ĐỒ ÁN ĐIỆN TỬ CÔNG SUẤT trung

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2/49SVTH: L XUN TRUNG 11D1CLC

CHNG I

TNG QUAN VCC PHNG PHP IU CHNH TC

A. NG CIN MT CHIU.

1.1. Khi qut chung:

ng c in mt chiu cho php iu chnh tc quay lin tc trng mt phmvi rng v trong nhiu trng hp cn c c tnh c c bit, thit bn gin hnv r tin hn cc thit biu khin ca ng cba pha.V mt su im nh

vy cho nn ng c in mt chiu c s dng rt ph bin trong cng nghip,trong giao thng vn ti.

2.1. Cu to ca n c in mt chiu.

ng c in mt chiu chia thnh 2 thnh phn chin:

- phn tnh( stato).

Gm cc b phn nh sau:

Cc chnh t: l b phn sinh ra ttrng, gm li st cc t v dy qun kch t.

+ Li st cc t lm bng l thp k thut in dy( 0,5-1)mm p li v tn cht.

+ Dy qun kch tc qun bng dy ng bc cch in.

Trong cc my c cng sut nh cc t chnh l mt nam chm vinh cu. Cntrong my c cng sut ln cc tl nam chm in.

Cc t pht gia cc tchnh dng ci thin trnh trng lm vic cu myin v i chiu.

+ li thp cc t ph c th l mt khi hoc c gp bng cc l thp ty theoch lm vic.



Xung quanh cc t phc t giy qun cc t ph, dy qun cc t phc ni vi dy qun phn ng.

Gng t: dng lm mch t ni lin cc cc t, ng thi lm v my.

- Phn quay(roto)Bao gm cc b phn sau:+ li thp phn ng: dng dn t, thng dng nhng tm thp k thut indy 0.5mm phcch in mng hai mt ri p cht li gim tn hao do dngin xoy gy ln.

Trong my in nh li thp phn ng c p trc tip vo trc cn tong myin ln gia trc v li st c gi roto.

Dy qun phn ng l phn sinh ra sc in ng v c dng in chy qua, n

thng c lm bng ng bc cch in.Cgp: dng i chiu dng in xoay chiu thnh mt chiu.

C cu ni than: dng a dng in t phn quay ra ngoi.2.2. phn loi ng c in mt chiu.

Cng nh my pht, ng c in cng c phn loi theo cch kch thch tthnh cc ng c sau:- ng c in kch tc lp:ng c in mt chiu kch t c lp c qun kch tc cp in t mtngun in ngoai c lp vi ngun in cp cho mch phn ng.- ng c kch t ni tip:ng c kch t ni tip c cun kch t mc ni tip vi cun dy phn ng.- ng c kch t hn hp:Gm 2 dy qun kch t: dy qun kch t song song v dy qun kch t ni tiptrong dy qun kch t song song l ch yu.2.3. Nguyn l lm vic ca ng cin mt chiu

Khi cho in p mt chiu U vo hai chi in, trong dy qun phn ng c

dng in I. Cc thanh dn c dng in nm trong t trng, s chu lc Ft

tc dng lm cho rto quay.

Khi phn ng quay c na vng, v tr cc thanh dn i ch

cho nhau, do c phin gp i chiu dng in, gi cho chiu

lc tc dng khng i, m bo ng cc chiu quay khng i.



Khi ng c quay, cc thanh dn ct t trng, s cm ng sc in ng E..

ng cin mt chiu sc in ng E ngc chiu vi dng in I nn

sc in ng Ecn c gi l sc phn din.

Phng trnh in l:

3.phng trnh c tnh c:

iu khin c tc ng c in mt chiu kch tc lp th ta phi phntch, tm cc mi quan h gia tc vi cc thng s khc ca ng c ta ra phng php iu khin. ng c in mt chiu kch tc lp th dngkch tc lp vi dng phn ng.V c nui bi hai ngun mt chiu c lpvi nhau.

Hnh 1. S ni dy ca ng cmt chiu kch tc lp

Theo hnh 1 ta vit phng trnh cn bng in p ca mch phn ng :

U=E+ ( R+ Rf)I (1)

Trong : U: in p t ln phn ng ng c(V).

E: Sc in ng phn ng (V)

R: in trca mch phn ng ()

Rf: in trph trong mch phn ng ()

I: Dng in mch phn ng ()

Vi R= r+ rcf+rb+ rct

Trong : r in trcun dy phn ng ()



Rcf: in trcun cc t ph()

Rb: in trcun b ( )

Rct: in trtip xc chi in ()

Sc in ng Eca phn ng ng c c xc nh theo biu thc sau:

trong : p : Si cc tchnh

N : S thanh dn tc dng ca cun dy phn ng

a : Si mch nhnh song song ca cun dy phn ng: T thng kch tdi mi cc t(wb)

: Tc gc (rad/s)

K =

: H s cu to ng c.Nu biu din sc in ng theo tc quay n (vng/ pht) th

= =

V vy : Vi : : H s sc in ng ca ng c. .



Thay (1) v (2) v bin i ta c : ; (4)Biu thc (4) l phng trnh c tnh c in ca ng c.Mt khc, mmen in

t: Mt= . (5)Nu b qua tn tht trong cc trc, tn tht t qut mt v tn tht trong thp thmmen ctrn trc ca ng cbng mmen in t, ta k hiu l M, tc l :

Mt= Mc=M

Vy phng trnh c tnh cca ng cl :

( 6)Biu thc (6) l phng trnh c tnh cca ng c in mt chiu kch tc

lp .

Khng xt n nh hng ca phn ng phn ng ngang trc lm gim t thng ca ng ctc l xem =const th quan h=f(M,I) l tuyn tnh.

