Copy of Nhap Da Trang Bi Dien Cua Ha

8/3/2019 Copy of Nhap Da Trang Bi Dien Cua Ha

1/86

TRNG HSP KT VINH N TRANG B INKHOA IN

SVTH: H Vn H - 1- Lp H in B_K3

LI NI U

Trong cng cuc xy dng v i mi t nc th vic pht trin khoa hc k thut

ng dng vo lnh vc cng nghip. nc ta nhp kh nhiu cc loi my mc, thit b

hin i. Do vy i hi qu tr nh ging dy cho hc sinh, sinh vin phi trang b nhng

kin thc c bn v nguyn l hot ng cng nh nguyn tc vn hnh ca trang thit b

nhm nm bt kp thi vi thc t ca x hi trong hin ti v trong nhng nm ti.

Trong qu trnh hc tp ti trng em c hc mn hc Trang B in, cng c

kin thc mn hc ny c nhiu ti n mn hc v cc loi my khc nhau giao

cho sinh vin. Em nhn ti: thit k h truyn ng chnh cho my tin. Thit k h

truyn ng chnh cho my tin l ti tng i kh, trong thi gian lm n va qua

vi s ni lc ca bn thn cng s gip ch o tn tnh ca cc thy, c gio trong b

mn trang b in c bit l s gip ch o tn tnh ca thy Trn Duy Trinh em hon thnh xong n ny.

n ca em gm 7 chng:

Chng 1: Tng quan v my tin

Chng 2: La chn phng n truyn ng

Chng 3: Tnh chn cc thit b mch ng lc

Chng 4: Thit k h thng iu khin

Chng 5: Xy dng s cu trc ca h truyn ng

Chng 6: Xt n nh v hiu chnh h thng

Chng 7: Thuyt minh s nguyn l ca h thng

Trong qu trnh thit k n, vi kin thc cn hn ch nn khng trnh khi nhng

sai st. V vy, em rt mong s ng gp kin ca cc thy c n mn hc ca em

c hon thin hn v bn thn em c kinh nghim b sung kin thc c tt hn. Em

xin chn thnh cm n!Vinh, ngy 18 thng 10 nm 2011

Sinh vin thc hin

H Vn H

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CHNG I

TNG QUAN V MY TIN

1. KHI NIM CHUNG

1.1c im cng ngh

Nhm my tin rt a dng, gm cc my tin n gin, Rvonve, my tin vn nng,

chuyn dng, my tin ct, my tin ng. Trn my tin c th thc hin c nhiu cng

ngh tin khc nhau: Tin tr ngoi, tin tr trong, tin mt u, tin cn, tin nh hnh.

Trn my tin c th thc hin c doa, khoan v tin ren, bng cc dao ct, dao doa, tar

ren. Kch thc gia cng trn my tin c th t c vi milimt n hng chc mt (Trn

my tin ng)

1 : Thn my

2 : trc

3 : Bn dao

4 : sau

Dng bn ngoi my tin

Chuyn ng chnh ca my tin lm vic ch di hn, l chuyn ng quay

ca mm cp, chuyn ng tnh tin lin tc ca bn dao. Cc chuyn ng ph gm

chuyn ng phanh cu dao v sau, ko phi,bm nc, nng h, kp v ni x v.v...

cc my c nh, ngi ta thng dng ng c lng sc ko cc truyn ng c

bn. Loi ng c ny c u im v mt kinh t ,n gin v c tnh c cng. iu chnh

tc bng phng php c kh, trong phm vi khng rng lm. Khi my yu cu phm vitc rng thng s dng ng c lng sc hai hay nhiu tc .

Mt trong nhng c im ca my tin c nng l yu cu iu chnh tc ng c

trong phm vi rng.V vy phn nhiu ngi ta dng ng c n mt chiu kt hp vi

tc 3- 4 cp. iu chnh tc in kh c thc hn bng cch thay i t thng ng

c, hoc bng phng php iu chnh 2 vng



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1.2 Cc thng s c trng cho ch ct gt ca my tin

a. Tc ct: L tc di chuyn tng i ca bn dao so vi chi tit ti im tip

xc. y l thng s c bn xc inh ch lm vic ca my v tnh ton ch ct

gt ca my, n ph thuc vo cc yu t nh vt liu lm dao v chi tit gia cng.

