Mc lc: CHNG 1.TNG QUAN V CNG NGH...................................................3 1.1.ng dung cua acquy....................................................................................3 1.2. Acquy axit..................................................................................................3 1.2.1. Cu tao cua binh acquy axit.................................................................3 1.2.2.Qua trinh hoa hoc trong acquy axit......................................................5 1.3.Acquy kim.................................................................................................6 1.3.1.Cu tao cua acquy kim........................................................................6 1.3.2.Qua trinh hoa hoc trong acquy kim ...................................................6 1.4.S khac nhau gia acquy axit va acquy kim.............................................7 1.5.Cac thng s c ban cua acquy...................................................................8 1.5.1.Sc in ng cua acquy......................................................................8 1.5.2.Dung lng cua acquy..........................................................................8 1.6.c tinh phong nap cua acquy.....................................................................9 1.6.1.c tinh phong cua acquy....................................................................9 1.6.2.c tinh nap cua acquy.......................................................................10 1.7.Cac phng phap nap acquy t ng........................................................11 1.7.1.Phng phap nap dong in...............................................................11 1.7.2.Phng phap nap in ap...................................................................12 1.7.3.Phng phap nap dong ap................................................................13 CHNG 2: TNH TON THIT K MCH CNG SUT.........................15 2.1.Cac mach chinh lu ..................................................................................15 2.1.1.Mach chinh lu hinh tia ba pha..........................................................15 2.1.2.Mach chinh lu c iu khin cu ba pha i xng...........................17 2.1.4.Mch chnh lu iu khin cu mt pha khng i xng..................20 2.2.Chn mch chnh lu ph hp..................................................................21 2.3.Tinh toan cac thng s vi mach a chon................................................21 2.4.Mach bao v Tiristor..................................................................................23 2.5.Tinh toan may bin ap...............................................................................24 CHNG 3. TNH TON THIT K CHO MCH IU KHIN ...............26 3.1.Yu cu v nguyn tc iu khin.............................................................26 3.1.1.Mc ch v yu cu...........................................................................26 3.1.2.Nguyn tc iu khin........................................................................27 3.2.Cc linh kin in t s dng trong mch.................................................29 3.3.S khi v chc nng...........................................................................31 3.3.1.Khu ng pha....................................................................................31 3.3.2 Khi to xung ng b.......................................................................31 3.3.3.Khi to in p rng ca...................................................................32 3.3.4.Khi phn hi dng in....................................................................33 3.3.5.Khi phn hi in p.........................................................................34 3.3.6.Khi chuyn mch np.......................................................................34 3.3.7.Khi to xung chm...........................................................................35 3.3.8. Khi khuych i xung v bin p xung...........................................36 3.4. Mch iu khin.......................................................................................391

3.4.1.Dng in p.......................................................................................40 3.4.2.Nguyn l hot ng ca s ..........................................................40 3.5.Khi ngun nui mch iu khin.............................................................40

2

CHNG 1.TNG QUAN V CNG NGH1.1.ng dung cua acquy Acquy l mt ngun in dc tch tr nng lng in di dng ha nng. Acquy l ngun in mt chiu cung cp cho cc thit b in trong cng nghip cng nh trong i sng hng ngy: nh ng c in mt chiu, bng n, c tch in, cc mch in t, Acquy l ngun cung cp cho cc ng c khi ng. Trong thc t c nhiu loi acquy nhng thng gp nht l hai loi sau: acquy axit v acquy kim. 1.2. Acquy axit 1.2.1. Cu tao cua binh acquy axit Bnh acquy thng thng gm v bnh, cc bn cc, cc tm ngn v dung dch in phn V bnh: V bnh acquy axit hin nay c ch to bng nha bnit hoc anphantonec hay caosu nha cng. tng bn v kh nng chu axit cho bnh, khi ch to ngi ta p vo bn trong bnh mt lp lt chu axit l polyclovinyl, lp lt ny dy khong 0,6mm. Nh lp lt ny tui th ca acquy tng ln t 2-3 ln. Pha trong v bnh ty theo in p danh nh ca acquy m chia thnh cc ngn ring bit v cc vch ngn ny c ngn cch bi cc ngn kn v chc. Mi ngn c gi l mt ngn acquy n. y cc ngn c cc sng khi bn cc to thnh khong trng gia y bnh v mt di ca khi bn cc, nh m trnh c hin tng chp mch gia cc bn cc do cht tc dng bong ra v ri xung y gy ln. Bn ngoi v bnh c c hnh dng gn chu lc tng bn c v c th c gn cc quai xch vic di chuyn c dng hn. Bn cc, phn khi bn cc v khi bn cc: Bn cc gm ct hnh li v cht tc dng. Ct c bng hp kim ch(Pb)antimion(Sb) vi t l (87 95)%Pb v (5 13)%Sb. Ph gia antimon thm vo c tc dng tng thm cng, gim han g v ci thin tnh c cho ct. Ct gi cht tc dng v phn phi dng in khp b mt cc. iu ny c ngha rt quan trng i vi cc bn cc dng v in tr ca cht tc dng (oxit ch) ln hn rt nhiu so vi in tr ca ch nguyn cht, do cng tng chiu dy ca ct th in tr trong ca acquy cng nh. Ct c dng khung bao quanh, c vu hn ni cc bn cc thnh phn khi bn cc v c hai chn t cc sng y bnh acquy. V in ct ca bn cc m khng phi l yu t quyt nh v chng cng t b han g nn ngi ta thng lm mng hn bn cc dng. c bit l hai tm bn ca phn khi bn cc m li cng mng v chng ch lm vic c mt pha vi cc bn cc dng. Cht tc dng c ch to t bt ch, axit sunfuric c v khong 3% cc mui axit hu c i vi bn cc m, cn i vi bn cc dng th cht tc dng c ch to t cc oxit ch Pb3O4, PbO v dung dch axit sunfuric c. Ph gia3

mui ca axit hu c trong bn cc m c tc dng tng xp, bn ca cht tc dng, nh m ci thin c thm su ca dung dch in phn vo trong lng bn cc ng thi in tch thc t tham gia phn ng ha hc cng c tng ln Cc bn sau khi c cht y cht tc dng c p li, sy kh v thc hin qu trnh to cc, tc l chng c ngm vo dung dch axit sunfuric long v np vi dng in mt chiu tr s nh. Sau qu trnh nh vy, cht tc dng cc bn cc dng hon ton tr thnh PbO2 (mu gch xm). Sau cc bn cc dng c em ra ra, sy kh v lp rp. Nhng phn khi bn cc cng tn trong mt acquy c hn vi nhau to thnh cc khi bn cc v c hn ni ra cc vu cc lm bng ch hnh cn ni ra ti tiu th. Vi ch rng, nu ta mun tng dung lng acquy th ta phi tng s bn cc mc song song trong mt acquy n. Thng ngi ta ly t 5 8 tm. Cn mun tng in p danh nh ca acquy th ta phi tng s bn cc mc ni tip. Tm ngn: Cc bn cc m v dng c lp xen k vi nhau v cch in vi nhau bi cc tm ngn v m bo cch in tt nht cc tm ngn c lm rng hn so vi cc bn cc. Cc tm ngn c tc dng chng chp mch gia cc bn cc m v dng, ng thi cc tm bn cc khi b bong ri ra khi s dng acquy. Cc tm ngn y phi l cht cch in tt, bn, do, chu c axit v c xp thch hp ngn cn cht in phn thm n cc bn cc. Cc tm ngn hin nay c ch to t vt liu polyvinyl xp, dy khong t 0,8-1,2mm v c dng mt phng hng v pha bn cc m cn mt mt c dng sng hoc g th hng v pha bn cc dng nhm to iu kin cho dung dch in phn d lun chuyn hn n cc bn cc dng v dung dch lu thng tt hn. Dung dch in phn Dung dch in phn trong bnh acquy l loi dung dch axit sunfuric c pha ch t axit nguyn cht vi nc ct theo nng quy nh ty thuc vo iu kin kh hu ma v vt kiu lm tm ngn. Nng dung dch axit sunfuric = (1,1 1.3)g/cm . Nng dung dch in phn c nh hng ln n sc in ng ca acquy. Nhit mi trng c nh hng ln n nng dung dch in phn. Vi cc nc trong vng xch o nng dung dch in phn quy nh khng qu 1,1 g/cm . Vi cc nc lnh, nng dung dch in phn cho php ti 1,3 g/cm . Trong iu kin kh hu nc ta th ma h nn chn nng dung dch khong (1,25-1.26) g/cm , ma ng nn chn nng khong 1,27g/cm . Cn nh rng: nng qu cao s lm chng hng tm ngn, chng hng bn cc, d b sunfat ha trong cc bn cc nn tui th ca acquy cng gim i rt nhanh. Nng qu thp th in dung v in p nh mc ca acquy gim v cc nc x lnh th dung dch vo ma ng d b ng bng.4