Hnh 2 thc tnh cca ng c in mt chiu kch tc lp

Tth ta c : khi I= 0 hoc M =0 ta c:



(7)0: c gi l tc khng ti l tng ca ng c, khi =0 ta c:

V M = k (9)Inm, Mnm: c gi l dng in ngn mch v mmen ngn mch.

Mt khc :phng thnh c tnh (4) v (6) cng c th vit di dng:

V : ta suy ra t (5)Trong : R = R

+ R

f,

: c gi l st tc ng vi gi trca M (hay I) , ta c th biu din c

tnh c in v c tnh ctrong hn vtng i vi iu kin tthng l nh

mc ( =m)

( Rcb=Um/ Imc gi l in trcbn )

T(4) v (6) ta vit c tnh c in v c tnh c n vtng i



*=1- R*I* (13)

= 1- R*M(14) cng ca c tnh c:

Cng sut (nng lng in)

Tphng trnh l tng : IU =(E+IR)I (16)

Ta c Pin=Pt+

Trong Pt = IE cng sut in t.

P =I2R tn hao cng sut ctrn in trphn ng .

Thc t :Pin=Pt+ P+ P0 (17)

Vi P0 tn hao ma st do s quay

T biu thc (4) hoc (6) ta thy l mt hm ph thuc R, ,U : =f(R,,U) do iu chnh tc ng c in mt chiu kch tc lpc ba phng php iu khin sau :

- iu khin in trph phn ng

- iu khin t thng kch t

- iu khin in p phn ng

Sau y ta xem xt tng phng php iu khin mt

2. CC PHNG PHP IU KHIN NG C IN MT CHIUKCH TC LP

2.1 Phng php iu khin bng in trph phn ng (Rf):

Nguyn l iu chnh:



Ni thm in trph Rfvo mch phn ng

Ta phn tch trn nn ta c = fRf, kt, U, gi thit rng :Nu gi=m=const ; U= Um= const; R=const th =f(Rf)

Mun thay i c gi tr Rfca mch phn ng bng cch ni tip mt in trph(Rf) thay i c gi trvo mch phn ng. Lc ny ta c : R = R+ Rf

Tphng trnh c tnh c :

Tphng trnh trn ta thy : khi tng gi tr ca Rf th tc ca ng cgim,khi gim gi tr ca Rf th tc ca ng ctng .

Lc ny ta c tc khng ti l tng:

Nh vy khi thay i Rf cho ta mt hc tnh nh sau:

Hnh 1-3 : c tnh cca ng c in mt chiu kch tc lp nhng intrph khc nhau

Nhn xt: Nu Rfcng ln th tc ng ccng gim, ng thi Inm v Mnmcng gim. Phng php ny c dng hn chdng in ng ckhi khing.



- u im : n gin , d thc hin.

-Nhc im : + cng c tnh cthp.

+ Tn tht nng lng trn in trln .

+ Phm vi iu chnh hp.

2.2. Phng php iu chnh bng tthng kch t:

Nguyn l iu chnh:

iu chnh t thng kch t ca ng c in mt chiu l iu chnh m men int ca ng cM = kIv sc in ng quay ca ng c

E=k

Hnh 1-4 :S ni dy iu chnh kch t ca ng cin mt chiu kch tc lp.

T biu thc (4) v (6) ta thy = f(U, kt, Rf), nu gi U=Um=const v in trphn ng R= const (Rf=0 ) lc ny = f(kt). thay i c tc ta cn

thay i kt, m t thng kch t do dng kch t sinh ra.

Vy iu chnh kt ta mc thm bin trRv vo mch kch t, khi iu chnhkt ta phi tun theo iu kin sau.

Khng thtng dng kch t Ikt ln hn dng nh mc ca cun dy kch t v nph hng cun kch t v khi kt = m bo ha ri, nu mun tng Ikt th ktcng khng tng ng knn ta iu chnh bng cch gim kt.

Trong trng hp ny ta c:



- Tc khng ti l tng

- cng ca c tnh c

Do cu to ca ng c in, thc thng iu chnh gim t thng. Nn khi tthng gim th x tng, cn s gim. Ta c thc tnh cvi x tng dn v cng ca c tnh gim dn khi gim t thng.

Hnh 1-5 : c tnh cv c in ca ng c in mt chiu kch tc lp khigim t thng.

T th ta nhn thy khi tthng thay i vi m>1 >2ta c:

- Dng in ngn mch : = - M men ngn mch : Mnm = kxInm = var (Mnm > Mnm1 > Mnm2)

T th c tnh ta thy 0



- Tc khng ti l tng: tng.- cng c tnh c :

gim.

- Di iu chnh khng rng. : D = 1 : 2+ Tc nh nht b chn bi c tnh t nhin (=m)

+ Tc ln nht maxb gii hn bi bn ckh v iu kin chuyn mch cang c

- u im : Cng sut mch iu chnh nh, tn tht nng lng nh

- Cng c th sn xut nhng ng cgii hn iu chnh 1:5 thm ch n 1:8nhng phi dng nhng phng php khng chc bit, do cu to v cngngh ch to phc tp khin cho gi thnh ca my tng ln.

2.3. Phng php iu khin ng c in mt chiu kch t c lp bngthay i in p phn ng:

Gi thit = m=const; R=const,(Rf=0), M = const. Lc ny = f(U)

Khi thay i in p phn ng ta c:

Nh vy khi thay i in p phn ng cho ta mt hcc c tnh sau.



Vi Um>U1>U2>U3>>Ui

Hnh 1-6 : thc tnh cca ng c in mt chiu kch tc lp khi inp phn ng thay i.

Nh vy khi thay i in p phn ng ng cta c mt hc tnh csongsong vi c tnh t nhin.

Nhn xt :

- u im : + khng gy n

+ khng gy tn hao phtrong ng c

+ di iu chnh rng D 10 : 1+ cng c tnh ckhng i trong tn di iu chnh

+ d tng ha.

- Nhc im : + phng php iu chnh ny cn mt b ngun c th thay itrn in p ra.