- Lng n dao: S (mm/vg)

- Chiu su ct : t (mm)

- Tui th ca dao : T

Tc ct c xc inh theo biu thc kinh nghim :

yVxVm

V

ZStT

CV

.. (m/ph)

Trong :

- t l chiu su ct

- T l tui th ( bn) ca dao

- S l lng n dao khi chi tit quay c mt vng.

- CV, xV, yV, m l h s m ph thuc vo chi tit gia cng, vt liu lm giao

v phng php gia cng. Vt liu gia cng l gang, thp CT5 vt liu lm giao bng thp

hp kim cng cao, nn chn

CV= 200; xV = 0,15 - 0,2 chn :xV= 0,2 ; yV= 0,35 - 0,8 chn :yV= 0,35 ; m = 0,1 -

0,2 chn: m = 0,1; T= 60 - 80 ph chn: T= 60 ph.

m bo nng sut cao nht, s dng my trit nht th trong qu trnh gia cng

phi lun t tc ti u, n c xc nh bi nhng thng s: su ct t, lng n dao

S v tc trc chnh ng vi ng knh chi tit xc nh. Khi tin ngang chi tit c ng

knh ln, trong qu trnh gia cng ng knh chi tit gim dn, duy tr tc ct (m/s)

ti u l hng s th phi tng lin tc tc gc ca trc chnh theo quan h.

VZ = 0,5.dct . ct .60.10-3 (m/ph)

Trong : - dct - l ng knh chi tit (mm),

- ct - tc gc ca chi tit (rad/s)

b. Lc ct: L lc tc ng ti im tip xc gia dao v chi tit, lc y ti im tip xc

gi l lc php tuyn chia lm ba thnh phn:



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- Lc tip tuyn Fz : chng li s quay ca chi tit,

- Lc dc trc FX : chng li s di chuyn ca bn dao.

- Lc hng knh FY : Chng li s t ca dao v chi tit.

T l cc thnh phn lc : Fz : FY : FX = 1 : 0,4 : 0,25

Lc ct l thng s quan trng xc nh t cc ch ct ca my.Thng thng lc

ct c xc nh theo cng thc kinh nghim:

FZ = 9,81.CF .txF

. SyF

.V nZ

Trong : CF , XF , YF , n l h s v m ph thuc vo vt liu lm dao, chi tit gia

cng v phng php gia cng.c. Cng sut ct:

L cng sut yu cu ca c cu chuyn ng chnh. Qu tr nh tin xy ra vi cng

sut ct l hng s v c xc nh.

PZ = 310.60

. ZZ VF (kW)

Bi v lc ct ln nht Fmax sinh ra khi lng n dao v su n dao ln, tng ng

vi tc ct nh Vzmin ; cn gi l lc ct nh nht Fmax ,xc nh bi t, s tng ng vitc ct Vzmin ; ngha l tng ng vi h thc :

Fmax . Vzmin = Fmin Vzmax .

S ph thuc ca lc ct vo tc nh hnh v:

d. Thi gian my:

L thi gian gia cng chi tit, n cn c gi l thi gian cng ngh hay thi gian

hu ch. tnh thi gian my phi cn c vo cc yu t ca ch ct gt v phng

php gia cng.



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tM =adV

L 310. (s)

tM =Sn

L

.(ph)

Trong : -L l chiu di gia cng

- Vadl tc n dao

- S l lng n dao

- n tc quay ca chi tit vg/ph

Nh vy gim thi gian gia cng, ta phi tng tc ct,lng n dao v nng sut

s tng

1.3 Ph ti ca c cu truyn ng c bn ca my tina. Trong truyn ng chnh ca my tin lc ct l lc hu ch ca my n ph thuc

vo ch ct ( t, S, V ) vt liu chi tit lm dao

Chuyn ng chnh ca my tin l chuyn ng quay c xc nh:

MZ =2

.dFZ (N.m)

Trong : - FZl lc ct (N)

- dl ng knh gia cng (m)

M men hu ch trn trc ng c:

Mhi =i

dFZ

.2

.(N.m)

i - l t s truyn t trc ng c n trc chnh ca my.

* i vi chuyn ng chnh l chuyn ng tnh tin:

Mhi = FZ . (N.m)

- l bn knh quy i lc ct v trc ng c.

* M men cn tnh trn trc ng c :

Mc =

hiM (N.m)



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- l hiu sut ca b truyn t trc ng c n trc chnh .

