ti :thit k v ch to b ngun np c quy t ng cho 10 c quy 12v dung lng 60ah

n tt nghip: Thit k v ch tao b ngun np c quy t ng

ti :thit k v ch to b ngun np c quy t ng cho 10 c quy 12v dung lng 60ahMc lc

ti............................................................................................................................1

Li gii thiu............................................................................................................4

Chng I : gii thiu chung v c quy...........................................................5

1 Cu to c quy ch-axit32 Yu cu cho vic np c quy...7

2.1 c quy...7

2.2 Qa trnh bin i nng lng trong c quy axit..7

2.3 Cc c tnh ca c quy....8

3. Cc phng php np c quy t ng..12

3.1 Np bng in dng in khng i......12

3.2Np bng in p khng i13

3.3Np bng dng p....13

Chng II : PHNG N CHNH LU...16

1. La chn b bin i ...16

2. Phn tch s chnh lu.............17

2.1Chnh lu c iu khin cu 1 pha i xng......17

2.1Chnh lu c iu khin cu 1 pha khng i xng.......20

Chng III : TNH TON thit k MCH LC...24

I. Chn s thit k27

II. Tnh ton chn van lc......29

III. Tnh ton bo v qu p. .29

Chng IV : Tnh ton thit k mch iu khin

I/ Nguyn tc iu khin trong h ng b.31

1. Nguyn tc iu khin ngang...31

2. Nguyn tc iu khin dc.......32

3. Khu ng b...32

4. Khu to in p ta ...33

5. in p ta dng rng ca........38

6. Khu so snh.....40

7. Khu dng xung .......42

8. Khu khch i xung (KX) ...43

II/ B iu khin ( BK ) ......43

CHNG V .THIT K SN PHM.49

I. Thit k mch ng lc v mch iu khin..49

1.Ngun cung cp cho mch iu khin...52

2. Tnh ton my bin p ngun....53

II . Tnh ton khi iu khin.....56

1. Khu ng b....57

2. Khi to in p rng ca.........57

3. Khi so snh..58

4. Khi to dng xung...58

5. Khi khuch i xung ..59

III. Lp rp sn phm.....62

LI GII THIU

Hin nay vi tin b khoa hc k thut v ang i mi cc phn t, cc mch iu khin c p dng rng ri vo trong cng nghip v i sng.Vi xu th pht trin ca khoa hc hin nay l ng dng khoa hc k thut in t,k thut tin hc,c kh chnh xc thc hin t ng ho c p dng cho tng my,t hp my cho tng dy truyn cng ngh,cc nh my tin ti t ng ho c ngnh sn xut. T ng ho lm gim nh sc lao ng chn tay ca con ngi lm cho t ng n tr thnh c trng ca nn sn xut cng nghip hin i.

Trong cng nghip ngoi ra cc mch iu khin ngi ta thng dng k thut s vi phn mm n gin,linh hot v d dng thay i cu trc tham s hoc cc lut iu khin.N lm tng tc tc ng nhanh v c chnh xc cao cho h thng.Nh vy n lm chun ho cc h thng truyn ng in v cc b iu khin t ng hin i v c nhng c tnh lm vic khc nhau.Trong nhng ng dng th vic p dng vo mch map c quy t ng ang c s dng rng ri v c nhng c tnh u vit.Bi c quy l ngun cp in mt chiu cho cc thit b trong cng nghip cng nh trong i sng hng ngy cung cp in cho cc ni cha c ngun in li phc v cho chiu sng,ti vi,thng tin lin lciu khin o lng ,cung cp cho cc thit b gin khoan ngoi bin.Chnh v vy vic nghin cu ch to mch np c quy l ht sc cn thit,n nh hng rt ln ti dung lng v bn ca c quy.

Di y em xin trnh by ton b ni dung ca n tt nghip.Thit k v ch to b ngun np c quy t ng do thy gio Lu c Dng ging vin trng i Hc Bch Khoa hng dn.Mc d trong thi gian qua em c gng tm hiu thc t,nghin cu ti liu thc hin n nhng khng th trnh khi nhng sai st , em mong tip tc nhn c s ng gp v ch bo ca cc qu thy c.

Em xin by t lng bit n ti s gip nhit tnh ca thy gio Lu c Dng ngi trc tip hng dn v ton b cc thy trong b mn T ng ho x nghip cng nghip Khoa in trng i hc Bch Khoa H Ni , gip v dy bo em trong nhng nm qua.

CHNG I

GII THIU CHUNG V C QUY

c quy l ngun cung cp nng lng in mt chiu cho cc thit b trong cng nghip v dn dng. C nhiu loi cquy nh:

+ c quy Axit (Acquy ch).

+ c quy Kim ( Acquy St- Niken, , acquy bc- km....)

Tuy nhin, trn thc t thng dng nht t trc n nay vn l acquy axit, v so vi cquy kim, cquy axit c nhiu tnh nng tt hn nh: S ca mi cp cc ln, c in tr trong nh, dung lng ca bnh c quy ln.

1/ Cu to ca bnh c quy ch - axit:

Bnh c quy axit gm v bnh c 6 ngn ring (hay 12V), trong mi ngn c t khi bn cc dng, khi bn cc m. Chng c ngn cch nhau bng cc tm ngn, mi ngn nh vy c coi l mt acquy n, cc acquy n ni tip vi nhau to thnh bnh c quy. Ngn u v ngn cui c u t do to thnh cc u cc dng (+), cc m (-) ca c quy, dung dch in phn (dung dch axit sunfuric) c vo trong tng ngn.

- V bnh:

V bnh c lm bng cc loi nha nh: ebnit, axphapec, cao su cngchng c kh nng chu axit, v c bn vng cao.

- Bn cc:

- Cu to ca mt bn cc trong cquy gm c phn khung xng v cht tc dng cht nn n.

+ Khung xng ca bn cc dng v m c cu to ging nhau. Chng c c bng hp kim ch v Stibi (Sb) v c to hnh dng mt li.

+ Cht tc dng c ch to t bt ch, dung dch axit sunfuric v khong 3% cht n nh mui ca cc axit hu c i vi bn cc m; cn i vi bn dng th cht tc dng c ch to t cc oxit ch Pb3O4, PbO v dung dch axit sunfuric, cht n trong bn cc m c tc dng tng xp, gim kh nng c v hin tng chng ho cng cho cc bn cc, lm tng in dung cho acquy.

