khng lm c. 2. Phn tch c im, yu cu truyn ng y ch phn tch yu cu truyn ng cho truyn ng quay chi tit mi. R rng, vic gia cng phi c thc hin vi nhng chi tit khc nhau, tc l c cc ch ct khc nhau. Cc ch ct khc nhau c thc hin bng cch iu chnh tc truyn ng chnh v truyn ng n dao. Khi gii quyt vn ny, ta cn phi quan tm n cc ch tiu: phm vi iu chnh tc , trn iu chnh, iu kin ph ti, ch lm vic, n nh tc v tnh kinh t ca h thng truyn ng. Sau y s ln lt cp n cc yu t : Phm vi iu chnh tc : trong my mi trn, di iu chnh tc c tnh bi t s gia tc quay ln nht v tc quay nh nht. Thng, cc my mi iu tc theo cp bng cch iu chnh s i cc nn di iu chnh b. C th m rng di iu chnh bng Nguyn Duy nh TH3 K49 -5-

cch dng b bin tn iu chnh trn tc cho ng c quay chi tit. Phm vi iu tc cng rng cng c nhiu ch gia cng. - trn iu chnh: nh ni trn, trn iu chnh ph thuc cch iu tc cho truyn ng n dao. iu chnh cng trn th cht lng b mt gia cng cng tt. - iu kin ph ti: c tnh c ph ti c cho bi phng trnh:M c = M co + (M m M co ). m x

Trong : Mm: Momen nh mc ca ng c. Mco: Momen cn = 0 : Tc ang xt. x: c trng cho ti: x = 0 ng vi chuyn ng n dao; x = -1 ng vi chuyn ng chnh. Nh vy, ni ring trong truyn ng n dao truyn ng quay trn cho chi tit, x = 0; suy ra phng trnh c tnh c tr thnh: Mc = Mm = const T , ta thy ni chung momen ti l khng i. Tuy nhin, trong vng tc thp, lng n dao nh, lc ct b hn ch bi chiu su ct ti hn. Trong vng ny, tc n dao gim lm cho lc n dao v momen cng gim theo. Vng tc cao th b gii hn bi cng sut ca ng c truyn ng nn ti , momen cng phi gim khng lm cng sut ca truyn ng qu ln. Tm li, ta c c tnh c ph ti truyn ng quay chi tit nh sau: ng c khng chuyn ng. m: Tc nh mc ca ng c.

Hnh 1.7: c tnh c ph ti ca truyn ng quay chi tit. Nguyn Duy nh TH3 K49 -6-

Nh vy, nhim v ca truyn ng ng c l phi lm c tnh iu chnh ca n ging c tnh c ca my ct. - Ch lm vic: Khi gia cng mi, chi tit quay lin tc cn mi di chuyn trn b mt vng cn gia cng. Do , ch lm vic ca truyn ng n dao l ch lm vic di hn v khng yu cu o chiu. - Ch ti: Khi h thng lm vic, chi tit c lp trn trc ca tang trng v quay vi vn tc t l vi tc ca trc ng c. Do , ng c mang ti ngay t u. Do ch gia cng khc nhau, cc chi tit khc nhau, nn khng qui i momen qun tnh ca chi tit v trc ng c m coi chi tit nh mt ti c sn trn trc ng c. - n nh tc : R rng, tc quay cng n nh th cht lng gia cng cng cao, b mt mi cng nhn, bng. Yu cu i vi truyn ng n dao my mi: % (510)% - Tnh kinh t: Thit b cho h truyn ng phi r, nhng vn cung cp hiu qu cao nht cho h. ng thi, thit b phi d kim v hot ng tin cy trong ch di hn. n y, ta s tm tng kt chng 1 rt ra yu cu c th cho vic thit k h thng iu khin ng c truyn ng cho chi tit. Cn c vo yu cu ra, ta phi thit k h thng t c nhng yu cu sau: Di iu chnh tc : D = max : min = 480 : 48 = 10 : 1. iu chnh v cp tc , khng yu cu o chiu. iu chnh gi mmen khng i v bm theo momen ti. iu chnh gi n nh tc . Lm vic di hn, tin cy. Gi thnh h. Phn sau s i su vo thc hin k thut cho yu cu ra.


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CHNG 2: TNH CHN PHNG N TRUYN NG1. Chn phng n truyn ng Nh trn ni, chuyn ng quay ca chi tit mi chnh l chuyn ng n dao. i vi my mi trn, cc my c nh, truyn ng quay chi tit (truyn ng n dao) thng dng ng c khng ng b nhiu cp tc , iu tc bng cch iu chnh s i cc. cc my ln th dng h thng b bin i - ng c mt chiu/ng c ng b. Cng sut m bi yu cu l nh, do y s dng ng c khng ng b ba pha rotor lng sc truyn ng quay chi tit. Trc y, ng c in mt chiu thng c a chung hn, k c trong di cng sut nh v tnh iu chnh n gin v tuyn tnh ca n. Tuy nhin, ngy nay, cng ngh in t v vi iu khin pht trin mnh m, vic iu khin ng c khng ng b khng cn l kh khn na, hn na ng c khng ng b ba pha lng sc r hn ng c mt chiu cng cng sut nhiu v rt ph bin trn th trng vi di cng sut rng, do , hon ton ph hp cho ng dng ca chng ta. Do yu cu iu chnh trn tc nn ta dng b bin tn cp ngun cho ng c. Hn na, vic dng bin tn cho ta d dng m rng di iu chnh, d dng p t cc k thut iu khin hin i, p t nhanh v chnh xc momen, iu chnh trn v n nh tc . Ngoi ra, bin tn hot ng tin cy v chc chn, d dng ci t tham s iu khin, c th dng 1 bin tn cho nhiu loi truyn ng. Do , vic s dng bin tn tr thnh mt chun cng nghip. C nhiu hng ln sn xut bin tn rt ni ting nh ABB, Siemens, vi cc sn phm rt ni ting trn th trng, tuy nhin y, ta s thit k li b bin tn phc v cho bi ton yu cu m khng s dng bin tn sn c. Bin tn c 2 loi: bin tn trc tip v bin tn gin tip phn bit nhau bi khu trung gian mt chiu gia b chnh lu v b nghch lu. Sau y s phn tch u nhc im ca tng loi chn ra loi bin tn thch hp nht vi ng dng ca ta. 1.1.Bin tn trc tip (Cycloconverter)


