00_thiet_ke_bo_nap_acquy_tu_dong_on_dong_va_on_ap_2612

8/7/2019 00_thiet_ke_bo_nap_acquy_tu_dong_on_dong_va_on_ap_2612

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n TCS Thit k b np c quy tng

SV thc hin: Khoa Tun - TH1_K48 Trang 1

TI 8THIT K B NP C QUY TNG

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 ngunnp :

Um = 24 -50 V

Im = 60 A

Imin = 40 A .


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LI NI U

Nc ta hin nay ang trn con ng Cng nghip ha - Hin i ha.

Bi vy tng ha ang pht trin mnh trong nhng nm gn y.Tng

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 tng 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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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 shai in cc c in th khc

nhau, n cung cp dng in mt chiu cho cc thit bin 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 nhintrong 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 trtrong 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 tng.

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 l p lt chu axit l


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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 cv

c thc 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 h p 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 trca

cht tc dng (xit ch PbO2) ln hn rt nhiu so vi in trca ch nguyn

cht, do cng tng chiu dy ca ct th in trtrong 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 cmt 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 ci 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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sunfuric c. Ph gia mui ca axit hu ctrong 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, dykhong 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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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 trv sc in ng ca acqui:

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 nhrng: 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

Eaq

in trdung dchi n hn

V/ngn /cm3

1.0 1.1 1.2 1.3 1.4 1.5 1.6

0

5

1

3

2

4

0.0

2.5

0.5

1.5

1.0


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nh mc ca acqui gim v cc nc x lnh th dung dch vo ma ng d

bng 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 mhoc 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.

Lc 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 n p m mch ngoi cc in t "e"

chuyn ng t cc bn cc dng n cc bn cc m - l dng in np In.


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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 trthnh 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+

( 2.1)

- cc m:

PbSO4 + 2e + 2H+

= Pb + H2SO4 (2.2)-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 trthnh 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. Sdngi 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 bnhacqui ta ch cn kho st mt bnh acqui n l .

3.1. Sc in ng ca acqui axit:


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* 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 + (V).

Trong :

E0: Sc in ng tnh ca acqui n, tnh bng vol.

: nng dung dch in phn khng ly theo n v g/cm3 m

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 trtrong 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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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 (Ah).

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) U,E (V)

Vng phng cho php

1,27

tgh

10

5

1,11

2,11

1,95

Khong ngh

1,0

1,5

2,0

0,5

Eaq Eo Up E Ip.raq

1,75

Cp=Ip.tp

tp (h)

(g/cm3)


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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 tri, 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 licho 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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T thc tnh np ta c nhn xt sau:

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

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 tiptc 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.

10

0

Khong nghVng

np no(23)h

Eaq Eo Un In.raq E

2,7V

2,11V

I (A) U,E (V)

Vng np hiu dng

1,27

ts

5

1,11

1,95

1,0

1,5

2,0

0,5

Cn

=In

.tn

tn (h)

(g/cm3)

Bt u si


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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 n p, sc in ng ca acqui tng dn, duy tr dng

in np khng i ta phi b tr trong mch np bin trR. Tr s gii hn ca

bin trc xc nh theo cng thc:

In

NaqUnR

0,2= .

Nhc im ca phng php np vi dng np khng i l thi gian np

ko di v yu cu cc acqui a vo n p phi c cng cdung 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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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 gin nh vi chnh xc n

3% v c theo di bng vol k.

Dng npRaq

EaqUnIn

= lc u s rt ln sau khi Eaq tng dn ln th In

gim i kh nhanh.

Phng php np vi in p np khng thay i c thi gian np ngn,

dng in np tng 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 tng 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 knhng 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 tng m ta thit k cn phi p ng nhng yu cu sau:

- Ban u tng np n dng vi dng np t trc

In = 0,05 .C20/1ngn cqui n.


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- Khi pht hin thy hiu in th trn cc cc ca acqui n tng ti 2,7 V th

tng 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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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 kthut 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 mtiristo c mtvan dn cho dng in chy qua .

in p p mch trong c hai na chu k vi tn sp mch bng hai

ln tn sin p xoay chiu . Hnh dng cc ng cong in p v dng in

ti (Ud,Id ) cho trn hnh v .