* Nhng ch khi pha ch dung dch in phn cho acquy : - Khng c dng axit c thnh phn tp cht cao nh loi axit k thut thng thng v nc khng phi l nc ct v dung dch nh vy s lm tng cng qu trnh t phng in ca acquy. - Cc dng c pha ch phi lm bng thy tinh, s hoc cht do chu axit. Chng phi sch, khng cha cc mui khong, du m hoc cht bn - m bo an ton trong khi pha ch, tuyt i khng c nc vo axit c m phi axit vo nc v dng u thy tinh khuy u. Np, nt v cu ni: Np lm bng nha ebonit hoc bakenit. Np c hai loi: - Tng np ring cho mi ngn - Np chung cho c bnh loi ny cu to phc tp nhng kn tt. Trn np c l dung dch in phn vo cc tm ngn v kim tra mc dung dch in phn, nhit v nng dung dch trong acquy. L c y kn bng nt c ren gi cho dung dch in phn trong bnh khi b bn v snh ra ngoi. nt c l nh thng kh t trong bnh ra ngoi lc np acquy. Np mt s loi acquy c l thng kh ring nm st l , kt cu nh vy rt thun tin cho vic iu chnh mc dung dch trong bnh acquy. Trong trng hp ny, nt khng ch l thng kh na. 1.2.2.Qua trinh hoa hoc trong acquy axit Trong acquy thng xy ra hai qu trnh ha hc thun nghch m c trng l l qua trnh np v phng in. Khi np in, nh ngun in np m mch ngoi cc in t chuyn ng t cc bn cc dng n cc bn cc m l dng in np In. Khi phng in, di tc dng ca sut in ng ring ca acquy, cc in t s chuyn ng theo hng ngc li v to thnh dng in phng Ip. Khi acquy np no, cht tc dng cc bn cc dng l PbO2 cn ti cc bn cc m l ch xp Pb. Khi phng in, cc cht tc dng c hai bn cc u tr thnh sunfat ch PbSO4 c dng tinh th nh. Khi np in cho acquy s xy ra phn ng: - cc dng:PbSO 4 2e + 2 H 2O = PbO2 + H 2 SO4 + 2 H +

(1.1)

- cc m: (1.2) - Ton b qu trnh xy ra trong c quy khi np in l:PbSO 4 + 2e + 2 H + = Pb + H 2 SO4

(1.3) Kt qu l to thnh mt in cc Pb v mt in cc PbO2. S phng in ca acquy sy ra khi ni hai in cc Pb v PbO2 va thu c vi ti, lc ny ha nng c d tr trong acquy s chuyn thnh in nng. y cc in cc xy ra cc phn ng ngc ca (1.1) v (1.2), ngha l trong2 PbSO 4 + 2 H 2O = Pb + PbO 2 + 2 H 2 SO4

5

acquy s xy ra phn ng ngc ca (1.3). Acquy s cung cp dong in cho n khi c hai in cc tr li thnh PbSO4 nhu ban u. Sau nu mun dng tip th ngi ta li np in cho acquy v c th tip din. 1.3.Acquy kim 1.3.1.Cu tao cua acquy kim Acquy kim l loi acquy m dung dch in phn c dng trong acquy l dung dch kim KOH hoc NaOH. Ty theo cu to ca bn cc acquy kim c chia lm ba loi: -Loi acquy St (Fe) Niken (Ni) -Loi acquy Cadimi (Cd) Niken (Ni) -Loi acquy Bc Ag) Km (Zn) Trong ba loi trn th loi th ba c h s hiu dng trn mt n v trng lng v mt n v th tch l ln hn, nhng gi thnh ca n li cao hn v phi s dng khi lng bc chim ti 30% khi lng ca cht tc dung, do loi ny t c dng. Acquy kim c cu to tng t acquy axit, tc n cng gm dung dch in phn, v bnh, cc bn cc, Bn cc ca acquy kim c ch to thnh dng thi hoc khng thi. Gia cc bn cc c ngn bi cc tm ebonit. Chm bn cc dng v chm bn cc m c hn ni nh chm bn cc ca acquy axit a ra cc vu cc cho caquy. Cc chm bn cc c t trong bnh in phn v c ngn cch vi v bnh bng lp nha vinhiplat. Loi acquy dng bn cc dng thi th mi thi l mt hp lm bng thp l trn b mt c khoan nhiu l: = 0.2 0.3mm cho dung dch thm qua. Nu l acquy st niken th trong hp bn cc m cha st c bit thun khit, cn trong bn cc dng l hn hp 75%NiO.OH v 25% bt than hot tnh. Loi acquy kim dng bn cc khng phn thi, th bn cc c ch to theo kiu khung xng, ri em cc cht tc dng c cu trc xp mn p vo cc l nh trn bn cc. 1.3.2.Qua trinh hoa hoc trong acquy kim Ging nh trong acquy axit, qu trnh ha trong acquy kim cng l qu trnh thun nghch. Nu bn cc trong acquy kim l st v niken th phn ng ha hc xy ra trong acquy nh sau: Trn bn cc dng:Ni (OH ) 2 + KOH + OH = Ni (OH ) 3 + KOH

Trn bn cc m:Fe(OH ) 2 + KOH = Fe + KOH + 2OH

(1.4)