+ iu khin phc tp

Kt lun : Vi s pht trin ca khoa hc k thut nh hin nay th phng phpny cng c s dng ph bin trong sn xut cng nhtrong cc lnh vc khc.

2.4. Nuyn l iu chnh:

c tnh iu chnh :



iu chnh in p phn ng, ta phi s dng mt b bin i, iu chnh cin p u ra cp cho mach phn ng ca ng c.

Hnh 1-7 : S ngyn l iu chnh ng c

B bin i dng bin i in p xoay chiu ca li in thnh mt chiuv iu chnh c gi trin p u ra theo yu cu. in trtrong ca b bini. Rbph thuc vo loi thit b, v thng thng cng sut ca b bn i vng cxp x bng nhau nn Rbcng c gi trng k so vi R ca ng c.

Ts nguyn l ta c s thay th

Hnh 1-8 : S thay thnguyn l iu chnh ng c

Ts thay thta c phng trnh cn bng in p.

E= Ub(Rb+ R)I (20)

Sc in ng ca ng c

E= km (21)

T biu thc (20) v (21) ta c



km= Ub( Rb+ R)I(22)

Tphng trnh (22) ta c phng trnh c tnh c in.

Phng trnh c tnh c:

Vi : M =kmI

Trong ml tthng nh mc ca ng c, m= const.Tc khng ti l tng

cng ca c tnh c

Ta thy tc khng ti l tng khng ph thuc vo M, I m ph thuc vo Ub.

Ub=kaUk

Trong : ka l h s khuych i ca b bin i

Ukl in p iu khin

Tphng trnh (23) v (24) ta c thc tnh c biu din nh sau.



Hnh 1-9: thc tnh c in khi Ub thay i

T th ta thy, khi Ubthay i th ta c nhng tc khng ti l tng khc

nhau, cn cng c tnh ckhng i v c tnh iu chnh dc hn c tnh tnhin.

Nhn xt :

- iu chnh tc trong bt k vng ti no, k c khng ti l tng , c tnhc iu chnh tuy mm hn c tnh tnhin nhng cng hn phngphp dngbin trv thay i kt

- iu khin phc tp, vn u t ln nhng n l phng php tt hn so vihai phng php trn

Kt lun : Sau khi phn tch ba phng php iu khin nu trn th phngphp iu khin bng thay i in p phn ng l tt hn c. Cho nn em chnphng php lm phng php nguyn cu cho ti ca em.

3. CC B BIN I IU CHNH TC NG C MT CHIUKCH TC LP BNG THAY I IN P PHN NG .

- Hin nay trong cng nghip ngi ta thng s dng ba loi b bin i sau:

+ B bin i my pht in mt chiu



+ B bin i xung p mt chiu

+ B bin i chnh lu c iu khin

- Tng ng vi vic s dng cc b bin i ta c cc h truyn ng sau:

+ H my phtng c( F )

+ Hiu chnh xung png c( XA )

+ H chnh lu Tiristo ng c( T )



Chng II

TNG QUAN V B CHNH LU THYRISTOR HNH CU 3 PHA.

THIT KS N NGHUYN L H THNG CHNH LUNG C IN 1 CHIU(h T- ) KCH TC LP.

A. TNG QUAN V B CHNH LU THYRISTOR HNH CU 3 PHA.I. TNG QUAN V B CHNH LU THYRISTOR HINH CU 3

PHA.1. 1 gii thiu v thyristor.

Thyristor l linh kin gm 4 lp bn dn P1-N

1-P

2-N

2lin tip to nn

ba cc: anot A, catot K v cc iu khin G(gate) Ti ba v tr tip xcnhau ca cc lp P1-N1-P2-N2 to nn cc lp tip gip J1,J2,J3.V l thuyt c hai loi thyristor:

- Thyristor kiu N hay thyristor c cc iu khin G ni vi vng N gn anot.

- Thyristor kiu P hay thyristor c cc G ni vi vng P gn catot.- Hot ng ca thyristor:- Thysristor lm vic ba trng thi: ng mkha

UBR: in p ngc nh thng

UBO: in p t mca thyristor

UTO: in p ri trn Thyristor

IL latching

IH dng duy tr.



+ thysristor kha nu UAK< 0 v s vn kho nu ta cho UAK> 0.+ thysristor chuyn qua trng thi t kha sang dn nu ng thi m bo 2iu kin: UAK> 0, v c dng iu khin IG mnh( v c cng sut vthi gian). Khi thyristor dn nu ngt dng iu khin i( cho IG=0) n svn dn chng nao dng in qua n ln hn mt gi tr gi l dng duy tr.

- Trong thc tngi ta s dng thyristor kiu N nhiu hn.II. CHNH LU THYRISTOR CU 3 PHA.

1.1 s chnh lu cu thyristor cu 3 pha.



1.2cc tham s chnh ca mch chnh lu cu 3 pha:- gi trtrung bnh in p chnh lu.

Khi th Ud0 = 2,34.U2- Tr strung bnh dng in qua van:

- in p ngc ln nht m van phi chu khi lm vic:

- Cng sut tnh ton mch MBA ngun: Sba = 1,05.Pd- Hs mch cu in p chn lu Km = 0,057 s chnh lu 3 pha( H

gm 6 thyristor chia than hai nhm)+ Nhm catot chung: T1,T3,T5+ Nhm anot chung: T2,T4,T6.



- in p pha:= =

= G m c tnh tgiao im ca cc na hnh sin ( hay thi gian chuynmch t nhin).

- Hot ng ca s .Gi thit T5 v T6 ang cho dng chy qua VF = U2C ,VG = U2b khi cho xung iu khin m T1. Thyristor mv U2a >0. S mcaT1 lm cho T5 b kha li mt cch t nhin v U2a > U2c, lc ny T6 v T1

cho dng chy qua. in p trn ti Ud = Uab =U2a = U2b.Khi cho sung iu khin mT2. Thyristor ny mv khiT6 dn dng, n t U2b nn anot T2 m U2b > U2c. s mca T2 lm cho T6b kha li mt cch t nhin U2b > U2c.

