V c bn, thay i tc ng c mt chiu l iu khin in p phn ng ca ng c

Trng i Hc Bch Khoa H Ni n tt nghip.

Li cam oan!

Em xin cam oan ti tt nghip ny l do em t thit k di s hng dn ca thy gio Nguyn Quang ch. Cc kt qu v s liu trong ti l hon ton trung thc.

hon thnh bn n ny, em ch s dng nhng ti liu tham kho c ghi trong bng cc ti liu tham kho, khng s dng cc ti liu khc m khng c lit k phn ti liu tham kho.

MC LC

li ni u

ChngI : Tm hiu chung v ng c in mt chiu.............................2

1.1Khi nim chung..........................................................................2

1.2 Cu to v nguyn l lm vic....................................................2

1.2.1 Cu to ca ng c in mt chiu..................................2

1.2.1 Nguyn l lm vic ca ng c in mt chiu...............6

1.3 c tnh c ca ng c in mt chiu kch t c lp............6 ChngII : Tm hiu h truyn ng cho ng c in mt chiu..........10 2.1 iu chnh tc cho ng c in mt chiu.........................10 2.1.1Nguyn l iu khin in p phn ng..................................10 2.1.2 Nguyn l iu chnh t thng.........................................15 2.2 La chn mch lc cho truyn ng ng c in mt chiu c o chiu quay..............................................................................17 2.2.1 Truyn ng T- c o chiu iu khin ring.............19

2.2.2 Truyn ng T- c o chiu iu khin chung...........22 2.3 Tm hiu s mch chnh lu 3 pha........................................27

2.3.1 s 3 pha c iu khin .................................................27

2.3.2 Tnh chn van ng lc......................................................29

Chng3 : Tm hiu v MentorII..........................................................32 3.1 Gii thiu v MentorII..............................................................32 3.1.1 Ngun cung cp................................................................32 3.1.2 u ra................................................................................32

3.1.3 Phn hi tc ..................................................................32 3.1.4 Phn hi dng in........................................................... 32 3.1.5iu khin..........................................................................33

3.1.6 Thc n...........................................................................33 3.2 Cu to v chc nng................................................................33 3.3 Cch ni mng ca MentorII....................................................38 3.4 Bng iu khin......................................................................39 3.5 Ngun tin ni tip...................................................................40 3.5.1 Kt ni............................................................................40 3.5.2 Cch iu chnh s b ....................................................41 3.5.3 Cc k t iu khin ca MentorII.................................42 3.5.4 a ch ni tip...............................................................42 3.5.5 Nhn dng tham s.........................................................42 3.5.6 Phn d liu ...................................................................42 3.5.7 Khi kim tra BCC.........................................................42 3.5.8 Gi d liu ti MentorII.................................................42 3.5.9 c d liu t MentorII ...............................................43 3.6 Cc tham s chnh ca MentorII............................................43 3.6.1 Menu1:Ci t tc .....................................................43 3.6.2 Menu2: tr................................................................48 3.6.3 Menu3: la chn phn hi v mch vng tc ...........51 3.6.4: Menu4 : la chn v gii hn dng in......56

3.6.5 : Mch vng tc ...................................................61

3.6.6:iu khin t thng......67

3.6.7 Menu10: tnh trng logic v chun on.......................71 3.6.8 Menu11:Hn hp .........................................................75 Chong4: Chng trnh phn mm ng dng ......78 4.1 t vn .............................................................................78 4.2 Phn mn MentorISoft ca MentorII....................................78

lI NI u

Ngy nay vi s pht trin khng ngng ca nn khoa hc k thut to ra nhng thnh tu to ln .Trong ngnh t ng ha cng gp phn khng nh vo thnh cng . Trong s nghip cng nghip ho, hin i ho t nc, c th ni mt trong nhng tiu ch nh gi s pht trin kinh t ca mi quc gia l mc t ng ho trong cc qu trnh sn xut m trc ht l nng sut sn xut v cht lng sn phm lm ra. S pht trin rt nhanh chng ca my tnh in t , cng ngh thng tin v nhng thnh tu ca l thuyt iu khin t ng lm c s v h tr cho s pht trin tng xng ca lnh vc t ng ho.

nc ta mc du l mt nc chm pht trin, nhng nhng nm gn y cng vi nhng i hi ca sn xut cng nh s hi nhp vo nn kinh t th gii th vic p dng cc tin b khoa hc k thut m c bit l s t ng ho cc qu trnh sn xut c bc pht trin mi to ra sn phm c hm lng cht xm cao tin ti hnh thnh mt nn kinh t tri thc.

Mt trong nhng vn quan trng trong ccdy truyn t ng ho sn xut hin i l vic iu chnh tc ng c .T trc n nay, ng c mt chiu vn lun l loi ng c c s dng rng ri k c trong nhng h thng yu cu cao .C nhiu phng php iu chnh iu chnh tc ng c mt chiu v d nh : thay i in p t vo phn ng ng ,thay i t thng ,thay i in tr ph trn mch phn ng .Da vo cc phng php c nhiu cc sn phm ra i ph t ng iu chnh tc ng c mt chiu . Mt v d tiu biu l MentorII ca Control techniques. MentorII c kh nng iu chnh tc ng c in mt chiu c o chiu quay.

MentorII c iu khin bi phn mm MentorSoft l mt phn mn kh mnh ca Control techniques.MentorSoft cho php hin th y tt c cc tham s bn trong ca MentorII.

Trong n ny bao gm c nhng ni dung chnh sau:

+ Tm hiu chung v ng c in mt chiu

+Tm hiu truyn ng cho ng c in mt chiu

+ Tm hiu chung v MentorII

+ Phn mm MentorSoft.

Trong thi gian lm n c s quan tm hng dn, gip nhit tnh ca thy NGUYN QUANG CH v cc thy c trong b mn em hc hi , tip thu c nhiu kinh nghim v ng c in mt chiu cng nh cch iu chnh tc ng c mt chiu v hon thnh bn n tt p. Tuy nhin, do thi gian v gii hn ca n cng vi phm vi nghin cu ti liu vi kinh nghim v kin thc cn hn ch nn bn n ny khng trnh khi nhnh thiu st rt mong s ng gp kin ca thy c cun n ca em c hon thin hn.Chng 1: tm hiu v ng c in mt chiu

1.1 Khi nim chung.

Trong nn sn xut hin i, my in mt chiu vn c coi l mt loi my quan trng. N c th dng lm ng c in, my pht in hay dng trong nhng iu kin lm vic khc.

ng c in mt chiu c c tnh iu chnh tc rt tt, v vy my c dng nhiu trong nhng ngnh cng nghip c yu cu cao v iu chnh tc nh cn thp, hm m, giao thng vn ti

ng c in c phn loi theo cch kch thch t, thnh cc ng c kch thch c lp, kch thch song song, kch thch ni tip v kch thch hn hp. Cn ch rng ng c kch thch c lp I= I; ng c kch thch song song v hn hp I = I + It; ng c in kch thch ni tip I = I = It.Trn thc t, c tnh c ca ng c kch thch c lp v kch thch song song hu nh ging nhau nhng khi cn cng sut ln ngi ta thng dng ng c in kch thch c lp iu chnh dng in kch thch c thun li v kinh t hn mc d loi ng c ny i hi phi c thm ngun in ph bn ngoi. Ngoi ra, khc vi trng hp my pht kch thch ni tip, ng c in ni tip c dng rt nhiu, ch yu trong ngnh ko ti bng in.

