dong co 1 chieu kich tu doc lap (do an)

8/8/2019 Dong Co 1 Chieu Kich Tu Doc Lap (Do an)

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n mn hc [email protected]

Li ni u

Vic s dng ng c in trong sn xut v i sng l rt rng ri ,c bit l ng c in chiu bi v ng c in 1 chiu c rt nhiu uim so vi ng c xoay chiu . Nhng gn lin vi vic s dng ng cin 1 chiu l qu trnh iu chnh, o chiu tc ca ng c sao cho

ph hp vi yu cu ca thc t

L 1 sinh vin ca khoa in v b mn T ng Ho c trang bvi nhng kin thc v nhiu mn hc trong c mn in t cng sut,qua cc bi ging ca cc thy c v qu trnh tm hiu em hon thnhbn n ny.

y l mng ti kh rng, vi khi lng cng vic ln v mi mi vi chng em cho nn em gp mt s kh khn trong qa trnh thitk, song c s hng dn tn tnh ca cc thy gio, c gio trong b mn,

c bit l thy gioPhm Quc Hiv s gip nhit tnh ca ccbn cng lp, nn em hon thnh bn n ny. Em xin chn thnh cmn s gip qu bu . Tuy nhin do hn ch v thi gian cng nh v trnh ca bn thn, nn khng trnh khi cn nhiu ch thiu st, em rtmong c s ch bo ca cc thy c gio v cc bn em hon thin hn

bn n ny.

Trong bn n ny em xin trnh by mt s ni dung sau :

Chng 1 : Gii thiu Chung v ng c in mt chiu.Chng 2 : Cc phng n tng th.Chng 3:Phn tch s hot ng ca mch thit k.Chng 4 : Thit k v tnh ton mch lcChng 5 : Tnh ton v thit k mch iu khin.

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Chng IGii thiu Chung v ng c in mt chiu

I.ng c in mt chiu

1. Tm quan trng ca ng c in mt chiuTrong nn sn xut hin i , ng c mt chiu vn c coi l mt loi

my quan trng mc d ngy nay c rt nhiu loi my mc hin i s dngngun in xoay chiu thng dng .

Do ng c in mt chiu c nhiu u im nh kh nng iu chnh tc rt tt , kh nng m my ln v c bit l kh nng qu ti . Chnh v vym ng c mt chiu c dng nhiu trong cc nghnh cng nghip c yucu cao v iu chnh tc nh cn thp , hm m, giao thng vn ti ...m iu quan trng l cc nghnh cng nghip hay i hi dng ngun in

mt chiu .Bn cnh , ng c in mt chiu cng c nhng nhc im nht nhca n nh so vi my in xoay chiu th gi thnh t hn ch to v boqun c gp in phc tp hn ( d pht sinh tia la in )... nhng do nhng -u im ca n nn ng c in mt chiu vn cn c mt tm quan trngnht nh trong sn sut .

Cng sut ln nht ca ng c in mt chiu hin nay vo khong10000 KW , in p vo khong vi trm cho n 1000 V . Hng pht trinhin nay l ci tin tnh nng ca vt liu , nng cao ch tiu kinh t ca ngc v ch to nhng ng c c cng sut ln hn ...

2.Cu to ca ng c in mt chiu.ng c in mt chiu c th phn thnh hai phn chnh : phn tnh v

phn ng.2.1.Phn tnh hay stato.

y l ng yn ca my , bao gm cc b phn chnh sau:a, Cc t chnh : l b phn sinh ra t trng gm c li st cc t v dy

qun kch t lng ngoi li st cc t . Li st cc t lm bng nhng l thpk thut in hay thp cacbon dy 0,5 n 1mm p li v tn cht . Trongng c in nh c th dng thp khi . Cc t c gn cht vo v my

nh cc bulng . Dy qun kch t c qun bng dy ng bc cch in vmi cun dy u c bc cch in k thnh mt khi tm sn cch in tr-c khi t trn cc cc t . Cc cun dy kch t c t trn cc cc t nyc ni tip vi nhau.

b, Cc t ph : Cc t ph c t trn cc cc t chnh v dng cithin i chiu . Li thp ca cc t ph thng lm bng thp khi v trn

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thn cc t ph c t dy qun m cu ro ging nh dy qun cc t chnh.Cc t ph c gn vo v my nh nhng bulng.

c, Gng t : Gng t dng lm mch t ni lin cc cc t, ng thilm v my. Trong ng c in nh v va thng dng thp dy un v hnli. Trong my in ln thng dng thp c. C khi trong ng c in nhdng gang lm v my.

d, Cc b phn khc.Bao gm:

- Np my : bo v my khi nhng vt ngoi ri vo lm h hng dyqun v an ton cho ngi khi chm vo in. Trong my in nh vva np my cn c tc dng lm gi bi. Trong trng hp ny npmy thng lm bng gang.

- C cu chi than : a dng in t phn quay ra ngoi. C cu chithan bao gm c chi than t trong hp chi than nh mt l xo t chyln c gp. Hp chi than c c nh trn gi chi than v cch invi gi. Gi chi than c th quay c iu chnh v tr chi than chong ch. Sau khi iu chnh xong th dng vt c nh li.

2.2 Phn quay hay rto.Bao gm nhng b phn chnh sau :

a, Li st phn ng : Dng dn t. Thng dng nhng tm thp kthut in dy 0,5mm ph cch in mng hai mt ri p cht li gimtn hao do dng in xoy gy nn. Trn l thp c dp hnh dng rnh sau khi p li th t dy qun vo.

Trong nhng ng c trung bnh tr ln ngi ta cn dp nhng l thnggi khi p l thnh li st c th to c nhng l thng gi dc trc.Trong nhng ng c in ln hn th li st thng chia thnh nhng on

nh, gia nhng on y c mt khe h gi l khe h thng gi. Khi mylm vic gi thi qua cc khe h lm ngui dy qun v li st.

Trong ng c in mt chiu nh, li st phn ng c p trc tip votrc. Trong ng c in ln, gia trc v li st c t gi rto. Dng girto c th tit kim thp k thut in v gim nh trng lng rto.

b, Dy qun phn ng.Dy qun phn ng l phn pht sinh ra sut in ng v c dng in

chy qua. Dy qun phn ng thng lm bng dy ng c bc cch in.Trong my in nh c cng sut di vi kW thng dng dy c tit dintrn. Trong my in va v ln thng dng dy tit din ch nht. Dy qunc cch in cn thn vi rnh ca li thp.

trnh khi quay b vng ra do lc li tm, ming rnh c dng nm cht hoc ai cht dy qun. Nm c lm bng tre, g hay bakelit.

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c, C gp : Dng i chiu dng in xoay chiu thnh mt chiu. Cgp gm nhiu phin ng c c m cch in vi nhau bng lp mica dyt 0,4 n 1,2mm v hp thnh mt hnh trc trn. Hai u trc trn dnghai hnh p hnh ch V p cht li. Gia vnh p v tr trn cng cch inbng mica. ui vnh gp c cao ln mt t hn cc u dy ca cc phnt dy qun v cc phin gp c d dng.

d, Cc b phn khc.- Cnh qut : Dng qut gi lm ngui my. My in mt chiu th-

ng ch to theo kiu bo v, hai u np my c l thng gi. Cnhqut lp trn trc my , khi ng c quay cnh qut ht gi t ngoi vong c. Gi i qua vnh gp, cc t li st v dy qun ri qua qut gira ngoi lm ngui my.

- Trc my : Trn t li st phn ng, c gp, cnh qut v bi. Trcmy thng lm bng thp cacbon tt.

3. Phn loi my inng c in mt chiu phn loi theo cch kch thch t thnh ng

c in kch hch c lp, ng c in kch thch song song ,kch thch nitip,kch thch hn hp

Trn thc t c tnh ca ng c kch t c lp v kch thch songsong l ging nhau nn khi cn cng sut ln ngi ta thng dng ng ckch t c lp c th iu chnh dng in kch thch c thun tin do m iu chnh tc d dng v kinh t hn mc d n i hi c dngbn ngoi.

