do an hoan thanh chinh thuc

8/8/2019 Do an Hoan Thanh Chinh Thuc

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http://www.ebook.edu.vn Trang 2

Li ni u

T ng ha pht trin v mang li nhng ng dng v

cng to ln cho s pht trin tt c cc ngnh k thut ca th

gii.Tuy nhin, nc ta n mi c ng dng v pht trin

mnh m trong nhng nm gn y.N gip nc ta pht trin

tin ti tr thnh mt nc Cng nghip ha - Hin i ha. Bi

vy t ng ha c nghin cu tt c cc ngnh k thut trong

trng ni chung v ngnh t ng ha ni ring

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 ,phn ln cc thit b u s dng in li ,Tuy nhin 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,hn na c qui c s dng nhiu trong cng ngh

t,xe my .

Do vy m vic c mt cng ngh np c qui ti u l rt cn

thit v quan trng .Trong n ny, em c giaothit k b np

c qui.Trong qu trnh lm chng em lun c s gip ,ch

bo ht sc tn tnh ca thy PHM QUC HI,nh c thy ch

dn m em hon thnh n mt cch tt nht.Tuy nhin do c hn

ch v mt kin thc nn em khng th trnh khi nhng thiu st

Em xin cm n tt c cc thy c trong ngnh t ng ha cho chng em c lm n y b ch ny ,c bit em xin

chn thnh cm n thy PHM QUC HI ,ngi lun tn tnh

gip v hng dn chng em.Em xin chn thnh cm n

H Ni ngy 20 thng 5 nm 2008

Sinh vin thc hin

Phm vn Luyn


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Chng I

Gii thiu chung v c qui

I. Khi qut chung v cu to v nguyn l lm vic ca c qui

c qui l ngun in ho hot ng trn c s hai in cc c

in th khc nhau,dng tch tr in nng , cung cp dng

mt chiu cho cc thit b in trong cng nghip cng nh trong

dn dng.

C nhiu loi c qui nhng ph bin l hai loi acqui: c qui

axit (c qui ch) v c qui kim.Tuy nhin c qui axt c s dng

rng ri v ph bin hn v so vi c qui kim th c qui axt c:

+ Sc in ng cao(2V),st p trong qu trnh phng nh

+ in tr trong nh

+ Gi thnh ca c qui axit r hn so vi acqui kim.

Trong n ny em c qui axit nghin cu cng ngh v thitk ngun np acqui t ng.

1. Cu to ca c qui axit

Bnh c qui axit gm v bnh, cc bn cc, cc tm ngn v

dung dch in phn.

1.1. V bnh

V bnh c qui axit c ch to bng nha bnit hocanphantpc hay cao su nha cng. Bn trong bnh lt mt lp

chu axit l polyclovinyl tng tui th ca bnh

Pha trong v bnh chia thnh cc ngn ring bit. Mi ngn

c gi l mt ngn c qui n, trong n ny, nghin cu c

qui ch vi in p danh nh l 14V c by ngn c qui n.


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1.2. Bn cc

Bn cc gm ct hnh li c c bng hp kim ch (Pb) -

antimon (Sb) vi t l (87 95)% Pb - (5 13)% Sb v cht tc

dng. 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 phn khi dng in khp b mt

bn cc ,c vu hn ni cc bn cc thnh phn thnh khi bn

cc

Bn cc m cht tc dng c ch to t bt ch, axit sunfuric

c v khong 3% cc mui ca axit hu c ,cc bn cc d

ngcht tc dng c ch to t cc xt ch Pb3O4, PbO v dung dch

axit sunfuric c. Ph gia mui ca axit hu c trong bn cc m

c tc dng tng xp, ci thin c thm su ca dung dch

in phn, gia tng phn ng ho hc trong bn cc

Nhng bn cc cng tn c hn vi nhau to thnh cc khi

bn cc v c hn ni ra ti tiu th. Nu mun tng dung lng

ca c qui th phi tng s tm bn cc mc song song ,mun t ng

in p danh nh ca c qui th tng s tm bn cc mc ni tip.

1.3. tm ngn

cc bn cc m v dng c lp xen k v cch in vi nhau

bi cc tm ngn

Cc tm ngn phi l cht cch in tt ,bn do,xp ,chi axt

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.