1. 2 Cu to v nguyn l lm vic.1. 2.1Cu to ca ng c in mt chiu.

Kt cu ch yu ca ng c in mt chiu nh hnh v 1.1 v c th chia lm hai phn chnh l phn tnh v phn quay.Cc thnh phn:

Bearing: Vng biCommutator : C gp

Armature core : Cun dy phn ng

Shaft: Trc quay.

Magnet:Nam chm

Hnh 1.1 S mt ct ngang v dc ca ng c mt chiu.

a). Phn tnh (stato).

y l phn ng yn ca my. Phn tnh gm c cc b phn sau:

Cc t chnh: l b phn sinh ra t trng gm c li st cc t v dy qun kch t lng ngoi li st cc t lm bng nhng l thp k thut in hay thp cacbon dy 0.5 n 1 mm p li v tn cht. Trong my in nh c th dng thp khi. Cc t c gn cht vo v my nh cc bulng. Dy qun kch t c qun bng dy ng bc cch in v mi cun dy u c bc cch in k thnh mt khi v tm sn cch in trc khi t trn cc cc t. Cc cun dy kch t t trn cc cc t ny v c ni ni tip vi nhau.

Cc t ph: c t gia cc cc t chnh v dng ci thin i chiu. Li thp ca cc t ph c t dy qun m cu to ging nh dy qun cc t chnh. Cc t ph c gn vo v my nh nhng bulng.

Gng t: dng lm mch t ni lin cc cc t, ng thi lm v my. Trong my in nh v va thng dng thp tm dy un v hn li. Trong my in ln thng dng thp dc. C khi trong my in nh dng gang lm v my.

Ngoi ra cn c cc b phn khc nh: Np my bo v my khi b nhng vt ngoi ri vo lm h hng dy qun hay an ton cho ngi khi chm vo in. C cu chi than a dng in t phn quay ra ngoi.

b). Phn quay (rto).

Gm c nhng b phn sau:

Li st phn ng: dng dn t. Thng dng nhng tm thp k thut in (thp hp kim silic) dy 0.5 mm ph cch in mng hai mt ri p cht li gim hao tn do dng in xoy gy nn. Trn l thp c dp hnh dng rnh sau khi p li th t dy qun vo. Trong nhng my c trung tr ln, ngi ta cn dp nhng l thng gi khi p li thnh li st c th to c nhng l thng gi dc trc.

Hnh 1.2 S cu to rto.

Trong nhng my in hi ln th li st thng c chia lm tng on nh. Gia cc on y c mt khe h gi l khe thng gi ngang trc. Khi my lm vic, gi thi qua cc khe lm ngui dy qun v li st. Trong my in nh li st phn ng c p trc tip vo trc. Trong my in ln, gia trc v li st c t gi rto. Dng gi rto c th tit kim thp k thut in v gim nh trng lng rto. Dy qun phn ng: l phn sinh ra s.. v c dng in chy qua. Dy qun phn ng thng lm bng dy ng c bc cch in. Trong my in nh (cng sut di vi kiloat ) thng dng dy c tit din trn. Trong my in va v ln, thng dng dy tit din hnh ch nht. Dy qun c cch in cn thn vi rnh ca li thp.

trnh khi quay b vng ra do sc ly tm, ming rnh c dng nm cht hoc phi ai cht dy qun. Nm c th lm bng tre, g hay bakilit.

C gp: dng i chiu dng in xoay chiu thnh mt chiu. Kt cu ca c gp gm nhiu phin ng c dui nhn cch in vi nhau bng lp mica dy 0.4 n 1.2 mm v hp thnh mt hnh tr trn. Hai u tr trn dng hai vnh p hnh ch V p cht li. Gia vnh p v tr trn cng cch in bng mica. ui vnh gp c cao hn mt t hn cc u dy ca cc phn t dy qun vo cc phin gp dc d dng.

Cc b phn khc nh: Cnh qut qut gi lm ngui my. Trc my t li st phn ng, c gp, cnh qut v bi.

1.2.2. Nguyn l lm vic ca ng c in mt chiu.

ng c in mt chiu thc cht l my in ng b trong s.. xoay chiu c chnh lu thnh s.. mt chiu. chnh lu s.. ta c hai u vng dy c ni vi hai phin gp trn c hai chi in lun t st vo chng. Khi rto quay, do chi in lun tip xc vi phin gp ni vi thanh dn. V vy s.. xoay chiu trong vng dy c chnh lu mch ngoi thnh s.. v dng in mt chiu nh h thng vnh gp v chi in. s.. mt chiu gia cc chi in c tr s ln v t p mch, dy qun rto thng c nhiu vng dy ni vi nhiu phin gp lm thnh dy qun phn ng v c c gp in (cn gi l c gp hoc vnh i chiu).

1.3. c tnh c ca ng c in mt chiu kch t c lp.Khi ngun in mt chiu c cng sut khng ln th mch in phn ng v mch kch t mc vo hai ngun in c lp vi nhau, lc ny ng c c gi l ng c kch t c lp.

Hnh 1.3 S ni dy ca ng c kch t c lp.

Theo s trn c th vit phng trnh cn bng in p mch phn ng nh sau:

U =E+(R + Rf)I (1-1).

Trong : U- in p phn ng, V.

E- sc in ng phn ng, V

R- in tr ca mch phn ng, ( Rf- in tr ph trong mch phn ng, ( I- dng in mch phn ng, A

Vi R = r + rcf + rb + rct r - in tr cun dy phn ng

rcf - in tr cun cc t ph

rb - in tr cun b

rct - in tr tip xc ca chi in

Sc in ng E ca phn ng ng c c xc nh theo biu thc:

E = (( = K(( (1-2).

Trong : p s i cc t chnh

N s thanh dn tc dng ca cun dy phn ng

a s i mch nhnh song song ca cun dy phn ng

( - t thng kch t di mt cc t, Wb ( - tc gc, rad/s

K = l h s cu to ca ng c

T (1-1) v (1-2) ta c:

( = - I (1-3).

Biu thc (1-3) l phng trnh c tnh c in ca ng c

Mt khc Mmen in t Mt ca ng c c xc nh bi:

M=K(I (1-4).

Suy ra: I =

Thay gi tr I vo (1-3) ta c:

(=- Mt (1-5).