Trong n ny ta xt n ng c in mt chiu kch t c lpS

Khi m ngun mt chiu c cng sut khng qu ln th mch phn

ng v mch t mc vo hai ngun mt chiu c lp nhau nn gi l ngc in mt chiu kch t c lp

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E

R

Rf

U

Ikt

Rkt

U

CKT

Ukt


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4 Nguyn l lm vic ng c in mt chiuKhi cho in p U vo hai chi than A,B trong dy qun phn ng sinh

ra dng in I .Cc thanh dn ab,cd c dng in nm trong t trng s chutc dng ca lc Ft tc dng lm cho Roto quay ,khi phn ng quay navng th v tr cc thanh dn ab,cd i ch cho nhau do cc phin gp ichiu dng in gi cho chiu tc dng khng i m bo ng c c chiuquay khng i,khi ng c quay cc thanh dn ct t trng s cm ng scin ng E ,chiu quay xc nh theo quy tc bn tay tri

Phng trnh phn ngU = E +R.I

II. iu chnh tc ng c v o chiu

1) Phng trnh c tnh c

U = E +(R +Rf)IU : in p phn ngE : Sut in ng phn ngR ,Rf: in tr phn ng,in tr ph trong mchphn ngI: Dng in mch phn ng

R =r +rct +rb +rtcr : in tr cun dy phn ngrct : in tr cc t ph

rb : in tr cun brtx : in tr tip xc chi in

E =

....2

.K

a

NP=

a

NPK

..2

.

=

P : S cc t chnh

N : S i mch nhnh song song ca cun dy phn ng

: T thng kch t di mt cc t Wb : Tc gc (Rad/s)

E = Ke . n.

u

fuufuu IK

RR

K

U

K

IRRU

K

E

)(

.

).(

.

+=

+==

Mmen in t Mt =K. .I

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E

I

Rf

U

Ikt

Rkt

U

CKT

Ukt


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Suy ra I =.K

Mdt

Nu b qua tn tht coi mmen in t bng mmen c u trcMt =Mc=M

MK

RR

K

U fuu.).(. 2

+

= (1)

Mmen ph thuc vo t thng v dng phn ngT phng trnh (1) suy ra : thay i tc ng c ta c th dng phngphp thay i in p phn ng U ,t thng tc l thay i dng kch tIkt v thay i in tr phn ng R ,Rf

M =K. .I .do mun o chiu ng c tc l o chiu mmen Mta c th dng phng php o chiu t thng (tc l o chiu dng kch tIkt ) hoc l o chiu dng in phn ng I

2. Cc phng php thay i tc a) Phng php thay i t thng ,thay i dng kch t Ikt .Vi mt phti Mc nht nh .Khi gim tc ng c tng ln

Khi kch thch dng khc nhau c tnh c nhn c khc nhau v dckhc nhau.Giao im mmen cn M =f(I) vi cc ng trn cho bit tc xc lp ng vi thng s khc nhau ca t thng .b) Thay i in p phn ng- iu chnh in p phn ng ng c in mt chiu cn c thit bngun nh my pht in mt chiu kch t c lp , cc b chnh lu iukhin cc thit b ny c chc nng bin i lng xoay chiu thnh mtchiu c sut in ng Eb iu chnh c l nh tn hiu Uk

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M(Iu)

n

Iu

dm

nodm

BB

LK


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- Phng trnh c tnh c ca h thng nh sau:

udm

bud

dm

b IK

RR

K

E.

+

=

- V t thng ca ng c c gi khng i nn cng ca c tnh

c cng khng thay i cn tc khng ti l tng tu thuc vo gitr in p Uk ca h thng do c th ni phng php iu chnhny l trit

- xc nh di iu chnh tc ta thy rng tc ln nht ca hthng b chn bi c tnh c bn l c tnh ng vi in p nh mcv t thng cng gi gi tr nh mc. Tc nh nht ca di iuchnh b gii hn bi yu cu v sai s tc v mmen khi ng ,khi mmen ti l nh mc th gi tr ln nht va nh nht ca tc l

dmM

= max0max

dm

M= min0min

tho mn kh nng qu ti th c tnh thp nht ca di iu chnhphi c mmen ngn mch l:

Mnmmin = Mcmax = KM.MmTrong : KM h s qu ti v mmen, do h c tnh c l nhng ngthng song song vi nhau nn theo nh ngha v cng c tnh c ta cth vit:

( ) ( )

( ) 1

1

1

11

max0max0

minmin

=

=

==

M

dm

dmM

dm

Mdmdmnm

K

M

MK

M

D

KM

MM

* Phm vi iu chnh ph thuc tuyn tnh vo

Trang 1

min

0min

max

omax

Mm

Mnmmin

k1

k2


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3.Vn o chiuChiu quay ng c ph thuc vo chiu quay mmen c th dng hai

phng php .Hoc thay i chiu dng phn ng I hoc i chiu t thng(i chiu dng kch t Ikt).

Nu dng phng php o chiu dng kch t .Khi my ang quay thdo h s in cm ca cun dy kch thch ln (do c nhiu vng dy) nnkhi thay i dng kch thch Ikt th xut hin sut in ng cm ng rt caogy ra in p lm nh thng cch in dy qun kch thch .

Do o chiu quay ng c ta chon phng php o chiu dngphn ng I

4) Mt s yu cu k thut khca) trn

i

i

1+=

Trong : 1, +ii l tc n nh ca ng c t c cp i ,i+1 1 tc l h truyn ng c th n nh mi v tr trong ton diiu chnh

b) Di iu chnh tc L phm vi iu chnh l t s gi tr ln nht v gi tr nh nht ca tc

lm vic ng vi mmen ti cho

min

max

=D

Trong :max

b hn ch bi bn ng c v bn ca vnh gp min b chn bi yu cu v mmen khi ng ,kh nng qu tiv sai s tc lm vic cho php.c) Chng mt kch tKhi m my phi m bo chng mt kch t m nguyn nhn l do ngn

mch kch thch

V khi E = 0 nn I =uu

u

R

U

R

EU=

Do U khng i v R rt nh (in tr cun dy phn ng) nn I rt ln lmchy dy qun v vnh gp

Cch khc phc iu ny l phi c b phn nhn bit c mt kch t( 0= v do I =0) th lp tc ngt ngun cp cho phn ng tc U = 0.

Khi I khng ln v trnh c s c trn.

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Chng 2Cc phng n tng th

la chn c mt phng n thch hp vi yu cu Cp in cho ng c

in 1 chiu kch t c lp c o chiu theo nguyn tc iu kin ring tacn phi la chn 1 phng n iu chnh tc v s mch lc ph hp

I .La chn phng n iu chnh tc ng cNi chung ta c th tin hnh iu chnh tc ng c theo 2 phng phpch yu nh nu trn : iu chnh in p phn ng v iu chnh tthng

Phng php iu chnh t thng .Phng php iu chnh t thng c th thay i tc c mt cchlin tc v kinh t nhng ch c th iu chnh c tc trong vng trn

tc nh mc v b hn ch bi cc iu kin c kh Phng php iu chnh in p phn ng .Phng php iu chnh tc bng cch thay i in p phn ng c thiu chnh tc 1 cch lin tc v d dng thc hin bng cch thay igc iu kin . Mt khc phng php ny trong qu trnh iu kinthay i tc th khng tiu hao nng lng v momen ca ng ckhng i .

Nhn xt:T nhng phn tch trn ta la chn phng n iu chnh tc ca ng

c bng cch thay i in p phn ng

II. La chn phng n mch lc. thay i v o chiu ca ng c in 1 chiu chng ta buc phi thitk cc b chnh lu o chiu vi s nguyn l nh sau .

Da vo cng sut cung cp cho ti m ta la chn mch chnh lu l mt phahay mch chnh lu 3 pha .Thng thng theo kinh nghim ta c :

+ Nu cng sut ca mch cung cp ln hn 5 7 kW ta s dngmch chnh lu 3 pha

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CL II

Ti

Rf


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+ Nu cng sut ca mch cn cung cp nh hn 5 7 kW ta s dngmch chnh lu 1 pha

- Cng sut mch phn ng :P = 600.5 = 3000W = 3kW < 5 7 kW nn ta s dng mch

chnh lu 1 pha cung cp cho phn ng.- Cng sut mch phn cm :Pcm = 400.0,6 = 240W = 0,24kW < 5 7 kW nn ta s dng

mch chnh lu 1 pha cung cp cho phn cm. Cc s chnh lu 1 pha

1. Chnh lu mt na chu k .

Vi s ny sng in p ra mt chiu b gin on trong mt nachu k khi in p ant ca van bn dn m , do khi s dng s chnhlu mt na chu k cht lng in p ti xu .

in p ti trung bnh ln nht trn tiUdo = 0,45U2

V cht lng in p xu nn hiu sut ca my bin p cng thp .