Cc sung iu khin lch nhauc ln lt a n cc iu khin ca

cc thyristor theo th t1,2,3,4,5,6,1,2..Trong mi nhm khi 1 Thyristor m n s kha ngay thyristor dn dng

trc n.Tm tt bng:

Gi trtrung bnh in p ti:



Nhn xt: Chnh lu cu thyristor cu 3 pha l loi c s dng nhiu nht trongthc t v c nhiu u im hn c. N cho php u thng vo li in 3 pha, p mch rt nh(5).Nu c dng bin p th gy mo li it hn cc loi trn ng thi cng sutMBA cng ch xp x bng cng sut ti. cng sut mch chnh lu c th rt lnln ti hng trm kw.

Nhc im ca n l st p trong mch van gp i s hnh tia nn khng phhp vi in p ra ti di 10V.

2.2.2. Hin tng trng dn .

Gi s T1 , T2ang dn dng . Khi = 1 cho xung iu khin mT3 , do Lc0 nn iT3 khng thtng t ngt t0 Id v dng iT1cng khng th gim tngt t Id 0 . C ba thyristor T1 , T3 , T5u dn dng , hai ngun ea , eb ningn mch .

Nu chuyn gc to t0 n 1 ta c :

ea = 2 . U2sin( +6

5 + ) (2-2)

eb = 2 . U2sin( +6

+ ) (2-3)

in p ngn mch Uc = UbUa = 6 . U2sin( + )Dng in ngn mch icc xc nh bi phng trnh :

Uc = 6 . U2sin( + ) = 2Xc. ddi

c ; ic =cX

U.2

.62 [ cos cos( + ) ] ; (2-5)



Hnh 2-6: S chnh lu cu ba pha c hin tng trng dn .

Hnh 2-7 : S dng sng chnh lu cu ba pha c trng dn .

Dng in chy trong T1 l iT1 = Td - Ic ;

Dng in chy trong T3 l iT3 = ic ;Gi thit qu trnh trng dn kt thc khi = 2 v k hiu = 21 l gc trngdn .

Khi = , iT1 = 0 , ta c biu thc sau :

2

ac ee

Ud

0

1 3 5

6 2 4

5;1 1;3 3;5

6;2 2;4 4;6

2

cb ee

2

ba ee

u2a u2b u2c

VG

VF

FG

T1T4

T3

T6

T5T2

Lc

I

eaebec

LcLc

ic



cos cos( + ) =2.6

..2

U

IX dc (2-6)

Hnh dng in p ti Udtrong giai on trng dn trong khong 1 2 . T2 dn

dng , T1 , T3 trng dn dng . Vy c th vit cc phng trnh sau :

ea - ec - 2Lcdt

diT1 = Ud (2-7)

ebec - 2Lc.dt

diT3 = Ud (2-8)

iT3 + iT1 = iT2 = Id =const (2-9)

T 3phng trnh ta rt ra :

Ud =2

ba ee - ec (2-10)

Do trng dn ( Lc 0) nn tr trung bnh ca in p ti b gim i mt U ,v c xc nh nh sau :

U =.2

6

0 0

).sin(.63

)2

( ddee

eba

b

=.2

63[cos - cos(+ ) ] (2-11)

T ta c thxc nh c : U =

cLX ...3 ; (2-12)

2.2.3. Nghch lu ph thuc.

Ta c ch chnh lu dng in trung bnh trn ti Idv in p trung bnh

Ud lun cng chiu . Cng sut tiu th trn ti P = Ud.Id lun dng v chiuca cng sut lun t pha ngun xoay chiu chuyn qua ti mt chiu , ta ni bbin i lm vic ch chnh lu.



Hnh 2-8 : a ) Ch chnh lu ; b ) Ch nghch lu .Khi tng gc m , gi tr trung bnh ca in p chnh lu Ud gim i nhng

vn cn dng . Tip tc tng > 2

, in p chnh lu trung bnh i du . Cc

M tr thnh m ( - ) v N thnh dng ( + ) , in pUd tng dn n khi = .V in p Udi chiu trong khi Id c chiu khng i nn cng sut P i du .iu ny c ngha l ti mt chiu trthnh ngun pht v cng sut truynngc t pha ti v ngun . Ta ni b bin i chuyn sang lm vic chnghch lu . Lu rng ch nghch lu , li xoay chiu nhn cng sut tcdng t pha ti nhng vn tip tc cung cp cng sut phn khng v nh hngn dng sng , tn s ca in p xoay chiu ta ni rng s ny lm vic ch

nghch lu ph thuc .

li c th nhn nng lng t pha ti th ti phi l mt ngun pht vin p ca ti phi ln hn in p ca ngun m bo cho van bn dn phncc thun

Hnh 2-9 : S dng sng ca nghch lu ph thuc .

Ti

a )

M

N

_

+

TiUd

Id

BB

Cng sut P = Ud.Id > 0

b )

M

N

+

_

Ud

Id

BB

Cng sut P = Ud.Id < 0

u2a u2b u2c

2 30

U VG

VF

t

1 3 5

6 2 4 6 2 4

1 3

Utb



Trong trng hp nghch lu ph thuc th ta phi c gc iu khin > /2ta c in p trung bnh ca nghch lu ph thuc l Utb< 0 nh hnh 2-9 .

B.S NGHUN L H H THNG CHNH LU NG C IN 1CHIU ( H T) KCH TC LP.