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Hnh 2.1: Bin tn trc tip s tia 3 pha. u im: Mch ch cn dng van Tiristor thng thng, qu trnh chuyn mch theo in p li. B bin tn khng s dng khu trung gian mt chiu nn hiu sut rt cao. C kh nng lm vic tn s thp thm ch ngay c khi c s c. Thng s dng cho di cng sut rt ln n vi chc MW. S dng nhiu van bn dn lm cho mch iu khin rt phc tp. H s cng sut thp.

Nhc im:

Tm li, vi ng dng l h truyn ng cho my mi, ta khng dng loi bin tn ny. 1.2.Bin tn gin tip: Bin tn gin tip khc bin tn trc tip ch n c khu trung gian mt chiu. Nh c khu trung gian mt chiu ny m khu chnh lu v khu nghch lu l cch ly nhau v iu chnh c lp vi nhau. Tn s u ra nh c th c iu chnh m khng ph thuc tn s u vo. Ty thuc vo khu trung gian mt chiu m phn ra thnh bin tn ngun dng v bin tn ngun p. 1.1.1. Bin tn ngun dng


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Hnh 2.2: Bin tn ngun dng Khu trung gian mt chiu l cun khng Lf, thc hin chc nng ngun dng cho b nghch lu. u im: C kh nng tr nng lng v li. Khng s ch ngn mch v dng in mt chiu c gi khng i. Ph hp cho di cng sut ln trn 100 kW Nhc im: Hiu sut km di cng sut nh. Cng knh v c cun khng. H s cng sut thp v ph thuc vo ph ti nht l khi ti nh. Do , vi ng dng my mi vi ti ch vo khong 2,2kW ca ta, bin tn ngun dng r rng l khng ph hp. 1.1.2. Bin tn ngun p:

Hnh 2.3: Bin tn ngun p.


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Khu trung gian mt chiu l t Cf, thc hin chc nng ngun p cho b nghch lu. u im: Ph hp vi ti nh, di 30kW H s cng sut ca mch ln (gn bng 1) Hnh dng v bin in p ra khng ph thuc ti, dng in cho ti qui nh. C th p dng k thut PWM gim tn hao do sng hi bc cao, kh p mch momen. Nhc im: Khng tr c nng lng v li, nu mun tr nng lng v li phi mc thm mt khu chnh lu mc song song ngc vi khu chnh lu ban u hoc dng chnh lu PWM hay bin tn 4 gc phn t. Nh vy, n y, ta thng nht chn b bin i l bin tn ngun p. Phn tip theo s chn phng php iu khin cho loi bin tn ny. 1.3.Phng php iu khin bin tn: C rt nhiu phng php iu khin cho bin tn ngun p. Ph bin trong cng nghip l iu khin theo lutU = const , iu khin theo h s trt, iu khin ta t thng rotor (FOC) f

v gn y iu khin trc tip momen (DTC) cng xut hin trong cc b bin tn cng nghip thay th cho FOC. n ny ch nh nghin cu ng dng phng php DTC cho iu khin b bin tn ngun p v mt s l do sau: Phng php DTC cho php p t rt nhanh momen do , hon ton ph hp vi ng dng my mi. Phng php DTC cho php c th iu chnh vi chnh xc l ty . M hnh n gin, khng ph thuc nhiu tham s, do , khng b nh hng bi sai lch do tham s ca ng c nh cc phng php khc. Khng phi thc hin php quay ta do , thi gian tnh ton nhanh. Tuy vy, DTC cng c nhc im: p ng tc thp rt km; p ng momen khng trn, nhp nh momen ph thuc di tr v kh c th khc phc s nhp nh momen ny.


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V mt l do na, l DTC d ra i lu nhng cha ph bin Vit Nam ( Vit Nam ph bin dng DTC v U/f) m ph bin cc nc chu u. Trong qu trnh hi nhp, cc b iu khin ca nc ngoi chc chn s trn vo Vit Nam, cho nn, em mun i su tm hiu, hc hi v th nghim phng php DTC nhm ng dng sau ny. Kt lun: phn tip theo, ta s i su vo thit k h truyn ng bin tn ngun p

ng c khng ng b ba pha rotor lng sc phng php iu khin trc tip momen. 2. Chn ng c Yu cu ca h thng: Momen cc i (Mmax): Tc quay chi tit (n): T s truyn (i): Hiu sut (): Momen qun tnh c cu (J): 25Nm 48 480 vng/pht 3 0,8 0,009 kg/s2

Phm vi iu chnh tc , quy i v trc ng c Tc b nht ca chi tit: nminct = 48 vng/pht minct =

48 = 5,03 rad/s 60 / 2

Tc quay ln nht ca chi tit: nmaxct = 480 vng/pht Di iu chnhD=

maxct =

480 = 50,3 rad/s 60 / 2

wmax = 10 : 1 wmin

Tnh quy i tc v trc ng c qua hp s c t s truyn i = 3 min = i . minct = 3 . 5,03 = 15,09 rad/s max = i . maxct = 3 . 50,3 = 150,9 rad/s Tnh Mmen quy i v trc ng cNguyn Duy nh TH3 K49 - 12 -

Momen cc i tang trng Mmax = 25Nm Mqd =

momen qui i:

M max 25 = = 10,42 Nm 0,8.3 .i

Cng sut c cc i yu cu ca ng c: Pmax = Mqd . max = 10,42 . 150,9 1,57 kW Chn ng c: ng c c chn cn c cng sut ln hn Pmax = 1,57 kW, c Momen nh mc 10,42Nm, c di iu chnh t nht l D = 10:1, v tc nh mc c 150,9 rad/s tc l c 1440 vng/pht. Trn c s , ta chn ng c khng ng b Rotor lng sc ca ABB c thng s nh sau:

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Tn: Cng sut nh mc: in p dy nh mc: Tn s nh mc: S i cc: Tc nh mc: Hiu sut: H s cng sut: Dng Stator nh mc:

M3AA 100LC 3GAA 102 313 - CG2. Pm = 2,2 kW U1m = 400V f = 50Hz p=2 nm = 1450 vng/pht = 86,8% cos = 0,77 I1dm = 4,8A sin = 0,638


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Dng Stator khi ng: Momen nh mc: Momen ti hn: Momen qun tnh: Trng lng:

Is = 4,8 . 8,5 = 40,8 A. Mdm = 14,5Nm Mth = 14,5 . 4,6 = 66,7 J = 0,009 kgm2 m = 25kg

Momen khi ng nh mc(s=1): Ms = 14,5 . 4 = 58 Nm

Tnh ton tham s ng c

Hnh 2.1: Mch in thay th mt pha ng c khng ng b Do yu cu xy dng b iu khin cho ng c, ta phi xy dng c m hnh ng c. xy dng c m hnh ng c, ta phi xc nh cc i lng sau: Rs, Rr, Ls, Lr, LM.. Cng sut nh mc a vo ng c:Pvdm = 3U 1m .I1m . cos = Pm

= 3.400.4,8.0,77 =

2200 = 2,53 kW 0,868

(2.2 1)

Tc ng b:nm = 60 f 60.50 = = 1500 vng/pht p 2

(2.2 2)

H s trt nh mc: s m = ns nm 1500 1450 = = 0,033 1500 ns (2.2 3)


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Tng tr mt pha:

Z in =

400 U1m = = 48,11 3I1m 3.4,8

(2.2 4)

Momen trn trc ng c:

M =

2.M th (1 + a .sth ) s sth + + 2 asth sth s2 4f 1 Rs + Rs2 + X nm

(2.2 5)

Trong :

- M th =

(

p .U 12m

) = 66 ,7 Nm

(2.2 6) (2.2 7) (2.2 8)

' - X nm = X s + X r

- a=

Rs ' Rr' Rr 2 Rs2 + X nm

- sth =

(2.2 9)

Khi s = sm = 0,033 ta c Momen nh mc:M m = 2.M th (1 + a .sth ) 2.M th (1 + a .sth ) = = 14 ,5 Nm sm sth 1 + 30.sth + 2 asth + + 2 asth sth sm 30.sth

(2.2 10)

Khi s = 1, ta c Momen khi ng:Ms = 2.M th (1 + a .sth ) = 58 Nm 1 sth + + 2 asth sth 1

(2.2 11)

Gii h 2 phng trnh (2.2 10), (2.2 11) vi n l a, sth ta c:a= Rs = 1,57 ' Rr' Rr 2 Rs2 + X nm

(2.2 12)

sth =

= 0 ,49

(2.2 13)

' Tip tc gii h 3 phng trnh (2.2 6), (2.2 12), (2.2 13) v gi thit X s X r ta c ngay:


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Rs = 3 ,32 ' Rr = 2 ,11 ' ' X s = X r = 7 ,64 Ls = Lr = 4 ,39 mH

(2.2 14)

Nhim v tip theo l phi tnh c in khng t ha XM. Xut pht t mch in thay th mt pha ng c khng ng b, ta c:' & = jX // R + Rr + j .X Z in s nm s

(2.2 15)

Suy ra:

' Rr Rs + s & Re{Z in } = Z in .cos = 2 ' 2 1 Rr X nm . Rs + + 1 + 2 X s X

(2.2 16)

ch nh mc: s = sm = 0,033 v Zin = 48,11 ; cos =0,77. Thay vo (2.2 16) ta tnh c:' Rr Rs + s m

Z in . cos =

1 2 X

R' Rs + r . s m

X + 1 + nm X

2

X = 74 ,55 = 48 ,11.0 ,77 2 L = 237 ,3mH

(2.2 17)

Cui cng, ta cn phi quy i momen qun tnh ca h thng v u trc ng c. cho n gin, ta gi thit cc bnh rng ca hp s l l tng (ngha l khng c mmen qun tnh) v chi tit cha c gn ln tang trng. Do , momen qun tnh qui i c tnh bi:J qd = J c + J ht 0,009 = 0,009 + 2 = 0,01 kg.m2 2 3 i

(2.2 20)

3. Chn thit b cho b bin i

Nhn xt, my mi dng trong cng nghip s dng in p 3 pha c Udy = 380V trong khi , ng c m ta chn dng in p 400V. Do , ta khng cn iu chnh in p DC bus ca u ra chnh lu. Do , mch chnh lu s dng diode v khng cn mch xung p thay i DC bus. Khi , theo hng dn m nh sn xut ABB a ra, dng in nh mc ca mi pha ng c s l: Im-380 = 1,05.Im 400 = 1,05.4,8 = 5,04 A. Trn c s , ta chn cc van cho ph hp.


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Mch nghch lu c u vo l khu trung gian mt chiu, u ra l in p ba pha ni vo ng c. Dng cc i mi pha ng c chnh l dng nh qua mi van. Bng thut ton iu khin, ta c th gii hn dng in khi ng l dng cc i qua mi pha ng c khng vt qu 2,5 ln dng in nh mc. Do , dng hiu dng ln nht qua vanImax = 2,5.Im = 2,5.5,04=12,6A.