1.4. Dung dch in phn

Dung dch in phn l dung dch axit sunfuric (H2SO4) c

pha ch t axit nguyn cht vi nc ct tu thuc vo iu kin

kh hu v vt liu lm tm ngn. Nng dung dch axit sunfuric


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Khi ni hai in cc Pb v PbO2 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 c qui

s xy ra phn ng ngc ca (2.3). Acqui s cung cp dng incho n khi c hai in cc li tr thnh PbSO4 nh ban u

3. cc thng s v c tnh ca c qui

3.1. Sc in ng ca c qui

* Sc in ng ca c qui axit ph thuc vo c tnh l ho

ca vt liu lm cc bn cc , dung dch in phn v

c xcnh bng cng thc thc nghim

E0 = 0,85 + (V).

Trong :

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

: nng dung dch in phn tnh bng vol quy v +150C.

*Sc in ng ca c qui khi phng in

Ep = Up + Ip. raq

Trong :

Ip : Dng in phng (A)

Up: in p o trn cc cc ca c qui khi phng in (A)

raq: in tr trong ca c qui, khi phng in hon ton th raq =

0,02 .

* sc in ng np En ca c qui

En = Un In.raq (V).

Trong :

In : dng in np (A).

Un: in p o trn cc cc ca c qui khi np in (V).

raq : in tr trong ca c qui khi np in. Khi np no th raq =

(0,0015 -0,001) .3.2.Dung lng ca c qui


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I (A) U,E (V)

Vng phng cho php

1,27

tgh

10

51,11

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)

Dung lng ca c qui l i lng nh gi kh nng cung

cp hoc tch tr nng lng ca c qui v c tnh theo cng

thc :

Ci = Ii.ti (Ah).Trong :

Ci dung lng thu c trong qu trnh phng np (Ah).

Ii Dng dn phng np n nh (A) tp(h).

3.3. c tnh phng in ca c qui

- th biu din mi quan h ph thuc ca sc in ng,

in p acqui v nng dung dch in phn theo thi gian phngkhi dng in phng khng thay i.

T th ta c cc nhn xt sau:

- Trong khong thi gian phng t tp =0 ti tp = tgh, sc in

ng, in p v nng dung dch in phn gim dn, tuy nhin

dc ca cc th l khng ln,y l giai on phng n nh

hay thi gian phng in cho php ca c qui

- T thi im tgh tr i,nu tip tc phng in th dc sc

in ng, in p ca acqui gim rt nhanh, mt khc cc tinh th


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10

0

Khong nghVngnp 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 si2,4V

sunfat ch (PbSO4) to thnh trong phn ng s c dng th, rn,

kh ho tan (bin i ho hc)

- Sau khi ngt mch phng mt khong thi gian, cc gi tr sc

in ng, in p v nng dung dch in phn ca c qui litng ln, y l thi gian hi phc hay khong ngh ca c qui

thi gian phc hi ny ph thuc vo ch phng in ca c qui

3.4. c tnh np ca c qui

- 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.

T th c 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 t

ng si ), lc ny in th gia cc cc ca acqui n tngti gi tr 2,4 V, tip tc np gi tr ny nhanh chng tng ti 2,7 V


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v gi nguyn,thi gian np ny gi l thi gian np no v thng

ko di t 2-3 h, lm tng thm dung lng phng in ca acqui.

Trong qu trnh sc in ng v nng dung dch in phn

l khng thay i..- Sau khi ngt mch np, in p, sc in ng v nng

dung dch in phn gim xung v n nh. y l khong ngh

ca c qui sau khi np.

- Dng in np nh mc i vi c qui qui nh bng 0,5.C20

(0,1.C10).

II. Cc phng php np in cho acqui

1. Phng php np vi dng np khng i

- Phng php np in vi dng np khng i cho php chn

dng in np thch hp , m bo cho acqui c np no

Cc c qui c mc ni tip vi nhau v phi tho mn

Un 2,7 Naq.

Trong :

Un: in p np (V).

Naq: S ngn c qui n mc trong mch np .

- Khi np sc in ng ca acqui tng dn, duy tr dng

in np khng i ta phi b tr trong mch np bin tr R vi tr

sIn

NaqUnR

0,2= .

- Nh

c im: thi gian np ko di- khc phc : s dng phng php np cng bc theo 2

nc. Dng n np nc th nht chn bng (0,3 - 0,5).C10, v khi

c qui bt u si th np nc th hai bng 0,1.C10.

2. Phng php np vi p khng i

- Phng php np p, c qui c mc song song vi ngun

np. Hiu in th cho mi ngn n c gi n nh v c gi tr

t 2,3 - 2,5 V vi chnh xc n 3%


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- Dng npRaq

EaqUnIn

= lc u s rt ln sau khi Eaq tng

dn ln th In gim i kh nhanh.

