Thiet Ke Bo Nap Ac Quy Tu Dong

Chng 1

n tt nghip

LI NI U

Ngun in mt chiu c ng dng rng ri trong nhiu lnh vc, t ng dng trong sinh hot n ng dng trong quc phng, cng nng nghip, cc ngnh khoa hc k thut v trong sn xut cng nghip hin i.

Nhng nm gn y, khoa hc k thut pht trin nhanh cha tng thy, nht l lnh vc in t, tin hc, t ng ho, o lng iu khin, v.v cc thit b in t trong cc dy chuyn sn xut cng nghip, cc dng c o lng t ng, cc h thng iu khin vi chnh xc cao, cc loi my tnh in t vi tc tnh ton nhanh c ng dng trong cng nghip, trong o lng v iu khin, v.v hu ht u cn mt ngun cung cp in mt chiu hot ng. Chnh v vy i hi c mt ngun in mt chiu tng ng; p ng c nhng nhu cu mi, nh c in th, dng in ln, kch thc gn nh, lm vic tin cy, chnh xc, n nh cao, kh nng lm vic lu di, chu c cc iu kin mi trng lm vickhc nhau. .. cc ngun in mt chiu ny c th l cc loi pin, acqy, nhng ph bin hn vn l ngun in mt chiu t acquy. Do vy vic thit k mt b ngun cung cp l rt quan trng v thit thc. Trong c vic thit k mt b np acquy t ng v lin tc cng c rt ch trng.

Vn ln ti ny l ch np cho acquy nh th no? cng ngh np ra sao? c rt nhiu phng php in cho acquy

+ Np vi dng in khng i

+ Np vi in p khng i

+ Np va dng, va p

Trong n ny ta dng phng php np hai cp dng in bi v khi np i hi tc nhanh, thi gian ngn v in p c quy phi y v .

H Ni, ngy 30 thng 5 nm 2010

Sinh vin thc hin Trn nh Ca

Chng 1ACQUY V CNG NGH NP ACQUY1.1. Gii thiu v acquy:Acquy l loi ngun in ho hc, c th bin in nng thnh ho nng v ngc li bin ho nng thnh in nng. Qu trnh bin ho nng thnh in nng gi l qu trnh phng in v qu trnh bin in nng thnh ho nng gi l qu trnh np in .Acquy l ngun in mt chiu c s dng rt rng ri v lm vic da trn hin tng in - ho hc. Acquy sn xut ra phi bo m cc tnh nng v in theo quy nh .

+ Sc in ng ln v t thay i khi phng, np in .+ Acquy phi lm vic thun nghch, ngha l hiu sut nng lng gn 100%.

+ in tr trong nh.+ Dung lng cho mt n v trng lng v mt n v th tch phi ln . + T phng in .1.2. Phn loi acquy: Trong iu kin hin nay c rt nhiu loi acquy khc nhau c sn xut tu thuc vo nhng iu kin, yu cu c th ca tng loi my mc, dng c, iu kin lm vic, cng nh nhng tnh nng kinh t k thut ca acquy. C th lit k mt s loi sau :

- Acquy ch ( hay acquy axt)

- Acquy kim

- Acquy khng lamen v acquy kn

- Acquy km - bc v acquy catmi- bc

Tuy nhin trn thc t th acquy axit v acquy kim c s dng nhiu hn. nhng thng dng nht t trc n nay l acquy axt; v so vi acquy kim n c mt vi c tnh tt hn nh : sc in ng ca mi bn " cp bn " cc cao hn v c in tr trong nh . V vy trong n ny ta chn loi acquy a xt nghin cu v thit k .

1.3. Cu to ca acquy axt:1.3.1.V bnh:V thng lm bng nhng nguyn liu cch in nh nha, cao su cng ( bnt) c thnh hnh hp, c chia thnh nhiu ngn, chu c kh hu nng, lnh va chm mnh v chu c axt.

y ca mi ngn c bn sng khi bn cc to thnh khong trn gia y bnh v mt di ca khi bn cc, trnh c hin tng chp mch gia cc bn cc do cht kt ta ri xung y bnh gy nn.

1.3.2.Tm ngn:Tm ngn c ghp gia cc bn cc m v cc bn cc dng trnh hin tng chp mch gia cc in cc khc du.

1.3.3.Phn phi bn cc m:Cc bn cc m ghp song song vi nhau to thnh khi bn cc m. Cht hot ng ca bn cc m l ch xp.

Hnh 1.1.cc bn cc acquy1.3.4. Phn khi bn cc dng:Cc bn cc dng cng c ghp song song nhau to thnh khi bn cc dng.

Cht hot ng bn cc dng l PbO2 ( bi xt ch )

tng dung lng v sc in ng ca acquy, ngi ta u ni nhiu bn cc ni tip nhau dng- m xen k nhau trong mt ngn gia (+) v (-) cch nhau bng mt l cch in.

1.3.5.Dung dch in phn:Dung dch in phn l axt sunfuric ( H2SO4 )

Nng dung dch in phn axt sunfuric P = 1,1 - 1,3g/cm3Nng dung dch in phn c nh hng ln n sc in ng ca acquy.

1.3.6.Cu ni:Cu ni bng ch ni tip cc u cc m ca ngn acquy ny vi u cc dng ca ngn acquy tip theo.

1.3.7.Np nt:Np y v bnh cng c lm bng nha hoc bng cao su cng, np c cc l dung dch vo bnh v u cc v nt y in dch khi ra.

1.4. Qu trnh ho hc trong cc acquy axt:Trong acquy axt thng xy ra hai qu trnh ho hc thun nghch 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 m n cc bn cc dng, l dng in np (In).Khi phng in, di tc dng ca sc in ng ring ca acquy cc in t s chuyn ng theo hng ngc li t (+) n (-) v to thnh dng in phng (Ip).Khi acquy c np no, cht tc dng cc bn cc dng l PbO2 , cn 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.