Vi my tin ng do c chuyn ng trt trn bng my nn c xut hin lc ma

st ni g trt ca my.

Fms = FN. = [ g.(mb + mct) + Fy ]. (N)

FN l lc y tc dng ln g trt

l h s ma st trt ph thuc vo tc mm cp tc nh mc

* ch xc lp lc ko ca cc chuyn ng mm cp c xc nh l tng lc

ct v lc ma st.

FK = FZ + Fms = FZ + [g (mb + mct ) + Fy ] . (N)

Khi m men trn trc ng c ng vi chuyn ng quayl:

Mc =.2

.

i

dFK (N.m)

i vi chuyn ng tnh tin l :

Mc =

.ZF (N.m)

b. C cu truyn ng n dao

- Trong lc truyn ng n dao ng c thc hin di chuyn bn dao hoc chi tit m bo qu trnh gia cng. H truyn ng n dao c thc hin bng nhiu phng n

khc nhau.

1. ng c truyn ng

2. Hp gim tc

3. Trc vt

4. Bnh vt

5. Bn dao

6. Bng my



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ng c truyn ng n dao s m bo mt lc cn thit di chuyn tnh tin bn

dao. Lc ny c xc nh bi lc cn chuyn ng khi di chuyn bn dao:

Fad = k.Fx+ Fms + Fd (N)

- k = (1,2 1,5) l h s d tr

- Fms - l lc ma st ca bn dao hng g trt

- Fd - l lc dnh.

Fms = .(g.mb + FY + FZ) (N)

- l h s ma st ca bn dao theo hng g trt

Lc dnh sinh ra khi khi ng bn dao:

Fd = .S (N)

- p sut dnh, thng bng 0,5 M/m2

S - din tch b mt tip xc g trt ca bn dao (cm2 )

Cc thnh phn lc n dao: Fx, Fms, Fdkhng ng thi trong qu trnh lm vic. Nn

khi xc nh ph ti truyn ng n dao phn ra thnh hai ch lm vic l khi ng lm

vic v n dao lm vic.

Khi khi ng, lc n dao xc nh bi 2 lc ma st do khi lng c b phn di

chuyn v lc dnh:

Fad.k = 0 .g.mb +Fd (N)

Vi 0 = 0,2 0,3 - h s ma st khi khi ng

Khi c cu n dao lm vic, lc n dao c tnh:

Fad.lv = k.Fx + .( g.mb +Fy + Fx ) (N)

Vi h s ma st khi lm vic, = 0,05 0,15.

* M men trc vt v tn c xc nh theo biu thc

Mtv = 0,5 . Fad . dtv . tg( ) (N.m)

Trong : dtv - ng knh trung bnh ca trc vt v tn (mm)

- gc lch ca ng ren trc vt ()

- gc ma st ca ng ren trc vt ( )



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* M men cn tnh trn trc ng c c xc nh bng cng thc:

Mc =.i

Mtv (N.m)

i, - l t s v hiu sut ca b truyn

Khi xc nh cng sut ng c truyn ng n dao ln lt chn t iu kin m men

ln nht trong hai tr s m men tng ng vi hai lc n dao khi khi ng v lm vic.

Bi v truyn ng n dao thng c phm vi iu chnh tc rng nn ng c cn c

kim tra theo iu kin mmen cn tnh tc nh nht c tnh n s gim m men ng

c do iu kin lm mt xu v kim tra theo iu kin m men khi ng.

th ph ti ca truyn ng

chnh

th ph ti truyn ng n dao

c. Ph ti ca c cu truyn ng chnh my tin:

Truyn ng chnh my tin ng c c th ring.Trn my tin ng chi tit gia

cng c ng knh ln v c t trn mm cp nm ngang.Do trng lng mm cp v

chi tit ln nn lc ma st g trt v hp tc kh ln. V vy ph ti trn trc ng c

truyn ng chnh l tng cc thnh phn lc ct, lc ma st g trt, lc ma st hp tc

. th biu din cc thnh phn cng sut ca truyn ng chnh my tin :



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- P1 l cng sut khc phc lc ct,

- P2 l cng sut khc phc lc ma st g trt,

- P3, P4 l cng sut khc phc lc ma st trong hp tc tng ng do lc ma

st v s quay ca mm cp.

- P5 l tng cng sut ca truyn ng chnh.