3.1 Tng quan v h truyn ng T- khng o chiu.

Khi dng cc b chnh lu c iu khin hay l b chnh lu dng thyristor lmb ngun mt chiu cung cp phn ng ng c in mt chiu, ta c in p chnhlu ca h:

Vy ta c phng trnh c in v phng trnh c tnh c ca h chnh lu T

khng o chiu.

l tc khng ti l tng, v lc vng dng in ginon h s c thm mt lng st p nn ng c tnh biu din dc hn. Tc khng ti l tng thc s ln hn tc khng ti gitng Vy khi thay i gc iu khin th thay i tn (- v

ta sc mt hc tnh song song vi nhau nm bn phi mt phng to [I] hoc [ M].

Trong : XBA: in khng my bin p. fl: Tn sli.

: in cm tng mch phn ng. m: s pha chnh lu



a. Ch dng lien tc.Dng chnh lu Id chnh l dng in phn ng ng c in. Da vo s thayth ta vit c phng trnh c tnh:

2.3. Tng quan h truyn ng T c o chiu.2.3.1. Nguyn tc c bn xy dng h truyn ng T-o chiu :

- Gi nguyn chiu dng in phn ng v o chiu dng kch t .



- Gi nguyn dng kch tv o chiu dng phn ng nhng c phn rabn s chnh :

+ Truyn ng dng mt b bin i cp cho phn ng v o chiu quay

bng cch o chiu dng kch t .+ Truyn ng dng mt b bin i cp cho phn ng v o chiu quay

bng cng tc t chuyn mch phn ng ( t thng gikhng i ) .

+ Truyn ng dng hai b bin i cp cho phn ng iu khin ring .

+ Truyn ng dng hai b bin i ni song song iu khin chung .

Tuy nhin , mi loi s u c u nhc im ring v thch hp vi tngloi ti , trong phn ny ta chn b truyn ng dng hai b bin i ni song songngc iu khin chung , bi n dng cho di cng sut va v ln c tn sochiu cao vthc hin o chiu m hn .Trongs ny ng c khng nhngo chiu c m cn c

2.3.2.Phng php iu khin chung :

S gm hai b bin i G1 v G2, u song song ngc vi nhau v cccun khng cn bng Lc . Tng b bin i c th lm vic ch chnh luhoc nghch lu .

Phng php iu khin kiu tuyn tnh : 1 + 2 =

Lc ny c hai mch chnh lu cng c pht xung iu khin , nhng lunkhc ch nhau : mt mch ch chnh lu ( xc nh du ca in p mtchiu ra ti cng l chiu quay ang cn c ) cn mch kia ch nghch lu . Vhai mch cng du cho mt ti nn gi tr trung bnh ca chng phi bng nhau .

Ut = Ud1 = U d2 ; (2-13)

Nu dng in lin tc ta c : Ud1 = Ud0.cos1 ;

Ud2 = Ud0.cos2 ;

Vy : Ud0.cos1 = Ud0.cos2 ;

Hay : cos1+ cos2 = 0 ; suy ra 1 + 2 = 1800 ;



Nu 1 l gc mi vi G1, 2 l gc mi vi G2 th s phi hp gi tr1 v 2 phi c thc hin theo quan h :

1+ 2 = 1800 ;

S phi hp ny gi l phi hp iu khin tuyn tnh (hnh 2-11) .

Hnh 2-11 : S phi hp tuyn tnh ca 1v 2 .

Gi s cn ng c quay thun , ta cho G1 lm vic ch chnh lu ,1= 0 90

0 , Ud1 > 0 , by gi2 > 900 , G2 lm vic ch nghch lu , Ui2 0 ;

Ui2 = U0cos 2 < 0 ;

Chai in p Ud1 v Ui2u c ln phn ng ca ng c M . ng c chc thnghe theo Ud1 v quay thun . ng c t chi Ui2 v cc thyristor khngth cho dng chy tcatt n ant .

Khi 1= 1 = 900

, th Ud1 = Ui2= 0 , ng c trng thi dng.th hm tisinh .

2 = 0 1 = 0

1=1800

2=1800

300

600

900

1500

1800

1200

0

ucuc1

uc2

uc4 uc3



Hnh 2-12 : S chnh lu cu ba pha dng phng php iu khin chung .

Gi s ucl in p iu khin biu khin cn khi ng M quay thunta cho uc = uc1 (hnh 2-11) .

1 = 300, 2 = 180

0 - 1 = 1500 , Ud1 =

2

3U0 , Ud2 = -

2

3U0 ;

G1 lm vic ch chnh lu cn G2 chun b sn sng lm vic chnghch lu . Nu by gicn gim tc ng c , ta cho uc = uc2 , cc gc m:

1 = 600, 2 = 180

0 - 1 = 1200 , Ud1 =

2

1 U0, Ud2 = -2

1 U0 .

T1T4

T3

T6

T5T2

Lc

FG

T4T1

T6

T3

T2T5

+ -Ckt

M

Ikt

L

Lc

Lc

Lc

u2au2bu2c

G1

G2



Lc ny , do qun tnh nn sc in ng E ca ng c vn cn gi nguyntr s ng vi trng thi trc , E > Ud1 b bin i G1 b kho li .

Mt khc E > |Ud2| nn b bin i G2 lm vic ch nghch lu ph

thuc , trnng lng tch lutrong ng c v ngun in xoay chiu . Dng inphn ng i du , chy t M vo G2ng c b hm ti sinh , tc gim xungn gi trng vi Ud1 .

Nu cho in p iu khin uc < 0 th G2 s lm vic ch chnh lu , cnG1 s lm vic ch nghch lu ph thuc .

Vy bng cch thay i in p iu khin uc ( uc > 0 hoc uc < 0 ) ta s thayi c gc m1v 2 :

+ Nu uc> 0 th 1 < 900, 2 > 90

0 dn n b chnh lu G1 lm vic chchnh lu , cn b bin i G2 lm vic ch nghch lu ph thuc kt qu llm cho ng c quay thun T .