Sba = 3,09.Ud.Id

+u im- L loi chnh lu c nguyn l n gin t van

+Nhc im- Cht lng in p xu- Hiu sut s dng my bin p thp

2. Chnh lu c chu k vi bin p c trung tnh

T2

U1R

U2

U2

T1

L

Trang 1

U2

RL

T

U1


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Dng dng ,p trn ti v in p ngc trn van T1 trong trng hpti in cm ln

Xt vi ti RL , in cm ln lc dng v p c cht lng tt- Bin p phi c hai cun dy th cp vi thng s ging ht nhau, mi

na chu k c mt van dn cho dng chy qua- Khi = cho pht xung m T1 , T1 dn do in p u anot dng v c

xung m . Khi = , in p trn anot = 0 nhng do c cun cm L nn vncn dng in Id nn T1 cha kho , T1 tip tc dn cho n khi = + ,pht xung m T2 th T1 b kho v T2 dn . T1 kho khng phi dng v 0m l do T2 dn .

T2 li dn cho n khi = 2 + , T1 c pht xung m , T1 dn v T2 b

kho li.Qu trnh c nh vy cho n khi in p ti p mch c tn s bng 2

ln tn s in p xoay chiu .Vi : l gc iu khin m .

+ in p trung bnh trn ti

Trang 1

IiT1

d

id Id

o

aUd

2?

Ud

o2 ?

iT1

Id


12/46


Ud =2

2+

.2 U2.sind =

22 .U2.cos = Udo.cos

Udo =

22 .U2 = 0,9.U2 :

in p chnh lu khng iu khin khi ti l thun tr .+ in p ngc trn van l ln

Unv = 2 2 U2Ti c in cm ln nn dng ti lin tc id = Id

Mi van dn thng mt na chu k+ Dng hiu dng qua van (chnh l dng hiu dng qua my bin p).

Ihd =

2

0

2.

2

1dId =

2

Id= 0,71.Id

+ H s p mch Km = 0,67.+ Cng sut my bin p Sba = 1,48.UdId

Nhn xt :

+ u im :- So vi chnh lu mt na chu k th chnh lu hnh tia c in

p vi cht lng tt hn- Dng qua van khng qu ln- iu khin van n gin

+ Nhc im

- Ch to my bin p phi c 2 cun ging nhau mi cun lmvic 1na chu k- Ch to bin p phc tp- Hiu sut s dng bin p xu hn- in p ngc trn van l ln .

3. Chnh lu cu mt pha

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Cc c tnh in p ti , dng qua ti , dng qua van v in p ngctrn van trong trng hp ti RL , R ln m bo dng qua ti l lintc c biu din trn hnh v

- Mch c T1 , T3 chung KatotT2 , T4 chung Anot

Na chu k u U2 > 0 , Anot ca T1 dng , Katot ca T2 m . Nu cxung iu khin m ng thi T1 v T2 th c hai van ny c m thng vt in p li ln ti Ud = U2 . in p ti mt chiu bng in p xoaychiu (Ud = U2) cho n khi no T1 , T2 cn dn .(Khong dn ca cc vanph thuc vo ti ) .

Na chu k sau, in p i du , anot ca T3 dng v katot T4 m ,nu c xung iu khin m ng thi T3,T4 th cc van ny c m thng v

Ud = - U2 , vi in p mt chiu c cng chiu vi na chu k trc+ Gi tr trung bnh ca ti

Ud = +

dU .sin.22

22 = cos.

2

222U = Udo cos.

Udo : in p trung bnh ti trong chnh lu cu khng iu khin .+ Dng qua my bin p cng bng dng qua van (khi van m) .

Trang 1

i2Id

UT1

o2 ?

I d

i T1T2I d

idI d

0

a

2 ?

i T3T4

Ud


14/46


+ Gi tr hiu dng dng th cp bin p

I2 = +

2

2

2dI = Id

+ in p ngc ln nht van phi chuUnv = 2 U2

+ Dng lm vic ca van tnh theo gi tr trung bnh

IT= +

2

2

1dI =

2

Id= 0.71 Id

+ H s p mch km = 0,67+ Cng sut bin p Sba = 1,23Pd

Nhn xt:Cht lng in p ca chnh lu cu mt pha ging nh chnh lu hnh tia .

Hnh dng ng cong in p , dng ti , dng qua van bn dn , h s pmch nh trong chnh lu hnh tia .+ u im :

- in p ngc trn van nh hn mt na so vi chnh lu hnh tiaUnv= 2 U2.

- My bin p ch tao n gin hn , v c hiu sut cao hn so vichnh lu hnh tia .

+ Nhc im :- S van nhiu hn- iu khin van T1 ,T2 v nhm T3 , T4 phi ng thi nn kh

khn hn.

Nhn xt chungT cc phn tch trn ta la chn mch lc l mch chnh lu iu khin

cu mt pha.Do yu cu cp in cho ng c c o chiu theo nguyn tc iu kinring nn y ta la chn s cu 1 pha i xng v s cu 1 phakhng i xng c h s cos cao hn nhng khng o chiu dng in rati .

Vy ta la chn s mch lc cung cp cho phn ng v phn cm

ca ng c nh sau:

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S mch lc cung cp cho phn ng .

S mch lc cung cp cho phn cm.

Trang 1

Ru

Eu

Lu

T1

T3

T2

T4

T1

T3

T4

T2

T1

T3

D2

D4

Rf

Rf


16/46


Chng 3Phn tch s hot ng ca mch thit k

i.nguyn l hot ng ca mch lc .Mch lc bao gm 2 s cu hot ng c lp ring r : b chnh luI chy th b chnh lu II khng chy , ngc li b chnh lu II chy th bchnh lu I ngh .

Do vy khi bnh thng 1 b hot ng ch chnh lu th s mch lc lc ny l s cu 1 pha .

1.S hot ng ca s cu 1 pha nh sau :Ti ta coi ti l RLE , cun cm L c thm vo lc in p v dng .

a) Mch hot ng khi La = o+ = pht xung m T1 , T2

T1T2 dn T3 , T4 khoin p ra ti Ud = U2

+ = + pht xung m T3 , T4T1 , T2 kho ; T3 , T4 dn (m)in p ra ti Ud = - U2

+ in p trung bnh ti

Ud =

22 U2cos

+ in p ngc ln nht trn vanUnv = 2.2 U

+ Dng lin tc nn id =Id+ Phng trnh mch ti

.2 2U sin =R.id +XL . dtdi

+E

Trang 1


17/46


++++

++=

dE

diXiRdU dLd .1

.1

.sin21

2

Ud =RId +E+ Dng hiu dng th cp my bin p

I2 = dd IdI =+

21

b) Trng dn

Trong my bin p c cun dy nn c in cm La nn trong mch s xyra hin tng trng dn.Gi s T1 ,T2 ang m cho dng chy qua iT1 =iT2 =IdKhi += pht xung m T3,T4V c La nn dng iT1,iT2 khng gim t ngt v 0 v dng iT3,iT4 cng

khng tng t ngt t 0 n IdLc ny c 4 van cng m thng cho dng chy qua ,ph ti b ngn mchud =0Ngun e2 b ngn mch sinh ra dng ngn mch ic

Phng trnh

).(sin2 2 aaa

a LXd

diXU

==

ic = ))cos((cos.2

)sin(.2 2

0

2

+=+cc X

U

X

U

t ic = ic1 + ic2 , ic1 = ic2 = 0,5.icic1 : lm tng dng trong T4 v lm gim dng trong T2Ic2 : lm tng dng trong T3 v lm gim dng trong T1

( )( ) += coscos.2

.2 24,3

cT

X

Ui

Trang 1

iT3iT1 Id

0

id

Ud

2

Rf


18/46


( )( ) += coscos.2

.2 22,1

cdT

X

UIi

Khi kt thc giai on trng dn tc khi , = th iT1,2 = 0

( )2.2

..2coscos

U

IX dc=+

Ta thy so vi khng trng dn c tnh in p st mt phn v l stp gy ra do trng dn U

( ) ( )( )

+=+= coscos..2

sin..2.2

2 2

0

2

UdUU

dc IXU..2

=

khi gi tr trung bnh in p trn ti l

Ud

= Ud - dcdc IX

U

IX ..2

cos..

