thiet ke bo nap acquy tu dong on dong va on ap 0726

n TCS

Thit k b np c quy t ng

TI 8 THIT K B NP C QUY T NG N DNG V N P bi:Thit k ngun np c quy . B ngun phi m bo hai ch np: np n nh dng in v np n in p . Khi c quy y phi ngt ngun np : Um = 24 -50 V Im = 60 A Imin = 40 A .

SV thc hin : Khoa Tun - TH1_K48

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n TCS

Thit k b np c quy t ng LI NI U

Nc ta hin nay ang trn con ng Cng nghip ha - Hin i ha. Bi vy t ng ha ang pht trin mnh trong nhng nm gn y.T ng ho iu khin cc qu trnh sn xut i su vo tng ng ngch, vo trong tt c cc qu trnh to ra sn phm. Ngy nay hu nh tt c cc my mc thit b trong cng nghip cng nh trong i sng hng u phi s dng in nng , c th l dng hon ton ngun nng lng in nng hoc mt phn nng lng in nng kt hp vi nng lng khc. Trn thc t c nhng lc rt cn nng lng in m ta khng th ly nng lng in t li in c. Do ta phi ly cc ngun in d tr nh c quy. Nh vy c th s dng c cc ngun cquy ta phi np in cho cquy. Bi b chnh lu np cquy t ng c s dng rng ri trong nhiu trng hp c th l rt quan trng , nu thiu n s khng c ngun in vn hnh , d tr cho cc my mc thit b m c th khng p ng c ch tiu kinh t k thut.


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n TCS

Thit k b np c quy t ng CHNG I GII THIU CHUNG V ACQUI

I. Tng qut chung cu to v nguyn l lm vic ca acqui: Acqui l ngun ho hot ng trn c s hai in cc c in th khc nhau, n cung cp dng in mt chiu cho cc thit b in trong cng nghip cng nh trong dn dng. Khi acqui phng ht dung lng ta tin hnh np in cho n v sau acqui li tip tc phng in c. Acqui c th thc hin nhiu chu k phng np nn ta c th s dng c lu di. Trong thc t k thut c nhiu loi acqui nhng ph bin v thng dng nht l hai loi acqui: acqui axit (acqui ch) v acqui kim. Tuy nhin trong thc t thng dng nht t trc ti nay vn l acqui axit v so vi acqui kim th acqui axt c mt vi tnh nng tt hn nh: + Sc in ng cao (vi cqui axit l 2V, cqui kim l 1,2V). + Trong qu trnh phng, s st p ca acqui axit nh hn so vi acqui kim. + Gi thnh ca acqui axit r hn so vi acqui kim. + in tr trong ca acqui axit nh hn so vi cqui kim. V vy trong n ny chng em chn loi acqui axit nghin cu cng ngh v thit k ngun np acqui t ng. 1. Cu to ca bnh acqui axit ( acqui ch ): Bnh acqui axit thng thng gm v bnh cc bn cc, cc tm ngn v dung dch in phn. 1.1. V bnh: V bnh acqui axit hin nay c ch to bng nha bnit hoc anphantpc hay cao su 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


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n TCS


polyclovinyl lp lt ny dy khong 0,6 mm. Nh lp lt ny m tui th ca bnh acqui tng ln t 2 3 ln. Pha trong v bnh tu theo in p danh nh ca acqui 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 acqui n, trong n ny, nhim v nghin cu l acqui ch vi in p danh nh l 12V nn ta c su ngn acqui 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 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 d dng hn. 1.2. 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) - antimon (Sb) vi t l (87 95)% Pb - (5 13)% Sb. Ph gia antimon thm vo c tc dng tng cng, gim han g v ci thin tnh c cho ct. Ct gia cht tc dng v phn khi dng in khp b mt bn cc. iu ny c ngha rt quan trng i vi cc bn cc dng v in tr ca cht tc dng (xit ch PbO2) ln hn rt nhiu so vi in tr ca ch nguyn cht, do cng tng chiu dy ca ct th in tr trong ca acqui s 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 ln cc sng y bnh acqui. V in ct ca bn cc m khng phi l yu t quyt nh v li 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 ca axit hu c i vi bn cc m, cn i vi cc bn cc dng th cht tc dng c ch to t cc xit ch Pb3O4, PbO v dung dch axit


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n TCS


sunfuric c. Ph gia 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 ho hc cng c tng ln. Cc bn sau khi c trt 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 vi tr s nh. Sau qu trnh nh vy cht tc dng cc bn cc dng hon ton tr thnh PbO2 (mu gch sm). Sau cc bn cc dng c em ra, sy kh v lp rp. Nhng phn khi bn cc cng tn trong mt acqui 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 ca cqui th ta phi tng s tm bn cc mc song song trong mt acqui n. Thng ngi ta ly t 5 8 tm. Cn mun tng in p danh nh ca acqui th ta phi tng s tm bn cc mc ni tip. 1.3. 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 acqui. Cc tm ngn y phi l cht cch in tt, bn, do, chu c axit v c xp thch hp d khng ngn cn cht in phn thm n cc bn cc. Cc tm ngn hin nay c ch to t vt liu polyvinyl xp, mn, dy khong t 0,8 1,2 mm v c dng mt phng hng v pha bn cc m cn mt mt c hnh sng hoc g hng v pha bn cc dng nhm to iu kn cho dung dch in phn d lun chuyn hn n cc bn cc dng v dung dch lu thng tt hn.