- u im: thi gian np ngn, dng in np t ng gim dntheo thi gian.

- Nhc: c qui khng c np no, v vy phng php np ny

ch dng 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.

3. Phng php np dng - p

- Ban u ta np vi dng np khng i In = 0,5.C10. Khi thy

c qui "si" th hiu in th gia cc cc ca ca cqui n 2,4V,

tip tc np th gi tr ny nhanh chng tng ti gi tr l 2,7 V.

Sau chuyn sang ch np n p vi gi tr in p np khng

i cho 1 ngn n l Un = 2,7Vv thng 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 ta chn phng php np dng - p

np cho c qui v b ngun np c qui t ng phi p ng nhng

yu cu sau:

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

In = 0,5 .C10/1 ngn c qui n.

- Khi pht hin thy hiu in th trn cc cc ca c qui n

tng ti 2,7 V th t ng chuyn t np n dng sang ch np

n p vi in p np t trc Un = 2,7V/ 1 ngn c qui n.

- Np n p cho ti khi dng in np tin v khng.


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chng II

La chn 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.

* bi : thit k b ngun np c quy c th np cho c quy

16-28V v dng np 10- 100A.

- V yu cu cn iu khin dng ,p nn ta chn phng n

chnh lu tiristor.- V ti yu cu cng sut nh v cht lng in p iu

chnh khng cn cao nn ta chn phng n chnh lu mt pha

nhm n gin ho vic thit k,gim gi thnh v tng linh

ng ca b chnh lu.

IICc 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 phi c hai cun dy th cp vi

thng s ging ht nhau , mi na chu k khi c xung ti iu

khin m tiristo c mt van dn cho dng in chy qua .

in p p mch trong c hai na chu k vi tn s p mch

bng hai ln tn s in p xoay chiu . Hnh dng cc ng cong

in p v dng in ti (Ud,Id ) cho trn hnh v .


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Trong na chu k u , khi U2>E th in p UAK ca T1dng, UAK ca T2 m.Vi vy T1 s dn nu c pht xung iukhin dng s chy qua T1-R-E, vi ngun l U2

Trong na chu k sau, khi U '2 > E th in p UAKca T2 dng,

ca T1 m, T2 s dn nu c pht xung iu khin, dng s chy

qua T2-R-E, vi ngun l U '2

Ch : Nu ta pht xung vo thi im U


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R

E

T2

T3T1

T4

U2

I2=I '2 =Ihdv=

d

R

EU22 )

sin2(

2

1

- Tr s dng hiu dng qua ti:

Ihd =

d

R

EU22 )

sin2(

1

Ta thy Ihdv=2

hdI

- in p ngc t ln van:

U ngcvan =2 2 U2

* Nhn xt : trong s ny , dng in chy qua van

khng qu ln . Khi van dn ,in p ri trn van nh.Vic iu

khin cc van bn dn y tng i n gin .Tuy vy ,vic ch

to bin p c hai cun dy th cp ging nhau , m mi cun ch

lm vic trong na chu k phc tp v hiu sut s dng bin p

xu hn , mt khc in p ngc ca cc van bn dn 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 k u , lc U2 > E in p anot ca tiristo T1

dng katot ca T2 m , nu c xung iu khin c hai van T1 ,T2


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E

R

T1

D1T2

D2

U2

ng thi ,th cc van ny s c m t in p li ln ti , T1 ,

T2 s dn n khi U2 < E.

Trong na chu k sau , khi U2


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- na chu k dng ca u2 khi


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Nhn xt: Mch iu khin s cu mt pha khng i xng

d iu khin,vic ch to bin p n gin.Tuy nhin tng st p

trn van l ln khng ph hp cho ti c p thp v n lm gim

hiu sut s dng b np

Kt lun:

T nhng phn tch trn ta chn s ''Chnh lu hnh tia hai

pha c im gia ''.v n c nhng u im sau

+ S lng van t nn mch iu khin s n gin

+ Tng st p trn cc van nh,hiu sut s dng thit bcao hn.Mt khc thit b gn nh .linh ng v gi thnh h


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N1

N2

N3

BAL

T1

T2

R C

R C

LRdAq

U1

Chng III

thit k v tnh ton mch lc

I.S mch lc

II.Cc phn t trn s mch lc :

1.Van lc:

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

m van phi chu.

*Ch tiu in p :- Van phi chu in p max khi cc acqui c np no,mi

ngn acqui c in p l 2V. c acqui 28V cn 14 ngn.