2.2..22 =

Ud = R.Id +E

2. S hot ng ca mch trong qu trnh ta o chiu ng cTrong qu trnh ta o chiu ng c ta cn 1 b iu kin lgic thc

hin cc bc o chiu nh sau :Gi s ta chuyn t s lm vic ca b chnh lu I sang b chnh lu II

Ta cn thc hin th t cc bc nh sau1. Ngt xung iu khin b chnh lu I2. o dng chy qua ti xc nh dng in v khng

3. Sau khi pht hin dng v khng phi d tr 1 thi gian tr > tph m bo thyristor kho chc khi mi pht xung iukhin van kia

4. Pht xung iu kin cho cc van b chnh lu IINu sut in ng Ed cha o chiu ,th khng c chy b chnh lu II

vi < 900 v lc sut in ng Ed v Id cng chiu nhau , trng hp nayd gy sp nghch lu do dng in rt ln .Do bt u phi iu khin b chnh lu II ln hn 900 , lc b chnh lu II lm vic ch nghch luph thuc .

a) Ch nghch lu ph thuc trong s cu 1 pha nh sau :Trong ng c in mt chiu c sc phn in ng ,trong iu kin

no n c th lm vic ch nghch lu - l ch bin i nng lngdng mt chiu pha ti thnh dng in xoay chiu cp tr li li in, ch lm vic nh vy chnh l khi ta hm ti sinh tit kim nng lng .

sc in ng E pht nng lng tr li li in th dng v p phingc chiu nhau Ud v Id ngc nhau

Trang 1


19/46


Do dng in ch chy theo mt chiu t A n K ca thyristor nn taiu chnh sao cho

- Chiu dng nh c- o chiu sc in ng Ed-

trong ch chnh lu vi gc m l

th Ud

=Ud0

.cos

.Nn nu 0

2 dU

tc l in p trn ti thay i cc tnh v ngc

chiu van .Do duy tr dng chy qua van t A n K ca van th ta phio chiu Ed v m bo dUE > vy ch chnh lu ,dng trong mchc duy tr bi Ud E > 0 th trong ch nghch lu ph thuc dng c duytr bi 0> dUE

Trong ch nghch lu li in nhn nng lng t pha tiiu kin hot ng ch nghch lu ph thuc

- 2

> - o chiu Ed

- m bo dUE >

Xt dng in p trn ti .

Trong mch c in cm bin p nn n c qu trnh trng dn vi gc trngdn

Xt T1,T2 ang dn , = pht xung m T3 ,T4 ,UA > UK nn c dngqua van v T1,T2 vn cn dn t 4 van cng dn trong khong .

Sau T1 ,T2 kho v T3,T4 min p ngc trn van T1 ,T2

Unv = ( ) 22 sin..2 UU =in p ngc ch xut hin trn van trong khoang t += n = vsau in p trn T1 ,T2 li l thun (UA > UK )Nh vy trong mt chu k lm vic mi van ch chu in p ngc trongkhong t += n = tc l trong khong thi gian )( += khong thi gian ny kh nh .

Nu off


20/46


Thyristor s t m tr li khng cn dng iu khin (m t nhin )Khi trong mch c hai ngun cng pht nng lng khi Id rt ln gys c sp nghch lu .Do an ton phi c off >+ )(

Chno

off 10= ( ng vi toff=555 s th c th kho van Thyristor thngthng ) )( off + /2 v sao cho dng in phn ng khng vt qu gi trcho php ng c c hm ti sinh. Nu nhp iu gim 2 ph hp vi qun

Trang 1

1

2L

iLd

b1

b2

b1

b2


21/46


tnh ca h th c th duy tr dng in hm v dng in khi ng ngckhng i , iu ny c thc hin bi cc mch vng iu chnh t ngdng in ca h thng trn s ca khi lgic LOG , iL , iL1 , iL2 l cc tnhiu lgic u vo b1,b2 l cc tn hiu lgic u ra kho cc b phtxung iu khin.

iL = 1 pht xung iu khin m B1.iL = 0 pht xung iu khin m B2.i1L (i2L) = 1 c dng in chy qua b B1 v B2b1(b2) = 1 kho b pht xung FX1 v FX2.

T mch lgic trn ta c:

LLLD

LLLD

iiib

iiib

1

__

22

2

__

1

__

1

.

.

+=

+=

Nhn xt:

H truyn ng van o chiu iu khin ring c u im l lm vican ton ,khng c dng in cn bng chy gia cc b bin i nnkhng cn thit k cun khng cn bng ,song cn mt khong thigian tr trong dng in ng c bng khng.

Do nguyn tc iu khin ring dng hai b bin i lm vic c lp,trong mt thi im th ch c mt b B lm vic cn b B kia phichc chn kho( c ngha l dng in qua b B ny phi bng 0). Tas dng xenx dng in nhn bit c dng in chy qua b B haykhng. xenxo dng ny ta dng in tr shunt .in tr ny khi c dngqua n th mt in p c 60 mV ly ra u ra .tn hiu ny qua bkhuych i thut ton khuych i in p ny

Ta dng mt khu thut ton so snh LM311: tn hiu dng in quaxenx c h trn mt in tr RS to ra tn hiu in p Vi.

Trang 1

-E

+E

R23

R24

R22 A5

UI


22/46


u ra chn 7 mc vi ngun nui +5V qua in tr ko ln 1k. t ca bso chn 1 c mc vo u ni t ca mch. Cch mc ny dn n u rac cc trng thi kh d l 0 v 5V .Lgic ca mch l:

V0 =5V i vi Vi > 0V0 =0V i vi Vi < 0

Nu V0 = 5V ngha l b B c dng in chy qua.Nu V0 = 0V c ngha l b B khng c dng in chy qua . Do mch iu khin ring cn c thi gian to tr nn ta chn b to

tr l mch 555

8

6

7

4

2 1

3

5

55 5

+ U

R1

R2

C1 C2

Ura

a.

Gc phc hi off =10o thi gian phc hi Toff=555 s

T = R1.C1 = 555.10-6 chn R1 = 470 v C1 = 1,5 F

Chng IVThit k v tnh ton mch lc

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23/46


I. Tnh chn TiristoKhi la chn van ta da vo 2 thng s c bn quan trng nht l dng

in qua van v in p ngc ln nht m van chu c , cc thng s c bnca van c tnh nh sau :

+ in p ngc t ln van :Ulv = Ung.max = knv . U2 = knv . Udm / ku

Trong :knv , ku ln lt l h s in p ngc v h s in p ca mch lci vi s cu 1 pha th : knv = 1,41 ; ku = 0,92Um l in p cung cp cho phn ng : Um = 600V

Do : Ung.max = 1,41.9,0

600= 940V

+ Dng in lm vic ca van cn c

Ilv = Ihd = khd.ImTrong :Ihd l dng in hiu dng qua vanKhd l h s hiu dng ng vi s cu 1 pha khd = 0,71Im l dng in qua ti ( cung cp cho phn ng )

Do : Ilv = 0,71.5 = 3,55 A s lm vic tin cy ta cn phi la chn cc van c d tr , do

ta chn thng s ca van :+ in p ngc ca van cn chn

Ung-v = kdt-U . Ung-max

Vi Kdt =1,6 2 l h s d tr in p , chn kdt -U = 1,6Do : Ungmax = 1,6.940 = 1504V 1,5 kV

+ Dng in hiu dng ca van c chnIhd-v = kdt-I .Ilv

Vi kdt-i l h s d tr dng in , chn kdt-I = 4Do : Ihd-v = 4.3,55 = 14,2 A- Da vo 2 thng s Ihd-v v Ung.max ta la chn van c s liu nh bng sau

K hiu Un

(V)Ihd(A)

Ig(A)

Ug(V)

Ih(A)

Ir(A)

U(V) dt

du(V/s) TC

(s)Tmax(0C)

T35-N-1800

BOF

1800 35 0,12 1,4 0,2 0,02 2 1000 120 125

Trong :

Trang 1


24/46


- in p ngc van : Un = 1800 (V)- Dng hiu dng van : Im = 35 (A)- Dng xung iu khin : Ik max = 120(mA)- in p xung iu khin : Uk max = 2 (V)- Dng in gi : Ih max = 200 (mA)- Dng in d : Ir max =20 (mA)- st p ln nht : )(2max VU =

- Tc thay i in p : )/(1000 sVdt

dU=

- Thi gian chuyn mch : tcm = 120( s )- Nhit lm vic cho php :Tmax =125 0C

II. Tnh ton MBA chnh lu

Mch in phn ng l mch chnh lu cu 1 pha nn ta cn thit kMBA 1 pha . Ta la chn MBA c mch t l hnh ch E ( c 1 tr ) clm mt bng cc l tn nhit .

1.Tnh ton thng s ca MBA :1. in p s cp ca MBA l in p li : U1 = 220 V2. in p th cp ca MBA- Phng trnh cn bng in p khi c ti :

Ud0.cosmin = Ud + 2Uv + Uba +UdnTrong

min = 100 gc d tr khi c suy gim in p li . Uv = 1,6(v) st p trn tiristor

Udn = 0 st p trn dy niUba = Ur + Ux : st p trn in tr v in khng MBA.