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n TCS 1.4. Dung dch in phn:


Dung dch in phn trong bnh acqui l loi dung dch axit sunfuric (H2SO4) c pha ch t axit nguyn cht vi nc ct theo nng qui nh tu thuc vo iu kin kh hu ma v vt liu lm tm ngn. Nng dung dch axit sunfuric = (1,1 1,3) g/cm3. Nng dung dch in phn c nh hng ln n sc in ng ca acqui. Hnh di trnh by nh hng ca dung dch in phn ti in tr v sc in ng ca acqui: V/ngn /cm3 5Eaq

2.5

4 1.5 1.0 0.5 0.0 3 2 1 0 1.0 1.1 1.2 1.3 1.4 1.5 1.6in tr dung dch in phn

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/cm3. Vi cc nc lnh (vng cc), nng dung dch in phn cho php ti 1,3 g/cm3. Trong iu kin kh hu nc ta th ma h nn chn nng dung dch khong (1,25 1,26) g/cm3, ma ng ta nn chn nng khong 1,27 g/cm3. Cn nh rng: nng qu cao s lm chng hng tm ngn, chng hng bn cc, d b sunfat ho trong cc bn cc nn tui th ca acqui cng gim i rt nhanh. Nng qu thp th in dung v in p


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n TCS


nh mc ca acqui gim v cc nc x lnh th dung dch vo ma ng d b ng bng. * Nhng ch khi pha ch dung dch in phn cho cqui: - 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 n ca acqui. - Cc dng c pha ch phi lm bng thu 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 axt c m phi t t axit vo nc v dng que thu tinh khuy u. 1.5. Np, nt v cu ni: Np lm bng nha bnit hoc bng baklit. Np c hai loi: - Tng np ring cho mi ngn - Np chung cho c bnh - loi ny kt cu phc tp nhng kn tt. Trn np c l dung dch in phn vo cc ngn v kim tra mc dung dich in phn, nhit d v nng dung dch trong acqui. 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 acqui. Np mt s loi acqui c l thng kh ring nm st l , kt cu nh vy rt thun tin cho vic iu chnh mc dung dch trong bnh acqui. Trong trng hp ny, nt khng c l thng kh na. Cu ni thng lm bng ch, dng ni cc ngn acquy n vi nhau. 2. Qu trnh bin i ho hc trong acqui axit: Trong acqui thng xy ra hai qu trnh ho hc thun nghch m c trng l qu trnh np v phng in. Khi np in, nh ngun in np m mch ngoi cc in t "e" chuyn ng t cc bn cc dng n cc bn cc m - l dng in np In. SV thc hin : Khoa Tun - TH1_K48 Trang 7

n TCS


Khi phng in, di tc dng ca sut in ng ring cu ca acqui, cc in t "e" s chuyn ng theo hng ngc li v to thnh dng in phng Ip. Khi acqui 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 hai bn cc u tr thnh sunfat ch PbSO4 c dng tinh th nh. Khi np in cho acqui s xy ra phn ng: - cc dng: PbSO4 2e + 2H2O = PbO2 + H2SO4 + 2H+ - cc m: PbSO4 + 2e + 2H+ = Pb + H2SO4 -Ton b qu trnh xy ra trong acqui khi np in l: 2PbSO4 + 2H2O = Pb + PbO2 + 2 H2SO4 (2.3) Kt qu l to thnh mt in cc Pb v mt in cc PbO2. S phng in ca acqui xy ra khi ni hai in cc Pb v PbO2 va thu c vi ti, lc ny ho nng c d tr trong acqui s chuyn thnh in nng. cc in cc s xy ra cc phn ng ngc ca (2.1) v (2.2), ngha l trong acqui s xy ra phn ng ngc ca (2.3). Acqui s cung cp dng in cho n khi c hai in cc li tr thnh PbSO4 nh ban u. Sau , nu mun dng tip ngi ta li np in cho acqui v c th qu trnh tip din. 3. Cc c tnh ca acqui axit : Mi ngn ca bnh acqui l mt acqui n c y cc tnh cht c trng cho c bnh. S d ngi ta ni tip nhiu ngn li thnh bnh acqui l tng in p nh mc ca bnh acqui. Do khi ngin cu c tnh ca bnh acqui ta ch cn kho st mt bnh acqui n l . 3.1. Sc in ng ca acqui axit: (2.2) ( 2.1)