17/57



Trong na chu ku , 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 thT1 s dn.Dng s chy qua T1-R-E, vi ngun l U2


18/57



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 c p xung vo thi im U


19/57



ca cc van bn dn fi chu c tr s rt ln.Thch hp vi mch chnh lu in

p thp nhng dng ln khng cn cht lng in p cao.

2. Chnh lu cu mt pha c iu khin i xng:

Trong na chu ku , lc U2 > E in p anod ca tiristo T1 dng lc

catod ca T2 m , nu c xung iu khin c hai van T1 ,T2 ng thi ,th

cc van ny sc m thng t in p li ln ti , T1 , T2 s dn n

khi U2 < E.

Trong na chu k sau , khi U2 > E , in p anod ca tiristo T3 dng lc catod ca T4 m , nu c xung iu khin c hai van T3 ,T4 ng thi ,thcc van ny sc mthng t in p li ln ti.

(vi iu kin 21


20/57



E

R

T1

D1T2

D2

U2

Dng hiu dng chy qua van :IhdV=2

hdI

in p ngc ln nht t ln van : 2max 2UUn =

* Nhn xt : So vi s trn ,s ny in p ngc ln nht t ln

van ch bng mt na,bin p d ch to v c hiu sut cao hn . Tuy nhin ,

s ny nhiu khi gp kh khn trong vic mcc van iu khin, nht l khi

cng sut xung khng ln .

3 . Chnh lu cu mt pha c iu khin khng i xng(thng hng) :

- na chu k dng ca u2 khi hay m cho xung iu

khin mT1 th T1v c D1u khng mc do trong mch c sc in ngE lm cho th UAKca tiristor m.

Khi


21/57



Gc dn dng ca it v ca tiristor trong s ny bng nhau v:

TD = = 2

V nguyn tc, c th thay i c trong khong (0;) nhng do

s c mt ca sc in ng E ca ti nn gc m c khng ch trong

khong ( ; ).

- Tr trung bnh ca in p trn ti:

Ud = ( )

dU

sin..21

2 + )(

+E

= )()]cos([cos2 2

++EU

- Tr trung bnh ca dng qua ti :

Id =

R

EUd= ])[(

.)]cos([cos

.

.2 2

++

R

E

R

U

- Tr trung bnh ca dng qua tiristor v it:

IT = ID =

dId.2

1

= )]([2

+dI

-Tr hiu dng dng qua van v dit:

Ihdv=2

hdI


22/57



T cc phn tch v cc strn , ta chn s chnh lu cu khng

i xng v so vi s cu i xng mc d di iu chnh v cht lng in

p chnh lu l nh nhau , nhng cu mt pha khng i xng ch s dng mt

na s van l tiristor, na cn li l it. T m gim c gi thnh thit b

bin i bi v it r hn rt nhiu so vi tiristor v siu khin cng tr

nn n gin hn do cn t knh iu khin. V vy ta chn mch cu khng i

xng np cho c quy.

Kt lun:

S la chn l : S ''Chnh lu cu mt pha khng i xng''.


23/57



Rs

Rf

R CR C

D1D2

T1T2

AT

220V50Hz

AQ

CHNG III

THIT K V TNH TON MCH LC

I.S mch lc :

II.Cc phn ttrn s mch lc :

1. Van lc:

chn van ta phi da vo ch lm vic nngn nht m van phi

chu.

Ch tiu in p :


24/57



- Van phi chu in p nng n khi cc acqui c np no:

Mi ngn acqui c in p l 2V. c acqui 50V ta cn2

50=25 ngn.

np no th cn in p np cho mi ngn l 2,7V. Khi :

dU =2,72

50=67,5 (V)

in p ngc ln nht trn van :

maxngU = 2 .U2

vi sdd kUU =2 cho s chnh lu cu mt pha ks = 0,9 thay vo ta c:

Ungmax = 2 .9,0

5,67=106,1 V

Do thc t in p li khng n nh v c php dao ng ,mt khc

c nhiu yu tnh hng ngu nhin trn mng in nn van c chn vi

mt h s d trin p nht nh:

max. nguv UKU >

vi Ku l h s d tr cho van. Ta chn : Ku =1,7

maxngU = 106,1.1,7 = 180,4(V).