(1.5) Nh vy qu trnh np in, st hidroxit trn bn cc m b phn tch thnh st nguyn t v anion OH . Cn bn cc dng, Ni (OH ) 2 c chuyn ha thnh Ni (OH ) 3 . Cht in phn KOH c th xem nh n khng tham gia vo phn ng ha hc m ch ng vai tr cht dn in, do sc in ng ca6

acquy hu nh khng ph thuc vo nng cht in phn. Sc in ng ca acquy ch c xc nh da trn trng thi ca cc cht tc dng cc tm cc. Thng thng acquy kim c np in hon ton sc in ng t c khong 1,7 n 1,85V. Khi acquy phng in hon ton sc in ng ca acquy l 1,2 n 1,4V. Nh vy in th phng in ca acquy kim thp hn acquy axit. Nu acquy axit in th phng in bnh qun l 2V th acquy kim ch l 1,2V. Hin nay cc nh thit k, ch to cha dng li nhng kt qu t c, ngi ta ch to c nhng acquy kim mi kh nh v nh, nhng vn c cc thng s k thut ca acquy axit. Nhng acquy ang hng ti vic thay th cc bn cc bng nhng hp kim c kh nng chng han g, gim kch thc v tng tnh bn vng. nhng tp cht mi c trn vo trong cht tc dng s ci thin c tnh phng in ca acquy ng k . Nhiu acquy mi khng c cu ni trn np v kt cu v bnh cng thay bng nhng vt liu rt nh nn gim c chiu dy thnh bnh, acquy cng t phi chm sc hn. 1.4.S khac nhau gia acquy axit va acquy kim C hai loi acquy ny u c mt c im chung l tnh cht ti thuc loi dung khng v sc phn in ng. Nhng chng cn c mt s c im khc bit sau : Acquy axit - Kh nng qu ti khng cao, dng np ln nht t c khi qu ti l Inmax = 20%C10 _Hin tng phng ln, do c qui nhanh ht in ngay c khi khng s dng. _S dng rng ri trong i sng, nhit cao va p ln nhng cng sut v qu ti va phi. _Dng trong t, xe my v cc ng c my n cng sut va v nh. _S dng nhng ni c yu cu c s dng vi cc thit b cng sut ln. _Dng ph bin trong cng nghip hng khng, hng hi v nhng ni nhit mi trng thp.7

acquy kim _Kh nng qu ti rt ln dng in np ln nht khi c th t ti: Inmax = 50%C10 _Hin t phng nh.

cng nghip c bit nhng ni c cng sut ln qu ti thng xuyn,

_Gi thnh thp _Gi thnh cao. 1.5.Cac thng s c ban cua acquy 1.5.1.Sc in ng cua acquy Sc in ng ca acquy kim v acquy axit ph thuc vo nng dung dch in phn. Ngi ta thng s dng cng thc kinh nghim Eo = 0,85 + Trong : (V) (1.6) Eo - sc in ng tnh ca acquy ( V ) - nng dung dch in phn 15 C ( g/cm3 ) Trong qu trnh phng in th sc in ng Ep ca acquy c tnh theo cng thc: Ep = Up + Ip.rb Trong : Ep - sc in ng ca acquy khi phng in ( V ) Ip - dng in phng ( A ) Up - in p o trn cc cc ca acquy khi phng in (V) rb - in tr trong ca acquy khi phng in ( ) Trong qu trnh np in th sc in ng En ca acquy c tnh theo cng thc: En = Un - In.rb Trong : En - sc in ng ca acquy khi np in ( V ) In - dng in np ( A ) Un - in p o trn cc cc ca acquy khi np in ( V ) rb - in tr trong ca acquy khi np in ( ) 1.5.2.Dung lng cua acquy -Dung lng phng ca acquy l i lng nh gi kh nng cung cp nng lng in ca acquy cho ph ti, v c tnh theo cng thc : Cp = Ip.tp Trong : Cp - dung dch thu c trong qu trnh phng ( Ah ) Ip - dng in phng n nh trong thi gian phng in tp ( A ) tp - thi gian phng in ( h ). -Dung lng np ca acquy l i lng nh gi kh nng tch tr nng lng ca c qui v c tnh theo cng thc : Cn = In.tn Trong : Cn - dung dch thu c trong qu trnh np ( Ah )8

(1.7)

(1.8)

In - dng in np n nh trong thi gian np tn ( A ) tn - thi gian np in ( h ). 1.6.c tinh phong nap cua acquy 1.6.1.c tinh phong cua acquy c tnh phng ca acquy l th biu din quan h ph thuc ca sc in ng, in p acquy v nng dung dch in phn theo thi gian phng khi dng in phng khng thay i .

Hnh 1.1: c tnh phng ca acqui T c tnh phng ca c qui nh trn hnh v ta c nhn xt sau: Trong khong thi gian phng t tp = 0 n tp = tgh, sc in ng in p, nng dung dch in phn gim dn, tuy nhin trong khong thi gian ny dc ca cc th khng ln, ta gi l giai on phng n nh hay thi gian phng in cho php tng ng vi mi ch phng in ca acquy ( dng in phng ). T thi gian tgh tr i dc ca th thay i t ngt .Nu ta tip tc cho acquy phng in sau tgh th sc in ng ,in p ca acquy s gim rt nhanh .Mt khc cc tinh th sun pht ch (PbSO4) to thnh trong phn ng s c dng th rn rt kh ho tan ( bin i ho hc) trong qu trnh np in tr li cho acquy sau ny. Thi im tgh gi l gii hn phng in cho9

php ca acquy, cc gi tr Ep, Up, ti tgh c gi l cc gi tr gii hn phng in ca c qui. c qui khng c phng in khi dung lng cn khong 80%. Sau khi ngt mch phng mt khong thi gian no, cc gi tr sc in ng, in p ca acquy, nng dung dch in phn li tng ln, ta gi y l thi gian hi phc hay khong ngh ca acquy. Thi gian hi phc ny ph thuc vo ch phng in ca acquy (dng in phng v thi gian phng ). 1.6.2.c tinh nap cua acquy c tnh np ca acquy l th biu din quan h ph thuc gia sc in ng , in p v nng dung dch in phn theo thi gian np khi tr s dng in np khng thay i .I (A) E,U (V)

10

2,11 1,95 1,75

E UP V ng phng n cho ph i p

Khong ngh

5

CP =I P.tP0 2 4 6 8 10

t

Hinhf1.2: c tnh np ca acqui T th c tnh np ta c cc nhn xt sau : Trong khong thi gian t tn = 0 n tn = tgh th sc in ng, in p, nng dung dch in phn tng dn. Ti thi im ts trn b mt cc bn cc m xut hin cc bt kh (cn gi l hin tng" si " ) lc ny hiu in th gia cc bn cc ca acquy n tng n 2,4 V . Nu vn tip tc np gi tr ny nhanh chng tng ti 2,7 V v gi10

nguyn. Thi gian ny gi l thi gian np no, n c tc dng cho phn cc cht tc dng su trong lng cc bn cc c bin i tun hon, nh s lm tng thm dung lng phng in ca acquy. Trong s dng thi gian np no cho acquy ko di t 2 3 h trong sut thi gian hiu in th trn cc bn cc ca acquy v nng dung dch in phn khng thay i . Nh vy dung lng thu c khi acquy phng in lun nh hn dung lng cn thit np no acquy. Sau khi ngt mch np, in p, sc in ng ca acquy, nng dung dch in phn gim xung v n nh. Thi gian ny cng gi l khong ngh ca acquy sau khi np. Tr s dng in np nh hng rt ln n cht lng v tui th ca acquy. Dng in np nh mc i vi c qui l In = 0,1C10 . Trong C10 l dung lng ca acquy m vi ch np vi dng in nh mc l In = 0,1C10 th sau 10 gi acquy s y. V d vi acquy C = 180 Ah th nu ta np n dng vi dng in bng 10% dung lng ( tc In = 18 A ) th sau 10 gi acquy s y. 1.7.Cac phng phap nap acquy t ng 1.7.1.Phng phap nap dong inA_ + _

V Un

D

D D+

A

R

A

R

_

....

+

_

....