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n TCS


* Sc in ng ca acqui axit ph thuc ch yu vo in th trn cc cc, tc l ph thuc vo c tnh l ho ca vt liu lm cc bn cc v dung dch in phn m khng ph thuc vo kch thc ca cc bn cc. Sc in ng ph thuc vo nng ca dung dch in phn v c th xc nh c mt cch kh chnh xc bng cng thc thc nghim sau: E0 = 0,85 + Trong : E0: Sc in ng tnh ca acqui n, tnh bng vol. : nng dung dch in phn khng ly theo n v g/cm m3

(V).

tnh bng vol quy v +150C. Ngoi ra sc in ng cn ph thuc vo nhit ca dung dch in phn na. * Trong qu trnh phng in, sc in ng ca acqui c tnh theo cng thc: Ep = Up + Ip. raq Trong : Ip : Dng in phng (A) Up: in p o trn cc cc ca acqui khi phng in (A) raq: in tr trong ca acqui khi phng in. Khi phng in hon ton th raq = 0,02 . * Trong qu trnh np in, sc in ng En ca acqui c tnh theo cng thc: En = Un In.raq (V). Trong : In : Dng in np (A). Un: in p o trn cc cc ca cqui khi np in (V). raq : in tr trong ca acqui khi np in. Khi np no th raq = (0,0015 0,001) .


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n TCS 3.2.Dung lng ca acqui:


* Dung lng phng ca phng ca acqui l i lng nh gi kh nng cung cp nng lng ca acqui cho ph ti v c tnh theo cng thc : Cp = Ip.tp Trong : Cp: L dung lng thu c trong qu trnh phng in (Ah). Ip: Dng dn phng n nh (A) trong thi gian phng in tp(h). * Dung lng np ca acqui l i lng nh gi kh nng tch tr nng lng ca acqui v c tnh theo cng thc: Cn = In.tn (Ah). Trong : Cn: L dung lng thu c trong qu trnh phng in (Ah). In: Dng in np n nh trong qu trnh np in (A). 3.3. c tnh phng ca acqui axit: c tnh phng ca acqui l th biu din mi quan h ph thuc ca sc in ng, in p acqui v nng dung dch in phn theo thi gian phng khi dng in phng khng thay i. I (A) 2,0 10 1,5 1,0 0,5 Cp=Ip.tp tgh tp (h) 2,11 Eaq 1,27 5 Eo U p E Ip.raq 1,11 U,E (V) Khong ngh 1,95 1,75 (g/cm3) (Ah).

Vng phng cho php


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n TCS


T th ta c cc nhn xt sau: Trong khong thi gian phng t tp =0 cho ti thi im tp = tgh, sc in ng, in p v nng dung dch in phn gim dn, tuy nhin trong khong thi gian ny dc ca cc th l khng ln, ta gi l giai on phng n nh hay thi gian phng in cho php tng ng vi mi ch phng in (dng in phng) ca acqui. T thi im tgh tr i, dc ca th thay i t ngt nu ta tip tc cho acqui phng in sau tgh th sc in ng, in p ca acqui s gim rt nhanh, mt khc cc tinh th sunfat ch (PbSO4) to thnh trong phn ng s c dng th, rn, kh ho tan (bin i ho hc) trong qu trnh np in tr li cho acqui sau ny. Thi im tgh gi l gii hn phng in cho php ca acqui, cc gi tr Ep, Up, ti tgh gi l cc gi tr gii hn phng in cho php ca acqui. Sau khi ngt mch phng mt khong thi gian, cc gi tr sc in ng, in p ca acqui, nng ca dung dch in phn li tng ln, ta gi l thi gian hi phc hay khong ngh ca acqui. thi gian phc hi ny ph thuc vo ch phng in ca cqui (dng in phng v thi gian phng ). nh gi kh nng cung cp in ca cc acqui c cng in p danh ngha, ngi ta quy nh so snh dung lng phng in thu c ca cc acqui khi tin hnh th nghim ch phng in cho php l 20h (10h). Dung lng phng trong trng hp ny c k hiu l C20 (C10). 3.4. c tnh np ca acqui: c tnh np ca cqui l th biu din quan h ph thuc ca sc in ng, in p cqui v nng dung dch in phn theo thi gian np khi tr s dng in np khng thay i.