Nu b qua tn tht c v tn tht thp th mmen c trn trc ng c bng mmen in t, ta k hiu l M. ngha l Mt = Mc = M.

(=- M (1-6).

y l phng trnh c tnh c ca ng c in mt chiu kch t c lp.

Gi thit phn ng c b , t thng ( = const, th cc phng trnh c tnh c in (1-3) v phng trnh c tnh c (1-6) l tuyn tnh. th ca chng c biu din trn l nhng ng thng hng.

Hnh 1.4 c tinh c in ca ng c Hnh 1.5 c tnh c ca in mt chiu kch t c lp. ng c in mt chiu kch

t c lpChng 2: tm hiu h truyn ng chong c in mt chiu.

I. iu chnh tc ng c in mt chiu. iu chnh tc ng c l dng cc bin php nhn to thay i cc thng s ngun nh in p hay cc thng s mch nh in tr ph, thay i t thng T to ra cc c tnh c mi c nhng tc lm vic mi ph hp vi yu cu.

Thc t c 2 phng php c bn iu chnh tc ng c in mt chiu l:

+iu chnh in p cp cho phn ng ng c.

+ iu chnh in p cp cho mch kch t ng c.

Cu trc phn mch lc ca h truyn ng iu chnh tc ng c in mt chiu bao gi cng c b bin i. Cc b bin i ny cp in p v dng in cho mch phn ng ng c hoc mch kch t ng c.

V phng din iu chnh tc , ng c in mt chiu c nhiu u vit hn so vi loi ng c khc, khng nhng c kh nng iu chnh tc d dng m cu trc mch lc, mch iu khin n gin hn, ng thi t cht lng iu chnh cao trong gii iu chnh tc rng.

1.Nguyn l iu chnh in p phn ng.

iu chnh in p phn ng ng c in mt chiu cn c thit b ngun nh my pht in mt chiu kch t c lp, cc b chnh lu iu khin Cc thit b ny c chc nng bin nng lng xoay chiu thnh mt chiu c sc in ng Eb iu chnh c nh tn hiu iu khin Uk. V l ngun c cng sut hu hn so vi ng c nn cc b bin i ny c in tr trong Rb v in cm Lb khc 0.

Hnh 2.1 S khi v s thay th ch xc lp.

ch xc lp c th vit c phng trnh c tnh ca h thng nh sau:

Eb + E = I ( Rb + R )

( = ( 2- 1 ).

( = (0( Uk) -

V t thng ca ng c c gi khng i nn cng c tnh c cng khng i, cn tc khng ti l tng th tu thuc vo gi tr in p iu khin Uk ca h thng, do c th ni phng php iu chnh ny l trit .

xc nh gii iu chnh tc ta rng tc ln nht ca h thng b chn bi ng c tnh c c bn, l c tnh ng vi in p phn ng nh mc v t thng cng c gi ga tr nh mc. Tc nh nht ca gii iu chnh b gii hn bi yu cu v sai s tc v v mmen khi ng. Khi mmen ti l nh mc th cc gi tr ln nht v nh nht ca tc l:

(max = (0 max -

( 2-2 ).

(min = (0 min -

tho mn kh nng qu ti th c tnh thp nht ca di iu chnh phi c mmen ngn mch l:

Mn.m min = Mc max = KM. MmTrong KM l h s qu ti v mmen. V h c tnh c l nhng ng thng song song nhau, nn theo nh ngha v cng c tnh c ta.

(min = ( Mn.m min Mm )

D = (2- 3).

Hnh 2-2 Xc nh phm vi iu chnh.

Vi mt c cu my c th th cc gi tr (0 max, Mm, KM l xc nh, v vy phm vi iu chnh D ph thuc tuyn tnh vo gi tr ca cng (. Khi iu chnh in p phn ng ng c bng cc thit b ngun iu chnh th in tr tng mch phn ng gp khong hai ln in tr phn ng ng c. Do c th tnh s b c:

(0 max . 10.

V th ti c c tnh mmen khng i th gi tr phm vi iu chnh tc cng khng vt qu 10. i vi cc my c yu cu cao v di iu chnh v chnh xc duy tr tc lm vic th vic s dng cc h thng h nh trn l khng tho mn c.

Trong phm vi ph ti cho php c th coi cc c tnh c tnh ca truyn ng mt chiu kch t c lp l tuyn tnh. Khi iu chnh in p phn ng th cng cc c tnh c trong ton gii iu chnh l nh nhau, do st tc tng i s t gi tr ln nht ti c tnh thp nht ca di iu chnh. Hay ni cch khc nu c tnh c thp nht ca gii iu chnh m sai s tc khng vt qu gi tr sai s tc cho php, th h truyn ng x lm vic vi sai s lun nh hn sai s cho php trong ton b di iu chnh. Sai s tng i ca tc c tnh c thp nht l:

S =

S =

( 2- 4 ).

Vi cc gi tr Mm, (0 min, Scp l xc nh ln c th tnh c gi tr ti thiu ca cng c tnh c sao cho sai s khng vt qu gi tr cho php. lm vic ny trong a s cc trng hp cn xy dng h truyn ng in kiu vng kn.

Trong sut qua trnh iu chnh in p phn ng th t thng kch t c gi nguyn, do mmen ti cho php ca h s l khng i:

Mc.cp = K(m.Im = Mm Phm vi iu chnh tc v mmen nm trong hnh ch nht bao bi cc ng thng ( = (m, M = Mm v cc trc to . Tn hao nng lng chnh l tn hao trong mch phn ng nu b qua cc tn hao khng i trong h.

Eb = E + I ( Rb + R )

I Eb = I E + I2 ( Rb + R )

Nu t Rb + R = R th hiu sut bin i nng lng ca h s l:

( =

( =

Khi lm vic ch xc lp ta c mmen do ng c sinh ra ng bng mmen ti trn trc M* = M*c v gn ng coi c tnh c ca ph ti l M* = ((*)x th:

( =

(2- 5).

Hnh 2-3: Quan h gia hiu sut truyn ngv tc vi cc loi ti khc nhau.

iu chnh tc bng cch thay i in p phn ng l rt thch hp trong trng hp mmen ti l hng s trong ton gii iu chnh. Cng thy rng khng nn ni thm in tr ph vo mch phn ng v nh vy s lm gim ng k hiu sut ca h.

u im: y l phng php iu chnh trit , v cp c ngha l c th iu chnh tc trong bt k vng ti no k c khng ti l tng.

Nhc im: Phi c b ngun c in p thay i nn vn u t c bn ln v chi ph vn hnh cao.