+

Phng php np dng in11

y l phng php np cho php chn c dng np thch hp vi mi loi acquy, bo m cho c qui c no. y l phng php s dng trong cc xng bo dng sa cha np in cho acquy hoc np s cha cho cc c qui b Sunfat ho. Vi phng php ny acquy c mc ni tip nhau v phi tho mn iu kin : Un 2,7.Naq (1.9) Trong : Un - in p np Naq - s ngn acquy n mc trong mch Trong qu trnh np sc in ng ca acquy tng dn ln, duy tr dng in np khng i ta phi b tr trong mch np bin tr R. Tr s gii hn ca bin tr c xc nh theo cng thc :R= U n 2,0 N aq In

(1.10)

Nhc im ca phng php np vi dng in khng i l thi gian np ko di v yu cu cc acquy a vo np c cng dung lng nh mc. khc phc nhc im thi gian np ko di, ngi ta s dng phng php np vi dng in np thay i hai hay nhiu nc. Trong trng hp hai nc, dng in np nc th nht chn bng ( 0,3 0,6 )C10 tc l np cng bc v kt thc nc mt khi c qui bt u si. Dng in np nc th hai l 0,1C10 1.7.2.Phng phap nap in ap Phng php ny yu cu cc acquy c mc song song vi ngun np. Hiu in th ca ngun np khng i v c tnh bng (2,3V 2,5V) cho mi ngn n. Phng php np vi in p khng i c thi gian np ngn, dng np t ng gim theo thi gian. Tuy nhin dng phng php ny acquy khng c np no. V vy np vi in p khng i ch l phng php np b xung cho acquy trong qu trnh s dng. Un V A

12

1.7.3.Phng phap nap dong ap _ A Un D +

A _ _ .... . .... . .... .

VR + +

_

+

Phng php np dng p y l phng php tng hp ca hai phng php trn. N tn dng c nhng u im ca mi phng php. i vi yu cu ca bi l np acquy t ng tc l trong qu trnh np mi qu trnh bin i v chuyn ho c t ng din ra theo mt trnh t t sn th ta chn phng n np acquy l phng php dng p. i vi acquy axit: bo m thi gian np cng nh hiu sut np th trong khon thi gian tn = 8h tng ng vi 75 80 % dung lng acquy ta np vi dng in khng i l In = 0,1C10. V theo c tnh np ca acquy trong on np chnh th khi dng in khng i th in p, sc in ng ti t thay i, do bo m tnh ng u v ti cho thit b np. Sau thi gian 8h acquy bt u si lc ta chuyn sang np ch n p. Khi thi gian np c 10h th acquy bt u no, ta np b xung thm 2 n 3h.

13

i vi acquy kim : Trnh t np cng ging nh acquy axit nhng do kh nng qu ti ca acquy kim ln nn lc n dng ta c th np vi dng np In= 0,2C10 hoc np cng bc tit kim thi gian vi dng np In = 0,5C10 . Cc qu trnh np c qui t ng kt thc khi b ct ngun np hoc khi np n p vi in p bng in p trn 2 cc ca c qui, lc dng np s t t gim v khng.

14

CHNG 2: TNH TON THIT K MCH CNG SUT2.1.Cac mach chinh lu 2.1.1.Mach chinh lu hinh tia ba phaUG

G1

G2

G3

G1t

U0 2 3

t

Ud I T1 t0 I T2 I T3 Id t1 t2 t3 t4t

t

t

t

t

Hnh 2.1: S v dng in p mch chnh lu hnh tia ba pha a.Nguyn l hot ng: t0 t1 : T3 thng Ud = Uc , Id = IT3 . t1 t2 : T1 thng Ud = Ua , Id = IT1 . t2 t3 : T2 thng Ud = Ub , Id = IT2 . t3 t4 : T3 thng Ud = Uc , Id = IT3 . b.Cc cng thc c bn: - in p trn ti Ud =

1 2

2

U0

d

(t )d (t ) =

3 2

5 + 6

+

2 .U 2 . sin(t )d (t )

6

=

3 2U 2 3 6 . 3. cos = U 2 cos 2 2

- Dng in trn ti: I d =

Ud RdId 315

- Dng in qua van: I T =

- Gi tr trung bnh ca in p chn lu: U d 0 = 1,17U 2 - in p ngc trn van: U ng = 2,45U 2 - Dng in pha th cp: I2 = 0,58Id - Dng in pha s cp:I1 = 0,47Id.Kba - Cng sut ti: Pd = Ud0.Id - Cng sut my bin p: Sba = 1,35Pd Nhn xt: Mch chnh lu c iu khin tia 3F c cu to phc tp, mun mch hot ng c cn mc bin p a im trung tnh ra ti, mi van ch lm vic trong 1/3 chu k v vy dng in trung bnh chy qua van nh. Mch dng ngun 3F nn cng sut tng ln rt nhiu, dng in ti n vi trm ampe.

16

2.1.2.Mach chinh lu c iu khin cu ba pha i xngUGUG1 UG2 UG3 UG4 UG5 UG6 t t t t t

U

T1

T3

T5

T1

t

t

Ud

T6

T2

T4

T6

T2

Id I T1,T4I T1 I T4 I T1

t

t

I T3,T6I T6 I T3 I T6

t

I T2,T5I T5 IT2 I T5 I T2

t

Ung

t

t

Hnh 2.2: S v dng in p chnh lu cu ba pha i xng a.Nguyn l hot ng. Mi Tiristor c pht 2 xung iu khin. Xung th nht xc nhgc m . Xung th 2 m bo thng mch ti. b.Mt s cng thc c bn. -in p trn ti: U = Ucos = 2,34Ucos . -Dng in trn ti: I = . -Dng in trung bnh qua van: I = . -in p ngc t ln van: U =2,45U -Dng in pha th cp: I = 0,816I . -Dng in pha s cp: I = 0,816I.K . -Cng sut my bin p: S = 1,05P . -Cng sut ti: P = UI .17

Nhn xt: Mch chnh lu iu khin i xng cu 3F thng c s dng rng ri trong thc t, mch cho ra cht lng in p bng phng, dng in chy qua ti lin tc trong sut qu trnh lm vic. Mch chnh lu ny thng c p dng vi nhng mch c cng sut ln v dng in chy qua mi van ch ch chy trong 1/3 chu k. 2.1.3.Mach chinh lu c iu khin cu ba pha khng i xng -Mt s cng thc c bn: -in p trn ti: U d = -Dng in trn ti: I = . -Dng in trung bnh qua van: I = . -in p ngc t ln van: U =2,45U -Cng sut my bin p: S = 1,05P . -Cng sut ti: P = UI3 6U 2 (1 + cos ) . 2

Nhn xt : Tuy in p chnh lu cha nhiu sng hi nhng chnh lu cu 3 pha khng i xng c qu trnh iu chnh n gin , kch thc gn nh hn.