Cc bn cc sau khi 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 bng dng in nh. Sau qu trnh nh vy, cht tc dng cc bn cc dng s tr thnh PbO2, bn cc m thnh Pb. Sau chng c lp rp vo bnh acquy to thnh cc khi bn cc, cc khi bn cc m v dng c lp xen k nhau v c cch in bng tm ngn.

- Tm ngn:

Tm ngn l cht cch in, c xp thch hp khng ngn cn dung dch in phn thm n cc bn cc. Tm ngn c lm t polyclovinyl, bng thu tinh ghp vi miplat c tc dng chng chp mch gia cc bn cc dng v m.

-Dung dch in phn:

Dung dch in phn trong c quy axit sunfuric (H2SO4) c pha ch t axit nguyn cht vi nc ct theo nng nht nh.

2/ Yu cu cho vic np c quy:

2.1/ cquy l ngun nng lng c tnh thun nghch: n tch tr nng lng di dng ho nng v gii phng nng lng di dng in nng, Qu trnh cquy cung cp in cho ph ti gi l qu trnh phng in; qu trnh c quy tch tr nng lng t mt ngun in m t chiu gi l qu trnh np in.

2.2/ Qu trnh bin i nng lng trong acquy axit:

a/ Qu trnh np:

Khi np, 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 sc in ng ring ca cquy cc in t s chuyn ng theo hng ngc li (t dng n m) v to thnh dng in phng Ip.

Khi cquy c np no cht tc dng cc bn cc dng 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.

- Qu trnh ho hc xy ra trong acquy axit c th vit nh sau:

+ Trn bn cc dng:

PbO2 + 3H + (HSO4)- + 2e ( PbSO4 + 2H2O

+ Trn bn cc m:

Pb + H2SO4 ( PbSO4 + 2e + 2H

b/ Qu trnh phng:

Khi phng, in axit sunfuric b hp th to thnh sunfurat, 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 axit sunfuric nn nng ca dung dch tng ln.

2.3/ Cc c tnh ca Acquy axit:

a/ Sc in ng ca Acquy axit (S)

S ca acquy ph thuc ch yu vo in th trn cc cc v ph thuc vo nng ca dung dch in phn c xc nh theo cng thc:

E0 = 0,85 + P (v)

Trong : E0 : Sc in ng tnh ca acquy (v)

P : Nng dung dch in phn 150C (g/cm3)

Thng thng, cc acquy axit c nng thay i trong khong 1,12 - 1,29 g/cm3. Do , S ca mt acquy n s thay i t 1,19 ( 2,14v, c ngha l nng ca dung dch in phn tng th S ca acquy cng tng.

b/ Cc c tnh phng v np ca acquy:

* Trong qu trnh phng in ca acquy th sc in ng c tnh theo cng thc:

EP=UP+IP.raq

Trong :

Ep: sc in ng ca acquy phng in (v)

IP: dng in phng (A)

Up: in p o trn cc cc ca acquy khi phng in (v)

raq: in tr ca acquy khi phng in (v)

- Dung lng phng ca acquy l i lng nh gi kh nng cung cp nng lng ca acquy cho ph ti, v c tnh theo cng thc:

CP=IP.tPTrong :CP: dung lng thu c trong qu trnh phng in. (A.h).

IP: dng in phng n nh trong thi gian phng in tP.

-c tnh phng ca acquy l th biu din mi 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.

Trong- khong thi gian phng 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 im 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 tinh th sunfat 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 cquy

c tnh phng ca c quy

sau ny, thi im tgh gi l gii hn phng in cho php ca cquy, cc gi tr EP, UP, P ti tgh gi l cc gi tr ti hn phng in cho acquy.

-Nu ngt mch vo thi im tgh th hiu in th UP s tng vt ln bng sc in ng ca acquy. Cn sau , nh khuych tn m nng dung dch cn bng dn, sc in ng cquy s tng dn ti E0 v bng 1,96 v i vi cquy c coi l phng ht in. on cui ca ng cong Eaq ng vi khong nghi ca cquy hay thi gian phc hi ca cquy, thi gian phc hi ny ph thuc vo ch phng in ca acquy ( dng in phng v thi gian phng.)

( Trong qu trnh np in ca cquy th S c tnh theo cng thc:

En=Un-In.raq

Trong :

In: dng in np (A)

Un: in p o trn cc acquy khi np.

raq: in tr trong ca acquy khi np in.

- Dung lng np ca cquy l i lng nh gi kh nng tch tr nng lng ca acquy v c tnh theo cng thc:

Cn=In.tnTrong Cn: dung lng thu c trong qu trnh np in(Ah)

In: dng in np n nh trong thi gian np in tn.

- c tnh np ca cquy l th biu din mi quan h ph thuc ca sc in ng, in p v nng dung dch in phn theo thi gian np khi tr s dng np in khng thay i (In = const).

Trong khong thi gian np t 0 n t = ts, 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 (gi l hin tng si), lc ny in th gia cc cc ca cquy tng ti 2,4v. Nu vn tip tc np, gi tr ny tng vt ln 2,7v v gi nguyn thi gian ny gi l thi gian np no, n c tc dng lm cho phn cc cht tc dng su trong lng cc bn cc c bin i hon ton, lm cho dung lng phng in ca cquy tng thm. Ta c th kt thc qu trnh np y. Nhng thng ngi ta phi tip tc np t 2h n 3h na, khi thy rng trong sut thi gian hiu in th v nng dung dch ca cquy khng thay i th mi tin chc acquy c np no.

+ Sau khi ngt dng in np th in p, sc in ng ca cquy, nng dung dch in phn gim xung v n nh, khi : E0 = 2,11(2,12v ng vi acquy c np no.

+ Tr s dng in np nh hng rt ln n cht lng v tui th ca cquy. Dng in np nh mc i vi cquy l In = 0,01 C10, trong C10 l in dung nh mc ng vi ch 10 gi phng.

3/ Cc phung php np c quy t ng:

Cc phng php np c quy t ng.C ba phng php np c quy l:

+ Phng php dng in.

+ Phng php in p.

+ Phng php dng p.

3.1 / Phng php np c quy vi dng in khng i.