2.1.2Nguyn l iu chnh t thng ng c.

iu chnh t thng kch t ca ng c in mt chiu l iu chnh mmen in t ca ng c M = K(I v sc in ng quay ca ng c E = K((. Mch kch t ca ng c l mch phi tuyn, v vy h iu chnh t thng cng l h phi tuyn:

iK = ( 2- 6 ).

trong : rK - in tr y qun kch thch

rb - in tr ca ngun in p kch thch

(K s vng dy ca dy qun kch thch

Trong ch xc lp ta c h:

iK = ; ( = f[iK]

Thng khi iu chnh t thng th in p phn ng c gi nguyn bng gi tr nh mc, do c tnh c thp nht trong vng iu chnh t thng chnh l c tnh c in p phn ng nh mc, t thng nh mc v c gi l c tnh c bn ( i khi chnh l c tnh c t nhin ca ng c ). Tc ln nht ca di iu chnh t thng b hn ch bi kh nng chuyn mch ca c gp in. Khi gim t thng tng tc quay ca ng c th ng thi iu kin chuyn mch ca c gp cng b xu i, v vy m bo iu kin chuyn mch bnh thng th cn phi gim dng in phn ng cho php, kt qu l mmen cho php trn trc ng c gim rt nhanh. Ngay c khi gi nguyn dng in phn ng th cng c tnh c cng gim rt nhanh khi gim t thng kch thch:

(( = hay (* = ( (* )2

Hnh 2- 4 S thay th (a) c tnh iu chnh khi iu chnh t thng ng c (b) Quan h ((ikt), (c) .

Do iu chnh tc bng cch gim t thng nn i vi cc ng c m t thng nh mc nm ch tip gip gia vng tuyn tnh v vng bo ho ca c tnh t ho th c th coi vic iu chnh l tuyn tnh v hng s C ph thuc vo thng s kt cu ca my in:

( = ciK =

Nhn xt: Phng php iu chnh bng cch thay i t thng c th diu chnh tc v cp v cho nhng tc ln hn tc c bn ncb. Phng php ny c dng iu chnh tc cho cc my mi vn nng hoc l my bo ging. Do qu trnh iu chnh tc c thc hin trn mch kch t nn tn tht nng lng t, mang tnh kinh t, thit b n gin.

Kt lun:

T nhng u, nhc im ca hai phng php iu chnh tc ng c in mt chiu ta va xt trn th ta thy phng php iu chnh tc ng c in mt chiu bng phng php iu chnh in p phn ng ng c l thch hp hn.

2.2. La chn mch lc cho truyn ng ng c in mt chiu c o chiu quay.

Chn truyn ng Tiristo - ng c in mt chiu (T- ) c o chiu quay.

Do chnh lu tiristo dn dng theo mt chiu v ch iu khin c khi m, cn kho theo in p li cho nn truyn ng van thc hin o chiu kh khn v phc tp hn truyn ng my pht ng c. Cu trc mch lc cng nh mch iu khin h truyn ng T- o chiu c yu cu o chiu cao v c logic iu khin cht ch.

C hai nguyn tc c bn xy dng h truyn ng (T- ) o chiu:

+ Gi nguyn chiu dng in phn ng v o chiu dng kch t ng c.

+ Gi nguyn chiu dng in kch t v o chiu dng in phn ng.

Trong thc t, cc s truyn ng (T- ) o chiu c nhiu song u thc hin theo mt trong hai nguyn tc trn v c phn ra thnh cc loi s chnh sau:

+ Truyn ng dng mt b bin i cp cho phn ng v o chiu bng cng tc t chuyn mch phn ng (( = const). H ny c u im dng cho cng sut nh, tn s o chiu thp:

Hnh 2- 4 S truyn ng dng mt b bin i cp cho phn ng v o chiu bng cng tc t chuyn mch phn ng.

+ Truyn ng dng mt b bin i cp cho phn ng v o chiu quay bng o chiu dng kch t. Loi ny dng cho cng sut ln, t thc hin o chiu:

Hnh 2- 5 S truyn ng dng mt b bin i cp cho phn ng v o chiu quay bng o chiu dng kch t.

+ Truyn ng dng hai b bin i, cp cho phn ng iu khin ring, hai b iu chnh lm vic ring r vi nhau. Ti mt thi im ch pht xung m mt b cn b kia kho hon ton. S ny dng cho mi gii cng sut v c kh nng o chiu vi tn s ln:

Hnh 2- 6 S truyn ng hai b bin i, cp cho phn ng iu khin ring, hai b iu chnh lm vic ring r vi nhau.

+ Truyn ng dng hai b bin i u ni song song ngc iu khin chung. S ny dng cho mi di cng sut va v ln, thc hin vic o chiu m nhng c nhc im kch thc cng knh, vn u t ln, tn tht ln.

Hnh 2- 7 S truyn ng hai b bin i u ni song song ngc iu khin chung.

+ Truyn ng dng hai b bin i ni theo s cho iu khin chung. Dng cho di cng sut va v ln c tn s o chiu cao.

Hnh 2- 8 S truyn ng dng hai b bin i ni theo s cho iu khin chung.

V nguyn tc xy dng mch iu khin, c th chia lm hai loi chnh: iu khin ring v iu khin chung.

1. Truyn ng T- o chiu iu khin ring.

Khi iu khin ring hai b bin i lm vic ring r nhau, ti mt thi im ch pht xung iu khin vo mt b bin i cn b kia b kho do khng c xung iu khin. H c hai b bin i l B1 v B2 vi cc mch pht xung iu khin tng ng l FX1 v FX2, trt t hot ng ca cc b pht xung ny c quy nh bi cc tn hiu logic b1 v b2. Qu trnh hm v o chiu c m t bng th thi gian. Trong khong thi gian 0 ( t1, B1 lm vic ch chnh lu vi gc (1< (/2 cn B2 kho. Ti t1 pht lnh o chiu bi iL, gc iu khin (1 tng t bin ln hn (/2, dng phn ng gim dn v khng, lc ny ct xung iu khin kho B1, thi im t2 c xc nh bi cm bin dng in khng SI1. Trong khong thi gian tr ( = t3( t2, B1 b kho hon ton, dng in phn ng b trit tiu. Ti t3, s.. ng c E vn cn dng, tn hiu logic b2 kch cho FX2 m B2 vi gc (2> (/2 v sao cho dng in phn ng khng vt qu gi tr cho php, ng c c ti sinh, nu nhp iu gim (2 ph hp vi qun tnh ca h th c th duy tr dng in hm v dng in khi ng ngc khng i, iu ny c thc hin bi cc mch vng dng iu chnh t ng ca h thng. Trn s khi logic LOG, iL, iL1, iL2, l cc tn hiu logic u vo; b1, b2 l cc tn hiu logic u ra kho cc b pht xung iu khin.

+ iL = 1 : pht xung iu khin m B1.

+ iL = 0 : pht xung iu khin m B2.

+ i1L(i2L) = 1 : c dng in chy qua B1 (B2).

+ b1, b2 = 1 : kho b pht xung FX1 (FX2).

Hnh 2- 8 S khi h truyn ng o chiu v cc tn hiu iu khin.