Phng trnh ho hc xy ra trong acquy axt:

Trn bn cc dng :

PbO2 + 3H+ HSO4+ 2e = PbSO4+ H2O

Trn bn cc m :

Pb + H2SO4 = PbSO4 + 2e + 2H

Trong trng hp tng qut c th c trng cc qu trnh bng bng sau:

Trng thi ca acquyBn cc dngDung dch in phnBn cc m

c np no

phng ht in

PbO2(xt ch)

PbSO4(sunfat ch tinh th nh)2 H2SO4(axt sunfuric)

2 H20

(nc)Pb

(ch xp nguyn cht)

PbSO4

(sunfat ch tinh th nh )

Nh vy, khi phng in axt sunfuric b hp th to thanh sunfat, cn nc th b phn ho ra , do nng ca dung dch gim i. Khi np in th ngc li, nh hp th nc v ti sinh ra axt sunfuric nn nng ca dung dch tng ln . S thay i nng ca dung dch in phn khi phng v np l mt trong nhng du hiu xc nh mc phng in ca acquy trong s dng.

1.5. Cc c tnh ca acquy axt:Trong phn ny ta ch nu mt vi c tnh ch yu ca acquy axt v n gin ta ch xt c tnh ca mt acquy n:

1.5.1.Sc in ng (S) ca acquy :

S ca acquy 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, khng ph thuc vo kch thc ca cc bn cc.

S ca acquy ph thuc vo nng ca dung dch in phn c xc nh bng cng thc thc nghim sau:

Eo= 0,85 + P ( 1.1 )

Vi E0 - S tnh ca acquy n. S tnh c o trong trng hp acquy khng phng in v bng vn k c bit.

P - nng dung dch in phn c tnh bng V quy v + 150C

Ngoi ra S ca acquy cn ph thuc vo nhit ca dung dch in phn. V d khi nhit thay i t 200C n 400C th S ca acquy n gim t 2,12 n 2,096V

1.5.2.Cc c tnh np v phng ca acquy:a.Phn tch qu trnh np Khi dung dch axit sunfuric vo cc ngn ca bnh th trn cc bn cc s sinh ra mt lp mng ch sunfat :

EMBED Equation.3 + ( +

em ni ngun in mt chiu vo hai u ca acquy th dng mt chiu s c khp kn qua mch acquy v dng i theo chiu: cc dng ngun mt chiu (u cc 1 acquy (chm bn cc 1(qua dung dch in fn(bn cc 2(u cc 2 ca acquy (cc m ngun mt chiu.

Dng in s lm cho dung dch in fn fn ly:

( +

Catin theo dng in i v fa chm bn cc ni vi m ngun in

v to ra fn ng ti :

2 + ( +

Cc anin chy v fa chm bn cc ni vi cc in dng ca ngun in to ra fn ng ti :

+ 2 + cc ( 2 +

Kt qu l cc chm bn cc c ni vi bn cc dng ca ngun in c ch ixit , chm bn cc kia c ch . Nh vy , hai loi chm cc c s khc nhau v cc tnh .

Khi np acquy ,lc u in th tng dn t 2V(2,4V .Nu vn tip tc np

gi tr ny nhanh chng tng ln 2,7V v gi nguyn.Thi gian ny gi l thi gian np no ,n c tc dng lm cho phn t cc cht tc dng su bn trong lng bn cc c bin i hon ton ,nh s lm tng thm dung lng phng in ca acquy .Trong s dng thi gian np no ca acquy

thng ko di khong 2h(3h , trong khong thi gian ny hiu in th ca acquy v nng dung dch in phn khng thay di.Sau khi ngt mch np , in p ,sc in ng , nng dung dch in phn ca acquy gim xung v n nh, dy gi l thi gian ngh ca acquy sau khi np .

C th np in cho acquy vi dng in c nh hoc np in th khng i . Np dng in c nh s nhanh nhng tn nng lng hn ch np in th khng i.

Hnh 1.2.Qu trnh np acquyTrn hnh v l c tnh np bng dng in khng i, nng dung dch khi np tng theo quy lut ng thng t 1,11 g/cm3 n 1,27g/cm3 cui qu trnh np.

Sc in ng E0 1,96V ng vi acquy coi l phng ht in.

Khi np in trong lng cc bn cc to thnh axt sunfuric v nng ca dung dch trong cc bn cc tr ln m c hn nng dung dch chung, do Eaq khi np ln E0 mt lng bng (E.

Th hiu ca acquy khi np :

Un = Eaq + In.Raq ( 1.2)

In : dng in np (A)

Un : Th hiu ca c quy trong qu trnh np

Raq : in tr trong ca c quy

(E : Mc chnh lch sc in ng trong qu trnh np.

cui qu trnh np S v th hiu Un tng ln kh nhanh cng vi cc bt kh to thnh trong acquy. Khi qu trnh np kt thc v cht tc dng cc bn cc tr li trng thi ban u, dng in In lc ny ch cn tc dng in phn nc thnh xi v hir v thot ra di dng cc bt kh. Hin tng ny c gi l s "si" ca acquy v l du hiu ca cui qu trnh np.

S si bt u trong acquy khi th hiu ca mi acquy n tng ti 2,4V ri ngay sau th hiu tng vt ln v n khi t gi tr tn cng 2,7V th ngng tng. im ny thc cht l im cui qu trnh np v c th kt thc np y, nhng thng ngi ta phi tip tc np khong 3 gi na, khi thy rng sut trong thi gian th hiu v nng dung dch ca acquy khng thay i th acquy mi c np no.

Sau khi ngt dng in np, th hiu ca acquy st hn xung bng Eaq v sau mt khong "ngh" (tc l sau khi cn bng nng dung dch v thot ht bt kh) gim n S tnh cho n gi tr E0= 2,11 ( 2,12V ng vi acquy c np no.

Nh vy nhng du hiu biu th mc cui cng ca qu trnh np:

Th hiu v nng dung dch ca acquy nng tng v chng phi khng thay i trong 3 gi lin.

in lng cung cp cho acquy khi np Qung ninh tnh bng :

Qn = In. Tn ( 1.3)

Tn : l thi gian np tnh n im cui qu trnh np.

Trong qu trnh acquy lm vic do tn tht v nhit v cho qu trnh phn ng ho hc khng hon ton li nn khi np phi cung cp cho acquy mt in lng ln hn in lng n c th sn sinh ra khi phng in. Ngoi ra, do phi tiu tn thm nng lng in cho vic in phn nc trong 3 gi lin nn khi np in lng cung cp cho acquy cn phi ln hn in dung Q thu c trong qu trnh phng khong 10 (15% na ( c trng bng phn gch vung trn hnh v )

b.Qu trnh phng

Trong qu trnh phng in ca acquy , xy ra cc fn ng ho hc sau:

Ti cc dng : + + 2 + 2e ( + 2

Ti cc m: + ( + 2e

Nh vy khi acquy phng in , ch sunfat li c hnh thnh hai chm bn cc , lm cho cc bn cc dn dn tr li ging nhau cn dung dch axit b phn tch thnh catin 2 v anin , ng thi qu trnh phng in cng to ra nc trong dung dch , do nng ca dung dch gim dn v sc in ng ca acquy gim dn .