18

Hnh 2.3: S v dng in p chnh lu cu ba pha khng i xng

19

2.1.4.Mch chnh lu iu khin cu mt pha khng i xng

Hnh 2.4: S v dng in p chnh lu cu mt pha khng i xng Trong s ny, gc dn dng chy ca Tiristor v ca it khng bng nhau. Gc dn ca it l : D = + Gc dn ca Tiristor l : T = Gi tr trung bnh ca in p ti:Ud = 1 2 U 2 sin d = 2U 2 (1 + cos )

U d max =Do :U2 =

U d max .64 ,8 = = 72V 2 2 2 2

2 2U 2

Gi tr trung bnh ca dng ti : U Id = d Zt Dng qua Tiristor:

1 IT = Id d = Id 2 2 Dng qua it:ID = 1 + + Id d = Id 2 2

20

Gi tr hiu dng ca dng chy qua s cp my bin p:I2 = 1 2 Id d = Id 1

Nhn xt: S chnh lu iu khin 1 pha khng i xng c cu to n gin, gn nh , d iu khin , tit kim van . Thch hp cho cc my c cng sut nh v va. 2.2.Chn mch chnh lu ph hp - C hai phng n dng s chnh lu i xng cu ba pha v chnh lu khng i xng cu ba pha u c nhiu knh iu khin, nhiu Tiristor nn gi thnh cao khng kinh t. - Do yu cu ca u bi, v s knh iu khin t nn ta chn s chnh lu iu khin cu 1 pha ng i xng. Chng c mt s u im: Hiu sut s dng my bin p cao hn mt s s nh cu 1 pha i xng. n gin hn v s lng Tiristor gim xung ch cn 2 nn mch iu khin c t knh iu khin hn, bo m kinh t hn. Cng mt di iu chnh in p mt chiu th cu khng i xng iu khin chnh xc hn. 2.3.Tinh toan cac thng s vi mach a chon Qua phn tch trn ta chn s chnh lu iu khin cu 1 pha khng i xng dng cho mch lc mch np c qui t ng . Phng n ny va p ng c yu cu k thut va bo m cho vic thit k. Nh phn tch trn: Ta chn phng n thit k cho mch np c qui l s chnh lu cu 1 pha khng i xng. C s nguyn l mch lc nh sau :

Hnh 2.5: S nguyn l mch lc S liu cho trc: in p ngun 3 pha : 220/380V ; f = 50 Hz Yu cu u ra (Ngun mt chiu t ng np acquy ):21

U = 16.2V I = 80A S liu tnh ton v chn la van. in p th cp ca bin p: T cng thc: U = .( 1 + cos ) in p U t max khi gc = 0. U = = 18V Dng in trung bnh chy qua van: I = = = 40A in p ngc t ln van l: U = .U = .18 25.5V Chn van m bo cho cc van hot ng tt chng ta chn tr v in p: k = 1,6 , h s d Do vy: - Chng ta chn van chu c in p ngc U = 25,5.1,6 = 40,8V - Dng in trung bnh chy qua van l: I = 40.1,2 = 48A Vi cc thng s v dng in, in p nh trn ta tin hnh tra s tay chn c cc van nh sau: - Chn Tiristor loi : T-50 Do Lin X (c) ch to: Sch in t cng sut ca Nguyn Bnh, Bng 1.3, tr 28 c cc thng s nh sau: + + + + + Dng in trung bnh qua van: I=50A. in p ngc cc i t nn van: U=0,1 1kV. Tn tht in p: U = 1,3V. Thi gian kha: t = 10,5 s. Tc tng in p: = 100V/ s22

T yu cu ca bi ta c:

van phi c h s d

tr v dng in: k = 1,2.

+ + +

Tc tng dng in: = 100A/ s Dng in iu khin: I = 0,9A. in p iu khin: U = 3V

- Chn Diot loi: B-50 Do Lin X (c) ch to: Sch in t cng sut ca Nguyn Bnh, Bng 1.1, tr 8 c cc thng s nh sau: + + + Dng in trung bnh qua van: I=50A. in p ngc cc i t nn van: U=0,1 1kV. Tn tht in p: U = 0,7V.

2.4.Mach bao v Tiristor. Tiristor v Dit cng rt nhy cm vi in p qu ln so vi in p nh mc, ta gi l qu in p, v vy cn mc thm mch bo v qu in p. Ngi ta chia ra 2 loi nguyn nhn gy nn qu in p: -Nguyn nhn ni ti: y l s tch t in tch trong cc lp bn dn. Khi kho Tiristor bng in p ngc, cc in tch ni trn i ngc hnh trnh, to ra dng in ngc trong khong thi gian rt ngn. S bin thin nhanh chng ca dng in ngc gy ra sc in ng cm ng rt ln trong cc in cm, lun lun c, ca ng dy ngun dn n cc Tiristor. V vy gia cc ant v catt ca Tiristor xut hin qu in p. -Nguyn nhn bn ngoi: nhng nguyn nhn ny thng xy ra ngu nhin nh khi ct ng ti mt my bin p trn ng dy, khi mt cu ch bo v chy, khi c sm st bo v mch qu p ngi ta thng dng mch L C, (xem hnh bn di):

23

Hnh 2.6: Mch bo v Tiristor Mch R C u song song vi Tiristor nhm bo v qu in p do tch t in tch khi chuyn mch gy nn Mch R C u gia cc pha th cp ca my bin p l bo v qu in p do ng ct ti ( dng in t ha ) my bin p gy nn. Thng s ca R C ph thuc o mc qu in p c th xy ra, tc bin thin ca dng in chuyn mch, in cm trn ng dy, dng in t ho my bin p .v.v Theo kinh nghim c ta chn: C = 0.22 F R = 50 2.5.Tinh toan may bin ap - Gi tr hiu dng in p th cp my bin p: U = 1,11.U 18V I = 1,11I 89A - Cng sut biu kin my bin p: S = U.I = 18.89 = 1602VA. -Chn mch t 3 tr , tit din tnh theo cng thc: Q=K . Trong : C - S tr mch t. f - Tn s ngun. K=5 6 Q = 5. 13cm . ng knh tr : d = = 4cm. Chn li thp c tit din 13 cm lm bng vt liu st t dy 0,2mm, l thp dp hnh ch E v ch I ghp li. Chn s b mt t cm trong tr B = 1,1T Chn ch s m = = 2,3 h = m.d = 2,3.4 = 9,2cm. Chn chiu cao tr h = 9cm.24

Tnh ton dy qun: S vng dy mi pha s cp my bin p: W = = = 693 vng S vng dy mi pha th cp my bin p: W = W = 693 = 57 vng. Chn s b mt dng in trong my bin p: J = J = 2,75 A/mm Dng in s cp ca my bin p l: I = = =7.28A Tit din dy dn s cp my bin p: S = = = 2.65mm Tit din dy dn th cp my bin p: S = = = 32,36mm. ng knh dy qun s cp: d= = = 1,84mm ng knh dy qun th cp: d = = = 6,42mm Theo sch in t cng sut ta chn dy tit din trn nh sau: d = 1,81 mm ; 22,9 gam/m d = 7 mm ; 220gam/m

25

CHNG 3. TNH TON THIT K CHO MCH IU KHIN3.1.Yu cu v nguyn tc iu khin 3.1.1.Mc ch v yu cu -Mun Tiristor m cho dng chy qua th phi c in p dng t trn anot v phi c xung p dng t nn cc iu khin. Sau khi Tiristir m yhif xung iu khin khng cn tc dng, lc ny dng in chy qua Tiristor do thng s mch ng lc quyt nh. Chc nng ca mch iu khin: -iu khin c v tr xung iu khin trong phm vi na chu k dng ca in p t nn anot-catot ca Tiristor. -To c cc xung c iu kin m c Tiristor, xung iu khin thng c bin t 2V n 10V, rng xung tx=20100s i vi thit b chnh lu. rng xung xc nh theo biu thc: tx= Trong : Idt : Dng duy tr ca Tiristor. : Tc tng trng ca dng ti. Mi quan h gia in p chnh lu vi vic thay i gc m Uc l=Uocos Trong : -Uc l : in p sau chnh lu. -Uo : in p chnh lu ln nht khi gc m =0 Cc yu cu i vi xung iu khin : -Pht xung iu khin chnh xc ng thi im do ngi thit k tnh ton. -Cc xung iu khin phi ln v bin v rng c th m c cc van -Cc xung iu khin phi c tnh i xng cao, m bo c phm vi iu chnh gc m. -C kh nng chng nhiu, tc ng nhanh. -m bo mch hot ng n nh v tin cy khi li in dao ng c v bin v tn s. Ngoi ra h thng iu khin phi c nhim v n nh dng in ti v bo v h thng khi xy ra s c qu ti hay ngn mch.