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

dU =2,7.28/2 =37,8 (V)

Mt khc c :Ud=0,9U2U2= 42,6 V

in p ngc ln nht trn van :

maxngU =2 2 .U2=119 V

in p ngc nh mc van l

Unmv=1,1.kdt.119= 235 V

Vi 1,1 l do thc t in p li khng n nh v c

php dao ng nn p li c th tng ln 10%

Kdt l h s d tr cho van. chn : Kdt =1,8

*Ch tiu dng in :


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- Dng in trung bnh qua van

Itbv=Id/2=100/2= 50 A

Do cng sut ti thp chn ch lm mt cho van t nhindng cnh tn nhit chun.

Itbv=(0,2-0.3)Imv Idmv=167 A

Vy iu kin chon van l Uv > 235 V v Iv >167A

***Chn loai van ST180S04P1V c cc thng s sau:

in p ngc Un cc i :400V

Dng lm vic nh mc ,dng in nh: Im

= 200A,Ipik

=5000A

Dng v p iu khin l Ig=150 mA,Ug=3V

Dng r Ir=30mA,st p U=1,8V,dU

dt=500V/s,tcm=100s

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 ,ngn mch tiristor, ngn mch u ra ca b bin i, ngnmch th cp my bin p.t Iatm=1,2Imv

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

Khi c s chuyn mch, cc in tch 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.V vy ta mc mch R C

mc song song vi thyristor nhm to ra vng phng in trong

qu trnh chuyn mch nn bo v c thyristor khng b qu

in p.

Mt khc mc RC cn lm gim tc tng p ca

Thyristor v khi bin thin in p vt qu du/dt cho php ca van


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th van s dn m khng cn dng iu khin . Theo kinh nghim

chn C=0,1 F , R=45 .

c) Hn ch tc tng dng

- 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

phi hn ch n bng cch mc ni vi ti mt cun cm

Nhng do s dng ngun bin p cho chnh lu nn in cm

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

3.Cun lc mt chiu

S p mch ca in p chnh lu lm cho dng in ti cnglm p mch theo .lm xu i cht lng in p mt chiu

Khi cng iu chnh su th cht lng p v dng cng xu,v

vy ta a cun lc nng cao cht lng in p

Gc iu khin max (iu khin su nht)

0,9.U2cosmax=Imin.Rtd max=84,3

H s p mch vo

Kmv=2

max

2

2 ( . ) 1

1

dm

dm

m tg

m

+

vi mm=2 km=13,8

Chn kmr=1,38 th ksb=10

L=2 1

.

sb

dm

k

m

=0,016H

5.Tnh ton my bin p

a) Tnh cc thng s c bn

*in p chnh lu khng ti :

Udo = Ud + UV + Uba + Udn

Trong :

Ud= 37,8 V - in p chnh lu

UV = 1,8 V - St p trn cc van

Uba =8% Ud = 3V -St p bn trong my bin p khi c ti .


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

Vy : Udo = 37,8+1,8+3=42,6V.

* Cng sut ti ti a:

Pdmax

= Udo

. Id

= 42,6.100= 4,26 kW

* Cng sut my bin p :

Sba = kP. Pdmax = 1,48.4,26 = 6,3k W

Vi s tia hai pha : kP=1,48

b) Tnh s b mch t

Tit din s b tr :

QFe = kQ.fm

Sba

.

trong

kQ l h s ph thuc phng thc lm mt

Vi my bin p du ta ly kQ = 5

m:s pha ca my bin p : m =2

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

Qfe=5.6300

2.50

=40cm2

c) Tnh ton dy qun

- in p cun dy s cp : U1 =220 V

- in p cun dy th cp : U2 =42,6 V

- H s my bin p : kba =2

1

U

U=220/42,6 = 5,16

* 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 t 1- 1,8 Tesla).

QFe- Tit din li thp(m2).

Chn thp lm my bin p l loi c m hiu l 330 dy0,5mm c B=1,1T.


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S vng dy cun s cp my bin p.

W1 =225 vng.

Svng dy cun th cp my bin p.

W2

= 44 vng.

*Dng in cc cun dy :

Dng th cp : I2 = k2 . Id = 0,58.100 = 58A

Dng s cp : I1 = I2/ kba = 11.24A

*Tit din dy dn :

Dy dn bng ng, chn J1 = J2 = 3(A/mm2)

Tit din dy qun s cp my bin p :

S1

=1

1

J

I= 11,24/3 = 3,75mm 2.

ng knh dy dn s cp

d 1=1

4S

=2,2 mm

Tit din dy qun th cp ca my bin p

S2 =2

2

J

I=58/3 = 19,33mm2.

ng knh dy dn th cpd2 =

S.4= 4,96mm.