Ch tiu dng in :

- Tnh dng in ca van

Dng in trung bnh thc t qua van:

AI

I dtbv 302

60

2===

Thc t phi chn van chu c h s qu dng KI = 1,2:

AIKI tbvIV 362,1.30. ===

Trong s ny, ch lm vic ca tiristor v it l ging nhau nn iu

kin chn van ging nhau.

V ti c cng sut nh nn ta chn iu kin lm mt cho van l lm mt

t nhin, dng cnh tn nhit chun vi i lu khng kh.

Vy iu kin chn van:


25/57



VUng 4,180max

AIV 36

La chn van : Diode : Loi C40-020R

Imax= 40A

Ungmax= 200V

U =1,1V

TCP= 2000C

Thyristor : Loi T10-40 do Lin X ch tocpI =40A

maxngU =200V

dkI =150mA

dkU =4V

U=1,75V

du/dt=100(V/s)

di/dt=40(A/ s)

2.Cc thit b bo v:

a) Bo v ngn mch, qu ti:S dng Aptmat (AT) ng ct mch lc, bo v khi qu ti v

ngn mch tiristor, ngn mch u ra ca b bin i, ngn mch th cp my

bin p.

b) Bo v qu p,tc tng in p cho van :

Bo v qu in p do qu trnh ng ct cc tiristorc thc hin

bng cch mc R C song song vi thyristor. Khi c s chuyn mch, cc in

tch t trong cc lp bn dn phng ra ngoi to dng in ngc trong khong

thi gian ngn. S bin thin nhanh chng ca dng in ngc gy ra sc in

ng cm ng rt ln trong cc in cm lm cho qu in p gia ant v katt

ca thyristor. Khi c mch R C mc song song vi thyristor n to ra vng


26/57



phng in trong qu trnh chuyn mch nn bo vc thyristor khng b

qu in p.

Nu tc bin thin in p vt qu du/dt cho php ca van th van

s dn m khng cn dng iu khin.Do ta phi mc thm R-C song song

vi thyristor , n s lm gim tc tng in p trn thyristor.Ta phi b tr

sao cho Thyristor phi nm st C. in trR c tc dng hn dng phng ca

t khi van dn.

Theo tnh ton kinh nghim ta chn C=0,3F , R=70 .

c) Hn ch tc tng dng :

V vi ti l c quy khng c tnh cm nn tc tng dng c th rt

ln c th gy hin tng t nng cc b trong van v vy ta phi c bin phphn ch n.

Bin php n gin nht l mc ni tip vi ti mt cun cm.

Tuy nhin v ta s dng ngun bin p cho chnh lu nn in cm trong

cun dy my bin p cng m bo iu kin trn.

3. Cc thit b ch th :

Ampe ko dng np: chn loi ampe k 100 A.

Vol ko in p np: chn loi vol k 100 V.

4. in trly tn hiu:

Rs: ly tn hiu phn hi dng v mch iu khin.

Tn hiu phn hi p ta ni trc tip vo hai u ca c quy.


27/57



5.Tnh ton my bin p :

a) Tnh cc thng scbn :

1. in p chnh lu khng ti :

Udo = Ud + UV + Uba + Udn

Trong :

Ud= 67,5 V - in p chnh luUV = 1,1 +1,75 =2,85 V - St p trn cc van

Uba =10% Ud = 6,75 V -St p bn trong my bin p khi c ti .

Udn 0 -St p trn dy dn (coi rt nh).

Vy : Udo = 67,5+2,85+6,75 =77,1 V.

2. Cng sut ti ti a:

Pdmax = Udo. Id = 77,1.60 = 4626 W

3. Cng sut my bin p :

Sba = kP. Pdmax = 1,23.4626 = 5690 W

Vi s cu mt pha : kP =1,23.

b) Tnh sb mch t(xc nh kch thc bn mch t):

Tit din sb tr :

U1 ,I1

Ud,Id

U2, I2

BA


28/57



QFe = kQ.fm

Sba

.

trong

kQ l h s ph thuc phng thc lm mt

Vi my bin p du ta ly kQ = 5m:s pha ca my bin p : m =1

f: l tn s dng in xoay chiu (y tn s l f =50Hz).