Np c quy vi dng in khng i

y l phng php np cho php chn c dng np thch hp vi mi loi c quy, bo m cho c quy c no. y l phng php s dng trong cc xng bo dng sa cha np in cho c quy hoc np s cha cho cc c quy b Sunfat ho. Vi phng php ny c quy c mc ni tip nhau v phi tho mn iu kin :

Un ( 2,7.Naq

Trong :

Un - in p np

Naq - s ngn c quy n mc trong mch

Trong qu trnh np sc in ng ca c quy 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 :

Nhc im ca phng php np vi dng in khng i l thi gian np ko di v yu cu cc c quy 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 ( 0,5 )C20 tc l np cng bc v kt thc nc mt khi c qui bt u si. Dng in np nc th hai l 0,05C203.2/ Phng php np c quy vi in p khng i

Phng php np c quy vi in p khng i yu cu cc c quy c mc song song vi ngun np . Hiu in th ca ngun np khng i v c tnh bng 2,3 -> 2,5 V cho 1 ngn c quy 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 lc u s rt ln sau khi tng ln th gim i kh nhanh.

Phng php np vi in p khng i c thi gian ngn dng in np t ng gim dn theo thi gian .Tuy nhin dng phng php ny c quy khng c np no, v vy phng php np vi in p khng i ch l phng php np b xung cho c quy trong qu trnh s dng.

- khc phc nhng nhc im v tn dng c hu ht cc u im ca cc phng php np trn ta kt hp 2 phng php np li thnh phng php np dng - p.

3.3/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 c quy 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 c qui l phng php dng p.

i vi c quy axit: bo m thi gian np cng nh hiu sut np th trong khon thi gian tn ( 16h tng ng vi 75 ( 80 % dung lng c quy ta np vi dng in khng i l In ( 0,1C20. V theo c tnh np ca c quy 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 16h c qui bt u si lc ta chuyn sang np ch n p. Khi thi gian np c 20h th c quy bt u no, ta np b xung thm 2 ( 3h.

i vi c quy kim : Trnh t np cng ging nh c quy axit nhng do kh nng qu ti ca c quy kim ln nn lc n dng ta c th np vi dng np In ( 0,1C20 hoc np cng bc tit kim thi gian vi dng np In ( 0,25C20 .

Cc qu trnh np c quy 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 quy, lc dng np s t t gim v khng.

Kt lun:

-V c quy l ti c tnh cht dung khng km theo sc phn in ng cho nn khi c quy i m ta np theo phng php in p th dng in trong c quy s t ng dng nn khng kim sot c s lm si c quy dn n hng hc nhanh chng. V vy trong vng np chnh ta phi tm cch n nh dng np cho c quy.

- Khi dung lng ca c quy dng ln n 80% lc nu ta c tip tc gi n nh dng np th c quy s si v lm cn nc. Do n giai on ny ta li phi chuyn ch np c quy sang ch n p. Ch n p c gi cho n khi c quy thc s no. Khi in p trn cc bn cc ca c quy bng vi in p np th lc dng np s t ng gim v khng, kt thc qu trnh np.

- Tu theo loi c qui m ta np vi cc dng in np khc nhau

+ c quy axit : - Dng np n nh In ( 0,1C20

- Dng np cng bc In ( ( 0,3 ( 0,5 )C20.

+ c quy kim :- Dng np n nh In ( 0,1C20

- Dng np cng bc In ( 0,25C20 .

* T cc phn tch trn ta tnh ton dng np v in p np theo yu cu u bi chng ta tin hnh np c quy vi dng in khng i.

+ Dng in np In ( 0,1x60 = 6A

T yu cu ca ti chng ta c: s lng c quy l 10 chic. Do vy chng ta c th c 3 cch mc np in cho c quy:

+ Mc 10 c quy ni tip vi nhau.

Dng in np nh In=0,1x60=6 A

in p np li rt ln Un=10x16,2=162 V

+ Mc 10 c quy song song vi nhau:

Dng in np nh In=0,1x60x10=60 A

in p np nh Un=16,2 V

+ Mc hn hp 10 c quy thnh 2 dy song song, mi dy c 5 c qui ni tip vi nhau

Dng in np In=0,1x60x2=12 A

in p np Un=16,2x5=81 V

Nhn xt:

Nh vy chng nu chng ta dng cch mc 10 c qui ni tip vi nhau th dng in np trong qu trnh n dng nh In=6A cn in p np khi np ch n p s rt ln Un=162 V.Phng php ny khng tho mn yu cu ca cng ngh v in p np ln. Cn vi cch mc 10 c quy thnh 10 chic song song vi nhau th dng in np rt ln (In= 60A) cn in p np nh( Un=16,2V). Phng php ny tho mn yu cu ca cng ngh nhng do dng in qu ln nn chng ta phi chn van chu c cng st ln, do vy s khng t c v vn kinh t.T chng ta thy :

Phng php ti u nht va p ng c yu cu ca cng ngh va t c hiu qu kinh t l phng php mc hn hp.

CHNG II

Cc Phng n chnh la

1 : La chn b bin i :

V ngun in li l ngun in xoay chiu nn mun np c dng in cho acquy th ta phi dng b chnh lu bin i dng in xoay chiu thnh dng in 1 chiu cung cp cho ti mt chiu ( y l ngun np in cho acquy).

- Thng s dng 2 loi chnh lu dng in l chnh lu 3 Pha v chnh lu 1 Pha.

1.1/ i vi chnh lu 3 Pha

Gm c chnh lu cu 3 pha i xng,chnh lu cu 3 pha khng i xng v chnh lu 3 pha hnh tia.

Ngi ta thng dng chnh lu 3 pha khi ti cng sut ln v c yu cu k thut cao nh p mch ca in p chnh lu thp, cht lng in p cao.S dng chnh lu 3 pha th khng lm lch pha ngun in li.

1.2/ i vi chnh lu 1 pha

-Gm s : +cu 1 pha i xng

+cu 1 pha khng i xng.

Ngi ta thng s dng s cu 1 pha khi ti cn cung cp c cng sut nh, nn n khng nh hng n ngun in cung cp.

- Trong n ny, em thit k khi ngun chnh lu vi yu cu cung cp cho ti mt in p t 12v ( 165v, dng in 1 chiu t 8(60A. Khi dng in np Inap = IC10 = 0,1 . 60 = 6A

Ta chn bnh acquy loi 12v, vi 6 ngn, in p mi ngn vo khong t 2(2,4v (khi c np no). Khi in p ngun np cho 1 bnh acquy phi tho mn: Un ( 2,7 Naq

( Unap ( 2,7 . 6 = 16,2 v

Un: in p np.