22/43


Chng IV

Mch iu khin

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

1. Mc ch v yu cu chung 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.

H iu khin s pht xung m hai tiristor T1,T2,Cc tiristor sm khi tho mn ng thi hai iu kin:

UAK >o, I >0

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.

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

+ 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 tr in p v tn s.

+ C kh nng chng nhiu cng nghip tt, tc ng nhanh

+m bo xung iu khin pht ti cc van ph hp m chc

chn cc van .Xung phi tha mn

cng sut (v in p v dng iu khin ).

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

ctrng thi dn .


23/43


2. Cu trc mch iu khin:

2.1Cc h iu khin chnh lu:

C hai h iu khin l h ng b v h khng ng b .

+ H ng b : trong h ny gc iu khin m van lun

c xc nh xut pht t mt thi im c nh ca in p lc

.V vy trong mch iu khin phi c 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 m van

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 nhng phi thc hin iu khin theo mch vng kn ,

khng th thc hin vi mch h.

2.2Nguyn tc iu khin:

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

(dch pha)v thng ng arccos(dc).

a) Nguyn tc iu khin thng ng tuyn tnh(dch pha)

- in p ng b (urc), ng b vi in p dt trn cc A - K

ca tiristor, thng t vo u o ca khu so snh.

0

us

2

Ucm

Usm

t


24/43


- in p iu khin (udk) - in p mt chiu iu chnh c

bin , thng t vo u khng o ca khu so snh .

hiu in th u vo ca khu so snh l:

Ud= udk urcMi khi udk=urc th khu so snh lt trng thi, to ra mt xung

iu khin.

Nh vy, lm bin i udk c th iu chnh c thi im xut

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

Gia v ucm c quan h:

=dkU

Um

Vi Udkmax=Um

b) Nguyn tc iu khin thng ng arcos(dc)

Theo nguyn tc ny dng b to in p rng ca v b so

snh .Tn hiu ng b Udk s ng b ha qu trnh lm vic ca

b phat xung rng ca

- Ti thi im Udk=Urc b so snh s lt trng thi .Nuurc= Um.cos t,chn thi im t=0 l thi im chuyn mch

t nhin th khi t= ta c

Umcos=Uk =arccosdk

m

U

U

Khi uk = Um th = 0.

Khi udk

= 0 th =2

.

Khi udk = - Um th = .

Nh vy, khi iu chnh udk t +Um n tr udk = -U m, ta iu

chnh c gc m t 0 n .


25/43


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 b iu khin:

Trong :

Ung: in p ngun

Uk: in p iu khin

1. Khu ng pha ( F ):

Ud

Uph

Uk

UngF Uta SS DX KK

B K

t

Ucm

0

2

usuAK


26/43


To in p trng pha vi in p th cp bin p mch lc.

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 rng ca c chu k lm vic theo nhp ca

in p ng pha.

3. Khu so snh( SS ):

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 m chc chn van

chnh lu, thng c s dng xung chm.

5. Khu khch i xung (KX):

Tin hnh khch i xung t mch dng xung a m bo m

chc chn tiristor. Khu ny cng thng lm nhim v cch ly

gia mch iu khin v mch lc.

6. 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 acqui t ng ,s n dng v n p phi

lun c m bo

III.Xy dng mch iu khin :

1. Khu ng pha :

a) S v nguyn l


27/43


N1

N2

N3

-

++

VR1 E

E

-EU1

OA1

R0

D1

D2

U2

U'2

R1

R3

Udf

Uo

Udb

T

Time (s)

0.00 10.00m 20.00m 30.00m 40.00m 50.00m

VF1

-20.00

20.00

- in p ng pha c so snh vi in p trn bin trVR1. Ti thi im Udf=U0 th i du ca in p ra khuch i

thut ton.- in p ti ca m:

Uo=. 1

1 3

EVR

VR R+

in p ca dng bng udf .in p vo bng:

Ura=K0.(u+-u-)=K0.(udf-u0)Khi udf> u0 th in p ra Ura=Ubh=E-1.5 VKhi uA< u

- th in p ra Ura=-Ubh=-( E-1.5)VKt qu ta c chui xung ch nht khng i xng.

b)Tnh ton

- in p ng pha qua it 1,2 c dng in p mt

chiu na hnh sin . chn in p xoay chiu ng pha UA=12(V)

in tr R1 c dng hn ch dng vo KTT.