T chng ta c :

QFe =5.50.1

5690= 53,34 cm

2.

c) Tnh ton dy qun:

- in p cun dy scp : U1 =220 V

- in p cun dy th cp : U2 =ku

Udo=

9,0

1,77=85,67 V

vi s cu mt pha : ku = 0,9

- H s my bin p : kba =2

1

U

U=

67,85

220= 2,57

S vng dy mi pha my bin p :

Ta c cng thc :

W =BQf

U

Fe ...44,4vng.

trong

W -S vng dy ca cun dy cn tnh.

U - in p ca cun dy cn tnh (V).

B - T cm (thng chn trong khong t 1 1,8

Tesla).

QFe- Tit din li thp(m2).

Ta chn thp lm my bin p l loi c m hiu l 330 dy 0,5mm tta c B=1,1T.

S vng dy cun scp my bin p.

W1 =170 vng.

Svng dy cun th cp my bin p.

W2 = 66 vng.


29/57



Dng in cc cun dy :

Dng th cp : I2 = k2 . Id = 1,11 . 60 = 66,6 A

Dng sc p : I1 = I2 / kba = 25,9 A

Tit din dy dn :

Chn sb mt dng in trong my bin p:Vi my bin p du v dy dn bng ng, chn J1 = J2 = 3(A/mm

2)

Tit din dy qun scp my bin p :

S1 =1

1

J

I=

3

9,25= 8,633 mm

2.

Tit din dy qun th cp ca my bin p :

S2 =2

2

J

I=

3

6,66= 22,2 mm

2.

ng knh dy dn :Do dy dn c tit din nh nn y chng ta chn dy dn trn.

ng knh ca dy dn th cp l :

d2 =

S.4=

14,3

2,22.4= 5,3 mm.

ng knh ca dy dn scp l :

d1 =

S.4

= 14,3

633,8.4

= 3,3 mm.


30/57



CHNG IV

MCH IU KHIN

I. Yu cu chung v cu trc mch iu khin :

1. Mc ch v yu cu chung vi mch iu khin:

* Mch iu khin l khu rt quan trng trong b bin i tiristor, n c vai

tr quyt nh n cht lng, tin cy ca b bin i. Mch iu khin rt

a dng nhng vi h thng mch lc c th ca mch np cn c mt hiu

khin thch ng. Vi mch ny, hiu khin s pht xung m hai tiristorT1,T2.

Cc tiristor s mkhi tho mn ng thi hai iu kin:

- Mt in p dng ln t ln hai cc ca tiristor theo hng t ant n

katt.

- Xung in p dng a vo cc iu khin ln v bin , rng.

lm thay i in p ra ti ch cn thay i thi im pht xung iu

khin, tc l thay i gc m ca cc van. u im ca tiristor l ch cn

dng v p iu khin nh nhng c th chu c p v dng rt ln chy qua.

* Mch iu khin phi thc hin cc nhim v chnh sau:

+ Pht xung iu khin (xung m van) n cc van lc theo ng

phng php iu khin cn thit.

+ m bo phm vi iu chnh gc iu khin min-max tng ng vi

phm vi thay i in p ra ca mch lc.

+ C i xng iu khin tt , khng vt qu 10-30 in ,tc l gc

iu khin vi mi van khng c qua lch gi tr trn .

+ m bo mch hot ng n nh v tin cy khi li in xoay chiu dao

ng c v gi trin p v tn s.


31/57



+ Cho php b chnh lu lm vic bnh thng vi cc ch khc nhau

do ti yu cu nh ch khi ng ,ch nghch lu , ch dng in lin

tc hay gin on , ch hm hay o chiu

+ C kh nng chng nhiu cn nghip tt .

+ tc ng ca mch iu khin nhanh ,di 1ms.

+m bo xung iu khin pht ti cc van ph hp mchc ch n cc

van ,c ngha l phi tho mn cc yu cu :

cng sut (vin p v dng iu khin ).