+ Nu uc< 0 th 1 > 900, 2 < 90

0 dn n b chnh lu G1 lm vic chnghch lu ph thuc , cn b bin i G2 lm vic ch chnh lu , kt qu llm cho ng c quay theo chiu ngc N .

c im ca cho dng ang xt l c mt dng in lc th chy tG1 vo G2 , lc th chy t G2 vo G1 m khng qua mch ti . Ngi ta gi dng

in ny l dng in tun hon .Dng in tun hon lm cho my bin p v cc thyristor lm vic nng n

hn . hn chdng in tun hon ngi ta dng bn in cm Lc(nh hnh 2-12) . Nh th slm tng cng sut t v gi thnh h thng . Tuy nhin phng

php iu khin chung cho php iu chnh nhanh ti a .

- Xc nh dng in tun hon icc .



s

Xt trung hp 1 = 30

0

, 2 = 180

0

- 1 = 150

0

, nh hnh 2-13 . Trong khong1n 3 : c T1 v T6 , T2 v T5 dn dng ,nhng ant T5 v catt T2 c cng mtin th , khng c dng chy t T5 sang T2 . Ch c dng tun hon chy t G1vo G2 qua T1 v T6. in p tun hon trong khong ny l :

ucc12 = u2au2b = 6 U2sin( + /6) ;

Nu chuyn to t O sang O2 , ta c :

ucc12 = -6

U2sin = 2.Xc ddicc12

;

icc12 =c

X

U

2

.6 2 .cos + C ;

Khi = 1 , icc12 = 0, ta c :

6 2 4

1 3 5

= 300

= 1500

1

2

2

u u u

1 2 3 4

6 2

3 5

1-1 0

0

0

0

icc21 icc12



icc12 = -cX

U

2

.62 .(cos - cos1) ; (2-14)

Tip tc xt cc khong khc , kt qu nhn c cho php ta kt lun :

hon icc12 chy t G1 vo G2, v ba xung dng in tun hon icc12 chy t G2 voG1 .

Tr trung bnh ca dng in tun hon :

Icc =2

3

1

1.2

.6 2

cX

U(cos - cos1)d =

cX

U

2

.63 2 .(sin11cos1) (2-15)

Phng php iu khin kiu phi tuyn : 1 + 2 = + .

y l kiu iu khin phi hp khng hon ton th lc ny s c thm h sphi tuyn v ta c :

1 + 2 = + ;

Gc ph thuc vo cc gi tr ca 1v 2 mt cch phi tuyn .

Hnh 2-14 : a ) S iu khin chung phi hp kiu tuyn tnh .

Id

0dm

1min

2max

Ic

1= /2

-0dm

1max

2min

a )

Id

1min

2max

0dm

-0dm

1max

2min

1= /2

Ic

b )



b ) S iu khin chung phi hp kiu phi tuyn .

2.3.2.Phng php iu khin ring :

Hai mch chnh lu hot ng ring bit . Mch ny hot ng (c pht

xung iu khin ) th mch kia hon ton ngh ( b ngt xung iu khin ) . V vyloi trc dng in tun hon v khng cn cun khng cn bng Lc . Songtrong qu trnh o chiu cn c thi gian cht ( nh nht l vi ms ) cho vanca mch phi ngng hot ng kp phc hi tnh cht kho ri mi bt u phtxung cho mch kia hot ng . V vy cn mt khi logic iu khin o chiu tincy v phc tp .

thay i trng thi lm vic ca cc b chnh lu th phi dng thit bcbit chuyn cc tn hiu iu khin t b chnh lu ny sang b chnh lu kia .Bi vy , khi iu khin ring , cc dc tnh c s bgin on ti trc tung . Nhvy , khi thc hin thay i ch lm vic ca h skh khn hn v h c tnhlinh hot km hn khi iu chnh tc .

Trong phng php iu khin ring cng c phi hp iu khin kiu tuytnh v phi tuyn .

T

T

T

T

T

T

FG

T4T1

T6

T3

T2T5

+ -Ckt

M

Ikt

L

u2

u2

u2

G1

G2



Hnh 2-15: S cu chnh lu cu ba pha c dng phng php iu khinring .

Do s ny dng phng php iu khin ring nn trong mch khng c

dng in tun hon dn n trong mch ta khng dng bn cun khng cn bngLc .

S nghuyn l h truyn ng T-:

B bin i van thyristor l mt loi ngun in p mt chiu, n trc tip bin idng soay chiu thnh dng mt chiu. vic iu chnh in p u ra ca b bini, c thc hin bng cch iu chnh gc m

ca van.

in p chnh lu Ud0( in p khng ti u ra) c dng p mch vi tn sp mch l n trong mt chu k ca in p s cp ca my bin p lc.

S chnh lu cu n= 2m.(vi m l spha).s hnh tia n=m.

Gi sin p cp cho b bin i van c dng.

K

n

Uk

C0

T1

T3

T5

T4

T6

T2

u2a u2b u2c

BA

U1, f1

+ -CK RK

CCSX

BKC



tTa bit sau mt chu kdng in v in p lp li nn ta ch cn xt cho mtchu kl , coi in trvan Rv=0.

S thay th mch.Mt b bin i van c th bao gm : My bin p lc, t van, khng lc, thit bbo v v h thng iu khin.

S thay th c dng nh sau:

S thay th chnh lu thyristor ng c in mt chiu

Khi van dn ta c phng trnh:

- Vi : =

= Nhn xt:- u im : H( T-) tc ng nhanh, tn tht nng lng t, kch thc v

trng lng nh, khng gy n v d tng ha do cc van bn dn c h

s khch ai ln, iu rt thun tin cho vic thit lp cc h thng tng iu chnh nhiu vng nng cao cht lng cc c tnh tnh v ccc tnh ng ca h thng.

- Nhc im: Do cc van bn dn c tnh phi tuyn, dng in p chnh lura c p mch cao, khnng linh hot v chuyn trng thi lm vickhng cao, gy tn tht ph v lm su iu kin chuyn mch trn c gp.