1.4 Kho st, ly thng s ca my tin

Trong thi gian nhn ti n mn hc Trang B in em tm hiu v mt s

loi my tin, trong c my tin 1K62 ca Lin X s dng loi ng c mt chiu

100 c cng sut 10kW, tc nh mc 1090 (vg/ph), v bo v, lm vic di hn,

pham vi iu chnh tc 1:2.



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CHNG II

LA CHN PHNG N TRUYN NG

2.1 KHI NIM CHUNG

Ngy nay vi s pht trin mnh m ca khoa hc k thut, cc my sn xut ngy

mt a dng dn n h thng trang b in ngy cng phc tp vi hi s chnh xc v

tin cy cao.Mt h thng truyn ng khng nhng phi m bo c yu cu cng ngh,

m cn phi n nh.Tu theo loi my cng tc m c nhng yu cu khc nhau, rt cn

thit cho gi n nh tc , m men vi chnh xc no trc s bin ng v ti v

cc thng s ngun. Do b bin i nng lng in xoay chiu thnh mt chiu v

ang c s dng rng ri.

B bin i ny c th s dng nhiu thit b khc nhau to ra nh h thng my

pht, khuych i t, h thng van.Chng c iu khin theo nhng nguyn tc khcnhau vi nhng u im khc nhau. Do c c mt phng n ph hp vi tng loi

cng ngh i hi cc nh thit k phi so snh nhng ch tiu kinh t - k thut a ra

phng n ti u.

2.1.1 Ni dung phng n:

Trn thc t, c rt nhiu phng n gii quyt. Tuy nhin mi phng n c nhng

u nhc im ca n. Nhim v ca nh thit k phi chn ra phng n ti u nht.

i vi nhng h thng truyn ng in n gin khng c nhng yu cu cao th chcn dng ng c in xoay chiu vi h thng truyn ng n gin.Vi h thng truyn

ng phc tp c yu cu cao v cng ngh, cht lng nh iu chnh trn, di iu chnh

rng th phi dng ng c in mt chiu. Cc h iu chnh km theo phi m bo cc

yu cu cng ngh v c kh nng t ng ha cao.

Nh vy, chn c h thng truyn ng ph hp th chng ta phi da vo cng

ngh ca my, cng sut lm vic a ra nhng phng n c th p ng yu cu

ca n. chn c phng n tt nht trong cc phng n t ra th cn phi so snh v

k thut v kinh t.

i vi truyn ng ca ng c in mt chiu th b bin i rt quan trng, n

quyt nh n cht lng h thng.Do vy vic la chn phng n v la chn b bin

i thng qua vic xt cc h thng.



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2.1.2 ngha ca vic la chn

Vic la chn phng n hp l c mt ngha c bit quan trng, n c th hin

qua cc mt.

+ m bo c yu cu cng ngh my sn xut.

+ m bo c s lm vic lu di, tin cy.

+ Gim gi thnh sn phm, tng nng sut.

+ D dng sa cha, thay th khi xy ra s c.

2.2 Cc phng n truyn ng

2.2.1H truyn ng my pht - ng c: (F - )

Trong h truyn ng my pht - ng c (F - ) ngun cung cp phn ng ng c l

b bin i my in (my pht iu khin kch t c lp).

S nguyn l:

ng c truyn ng cho my sn xut, my sn xut c cp in phn ng t

my pht F. ng c s cp ko my pht F v ng c mt chiu KB K, ng c Kcng ko my pht t kch t K cp in kch t cho ng c v my pht F.

Bin tr RKK dng iu chnh dng in kch t ca my pht t kch t F. Ngha

l iu chnh in p pht ra cp cho cc cun kch t my pht KTF v cun dy ng

c KT . Bin tr RKF dng iu chnh dng kch t my pht F, do in p pht ra

ca my pht F t vo phn ng ng c . Bin tr RK dng iu chnh dng kch

t ng c, do thay i tc ng c nh thay i t thng.



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Phng trnh c tnh c ca ng c ca ng c

d

uu

d k

RI

k

U

.

Vi U = UF - R .I hay ud

uEuD

d

Ik

RR

k

FE.

.

T phng trnh c tnh c ca

h F - ta c h c tnh c ca h l

nhng ng thng song song nm c

bn gc phn t ca mt phng ta

vi c tnh cng

* nh gi cht lng ca h thng:

- u im:

+ Phm vi iu chnh d dng v ln

+ C kh nng iu chnh rt bng phng

+ Tn hao khi m my, o chiu quay v khi iu chnh tc b, v qu trnh nuc thc hin trn mt kch t.