Dng mt chiu ca khu trung gian mt chiu:Id = I max 12 ,6 = = 14 ,6 A 0 ,861 0 ,861

Dng hiu dng qua van ca mch chnh lu:I lv = Id = 4 ,9 A 3

in p mt chiu DC Bus:U d = 2 ,34. U1 380 = 2 ,34. = 514 ,8 V 3 3 U1 380 = 2 ,45. = 539 V 3 3

in p ngc ln nht m cc van ca mch chnh lu cng nh nghch lu phi chu:U ng max = 2 ,45.

3.1.Chn van cho mch nghch lu Van nghch lu lm mt bng tn nhit nhm v qut gi, chn h s d tr v dng ki = 2,5. Suy ra dng Collector ca van bng:Icmax = 12,6.2,5 = 31,5 A

in p nh thng (VBreakDown) ca van nghch lu phi ln hn Udcbus = 514,8V. ch chnh lu, van cn phi chu c in p ngc Ungmax = 539V. Chn h s d tr v p ku = 1,1 suy ra:Vbreakdown = 1,1 . 514,8 = 566,3V Vngmax = 1,1.539 = 592,9

Trn c s , chn ra 6 van IGBT 1MBH30D-060 ca hng Fuji Electric c c tnh nh sau:


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Thuc tnh - VCEMax - VGEMax - ICMax (25oC) - ICMax (100oC) - Cng sut tiu th ln nht - C sn diode ngc, IDmax - ton/toff S mch nghch lu nh sau:Q1 1MBH30D-060

Gi tr 600V 20V 58A 30A 220W 30A 1,2s/1,0s

Q3 1MBH30D-060

Q5 1MBH30D-060

C1 3300uF - 630V

Q4 1MBH30D-060

Q6 1MBH30D-060

Q2 1MBH30D-060

Hnh 2.2: S mch nghch lu.

3.2.Chn van cho mch chnh lu: Mch chnh lu nh trn phn tch dng chnh lu diode mc theo s cu 3 pha in p ra t nhp nh nh nht (p mch 6 ln trong 1 chu k). Chn h s d tr v dng ki = 2,5 suy ra dng qua van:Iv = 2,5.Ilv = 2,5.4,9 = 12,25 A

in p ngc trn diode Ungmax = 539V. Chn h s d tr v p ku = 1,1 suy ra:Ungmax = 1,1.539 = 592,9 V

Vy chn ra 6 van YG963S6R ca hng Fuji c cc thng s:


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3.3.Chn t cho khu trung gian mt chiu Khu trung gian mt chiu l t c dung lng ln. Thng chn theo kinh nghim v theo cng sut ti: c 1W ti th ng vi 1 2 dung lng t. Cng sut m nghch lu cn cp cho ng c tnh trn: Pvdm = 2,53kW do , t cn chn c dung lng c khong 2530 5060. Tuy nhin, t thng c ch to theo cp 1000, 2200, 3300, do y, ta chn t c dung lng 3300 l . Nh trn tnh, EDCBus = 513,4 V. Cp in p ca t phi ln hn hoc bng 513,4V. Do t c ch to theo cc cp 385V, 400V, 630V, 1000V, Nn y, ta ch cn chn t c cp in p 630V l .Kt lun: nh vy, n y, ta chn xong ng c v cc thit b cn thit ca bin tn cp cho ng c. Cng vic tip theo l iu khin bin tn, cp in cho ng c mt cch hp l t c yu cu cng ngh phn tch chng mt. Vic tng hp b iu khin c thc hin chng tip theo.


- 19 -

CHNG 3: TNG HP CC B IU CHNHI. Tng hp b iu chnh DTC:

1. Nguyn l ca phng php DTC DTC l t vit tt ca Direct Torque Control, tc l phng php iu khin trc tip mmen. y l mt phng php mi xut hin vo gia nhng nm 80, do Depenbrock v Takahashi c lp xut trong hai ti liu : Mirekte Selbstregelung (DSR) fur hochdynamische Drehfeldantriebe mit Stromrichterspeisung Depenbrock 1988 v A new quick Response and High-Efficiency control sreategy of an induction motor Takahashi 1986. ng c khng ng b u tin c iu khin bng phng php DTC c ra mt vo nm 1995 do hng ABB ch to. y l mt phng php mi, trong vic phi hp iu khin b bin tn v ng c khng ng b l rt cht ch. Logic chuyn mch ca bin tn da trn trng thi in t ca ng c khng ng b m khng cn n iu ch rng xung p ca bin tn. Do s dng cng ngh bn dn tin tin vi cc phn t tnh ton c tc cao m phng php iu chnh trc tip mmen cho cc p ng u ra thay i rt nhanh, c vi phn nghn giy. Phng php iu khin DTC cho php iu khin ng c khng ng b theo mt nguyn l n gin :M= Lm Lm 3 p' 3 p' 3 p' s .is = . . r s = . . . . r s sin . 2 2 2 Ls Lr Lm 2 Ls Lr L2 m

(3.1 1)

y l biu thc tnh mmen trong h ta tnh, gn cht vi trc dy qun stator ( , ) .Vc t t thng rotor thng bin thin chm hn vc t t thng stator, do c th t c gi tr mmen theo yu cu bng cch quay vct t thng stato cng nhanh cng tt theo hng no , lm thay i gc mmen . Bin i (3.1 1) ta c:M = Lm 3 p' . s + s . j 2 2 Ls Lr L2 md s dtd s hay s = U s t dt

[(

)

s

]

(3.1 2)

Trong phng trnh in p stator: U s = Rs i s +

Nu b qua st p trn in tr thun Rs th: U s =


- 20 -

Vct U s c th ly cc gi tr gin on: U s =

2U dc S a + e j 2 / 3 S b + e j 2 / 3 S c 3

[

]

Trong Udc l in p mt chiu . Sa,Sb v Sc l cc hm ng ct ca cc van tng ng ca mch lc ca bin tn Mi kha Sa,Sb, Sc ly gi tr 1 khi ni vo +Udc hoc 0 khi ni vo Udc . Ta tng tng chia khng gian in p thnh 6 sectors (S1, S2, S3, S4, S5, S6) vi hai trng thi ca ba kha Sa, Sb, Sc ta c 8 trng hp V1, V2, V3, V4, V5, V6 tng ng vi gi tr trn, cn V0 = 111 v V7 = 000.