Qu trnh phng in ca acquy cng c th chia lm hai giai on : giai on u in p ,sc in ng , nng dung dch in phn ca acquy gim chm,y gi l giai on phng n nh hay thi gian phng in cho php ca acquy .Trong giai on tip theo ,in p acquy s gim rt nhanh .

Hnh 1.3.Qu trnh phng acquy

Trn hnh l s phng v c tnh phng ca acquy axt qu trnh phng cng c phn tch tng t nh trn, trong n ny ta ch quan tm n phng php np in cho acquy v vy y ta khng cn nu chi tit v c tnh phng ca acquy axt.

1.6. Cc phng php np in cho acquy axt: np in cho acquy ngi ta s dng ba phng php :

- Np bng dng in khng i In = const

- Np bng th hiu khng i Un = const - Np va dng va p n ny ta s dng phng php np bng dng khng i vi hai cp dng in

1.6.1. Np bng dng in khng i:Theo phng php ny, dng in np c gi nguyn mt tr s khng i trong sut thi gian np ( np mt nc) hoc trong nhng trng hp np vi cho php np hai nc tc l thay i cng dng in np mt lan.

V dng in np : (1.4)

M Eaq trong khi np tng dn, nn mun gi cho In = const trong qu trnh np th phi tng dn Un. thc hin c vic ny ngun np phi c nhiu nc in th , nu khng phi mc thm mt bin tr ni tip vi acquy.

Trong trng hp np hai nc th nc th nht kt thc khi th hiu mi acquy n t 2,4V (bt u si bt kh trong acquy). Sau chuyn sang nc th hai vi cng dng in np gim i v kt thc qu trnh np cui nc ny.

Theo cch ny tt c cc acquy khng l thuc vo th hiu nh mc c mc ni tip vi nhau v cn m bo iu kin :

Un > 2,7. Naq ( 1.5 )

Trong :Un : in p np

Naq : S ngn acquy n mc trong mch np.

Tt c cc acquy phi c in dung nh nhau, nu khng s phi chn cng dng in np theo acquy c in dung nh nht v nh vy acquy c in dung ln s phi np rt lu

Trong qu trnh np sc in ng ca acquy 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 (1.6)

* u im ca phng php :

Np bng dng in khng i l phng php np ch yu v tng hp nht , trong np mt nc l c bn, cn np hai nc ch p dng khi cn rt ngn thi gian np. phng php ny cho php chn tu cng dng in np cho thch hp vi tng loi acquy. Tt c cc acquy mi trc khi em s dng ni chung u phi np qua cch ny.

Phng php np in mt nc p dng trong trng hp np ln ln c acquy mi, acquy c v np cha cc acquy b sunfat ho nh.

* Nhc im :

Thi gian np ko di ( thng 25 n 50 gi, ring acquy np kh th ngn hn ) v thng xuyn phi theo di, iu chnh cng dng in np.

1.6.2. Phng php np vi in p khng i:Theo phng php ny tt c cc acquy c mc song song vi ngun in np v m bo th hiu ca ngun bng ( 2,3 ( 2,5)V trn mt acquy n. Th hiu ca ngun np phi c gi n nh vi chnh xc n 3% v c theo di bng vn kDng in np : (1.7) Lc u In s rt ln, sau khi Eaq tng dn th In gim i kh nhanh. Do dng in np ban u rt ln nn thi gian np gim i nhiu. Trong khong 3 gi u acquy nhn c 80% in lng yu cu. Qu trnh np kt thc khi dng np rt nh, gn bng khng, cn th hiu np t ( 2,3-2,4)V trn mt acquy n, nn qu trnh np thc ra mi ch n im bt u si kt thc, do khng th np no cho acquy bng phng php ny. Nh vy phng php np Un = const khng th thay th cho phng php np ch yu vi In= const ni trn m ch c th coi l phng php ph.

* u im ca phng php

Phng php np vi Un = const c thi gian np ngn hn v t tn cng, v dng in np t ng gim theo thi gian khng cn phi theo di v iu chnh, thch hp vi vic np b sung cho cc acquy ang s dng.

* Nhc im ca phng php

Khng th cng lc np c acquy c v mi v np cha cc acquy b sunfat ho; cng dng in np ban u rt ln, tuy khng lm hng acquy nhng c hi cho tui th v in dung ca acquy v gy qu ti cho thit b np nu khng c c cu hn ch dng in.

1.6.3.Np hai cp dng in:Ch np hai cp dng in:Dual Current Step Charge.

Trong ch ny b np s thc hin hai ch np vi hai mc dng in khc nhau l (Imax+IH) v IH.

Ch ny thng dng khi np ng thi cho nhiu nhnh acquy mc ni tip nhau,lc giai on np vi dng IH (holding current thng ly bng 10%Imax) s m bo cho cc acquy c t chnh t trng thi y nh nhau.

Nguyn l hot ng th hin c tnh v th np.ch np vi mc dng ln bng (Imax+IH) c gi l Bulk charge,nhng khc vi kiu np dng p ch khi in p acquy t tr s V21 th khng chuyn sang ch qu np (Over charge),tc l chn s 8 ca IC khng chuyn v mc thp m lun gi mc cao.Khu khuych i dng np c s dng lm b iu chnh dng gi dng np bng IH.

VIN

V12

STAGE2 VF

V21

STAGE1:BULK CHARGE STAGE1 STAGE 2:HOLDING VOLTAGE DONG IN NP

IH Imax+IH c tnh VA qu trnh np

INPUTSUPPLY

VOLTAGE

V12

VF

B C V21

CHARGE A D E

VOLTAGE

Imax+IH

CHARGE

CURRENT IH

OFF

STATE ON

LEVEL

STAGE1 STAGE 2 STAGE 1 Hnh 1.4. th trng thi

Cc im c trng ca qu trnh np l:

A Ngun vo c cp v b np lp tc hot ng vi ch np dng ln (Imax+IH).