28/43


-

++

VR2

E+E

-ER5

R4

C

Udz

Udf

OA2

D3

Urc

T

Time (s)

0.00 10.00m 20.00m 30.00m 40.00m

U2

1.00

10.00

Chn R2=R1=10(K )

Chn gc duy tr v kho nng lng l 5o th in p t vo ca

dng ca b so snh l:

Ud= 2 Usin5o= 2 *12*sin5o=1.48(V)

. 1

1 3

EVR

VR R+=1,48R3=7VR1

Chn R1=35(K ) , VR1 loi 100 K ,t gi tr 5 K

Chn Khuch i thut ton l loi TL084 c:

Ngun cung cp E= 12V

Nhit lm vic : t=-25 850C

Cng sut tiu th: P=680 mW

Tng tr u vo : Rin=106 M

Dng in ra : Ira=30pA

2. Khu to in p rng caa) S v nguyn l

- in p ng b c a vo ca o ca khu to in prng ca.

- Khi Udp


29/43


- in p trn t C c np tng tuyn tnh n tr s ngng

ca it n p DZ v gi in p ra tr s ny.

- na chu k sau, khi Udb>0 th 3 kho nn dng qua 3bng 0,t phng in,in p trn t C (in p ra) gim tuyn

tnh. Khi in p gim n khng ri m th t DZ dn nh it

bnh thng gi cho in p gi tr 0.

b) Tnh ton

*Khi Udp0 (Udp=+Ubh) th 3 kho , t C phng in

- Dng phng in : Ip=2 4

E

VR R+


30/43


T

Time (s)

0.00 25.00m 50.00m 75.00m 100.00m

VF1

-20.00

20.00

-

++

E

-E

Udk

Urc

Uss

R6

OA3

- in p trn t C gim dn theo thi gian:

uc(t)=UC(0)+1

Cci dt =Udz-

1

C ci dt =10- 2 4

E

VR R+.t.

1

C

- thi gian phng in t=tp =9,5ms,ti thi im tp in p

trn t bng bng 010-2 4

E

VR R+1

C.tp =0 VR2+R4=57K

Chn R4=10(K ) , VR2 loi 100 (K ),t gi tr 50 K

3. Khu so snh:

a) S v nguyn l

So snh in p ta v in p iu khin, im cn bng ca

hai in p ny l thi im m tiristor.

th in p Uss

Khi Udk>Urc th in p ra ca khu so snh l Ura=+Ubh=E-1,5 V

Khi Udk


31/43


-

++

E

-E

R7

R8

R9

OA3

C1

Udx

T

Time (s)

0.00 5.00m 10.00m 15.00m 20.00m

Voltage(V)

-10.00

0.00

10.00

20.00

a) S v nguyn l

- Khu to xung chm l mch dao ng hot ng ch tdao ng.in p trn t C1 c so snh vi in p Vp

Vp=7

7 8

.raU R

R R+= 7

7 8

1,5.

ER

R R

+

- T c np t gi tr Vp-n Vp+ v phng t gi tr Vp+ n

Vp- vi hng s phng np =R9 .C1

- Thi gian phng tp v thi gian np tn bng nhau,chu k phngnp T=tp +tn ,nu chn R7 =R8 th chu k phng np

T=2,2.R9.C1

b,tnh ton

chn R7=R8 =1k ,


32/43


N1 N2

&

+

D4

D5

D6

R10

R11

R12

C2

C3 R13

T1

T2

EUdf

Uss

Udx

- chn tn s lm vic ca mch dao ng l 10 kHz,C 1 =20nF

th ta c T=10-4=2,2. 20.10-9.R9R9 =2.5 k , chn R9 =3 k

kt qu ta c dy xung hnh ch nht

5,khu tch xung ,bin p xung v khuch i xung

a. s v nguyn l

* sau khu dng xung, ta c hai xung iu khin trong mt

chu k in p xoay chiu,iu ny l khng mong mun.v vy ta

s dng khu tch xung xc nh chu k dng (m) pht xung

iu khin cho Tirirtor khi in p trn n l dng( UAK >0)

- Ta ly in p t khu ng pha ,sau khu so snh v in p

sau khu to xung vo chn cng logic AND,cc in p dng s

c gi tr 1 logic.Cng AND s cho xung iu khin khi in p

trn tiristor l dng,in p ny c a vo khu khuch i

xung

*Khu khuch i xung

- Khi khuch i xung gm hai tranzitor T1,T2 mc theo kiu

Darlington

- D4 bo v Tranzitor khi c dng in ng

c


33/43


T

Time (s)