Hnh 3.1 : Cc hm ng ct v thc hin cc vector p.


- 21 -

C th iu khin phn ly bin t thng stator v mmen in t M bng cch tc ng vo cc thnh phn hng knh v thnh phn tip tuyn ca vt t thng mc vng stator trong qu o ca n. Pha a b c V0 0 0 0 V1 1 0 0 V2 1 1 0 V3 0 1 0 V4 0 1 1 V5 0 0 1 V6 1 0 1 V7 1 1 1

Bng 3.1: Vct in p ng vi trng thi kha Sabc

T phng trnh (3.1 2) ta thy, gi s, vector t thng Stator ang mt sector no . Bng vic thc hin mt vertor in p thch hp s lm cho modul vector t thng Stator, gc momen v do , momen thay i.

Hnh 3.2: S thay i ca vector t thng theo vector in p.

Nh vy, bng vic thay i vector in p mt cch thch hp, vector t thng s quay trn trong khng gian vi modul nm trong mt di tr no . Bng vic phn tch yu cu tng gim t thng v momen trong mi sector, ta thu c bng 3.2 l chin lc ng ct ti u cho nghch lu p thc hin DTC.


- 22 -

Hnh 3.3: Qu o t thng.

T

S1

S2

S3

S4

S5

S6

TI FI T= TD TI FD T= TD

V2 V0 V6 V3 V7 V5

V3 V7 V1 V4 V0 V6

V4 V0 V2 V5 V7 V1

V5 V7 V3 V6 V0 V2

V6 V0 V4 V1 V7 V3

V1 V7 V5 V2 V0 V4

Bng 3.2: Bng chn vector in p ti u.

Bng vic thc hin iu chnh Momen v t thng bng cc khu Relay, p ng ca momen s c dng nh sau:

Hnh 3.4:p ng kiu Relay ca momen.


- 23 -

2. M hnh h thng DTC:

Hnh 3.5: S nguyn l phng php DTC.

Hnh 3.5 l s n gin nht ca phng php DTC. Sai lch gia t thng c lng v t thng t (u ra ca b iu chnh t thng) c a vo khu Relay 2 v tr, sai lch gia mmen c lng v mmen t (u ra ca b iu chnh tc ) c a vo khu Relay 3 v tr. u ra ca cc khu Relay mang thng tin v yu cu tng hay gim momen v t thng cng vi thng tin v gc t thng Stator (cho ta bit v tr ca vector t thng Stator ang trong Sector no) cng c c lng v a vo bng chn 3.2. Kt qu, u ra bng chn, ta c vector in p cn phi thc hin thc hin yu cu tng gim momen v t thng ni trn. T nguyn l trn, d dng rt ra cc nhn xt sau y v chnh xc ca phng php iu khin: chnh xc iu khin hon ton ph thuc vo chnh xc ca php c lng momen, t thng v v tr. chnh xc iu khin ph thuc vo di tr momen v t thng. R rng, di tr momen cng hp th p ng momen cng mn, di tr t thng cng hp qu o ca t thng cng trn tc l dng Stator cng sin. Tuy nhin vic thu hp di tr s ko theo cc van IGBT phi tng tn s chuyn mch. 3. Cc vn v c lng v nng cao chnh xc: Nguyn Duy nh TH3 K49 - 24 -

Nh trn nhn xt, vic c lng ng cc tham s ca h thng s nng cao cht lng iu chnh. Xut pht t phng trnh cn bng p trong h ta :u s = Rs .is + d s dt

(3.1 3)

Suy ra:

s = ( us Rs .is )dt0

t

(3.1 4)

Phng trnh (3.1 4) c dng c lng t thng Stator. C th nh sau:t s = ( us Rs .is )dt 0 t = ( u R .i )dt s s s s 0

(3.1 5)

Cc gi tr u s ,u s ,is ,is c tnh t cc gi tr ia, ib, ua, ub o c t pha u ra ca nghch lu theo qui tc chuyn i abc . (ic v uc khng cn tnh do gi thit ti ng c l i xng nn hin nhin: ia + ib + ic = 0 v ua + ub + uc = 0) 2 u a = u 1 1 3 u = ua 2 ub 2 uc 1 1 u ub = u + 3 3 u = 3 (u u ) b c 1 1 2 u uc = u 3 3

(3.1 - 6)

T cng thc (3.1 1) c th vit li cng thc tnh momen nh sau:M = 3 p' 3 p' ( s is s is ) s .is = . 2 2

(3.1 7)

Gi tr is ,is c xc nh bng phng trnh (3.1 6); s , s c xc nh bng phng trnh (3.1 5). Cui cng, ta cn phi xc nh v tr v modul vector t thng s . Ta c:


- 25 -

s = s + j s = s . 2 2 s = s + s = arctan s s

(3.1 8)