B in p acquy t mc V12 v mch vng in p chuyn sang mc thp hn l VF,acquy c np bng dong IH.

C Acquy c ti v bt u phng in,b np vn ch cp dng IH.

D Khi in p acquy gim n tr s VF th b np chuyn sang cp dng ln (Imax+IH).

E Acquy tip tc phng in n tr s V21 th b np chuyn v trng thi ban u

1.7. Cch bo qun acquy:* Bo qun thng xuyn:Chm sc v s dng acquy ng k thut s nng cao hiu sut s dng, ko di tui th ca acquy v m bo an ton cho ngi, phng tin s dng.Lun m bo mc dung dch in phn, khi thiu phi b sung bng nc ct cho .

Bi m vo cc u bc acquy chng g, thng xuyn lau chi sch s, b mt acquy phi lun lun kh trnh hin tng phng in trn b mt acquy.

Kim tra cc nt v khng lm bp l thng hi

* Bo qun trong kho:- Kho cha acquy phi thong, nhit trung bnh phi nh hn 350C

- Kho khng acquy chung, acquy axt ring, acquy kim ring.

- Nn nh phi di nha ng v b tr gn gng tin vn chuyn ra vo trong kho.

- Trc khi ct acquy phi np qu lng, lau sch s cc nt v v bnh.

- Trong thi gian bo qun mi thng phi np mt ln v phi y vi ct.

Chng 2THIT K CH TO B NP ACQUY T NGLa chn s chnh lu

- Chnh lu iu khin mt pha na chu k

- Chnh lu iu khin mt pha hai na chu k

- Chnh lu iu khin cu mt pha

- Chnh lu iu khin ba pha

- Chnh lu iu khin cu mt pha khng i xng

- Chnh lu 6 pha dng in khng cn bng1.1.Chnh lu iu khin mt pha, na chu k:

Xc nh gi tr trung bnh Ud,Id

khi gc ( m , ( s l gc tt

Ud = .U2 sint

- id ko di ( ( ( na chu k m

( U2 = Ud = Rid + L .U2 sint

( .U2 sin ( = Rid + x

Khi id = 0 ( = (( = (Theo nh ngha v gi tr trung bnh ca hm s

Ta c : Ud = Rid

Ud = ( cos ( - cos ( )

Id = ( cos ( - cos ( )1.2.Chnh lu iu khin mt pha 2 na chu k :U1 = U2sint

U2 = -U2sint

Ta c id l dng lin tc, v vy khi bit c gc m ( th xc nh c gc tt (Khi T1 m ( .U2 sin ( = Rid + x

Ta c Ud = RId

Ud = . Cos (Id = . Cos (1.3.Chnh lu iu khin cu 1 pha:

Khi Id l dng gin on

Id =

IT =

Khi id l dng lin tc

Id = Id

Ud = Rid ( Ud =

1.3.1.Chnh lu cu 1 pha khng i xng

Id =

IT =

Cho thm 2 i t

ID =

Id ( ( + ()/2 (a.S chnh lu cu mt pha khng i xng

b.Nguyn l hot ng ca s :

Khi ( = (1 cho xung iu khin m T1. Trong khong (1, (2 T1 v D2 cho dng chy qua. Khi U2 bt u i du, D1 m ngay, T1 t nhin kho li dng Id = Id chuyn t T1 sang D1Khi D1 v D2 cng cho dng chy qua Ud = 0

Khi ( = (3= ( + ( cho xung m T2, dng ti id = Id chy qua D1 v T2 i t D2 b kho li.

Trong s ny, gc dn ca tiristor v i t khng bng nhau.

Gc dn ca i t l : ( D = ( + ( , cn gc dn ca tisristor

(T = ( - (.

Gi tr trung bnh ca in p ti:

Ud =

Gi tr trung bnh ca dng ti :

Id =

Gi tr dng trung bnh qua van :

IT =

Qua dng i t : ID =

Gi tr hiu dng ca dng chy trong cun dy th cp my bin ap

I2 =

in p ngc cc i t ln van

Uim = .U2

Cng sut my bin p :

Sba = 1,93Pd

1.4.Chnh lu iu khin 3 pha:a.S 3 pha hnh tia

Ua = U2 m sin (Ub= U2 m sin ( ( ( 1200)

Uc= U2 m sin ( ( ( 1200)

Trong mch c L nn id l dng lin tc

id = Id

Gc m ( > 0

Ud =

Id =

b.S cu 3 pha:Gm 3 tiristor u catt chung : T1, T3 , T5

3 tiristor u ant chung : T2 , T4 , T6

Ud = Id.R

Id = ( IT = ID =

1.5.Chnh lu 6 pha

Trn s c 1 my bin p 3 pha, 6 tiristor

1- Cp ngun cho T1 T3 T5


( Lm vic c lp

i vi s c 1 my bin p 3 pha, 6 tiristor



( S h lm vic c lp

Nhn xt

Ta phi dng 2 my bin p xung iu khin T1, T2 . Tuy nhin ta li cch ly c mch lc v mch iu khin v in

S ch dng mt na van iu khin nn gim gi thnh h thng iu khin n gin hn.1.6.Chn phng nQua phn tch trn, vi cc s chnh lu khc nhau ta thy s chnh lu cu mt pha khng i xng c:

- S van iu khin t, nn t knh iu khin, vn u t gim h thng iu khin n gin.

- Mch lc n gin, m bo kinh t

- Cng mt ri iu chnh th cu 1 pha khng i xng iu chnh chnh xc hn.

Nhng u im m bo v yu cu k thut v thit k ngoi ra n cn c u im kinh t.

T ta i n chn phng dng s mch chnh lu cn 1 pha khng i xng.

Chng III

Tnh chn mch lc

1.1.Tnh chn van:

S liu: Thit k b np acquy t ng

Cho 9 bnh acquy, mi bnh 12V mc ni tipu vo: 220 (V) , f= 50 (Hz)u ra : Ud = 108 (V)Id = 16 (A)Do in p cao nht Voc=1,25E nn ta phi chn:Ud=108.1,25=135(V)

Gi tr hiu dng ca in p pha th cp my bin p c phm vi iu chnh khi u vo dao ng, ta chn ( = 300

C : Udo =

U2 = 160,63(V)in tr np:

EMBED Equation.3 .