26

3.1.2.Nguyn tc iu khin Ngi ta thng dng hai nguyn tc iu khin thay i gc m ca cc Tiristor l: Nguyn tc iu khin thng ng tuyn tnh v nguyn tc iu khin thng ng arccos. a.Nguyn tc iu khin thng ng tuyn tnh. Theo nguyn tc ny ngi ta dng hai in p: -in p ng b, k hiu l Ur c dng rng ca, ng b vi din p t yteen anot-catot ca Tiristor. -in p iu khin, k hiu l Uc , l in p mt chiu, c th iu chnh c bin .

Hnh 3.1: Nguyn tc iu khin thng ng tuyn tnh Tng i s Ur + Uc c a n u vo ca mt khu so snh. Bng cch lm bin i Uc ta c th iu chnh c thi im xut hin xung ra, tc l thi im iu chnh gc m . Khi: Uc = 0 ta c = 0 Uc < 0 ta c > 0 Quan h gia v Uc c biu din qua cng thc sau: = Ngi ta thng ly Urmax = Ucmax . b.Nguyn tc iu khin thng ng arccos.27

Theo nguyn tc ny ngi ta cng dng hai in p: -in p iu khin Uc l in p mt chiu c th iu chnh c bin theo c hai hng ( m v dng). -in p ng b Ur vt trc in p anot-catot ca Tiristor mt g bng /2 (nu uAK=Asint th ur=Bsint).

Hnh 3.2: Nguyn tc iu khin thng ng ARCCOS Trn hnh v ng nt t l in p anot-catot ca Tiristor. T in p ny ngi ta to ra ur . Tng i s ur+uc c a ti u vo ca khu so snh. Khi ur+uc=0 th ta nhn c c mt xung u ra ca khu so snh. uc+Bcos=0 Do =arccos Ngi ta ly B = Ucmax Khi uc = 0 th = /2 Khi uc = UCmax th = Khi uc = -UCmax th = 0 Nh vy khi bin thin t -UCmax n +UCmax th bin thin t 0 n .28

Nguyn tc iu khin thng ng arccos c s dng trong cc thit b bin i i hi cht lng cao. 3.2.Cc linh kin in t s dng trong mch. - Ton b mch in phi dng 2 cng AND nn ta chn mt con IC 7415. Mi con IC7415 c 3 cng AND.

Hnh 3.3: S chn IC7415. -Ta dng mt con HCF4066 phc v cho vic chuyn mch np: Vcc

Gnd

Hinh 3.4: S chn IC HCF4066 *Thng s ca HCF4066: in p ngun nui : VDD = -0,518 (V) chn VDD = +12 (V) in p u vo: VIN = -0,5VDD+0,5(V) Nhit lm vic : T = -40 850 C Cng sut tiu th: P = 200 (mW) = 0,2 (W) -Ta s dng mt con IC 7404 c s chn nh sau:

29

Hnh 3.5: S chn IC7404 - Mch s dng 11 khuych i thut ton (OA1OA11) do vy chng ta cn 3 con IC TL084. mi con c s b tr chn nh hnh bn di:

Hnh 3.6: S chn ICTL084. *Thng s ca TL084 : in p ngun nui : Vcc = 18 (V) chn Vcc = 12 (V) Hiu in th gia hai u vo: 30 (V) Nhit lm vic : T = -25 850 C Cng sut tiu th: P = 680 (mW) = 0,68 (W) Tng tr u vo : Rin= 106 ( M ) Dng in u ra : Ira = 30 ( pA). Tc bin thin in p cho php : du/dt = 13 (V/ s).

30

3.3.S khi v chc nng 3.3.1.Khu ng pha.

D2

1

Ung

R2 D3

Hnh 3.7: S khi ng pha. Tn hiu ng b c th ly t bin p lc cng c th ly t mt bin p khc. Do trong mch iu khin c nhiu khu s dng ngun in p thp nn chng ta dng mt bin p c qun nhiu cun dy th cp, mi cun c mt chc nng ring bit, trong s dng cun c in p 0V-12V-24V dng cho khu ng b. Mch to xung ng b c ly t in p li U = 220V, f=50Hz, trng pha vi in p t nn cun s cp ca bin p ng lc. Hai it D2 v D3 lm nhim v chnh lu to ra tn hiu U1 lm ngng so snh vi tn hiu mt chiu. Ta chn 2 it D1 v D2 l it IN4004. 3.3.2 Khi to xung ng b. + E1

R3 _ OA2 R5 2 + R4 D11

U0

Hnh3.8: Khi to xung ng b.U Uo t Urss

t U2

t

31

in p U1 c so snh vi in p U0 to ra cc tn hiu tng ng vi thi im in p ngun i qua im khng. U0 cng nh th xung U2 cng hp v phm vi iu chnh cng ln la chn max = 1750 th U0 = 2U 2 sin 5 0 (4.1) T phng trnh 4.1 ta c U 0 = 2 .12. sin 5 0 = 2 .12.0,087 = 1,48V .0 Ta c R = R + R 4 3 4

U

E

1,48 12 = R4 R3 + R4

1,48( R3 + R4 ) = 12 R4 1,48 R3 = 10,52 R4

R3 = 7,1 R4

tn tht trn in tr nh chng ta chn R4 = 4,7K, R3 = 33K,R5= 2,2K 3.3.3.Khi to in p rng ca.

Hnh 3.9: Khi to xung rng ca. Nguyn l c bn ca khu ny l dng mch tch phn v kha in t T1, T1 l transitor ngc C828 (c thng s:30V 0,05A 0,25W). Khi U2=0,T1 kha, t C1 c np bi dng in:

-Ti thi im in p U2 chuyn t 10 t C1 phng ht in (UC1=0) v bt u c np in. Khi U2 chuyn t 01 transitor T1 thng, C1 bt u phng in cho ti khi in p U2 chuyn t 10. T C1 phng in trong sut rng ca xung. Khi U2 chuyn trng thi t 01 t C1 c np in tr li.

-in p t c hnh thnh do s np ca t C1, mt khc bo m in p t c trong mt na chu k in p li l tuyn tnh th hng s thi gian t np c Tr = R6. C1 = 0,005 (s)3 r Chn t C1 = 0,1 ( F) th in tr R6 = C = 0,1.10 6 = 50.10 1 Vy: R3 = 50 k . R7=10 k. 3.3.4.Khi phn hi dng in.

T

0,005

Hinh 3.10: Khi phn hi dng in. in p phn hi c ly trn in tr Rs ca mch lc. Tn hiu qua khuch i thut ton OA6 c lt trng thi, sau c cng vi tn hiu ch o ly trn chit p VR1 U4 = Ufh1 + Ucd1 (Ucd1 ly trn trit p VR1). Ban u khi cha ni ti vo mch, in p ca b chnh lu l Ud=U0, dng in Id=0. Khi ni ti vo mch, dng in s tng ln, do ni tr ca acqui nh nn dng in s tng ln rt ln lm gim tui th ca acqui. hn tr tc tng ca dng in chng ta s dng khu phn hi dng in lun lun n nh gi tr t. Khi bt u np dng in trong mch tng ln, lm cho in p ly trn in tr RS tng ln, in p U4 tng, qua khuch i thut ton OA7 tn hiu c lt trng thi, in p U5 tng, Udk tng. in p iu khin tng, lm tng gc m . Do in p trn mch lc gim xung, in p gim lm cho dng in gim xung bng gi tr t chnh l dng in np cho acqui. Ngc li khi dng in trong mch lc gim xung, th s lm cho in p iu khin gim, gc m tng ln, in p trn mch lc tng ln dn n dng in np tng ti gi tr t.