C sn dc ng mvan chinh xc vo thi im

quy nh ,thng tc tng p iu khin phi t

10V/us ,tc tng dng iu khin t 0,1A/us . rng xung iu khin cho dng qua van vt tr s

dng in duy tr Idt ca n , khi ngt xung van vn gi

c trng thi dn .

C dng ph hp vi s chnh lu v tnh cht ti.

+ Ngoi ra h thng iu khin phi c nhim vn nh dng in ra ti

v bo v h thng khi xy ra s c qu dng hay ngn mch ti.

2. Cu trc mch iu khin:

Cc hiu khin chnh lu:

C hai hiu khin cbn l hng b v h khng ng b .

+ Hng b : trong h ny gc iu khin mvan lun c xc nh

xut pht t mt thi im cnh ca in p lc .V vy trong mch iu

khin phi c mt khu thc hin nhim v ny gi l khu ng pha m

bo mch iu khin hot ng theo nhp ca in p lc .

+ H khng ng b : trong h ny gc iu khin mvan khng c

xc nh theo in p lc m c tnh da vo trng thi ca ti chnh lu v

vo gc iu khin ca ln pht xung mvan ngay trc y .Do , mch

iu khin ny khng cn khu ng pha ,tuy nhin b chnh lu hot ng


32/57



bnh thng bt buc phi thc hin iu khin theo mch vng kn , khng th

thc hin vi mch h.

Nguyn tc iu khin:

iu chnh gc mca cc tiristor trong na chu k in p dng ta

thng dng hai nguyn tc iu khin: thng ng tuyn tnh v thng ng

arccos.

a) Nguyn tc iu khin thng ng tuyn tnh:

Theo nguyn tc ny ngi ta dng hai in p :

- in p ng b (us), ng b vi in p dt trn cc A - K ca tiristor,thng t vo u o ca khu so snh.

- in p iu khin (ucm) - in p mt chiu c thiu chnh c bin ,

thng t vo u khng o ca khu so snh .

By gihiu in thu vo ca khu so snh l:

Ud= ucm - us

Mi khi ucm=us th khu so snh lt trng thi, ta nhn c "sn xung"

ca in p u ra ca khu so snh. "Sn xung" ny thng qua a hi mt

trng thi n nh to ra mt xung iu khin.

Nh vy, bng cch lm bin i ucm ngi ta c thiu chnh c thi

im xut hin xung ra, tc l iu chnh c gc mca tiristor.

Gia v ucm c quan h:

0

us

2

Ucm

Usm

t


33/57



sm

cm

U

u =

Ngi ta ly Ucmmax=Usm

b) Nguyn tc iu khin thng ng "arccos":

Theo nguyn tc ny ngi ta dng hai in p :

- in p ng b (us),vt trc uAK=Um.sin t ca tiristor mt gc l2

:

us= Um.cos t

- in p iu khin (ucm) - in p mt chiu c thiu chnh c bin

(theo hai chiu dng v m)

Nu t us vo cng o v ucm vo cng khng o ca khu so snh th

khi us=ucm ta s nhn c mt xung rt mnh u ra ca khu so snh khi

khu ny lt trng thi:

Um.cos = ucm .

Do : = arccos

m

cm

U

u.

Khi ucm = Um th = 0.

Khi ucm = 0 th =2

.

Khi ucm = - Um th = .

t

Ucm

02

us uAK


34/57



Nh vy, khi iu chnh ucm t tr ucm = +Umn tr ucm = -U m, ta c th

iu chnh c gc m t 0 n .

Nguyn tc iu khin ny c s dng trong cc thit b chnh lu i

hi cht lng cao.

II. S khi v chc nng:

Da vo nguyn tc iu khin v yu cu ca cng ngh ta thit lp c

s khi ca biu khin:

Trong :

Ung: in p ngun

Uk: in p iu khin

1. Khu ng pha ( F ):

C nhim v to in p trng pha vi in p th cp bin p mch lc.

Khu ny c chc nng xc nh im gc tnh gc iu khin . V vy n

c gc pha lin h cht ch vi in p mch lc. Thng thng khu ng pha

cn lm nhim v cch ly gia mch lc in p cao vi mch iu khin in

p thp.