Chn s bUba = 6%.Ud = 6% . 600 = 36(V)

T phng trnh cn bng in p ti ta c:

)(87,64910cos

3602.2600

cos

.2

0

min

VU

UUUUU

do

badnvddo

=+++=

+++=

3. Cng sut biu kin ca MBASba = ks.Pd-max

Trong :+ ks : H s cng sut theo s , vi s cu 1 pha : ks = 1,23+ Pd-max Cng sut ca ph ti lc cc i

Trang 1


25/46


Pd-max = Udo.Id = 649,87.5=3249,35 WDo : Sba = 1,23.3249,35 =3996,885VA

in p pha th cp :

)(07,7229,0

87,6492 V

k

UU

u

do ===

Dng in hiu dung th cp MBA:)(52 AIII dmd ===

Dng in hiu dng s cp MBA:

)(41,165.220

07,722.. 2

1

221 AI

U

UIkI ba ====

- Tnh ton t mchChn mch t hnh chc ghp t nhng l tn Silic loi 310 c

B dy tn : 0,35mmTn hao l : 1,7 W/kgT trng : d = 7,8kg/dm3

- Chn loi thp 330 cc l thp c dy 0,5 mm- Chn s b mt t cm trong tr B =1(T)* Tnh trTit din tr s b ca tr T c tnh theo cng thc:

QFe = kQ.fm

Sba

.

Trong :

Sba l cng sut pha ca my bin pKQ H s ph thuc vo phng php lm mt vi MBA lm mtkhng kh : kQ = 6m la s tr ca MBA :m =1

Thay s : QFe = 6.50.1

3996,885= 53,64cm2

Ta c : QFe = a.b = 53,64

Trong qu trnh thit k ta thng chn : 1< 5,1 c = 0,5a = 3cm

m =a

h= 2,5 => h = 2,5a = 15cm

Chn chiu cao tr h = 15(cm)

Trang 1


26/46


- Tnh ton dy qunS vng dy mi pha s cp MBA:

74,1841.10.64,53.50.44,4

220

...44,4 41

11 ==

TFe BQf

Uw (vng)

Ly w1 = 185 (vng )S vng dy mi pha th cp MBA:

19,607185.220

07,722. 1

1

22 === w

U

Uw (vng)

Ly w2 = 608 (vng)- Chn s b mt dng in trong MBA vi dy dn bng ng MBA

kh chn :J1 = J2 = 2,75(A/mm2)

Tit din dy dn s cp MBA:)(97,5

75,2

41,16 2

1

11 mm

J

IS ===

Chn dy dn tit din hnh trn , cch in cp B

- Chun ho tit din theo tiu chun: S1 = 6,29(mm2).- ng knh dy dn d1 = 2,83mm- ng knh dy dn k c cch in dn1 = 2,95mm

Tit din dy dn th cp MBA:

)(818,175,2

52

2

22 mmJ

IS ===

Chn dy dn tit din ch nht, cch in cp B .- Chun ho tit din theo tiu chun: S2 = 1,9113(mm2).- ng knh dy dn d2 = 1,56mm- ng knh dy dn k c cch in dn2 = 1,65mm- Kt cu dy qun s cp :

Trang 1

b

hH

c

C

a c

Hnh 1 s kt cu mybin p


27/46


Thc hin dy qun theo kiu ng tm b tr theo chiu dc trTnh s b s vng dy trn mt lp ca cun s cp :

4,4695,0.295,0

59,015.

.2

1

11 =

=

= eg

kb

hhw (vng)

Ly w1l = 46 vngTrong : ke = 0,95 h s p chth : chiu cao trhg : khong cch t gng n cun dy s cphg = 2.dn1 = 2.2,95 = 0,59cm

- Tnh s b lp dy cun s cp:

02,446

185

11

111 ===

w

wn (lp)

Chn lp n11 = 4lp.Nh vy c 185 vng chia lm 4 lp , chn 3 lp u 46 vng , lp th 4 c

197 3 . 46 = 47 (vng)Chiu cao thc t ca cun s cp :

)(28,1495,0

46.295,0.111 cm

k

bwh

e

===

Chn ng qun lm bng vt liu cch in lt gia tr v dy qun c bdy : d01=0,3 (mm)Chn lp cch in gia cc lp dy qun c b dy cd1 = 0,1mm

- ng knh ca cun dy trong cng

Dt1 = 22 ba + + 2.d01= 22 94,86 + + 2. 0,03= 10,82 cm

- ng knh ca cun dy ngoi cngDn1 + 2( dn1 + cd1 )n1 = 10,82 + 2( 0,295+0,01).4 = 13,26cm- ng knh ca trung bnh ca cun dy s cp

Dtb-1 =2

11 nt DD + =2

26,1382,10 += 12,04cm

- Chiu di dy qun s cpL1 = II.Dtb-1.w1 = 3,14.12,04.185 =6994,036cm

2.Kt cu dy qun th cp- Chn chiu di cao ca cun dy th cp h2 = h1 = 14,28cm

Tnh s b s vng dy trn mt lp ca cun s cp :

3,8095,0.165,0

33,028,14.

.2

2

2

2 =

=

= eg

l kb

hhw (vng)

Ly w1l = 80 vngTrong : ke = 0,95 h s p cht

h : chiu cao tr

Trang 1


28/46


hg : khong cch t gng n cun dy s cphg = 2.dn2 = 2.0,165 = 0,33mm

- Tnh s b lp dy cun s cp:

6,780

608

21

22 ===

w

wn l (lp)

Chn lp n21 = 8 lp.Nh vy c 608 vng chia lm 8 lp , chn 7 lp u 560 vng , lp th 8 c608 7 . 80 = 49 (vng)Chn ng qun lm bng vt liu cch in lt gia tr v dy qun c bdy : d02 = 0,3 (mm)Chn lp cch in gia cc lp dy qun c b dy cd2 = 0,1mm

- ng knh ca cun dy trong cngDt2 = Dn1+ 2.d02 = 13,32m

- ng knh ca cun dy ngoi cng

Dn2=Dt2 + 2( dn2 + cd2 )n2 = 13,32+ 2(0,165+0,01).8 = 16,12cm- ng knh ca trung bnh ca cun dy s cp

Dtb-2=2

22 nt DD + =2

12,1632,13 += 14,72cm

- Chiu di dy qun s cpL2 = II.Dtb-2.w2 = 3,14.14,72.608 = 281022,24cm = 2810m

Kim tra iu kin ca ca s- Chiu dy ca dy qun s cp v th cp khi qun ln tr :

D = ( d01+ d02 ) + (dn1 + cd1)w1+ ( d02 + cd02).w2= 0,3+ 0,3 + ( 2,95+ 0,1).4 + ( 1,65 + 0,1).8 = 26,8mm

- Khe h khng kh gia dy qun v mch tE = C- D = 3- 2,505 = 0,495 cm => lm mt v cch in

Nh vy kch thc ca li thp m ta thit k l ph hp..

III. Tnh chn cc thit b bo v cho mch ng lc1. Bo v qu nhit cho cc van bn dn

Khi lm vic vi dng in chy qua trn van c st p , do c tnhao cng sut p ,tn hao ny sinh ra nhit t nng van bn dn .

Mt khc van bn dn ch c php lm vic di nhit cho php Tcpno nu qu nhit cho php th cc van bn dn s b ph hng . cho van bn dn lm vic an ton, khng b chc thng v nhit, ta phi chnv thit k h thng to nhit hp l .