3.3.5.Khi phn hi in p.

Hnh 3.11: Khi phn hi in p. Tn hiu phn hi in p c ly trn in tr phn hi Rf . Khuch i thut ton OA8 ng vai tr l khu lp tn hiu, vi h s khuch i l .

Mch phn hi in p lm nhim v n nh in p khi dung lng acqui t ti 80% nh mc. Bin tr VR2 l bin tr ly in p ch o, Ucd np ln nht khi mi ngn acqui t ti 2,7V. UdkU = U7 = UphU - Ucd Ucd : in p ch o ly trn bin tr VR2. UphU : in p ti u ra ca khuch i thut ton OA8 (U6). Khi in p np tng ln ln hn gi tr in p t cho mi ngn acqui n l 2,7V lm cho Uf tng, UfhU tng, lm cho UdkU tng ln. in p iu khin tng lm cho gc m tng, do vy in p acqui gim xung bng gi tr t. 3.3.6.Khi chuyn mch np.

Hnh 3.12: Khi chuyn mch np.

Khi dung lng ca acqui t ti 80% gi tr nh mc mch s t ng chuyn t ch np dng in sang ch np bng in p. Bin tr VR3 l bin tr t gi tr ch o, tng ng vi in p trn mi ngn acqui l 2,4V. Khi in p cho mi ngn acqui di 2,4V, in p U6 nh hn in p ch o ly trn bin tr VR3, in p ti u ra ca khuch i thut ton OA10 (U8) m. Tn hiu ny kha ch np p v cho ch np dng hot ng. Khi in p trn mi ngn acqui t ti gi tr 2,7V , th in p U6 ln hn in p ly trn bin tr VR3, in p ti u ra ca khuch i thut ton OA10 chuyn trng thi (01). Tn hiu ny lm m ch np bng in p ng thi qua phn t o kha ch np bng dng in. 3.3.7.Khi to xung chm.

Hnh 3.13: Khi to xung chm. B OA11 l mt a hi dao ng to ra cc xung vung c tn s cao lp i lp li theo chu k, vi mc ch lm gim kch thc ca my bin p xung. T in C2 v bin tr VR to thnh mch tch phn. Mch R28 v R29 l mch phn hi. Nguyn l lm vic ca mch nh sau: gi s ti thi im 0 in p in p ra ca khuch i thut ton t gi tr cc i Ur +E. thng qua mch phn hi R28, R29 u vo + ca khuch i thut ton s c tn hiu phn hi duy tr khuch i thut ton nm ch bo

ha dng. Lc ny t C2 c np thng qua in tr R2. Khi t=tt , in p UC t gi tr U0 , khuch i thut ton lt trng thi v Ur = -Urmax -E. in p trn t C2 khng th thay i gi tr t ngt v lc ny t C2 li phng in qua

R1. thi im t = t2, khi U C = U 0 = R + R R28 , khuych i thut ton lt 28 29 trng thi Ur = Urmax +E v sau qu trnh li c lp li. Thi gian np ca t C2 l: Tnap = 1,1C2R2 . Thi gian phng ca t l: Tphong = 1,1C2R1 . Nu chn thi gian phng bng thi gian np th R1=R2 = VR/2 Mch to chm xung c tn s f= 1/2fx = 3 ( kHz) hay chu k ca xung chm T= 1/f = Tnap+Tphong = 333 ( s) Chn R1= R2 = VR/2 th T= 1,1 VR.C2 = 333 ( s) vy : VR. C2 = 302,73 ( s) Chn t C2 = 0,1 F c in p U = 16 (V) ; VR= 3027,3 ( ). thun tin cho vic iu chnh khi lp mch th ta chn VR l bin tr 3 K . 3.3.8. Khi khuych i xung v bin p xung.

E

Hnh3.14: Khi khuych i xung v bin p xung. Tnh BAX Theo phn tnh ton mch lc ta chn van Tiristor loi 25RIA102M. Van c cc thng s: Ug = 3 V Ig = 0,9A Gi tr ny l gi tr dng v p th cp my bin p. Chn vt liu st t 330, li st t c dng hnh ch lm trn mt phn t ca c tnh t ho B = 0,7 Tesla, H = 50 A/m, c khe h. + Chn t s ca my bin p: m = 3. + in p cun th cp BAX U2 = Uk = 3V + in p t ln cun s cp BAX : U1 = m.U2 = 3.3 = 9V

+ Dng in th cp BAX: I2 = Ik = 0,9 A + Dng in s cp BAX: I 0.9 I1 = 2 = = 0.3 m 3 + t thm ca li st t:3 tb = 0 H = 10 6.50 = 14.10 F + V mch c khe h nn phi tnh t thm trung bnh. S b chn: chiu di trung bnh ca ng sc l = 0,1mm, khe h lkh = 10-5m.

B

0,7

tb =

l l kh + 1

=

0,1 10 5 + 0,1 14.10 3

= 5,8.10 3

Th tch li st t:

V=

Trong : - tb : t thm trung bnh ca li st - 0 : t thm ca khng kh - tx : chiu di xung truyn qua BAX c gi tr t 10 600 s, y chn tx = 100 s - Sx : st bin xung ly Sx = 0,15 - U1 : in p s cp - I1 : dng in s cp Thay s vo ta c : 5,8.103.106.100.106 .0,15.9.0,3 V= =0.479.106 m3 2 0, 7 - Chn mch t c th tnh V = 1,4 cm3 vi th tch ta c cc kch thc mch t: a = 4,5 mm b = 6 mm d = 12 mm D = 21 mm Q = 0,27 cm2 = 27 mm2 Chiu di trung bnh mch t : l = 5,2 cm S vng qun dy s cp BAX: - Theo lut cm ng in t : U 1 .t x w1 = (vi k = 0,76 l h s cht y). B.Q.k

tb . 0 .t x .S.U 1 .I 1 B 2

9.100.106 W1 = = 63 0, 7.27.106 .0, 76

- S vng dy th cp : W 63 W2 = 2 = = 21 3 3 - Tit din dy qun th cpS1 = I1 J1 0,3 = 0, 05mm 2 6

- Chn mt dng in J1 = 6 A/mm2S1 =

ng knh dy qun s cp :d1 = 4S1 4.0, 05 = = 0, 25 mm

Chn dy dn c tit din trn ( Bng II.3 - TCS - Nguyn Bnh). Tit din S1 =0,04909(mm 2 ), ng knh d1 = 0,25(mm), trng lng 0,439gam/m, tr sut 0,366 ohm/m. Tit din dy qun th cp:S2 = I 2 0, 9 = = 0,18mm 2 J2 5

Chn mt dng in J2 = 5 A/mm2 - ng knh dy qun th cp:d2 = 4 S2 4.0,18 = 0, 4787 mm

Chn dy dn c tit din trn ( Bng II.3 - TCS Nguyn Bnh). Tit din S2 = 0,1886 (mm 2 ), ng knh d2 =0,49(mm), trng lng 1,68gam/m, tr sut 0,0914 ohm/m. - Kim tra h s lp y:kld = S1w1 + S 2 w 2 d12 w1 + d 22 w 2 0, 252.63 + 0, 492.21 = = = 0, 06 d2 d2 122 4

kl =0,06 < 1:nh vy ca s din tch cn thit. Tnh ton khu K cui cng T1, T2: chn transistor cng sut loi 2SC911 lm vic ch xung c cc thng s: + Transistor loi npn, vt liu bn dn l Si + in p gia collector v baz l khi h mch Emito : UCB0 = 40 V + in p gia Emito v Baz khi h mch Colecto : UEB0 = 4 V + Dng in ln nht Colecto c th chu ng c : ICmax = 500 mA + Cng sut tiu tn Colecto : PC = 1,7 W + Nhit ln nht mt tip gip T1 =1750 C. + H s khuych i = 50.