Naq: s ngn acquy n.

( Ta c 2 cch np cho acquy:

Cch 1: Mc lin 10 bnh acquy ni tip nhau.

(Ud = 16,2 . 10 = 162 v ( in p qu ln.

Id = 6A ( dng in qu nh.

Cch ny khng kh thi do dng np qu thp m in p np qu ln.

Cch 2: ta mc hn hp ni tip v song song cc bnh acquy vi nhau, chia lm 2 cp mc song song vi nhau, mi cp gm 5 bnh mc ni tip nhau.

Cch mc hn hp cc bnh c quy

Do Udmax = 16,2 .5 =81 V

Idmax = 6 .2 =12A

( cng sut ca mch : P = U .I = 81 . 12 =972 (w) = 0.972 (Kw)

V theo u bi l thit k mch ngun np c quy t ng ,tc l khi c quy i th phi t ng np y cho c quy, cn khi c quy no th s t ng ngt.Tt c iu ny u phi t ng nn ta chn mch chnh lu c iu khin.

Nh ta tnh trn P =0.972 Kw < 5Kw nn ta chon phung n mch chnh lu cu mt pha, c hai phng n chnh lu :

+ chnh lu c iu khin cu mt pha i xng

+ chnh lu c iu khin cu mt pha khng i xng

2 : Phn tch s chnh lu :

2.1: Chnh lu c iu khin cu 1 pha i xnga. S nguyn l

Trong s cu 1 pha i xng c tiristor : + T1 ,T2 l nhm u katot chung

+ T3 ,T4 l nhm u ant chung

b. Nguyn l hot ng ca s

Trong na chu k u ,khi U2 >E in p anot ca tiristo T1 dng,lc catot ca T2 m,nu c xung iu khin c 2 van T1 , T2 ng thi th cc van ny s c m thng t in p li ln ti T1 , T2 s dn n khi U2 E nu c xung iu khin m tiristo T2 th T2 ,D2 thng cho php dng qua ti.

in p trung bnh t ln ti : Vi

c.Dng in p:

d.Nhn xt u nhc im ca s chnh lu 1 pha

-Vi vic thit k ch to b ngun chnh lu iu khin tiristor cho mch ngun np c quy t ng . Qua vic phn tch cc s chnh lu ta thy s chnh lu cu mt pha khng i xng c nhng u im sau:

+tit kim nng lng hn(h s cos cao hn chnh la iu khin)

+s van iu khin t hn mt na so vi s cu i xng nn gim c gi thnh thit b bin i, s iu khin cng n gin hn, s knh iu khin van gim .V cng gim cc phn t bo v cho tiristor do dng mt na s van l iot

+ S cu i xng v s cu khng i xng c cht lng in p v di iu chnh l nh nhau .

+Th cp mybin p ch c mt cun dy so vi 2 cun dy trong s c im trung tnh nn vic thit k ch to bin p ngun tr nn n gin hn .

KL: Vi nhng u im ca cc mch chnh lu v cc nhn xt trn ta i n chn phng n chnh lu iu khin cu mt pha khng i xng .S ny p ng y yu cu k thut v in p chnh lu , bn cnh chi ph gi thnh cho vic la chn thit b li r hn. H s cos ca s khng i xng cao hn s i xng.

CHNG III

tnh ton v Thit k mch lc

Chn dng np l In=10%=6A(trong 10h)

Chn phng n mc 10 bnh thnh 2 nhnh song song nhau mi nhnh gm 5 bnh ni tip nhau vy:

in p np: Um = 2,7*6*5=81 V (1bnh 12v gm 6 ngn)

dng in np: Im =10%60.2= 12 A

la chn mch chnh lu chi tit cn bit kh nhiu yu cu k thut v cc chi tit c th.Tuy nhin nhng yu cu ny thng kh nm bt c y ,v vy khi thit k ch cn da vo mt s yu cu tt thiu a ra mt vi phng n mch lc cho thit k s b ban u.sau tin hnh thit k cc chi tit khc. Vy ta chn phng n chnh lu cu 1 pha bn iu khin.

Cc yu cu ti thiu:

in p ra ti nh mc :Udm=81 V

Dng in nh mc : Im =12A

Nhit mi trng lm vic: tmt =300 1/ Chn s thit k :

T yu cu ca ti v nhng phn tch trn th ta chn mch chnh lu cu 3 pha bn iu khin dng tiristo.s mch lc nh H 3.1

T s trn ta c Ud = 2.34U2()

Gi s vi gc l nh nht = 300

Ud = 2,34.220() = 480.3V

Vy vi mch chnh lu cu 3 pha bn iu nh trn la ph hp.

th ca mch chnh lu cu 3 pha bn iu ng vi cc gc m c dng nh sau:

S nguyn l mch np v th ca mch chnh lu cu 3 pha bn iu ng vi cc gc m

2. Bo v thit b tuyt i:

-Cc phn t bn dn cng xut c s dng rng di vi nhng u im nh gn nh , lm vic tin cy ,tc ng nhanh hiu xut cao ,d thc hin t

ng ho .. Vic s dng cc phn t ny phi ch n nhng ch s gii hn s dng do nh sn xut quy nh:

+ Gi tr in p ngc ln nht

+ gi tr trung bnh cho php vi dng in

+ Nhit ln nht ca mt ghp

+ Tc tng trng ln nht ca dng in di / dt

+ Tc tng trng ln nht ca in p du/ dt

+ Thi gian kho tof2.1: Bo v khi tc tng trng ca dng in qu ln

Tiristor lm vic trong mch khng c cm khng hay mch xung th thng chiu tng trng dng in t vo anot ca tiristor IA tng qu nhanh ( tc tng dng anot di/ dt )vt qu mt gi tr gii hn th tiristor x b ph hng ( cc lp bn dn b dnh thng do hiu ng knh) .V vy cn hn ch tc tng dng thng dng cun dy li khng kh ( cun khng xoay chiu )mc ni tip vi tiristor nh hnh v:

2.2: Bo v qu in p

nguyn nhn gy qu in p

+nguyn nhn ni ti do s tch t in trong cc lp bn dn khi khoa tiristor bng in p ngc , cc in tch bin i ngc li hnh trnh to ra dng in ngc trong khong thi gian rt ngn gy ra mt sc in ngcm ng rt ln,lm cho lp tip gip Anot- Katot xut hin qu in p .Lm cho tiristor x t m khng cn tn hiu iu khin .