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n TCS


I (A)

U,E (V)

Bt u si 2,4V

2,7V 2,11V

2,0 1,95 Eaq Eo Un In.raq E 1,5 10 5 1,0 0,5 Cn=In.tn 0 T th c tnh np ta c nhn xt sau: dung dch in phn tng dn ln. - Ti thi im tn = ts trn b mt cc bn cc xut hin cc bt kh do dng in in phn nc thnh xy v hyr (cn gi l hin tng si ), lc ny trn in th gia cc cc ca acqui n tng ti gi tr 2,4 V. Nu ta vn tip tc np gi tr ny nhanh chng tng ti 2,7 V v gi nguyn. Thi gian np ny gi l thi gian np no, c tc dng lm cho cc phn cht tc dng su trong lng cc bn cc c bin i hon ton, nh s lm tng thm dung lng phng in ca acqui. Trong s dng, thi gian np no cho acqui thng ko di t 23 gi, trong sut thi gian , hiu in th trn cc cc ca acqui v nng dung dch in phn l khng thay i. Nh vy dung lng thu c khi acqui phng in lun nh hn dung lng cn thit np no acqui. Sau khi ngt mch np, in p, sc in ng ca acqui, nng dung dch in phn gim xung v n nh. Thi gian ny cng gi l khong ngh ca acqui sau khi np. ts 1,11 (g/cm3) 1,27 Vng np no Khong ngh (23)h tn (h)

Vng np hiu dng

- Trong khong thi gian np t tn = 0 n tn = ts, sc in ng, in p, nng


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n TCS


Tr s dng in np nh hng rt ln n cht lng v tui th ca acqui. Dng in np nh mc i vi acqui qui nh bng 0,05.C20 (0,01.C10). II. Cc phng php np in cho acqui: 1. Phng php np acqui vi dng np khng i : Phng php np in vi dng np khng i cho php chon dng in np thch hp i vi tng loi acqui, m bo cho acqui c np no. y l phng php s dng trong cc xng bo dng, sa cha np in cho cc acqui mi hoc np in cho cc acqui b sunfat ho. Vi phng php np ny cc acqui c mc ni tip vi nhau v phi tho mn iu kin: Un 2,7 Naq. Trong : Un: in p np (V). Naq: S ngn acqui n mc trong mch np . Trong qu trnh np, sc in ng ca acqui tng dn, 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= Un 2,0 Naq . In

Nhc im ca phng php np vi dng np khng i l thi gian np ko di v yu cu cc acqui a vo np phi c cng c 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 np hai nc th dng n np nc th nht chn bng (0,3 0,5).C20, v kt thc np nc mt khi acqui bt u si. Dng in np nc th hai bng 0,05.C20. 2. Phng php np acqui vi in p np khng thay i:


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n TCS


Phng php np acqui vi in p np khng thay i yu cu cc acqui c mc song song vi ngun np. Hiu in th ca ngun np khng thay i v c tnh bng t 2,3 2,5 V cho mt ngn acqui n. Hiu in th ca ngun np phi c gi n nh vi chnh xc n 3% v c theo di bng vol k. Dng np In = gim i kh nhanh. Phng php np vi in p np khng thay i c thi gian np ngn, dng in np t ng gim dn theo thi gian. Tuy nhin dng phng php ny acqui khng c np no, v vy phng php np vi in p khng i ch l phng php np b xung cho acqui trong qu trnh s dng. khc phc nhng nhc im v tn dng c ht nhng u in ca cc phng php np trn, ta kt hp hai phng php np li thnh phng php dng - p. y cng chnh l phng php np m chng ta chn thit k mch iu khin cho ngun np acqui t ng trong n ny. 3. Phng php np dng - p: Ban u ta np acqui vi dng np khng i vi tr s qui nh l In = 0,05.C20. Ti khi thy acqui "si" - ng vi thi im hiu in th gia cc cc ca ca cqui n tng ti gi tr 2,4V - tip tc np th gi tr ny nhanh chng tng ti gi tr l 2,7 V. n y ta chuyn sang ch np n p vi gi tr in p np khng i l Un = 2,7V. Giai on np n p ko di t 2 n 3 gi, hoc khi dng np tin ti khng (In = 0) th kt thc qu trnh np. Kt lun: Qua phn tch k nhng c tnh ca acqui, c bit l c tnh np, ta chn phng php np dng - p np cho acqui. Nh vy b ngun np acqui t ng m ta thit k cn phi p ng nhng yu cu sau: - Ban u t ng np n dng vi dng np t trc In = 0,05 .C20/1ngn cqui n. SV thc hin : Khoa Tun - TH1_K48 Trang 14Un Eaq lc u s rt ln sau khi Eaq tng dn ln th In Raq

n TCS


- Khi pht hin thy hiu in th trn cc cc ca acqui n tng ti 2,7 V th t ng chuyn t np n dng sang ch np n p vi in p np t trc Un = 2,7V/ 1 ngn acqui n. - Np n p cho ti khi dng in np tin v khng.