Trn hnh v 2- 8 cho mt v d mch iu khin qu trnh o chiu. th thi gian ca cc tn hiu m t hnh v trn.

b1 = ; b2 =

Khong thi gian tr c m bo bi cc mch xung c rng khng i T.

H truyn ng van o chiu iu khin ring c u im l lm vic an ton, khng c dng in cn bng chy gia cc b bin i, song cn mt khong thi gian tr trong dng in ng c bng khng.

Hnh 2- 9 S mch lgc LOG.

2. Truyn ng (T- ) o chiu iu khin chung.

Trn H 2- 9 m t v d v h T - o chiu iu khin chung, ti mt thi im c hai b bin i u nhn c xung m, nhng ch c mt b bin i cp dng cho nghch lu, cn b bin i kia lm vic ch i.

c tnh iu khin ca B1 l ng I, c tnh iu khin ca B2 l ng II. Gi thit (1 < (/2; (2 > (/2 sao cho th dng in ch c th chy t B1 sang ng c m khng th chy t B1 sang B2 c. t c trng thi ny th cc gc iu khin phi tho mn iu kin:

hay

Nu tnh n gc chuyn mch ( v gc kho ( th gi tr ln nht ca gc iu khin ca b bin i ang ch nghch lu i phi l:

(max = ( - ( (max + ( ).

V gi tr nh nht ca gc iu khin ca b bin i ang lm vic ch nghch lu l:

Nu chn | Ed1| = | Ed2| th (1 + (2 = ( v ta c phng php iu khin chung i xng, khi ny s tng trong mch vng gia hai b bin i s trit tiu v dng in trung bnh chy vng qua hai b bin i cng trit tiu:

trong Rcb l tng in tr trong mch vng cn bng.

Trong thc t iu khin thng dng phng php iu khin chung khng i xng, tc l (2 > ( - (1, khi | Ed2 | > | Ed1 | v khng c dng in cn bng.

Hnh 2-10 S nguyn l v c tnh chnh lu o chiu iu khin chung.

Trong cc phng php iu khin chung, mc d m bo , tc l khng xut hin dng in trung bnh ca dng in cn bng, song gi tr tc thi ca s cc b chnh lu e1(t),ed2(t) lun khc nhau, do vn xut hin thnh phn xoay chiu ca dng in cn bng. hn ch bin dng in cn bng thng dng cc cun khng cn bng Lcb. Trong s chnh lu cu ba pha dng in cn bng chy trong hai vng c lp mi vng to thnh mt chnh lu ba pha hnh tia.

Trn cc hnh 2- 11 v 2- 12 gii thiu qu trnh in p cn bng Ucb, dng in cn bng icb trn vng I. Cc in p Uk1 v Uk2 c o gia cc im K1 v A2 ca chnh lu vi im trung tnh ca ngun xoay chiu ba pha,in p chnh lu Ed1 c o gia im 1 v trung tnh ngun. Trn hnh 2- 11 m t qu trnh khi iu khin chung i xng, hnh 2- 12 m t qu trnh khi iu khin khng i xng, c th thy r tc dng gim bin dng cn bng khi iu khin chung khng i xng. Dng in p chnh lu Ed hi c bit do c tnh n st p trn cc in khng cn bng:

Bng cch tng t c th xy dng c cc th Uk, UdA1 v Ed2, cc th ny c dng tng t trn. in p chnh lu ca c b bin i s bng:

Ud = Ed1 Ed2.3. Nhn xt chung.

u im ni bt ca h T - l tc tc ng nhanh cao, khng gy n v d t ng ho do cc van bn dn c h s khuych i cng sut rt cao, iu rt thun tin cho vic thit lp cc h thng t ng iu chnh nhiu vng nng cao cht lng cc c tnh tnh v cc c tnh ng ca h thng.

Nhc im ch yu l do cc van bn dn c tnh phi tuyn, dng in p chnh lu ra c bin p mch cao, gy tn tht ph trong my in, v cc truyn ng c cng sut ln cn lm xu dng in p ca ngun v li xoay chiu. H s cng sut cos ( ca h ni chung l thp.

Ngoi ra trong h truyn ng van o chiu iu khin ring c u im l lm vic an ton, khng c dng in cn bng chy gia cc b bin i, song cn mt khong thi gian tr trong dng in ng c bng khng.

T nhng u im ta chon h truyn ng T - o chiu iu khin ring.

III. Tm hiu mch chnh lu cu 3 pha.

1. S chnh lu cu 3 pha c iu khin.

Hnh 2-13. Chnh lu cu ba pha iu khin i xng. a- s ng lc, b- gin cc ng cong c bn, c, d - in p ti khi gc m (= 600 (= 600.

y l chnh lu ba pha hai na chu k vi hai nhm: T1, T3, T5 hnh thnh nhm catt chung; cn T2, T4, T6 l nhm ant chung.

Theo dng sng in p th in p tng p mch bc su v tr s nh ca n bng in p dy. Gc m ( c tnh t giao im ca cc na hnh sin.

Gi thit T5 v T6 ang dn nn VF = Uc , VG = Ub .

Ti (t1 = (/6 + ( cho xung iu khin m T1. Tiristor ny s m v Ua >0 . S m ca T1 lm cho T3 b kho mt cch t nhin v Ua > Uc. Lc ny T6 v T1 dn v in p trn ti l: UL = Ud = Ua - Ub .

Ti (t2 = 3(/6 + ( cho xung mi m T2. Tiristor ny s m v khi T6 dn c in p Ub t nn ant ca T2 m Ub > Uc. S m ca T2 lm cho T6 b kho li mt cch t nhin.

Cc xung iu khin lch nhau (/3 ln lt c a n cc cc iu khin theo th t nh sau:

Thi imMKho

(/6 + (T1T5

3(/6 + (T2T6

5(/6 + (T3T1

7(/6 + (T4T2

9(/6 + (T5T3

11(/6 + (T6T4

in p trung bnh trn ti c tnh theo cng thc:

Ud = UL = ( 2- 7 ).

= ( 2- 8 ).

Trong Uf. N max l in p pha cc i, Uf. f max l in p dy cc i.

Khi gc m ( nh dng sng biu din trn hnh (2- 13) cho thy in p Ud p mch bc su, nhng khi ( ln, in p trn ti s c phn m, dng in trong cc tiristor c dng ch nht nhng dng in qua th cp my bin p l hon ton i xng v khng c thnh phn mt chiu trnh cho li st b bo ho. S cu ba pha cn gi l cu Graetz c s dng rng ri bi dng in trong cc dy qun v dy ngun hon ton i xng.

Cng sut nh mc ca my bin p:

S1 = S2 = 1,05 Pd (2- 9).