+ Nguyn nhn bn ngoi : nh ct khng ti mt my bin p trn ng dy ,khi cu ch bo v chy , sm st...

Vy hn ch tc tng p ,tnng dng cp in tr R v t in C mc ni tip vi nhau v ni song song vi hai cc anot,katot ca tiristor nh hnh v:

3/ Tnh ton chn van lc

A/ Chn cp in p cho van.

- chn van ta phi da vo ch lm vic nng n nht

M van phi chu.

- Van phi chu in p max khi cc cquy c np no,mi

ngn cquy c in p l 2V. c acqui 12V cn 6 ngn.

np no th in p np cho mi ngn phi l 2,7V

in p np : U = 2,7.6.10 = 162V

Dng in np : In = 10%Im = 10.16 = 160A

Theo bng 1.1 bin in p ngc ln nht t ln van khi lm vic bng:

U2(max) =2,45U2=2,45.220=539V

chn h s d tr v p cho van l 1,7 th in p cn cho van tng ng s l :

U2=1,7Ungmax=1,7.539=916.3V

Vy chn loi van ?????? c ch to vi phm vi cp in p (???????)v vy ta chn van cp in p l ???? tng ng vi in p thc l ??????? V

IV/ TNH TON BO V QU P

=0 th dng qua van l ln nht lc gc dn qua mi van trong s cu l = vy ta c

Kdd h s dng in qua van,l t s gia gi tr hiu dngv gi tr trung bnh ca dng in i qua van

Tra cm nang v T-250 c cc s liu :U0=1,44V,R=112.10-5

Rt=0,220C/W vi V=6m/s,t0pn=1250C

U0 in p ngng ca van [V]

R in tr ng ca van []

RT nhit tr ca van c km theo tn nhit [0C/W]

Tpn nhit cho php qu bn dn p-n [0C]

Kdd h s dng din qua van,l t s gia gi tr hiu dngv gi tr trung bnh ca dng in i qua van

Vy qua phn tch v tnh ton trn ta quyt nh chn mch iu khin v mch ng lc nh h3.14

h3.14 mch iu khin v mch ng lc

CHNG IV

TNH TON THIT K MCH IU KHIN

Nguyn l chung mch iu khin

Trong cc h chnh lu th c 2 h iu kin c bn l h ng b v h khng ng b do h ng b c nhc im l nhy vi nhiu li in cao v c khu ng b lin quan n in p lc , nhng c u im hot ng n nh v d dng thc hin. Ngc li h khng ng b chng nhiu li in tt hn nhng km n nh . hin nay a s cc mch iu kin chnh lu thc hin theo h ng b

I / CC NGUYN TC IU KHIN TRONG H NG B

C hai nguyn tc iu khin.

1/ nguyn tc iu khin ngang

H3.1 l s cu trc v th minh ho ca nguyn tc iu khin ngang

H3.1a l s cu trc iu khin ngang

B : Khu ng b to ra in p hnh sin c gc lch pha c nh so vi in p lc

DF : khu dch pha c nhim v thay i gc pha ca in p ra di tac ng ca in p iu khin Uk.

TX : Xung iu khin to thnh khu to xung vo thi im khi in p dch pha Udf qua im 0.

DX: Nhm to ra cc xung c dng ph hp m chc chn van

chnh lu, thng c s dng xung chm.

KX: khuch i xung khuch i tn hiu gi ti cc iu khin ca van

H3.1b th minh ho ca nguyn tc iu khin ngang

2/ Nguyn tc iu khin dc.

H3.2 l s cu trc v th minh ho ca nguyn tc iu khin ngang

H3.2a l s cu trc iu khin dc

Uta : To ra in p ta c dng c nh (thng dng rng ca)

SS : Khu so snh xc nh im cn bng ca U ta v Uk

H3.2b th minh iu khin dc

3/ Khu ng b

Theo s cu trc mch ny c 2 chc nng:

m bo quan h gc pha c nh vi in p ca van lc nhm xc nh im gc tnh gc iu khin ,mch c tn gi l gc iu khin

Hnh thnh in p c dng ph hp lm xung nhp cho hot ng ca khu in p ta phi sau n

a/ ng pha bng my bin p

H3.3 bin p ng pha cho chnh lu 3 pha

H3.4 th bin p ng pha cho chnh lu 3 pha

Khi cun s cp u cun th cp u Y (kiu /Y) ta s t phm vi iu chnh =01800b/ 11 ng pha bng phn t quang

S dng phn t quang di dng IC chuyn dng cho php thc hin chc nng ng pha m vn m bo cch ly tt v in vi mch lc ng thi trnh ch to bin p ng pha,do lm gim kch thc mch . S hnh 3.4

H3.4 S v th ng pha bng phn t quang

Khi in p lc dng th dng s qua it pht quang LED, n pht sng lm m tranzitor quang , con na chu k m bng ny tng ng s kho.vy in p ra s c xung ch nht mt na chu k m bng ny tng ng s kho.R1 hn ch dng cho LED.

c/ ng pha kt hp vi khuch i thut ton

Mch chnh u hai na chu k c im gia(tia hai pha) dng it D1 v D2. in p chnh lu c a ti ca (+) ca khuch i thut ton OA so snh vi in p ngng Ung ly t bin tr P1,in p ng b s tun theo quan h sau

Udb = A0(U+-U-) = A0(Ucl - Ung)

Do nu Ucl > Ung th Udb dng th Udb = +Ubh nu Ucl < Ung thi Udb m v Udb = - Ung Vy in p ng b c dng nh hnh3.5 H3.5 S v th ng b v OA

4/ Khu to in p ta (Uta):

To in p c dng rng ca c chu k lm vic theo nhp ca

in p ng pha.