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n TCS

Thit k b np c quy t ng CHNG II PHNG N CHNH LU

I. Nhn xt chung: B chnh lu l thit b dng bin i ngun in xoay chiu thnh ngun in mt chiu nhm cung cp cho ph ti in mt chiu. Trong k thut c nhiu phng n chnh lu nh: chnh lu khng iu khin (chnh lu it); chnh lu iu khin (chnh lu tiristor); chnh lu mt pha; ba pha; su pha. Tu thuc vo yu cu c th m ta la chn phng n chnh lu thch hp nht nhm p ng c cc ch tiu v mt k thut v kinh t. II.Yu cu c th : Trong n ny ,vi yu cu c th l: thit k b ngun np c quy c th np cho c quy 24-50V v dng np 40- 60A. - V yu cu ca dng chnh lu iu khin nn ta chn phng n chnh lu tiristor. - V ti yu cu cng sut v cht lng in p iu chnh khng cao nn ta chn phng n chnh lu mt pha nhm lm gim gi thnh u t thit b v n gin ho vic thit k tnh ton. T nhng nhn xt trn ta cn phn tch cc s chnh lu iu khin mt pha tm ra phng n thch hp nht. III.Cc phng n thit k mch chnh lu : 1. Chnh lu mt pha 2 na chu k c iu khin: Trong s ny ,my bin p fi c hai cun dy th cp vi thng s ging ht nhau , mi na chu k khi c xung ti iu khin m tiristo c mt van dn cho dng in chy qua . in p p mch trong c hai na chu k vi tn s p mch bng hai ln tn s in p xoay chiu . Hnh dng cc ng cong in p v dng in ti (Ud,Id ) cho trn hnh v .


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n TCS


Trong na chu k u , khi U2>E th in p anot T1 dng, in p Katot T1 m, T1 sn sng dn.Nu cp xung iu khin cho T1 vo lc ny th T1 s dn.Dng s chy qua T1-R-E, vi ngun l U2


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n TCS


Trong na chu k sau, khi U '2 > E th in p anot T2 dng, in p Katot ca T2 m, T2 sn sng dn.Nu cp xung iu khin cho T2 vo lc ny th T2 s dn.Dng s chy qua T2-R-E, vi ngun l U '2 Ch : Nu ta cp xung vo thi im U u- th in p ra Ura=Ubh Khi uA< u- th in p ra Ura=-Ubh Kt qu ta c chui xung ch nht khng i xng.

UA

Uref

UB

2

1 Trang 36


n TCS


b)Tnh ton : in p sau khi t u ra ca bin p ng pha qua it 1,2 c dng in p mt chiu na hnh sin . chn in p xoay chiu ng pha UA=9(V) in tr R2,R1 c dng hn ch dng vo KTT. Thng chn R2,R1 sao cho dng vo KTT nh hn 1(mA) do : R2> Chn R2=R1=10(K ) Chn gc duy tr v kho nng lng l 5o th in p t vo ca dng ca b so snh l: Ud= 2 Usin5o= 2 *12*sin5o=1.48(V) Ta c :E R = 1.48 VR + R 2 3 UA 9 = 3 = 9000() I v 10

Do ta c: VR+R1=90(K ) Chn R1=10(K ) , VR=100(K ) Chn Khuch i thut ton l loi TL084 c: Ngun cung cp Vcc= 12V Nhit lm vic : t=-25 850C Cng sut tiu th: P=680 mW Tng tr u vo : Rin=106 M Dng in ra : Ira=30pA


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n TCS 2. Khu to in p rng ca: a) S v nguyn l :DZ


C1

D3

R4 + -

Urc OA2

VR2

R5

in p ca b pht xung ch nht c a vo ca o ca khu to in p rng ca. Khi Udp0 th 3 kho nn dng qua 3 bng 0 lc ny dng qua t C1 bng dng qua in tr R4 , dng ny ngc chiu vi dng qua t C1 na trc ngha l t C1 phng in do in p trn t C1 cng nh in p ra gim tuyn tnh. Khi in p gim n khng ri m th t DZ1 dn theo ch nh it bnh thng gi cho in p gi tr 0. b) Tnh ton : Khi Udp0 (Udp=+Ubh) th 3 kho , t C phng in Dng phng in : Ip=E R 4 + VR

in p trn t C gim dn theo thi gian: uc(t)=UC(0)+I R4 1 E ic dt = U OA C t = U OA (R4 + VR )C t C

gi tp l thi gian phng ca t in . Ta chn tp=9(ms) Chn R4 , VR sao cho t phng v 0 V trong 9 (ms)


Trang 39

n TCS ta c : 0 = UOAE t (R4 + VR )C p


tp =

(R4 + VR )CU OA (R4 + VR )0.22 * 10 6 * 6.2E = 12

= 9 * 10 3

suy ra

R4+VR=88000( ) hay R4+VR=88(K )