Dng in chy qua Thyristor:

.

Dng in chy qua it:

.

T s my bin p:

K = 0,73in p ngc ln nht mi van phi chu:

Ungm = .U2 = .160,63=227,2(V)Dng in chnh lu nh mc :

Id = 16(A )

Dng in trung bnh chy trn mi van

Iv= = 8(A)

Mch c cng sut nh nn s dng phng php lm mt t nhin , cnh tn nhit gn vo van kt hp vi i lu khng kh

Khi : Ilv = 25% IvIv = = 32(A)Gi tr hiu dng ca dng chy qua mi pha th cp bin p

I2 = Id. = 3214,6(A)Gi tr hiu dng ca dng chy qua mi pha s cp bin p :

I1 = K.I2 = 0,73.14,6=11(A)Dng in trung bnh chy qua Thyristor: ID=1,2.9,33=11,2(A)Dng in trung bnh chy qua it:

IT=1,2.6,67=8(A)Chn van :

H s d tr v in p : Ku = 1,6

H s d tr v dng in : Ki = 1,2

Phi chn van chu c in p ngc : 1,6 . 227,2=364 (V)

Phi chn van chu c dng in : 1,2 . 32=38,4 (A)* Chn 2 thyristor : T12N400COE trong gio trnh Tnh ton thit k thit b in t cng sut tc gi T.S Trn Vn Thnh- Dng trung bnh ln nht: Itbmax = 15(A)- in p ngc cc i : Uim = 400 V

- Tn tht in p : (U = 2,8 (V)

- Gi tr in p iu khin : Ug = 2(v)

- Gi tr dng in iu khin :Ig=40(mA)

- Thi gian chuyn mch :tcm=50

- Nhit lm vic ti a ca van: tmax=1250C - Tc tng in p: du/dt = 200(V/(s)* Chn 2 i t :it SKN20/04- Dng trung bnh ln nht: Itbmax =20 (A)

- in p ngc cc i : Ungm = 400 (V)

- Tn tht in p trng thi m: (U = 1,55(V) - Nhit cho php:Tmax=1800C

1.2. Tnh ton my bin p:Cng sut biu kn ca my bin p :

S = U2. I2 = 160,63.14,6=2345,2(VA)Chn mch t ch E v ch I ghp li, tit din tr c theo cng thc kinh nghim Q = K. (cm2)

Trong : K = 4 ( 5 nu l my bin p du

K = 5 ( 6 nu l my bin p kh

S: Cng sut biu kin ca my bin p KVA ; VA

C : S tr

f : Tn s ngun in xoay chiu

Ta s dng my bin p kh, ly K = 5 c

Q = 5. = 19,77(cm2)

Chn li thp c tit din 20m2 lm bng vt liu st t dy 0,6mm, l thp dp hnh ch v ch I ghp li. Tu thuc vo vic chn l thp m c kch thc li.

S vng/vn c tnh theo cng thc:

wo= 0,25( vng/vn)

S vng dy cun s cp :

w1 = U1.wo = 220. 0,25=55 (vng)S vng dy cun th cp:

w2 = U2.wo = 160,63.0,25 = 40 (vng)

Tit din dy qun :

I1 = 11 (A)

I2= 14,6 (A)

Chn mt dng in : J1 = J2 = 2,75 A/mm2C q1 = 4 (mm2)q2= 5,31(mm2)ng knh dy qun s cp :

d1 = 2,26(mm)ng knh dy qun th cp :

d2 = 2,6(mm)Tra sch "in t cng sut" chn dy :

+ Dy qun s cp chn :

- ng knh d1= 2,44 mm

- Trng lng ring 1 mt di: mcu=41,6 g/m- in tr dy 1 mt di : R/m=0,00375 ((/m)

- ng knh dy k c cch in dy:dng=2,57mm

- Tit din li ng:Scu=4,676mm2+ Dy qun th cp :

- ng knh ngoi: d2 = 2,83 (mm)

- Trng lng ring 1 mt di:mcu = 55,9g/m

- in tr dy 1 mt di : R/m = 0,00278 ((/m)

- ng knh dy k c cch in li dy:dng=3,16mm

- Tit din li ng:Scu=6,29mm21.3. Tnh ton bo v qu p:Trong cc b bin i dng van bn dn cn phi c cc phn t bo v s tng trng dng v p tiu van , trong s ca ta ang xt dng my bin p nn thnh phn cm khng ca my bin p gip ta bo v hn ch s tng trng dng in. V vy ta ch tnh ton bo v qu in p

Thyristor rt nhy cm vi in p qu ln so vi in p nh mc ngi ta chia lm hai 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 Thyristor bng in p ngc, cc in tch ni trn i ngc li 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 Thyristor . V vy gia a nt v ca tt ca Thyristor xut hin qu in p.

- Nguyn nhn bn ngoi: Nhng nguyn nhn ny thng xy ra ngu nhin nh khi ct khng ti mt my bin p trn ng dy, khi mt cu ch bo v chy, khi c sm st

bo v qu in p ngi ta thng dng mch Rc

Mch RC u song song vi Thyristor nhm bo v qu in p do tch t in tch khi chuyn mch gy ln.

Mch RC ta s dng cc ch vit tt sau y :

Ump , Uimp: Gi tr cc i cho php ca in p thun - ngc t ln trn it hoc Thyristor mt cch chu k, cho trong s tay tra cu

+ Tnh R - C

Xc nh h s qu in p theo cng thc :

K=

Trong :

UnmP - l gi tr in p cc i cho php t ln Thyristor mt cch khng chu k cho trong s tay tra cu.

Unm- l gi tr in p ngc cc i thc t t ln Thyristor

b- H s d tr v in p : b = 1 ( 2

Chn b = 1,6 v thay gi tr UnmP = 400 (V) , Uim = 227,2 (V)

Vo cng thc trn ta c gi tr ca K = = 1,1Theo kinh nghim ta chn :

C= 0,1 (F

R= 50 (Chng 4Tnh ton v thit k mch iu khin

1.1.Mc ch v yu cu:Mch iu khin l khu rt quan trng trong b bin i Thyristor, c vai tr quyt nh n cht lng, tin cy ca b bin i.

Thyristor ch m cho dng in chy qua khi c in p dng t trn ant v c xung p dng t vo cc iu khin. Sau khi Thyristor m th xung iu khin khng cn c tc dng g na, dng in chy qua Thyristor do thng s ca mch lc quyt nh.