2. Khu to in p ta (Uta):

To in p c dng cnh ( tam gic, rng ca, cosin ) c chu k lm

vic theo nhp ca in p ng pha.

Ud

Uph

Uk

UngF Uta SS DX KK

B K


35/57



3. Khu so snh( SS ):

Nhn tn hiu in p ta(Uta)v in p iu khin(Uk)v tinhnh 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.

4. Khu dng xung ( DX):

Nhm to ra cc xung c dng ph hp mchc chn van chnh lu.

mi ch lm vic cc xung ny c khi ng nhmch so snh, thng

c s dng xung chm.

5. Khu khch i xung (KX):

Tin hnh khch i xung t mch dng xung a ln sao cho c cng sut

( U, I ) mchc chn tiristor. Khu ny cng thng lm nhim v cchly gia mch iu khin v mch lc.

Trong trng hp mch lc chy in p thp th chng ta c th b cch ly.

6. Biu khin ( BK ):

Khu ny c nhim v nhn cc tn hiu t cng ngha ti v cc tn hiu

phn hi ly t ti v x l theo nhng qui lut iu khin nht nh quyt

nh a ra Uk tc ng n gc iu khin khng ch ngun nng lng ra ti

cho ph hp nht.

Trong n ny p ng nhng yu cu iu khin, ta s dng "l

thuyt iu khin theo lch" n nh dng in v n p trong tng giai

on np ca qu trnh np acqui tng. 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 n p acqui tng sn dng v n p c thc hin

theo s sau:

In - Uph - Uk- - Ud - In.

Un - Uph - Uk - - Ud - Un.


36/57



III.Xy dng mch iu khin :

1. Khu ng pha :

a) S v nguyn l :

u2'

in p ng pha c so snh vi in p trn bin trVR1. Ti thiim UA=UVR1 th i du ca in p ra khuch i thut ton.

in p ti ca m:

2

3

*1

RVRR

Eu

+=

in p ra ca dng bng uA .in p ra bng:

Ura=K0*(u+-u

-)=K0*(uA-u

-)

Khi uA> 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

UB

Uref

12

D2

GND

-

+

OA1Uv Udp

+E

GND

R1

R3

D1

R2

VR1


37/57



b)Tnh ton :

in p sau khi tu ra ca bin p ng pha qua it 1,2c dng

in p mt chiu na hnh sin . chn in p xoay chiu ng pha UA=9(V)

in trR2,R1c dng hn ch dng vo KTT. Thng chn R2,R1

sao cho dng vo KTT nh hn 1(mA) do : R2> )(900010

93

==

v

A

I

U

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*sin5

o=1.48(V)

Ta c : 48.12

3

=+

RRVR

E

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 tru vo : Rin=106

M Dng in ra : Ira=30pA


38/57



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

in p ca b pht xung ch nht c a vo ca o ca khu toin p rng ca.

Khi Udp0 th 3 kho nn dng qua 3 bng 0 lc ny dng

qua t C1 bng dng qua in trR4 , dng ny ngc chiu vi dng qua t C1

na trc ngha l t C1 phng in do in p trn t C1 cng nhin pra gim tuyn tnh. Khi in p gim n khng ri m th t DZ1 dn theo

ch nhit bnh thng gi cho in p gi tr 0.

b) Tnh ton :

Khi Udp


39/57


40/57



ta c : 0 = UOA-( ) p

tCVRR

E

+4

( ) ( ) 36

44 10*912

2.6*10*22.0

=+

=+

=VRR

E

CUVRRt OAp

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 mtiristor. Ta so snh in p tav 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 trvo 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:

Udk

R7

-

+

OA3

R6

Urc

Uss


41/57



Khi Udk>Urc th in p ra ca khu so snh l Ura=+Ubh

Khi Udk


42/57



Khu dng xung :

y l khu nhm to ra dng xung ph hp tha mn yu cu hot

ng ca mch lc .

Ta s dng to xung n bng mch vi phn RC .

Khi Uss = -Ubh : t C c np bng ngun m theo ng : 0RCUss

.