2. Tnh chn van ng lc. Thng s ca ng c: Pm = 1 KW Um = 220 V

nm = 1000 vng/pht

= 81 %

Im = 5,6 A

in p ngc ln nht tiristor phi chu:

Un max = Knv . U2 = Knv .

trong : Knv = ; Ku =

Un max = . = = 230,38 (v)

in p ngc ca van cn chn:

Unv = Kdt U . Un. maxtrong : Kdt U h s d tr in p, chn Kdt U = 1,8

Unv = 1,8 . 230,38 = 424,69 (v)

Dng in lm vic ca van c tnh theo dng hiu dng:

Ilv = Ihd = khd. Id

trong s cu ba pha, h s dng in hiu dng: khd =

Ilv = Ihd =

Chn iu kin lm vic ca van l c cnh tn nhit v din tch tn nhit; khng c qut i lu khng kh vi iu kin dng in nh mc ca van cn chn l:

Im = ki. Ilv = 3,2. 3,23 = 10,34 10 (A)

m ki h s d tr dng in, chn ki = 3,2

T cc thng s Unv, Im ta chn 6 tiristor loi BTW 42 60 RC c cc thng s sau:

in p ngc cc i ca van : Un = 600 (v)

Dng in nh mc ca van : Im = 10 (A)

nh xung dng in : Ipik = 150 (A)

Dng in ca xung iu khin : Ik = 0,05 (A)

in p ca xung iu khin : Uk = 1,5 (V)

Dng in r : Ir = 0,003 (A)

St p ln nht ca tiristo trng thi dn : (U = 2 (V)

Tc bin thin in p :

Thi gian chuyn mch : tcm = 35 ((s)

Nhit lm vic cc i cho php : Tmax =

Tnh ton chn my bin p chnh lu

Cng sut biu kin ca my bin p:

in p ngun chn :U1=220 V

Tnh in p th cp my bin p:

Vi Ud :in p ti

:St p trn van

:St p trn dy ni, coi nh bng 0.

: St p trn bin p,ly d tr st p ca my bin p l 6%Chn gc m nh nht ca van l 100Chng 3: Tm hiu v mentor II

3.1. Gii thiu v Mentor II

MentorII l mt phin bn mi nht ca Control Techniques. Mentor II. c ng dng trong nhng k thut tin tin c tnh linh hot cao. y l mt sn phm rt cn cho mt h thng i hi s chnh xc v yu cu s ti sinh. V d nh trong h thng my cun, my v, my dn giy, cu trc. MentorII c b vi s l cng nghip iu khin ng c in mt chiu. Phm vi u ra ca dng in l 25A n 1850 A. Thit b ny c iu khin ng c mt chiu ch mt gc phn t hoc bn gc phn t. iu khin mt gc phn t l iu khin ng c ch quay theo chiu thun. iu khin bn gc phn t l iu khin ng c c o chiu quay. C hai kiu iu khin trn u iu khin tc ng c, c th thm iu khin mmen ng c. Nhng thng s ca MentorII c la chn v thay i ti bng iu khin, MentorII hay mt giao din qua truyn thng ni tip. Sau y l mt s c tnh ca MentorII.

3.1.1. Ngun cung cp

S c mt 1 hay nhiu pha u vo c t ng pht hin. Thit b s chy m khng ti

3.2.1. u ra:

6 xung u vo SCRR to ra 12 xung u ra.

3.1.3. Phn hi tc :

in p phn ng dng ng c hoc my pht tc hoc phn hi s. C PID trong mch vng tc .

3.1.4. Phn hi dng in:

L 0.1%

Mch vng dng in tuyn tnh, tn s 80Hz.

p l mi gi tr ca dng in.

3.1.5. iu khin

Tt c cc tn hiu tng t v hu ht cc tn hiu s nhp vo u c th do ngi s dng to ra cho cc ng dng c bit.

PID mch vng tc

B tn hiu s cho iu khin v tr

B pht tc cho o lng

Chng trnh iu khin gim t thng.

Pht hin t ng tn hiu ni tip v s c mt pha.

H thng thc n thit lp tham s.

C th thit lp li thng s cui trong mi thc n.

Thc n thit lp phc v cho vic truy cp nhanh ti tham s.

Cho d iu khin n hay iu khin hon ton, v cn bn l mt hm in p ra, l hm gc m ca SCR c th kim sot chnh xc.

Cht lng ca thng tin p li t ng c tu thuc vo kh nng nhn ca thit b. Mt s d liu c th l ngui ngoi nh tc t, mmen t, tc phn hi ca ng c. Mt s bn trong nh in p v dng in u ra, v iu kin ca h thng ti mi giai on.

MentorII trang b mt b vi x l v phn mm c nh hnh bi nhng tham s ci t bi ngi s dng. Nhng tham s l nhn t quan trng lin qua ti hot ng ca ng c. Xa hn na nhng tham s c cung cp cho truyn thng, bo mt v hm thao tc khc.

3.1.6. Thc n.

S lng tham s ln, tuy nhin vic hiu chng v truy cp chng c lm d dng bi vic thu xp chng trong nhng thc n, mi thc n gm mt nhm logic hoc hm c bit.

3.2. Cu to v chc nng.

MentorII c nhiu chc nng nn cu to tng i phc tp. Trong bn n ny chng ta i su vo tm hiu MentorII M25 v M25(R).

Cc hm iu khin ng c mt chiu l iu khin tc , mmen, phng hng quay. Tc t l thun vi thnh phn ng v t l nghch vi t thng Mmen t l thun vi dng in phn ng v t thng. Hng quay lin quan ti cc tnh ca in p phn ng v kch t:

Hnh 3.1. S ni in p phn ng v kch t vo MentorII.

a. in p phn ng: back emf l mt thnh phn ca in p phn ng. Nh vy gi thit t thng khng i, c th iu khin tc ti im ni in p cc i. Dng in phn ng cng lm mt hm ca in p phn ng, do vy tc s ph thuc vo in p v mmen cc i t tc c s (ti in p phn ng cc i).

b, in p kch t: n xc nh dng in kch t, t thng. Nu in p kch t l c lp vi in p phn ng th tc tng n tc c s v lc dng in max. Khi mmen t l vi t thng, mmen cc i s gim nu tc c tng bng cch gim t thng.

V c bn, thay i tc ng c mt chiu l iu khin in p phn ng ca ng c. MentorII c trang b c kh nng iu khin t thng nu tc ln hn tc c bn c yu cu. iu khin ring t thng ng c t n tc v mmen cng c ng dng. Ngoi ra ta la chn mt phng thc phn hi ca MentorII c mt mch vng khp kn.

Mt ngun in p mt pha c cung cp cho cu thyriostor v mt tr khng c mc song song vi n sinh ra mt dng in gin on dng m gc m thyristorr, v dng ngun in khi qua im khng na chu trnh. in p cc i khi thyristorr m, l lc f trong hnh 2.1 tr v khng. Khi lm chm gc m lm gim dng in ra. Khi ti lm cm ng, nh kch t ca mt ng c chng hn dng in tr thnh lin tc. th dng in chm pha hn in p do cm ng ca ti v mt phn v s tr ca gc m.