- Khc phc: thit k truyn ng van c gng lm hp vng dng gin onbng cch ni khng lc ln, tng s ln p mch, ni van m.



Chng III

TNH TON MCH NG LC

Vi cc s liu:

- Ngun in li xoay chiu 3 pha: 220V/380V.- Ti l ng c in mt chiu kch tc lp c:+ Pm = 10KW ; Um = 220V ; Im = 63A; R = 0,3.+ H s d trin p Ku =1,5 1,8+ H s d trdng in Kd = 1,1 1,4

I. Tnh ton cc thng s cn li c bn ca ng c.ta c:

=

= 0,86 =0,145

ta chn = 0,86 l hiu sut cu ng c in 1 chiu.II. Tnh ton chn van ng lc.

Van n lc c la chn da vo cc yu tc bn: dng ti, s chn,iu kin tn nhit, in p lm vic.

1 . tnh ton v chn thyristor.a.in p ngc ca van.

(1.1)Vi = thay vo (1.1) lc c th tnh:

= = = 230.3 (V).Trong : in p ti, ngun xoay chiu, ngc ca van.

Ud=220, = cc h sin p ngc v in pti.



c th chn van theo in p hp l, in p ngc ca van cn chn phi lnhn in p lm vic Ulv qua h s d trin p Kdtu = ( 1,5 ta chn :

Kdtu = 1,8:

= 1,8. 230,3 =414,45 (v).b. Dng in lm vic ca van.

Chn dng in qua van.Vi s cu 3 pha ta c:

= 36,4 (A).Chn h s d tr dng l ( 1,1 ) ta chn =1,4

= 1,4 . 36,4 = 50,96 (A).Van chu c dng in nh mc l: = 50,96 (A).Sau khi tnh ton ta phi chn thyristor vi cc thng s: Idmv = 50,96 (A),

Unv = 414,54 (V).ta chn c 6 thyritor loi : T60N600BOC.Vi cc thng s:+ Dng in nh mc ca van : =60 (A)+ in p ngc t ln van: = 600 (V).+ st p trn van: + Dng in r : 25 (mA).+ Dng in nh cc i : .+Dng in xung iu khin: = 150(mA).+ in p xung iu khi: = 1,4 (V).+ Thi gian chuyn mch ( mv kha) : + Nhit lm vic cc i 125 (0C ).



+ Dng duy tr: + o hm in p:

= 400( V/s).

2. Tnh ton my bin p( MBA).T cc thng sc bn :

+ in p trung bnh trn ti: Ud = 220 (V).

+ Dng lm vic trung bnh trn ti( A- DC) : Id = 63( A).

Ta chn my bin p 3 pha hnh tr.

im p pha s cp MBA: in p chnh lu khng ti.

Ta c: = 235 (V).Ud = 220 (V).

in p st p trn van.

.St p bn trong my bin p khi c ti. Bao gm cc st p

trn in tr v st p trn in cm cc ai lng ny khi chon s bvo khong ( 5 chn = 13,2 ( V).st p trn dy ni: . = . = 0 Cng sut biu kin cu my bin p.

. = 1,05. = 12209,3 ( W).

- cng sut biu kin my bin p.

- h s cng sut ca s mach ng lc cu 3 pha iukhin i xng.

in p cudy s cp : in p cu cc cun dy th cp.



= = 100,43 ( V).in p chnh lu khng ti.

H sin p chnh lu mch cu 3 pha iu khin i xng.

+ Vi in p cun s cp : U1 = 380(V).+ T s bin p: n = = 100,6/ 380 = 0,26.

Dng in ca cun dy:- Dng in chy trong mi pha cun th cp:

= 63 . = 51,44 (A)- Dng in chy trong mi pha cun s cp.:

= . n = 51,44 . (A).a. Tnh ton s b mch t. Tit din s b tr.

Ta c : . = 6. = 54,13 (cm2)

Cng sut my bin p tnh bng W.

- H s ph thuc vo hnh thc lm mt. Ta chon lm mt my bin pkh .ms tr ca my bin ap( v my bin p 3 pha chon m= 3).f - tn s ngun in xoay chiu. vi li in Vit Nam ta chn f= 50(H).

ng knh tr: = 8,3 ( cm).Chn ha ng knh tr theo tiu chun: d = 8(cm).( theo bng 41a trang213 sch thit k my bin p in lcPHAN TTH).

Chn loi thp 330 cc l thp c dy 0,5mm chn s b mt t cmtrong tr BT = 1 T.

Chn t s m = h/d = 2,3. Suy ra h = 2,3 . d = 18,4 cm.( thng thng m =2 2,5 ).



- Chn chiu cao tr h = 18 cm. S vng dy ca cun s cp:

W1 =

0,1.10.1,61.50.44,4

380

...44,44

1

BQf

U

Fe

= 280 ( vng)

Chn W1= 280 (vng)

S vn dy ca cun th cp:W2 =W1 .

U

U.

1

2= 280.

380

43,100=72,8 ( vng).

Chn W2 = 73 (vng).

- Vi f = 50 nn ta chn B= 1T( Tesla). Tit din dy dn cun s cp my bin p:

Ban u chn : ( 2 ). .

- Chn dy dn c tit din hnh ch nht chun ha theo tiu chun cchin l :

.

- Kch thc ca dy c kn cch in l: .- Tnh li mt dng in trong cun s cp:

= 2,76 A/. Tit din dy dn qun th cp my bin p:

Ban u chn : ( 2 ).

.



- Chn dy dn c tit din hnh ch nht chun ha theo tiu chun cchin l : .

- Kch thc ca dy c kn cch in l:

.

- Tnh li mt dng in trong cun scp: = 2,722 A/

Kt cu dy qun qun s cpThc hin dy qun kiu ng tm b tr theo chiu dc trc.