Khi Uss = +Ubh : s xut hin xung in p trn R c gi tr bng in p c

sn trn t cng vi in p u ra ca So snh. Do tng s l 2Ubh. Sau t

C bt u qu trnh np o cui cng li n tr s Ubh nhng ngc du ban

u .

in p trn t :

).21.()( t

bhc eUtu

= ,vi CR.=

in p u ra mch vi phn chnh l in p trn in trR:

t

bh

t

bhbhcss eUeUUuuu

=== ..2).21.(

suy ra dng in c quy lut :

t

bh eR

Uti

= ..

.2)(

Nh vy in p suy gim theo hm m vi hng s thi gian ,do sau

thi gian khong 3 th c th cho rng in p ra v khng.Vy rng

xung n to ra theo phng php ny l : tx=3


43/57



Khu khuch i xung :

y l khu khuch i cng sut xung t khu dng xung a n m

chc van , cch ly mch lc v mch iu khin . Ta s dng KX dng bin

p xung.

Khu tch xung :

Sau khu to dng xung ta nhn c 2 xung iu khin do trong mt chu

k in p xoay chiu mi van s nhn c 2 xung iu khin c hai na chu

k. Vic pht xung iu khin cho van khi in p trn van m l c thc

,song khng mong mun. Ta s s dng Khu tch xung xc nh c

chu k dng (m) s pht xung cho Thyristor no.Lc van lc nhn xung

iu khin chgiai on in p trn n l dng.in p U'2c ly t khu ng pha: khi U'2 > 0 qua KTT cho in p ra

dng v chn cng AND vi logic 1 , kt hp vi Udxc a vo chn kia

ca cng AND s cho xung ch c khi in p trn thyristor dng .Sau chn

ra cng AND a vo khu Khuch i xung.

b) Tnh ton :

Khu khuch i xung :

Thyristor c : IG = 0,15 A v UG = 4V.

My bin p xung c t s cc cun dy l k=2. in p v dng in cun

sc p : U1 = UG . k = 8V

I1 = Ic = IG/k = 0,075 A

Chn E = 12 V


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C hai van T1 v T2u chn theo iu kin in p nh nhau l chic tr

s ngun Ecs.

V dng in , bng T2 chn theo dng in qua cun s cp ca bin p

xung:

IT2=I1=0,075 A

Vy chn bng T2 loi BD135 C tham s UCE=45V; ICmax=1,5A; min=40

Dng qua colector ca T1 chnh l dng qua bazT2

IT1=1,5/40=0,0375A

Chn T1 loi BC107 c UCE=45V; ICmax=0,1A;min=110

= kIs

ER CS 3,29

5,1.2,1

12.110.40

.

.

max1

21

11

chn R11=30 k

Sau khi chn c cc phn t ca mch khuch i xung c th tnh ton

cc phn t ca mch to xung vi s liu cn thit nh sau :

rng xung tx=2.tm=2.45=90 s

Khu dng xung :

Dng qua t : t

bh e

R

Uti

= ..

.2)(

Dng xung nhn vi gi trnh : Imax = 2Ubh/R

Chn gi trnh khng qu 8 mA .in p bo ha : Ubh = E-1,5 = 10,5 V

Vy ta c : R10 > 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


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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 hkhng kh.

* Tnh th tch li thp cn c :

2

20 ...... B

IUstlQV

xxtb

==

Trong : tb - t thm trung bnh

tb=H

B

.0

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 scp 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 scp bin p xung:


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I1 = I2 /m =0,15/2=0,075(A)

+ t thm trung bnh tng i ca li st:

tb =B/0 . H = 36 10.830.10.25,13,0

=

(H/m)

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= 3362

663

6,010.6,03,0

075,0.8.1,0.10.90.10.25,1.10.8cmm ==

Chn li hnh tr k hiu 1811 c V=1,12 cm 3 , ng knh ngoI

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

tch ta c kch thc mch t nh sau:

a = 3,5 mm

Q = 0,443 cm2

= 44,3 mm2

d = 11 mm

D = 18 mm

+ S vng qun dy scp bin p xung:


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w1 = U1 tx / B.Q = =

4

6

10.443,0.3,0

10.90.854 ( vng )

+ S vng dy th cp :

w2 = w1 / m = 54/2 = 27 (vng )

Chn mt dng in : j1 =6 ( A/mm2) , j2 = 4 (A/mm

2)

+ Tit din dy qun th cp:

s1 = I1 /J1 = 0,075/6 = 0,0125 (mm

2).