+ C th o chiu ng c mt cch d dng.

+ C kh nng qu ti cao.

+ c tnh qu tt, thi gian qu ngn

+ in p u ra ca my pht bng phng c li cho ng c

+ C kh nng gi cho c tnh c ca ng c cao v khng i trong qu trnh lmvic.

- Nhc im:

+ H thng s dng nhiu my in quay cho nn gy n, kt cu c kh cng knh

chim nhiu din tch

+ Tng cng sut t ln.



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+ Vn u t ban u ln.

+ My in mt chiu thng c t d ln, c tnh t ha c tr nn kh iu chnh

su tc .

2.2.2H truyn ng tiristo - ng c (T - ):

S gm:- FT: My pht tc dng phn hi m tc phn ng ca ng c.

- BB: B bin i dng tiristor bin i in p xoay chiu thnh in p mt

chiu cp cho ng c.

- : ng c mt chiu kch t c lp ko my sn xut

-TH&K: Khu tng hp v khuych i tn hiu.

- UCd: tn hiu t vo.

- .n: tn hiu phn hi m tc

* Nguyn l lm vic ca h thng:

Gi thit ban u h thng c ng vo li in vi in p thch hp, lc ny

ng c vn cha lm vic. Khi t vo h thng mt in p ng vi mt tc no

ca ng c thng qua khu tng hp khuych i v mch pht xung (FX) s xut hin cc

xung a ti cc iu khin ca cc van b bin i. Nn lc ny cc van ang t in

p thun th van s m. u ra ca BB c in p UCd t ln phn ng ca ng cdn n ng c quay, tc ca n ng vi UCd ban u.

Trong qu trnh lm vic, nu mt nguyn nhn no lm cho tc ng c gim

th ta thy: Uk = Ucd - .n, nn khi n gim Uk tng gim U tng n tng

ti im lm vic yu cu. Khi n tng qu mc cho php th qu trnh xy ra ngc li,

chnh l qu trnh n nh tc .



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* H c tnh ca h thng

Sc in ng ca BB

Eb = Ebm .cos = Ub (Ub = U u ra ca b bin i)

Eb = Kk= Uc. Kb (U - .n) = arc cos.bm

dbdk

E

nUKK).(

Phng trnh c tnh c ca h thng:

dd

ub

dd

dbdk

dd

ub

dd

d

K

RR

K

nUKK

K

RR

K

U

..

).(.

.

u

bdk

ub

bdk

dbdk IKK

RR

KK

UKK

..1..1

..

* nh gi cht lng ca h thng:

- u im:

+ Tc ng nhanh khng gy n v d t ng ha do cc van bn dn c h s

khuych i cng sut cao.

+ Cng sut tn hao nh, kch thc v trng lng nh.Gi thnh h d bo dng

sa cha.

- Nhc im:

+ Mch iu khin phc tp, in p chnh lu c biu p mch cao, gy n tn

tht ph ng k trong ng c v h thng.

+ Chuyn i lm vic kh khn hn do ng c tnh nm trong mt phng ta

.



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+ Trong thnh phn ca h bin i c my bin p nn h s cos thp.

+ Do vai tr ch dn dng mt chiu nn vic chuyn i ch lm vic kh khn

i vi cc h thng o chiu.

+ Do c vng lm vic gin on ca c tnh nn khng ph hp truyn ng ng

c ti nh.

a. Cc s ni dy ca h truyn ng tiritor ng c (T )

Trong k thut in hin nay c nhiu trng hp phi s dng ngun in p mt

chiu c tr s thay i c cung cp cho cc ph ti khc nhau tu thuc mc ch s

dng.Cc ngun in p mt chiu nh my pht in mt chiu, cc b bin i tnh

(Khuych i t) c kh nhiu nhc im, trong c nhc im c bn l tn tht ring

kh ln. Cng vi s pht trin ca k thut bn dn v vi mch in t th vic s dng cc

b chnh lu bn dn c iu khin ngy cng c ph bin v c nhiu u vit.