Nhn xt: Phng php DTC c thc hin da trn cc phng trnh t (3.1 5) n (3.1

8). Ta nhn thy, ton b phng php ch dng ng mt tham s ca ng c l in tr Stator Rs, cc thng s khc u c tnh ton c lng. in tr Stator l thng s d dng o lng vi chnh xc cao, d dng b nhit . Ngoi ra, m hnh khng h dng n cc php quay h trc ta (nh phng php FOC) mt php ton tn nhiu ti nguyn v thi gian thc hin vi chnh xc ph thuc ln vo kh nng ca vi x l. Do , c th ni phng php DTC tin cy, n gin v chnh xc hn cc phng php khc nhiu. 4. Cc vn v nng cao chnh xc ca phng php: Theo nhn xt ca mc trn, chnh xc ca phng php ph thuc vo chnh xc ca php c ng cc tham s momen, t thng, v tr. Vic nng cao phn gii v kh nng tnh ton ca vi x l s lm tng chnh xc ca cc php c lng, qua ci thin tnh chnh xc ca phng php. mc n mn hc, vn nng cao chnh xc ca php c lng s khng c trnh by m c th c tham kho thm trong cc ti liu [13], [14]. chnh xc cn ph thuc vo tnh ng n ca vic nhn dng in tr Stator tham s duy nht ph thuc m hnh ng c. Cng vic ny cng khng qu kh khn v c cp trong ti liu [17] v cng khng c nu ra trong phm vi ca n. nng cao chnh xc ca phng php, cn mt cch na l ci tin bng chn vector in p. Ti liu [15] a ra phng php s dng 12 sectors 12 vectors, ti liu [14] a ra phng php s dng 12 sectors 6 vectors, ngoi ra cn c phng php 6 sectors 12 vectors, 24 sectors 6 hoc 12 vector, mc n mn hc, n vn s trnh by phng php s dng 6 sectors 6 vectors truyn thng. Cc ci tin khc ch thay i phn bng chn, tng ng vi vic thay i mt cht cch tnh ton v tr vector t thng Stator, tuy nhin s khng c trnh by y.II. Tng hp vng iu chnh tc

1. Phn tch: Nguyn Duy nh TH3 K49 - 26 -

Hnh 3.6: Vng iu chnh tc .

Nh ni, b DTC nh mt khu Deadbeat tc ng rt nhanh, do , khi tng hp vng iu chnh tc , ta c th coi khu DTC l mt khu qun tnh bc nht vi hng s thi gian Tm. M phng b DTC vi tn hiu t dng bc thang ch khng ti, ta thy, Momen p ng ngay ch sau khong 1-2ms ty thuc tc t. Mc d khi tng hp b iu chnh, ta phi tm cch b nh hng ca Tm, tuy nhin nu Tm qu nh s gy ra qu iu chnh ln i vi momen, vy cho Tm = 2ms c mt p ng va phi. Do , ta c hm truyn xp x ca khu DTC:GDTC = 1 1 = 1 + s .Tm 1 + 0 ,002.s

Vng iu chnh tc cn khp kn bng ng phn hi m tc . Phn hi tc dng my pht tc c hm truyn coi nh mt khu qun tnh bc nht c hng s thi gian rt nh, chng hn T = 0,75ms, h s khuych i k = 1. Khu pht tc c hm truyn:Gw = 1 1 = 1 + s .T 1 + 0 ,00075.s

Coi l u ra th i tng ca vng iu chnh tc khi khng ti l:

Go =

J s (1 + sTm )(1 + sT )

1

Bin i Go ta c:1 J J = Go = s (1 + sTm )(1 + sT ) s 1 + (Tm + T )s + Tm .T .s 2 1

[

]

D thy: T.T = 2.10-3.7,5.10-4 = 15.10-7 0, do c th b qua thnh phn T.T.s2 trong hm truyn i tng. Nguyn Duy nh TH3 K49 - 27 -

t T = Tm + T ta c i tng ca vng iu chnh tc tr thnh khu tch phn qun tnh c hm truyn xp x:Go = J s (1 + Ts ) 1

Gi s h kn c hm truyn: Hm truyn sai lch tc : Fe = B iu chnh Rw:

1 + a1 s + a2 s 2 + ... Fk = * = 1 + b1 s + b2 s 2 + b3 s 3 + ... ( b a1 )s + ( b2 a2 )s 2 + ... * = 1 Fk = 1 * 1 + b1 s + b2 s 2 + b3 s 3 + ...

(3.2 1) (3.2 2)

R =

1 + a1 s + a2 s 2 + a3 s 3 ... 1 Fk . = J .s.(1 + (T + T )s ). ( b1 a1 )s + (b2 a2 )s 2 + ( b3 a3 )s 3 + ... Go Fec1 = lim Fe ( s ) = 0 s 0 Fe ( s ) = b1 a1 c2 = lim s 0 s

(3.2 3)

t:

ta c:

Do c1 = 0, hm truyn h kn khu v sai cp 1. Tuy nhin, khi c ti, nu ch dng h v sai cp 1 s mc phi sai lch tnh. khc phc iu ny, h phi v sai n t nht l cp 2. Khi , c2 = 0, tc l b1 = a1 = a. t tip:c2 = lims 0

Fe (s ) = b2 a2 . s2

D thy, nu c2 = 0 th h v sai n cp 3. Khi , b iu chnh ph thuc ti 3 tham s, gy kh khn cho vic chnh nh. Vy, ta ch dng li xt h kn v sai n cp 2, ngha l c2 0. cho n gin, chn a2 = 0 v ct cc hng t bc cao. Khi , t b2 = b, b3 = c, hm truyn cc khu a v:


- 28 -

1 + as Fk = 1 + as + bs 2 + cs 3 bs 2 + cs 3 Fe = 1 + as + bs 2 + cs 3 1 + as 1 + Ts Ja 1 = . 1 + R = J .(1 + Ts ). c c b as bs 1 + s 1 + s b b

(3.2 4)

T hm truyn ca b iu chnh ta thy ngay, nu cho

c = T th b iu chnh tr thnh b Ja . H (3.2 4) tr thnh: b

khu PI c hng s thi gian tch phn Ti = a, h s khuch i kp =1 + as Fk = 1 + as + bs 2 ( 1 + Ts ) R = Ja 1 + 1 b as

(3.2 4a)

Phn tip theo s chnh nh tham s ca b R da trn tnh chn tham s a, b cho ph hp. 2. Chnh nh tham s PID theo chun ti u i xng: Mt iu d nhn thy l nu cho a = 4T ; b = 8T2; c = 8T3, t h (3.2 4a) ta c ngay hm chun ca tiu chun ti u i xng:1 + 4Ts Fk = 1 + 4Ts + 8T 2 s 2 + 8T 3 s 3 R = J 1 + 1 2T 4Ts