+ ng knh dy qun scp :

d1 =

14s = 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 =

24s = 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 sn

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

tin 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 bu 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 bin t .


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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:

Cc phn tthc 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 tin ( 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).

Sn p dng IC n p

Tnh chn cc phn ttrn s:

- UA 7812 c in p u vo : 7 35V

UrUVUUU H p

cch ly

Chnhlu

Mchlc

Mch nnh in


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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 1Ain 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

ttrong mch iu khin:

Chn my bin p mt pha c mt cun scp 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 l20/ 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 c p 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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+ 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 th8 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 nuiPN=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. 2,35.2,1=baS = 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 scp:

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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- Dng in trong cun dy scp 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:

Scp: S1 = I1/ J = 0,16/ 3 = 0.053 (mm2) .

- ng knh dy qun scp:

d1 = 26,014,3

053,0.441 ==

s(mm).

ng knh cc cun th cp ta chn bng 0,26 mm .

S nguyn l:

7.Khu phn hi:

a) S nguyn l:

m

A Tr.t

GND a'asrvun


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b) Nguyn tc hot ng:

Cc tn hiu phn hi dng UphI v p UphUc ly t mch lc ri a v

cc khu phn hi to ra Ukiu khin gc m nhm n nh cc gi tr

dng hoc p t trc theo nguyn tc:

IUphIUkUclI

IUphIUkUclI.

UnUphUUkUclUn

UnUphUUkUclUn.

c) Tnh chn cc phn ttrn s:

Cc b khuch i thut ton ta s dng IC LM348. S ni cc chn

nh hnh v.

-Khu phn hi dngin:

Theo nh trnh by trn, dng in phn hi c ly trn Rsun

, ta chn

Rsun loi 50A/60mV.

in p ri trn Rsunng vi gi tr dng Id = 60A l :

UphI =50

60.60 = 72 mV = 0.072

V.


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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 kin p ln nht trn C8 l 1V nn khng thdn qua it c d l phn cc thun)

Ta c :

UC8=Ubh.(1- 816CR

t

e

) + UC8(0)

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

8161CR

t

e

=10

1= 0,1

816CR

t = 0,105

tc p ng mt cch hp l th ta chn thi gian t=10s

R16.C8= 100105,0

10

Chn C8=1000F R16=100K

Tip theo l b khuch i o:

U2=-(1819

28

R

U

R

U VRC + ).R17

Ta nhn thy vi mch lc nh trn v ti l ngun E nn van mchc

lc cm ti vo (I=0) th Udk=-10V

Udk=U2=-UVR2.18

17

R

R=-10V


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Chnh UVR2=1V, R17=20K ; R18=2K

R19 = R18 =2K;

iu chnh chit p VR6 ta siu chnh c dng vo ti.

- Khu phn hiin p:

Ta ly UphUhai 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 , thayi 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

CM1ng 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 CM2ng 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 trin p chuyn mch

tng ng vi in p np.


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CC TI LIU THAM KHO

Ti liu Tc gi

in t cng sut V Minh Chnh, Phm Quc Hi

Trn Trng Minh

in t cng sut Nguyn Bnh

Hng dn thit k mch in t cng sut Phm Quc Hi

Tnh ton thit k thit bin t cng sut Trn Vn Thnh

Phn tch v gii mch in t cng sut Phm Quc Hi, Dng Vn

Nghi

Kthut mch in t Phm Minh H

Cc ti liu vc quy .

H Ni , ngy , thng , nm

Sinh vin thc hin

Khoa Tun


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Mc lc

Chng I : Gii thiu chung vc quy 3 - 14

Chng II : Phng n chnh lu 15 - 21

Chng III : Thit k v tnh ton mch lc 22 - 28

Chng IV : Mch iu khin 29 - 53

00_thiet_ke_bo_nap_acquy_tu_dong_on_dong_va_on_ap_2612

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