* S ni dy hnh tia:

Hnh 2 1: S nguyn l h thng CL - hnh tia 3 pha v s thay th

c im ca s ni dy hnh tia:

- S van chnh lu bng s pha ca ngun cung cp

- Cc van c mt in cc cng tn ni chung, in cc cn li ni vi ngun xoay

chiu. Nu in cc ni chung l katt, ta c s catt chung, nu in cc ni chung l

ant, ta c s ni ant chung.

- H thng in p ngun xoay chiu m pha phi c im trung tnh. trung tnh ngun

l in cc cn li ca in p chnh lu



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* S hnh cu:

c im ca s chnh lu cu:

- S van chnh lu bng 2 ln s pha ca in p ngun cung cp, trong c m van c

katt ni chung (cc van 1, 3, 5) to thnh cc dng ca in p ngun ; m van c ant

chung ( 2, 4, 6) to thnh cc m ca in p chnh lu

- Mi pha ca in p ngun ni vi 2 van, 1 nhm ant chung, 1 nhm katt

chung.

Hnh 2 - 2: S nguyn l h thng CL- hnh cu 3 pha v s thay th

b. Nguyn l lm vic ca BB xoay chiu - mt chiu

* S tia:

Xt s tia 3 pha katt ni chung

mt Thyristor m cn c 2 iu kin

- in p Ant - Katt phi dng ( UA > 0)

- C tn hiu iu khin t vo in cc

iu khin v Katt ca van

Do c im va nu m trong s tia 3 pha



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cc van ch m trong mt gii hn nht nh.

V d: pha A, trong khong t = 0 . uA > 0 Tuy nhin cc khong t = 0 / 6

uC > uA v t = 5/6 ub > uA

Nh vy van T1 ni vo pha A ch c th m trong khong t = /6 - 5/6. Trong

khong ny nu tn hiu n cc iu khin ca T1 th T1 m.Tng t vi T2 v T3.

Thi im 0 = t = /6 c gi l thi im m t nhin ca s chnh lu 3

pha.Nu truyn tn hiu m van chm hn thi im m t nhin mt gc in th

khong dn dng cu van s thay i (nh hn 2/3) dn n tr s trung bnh ca in p

chnh lu s gim i. Khi gc m cng ln th Ud cng nh

* S cu:

T kt cu ca s chnh lu cu ta c nhn xt: c dng qua ph ti th trong s

phi c t nht 2 van cng thng, mt nhm ant chung, mt nhm katt chung. Vy

vi gi thit l s lm vic ch dng lin tc v b qua qu trnh chuyn mch th

khi b chnh lu cu m pha lm vic, mt thi im bt k trong s lun c 2 van c

th dn dng khi c xung iu khin: Van nhm katt chung ni vi pha c in p dng

nht v van nhm ant chung ni vi pha c in p m nht. Thi im m t nhin ca

s cu cng c xc nh nh i vi s tia c s pha tng ng:

- iu khin in p chnh lu trn ph ti mt chiu ta thay i thi im a

xung iu khin n cc cc iu khin ca cc van, lm thay i khong dn dng ca van

lm in p trung bnh ca chnh lu thay i.

c im ca cc s hnh tia l ngoi cc thi gian chuyn mch cc van ng vi

(l khong thi gian khi mt van no ang ngng lm vic v van tip sau ang bt u



18/86



lm vic )dng in ph ti id bng dng in trong van ang m. Do dng in trong

mch ph ti c xc nh bi sc in ng pha lm vic ca my bin p, cn st p

trong b bin i th c xc nh bi st p trn pha .

s cu, bn ngoi chu k chuyn mch vn c hai van lm vic ng thi .Dng

in ph ti chy lin tip qua hai van v hai pha ca my bin p di tc dng ca hiu ssc in ng ca cc van tng ng, ngha l di tc dng ca sc in ng dy. Sau

mt chu k bin thin ca in p xoay chiu c su van ca b bin i u tham gia lm

vic.