+ Dng in lm vic ca colecto IC=I1=50 mA. +Dng in lm vic ca Bazo IB =I C 3 50 = = 1(mA ) 50

Ta thy rng loi thyristor chn c : + in p iu khin Uk=3V + dng iu khin Ik= 0.9A. Ta c: I c = I1 = 0,3 A3 C Vy th: I B = = 50 = 6.10 ( A) = 6( mA)

I

0.3

Ta chn: R 32 = R 34 =R31 = R33 =

+ E U1 12 9 = = 10 I1 0.3

Uv 5 = = 694.44 k .I B 1, 2.6.10 3

Trong : k: L h s d tr, chn k=1,2 Uv : L in p ra ca 7415 chn l 5V Vy thun tin cho vic mua linh kin ta chon R31 v R33 l 700 Tt c cc it trong mch iu khin dng loi 1N4009 c cc tham s: - Dng in nh mc : Im = 10 (mA) - in p ngc ln nht : Ung = 25 (V) - in p cho Diot m thng : Um =1(V) 3.4. Mch iu khin -Xem hnh nh km-

3.4.1.Dng in pU U o Us rs 0 2 3

t

2

t

U 2

t U 3 Uk t U2 1

t U1 1

t U0 1

t U 9

t U4 1

t U3 1

t

Hnh 3.15: Dng in p ra ca mc iu khin. 3.4.2.Nguyn l hot ng ca s Tn hiu xoay chiu c chnh lu bi 2 dit D11, D12, s c so snh vi in p U0 to ra tn hiu ng b U2 trng vi thi im in p li i qua im 0. Tn hiu ng b ny s m kho in t bng thng Q1 gim in p trn t v 0, t C1 c np in theo cng thc UC = E.t/R7 v u ra ca khuych i thut ton OA2 s c tn hiu rng ca. Sau tn hiu ny c so snh vi tn hiu iu khin nh b so snh bng khuych i thut ton OA3. B OA11 l mt a hi dao ng xung c tn s cao U9 vi mc ch gim kch thc ca my bin p xung. Tn hiu cao tn trn ln vi tn hiu iu khin U12 cng cc tn hiu phn phi U10, U11 thnh tn hiu U14, U13. Nhng tn hiu ny c khuych i thng qua my bin p xung a trc tip cc iu khin ca Tiristo. 3.5.Khi ngun nui mch iu khin Bin p ngun nui v bin p ng pha dng chung cun s cp. Do ta s dng mt my bin p vi mt cun s cp v nhiu cun th cp, mi cun thc hin mt chc nng ring. Cun 0V-12V-24V s

dng lm cum ng pha vi tn hiu ngun, cun 0V-18V-36V s dng lm ngun nui mch iu khin. + 12V 7812C1 D1 D1 C3

D1

D1

C2

C4

-12V 7912 Hnh 3.16: Khi ngun nui mch iu khin. - Cc linh kin s dng trong mch: + Chnh lu cu 5A. + T lc ngun trc v sau n p C1 = C2 = C3 = C4 =220F/50V. + Vi mch n p 78L12, 79L12 l loi vi mch n p c cng sut nh. Dng in ti khng vt qu 100mA. Chng c bao gi di 2 dng: v st l hiu bng ch H, v bng cht do k hiu bng ch Z.

78L

79L

V CC R

CC V

R

Hnh 3.17:S b tr chn. Tnh ton my bin p ngun:

0 VU 2 1

1 2 VU 2 2

U n g

2 4 V 0 VU 2 3

1 8 VU 2 4

3 6 V

- Khi ngun 12 cp cho khuych i thut ton, I1 = 500mA. Cng sut ca ngun nui l: P1 =U1.I1 = 36.0,5 = 18W - Khi ngun ng pha 0V 12V 24V, I2 = 500mA. Cng sut ca ngun ng pha l: P2 =U2.I2 = 24.0,5 = 12W - Cng sut ca my bin p l: P = P1 + P2 =18 +12 = 30W - Dng in s cp my bin p l:I1 = P 30 = = 0,136 A U 1 220 k P = 1,2 30 = 0,22cm 2

- Tit din li thp mch t:S=

Ta chn li thp c tit din S = 0,92cm2, lm bng thp k thut in dy 0,2mm, gm cc l thp hnh ch v ch I ghp li vi nhau: Theo cng thc kinh nghim chng ta tnh s vng/vn: n0 =50 k (vi S

k = 4060 l h s ca my bin p, ly k = 50) n0 = 0,92 = 54 vng/vn. - S vng dy cun s cp l: W1 = n0.U1 = 54.220 = 11880 vng. - S vng dy cun th cp l: Cun 12V: W21 = W22 = n0.U = 54.12 = 648 vng. Cun 18V: W23 = W24 = n0.U = 54.18 = 972 vng. - Dng in trong cc cun th cp:I 21 = I 22 = I 23 = I 24 W1 W 11880 .I 1 = 1 .I 1 = .0,136 = 2,5 A W21 W22 648 W W 11880 = 1 .I 1 = 1 .I 1 = .0,136 = 1,66 A W23 W24 972

- Tit din dy qun:

I 1 1,36 = = 0,272mm 2 (chn J = 5A/mm2) J 5 I I 2,5 = 0,5mm 2 (chn J = 5A/mm2) + Cun 12V: S 21 = S 22 = 21 = 22 = J J 5 I I 1,66 = 0,33m 2 (chn J = 5A/mm2) + Cun 18V: S 23 = S 24 = 23 = 24 = J J 5

+ Cun s cp: S1 =

- ng knh dy th cp l:4.S1 4.0,272 = = 0,59mm 4.S 21 4.S 22 4.0,5 = = = 0,8mm . + Cun 12V: d 21 = d 22 = 4.S 23 4.S 24 4.0,33 = = = 0,65mm . + Cun 18V: d 23 = d 24 =

+ Cun s cp: d1 =

- Tra s tay|Thng s dy dn tit din trn (sch in t cng sut NXB Khoa hc k thut 1996, Nguyn Bnh), ta chn c dy: + Dy s cp: d1 = 0,59mm, S1 = 0,2734mm2, R=0,21 /m. + Dy s cp: d21 = d22 = 0,8mm, S1 = 0,5027mm2, R= 0,0342 /m. + Dy s cp: d23 = d24 = 0,67mm, S1 = 0,3526mm2, R=0,0488 /m. Tm tt thng s trn mch iu khin: T: in tr: Ring : C3 , C4 = 0,1F C1 , C2 = 0,22F Cc in tr mch iu khin sao cho ph hp vi dng vo ca R1 , R2 = 50 R7 = 50k R4 = 4,7k, R5 = 33k Chn cc bin tr VR1 VR4 = 2k, VR5 = 3k , Rs =Rf=Rf1=3k Transitor loi: 2SC911 it: D11 ,D12 chn loi 1N4004 Cc it cn li chn loi 1N4009 cc IC, nhng do cc thng s ny rt kh tra nn ta thng chn R =1050k