Mch iu khin c cc chc nng sau :

- iu khin c v tr xung iu khin trong phm vi na chu k dng ca in p t trn ant, ca tt, Thyristor.

- To ra c cc xung iu kin m c Thyristor ( xung iu khin thng c bin t 2 - 10V, rng xung Tx= 20 - 100 (s i vi thit b chnh lu tx ( 10 (s i vi thit b bin i tn s cao)

rng xung c xc nh theo biu thc :

tx=

Trong : Idt - dng duy tr ca Thyristor

di - tc tng trng ca dng ti

dt

Cu trc mch iu khin

Khu 1 - L khu ng pha

Khu 2- Khu to in p rng ca

Khu 3- Khu so snh

Khu 4- Khu khuych i

Bng cch tc ng vo Uc, c th iu chnh c v tr xung iu khin, cng l iu chnh c gc (.

1.2. Nguyn tc iu khin: iu khin gc m ca cc Thyristor trong thc t ngi ta thng dng 2 nguyn tc iu khin

- Nguyn tc iu khin thng ng "ARCCOS"

- Nguyn tc iu khin tuyn tnh

thc hin vic iu chnh v tr xung trong na chu k dng ca in p t trn Thyristor1.2.1. Nguyn tc iu khin thng ng tuyn tnhi vi nguyn tc ny, ta dng 2 in p

- in p ng b L/s , c dng rng ca, ng b vi in p t trn ant, ca nt.

- in p iu khin Uc l in p mt chiu c th iu chnh c bin . Tng i s Uc + Ui c a n u vo ca khu so snh.

Nh vy bng cch lm bin i Uc ta c th iu chnh c thi im xut hin xung ra, tc l iu chnh gc (Khi Uc = 0 ( ( = 0

Uc < 0 ( ( > 0

( ( v Uc c quan h :

( = (.

ly Ucmax = Urmax

1.2.2.Nguyn tc iu khin thng ng "Arccos"

i vi nguyn tc ny ngi ta cng dng 2 in p

- in p ng b Ur vt trc in p a nt, ca tt , Thyristor mt gc (/2 ( nu Uak= A sin wt th Ur= Bcoswt )

- in p iu khin Uc l in p mt chiu, c th iu chnh c bin theo 2 hng dng - m )

Trn hnh v, ng nt t l in p a nt, ca tt, Thyristor t to ra Ur

M Ur + Uc c a n khu so snh

Khi Ur + Uc = 0 ta nhn c mt xung u ra ca khu so snh.

Uc + Bcos ( = 0

( ( = Ur cos ( - )

Do B = Uc max

Khi Uc = 0 ( ( = (/2

Uc = Ucmax ( ( = (Uc = - Ucmax ( (= 0

Vy khi cho Uc bin thin t - Ucmax n + Uc max th ( bin thin t 0 ((1.2.3. Khu to in p ng pha

Mch ly xung ng pha c ly t ngun 220V, tn s f = 50Hz pha th cp ly 12V. Bin p th cp c im gia c qun v ly im gia lm im 2 u cn li ca mch u vo 2 it to in p p mch (+) lin tc, im gia u mt.

Gi tr Ud =

= 0,9 . 12 = 10,8 (V)

1.2.4. Khu to in p rng ca ng b

S dng 1 tranfistri v 1 t in

Ub - xung ng b im khng ca in p ngun xoay chiu

Ur - in p rng ca ng b

Ur = E ( 1-e-t/rc)

Khi transistor kho ( t C c np in theo chn ra (+) ca IC v m ngun .

1.2.5.Khuych i xung, bin p xung

Hot ng ca s :

Tn hiu vo Uc, l mt tn hiu logic. Khi Uc= 1 th T m bo ho, khi Ue= 0 ( T kho li.

in tr R1 hn ch dng coletoc, i t Dr hn ch in p trn cc cc coletot- emit ca T. i t D2 ngn chn xung p m c th c khi T kho. Rg hn ch dng iu khin. R2 l in tr nh hng n bin v sn xung ra.

Khi Ur= 0 ( il= ic = 0 ( Uce = E , id1 = 0

Gi thit khi t = 0 + Uc = 1, T m th in cm L khng cho ic t ngay tr s bo ho ca n.

Ic =

Dng ic ch c th tng trng t t theo lut hm m

iL = ic = (1 - e-t/t)

Vi t =

Sau khong thi gian 5t ( ic =

Bn BAX ca phn th cp xut hin mt xung in p trn R2 m Thyristor, khi t = T1, Ue = 0 lc iL t gi tr :

iL (T1)= ic .T1)= (1 - e-T1/t)= Io <

T b kho ( ic = 0

Nu khng c i t Dr th nng lng

W = LI20 s sinh ra qu in p trn cc cc C, E ca T. Qu in p ny c th t ti 100V ph hu T = 0,8V th Dr m cho dng chy qua ( ngn mch 2 im C v E. Do

Uce= E + 0,8V

( t = Tq ( iL gim

iL = I. e-t/t vi t'=

Trong R' l in tr ca cun cm L v in tr thun ca Dr.

1.2.6.Khu so snh

So snh 2 tn hiu cng du :

Ud = Ui - U1 khi

U1 < Ui ( Ud> 0 ( U2 = Usat

U1 > Ui ( Ud < 0 ( U2 = - Usat

- So snh 2 tn hiu khc du :

Khi Ur < Uc ( U2 = - U sat

Ur > Uc ( U2 = Usat

1.3. Tnh ton v chn linh kin cho mch iu khin

1.3.1.Tnh chn khi ngun

Khi ngun to in p ( 12V cung cp cho cc OA

D1 ( D4 Chn loi 1N c I = 2A , U = 70V

C1C2 Lc ngun in p trc n p 7812 v 7912

Chn loi 2200 (F - 35 V

C3C4 lc ngun sau n p chn loi 470 (F - 25V

C5C6 lc sng hi v nhiu cao tn chn t xoay chiu 100 (F- 40V

Khi ngun to En= 24V

D5 (D8- chn loi B-10 c I = 10A , U = 100v

C7 t lc ngun cng sut chn loi 2500 (F - 40V

Khi ngun to in p ng pha

UF= ( 16 sin Wt

D9 ( D10 ly tn hiu ng pha chn loi 1N5402

C I =2A , U = 70V

- Tnh my bin p ngun :