4.1/ in p ta dng rng ca

a s cc in p ta trong mch iu khin chnh lu hin thi u dng dng rng ca v n khc phc c nhc im ca dng hnh sin c ngha l n t b nh hng in p v tn s ngun xoay chiu. Tuy nhin nhc im ca n l khng t c quan h tuyn tnh gia in p iu khin v in p chnh lu nn kh khn khi tin hnhqu trnh t ng iu chnh v n nh cc thng s ca mch.C th chia lm 2 loi rng ca chinh

Rng ca phi tuyn

Rng ca tuyn tnh

a/ To rng ca phi tuyn

Dng rng ca phi tuyn c to bng tranzitor v t

H3.6 S v th to in p rng ca phi tuyn hai na chu k

Trong mch nay tranzitor hot ng nh mt cng tc ng ngt theo nhp ca in p ng pha. thi im 12 in p ng pha m in p ny li t vo baz-mite bng tranzitor T lm n ng cc cc colect v mite khng thng vi nhau tng ng nh mt cng tc m v vy t s np.t ngun 1 chiu E theo quy lut:

Uc(t) = E(1- e-1/RC)

Trong khong 23 th tranzitor T m ng nh mt cng tc ng.V vy t s phng

To rng ca tuyn tnh

Nhn xt: Nhc im chung ca cc s to in p rng ca dng tranzitor l s ph thuc kh r vo thi im m v kho cc bng v in p ng pha ,do vy in p rng ca cng t nhiu b bin ng theo in p li in xoay chiu . iu ny lm nh hng ti gc m cng nh phm vi iu chnh . mt khc tuyn tnh ca rng ca cng khng cao.

H3.7a S to rng ca tuyn tnh

H3.7b th to rng ca tuyn tnh

b/ Mch dng khuch i thut ton(OA)

Hin nay cc mch to rng ca s dng OA ngy cng ng dng nhiu hn do khc phc c cc nhc im trn ,v do gi thnh ca OA tng i r.S v th nh h3.7a,b

H3.7a S mch khuch i dng thu ton OA

H3.7b th mch khuch i dng thu ton OA

Rng ca tuyn tnh 1 na chu k .h3.7b l s to rng ca c dng i xung dng hai OA . khuch i thut ton OA1 u theo s so snh c nhim v xc nh im di qua im O ca in p ng pha vi sai s ch vi phn nghn vn v gi l in p ng b Udb. in p u ra ca OA1 c dng xung ch nht ch c hai tri s m v dng bng gi tr bo ho Ubh.Tng OA2 c nhim v to in p rng ca .nguyn l hot ng nh sau:

na chu k khi in p Udb < 0 (A1 bo ho m Udb = -Ubh ) it 3 dn .S dng c im ca OA l in th gia hai ca (+) v (-) ca n bng nhau , ta c in th im (-) ca OA2 bng 0 do im (+) ni vi 0v . lc Uc = Urc.

na chu k in p Udb > 0 (A1 bo ho m Udb = +Ubh ) it 3 kho.Nn dng qua R2 = 0 .Lc ny dng qua t bng dng qua in tr R3 dng in ngc chiu vi dng qua t C .C ngha l t C phng in.

5/ Khu so snh

So snh gia in p ta Uta v in p iu khin Uk, tm thi

im hai in p ny bng nhau ( Uk = Uta) pht xung iu

khin tc l xc nh gc m .

a) so snh dng tranzitor:

s h3.7 l s dng tranzitor

H3.7 s Khu so snh dng tranzitor

Nguyn l hot ng nh sau :

khi Urc < Uk th tranzitor kho v c UBE = Urc Uk 0 .Do in p ra Ura = 0.

b/ so snh dng khuch i thut ton

Trong kiu ny hai in p cn so snh c a ti hai cc khc nhau ca OA .in p ra s tun theo quy lut :

Ura = K0U = K0(U+ - U-), K0 l h s khuch ai ca OA

Tu thuc vo in p ta v in p iu khin a vo ca no m in p ra xut hin sn xung m hoc xung dng . thi im cn bng gia chng

nu in p iu khin a vo ca (+) cn in p ta a vo ca (-) c ngha l U+Uta nh hnh 3.8a th in p ra l:

Ura = K0(U+ - U-) = K0(Uk - Uta)

H3.8 So sanh 2 ca dng khuch i thut toan OA

nhn xt :khuch i thut ton OA l phn t so snh l tng v:

Tng tr vo OA rt ln nn khng gy nh hng n cc in p a vo so snh n c th tch bit hon ton khng chng nh hng ln nhau.

Tng vo OA cng l khuch i vi sai , mt khc tn s nhiu nn h s khuch i rt ln v th so snh rt cao.

Sn xung dc ng nu so vi tn s 50Hz

6./ Khu dng xung ( DX)

Nhm to ra cc xung c dng ph hp m chc chn van

chnh lu, thng c s dng xung chm.. To xung n bng vi mch RC :l vi mch n gin gm 1 t v mt in tr mc nh h4.9.khi in p a t khu so snh Uss mc thp Ubh th t C c np bng ngun m theo ng 0RCOAss(-E)0 n tr s bng Ubh vi du in p hnh 3.9 .Khi Uss chuyn ln mc cao dng Ubh thi im ban u trn in tr R xut hin 1 xung in p c tr s bng tng in p c sn trn t cng thm in p ca khu so snh do chung mc ni tip nhau nn chng l +2Ubh sau t C li np o cui cng li n tr s Ubh nhng ngc du ban u.

H3.9 to xung n bng mch vi phn RC

7/ Khu khch i xung (KX):

Tin hnh khch i xung t mch dng xung a m bo m chc chn tiristo. Khu ny cng thng lm nhim v cch ly gia mch iu khin v mch lc.

khuch i xung ghp trc tip :kiu ghp trc tip cho php ai van dng xung iu khin ti u(nh h3.10 )nhng c nhc im l khng cch ly mch lc v mch iu khin do ch ng dng vi nhng mch chnh lu in ti di 40v.