0.00 10.00m 20.00m 30.00m 40.00m

U3

-20.00

20.00

U4

-10.00

20.00

U5

-1.00

7.00

U6

0.00

4.00

- D6 ngn chn xung p m c th c khi van kha

-D5 hn ch qu p trn cc cc colector v emito ca tranzitor

T1,T2

- R11 to mt st p 0,4 v,iu khin T2 lc dng ra ln v

chuyn t m sang kha nhanh hn

- R12 hn ch dng ngn mch ca T2

- R13 va C3 dng lc

-C2 np in khi cha c xung iu khin ,khi c xung iu

Khin ,C2 phng in cung cp in cho bin p xung

th p nhn c sau AND

b.tnh ton

Ug=3v,Ig=150mAdng v p th cp bin p xung l :

I2=150mA

U2=3+0,6=3,6v (0,6 l st p trn D6 khi n dn)

Chn my bin p c t s bin i k=2,suy ra in p v dng

s cp bin p xung

I1=75mA, U1=3,6.2=7,2 v


34/43


R12=(E-7,2)/0,075=64

Chn R12= 70

Chn tranzitor T1 loi ST 603 c

Uce=30 v,Ice=800 mA ,= 30-100

Chn tranzitor T2 loi C828 c

Uce= 30v,Ice=300mA, = 30-100

R11= 32

ce

ce

U

I= 3

3

3

ce

ce

U

I

=0.4.30/75=160 ,chn R11 bng 165

Diode D4,D5,D6 chn loi :D-1001 c

Imax =1A,Ung =200v, U=0,6 v

6.Ngun cung cp cho mch iu khin :

a) 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 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).

S n p dng IC n p

UrUVUdU2U1 H p cchly

Chnhlu

Mchlc

Mch n nhinp


35/43


Tnh chn cc phn t trn s

- IC 7812 c in p u vo : 7 35VDng in u ra :0 1A

in p ra E=12V

IC 7912 c in p u vo : 7 35V

Dng in u ra : 0 1A

in p ra E=-12V

- Chn t lc phng C1=C2=1000F, C3=C4=100 FChn t lc nhiu C5=C6=0,1F .

- Chn cc cu chnh lu c I=1A; U=40V(khng c tn nhit)

* chn my bin p

* Chn my bin p mt pha c mt cun s cp v nhiu cun

th cp

- in p s cp 220v

-Cun dy ng pha :ly in p ng pha l U=14v

-Cun th cp bin p cung cp in p cho chnh lu ca b

to ngun nui,in p th cp l :U=40 v

-Cng sut my bin p bng tng cng sut tiu th khu

ng pha,khu my bin p xung,khu to ngun nuI, 8 IC

TL084 v 2 cng AND


36/43


-

++

-

++

-

++

-

++

1

1

1

Iph

Up

E

Uphu

VR5

VR6E

VR4

R20

R19

CM2

CM1

NOT

Uphi

SUN

-E

R28

R30

R24

R26

R27R23

R22

R16

R17

R18OA5

OA6

R21

R29

OA8

E

-E

E

-E

E

-E

OA7

R25

KHAU PHAN HOI

Khau on dong

khau on ap

Khau chuyen mach

Udk

S=Sdf+SIC +SSND +Sbax

=1,48.0,9.14.1+8.0,68+2.2,5.10-9 +2.3.0,15=25 w

-Chn hiu sut ca mach to ngun nui cho mch iu khin l

75% v vy chn cng sut my p l

Sba = 25/0,75=34 VA,suy ra Iscp=0,15A

7.Khu phn hi

7.1) S v nguyn tc hot ng

- Cc tn hiu phn hi dng UphI v p UphU c ly t mchlc ri a v cc khu phn hi to ra Uk iu khin gc m nhm n nh cc gi tr dng hoc p t trc

Do p ,dng ra ti ( dng, p np)l lin tc nn ta c

Ud = 0,9U2 cos

Id =d

t

U E

R

= 2

0,9. .cos

t

U E

R

- Do vy thc hin n dng v p theo nguyn tc sau:


37/43


38/43


- c qui c np no th in p t cho mi ngn n l

2,7v.V vy in p np khi l

Ud = Un =2,7.28/2= 37.8 V

0,9 .U2 cos =37,8 =9,63Uk =Urc . /180 =0,54 V

- OA6 l mch cng o ,chon R24=R25=R26=5k

Suy ra Uk = -Uphu -UtVR6

Chn Uphu =6 VUp.23

22 23

R

R R+=6 vi (Up=Ud=Un)