+Tn tht cng sut trn 1 TiristorP=U.Ilv = 2.3,55 = 7,1(W)

Trong : U St p trn van : U = 2 (V)

Trang 1


29/46


Ilv Dng lm vic chy qua van Ilv = 3,55A- Ta bit rng nu P < 20W th van c th lm vic vi dng in ti

a l 10%Im m khng cn cnh tn nhit .Nhng v I lv >10%Im ( 3,55> 10%.35) nn ta phi thit k cnh tn nhit gn ln van to nhitcho cc van

- Tng din tch b mt to nhit :

.mk

PS

=

Trong :p : tn hao cng sut (W)T: chnh lch so vi mi trngTmt = 400CNhit lm vic cho php ca Tcp=1250C

Chn nhit trn cnh to nhit Tlv=800

CT=Tlv-Tmt =80 40 = 400CKm :H s to nhit bng i lu v bc xChn Km=8 (w/m2. 0C)

Vy:22

875,221)(0221875,040.8

1,7cmmSm ===

Chn loi cnh to nhit c 10 cnhKch thc mi cnh a.b= 5.5(cm2)Tng din tch to nhit ca cnh

S=10.2.5.5 = 250(cm2)

2.Bo v qu dng in ca van

Dng 1 aptmt ng ngt mch ng lc, t ng bo v khi quti v ngt mch Tiristor, ngt mch u ra bin i, ngt mch th cpMBA ngt mch ch nghch lu.Chn 1 ptmt da vo cc thng ssau:

+ in p nh mc : Udm =220V+ Dng in lm vic nh mc : Im = 1,1.I1 = 1,1. 16,41 = 18,051A

+ Chnh nh dng ngn mch : Inm = 2,5. I1 = 1,5.16.41 = 41,025A+ Dng qu ti : Iqt = 1,5. Im = 1,5. 16,41 = 24,615A

Vy ta chn 1 ptmt c :

Loi Um (A) Im( A) Inm (A) Iqr (A)

Trang 1


30/46


LK 220 20 45 25

- Chn 1 cu giao to khe h an ton khi s cha h truyn ng.- Dng dy chy tc ng nhanh bo v ngn mch cc Tiristor ngn

mch u ra ca b CL.+ Nhm 1CC : Dng in nh mc dy chy nhm 1cc

)(5,55.1,1.1,11 AII lvCC ===

+ Nhm 2CC : Dng in nh mc dy chy nhm 2cc)(095,355,3.1,1.1,12 AII hdCC ===

+ Nhm 3CC : Dng in nh mc dy chy nhm 3cc , dng bov ng c trong trng hp m my :

)(255.5.3 AIkI DCdmmmCC ===

Vy chn cu chy nhm 1CC, 2CC v 3CC c Iday chy theo bng sau :

Nhm 1CC 2CC 3CCIdc (A) 6 4 25

3.Bo v qu in p cho vanBo v qu in p : do qu trnh nh ct cc Tiristor c thc hin

bng cch mc R-C song song vi Tiristor .Khi c s chuyn mch cc intch tch t trong cc lp bn dn phng ra ngoi to ra dng in ngc trongkhong thi gian ngn ,s bin thin nhanh chng ca dng in ngc gy rasc in ng cm ng rt ln trong cc in cm lm cho qu in p giaanod v catod ca Tiristor .

Khi c mch R-C mc song song vi Tiristor to ra mch vng phngin tch trong qu trnh chuyn mch nn Tiristor khng b qu in p .Theo kinh nghim R1 = ( 20 -> 100 )

C1 = ( 0,1-> 2) F

Bo v xung in p t li in . ta mc mch R-C nh mch lc nym nh xung gn nh nm li hon ton trn in tr ng dy .

Chn R1= 40 ()v C1= 0,8F

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31/46


Chng VTnh ton v thit k mch iu khin

I.Yu cu i vi mch iu khin

Mch iu khin l khu rt quan trng trong b bin i Tiristo v nng vai tr ch o trong vic quyt nh cht lng v tin cy ca BB.Yu cu ca mch iu khin c th tm tt trong 6 im chnh sau:

+ Yu cu v rng xung iu khin.+ Yu cu v ln xung iu khin.+ Yu cu v dc sn trc ca xung (cng cao th vic m cng tt

thng thng 0,1A/,dt

diDK ).

+ Yu cu v s i xng ca xung trong cc knh iu khin.+ Yu cu v tin cy.

. in tr knh iu khin phi nh Tiristor khng t m khi dngr tng.

.Xung iu khin t ph thuc vo dao ng nhit , dao ng inp ngun.

. Cn kh c nhiu cm ng trnh m nhm.+ Yu cu v lp rp vn hnh.. Thit b thay th d lp rp v iu chnh.. D lp ln v mi khi c kh nng lm vic c lp.

II. Nguyn l chung ca mch iu khin.

1. Nhim v ca mch iu khinL to ra cc xung vo nhng thi im mong mun m cc van nglc ca b chnh lu.

- Tiristor ch m cho dng in chy qua khi c in p dng t trn Antv c xung p dng t vo cc iu khin khng cn tc dng g na.

- Chc nng ca mch iu khin :+ iu chnh c v tr xung iu khin trong phm vi na chu k d-

ng ca in p t trn ant katt ca tiristor.+ To ra c cc xung iu kin m tiristor rng xung tx < 10s.

Biu thc rng xung:

dt

di

It dtx =

Trong : It l dng duy tr ca tiristor.di/dt : tc tng trng ca dng ti.

i tng cn iu khin c c trng bi i lng iu khin l gc .

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2. Nguyn tc iu khin.Mch iu khin tiristor c th phn loi theo nhiu cch. Song cc mch

iu khin u da theo nguyn l thay i gc pha v theo ta c hainguyn l khng ch ngang v khng ch ng.

- Khng ch ngang l phng php to gc thay i bng cch dchchuyn in p ra hnh sin theo phng ngang so vi in p ta.

+ Nhc im ca phng php khng ch ny l gc ph thuc vodng in p v tn s li, do chnh xc ca gc iu khin thp.- Khng ch ng l phng php to gc thay i bng cch dchchuyn in p ch o theo phng thng ng so vi in p ta rng ca.

+ Phng php khng ch ng c chnh xc cao v khong iukhin rng ( t 0 -> 1800 ).

+ C hai phng php iu khin ng :. across. tuyn tnh

Nguyn tc iu khin thng ng tuyn tnhTng i s ca Ur + Uk a n u vo ca mt khu so snh. Bng cchlm bin i Uk ta c th iu chnh c thi im xut hin xung ra tc liu chnh c gc .

Khi Uk = 0 ta c = 0.Khi Uk < 0 ta c > 0.Quan h gia v Uk nh sau:

rmax

dk

U

U. =

Nguyn tc iu khin thng ng acrcoss.Nguyn tc ny dng hai in p:

+ in p ng b Ur vt trc in p ant catt mt gc bng /2(Nu UAK = A.sinwt th Ur = B.coswt )

+ in p iu khin c Uk l in p mt chiu c th iu chnhc bin theo hai hng (dng v m).

Trn hnh v ng nt t l in p ant catt tiristor, t in p ny ng-i ta to ra Ur . Tng i s Ur + Uk c a n u vo ca khu so snh .

Khi Ur + Uk = 0 ta nhn c mt xung u ra ca khu so snh :Uk + B.cos = 0

Do = arccos(-Uk/B)Thng ly B = Uk maxKhi Uk = 0 th =/2

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Khi Uk = - Uk max th = 0Nh vy khi cho Uk bin thin t - Uk max n + Ukmax th bin thin t

0 n .Nguyn tc ny c s dng trong cscs thit b chnh lu i hi cht lng

cao. Nhn xt: Theo yu cu thit k mch iu khin ta thy nguyn tciu khin thng ng tuyn tnh l ph hp, ta chn nguyn tc iu khinny.

Phng php iu khin thng ng tuyn tnh

Khi in p xoay chiu hnh sin t vo anod ca Tiristo, c th iukhin c gc m ca Tiristo trong vng in p + anod, ta cn to mtin p ta dng tam gic, ta thng gi l in p ta l in p rng ca Urc.Nh vy in p ta cn c trong vng in p dng anod.

Dng mt in p mt chiu Uk so snh vi in p ta. Ti thi im(t1,t4) in p ta bng in p iu khin (Urc = Uk), trong vng in pdng anod, th pht xung iu khin Xk. Tiristo c m t thi im cxung iu khin (t1,t4) cho ti cui bn k (hoc ti khi dng in bng 0)

S nguyn l iu khin chnh lu

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Udf

Urc

Udk

Ud

Xdk

t1 t2 t3 t4 t5

t

t

t

t


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S khi mch iu khin.

thc hin c nu trong phn nguyn l iu khin trn,mch iu khin bao gm ba khu c bn trn hnh 3

Nhim v ca cc khu trong s khi hnh 3 nh sau:

+ Khu ng pha c nhim v to in p ta Urc (thng gp l in pdng rng ca tuyn tnh) trng pha vi in p anod ca Tiristo

+ Khu so snh c nhim v so snh gia in p ta vi in p iukhin Uk, tm thi im hai in p ny bng nhau (Uk = Urc). Ti thiim hai in p ny bng nhau, th pht xung u ra gi sang tngkhuch i.