(3.2 5)

gim bt qu iu chnh, ta mc thm mt khu qun tnh bc nht c hm truyn1 ngay sau tn hiu t lm suy gim tc bin thin ca n. Vi tc t 100 1 + 4Ts

rad/s, p ng i vi tiu chun ti u i xng nh sau:


- 29 -

Hnh 3.7: Qu o t thng

Hnh 3.9: p ng tc v momen

Hnh 3.8: c tnh c

Nhn xt, khi c ti (dao ct n vo chi tit), qu iu chnh v tc khong 5% tha mn yu cu cng ngh nu. D vy, c tnh khi ng xu, qu iu chnh ln n 43%, thi gian qu khong 65ms. Do , ta th i tm mt tiu chun c p ng tt hn. 3. Chnh nh tham s PID da trn dao ng ti hn chun Tyreus & Luyben: Quay tr li vi h (3.2 4a), ta , nu cho a = T th h (3.2 4a) tr thnh:


- 30 -

Fk =

1 + Ts 1 1 . = = 2 2 1 + Ts + bs ( 1 + Ts ) 1 + bs b 1 + s2 b

1 b

(3.2 6)

Phng trnh (3.2 6) l hm chun ca h mi, d thy y chnh l mt khu dao ng iu ha vi tn s gc u = chnh ku =1 hay chu k Tu = 2 b ; khi , h s khuch i ca b iu b

Ja JT = . Tu v ku, gi l tn s dao ng ti hn v h s khuych i ti hn. D b b2 2

Tu = 2kT thy, nu t b = k .T th ta c: J k u = k 2 T

Ti liu [] a ra mt phng php chnh nh PID phng php chnh nh ca Tyreus & Luyben da trn dao ng ti hn nh sau. 1 R = k p 1 + sTi a = Ti = 4 ,4kT Ja 0 ,35 J = 0 ,35k u = 2 k p = b k T 4 ,4k 3T 2 b= T = a = 2 ,2T u 0 ,35 i

(3.2 7)

Hm truyn h kn tr thnh:Fk = 1 + as = 1 + as + bs 2 ( 1 + Ts ) 1 + 4 ,4kTs 4 ,4k 3T 2 (1 + Ts ) 1 + 4 ,4kTs + 0 ,35

(3.2 7)

Nh vy, cht lng ng hc ca h hon ton ph thuc tham s k. Theo kt qu m phng, gi tr ti u ca k = 0,45. Khi , p ng ca h thng:


- 31 -

Hnh 3.10: Qu o t thng


Hnh 3.11: c tnh c

Nhn xt : Tc gn nh phng, qu iu chnh khong 2% khi c ti. p ng vi tc t nhanh hn so vi dng chun ti u i xng, khi ng ch mt khong 30ms v hu nh khng c qu iu chnh v tc . R rng, p ng theo Tyreus & Luyben tt hn nhiu so vi p ng ca h thng theo chun ti u i xng, m bo c cht lng b mt chi tit khi gia cng mi. 5. Vn iu khin vng tc cao: Vng tc cao phi gim t thng m bo cn bng v mt cng sut. mc n mn hc, vn ny cng khng c xt. Nguyn Duy nh TH3 K49 - 32 -

Kt lun: n y, ta xy dng xong phng php iu khin cho h thng. Phn tip

theo s tin hnh thc hin k thut phn cng v m phng h thng trn Matlab & Simulink.


- 33 -

CHNG 4: XY DNG MCH IU KHINI. II. III. IV. V. Khi mch ng lc Khi mch sensor Khi mch ch th Khi mch ngun Khi mch vi x l


- 34 -

CHNG 5: M PHNG H THNG1. M hnh h thng m phng trn Matlab & Simulink.

H thng c m phng vi m hnh ng c khng ng b ca Matlab, vi cc thng s tnh chng 2, mc 2.

Hnh 5.1: M hnh h thng.

Hnh 5.2: Khu o lng in p v dng in


- 35 -

Hnh 5.3 : c lng t thng, momen v v tr.

c lng t thng c thc hin bng phng trnh (3.1 4), c lng Momen dng phng trnh (3.1 7). V tr c c lng bng cch dng cc khi chuyn i Real Imag v Magnitude Angle sn c ca Simulink. Bng cch ny ta trnh c vic x l t thng bng 0 khi khi ng.

Hnh 5.4: Khi bng chn vector in p.

Sai lch t thng v momen c a vo mt lookup table 3D. Lookup Table 3D s chn ra vector in p tng ng, lookup table 2D a ra cc thnh phn ca vector in p . Ni dung ca bng chn c khi to trong file Parameter_inital.m c ni dung c cp trong phn ph lc:


- 36 -

Hnh 5.5: Chuyn i h trc ta abc .

Hnh 5.6: M hnh khi bin tn.2. Kt qu m phng.

H thng c m phng vi cc u vo tc t khc nhau, momen ti cc i Tmax = 10,5 Nm. Di tr momen T = 1 ; di tr t thng = 0 ,005 ; thi gian m phng t = 0,15s. Kt qu m phng nh sau:


- 37 -

Tc cc tiu: * = 15,09 rad/s

Hnh 5.7 Tc v momen.

Hnh 5.8: Dng Rotor v dng Stator

Hnh 5.9: Qu o vector t thng Stator

Hnh 5.10: p ng t thng Stator


- 38 -

* = 100 rad/s


Hnh 5.12 : p ng dng Rotor v Stator.




- 39 -

Tc cc i * = 150 rad/s

Hnh 5.15 Tc v momen.