R22=5,3 R23 chn R23= 5k ,R22=27k

UtVR6= -6,54 V

Suy ra E.6

6 27

VR

VR R+=6,54VR6=1,2 R27

Chn R27= 1k ,VR6 loi 20k t gi tr 12k

*Vy t gi tr in p ban u ti VR6 l UtVR6 =-6,54 V

c, Khu chuyn mch

- Ban u c qui c mc vo mch np th dng np tng v

in p c qui tng dn ln, tc l dng phn hi v p phn hi

tng dn ln. Lc ny do p phn hi nh hn UVR5(p t VR5 )

nn p u ra OA8 UraOA8 =-E ( mc thp), do chuyn mch

CM2 ngt cc ng phn hi p ra khi mch. ng thi do c

cng NO nn chuyn mch CM1 ng ng phn hi dng vi

mch thc hin qu trnh n nh dng.- Khi p phn hi UphU bng UVR5 th UraOA8 =E do CM2

ng cn CM1 ngt nn mch thc hin qu trnh n p.

t UtOA5 =6 V

Suy ra E.5

5 28

VR

VR R+=6V VR5=R28

Chn VR5 loi 5k ,t gi tr 4k ,chn R28 =4k

Chn R29=R30 =k


39/43


-

++

R32

R33

R31 VR7

OA9

E

-E

Uphu

RH

T3

D7

D8E

E

Vy din p t VR5 l 6 V

8.Khu bo v qu in p

Nguyn l lm vic ca mch

- Khu bo v qu p s ngng qu trnh np cho c qui khiin p np cho c qui ln hn in p np nh mc 10%

- in p phn hi Uphu ly t in tr R23,c so snh viin p UtVR7 t VR7

- Khi in p np tng qu gi tr nh mc 10% th in p

Uphu cng tng tng ng 10%Nn ta t UtVR7 =1,1.6= 6,6 V

- Khi UphuUtVR7 (tc l Un110%Unm ) u ra ca OA9 ldng (+) t ln bazo ca T3 ,lm T3thng .Dng in t ngun Equa rle RH qua T3 v t , lm rle tc ng ,tip im thngng c m lm cho mch iu khin tc ng ,lm c qui cct khi mch lc

Khi Uphu


40/43


T

Time (s)

0.00 10.00m 20.00m 30.00m 40.00m

Axis label

-50.00

50.00

U1-20.00

20.00

U2

1.00

10.00

U3

-20.00

20.00

U4

-10.00

20.00

U5

-1.00

7.00

U6

0.00

4.00

U7

-300.00m

900.00m

S hot ng ca mch iu khin


41/43


-+

+

+

N1

N2

N3

- +

+

-+

+

-+

+

&

N1

N2

N1

N2

N3

N1

N2

N3

N1

N2

&

- +

+

- +

+

- +

+

- +

+

1

tt

-

+

- +

+

VR2

E

+E

-E

R4

+E

-E

E

OA3

D4

D5

D6

R10

R11

R12

C2

C3

R13

T1

T2

E

+E

-E

R3 O

A1

OA2

DZ

R5

D3

D1

D2

OA4

R1

VR1

Udf

U0

R15

Z2

E-E

C1

C

R9

R0

R6

R7

R8

U4

Udb

Uss

Urc

Udk

D7

D8

Uand

Udx

BA

Ug

BAL

T1

T2

R

C

R

C

U1

Rd

Aq

D4

D5

D6

R10

R11

R12

C2

C3

R13

T1

T2

E

Uand

Ug

E

Uphu

VR5

VR6

E

VR4

R20

R19

CM2

CM1

NOT

Uphi

SUN

-E

R28

R30

R24

R26

R27

R23

R22

R16

R17

R18

OA5

OA6

R21

R29

OA8

E

-E

E

-E

E

-EOA7

R25

R15

Z2

R32

R33

R31

VR7

OA9

E-E

Uphu

RH

T3

D7

E

E

D8


42/43


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

Tnh ton thit k thit b in t cng sut Trn Vn Thnh

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

Dng vn Nghi

K thut mch in t Phm Minh H

Cc ti liu v c quy .

H Ni , ngy 20 thng 5 nm 2008

Sinh vin thc hin

Phm vn Luyn


43/43

do an hoan thanh chinh thuc

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