Chn R4=10(K ) , VR=100 (K ). 3. Khu so snh: a) S v nguyn l: y l khu dng xc nh thi im m tiristor. Ta so snh in p ta v in p iu khin im cn bng ca hai in p ny l thi im m tiristor. so snh hai tn hiu tng t ngi ta c th dng KTT hoc dng transistor nhng trong thc t ngi ta thng dng KTT do cc u im sau : - Tng tr vo ca Opam rt ln nn khng gy nh hng n in p a vo so snh , n c th tch bit hon ton chng khng gy tc ng sang nhau . - Tng vo ca Opam thng l loi khuych i vi sai , mt khc c nhiu tng nn h s khuych i rt ln. V th chnh xc so snh rt cao , tr khng qu vi micro giy. - Sn xung dc ng nu so vi tn s 50 Hz. Thc t khi chnh lch gia Urcv Udk ch khong vi milivn th in p u ra ca n thay i hon ton t trng thi bo ho m sang trng thi bo ho dng v ngc li. Vi nhng u im ta dng KTT so snh, ta dng khu so snh kiu hai ca, s nh hnh v:R6 R7 + OA3

Urc Udk

Uss


Trang 40

n TCS


Khi Udk>Urc th in p ra ca khu so snh l Ura=+Ubh Khi Udk 2U/Imax =2,6k ; ta chn R10 = 3k. Chn to xung kim vi tx = 90.10-6 s nn R10.C =tx/3 = 30.10-6 s Suy ra ta chn C = 10nF Khu tch xung: Chn KTT l loi TL084, cng AND l loi IC 4081 c 4 cng AND trong mt v v c cc thng s: Ngun nui: Vcc=3 15 (V) .Chn Vcc=12(V) Nhit lm vic :-40 80 oC SV thc hin : Khoa Tun - TH1_K48 Trang 44

n TCS


in p ng vi mc logic cao :2 4.5(V) ,dng 1 (mA) Cng sut tiu th :P=2.5 (nW\1cng) 5. Tnh ton bin p xung : * Bin p xung thng phi lm vic vi tn s cao nn li thp cho tn s li in 50Hz khng p ng c , Chn vt liu lm li l st Ferit HM. Li c dng hnh xuyn, lm vic trn mt phn ca c tnh t ho c: B = 0,3 (T), H = 30 ( A/m ) khng c khe h khng kh. * Tnh th tch li thp cn c :V = Q.l =

tb . 0 .t x .s x .U .I 2B 2

Trong : tb - t thm trung bnh tb=B 0 .H

o = 1,25 . 10-6 (H/m); Q - tit din li st; l - chiu di trung bnh ng sc t; tx- rng mt xung ,(s) sx- st p xung cho php , thng ly bng 0,10,2 vi tx= 90 s + T s bin p xung : thng m = 23, chn m= 2 + in p cun th cp my bin p xung: U2 = Udk =5V + in p t ln cun s cp my bin p xung: U1 = m. U2 = 2.4 = 8 (V) + Dng in th cp bin p xung: I2 = Idk =0,15 (A) + Dng in s cp bin p xung: SV thc hin : Khoa Tun - TH1_K48 Trang 45

n TCS


I1 = I2 /m =0,15/2=0,075(A) + t thm trung bnh tng i ca li st: tb =B/0 . H = trong : 0=1,25.10-6 (H/ m) l t thm ca khng kh Th tch ca li thp ca li thp cn c: V= Q.l = (tb . 0 . tx . sx . Ul . Il )/ B2 Thay s V=8.10 3.1,25.10 6.90.10 6.0,1.8.0,075 = 0,6.10 6 m 3 = 0,6cm 3 0,3 23

0,3 = 8.10 3 (H/m) 6 1,25.10 .30

Chn li hnh tr k hiu 1811 c V=1,12 cm tch ta c kch thc mch t nh sau:

, ng knh ngoI

18mm , ng knh trong 11 mm, tit din li tng ng 0,443 cm 2 ,vi th

a = 3,5 mm Q = 0,443 cm2 = 44,3 mm2 d = 11 mm D = 18 mm + S vng qun dy s cp bin p xung:


Trang 46

n TCS w1 = U1 tx / B.Q =

Thit k b np c quy t ng8.90.10 6 = 54 ( vng ) 0,3.0,443.10 4

+ S vng dy th cp : w2 = w1 / m = 54/2 = 27 (vng ) Chn mt dng in : j1 =6 ( A/mm2 ) , j2 = 4 (A/mm2) + Tit din dy qun th cp: s1 = I1 /J1 = 0,075 /6 = 0,0125 (mm2 ). + ng knh dy qun s cp : d1 =4s1

= 0,13 (mm)

+ Tit din dy qun th cp: s2 = I2 / J2 = 0,15/4 = 0,0375 (mm2 ). + ng knh dy qun th cp: d2 =4s 2

= 0,22(mm).

6.Ngun cung cp cho mch iu khin : Mch iu khin trn i hi ngun cung cp l in p mt chiu , tr s n p v n nh tu thuc vo tng khu trong mch .Cn thit k cc loi ngun sau : - Ngun khng i hi n nh cao s dng mch chnh lu ch lc bng t in v khng cn n p cung cp cho khu ng pha , khu khuch i cng sut . - Ngun mt chiu n p dng IC n p cp ngun cho cc vi mch nh khuch i thut ton , IC logic . a) Ngun nui n p dng IC n p 7812 ,IC7912: Hu ht cc thit b u dng ngun mt chiu. Ngun mt chiu ny c to ra bng cch bn i in p li 220V xoay chiu sau n nh in p mt chiu ny v cung cp cho cc thit b in t .