- S vng dy trn mt lp ca qun s cp: = 39,1 (vng)Ly = 39 ( vng).

Trong : H s p cht.h- Chiu cao tr.

khong cch tgng n cun dy s cp. chn s b 1,5 cm- S lp dy qun s cp: 2,7

39

280

11

111

W

Wn (lp).

- Chn s lp = 7 ( lp).- Nh vy c 280 vng chia thnh 7 lp. mt lp c 40 vng.- Chiu cao thc t ca cun s cp:

= = 18,41(cm).- Chn ng qun dy lm bng vt liu cch in c b dy:

- Khong cch t tr ti cum s cp .- ng knh trong ca ng cch in: 2 = 9 + 2.1 2. 0,1 = 10,8 (cm).

- ng knh trong ca ca qun s cp: + 2 = 10,8 + 2.0,1 = 11 (cm).

- Chn bdy cch in gia cc lp dy cu qun s cp l:



.- B dy cun s cp:

= ( ). = ( 1,23 + 0,1).7 = 9,31 (mm).= 0,931(cm).- ng knh ngoi ca cun s cp:

+ 2 = 11 + 0,931. 2 = 12,862(cm).- ng knh trung bnh cun s cp: = = = 11,931 (cm).

- Chiu di dy qun s cp: =. = 11,931 . . 280 =10495,1 (cm) = 104,951 (m).

- Chn bdy cch in gia cun s cp v th cp: .

Kt cu dy qun th cp:- Chn s b chiu cao cun th cp: h1 = h2 = 18,41 (cm).Tnh s b vng dy trn mi lp :

= 19( vng).- Tnh ton s b lp dy qun th cp: = = = 3,8 ( lp).- Chn s lp dy qun th cp:

= 4 (lp).Vy 73 vng chia thnh 4 lp, 3 lp u mi lp 24 vng, 1 lp sau c 1vng.

- Chiu cao thc tc ca cun th cp: = (cm).

- ng knh trong ca cun th cp: 1 = 14, 862 (cm).Vi a12 : l bdy cch in cc cun dy vi nhau.(vi my bin p cdng in nh th a12 thc(0,3

chn = 1mm.)

- Chn bdy cch in gia cc lp dy cun th cp: - B dy cun th cp:- = ( ). = ( 0,224 + 0,01).4 = 0,936 (cm).- ng knh ngoi ca cun th cp:

+ 2 = 14,628 + 2.0,936 = 16,5 (cm)



- ng knh trung bnh ca cun th cp: = = = 15,653 (cm).

- Chiu di dy qun th cp: =. = 15,653. . 73 = 3569(cm) = 35,69 (m).- ng knh trung bnh cc cun dy: = = = 13,75 (cm).

- Chn khong cch gia hai cun th cp:

= 2( cm).

Tnh kch thc mch t.Vng knh tr l d = 9 (cm) ta c s bc l 5 trong na titdin tr.( theo bng 41a trang 213 sch thit k my bin p in lc PHAN TTH).

- Tnh ton b tit din bc thang ca tr: 56,7(). ( theo bng 41a trang 213 sch thit k my bin

p in lcPHAN TTH).

- Tit din hiu qu cu tr: = 56,7 . 0,891 = 55,63( cm2).- Tng chiu dy ca cc bc thang hnh tr:

- S l thp trong cc bc:

+ Bc 1:

+ Bc 2: + Bc 3:

+ Bc 4:



+ Bc 5: - n gin trong ch to to ta chn gng c tit din hnh ch nht

c cc kch thc sau:+Chiu dy ca gng bng chiu dy cu tr: b = dt =8 (cm).+Chiu cao ca gng bng chiu rng tp l thp th nht cu tr: a =8,5 (cm).+Tit din gng: Qbg = a

- Tit din hiu qu ca gng: .

- S l thp dng trong mt gng: Tnh chnh xc mt in cm trong tr:

---

- Mt t cm trong gng:

- Chiu rng ca s:c = 2( + + + ) + = 2( 1 +0,931+ 1 + 0,936) + 2=9,734 (cm).

- Khong cch gia hai tm truc:

- Chiu rng mch t:L = 2c + 3d = 9,734. 2 + 3. 9 = 44,468 (cm).

- Chiu cao mch t: H = h + 2a = 21 + 2 .8,5 = 38 ( cm). Tnh khi lng cu st:

- Th tch ca tr:

- Th tch ca gng: 5388

)(099,110.63,55.280.50.44,4

380

...44,4 41

1 TQf

U

WB

Tg

)(009,1099,1588,60

63,55..

TQ

QBB T

g

T

g



- Khi lng ca tr: = 3,541 . 7,85 = 27,797 (kg)

- Khi lng ca tr: = 3,541 . 7,85 = 27,797 (kg)

- Khi lng ca gng:

- Khi lng cu st:

- Th tch ca ng:

104,951.10. .+35,69.10.18,9. )= 3,573 (dm3)

- Khi lng cu ng: 8,9 . 3,573

Tnh ton cc thng s ca my bin p:- in trtng ng cun s cp my bin ap 75 o C:

. 0,02133 =0, 455().Trong ).

- in trtng ng cun s cp my bin ap 75 o C: . 0,02133 = 0,04().

Trong ).- in tr my bin p quy i v th cp:

= 0,04 + 0,455( = 0,071().- St p trn in trmy bin p:



- in khng my bin p quy i v th cp: (

)

0,087().Trong : +

+ r = = = 7,483 (cm) bn knh trong dy qun th cp.- in cm my bin p quy i v th cp:

- St p trn in khng my bin p:

- St p trn my bin ap: =

- Tng trngn mch quy i v th cp: =

- Tn hao ngn mch trong my bin ap: . = 3.0,071. = 564(W).

-


49/49

+ Attomat: dng ng ct mch ng lc, tng bo v khi qu ti v ngnmch.

Chn attomat:

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