Hnh v 3.2. Ngun cung cp cho mch kch t

o chiu quay ng c in theo hai cch, tu thuc vo kiu cu ca thit b. Cch iu khin n gin nht l dng mt cu ba pha iu khin ng c. Lc ny ng c khng o c chiu quay. V vy, mun o chiu ta phi c kho chuyn i nh trong hnh v 3.3

Hnh 3.3. S mc mt cu 3 pha dng cng tc chuyn i o chiu.

Tuy nhin thc t yu cu iu khin y hai chiu ca ng c. Vi kh nng o mmen nhanh chng v lin tc. Ta mc hai cu song song ngc nh hnh v 3.4. S ny c th iu khin y o chiu v hm m khng cn kho chuyn i.

Hnh 3.4. S mc hai cu 3 pha song song ngc

Khi hm bng phng php hm ng nng hnh v 2.5. Lc ny s gim tc khng kim sot c v cng khng tuyn tnh.

Hnh 3.5. Hm ng nng.

D s dng mt gc phn t hay bn gc phn t, ng c in vn lun ph thuc vo in p. M in p ta c th kim sot c chnh xc thng qua gc m ca thyristor ca cu 3 pha.

Nh ni phn trn, thay i tc ng c ta c th thay i in p phn ng. lm c iu ny ta iu khin gc m ca cc thyristor. Mentor II cho php ngi s dng iu khin t ng gc m cho thyristor. Ngi s dng ch cn t gi tr tc yu cu vo v truy nhp cc tham s ca MentorII sao cho h thng lm vic ti u nht. Trong s hnh 3.6 ta thy c hai mch vng khp kn l mch vng tc v mch vng dng inh. mch vng tc , c tn hiu t u vo. Tn hiu ny c s dng t tc vo iu khin ng c. Trn MentorII ta c th t tc bng bin tr hoc phn mn. Tn hiu phn hi tc c ly t my pht tc so snh vi tn hiu t. mch vng dng in, tn hiu phn hi v ly t bin dng ba pha ca ngun in vo MentorII.

3.3. Cch ni mch ca MentorII.

MentorII c th chy c ch mt gc phn t v bn gc phn t nn hai kiu ni dy cho MentorII. Trong bn thuyt minh ny ta ch xt cch ni dy ch bn gc phn t.

Hnh 3.8. S ni dy 4 gc phn t ca MentorII

Trc tin ta ni hai cng tc t LC v RR. Cng tc t LC l cng tc chnh ng ngun ba pha vo Mentor II ti ba im L1-L3 ng thi ng t L11 v L12. Ngoi ra cn c ba tip im na ca cng tc t LC. cng tc t RR cng c ba tip im lin ng vi tip im ca LC. Phn ng ca ng c c ni vo hai u A1 v A2; phn kch t c ni vo hai u F1 v F2. Nu iu khin tc ng c bng bin tr vo u vo s 1 3 ca MentorII. V cui cng ta mc ngun iu khin vo ba im E1 E3.

Lu trong qu trnh ni dy, E1 - E3 phi trng pha vi L1 L3.

3.4. Bng iu khin

Bng iu khin ca Mentor II l ni iu khin v truy nhp cc tham s ca qua iu khin ng c

.

Hnh 3.9. Bng iu khin ca MentorII.

Bn phm ca MentorII phc v 2 mc ch chnh l:

Ci t li cc tham s theo yu cu s dng.

Thao tc n cc tham s cn hin th.

Bn phm gm c mt nt Reset, mt nt Mode, hai nt la chn thc n v hai nt la chn tham s. Bm nt Mode mt ln iu chnh tham s (nu hiu th nhp nhy th cho php iu chnh). Lc ny ta c th dng hai nt la chn tham s iu chnh, c th iu chnh nhanh bng cch n v gi phm . Nhn nt Mode ln na thot khi s iu chnh. Lu gi tr ca tham s mi iu chnh s b mt i khi tt ngun ca thit b. Do ta phi truy nhp n thc n v t tham s 00 bng 1.

Mn hnh ca MentorII hin th thc n (bn tri du thp phn), tham s (bn phi du thp phn) v d liu tham s c chn.

Ngoi ra cn c 6 n led hin th tnh trng lm vic ca MentorII. Lu rng 2 n led cu 1 v cu 2 sng th khng nht thit lc cu ang hot ng m c th do s truyn dn ph thuc vo gc m hay iu kin hot ng.

3.5. Truyn tin ni tip.

Giao tip ni tip vi MentorII l mt c tnh quan trng trong giao tip vi thit b ngoi vi trong cng nghip. Thit b ngoi vi c th ci t ton hoc tng phn. C kh nng bin i cc tham s ngay lp tc tho mn cc trng thi ca mt chu trnh nhim v hoc iu kin hot ng khc nhau trong qu trnh hot ng.

Phng tin ny gip ta theo di lin tc hot ng ca thit b phc v cho iu khin hoc mc ch phn tch.

Mt phng thc truyn tin chun cho tt c cc MentorII. N l gia din my my, cho php mt hoc nhiu thit b c s dng trong h thng iu khin bi PLC hoc my tnh.

MentorII c th iu khin trc tip, hot ng ca chng c th thay i, v trng thi ca chng c kim tra bi mt h thng iu khin c th giao tip khong 15 MentorII, v c th ln n 99 nu c s dng b m hng.

Cng truyn tin ca thit b l chn PL2. Ni theo chun RS422. Nghi thc l ANSI x3.28 2.5 A4, nh tiu chun cho nhng giao din cng nghip.

3.5.1. Kt ni.

Nhng phng thc truyn tin ni tip 9 chn loi D ni vi PL2 trn MDA 2B. Chn ni ny cung cp chun RS422 (ghi ch: RS422 thc t cng ging nh RS485 cho php nhiu h thng gim st.)

Ch : Kt ni RS232 c th thay th mt phn ca RS422.

Nhng yu t ca thng tin gia h thng iu khin v MentorII l k t ASCII.

3.5.2. iu chnh s b:

Mi thit b yu cu mt s nhn dng, hoc a ch t bi tham s 11.11. Baud 11.12 i hi s c t ph hp vi h thng iu khin. D liu, trng thi thit b, t tham s c th ly t thit b theo mt vi cch

Nhng yu t ca thng tin gia h thng iu khin. D liu, trng thi thit b, t tham s c th ly t thit b theo mt vi cch.Chn sRS232RS422

1NC0V

2TXDTXD

3RXDRXD

4

5

60VTXD

70VRXD

8

9

Dy cp truyn tin khng c chy song song vi dy cp in no c bit l dy ni thit b vi ng c. Nu khng trnh c th phi m bo khong cch cc tiu l 300mm. Chiu di cc i ca RS422 khong 1 mt.