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n TCS


Ngun n p l ngun lun n nh in p ra khi thay i in p vo hoc thay i ti . S khi ca b ngun mt chiu n p:

U

H p cch ly

U

Chnh lu

U

Mch lc

UV

Mch n nh in

Ur

Cc phn t thc hin khi chc nng: - Khi h p v cch ly dng my bin p thc hin. - Khi chnh lu dng it ( hoc cu chnh lu ) thc hin. - Mch lc dng t in ( t ho ) c in dung ln thc hin . - Mch n nh in p dng IC chuyn dng thc hin. IC n p chuyn dng c gi thnh r v tham s tt nn phn ln ngun n p dng cho mch iu khin dng IC n p ch to sn, trong IC n p 78xx l thng dng nht hin nay. IC ny c ch to cng nghip vi cc cp in p ra chun v c th hin bng hai s xx. Dng ti cho php IC ny l 1A( khi c tn nhit tt). S n p dng IC n p

Tnh chn cc phn t trn s : - UA 7812 c in p u vo : 7 35V Trang 48


n TCS

Thit k b np c quy t ng Dng in u ra :0 1A in p ra E=12V

UA 7912 c in p u vo : 7 35V Dng in u ra : 0 1A in p ra E=-12V - Chn t lc phng C3=C5=1000F, C3=C5=100 F Chn t lc nhiu C4=C6=0,1F . - Chn cc cu chnh lu c I=1A; U=50V(khng c tn nhit) b) Tnh chn my bin p cp cho ngun nui n p v cc linh kin in t trong mch iu khin: Chn my bin p mt pha c mt cun s cp v nhiu cun th cp + Hai cun chung 0V-6V-12V to in p ng pha . + Hai cun th cp ring dng cho ngun nui n p . Hai chnh lu cu mt pha to in p ngun nui i xng cho IC . in p u vo ca IC n p chn 20V. in p th cp cc cun dy ny l 20/ 2 =14,18V Chn in p ca hai cun th cp ny l 14V + Mt cun th cp to ngun nui cho bin p xung ,cp xung iu khin cho cc tiristor(+12V). Mi khi pht xung iu khin cng sut xung ng k , nn cn ch to cun dy ny ring r vi cun dy cp ngun IC , trnh gy st p ngun nui IC in p pha th cp cun dy ngun nui bin p xung l 12/ 2 =8,485V chn 9V * Tnh ton my bin p:


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n TCS + in p li: U1 =220V. + Cng sut cun dy ng pha:


- in p ly ra mi cun ng pha l 9V - Dng in chy qua cc cun dy ng pha l 1A cng sut Pp=2.9.1=18 (W) + Cng sut tiu th 8 IC TL084 v 2 cng AND l PIC=8.0,68 +2.2,5.10-9=5,44 (W) + Cng sut bin p xung cung cp cho cc iu khin Tiristor PT= 2.Udk.Idk=2.4.0,15=1,2(W) + Cng sut s dng cho vic to ngun nui PN=Pdp + PIC + PT =18+5,44 +1,2 = 24,64(W) - H s cng sut my bin p = 0,7, ta c cng sut my bin p l: Sba = PN/ . Sba = 24,64/ 0,7 = 35,2 (VA). - Chn my bin p mt pha mt tr c li st lm bng tn silic dp hnh ch E,I dy 0,35 mm ghp li. Khi tit din li st c tnh bi: S = 1,2. Sba = 1,2. 35,2 = 7,12 (cm2), ta chn S = 8(cm2). - H s dy qun: N0 = (40 60)/ S = (40 60)/8 = (5 7,5) (vng/ vol) Ta chn N0 = 6 ( vng / vol). S vng dy qun s cp: W1 = 6.220 = 1320 (vng ) S vng dy qun th cp: W2 = N0.U2 2 cun cho ngun : Wmn = 6.14 = 84(vng) 2 cun uv,rs : Wuv = Wrs = 4.10 = 40 (vng). Cun 0V 9V 18V: Wa = Wa' = 6.9 = 54 (vng)


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n TCS


- Dng in trong cun dy s cp my bin p: I1=Sba/U1=35,2/220=0,16(A) - Tit din dy: Ta chn mt dng in J =3 A/ mm2, ta s c tit din cun dy: S cp: S1 = I1/ J = 0,16/ 3 = 0.053 (mm2) . - ng knh dy qun s cp: d1 =4 s1

=

4.0,053 = 0,26 (mm). 3,14

ng knh cc cun th cp ta chn bng 0,26 mm . S nguyn l: A Tr.t

m 7.Khu phn hi:

n

u

v

r

s

a

GND a'

a) S nguyn l:


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n TCS


b) Nguyn tc hot ng: Cc tn hiu phn hi dng UphI v p UphU c ly t mch lc ri a v cc khu phn hi to ra Uk iu khin gc m nhm n nh cc gi tr dng hoc p t trc theo nguyn tc: I I Un Un UphI UphI UphU UphU Uk Uk Uk Uk Ucl Ucl Ucl Ucl I I . Un Un .

c) Tnh chn cc phn t trn s : Cc b khuch i thut ton ta s dng IC LM348. S ni cc chn nh hnh v. - Khu phn hi dng in: Theo nh trnh by trn, dng in phn hi c ly trn Rsun , ta chn Rsun loi 50A/60mV. in p ri trn Rsun ng vi gi tr dng Id = 60A l : UphI = 60.60 = 72 mV = 0.072 V. 50


Trang 52

n TCS


Ta cho tn hiu ny so snh vi in p trn trit p VR6, n c s dng iu chnh dng np. R15=32K ,VR6=1K, R15=1K. Tn hiu ra b so snh U1 ch c 3 trng thi l (+Ubh, 0, -Ubh) Ta cho tn hiu ny qua it D11,D12 v R16,C8 nh hnh v. Chn D11 v D12 c in p thun 1,5V, khi dng qua c it ny cn phi c in p ti thiu t ln it l 1,5V. Khi U1=Ubh th C8 c np,in p tng dn. Khi U1=U-bh th C8 c np , in p gim dn Khi U1=0 th t C8 khng c np nhng chng cng khng b phng v c D11 v D12 cn.(ta thit k in p ln nht trn C8 l 1V nn khng th dn qua it c d l phn cc thun) Ta c : UC8=Ubh.(1- et R16 C8 t R16 C8

) + UC8(0)

Ubh=10V, gi s ban u UC8(0)=0V1 e

=

1 = 0,1 10

t = 0,105 R16C8

tc p ng mt cch hp l th ta chn thi gian t=10s R16.C8=10 100 0,105

Chn C8=1000 F R16=100K Tip theo l b khuch i o: U2=-(U C 8 U VR 2 + ).R17 R19 R18

Ta nhn thy vi mch lc nh trn v ti l ngun E nn van m chc lc cm ti vo (I=0) th Udk=-10V Udk=U2=-UVR2.R17 =-10V R18


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n TCS


Chnh UVR2=1V, R17=20K ; R18=2K R19 = R18 =2K; iu chnh chit p VR6 ta s iu chnh c dng vo ti. - Khu phn hi in p: Ta ly UphU hai u ra ca mch chnh lu V mch in ta thit k dng np cho c quy t 24 n 50V nn trc khi phn hi ti mch iu khin ta cn gim p . Ta ly VR3 in p a vo mch n p. Ta chn R12=90K; c th thay i c in p np ta chnh trit p VR3 Vi chit p ny ta c th thay i in p vo b khuch i o , thay i c rng ta chn h s khuych i ca b khuch i o l 2. Chn R20 = R21 = R22 = 10K; R23 = 20K VR3 chn loi 10K -> Udk=-2UphU Thay i v tr ca chit p ta thay i in p np. - Khu chuyn mch: Ban u acqui c mc vo mch np th dng np tng v in p acqui tng dn ln, tc l dng phn hi v p phn hi tng dn ln. Lc ny do p phn hi nh hn UVR1 nn u ra ca thp, do chuyn mch CM2 ngt cc ng phn hi p ra khi mch. ng thi do c cng NO nn chuyn mch CM1 ng ng phn hi dng vi mch thc hin qu trnh n nh dng. Khi p phn hi UphU bng UVR1 th U3 o du do CM2 ng cn CM1 ngt nn mch thc hin qu trnh n p. Chn: VR1 =100K. Ta gn VR1 v VR3 cng 1 trc iu chnh, khi ta ch cn vn 1 nm iu chnh in p np th trc ny cng chnh lun gi tr in p chuyn mch tng ng vi in p np.


Trang 54

n TCS



Trang 55

n TCS

Thit k b np c quy t ng CC TI LIU THAM KHO

Ti liu in t cng sut in t cng sut Hng dn thit k mch in t cng sut Tnh ton thit k thit b in t cng sut Phn tch v gii mch in t cng sut K thut mch in t Cc ti liu v c quy .

Tc gi V Minh Chnh, Phm Quc Hi Trn Trng Minh Nguyn Bnh Phm Quc Hi Trn Vn Thnh Phm Quc Hi, Dng Vn Nghi Phm Minh H

H Ni , ngy

, thng

, nm

Sinh vin thc hin

Khoa Tun


Trang 56

n TCS


Mc lcChng I : Gii thiu chung v c quy Chng II : Phng n chnh lu Chng III : Thit k v tnh ton mch lc Chng IV : Mch iu khin 3 - 14 15 - 21 22 - 28 29 - 53


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thiet ke bo nap acquy tu dong on dong va on ap 0726

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