Hnh 5.16: Dng Rotor v dng Stator



- 40 -


CHNG 6: KT LUNH thng c xy dng trn hon ton p ng c yu cu cng ngh nu, ng thi thc hin thnh cng thut ton DTC iu chnh trc tip momen vi cht lng cao. p ng tc lc khi ng hu nh khng c qu iu chnh, khi c ti, tc ch mc phi qu iu chnh n 2%. H thng c ng hc cao v n gin, do , hon ton thch hp vi ng dng truyn ng cho ng c quay trn chi tit mi. H thng ch s dng Rs - in tr Stator - l tham s d dng nhn dng v d dng b s thay i do nhit , nn khng b nh hng ln bi sai lch ca m hnh ng c, do , nng cao tnh khng nhiu, hot ng tin cy. Tuy nhin, p ng tc thp cha tt. Khng kim sot c dng Stator. p ng momen cha c mn. y cng chnh l nhc im c bn ca phng php DTC v cn c nhiu thi gian nghin cu khc phc. mc n mn hc, n phn tch, a ra yu cu v truyn ng v thit k h thng thc hin thnh cng yu cu ra. Em knh mong nhn c nhng li nhn xt ch bo ca cc thy c lm cho n hon thin hn, khc phc c nhng nhc im, pht huy nhng u im ca n. Mt ln na em xin chn thnh cm n thy Nguyn Minh H cng cc thy c trong b mn T ng Ha XNCN i hc Bch Khoa H Ni n cn hng dn em hon thnh bn n.


- 41 -

CHNG 7: TI LIU THAM KHO1. 2. 3. 4. 5. 6. 7. 8. 9. Trang b in in t my gia cng kim loi Nguyn Mnh Tin, V Quang Hi Gio trnh gia cng c kh Nguyn Th Cng, Phm Ngc Lin, Nguyn c L NXB iu chinh t ng truyn ng in Bi Quc Khnh, Phm Quc Hi, Nguyn Vn iu khin t ng truyn ng in xoay chiu ba pha PGS, T.S Nguyn Phng Truyn ng in thng minh PGS, TS Nguyn Phng Quang NXB Khoa hc v K C s truyn ng in. - PGS TS Bi Quc Khnh, PGS TS Nguyn Vn Lin, NXB in t cng sut V Minh Chnh, Phm Quc Hi, Trn Trng Minh NXB Khoa hc Hng dn thit k mch in t cng sut Phm Quc Hi 2000. L thuyt iu khin t ng h tuyn tnh Nguyn Thng Ng NXB Khoa hc v

NXB Gio Dc 2006. H Ni 2005. Lin, Dng Vn Nghi NXB Khoa hc v K thut 1999. Quang NXB Gio Dc 1998. thut 2004. Khoa Hc v K Thut 2007. v K thut.

K thut 2000.10. C s h thng iu khin qu trnh Hong Minh Sn NXB Bch Khoa H Ni

2006.11. Matlab & Simulink dnh cho k s iu khin t dng PGS, TS Nguyn Phng Quang

NXB Khoa hc v K thut 2006.12. Tng hp h thng iu khin hai mch vng v tr v tc trong truyn ng ng c khng ng b - Nguyn Th Cng, Trng Minh Tn, BC Hi ngh khoa hc ton quc ln th

20 H BKHN.13. Modern power electronics and AC drives. - B.K Bose - NXB Prentice Hall. 14. Improvements in Direct Torque Control of Induction Motors - Jos Luis Romeral

Martinez, Antoni Arias Pujol - UNIVERSITAT POLITCNICA DE CATALUNYA


- 42 -

15. 12-Sector Methodology of Torque Ripple Reduction in a Direct Torque Controlled Induction Motor Drive - Borra Suresh Kumar, R. A. Gupta, Rajesh Kumar, SICE-ICASE

International Joint Conference 2006.16. A completely vectored direct torque control scheme for induction motor - Dhifaoui

Rachid, Hasnaoui Othman, Bacha Fouzi - Systems, Man and Cybernetics, 2002 IEEE International Conference.17. Adaptive stator resistance compensator for high performance direct torque controlled induction motor drives - Byeong-Seok Lee and R. Krishnm, Industry Applications Conference,

1998. Thirty-Third IAS Annual Meeting. The 1998 IEEE


- 43 -

PH LC:M ngun chng trnh khi to tham s h thng m phng%Parameters_inital.m clear DeltaT = 1; DeltaF = 0.005;

look_up_table(2,3,:) = [2, 3, 4, 5, 6, 1]; look_up_table(2,2,:) = [0, 7, 0, 7, 0, 7]; look_up_table(2,1,:) = [6, 1, 2, 3, 4, 5]; look_up_table(1,3,:) = [3, 4, 5, 6, 1, 2]; look_up_table(1,2,:) = [7, 0, 7, 0, 7, 0]; look_up_table(1,1,:) = [5, 6, 1, 2, 3, 4]; look_up_table;

S0 = [0; 0; 0]; S1 = [1; 0; 0]; S2 = [1; 1; 0]; S3 = [0; 1; 0]; S4 = [0; 1; 1]; S5 = [0; 0; 1]; S6 = [1; 0; 1]; S7 = [1; 1; 1]; valve_state = [S0, S1, S2, S3, S4, S5, S6, S7];

Rs = 3.32; p = 2;

%Tong hop vong toc do J = 0.01;


- 44 -

T = 0.002; Tw = 7.5e-4;

%Toi uu doi xung Kp = J/(T+Tw)/2; %He so khuech dai Ki = J/8/(T+Tw)^2; %He so tich phan a = Kp/Ki; %Hang so thoi gian quan tinh khau loc dau vao

%Tyreus Luyben k = 0.45; %tham so quyet dinh tan so dao dong toi han Kp = 0.35*J/(T+Tw)/k^2; %He so khuech dai Ki = 0.35*J/2/pi/(T+Tw)^2/2.2/k^3; %He so tich phan a = 4.4*pi*(T+Tw)*k;%Hang so thoi gian quan tinh khau loc dau vao

'ready'


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