K t nghaM ASCI

HEXPhm iu khin

EOTBit u tin ca cu lnh gi cho MentorII04D

ENQBit kt thc ca lnh c d liu05E

STXBit u tin ca cu tr li ca MentorII02B

ETXBit kt thc ca cu tr li ca Mentor II03C

ACKTn hiu thng bo MentorII nhn c lnh06F

BSLi li tham s trc tham s hin hnh08H

NAKTn hiu thng bo MentorII khng hiu cu lnh15U

3.5.3. Cc k t iu khin ca Mentor II.

3.5.4. a ch ni tip.

Mi thit b c mt nhn dng hay a ch (tham s 11.11) v vy ch c mt thit b c ni l tr li. Cho an ton, mi s 2 k t a ch ca thit b c lp li, nh vy a ch ca thit b s 14 c gi 4 k t: 1 1 4 4

3.5.5. Nhn dng tham s.

Truyn tin bi giao din ni tip, tham s c xc nh bi 4 ch s ch thc n v s tham s, nhng khng c thp phn. V d thc n 01 tham s 01 c vit l 0 1 0 1.

3.5.6. Phn d liu.

D liu chim 5 c tnh tip theo sau tham s. Khng s dng du thp phn.

3.5.7. Khi kim tra BCC.

Cho php thit b v h thng iu khin m bo thng tin truyn i khng b li tt c cc lnh v tr li u phi c kt thc bi mt khi kim tra.

3.5.8. Gi d liu.

d dng ta ly mt v d c th minh ho. Gi mnh lnh gim i 47.6 % gi tr ca tham s 01.17 n MentorII c a ch l 14.

Khi mun gi d liu n MentorII thng qua cng Com phi ng theo cu trc sau:

CONTROL

EOT

Control

-DADDRESS

1144CONTROL

STX

Control

-BPAR

0117DATA

-0476CONTROL

ETX

Control

-CBCC

Lu : Mc d liu c th t mt n nm k t u c.

Khi xc nhn c tn hiu gi n MentorII s tr li thng ip:

M ACK nu MentorII hiu v thc hin c mnh lnh gi n.

M NAK nu MentorII bo mnh lnh sai, d liu di qu hoc BCC sai.

32.5.9. c d liu t MentorII.

d dng ta ly mt v d c th minh ha: Gi mnh lnh c gi tr ca tham s 01.17 n MentorII c a ch l 12.

Ta c th c cc gi tr ca tham s trn MentorII qua mnh lnh c d liu c cu trc sau:

CONTROL

EOT

Control

-D

ADDRESS

1122PAR

0117CONTROL

ENQ

Control

-E

Khi nhn c mnh lnh nh trn MentorlII s tr li vi cu trc nh sau:

CONTROL

STYX

Control

-B

PARAM

0117DATA

-0476CONTROL

EXT

Control

-CBCC

3.6. Cc tham s chnh ca MentorII.

3.6.1 Menu1:Ci t tc Cc tham s ca MentorII c tch thnh cc nhm thun tin cho vic tra cu v truy nhp. Cc nhm gi l cc thc n, mi thc n s c cc chc nng khc nhau.

3.6.1. Mentor1: Ci t tc

C 4 kin t tc 01.17, 01.18, 01.19 v 10.20. Mt trong bn kiu c th t tc t 1000. n 1000. V c th truy cp qua bn phm, chng trnh hoc truyn ni tip ti bt k thi im no. Bn tham s ny gip cho MentorII c tnh linh hot cao khi giao tip vi cc thit b khc. Hai la chn 01.14 v 01.15 la chn mt trong bn kiu t tc trn.

Vic thay i cc tham s la chn lng cc hay n cc o cc, v tc ln nht v nh nht ca quay thun, quay ngc. Tham s 01.11 t ON nu 01.11 = 0 th 10.03 = 0. Tham s 01.12 o cc tnh. Tham s 01.13 la chn 01.05 hay khng.

a. Tham s 1.1: RO tc t trc khi b.

Theo di gi tr ca tc t lin tc. Tham s 1.1 cng c s dng bt u khi ng cng vi 1.6.

b. Tham gia 1.2: RO tc t sau khi b.

Theo di gi tr ca tc t sau khi c thm 1.4

c. Tham s 1.3: RO t trc khi tr:

Tc t trc khi c tr (tham kho thc n 2)

d. Tham s 1.4: RW t b:

gi tr t (t 1000 n +1000) c cng vo gi tr tc t 1.1.

e. Tham s 1.5: RW t inch.

L ngun ca tc t khi chn bi 1.13 (iu khin bi chn TB3 22 v TB3-23). Cung cp phng tin tin li t cc tc yu cu khc nhau. Phi nh hn tc cc i t bi 1.6 v 1.9.

f. Tham s 1.6. RW tc quay thun cc i.

t gii hn di ca tc quay thun. Tc ny khng c ngha nu 1.10 1 s ngn s chnh lch gia tc cc tiu quay thun v quay ngc khi tc t vo l 0.

h. Tham s 1.8: RW tc quay ngc cc tiu:

t gii hn di ca tc quay thun. Tc ny khng c ngha nu 1.10=1 s ngn s chnh lch gia tc cc tiu quay ngc khi tc t vo l 0.

i. Tham s 1.9:RW tc quay ngc cc i.

t gii hn trn ca tc quay ngc.

j. Tham s 1.10: RW la chn lng cc:

Trong trng thi bnh thng (=1) cho php thit b tr li tn hiu tc t 1.2 trong trng hp hng quay c xc nh bi tn hiu lng cc. Cc tnh dng gy ra quay thun, cc m gy ra quay ngc. Khi 1.10=0 thit b tr li tn hiu theo kiu n cc, cc tnh m xem nh tc 0. Khi o chiu c xc nh bi 1.12 (4 gc phn t)

k. Tham s 1.11RWt ON

Mc nh l 0 nu TB3-21 khng kch hot . Khng th t l 1 tr phi TB3-21 c kch hot. Tham kho menu 8. iu khin bi TB3 25, 22, 23, 24.

l. Tham s 1.12 RW la chn quay ngc.

Quay ngc khi o cc tnh ca tc t. N c hiu ng (trong 4 gc phn t) khi quay ngc tn hiu tc m khng quan tm n hng quay ca ng c. Mc nh bi TB3-25,22,23,24.

m. Tham s 1.13RW la chn inch.

Thay th tt c cc tc yu cu t bi 1.5. Mc nh 1.13, tc t bnh thng. iu khin bi TB3-22,23.

n. Tham s 1.14RW t Selector1.

Chn 1 v 3 hay 2 v 4. Bn gi tr ca 1.14 v 1.5 s cho php la chn mt trong bn gi tr ca 1.17 v 1.20.

o. Tham s 1.15 RW t Selector2.

Chn 1/2 hay 3/4 bn gi tr ca 1.14 v 1.15 s cho php la chn mt trong bn gi tr ca 1.17 v 1.20.

p. Tham s 1.16 RW t ng b.

Khng cho thit b chy cho n khi c tn hiu tc t.

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