H3.10 Khuch i xung tc tip

Khuch i xung ghp qua phn t quang :cc phn t quang m ta c th s dng ghp ni gia Khuch i xung v van lc nh h3.11a

H3.11a cc phn t quang

H3.11b S ghp ni khuch i bng ghp phn t quang

Chng c u im ni bt : m bo cch ly gia iu khin v lc v truyn c cc xung c rng tu . hin nay cng nghip ch to c cc phn t opto dng IC rt thun tin cho mch iu khin .Tuy nhin do dng in ti m n chu c ch vi chc mAnn khng cng sut m van lc ,v vy v tr ca n trong mch iu khin phi trc tng khuch i .Lc tng khuch i cui u trc tip nh h3.11b

Khuch i xung ghp bng bin p xung: L phng php ghp thng dng nht hin nay v d dng cch ly mch iu khin v mch lc.Tuy nhin do tnh cht vi phn ca bin p nn khng cho php truyn cc xung rng vi ms .chnh v vy ngi ta phi truyn xung rng di dng xung chm bin p xung hot ng bnh thng. n gin mch ng thi vn m bo h s khuch i dng cn thit ,tng khuch i hay u theo kiu Dalintn

a/ Xung iu khin dng xung n v xung kp

H3.12 l s tt gin cho loi ny .C hai bng T1,T2u chn theo iu kin in p nh nhaul chu c tr s ngun E

H3.12a khuch i xung ghp bin p xung dng n v kp

b/ xung iu khin dng xung chm

H3.12b khuch i xung ghp bin p xung dng xung chm

9/ B iu khin ( BK ):

Khu ny c nhim v nhn cc tn hiu t cng ngh a ti v cc tn hiu phn hi ly t ti v x l theo nhng qui lut iu khin nht nh a ra Uk tc ng n gc iu khin khng ch ngun nng lng ra ti cho ph hp nht. n nh dng in ta phi phn hi m dng in. n nh in p ta phi phn hi m in p.Trong qu trnh np acquy t ng ,s n dng v n p philun c m bo. iu chnh gc m tiristo ta c nhng phng php sau:

a/ iu chnh gc m tiristo theo phng php thng ng tuyn tnh dng tranzitor s nh h3.13

H3.13a iu chnh gc m tiristo theo nguyn tc thng ng tuyn tnh

Nguyn l hot ng nh sau:

Gi s na chu k u a m khi 1 dn.Na chu k sau a m lc 2 dn.in p ti A (UA) c dng nh (h3.14b.)

Khi UR2 UA >0 Tr1 dn Uc = 0 Tr2 kho lc ny t C1 np in in p UD c dng nh (h4.13b).

Khi UR2 UA UD th u ra ca KTT1 c in p =+Ubh do UD c ni vo cng (-)ca KTT1. Udk c ni vo cng (+)ca KT. 3 c nhim v ch ly phm dng ca KTT1 nn in p ti E (UE=+Ubh)khi UF,Ug1,Ug2 c dng nhu trn th 4.13b

H3.13b Dng in p ca iu chnh gc m tiristo theo nguyn tc thng ng tuyn tnh

b/ iu chnh gc m tiristo bmg vi mch 555 H3.13 nguyn tc iu chnh gc m bng vi mch 555

Nguyn tc hot ng nh sau: khi ng pha s bin i in p ng pha Uf ( l in p hnh sin cng tn s vi in p cn chnh lu )thnh mt dy xung hep ln cn im 0 ca in p ng pha ()gi l dy xung nhp) .a dy xung vo chn 2 ca vi mch 555 mc theo s h4.13 th ca ra chn 3 s xut hin 1 dy xung tng ng.Thi gian tn ti ca xung ny = VR.C2 v thay i c nh bin tr VR do thay i c thi im xut hin sn xung.Ly sn xung ca xung ny bin thnh xung m ca tiristo.

Qua nhng phn tch trn ta quyt nh chn phng n thit k mch ng lc v mch iu kin nh s h3.14.Cc mch chnh lu ba pha chnh lu cu khng i xng gm 3 Tiristo thuc 3 pha v vy mch iu khin gm 3 knh mi knh iu khin cho mt van ca mt pha. Hay ni cch khc mch iu khin chnh lu cu ba pha khng i xng gm 3 mch iu khin mt pha ghp li , s nguyn l di y v cho mt pha ,hai pha cn li tng t. th song u ra nh h 3.15

H3.14S ton nguyn l mch ton b mch iu khin

H3.15 th dng in p ca mch ng lc v mch iu khin

CHNG V

THIT K SN PHM

I/ THIT K MCH NG LC V MCH IU KHINQua nhng phn tch trn ta quyt nh chn phng n thit k mch ng lc v mch iu kin nh s h4.1

H5.1 S mch ng lc v mch iu khin.

H 5.2 Biu dng xung ti cc im trn mch iu khin

1 .NGUN CUNG CP CHO MCH IU KHIN:

S khi ca b ngun mt chiu n p

Cc phn t thc hin khi chc nng:

- Khi h p v cch ly dng my bin p

- Khi chnh lu dng chnh lu cu

- Mch lc dng t in c in dung ln.

- Mch n nh in p dng IC n p 78xx 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).

Khi h p v cch ly dng my bin p

Ngun nui n p dng IC n p 7812 ,IC7912:Ngun n p l ngun lun n nh in p ra khi thay i in p vo hoc thay i ti .

S n p dng IC n p 78xx,79xx

H5.3 S n p dng IC n p

Tnh chn cc phn t trn s

- IC 7812 c in p u vo : 17 35V

- Dng in u ra :0 1A

-in p ra E=12V

-IC 7912 c in p u vo : 17 35V

-Dng in u ra : 0 1A

-in p ra E=-12V

- Chn t lc phng C1=C2=1000F, C3=C4=100 F

-Chn t lc nhiu C5=C6=0,1F .

- Chn cc cu chnh lu c I=1A; U=24V(khng c tn nhit)

2/ TNH TON MY BIN P NGUN

- Khi ngun 24V cp cho bin p xung, dng I = 420 mA

# P21 = U21 . I21 = 240,42 = 10 W

- Khi ngun to in p ng pha 0V- 6V 12V, dong I = 500 mA

#P22 = U22 . I22 = 12 .0,5 = 6 W

-Khi ngun 12 cp cho khuych i thut ton ,dng I = 500 mA

# P23 = U23 . I23 = 24. 0,5 = 12 W

- Khi ngun 10V cp cho cch ly quang ,dng I = 1A

# P24 = U24 . I24 = 10 . 1 = 10W

# Cng sut my bin p ngun l

P = 2 .P21 + P22 + 5 .P23 + P24 = 2.10 + 6 + 5.12 + 2.10 =106 W

# Dng s cp bin p ngun

-Tit din li thp mch t

-Ta chn li thp c tit din 25 cm2 , lm bng l thp k thut in dy 0,35 mm , gm cc l hnh ch E v ch I ghp li vi nhau .