+ Khu to xung c nhim v to xung ph hp m Tiristo. Xung m Tiristor c yu cu: sn trc dc thng ng, m bo yu cu Tiristom tc thi khi c xung iu khin (thng gp loi xung ny l xung kimhoc xung ch nht); rng vi rng xung ln hn thi gian m caTiristo; cng sut; cch ly gia mch iu khin vi mch ng lc (nu

in p ng lc qu ln)1) Khi ng pha

Ngy nay cc vi mch c ch to ngy cng nhiu, cht lng ngycng cao, kch thc ngy cng gn, ng dng cc vi mch vo thit k mchng pha c th cho ta cht lng in p ta tt, iu khin gc m ln gc

Trang 1

ng pha So snh To xung

A1 A2

00

11

22

33

44

55

R1A

00

11

22

33

44

55

R2

Ur

R3

C1

CD1

B

Tr

U1


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m c th t ti 1800. Trn s m t s to in p ta dng khuchi thut ton (KTT).

2) Khu so snh

S khu so snh

xc nh c thi im cn m Tiristo chng ta cn so snh hai tn

hiu Uk v Urc. Vic so snh cc tn hiu c th c thc hin bngTranzitor nhng Tranzitor khng lm vic ch ng ct nh ta mongmun, do nhiu khi lm thi im m Tiristo b lch kh xa so vi imcn m ti Uk = Urc. khc phc iu ny ta dng khuych i thut ton.KTT c h s khuych i rt ln nn ch cn mt tn hiu rt nh uvo ta c in p u ra ngun nui .u im ca n l c th phtxung chnh xc ti Uk=Ura

3) Khu to xung

Vi nhim v to xung ph hp m Tiristo nh nu trn, tngkhuch i cui cng thng c thit k bng Tranzitor cng sut, nh m ttrn hnh 6a. c xung dng kim gi ti Tiristo, ta dng bin p xung(BAX), c th khuch i cng sut ta dng Tr, it D bo v Tr v cundy s cp bin p xung khi Tr kho t ngt. Mc d vi u im n gin,nhng s ny c dng khng rng ri, bi l h s khuch i catranzitor loi ny nhiu khi khng ln, khuch i c tn hiu t khuso snh a sang.

Tng khuch i cui cng bng s darlington nh trn hnh 6b thnghay c dng trong thc t. s ny hon ton c th p ng c yucu v khuch i cng sut, khi h s khuch i c nhn ln theo thng

s ca cc tranzitor.Trong thc t xung iu khin ch cn c rng b (c khong (10

200) s), m thi gian m thng cc tranzitor cng sut di (ti a ti mtna chu k - 0.01s), lm cho cng sut to nhit d ca Tr qu ln v kch th-c dy qun s cp bin p d ln. gim nh cng sut to nhit Tr v kchthc dy s cp BAX chng ta c th thm t ni tng nh hnh 6c. Theo s

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A33

Ura

R4

Udk

R5Urc

b.

A3

Ura

Urc

Udk

c.

R5

R4


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ny, Tr ch m cho dng in chy qua trong khong thi gian np t, nndng hiu dng ca chng b hn nhiu ln.

3 .S mch iu kin- T nhng phn tch trn ta c s ca mch iu kin :

4. Nguyn l hot ng ca s iu khin:

Trang 1

S cc khu khuch i.a- bng tranzitor cng sut; b- bng s darlington;c- s c t ni tng.

R

A1 A2

R1A

R2

R3

C1

D1B

Tr1

U1

Ud

Urc

+E

A3

Ura

R4

R5

BAX

D2

R6

R7

C2

Tr2

R

Uv Tr2

BAX+E

D2

Tr1

b.

R6

Uv

Tr1

BAX+E

D2

a.

R6

R

R6

Uv Tr2

BAX+E

D2

Tr1C2

D

c.R

f


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in p vo ti im A (UA) c dng hnh sin, trng pha vi in panod ca Tiristo T, qua khuch i thut ton (KTT) A1 cho ta chui xungch nht i xng UB. Phn p dng ca in p ch nht UB qua it D1ti A2 tch phn thnh in p ta Urc. in p m ca in p UB lm mthng tranzitor Tr1, kt qa l A2 b ngn mch (vi Urc = 0) trong vng UBm. Trn u ra ca A2 chng ta c chui in p rng ca Urc gin on.

in p Urc c so snh vi in p iu khin Uk ti u vo caA3. Tng i s Urc + Uk quyt nh du in p u ra ca KTT A3.Trong khong 0 t1 vi Uk > Urc in p UD c in p m. Trongkhong t1 t2 in p Uk v Urc i ngc li, lm cho UD lt ln dng.Cc khong thi gian tip theo gii thch in p UD tng t.

Cc xung ra UF lm m thng cc tranzitor, kt qu l chng ta nhnc chui xung nhn Xdk trn bin p xung, a ti m Tiristo T.

III. Tnh ton cc thng s ca mch iu khinMch iu kin c tnh xut pht t yu cu v xung m Tiristor , do

ta c cc thng s c bn tnh mch iu kin :+ in p iu kin Tiristor : Uk = 1,4V+ Dng in iu kin Tiristor : Ik = Ig = 0,12A+ Thi gian m Tiristor : tm = 120ms+ rng ca xung iu kin :tx = 167ms+ Tn s xung iu kin fx = 3kHz

+ mt i xng cho php : = 40

+ in p nui mch iu khin U = 12V+ Mc st bin xung : sx = 0,15

1. Tnh my bin p xungChn vt liu lm li l Fit HM , li c dng hnh xuyn lm vic trn q

fn ca c tnh t ha c B = 0,3T ;H = 30A/m khng c khe h khngkh

+ T s bin p xung chn m 3+ in p th cp ca MBA xung : U2 = Uk = 1,4V

+ in p t ln cun th cp MBA xung :U1 = mU2 = 3,14 = 4,2V

+Dng in th cp ca MBA xung : I2 = Ik = 0,12A+ Dng in s cp ca MBA xung : I1 = 2.I2/m = 2. 0,12/3 = 0,08A+ t thm trung bnh tng i ca li thp

tb = 60 10.25,1.30

3,0

. =

H

B= 8.103

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Vi 0 = 1,25.10-6 (H/m) l t thm ca khng khTh tch ca li thp cn c :

V = Q.l = (tb.0.tx.sx.U1.I1)/B2

Thay s :

V= 8.10

3

.1,25.10

-6

.0,15.167.4,2.0,08)/0,3

2

= 0,4676.10-6m3 = 0,4676.10-3cm3

Da vo bng tra cu ta la chn c mch t c th tch V= 0,645cm 3 vkch thc c th nh sau: Q =0,1cm2, l = 6,45cm ; a = 2,5mm ; b = 4mm ; d=20mm ;

- S vng dy s cp my bin p xung :- Theo nh lut cm ng in t

U1 = W1Qtx

BQW

dt

dB = 1

W1 = 8,23310.1,0.3,0

10.167.2,46

6

1 ==

BQ

txU(vng)

Chn w1 = 234 vng- S vng dy th cp

W2 = 7832341

==m

W

(vng)Chn mt dng in J1= 6 (A/mm2)- Tit din dy cun th cp

S1 = 0067,06

08,0

1

1 ==J

I(mm2)

+ ng knh dy cun th cp

Trang 1

d D

a

b

Rf


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d2 = 092,00067,0.44 1 ==

S(mm)

Chn d = 0,1 (mm)- Tit din dy cun th cp

S2 = 03,0412,0

2

2 ==J

I

(mm2)

Chn mt dng in J2 = 4(A/mm2)+ ng knh dy cun th cp

d2 = 95,103,0.44 2 ==

S(mm)

Chn dy c ng knh d2 = 1,95 (mm)- Kim tra li h s lp y

Kl =1747,0

20

78.95,1234.1,0..

)4.(

..2

22

2

2

21

2

1

2

2211


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Uk = 2,0 (V) Ik = 0,1 (A)Do ta ch cn 1 tng khuych i cng cng sut iu khin Tranzito

- Chn ngun cp cho bin p xung E = 12(V) , vi ngun E =12 (V) taphi mc thm in tr R10 ni tip vi cc Emito ca Ir3 , R1 .

R7 = 5,9208,02,412

1

1

=

=

I

UE

()Tt c cc Diod trong mch iu khin u dng loi 1N4009 c thams .