+ Khi ngun to in p cung cp cho OA: - 17V- 0V- + 17V

U1 = 220 V , U2 = 30V , I2 = 500mA

C P2.1 = U2. I2 = 34 . 0,5 = 17 (w)

+ Khi ngun to in p ng pha - 12V - 0 - + 12V dng 150 mA

P2.3= 24. 0,15 = 3,6 (w)

Cng sut ca my bin p ngun :

P = P2.1 + P2.2 + P2.3 = 17 + 24 + 3,6 = 44,6 (w)

Tit din ca li :

Q= K. = 1,2. = 7 (cm2)

Chn li thp c tit din Q= 8 cm2 lm bng l thp k thut E310 ghp li, dy 0,35 mm, v I

+ S vng dy , dng/vn

Wo= = 6 vng/v

Vng dy cun s cp :

W1 = U1. Wo = 220.6 = 1320 vng

S vng dy th cp :

W2.1.1 = W2.1.2 = 17.6 = 102 ( vng)

W2.2 = 24.6 = 144 vng

W2.3.1 = W2.3.2 = 12.6 = (2 ( vng)

Tit din dy chn J = 2A/mm2Dng th cp : ( mm2)

Tit din dy th cp : S2.1 = ( mm2)

S2.2 = ( mm2)

S2.3 = ( mm2)

- ng knh dy:

ng knh dy s cp :

d1 = (mm)

d2 = (mm)

d3 = (mm)

Tra s tay:

d1= 0,27mm ; 0,509g/m

;0,315 ohm/m

d2.1 = 0,57mm ; 2,27g/m ; 0,0675 ohm/m

d2.2= 0,8mm , 4,47 g/m ; 0,0342 ohm/m

d2.3= 0,31mm ,0,671 g/m;0,239 ohm/m

1.3.2.Tnh chn cc phn t khu to in p ng pha

Chn OA1, OA2 OA10 loi TL 084 c cc thng s :

Ecc= ( 12V

Zv = 1012 (Ud = ( 30v

Zra = 60 (Ur + ( 25v

K = S.104du/dt = 0,3v/(s

T0 = -25 + 850Khi Ud < Ub tn hiu ra ca OA1 l tn hiu xung m lc transistor T1 kho, t C1 c np theo ng ra ca OA2 vi dng np

In = -

Khi UB< Ud , u ra ca OA1 l xung dng lc T1 m t C1 phng in t ( phn dng ca t ) qua T1

- Chn chu k phng np ca t C1 l 10ms

Thi gian np tn =

Thi gian phng tp= 0,97ms

Chn Ucmax = 9v ta c

Uc =

Vi R4 + UR2 = R

Chn C1 = 470 nF = 47.10-8F

( R =

Chn UR2 = 0,5K ( ; R4= 100 (Ta c Uo= vi R' = R3+ UR1

t Uo= 0,5 (1 (V) ( UR1 = 1 K ( ; R3 = 100 (1.3.3.Khu phn hi p

Chn Rf >> Rt, do vy dng qua R7 chn c 10mA nn

Rf = (K ()

Ngun np phi tho mn yu cu l khi in p c quy nh hn 90% in p nh mc c quy ( UnAQ < 90% Um) th mch c np theo ch dng.

Khi UnAQ > 90% Um th mch t ng chuyn sang ch np ( np no).

Ta t ngng cho OAo l 5v ta chn in p trn UR3= 5v phi t v tr VR5 sao cho in p trn c quy t ti 108V th in p phn hi l 5V, ta c : RP = = 0,5 (K ()

Chn R18= 0,5 k(R16= R19 = 10 k(H s khuych i ca OA7 l K = 5.104Ta c R41= K. R19= 5.104.104 = 5.102 M (in tr R22 hn ch u vo b khuych i chn R22 = 10 k(, R21, R23 phn p t gi tr ngng so snh vi in p phn hi, u ra khuych i cp cho kho chuyn mch CM1

Chn R21 = R23 = 15k(in tr R17 , R20 hn ch dng vo CM1

Chn R20 = R17 = 20 k(1.3.4.Khu phn hi dng

in tr RS ly tn hiu phn hi dng chn loi 75 (A

Cc in tr R24, R26, R27 hn ch dng vo b khuych i v phi hp tr khng

Ta chn : R24 = R26 = R27 = 10k(Nn R25 = K.R24 = 5.104.104 = 5.102 ( (in tr R28 = R30 = 10 k (R29, R31 t nghing so snh vi dng phn hi c khuych i a vo kho chuyn mch CM2.

Chn R29 = R31 = 15 k (in tr R32 hn ch dng vo kho chuyn mch CM1 , chn R32 = 20k (Cc kho chuyn mch CM1 v CM2 lm vic theo nguyn tc: khi c in p (+) t vo th kho chuyn mch ( CM thng cho dng chy qua. Khi c in p (-) t vo th CM kho ngt dng.

1.3.5.Khu bo v qu p trn c quy

Khi ch np ( np no) nu in p vt qu in p nh mc ca c quy s dn n nh thng cc bn cc lm hng c quy. V vy cn c b bo v qu p cho c quy.

t ngng cho OA11 l 7V

Chn h s np ca c quy l 1,2 ta c in p c quy ln hn I' l

UAQmax = 120.1,2 = 144 (V)

Theo t s t chit p v tr sao cho khi in p c quy tng n 144v th in p phn hi trn Rp l 7v , khi in p ti (A) t 7V th u ra OA11 lt trng (-) sang (+) lm cho T6 khng, dn n rle Rl tc ng m cc tip im ca n trn mch iu khin ngt mch bo m an ton cho c quy.

Chn chit p VR4 = 10k (R32 hn ch u vo v phi hp tr khng chn R32 = 5K (R33, R34 hn ch v to ch lm vic cho T6.

