do an hoan thanh chinh thuc
TRANSCRIPT
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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.
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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 .
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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
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- 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 .
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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
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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
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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.
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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
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- 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+
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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
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-
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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 ,
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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
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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
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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
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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
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-
++
-
++
-
++
-
++
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:
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- 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
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-
++
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
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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
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-+
+
+
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
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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 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
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8/8/2019 Do an Hoan Thanh Chinh Thuc
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