+ Theo cng thc kinh nghim ta c no l s vng/vn:

, vi K l h s my bin p { 40 60 }

Ta chn K = 50 , no = 50/ 20,6 = 2,4 vng/v

-S vng dy cun s cp l :

W1 = no . U1 = 2,4 . 220 = 528 vng

S vng dy cc cun th cp l :

+ Cun 24V : W21 = no . U21 = 2,4 .24 = 58 vng

+ Cun 18V : W22 = no . U22 = 2,4 . 18 = 43 vng

+ Cun 6V : W23 = no . U23 = 2,4 . 6 = 14,5 vng

+ Cun 10V : W24 = no . U24 =2,4 .10 = 24 vng

-Chn mt dng in J = 5 A/mm2

-Tit din dy s cp :

-ng knh dy s cp:

- Dng in trong cc cun th cp

+ Cun 24V:

+ Cun 18V:

+ Cun 6V:

+ Cun 10V:

-Tit din dy th cp :

+ Cun 24V:

+ Cun 18V:

+ Cun 6V:

+ Cun 10V:

-ng knh dy th cp l

+ Cun 24V:

+ Cun 18V:

+Cun 6V :

+ Cun 10V:

-Tra s tay Thng s tit din dy trn sch in t cng sut Nguyn Bnh ta chn c dy :

+ dy s cp d1 = 0,35 mm; S = 0,096mm2 ; 0,855 gam/m ; 0,187 ohm/m

+ dy th cp :

d21 =1,04 mm; S21 = 0,874 mm2 ; 7,55 gam/m ; 0,0202 ohm/m

d22 = 1,22 mm; S22 = 1,178 mm2 ; 9,4gam/m ; 0,0163ohm/m

d23 = 2,1 mm ; S23 = 3,494 mm2 ; 30,8 gam/m ; 0,00506 ohm/m

d24 =1,64 mm ; S24 2,112 mm2 ; 18,3 gam/m ; 0,00850 ohm/m

II/ TNH TON KHI IU KHIN

1/ Khu ng b

Ta s dng khuch i thut ton mc theo

s so snh ( hnh 4.4 )

Chn in tr R3, R4 bng 15 k,

chn khuch i thut ton OA loi TL 081

Hnh 5.4 s nguyn l khu to xung ng b

2/ Khi to in p rng ca

Khi to in p rng ca tuyn tnh mt na chu k hnh 4.5.in p trn t C bng in p u ra ca 0A3 Uc2 = UD , in p trn in tr R5 bng Uc b qua st p trn 1. Trong s ny khng khng ch thi gian np cho t nn ta chn R5 E1 / Icmax = 12/ 1,5 = 8 chn R15 = 10 , chn cng sut ca in tr R15 bng 5W .

Kim tra st p trn in tr R15 khi Tr1 dn dng :

UR15 = I1 . R15 = 0,075 . 10 = 0,75 V

vy in p

U1 = E1 - UR15 = 12 0, 75 = 11,25 V

ln hn 8 V nn t yu cu . Bng Tr1 chn loi BC107 c Uce = 45V , Icmax=0,1A,

tra bng c min = 110 . Vy tr s in tr R14 k(Chn R14 = 20 k, chn it 3, 4, 5,6,7 loi 1N4001 c tham s Itb = 1A Ungmax= 400V) Khu phn hi

H5.9 S nguyn l phn hi dng in

Uk = K1 ( Ut Uph ) chn R18 = 10 k suy ra R19 = 500 k.Chn R16 = 5k , R17 = 10k, R20 = R21 = 10 k, OA6 loi A741

III/ LP RP SM PHM

1 Thit k mt ngoi ca t

S H5.10 l s b tr mt ngoi t gm: 1 ng h ampe,3 m bo mt n bo trong qu trnh np hiu dng mt n bo khi np no mt cu tr CC mt cng tc CT v n bo ngun(B ngun).

H5.10 S b tr mt ngoi t

H5.11 S u ni2 /Thit k mch iu khin

Sau khi chn c phng n thit k mch iu khin nh h5.11ta tin hnh thit k mch in.Cc bc lm nh sau Ly tm phit ng ct thnh hnh ch nht c kch thc 15x17 cm

Sau in mch in ln tm pht

em ngm vo dung dich FeCl3.Sau 1 thi gian ta c mch in nh h5.12

H5.12 S mch in

Ly linh kin lp ln mch in theo nh h5.13

H 5.13 S b tr linh kin

TI LIU THAM KHO

1. Gio trnh Trang b in t

Tc gi : Nguyn Vn Cht

2. Ti liu hng dn s dng c quy axit

Ngi bin son : Pham Hong Kin

3. Ti liu hng dn thit k thit b in t cng sut

Tc gi : Trn Vn Thnh

4. Hng dn thit k mch in t cng sut

Tc gi : Phm Quc Hi

5. S tay linh kin in t

Tc gi : K s on Thanh Hu

6. Kh c in kt cu s dng v sa cha

Tc gi : Nguyn Xun Ph T ng

7. in t cng sut l thuyt thit k ng dng

L Vn Doanh (ch bin)

Nguyn Th Cng

Trn Vn Thnh

8. Gio trnh in t cng nghip

Tc gi : V Quang Hi

9. K thut in t

Tc gi : Xun Th

+

-

Cu to bnh c quy

Vu cc

Ngn c quy n

nt

npp

V

bnh

b) Cu to bn cc

Khung xung

bn cc

Cht tc dng

Vu bn cc

a ) phn khi cc bn

cc v tm ngn

Tm ngn

Bn cc dung

Bn cc

m

cc dung

Cc m

c) kt cu bnh c quy

C

P

= I

P

.t

P

Vng phng in cho php

4

0

5

10

1,75

1,95

2,11

I (A)

U (V)

20

12

8

8

t

E

U

P

Khong ngh

E

Khong ngh

1,95V

C

n

= I

n

.t

n

Vng np chnh

5

10

0

10

1

ts

s

20

(2(3) h

Vng np no

t

I (A)

U,E (V)

2,4V

2

2,7V

un

Bt u si 2,4V

2,1V

Eaq

E

c tnh np ca c quy

A

V

_

+

A

.

.

.

.

.

.

.

.

A

_

_

+

+

D

DD

R

R

UN

+

U

T2

U

2

R

L

E

d

d

i

i

i

i

i

T3

T4

T1

12