Ta chn R33 = R34 = 170 (i t D18 bo v lp tip gip BE khi transistor kho.

i t D19 bo v cun dy R le khi T kho

Chn D18 , D19 cng loi 1 N5401 c I = 2A, U = 70V

T chn loi B4 x 47 c thng s :

P = 125 w

Ic = 9A

Uce = 200V IB = 1,2A

1.3.6.Mch pht xung chm

Ta s dng b pht xung ch nht tn s 8KHz

Chn OA5 loi IC (A 741

Cc in tr R9, R10, R11 cng tr s 10 k (i t D7 loi 1N5402

Tn s giao ng ph thuc R8 v C2 chn C2 = 10 (F

Ta c : T = 2R8C2 ln ( 1 + ) =

T = 2,2 . R8. C2 ( R8=

Chn R8= 5,8 k (1.3.7.Mch so snh

Mch thc hin so snh 2 tn hiu phn hi v tn hiu rng ca, ng thi to xung ch nht.

in tr vo ca Ic: Zv = (Ta chn: R5 = R6 = 10 k (OA3 chn loi TL 084

R7= 15 k (i t D8 chn loi 1N 5402 ,I= 2A , U = 70V

1.3.8.Mch khuych i xung

Cc tn hiu Uss + Ucc c a n cng phn t s vo khu khuych i trong b bin i chn transistor loi

T (- 100 c Ug= 8 (v)

Ig = 0,4 A

T s bin p xung chn m = 1,5

Dng in s cp BAX c biu thc

I1=

(b qua dng t ho I ()

Chn T3 loi D163 lm vic ch xung c :

Uclo= 85 V

f = 15 MHz

Ic max = 6A

( = 10 ( 320

Vbco= 6 V

Chn IC = 0,4A ;( =40

Ta tnh : I ( T3 =

Dng IcT2 = IbT3 do chn T2 loi C28 c

Uce = 30V

Ic= 50mA

( = 100 ( 220

V bin p lm vic ch xung nn s c thi im BAX b bo ho, hn ch dng qua cuc s cp BAX ta mc thm R36in p th cp : BAX : U2 = f (v)

( U1 = mU2 = 1,5.7 = 10,5 (v)

( R36 = (()

in tr R36 l in tr cng sut

R36 = 39 ( ; P = 2w

i t D11 ngn chn xung m t vo cc G ca Thyristor

D12 bo v lp tip gip K - G lc kho

Chn D11 - D12 loi 1N5402

in tr R35 hn ch dng Ig ca Thyristor

Ig =

Rg = 10 ( ( R35 = (()

B pht xung cho knh iu khin Thyristor T2 cng c thng s tng t.

1.3.9.Tnh ton BAX

Chn vt liu st t f 330 li thp dng hnh ch I lm vic trn mt phn c tnh t ho (B = 0,7 T

(H = 50A/m c khe h

t thm ca li st

( =

Do mch t c khe h nn phi tnh t thm trung bnh chn s b

Chiu di trung bnh ng sc : l = 0,1

( = 10-5m

Do : Mtb =

Th tch li st t

V = Q1 =

V =

V= 2,5 ( cm2 )

Thc t ta thng dng BAX bng vt liu ferit c kch thc v ngoi v gng hnh tr l 30 x 25 (mm2)

S vng cun s cp n1 = 80 vng

S vng cun th cp n2 = 60 vng

1.4. Mch iu khin:

T yu cu ca cng ngh chng ta xy dng c mch iu khin c s sau:

S :Dng in p:

Nguyn l lm vic ca mch:

Khi Ub > U tn hiu ra ca OA1 l tn hiu xung m ( Uc < 0 ) khi transistor T1 kho "khng m", t C1 c np in theo a u ra ca OA2.

Khi UB < U tn hiu ra ca OA1 l xung dng t vo cc baz ca transistor T1, lc ny transistor T1 m bo ho, t C1 phng in t ( +t ) qua CE ca T1 to ra c in p rng ca.

Lc bt u np in cho c quy , in p c quy c np thp in p phn hi ly trn Rf thp hn gi tr t ca UR3 ( UA < 50) u ra ca OA6 m dn n kho chuyn mch CM1 kho v CM2 m. Lc ny tn hiu phn hi dng ly trn Rs c a vo b khuych i chy qua CM2 a ti u cng ca khu so snh, lc ny c quy c np theo ch dng.

in p c quy c np dng ln n khi in p trn Rf tng ln v ga tr in p Ua = 5V th u ra ca OA6 lt trng thi sang dng ( ng kho chuyn mch CM2 li v m kho chuyn mch CM1. in p phn hi ly trn R7 c khuych i v a ti u cng ca khu so snh. Lc ny c quy c np theo ch A1.

B so snh OA3 thc hin vic so snh hai tn hiu Urc v UCM1 hoc Ucm2, nu Urc < Ucm th transistor T2 v transistor T1 thp dn n pht xung vo m Thyristor T1.

Nu Urc > Ucm th kho cc knh iu khin cc Thyristor trn mch lc.

Nu in p c quy vt qua gi tr t 144v th in p ti im A l UA > U= 7v , u ra ca OA11 dng, lm cho transistor T6 thi dn n c dng chy qua cun rle, lm cho rle tc ng m tip im thng ng ca r le trn mch iu khin cho nn mch .

KT LUN hon thnh n ny, ngoi s c gng ca bn thn, cn c s hng dn tn tnh ca thy c trong b mn "T ng ho x nghip - cng nghip" c bit l thy Phm Quc Hi trc tip hng dn cho em trong sut thi gian lm n. n nay n c hon thnh, nhng vn cn nhng thiu st, em rt mong nhn c s ph bnh ca cc thy c cho cun n ca em c hon chnh hn. Em xin chn thnh cm n !H Ni, ngy 30 thng 5 nm 2010

Sinh vin

EMBED Unknown

EMBED Unknown

EMBED Unknown

Rf

+E

OA1

OA5

R28

D6

C2

R29

R27

R30

+E

D9

T4

R34

R36

T5

R35

D10

AND

Rs

7404

R20

OA7

R16

R18

OA9

OA8

4066

4066

D4

R8

-E

R3

R4

+E

R5

D11

D3

R2

D2

R6

OA2

R7

D1

D5

OA3

OA4

T1

C1

Udk

R9

R1

R17

D14

OA10

R23

+E

VR3

+E

VR2

+E

VR1

OA6

R15

R10

R11

R33

AND

D7

T2

R31

T3

R32

D8

R12

R22

R14

R21

R19

R13

PAGE 57 ti: Thit k b np c quy t ng H Bch khoa H Ni

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(-)

A

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+

-

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0

2

4

8

12

18

20

0,5

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1,27

1,11

Q

p

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p

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p,

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