+ Dng in nh mc Im = 10 (mA)+ in p ngc ln nht Un = 25 (V)+ in p cho Diod m thng Um = 1(V)

3.Chn t C2 v R6- in tr R6 dng hn ch dng in a vo Baz ca Transistor I r2

ta chn R9 tho mn iu kin:R6 === kI

U

B

8,225,28110.2.8,0

5,43

3

Chn R6 = 3kChn C2 sao cho C2.R6 tx

=> C2 nFFRtx 6,55)(10.6,55

10.3

167 93

6

===

Chn C2 = 33nF

4.Chn cc khut thut ton- Mi knh iu kin phi dng 4 khuych i thut ton , do ta chn8 IC loi TL084 do hng texas instrument , mi IC ny c 4 khuchi thut ton

- Thng s ca IC TL084 :+ in p ngun nui : VCC = 18V, Chn VCC = 12V+ Hiu in th gia 2 u vo : 30V+ Nhit lm vic : T = - 2585 C0

+ Cng sut tiu th P = 0,68W+ Tng tr u vo : Rmin = 106+ Dng in u ra : Ira = 30pA+ Dng in u vo : Iv = 1mA

+ Tc bin thin in p cho php : )/(13 sVdt

du=

S chn IC TL084 :

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.S chn IC TL084

13

1

14

2 3 4 5 6 7

12 11 10 9 8- -

--

+ +++

Ucc

5.Tnh tng so snh

- Mi knh iu kin c 1 khuych i thun ton ng vai tr tng sosnh ta chn loi IC TL084 nh trn

- Chn R4 = R5 > == kI

U

r

dk 12

10.1

123

- Trong ngun nui VCC = 12V , th in p vo A3 Ur = 12V dngin vo hn ch Ilv < 10-3A do ta chn R4 = R5 = 15k khi dngvo A3 :

Ilv-max = mA8,010.15

123=

6.Chn khu ng pha

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A33

Ura

R4

Udk

R5Urc

A1 A2

R1A

00

11

22

33

44

55

R2

Ur

R3

C1

CD1

B

Tr

U1


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in p t c hnh thnh do s np ca t C1 , mt khc m bo phmvi iu kin rng th gc iu kin = 0 1800 th hng s th gian t npc : Tr = R3.C1 = T/2 = 1/2f = 0,01s

Chn C1 = 1F th in tr R3 = == kCTr 10

10.1

01,06

1

thun tin cho vic iu chnh khi lp rp mch R 3 thngg chn bin trR3 ln hn 10k iu chnh

- Chon Tranzitor T1 loi A564 c cc thng s:+ TRanzitor loai pnp lm bvng Si+ in p gia Emitor v Baz lc h mch colector :UEB0 =7V+ Dng in ln nht m colector c th chu c : IC-max = 100mA+ Nhit ln nht mt tip gip : TCP= 1500

+ H s khuych i = 250

+ Dng cc i ca Baz : IB3 =A

IC 4,0250

100==

+ in tr hn ch dng in i vo baz Tranzitor Tr1 c chn nhsau :Chn R2 tho mn iu kin :

R2 == k

I

U

B

N 3010.4,0

123

max

Chn R2 = 30kChn in p ng pha : UA= 9Vin tr R1 hn ch dng in i vo khuych ddaij thut ton A 1 th-

ng chn R1 sao cho dng vo khuych i thut ton : Ir < 1mA. Do R1 > == kI

U

r

A 910.1

93

Chn R1 = 10 k

7.Tao ngun nuiTa cn chn ngun nui 12V cp cho BAX nui IC, cc b iu

chnh dng in tc , in p t tc . Ta dng mch chnh lu cu bapha dng it, in p t cp MBA ngun nui :

n nh in p ra ca ngun nui ta dng hai vi mch n p

7812 v 7912, cc thng s chung ca vi mch ny:in p u vo : UV = 7 35V.in p u ra : Ura = 12V vi IC 7812

Ura = - 12V vi IC 7912.Dng in u ra : IR = 0 1A.St p nh nhttrn IC 7812 l U = 4V

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Ud= 12- 4 = 8V

)(88,88,0

82 VU ==

Ta chn U2 = 9V.T in C4 , C5 dng lc thnh phn sng hi bc cao.Chn C4 = C5 = C6 = C7 =470F.

8.Tnh ton my BA ngun nui v ng phaTa thit k MBA dng cho c 3 vic to in p ng pha v tongun nui cho tt c cc mch iu kin bao gm cng sut cung cp cho 8mch ng pha v cung cp ngun nui cho 8 IC TL084 to ra 32 khuychi thut ton. Chn kiu MBA 1 pha 1 r, trn mi tr c 2 cun dy : 1cun s cp v 1 cun th cp.

in p ly ra th cp MBA lm in p ng pha , ly ra th cpca ngun nui :

U2 = U2p = UN = 9(V).- Dng in th cp MBA ng pha :

I2p = IA1-max = 1(mA)- Cng sut ca MBA cung cp cho vic to nn p ng pha

Pp = 8.I2-dp .U2-p = 8.9.1.10-3 = 0,072( W)- Cng sut tiu th 8IC TL 084 s dng lm khuych thut ton .

P8IC = 8.0,68 = 5,44W- Cng sut BA xung cp cho cc 8 iu khin Tiristor

Px = 8. Uk . Ik = 8.1,4.0,12 = 1,344(W)

Trang 1

+12V

0

7912

.

C7

C6

-12V

C5

C4

.

7812

220~


44/46


- Tng cng sut ca MBA cung cp :P = Pp + P8IC + PXP = 0,072 + 5,44 +1,344 = 6,856(W)

- Cng sut ca MBA c k n 5% tn tht trong my :

S = 1,05.P = 1,05 .6,856 = 7,1988 (VA)- Dng in th cp MBA :

I2 = 9.61988,7

.6 2=

U

S= 0,133(A)

- Dng in s cp MBA:

I1 = 220.31988,7

.3 1=

U

S= 0,011(A)

- Tit din tr ca MBA c tnh theo cng thc kinh nghim:

)(32,150.3

1988,7.6.

.2cm

fmSkQ QT ===

Trong : kQ = 6 : h s ph thuc phng thc lm mt.m = 3 : s tr ca MBAf = 50 : tn s in p li.

- Ta chn li thp hnh ch E , c t cm B = 1T c kch thc nhhnh v :

Ta c : QFe = a.b = 1,32cm2

Trong qu trnh thit k ta thng chn 1 h = 2,5.1 = 2,5 cm

n = 5,0=a

c=> c = 0,5.1 = 0,5 cm

- Chn mt t cm B = 1T trong tr ta c s vng dy s cp :

Trang 1

b

hH

c

C

a c


45/46


07,5751,32.104,44.50.1.

220

4,44.f.B.Q

Uw

4

Fe

11 === (vng)

Chn w1 = 7508 ( vng)- Chn mt dng in J1 = J2 = 2,75A/mm2.

- Ta c tit din dy qun s cp :)0,004(mm

2,75

0,011

J

IS 2

1

11 ===

- Chn dy dn trn , tit din theo qui chun : S1 = 0,00785mm2

- ng knh thc ca li dy qun s cp : d1 = 0,1mm- ng knh c k cch in l D1 = 0,12mm.- S vng dy th cp :

07,13220

9.7508

U

U.w

1

212 === w (vng)

Chn w1 = 308 ( vng)- Chn mt dng in J1 = J2 = 2,75A/mm2.- Ta c tit din dy qun th cp :

)0,048(mm2,75

0,133

J

IS 2

2

22 ===

- Chn dy dn trn , tit din theo qui chun : S2 = 0,095mm2

- ng knh thc ca li dy qun s cp : d2 = 0,11mm- ng knh c k cch in l D2 = 0,14mm.- Chiu di mch t :

L = 2c +3a =2.0,5 +3.1 = 4cm

- Chiu cao mch tH = 2a + h =2.1 + 2,5 = 4,5 cm- Chn cc it cho b chnh lu- Dng in hiu dng qua it

I =2

2I = 094,041,1

133,0= ( A)

- Chn it c dng nh mcI-m = ki.I = 10.0,094 = 0,94 A ( ki = 10 h s d tr v dng )

- in p ngc ln nht m it phi chuUng-max = )(229.6.6 2 VU ==

- Chn it c in p ngc ln nhtU ng-max = ku.Ung-max = 2.22 = 44 (V)Vi ku = 2 h s d tr v in pChn it loi : kY Z70 c Ung-max = 50 (V) ; Im = 20 (A)

Trang 1


46/46


Ti Liu tham kho

1. in t cng sut - Nguyn Bnh ( Nh xut bn KHKT)2. Truyn ng in - Nguyn Vn Lin ( Nh xut bn KHKT )3. My in Tp 1,2,3 Nguyn Khnh H - V Gia Hanh

( Nh xut bn KHKT )4. Ti Liu Hng Dn Thit K TCS - Phm Quc Hi

dong co 1 chieu kich tu doc lap (do an)

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