Tr s trung bnh ca sc in ng chnh lu Ed trng thi dng in lin tc c

xc nh nh sau :

Ed = Emcos

Trong Em l tr s cch i ca sc in ng chnh lu ng vi trng hp 0

Vi s 3 pha hnh tia tr s cc i ca sc in ng chnh lu l: E m1 =1,17E2f

.Vi s cu l Em2 =2,34E2f

Trong E2f l tr s hiu dng ca s.. pha th cp my bin p

Kt lun: ph hp vi yu cu ca ti th ta chn b chnh lu cu 3 pha.

c. Dng in chnh lu trn ph ti mt chiu:

Do in p chnh lu lp i lp li 2m (hoc m) ln trong mt chu k ca in pngun nn ch xc lp th dng qua ti cng lp i lp li nh vy (tu thuc s

chnh lu l tia hay cu, s pha chn hay l). Nh vy ch cn bit dng v p trn ti trong

khong thi gian l 1/m chu k hay l tng ng gc in 2 / q ( q = 2m hoc q =

m). xc nh dng v p trn ti ta da vo s thay th ca chnh lu trong mt

khong thi gian lm vic ca mt van.

Hnh 2 - 3: S thay th ca chnh lu trong khong thi gian lm vic ca

van



19/86



- U: tng i s in p ngun xoay chiu tc ng trong mch vng ni vi cc van

ang dn dng trong s thi gian ang xt.

- Nu l s tia th ch c 1 van m, u = uf.

- Nu l s cu th c 2 van 2 pha khc nhau cng lm vic, u = ud.

- Nu chn mc thi gian xt t = 0 l thi im bt u m mt van trong s th

u = Um.sin(t + )

+ Um - Bin in p ngun (pha hoc dy)

+ - gc pha u

T c trng cho van ang dn dng, s tia l 1 van, s cu l 2 van ni tip

nhau, b qua st p trn van

- Ed, Rd, Ld l cc phn t ca ph ti

- Ud, Id - dng v p trn ti.

Phng trnh cn bng in p t s thay th:

Rd.id + Ld.dt

did = Um.sin( t ) - Ed

Gii phng trnh ny ta nhn c biu thc ca dng in chnh lu:

id = i0 + Im.[1-

/2

.)(1

)sin( tearctg

] Im.[1-

2)(1

)sin(

arctgt ]

Tu thuc c tnh ph ti, dng s , gi tr gc iu khin m c th c cc ch

lm vic khc nhau:

- Nu trong ton b thi gian lm vic id >0 ta c ch dng ti lin tc

- Nu trong mt chu k lm vic m dng ti c q khong bng khng v q khong

khc khng ( q = m nu l s tia, q = 2m nu l s cu ) ta c ch dng ti gin

on.

- Ch gii hn gia 2 ch nu trn c gi l ch dng bin lin tc.

d. o chiu trong h thng T

Trong nhiu trng hp cn phi thay i c chiu dng in qua ph ti ca b

chnh lu. Do tnh dn dng mt chiu ca cc van nn phi o chiu bng cng tc t

hoc s dng cc s c bit gm 2 b chnh lu, mi b dn dng theo mt chiu.



20/86



C 2 b chnh lu iu khin l s u cho v s song song ngc.V mt

nguyn l th s u cho hoc s song song ngc hot ng tng t nh nhau. Khi

BB ny lm vic th BB kia ngh, khi i ch ca BB th dng in qua ti c i

chiu. Thc t ngi ta hay s dng s u song song ngc vi cc phng php iu

khin khc nhau. iu khin 2 BB song song ngc c th s dng 2 phng php iu chnh:

- iu khin ring r (iu khin c lp): L s dng 2 b pht xung c lp nhau.

Khi b pht xung ny lm vic (pht xung m cho BB) th b kia ngh, do cc van

trong BB cn li khng th m c.Khi cn o chiu th cho b ny ngh, sau cho b

th 2 pht xung m cc van ca BB th 2.Phng php iu khin ny c nhc im

l tn s o chiu khng cao v cc van Thyristor cn c thi gian khi phc c tnh

kho ca n.

- iu khin ph thuc: C 2 b pht xung cng pht xung n cc BB, trong

mt b lm vic ch chnh lu, b cn li lm vic ch nghch lu ch.Khi s

dng phng php ny, s c dng in khng cn bng chy trong cc BB. hn ch

dng ny ngi ta s dng cc cun khng cn bng

Hnh 2 - 4: S ni song song ngc ca h thng CL - c o chiu quay.

Chn phng n iu khin chung

- c im ca phng php ny l hai mch chnh lu cng hot ng, tc l cng

c pht xung iu khin.Tuy nhin mt b phn lm vic ch chnh lu, l b xcnh du ca in p mt chiu hoc xoay chiu ca ng c, cn b kia chy ch

nghc lu v sn sng chuyn sang ch chnh lu.

Hnh v l th d v b chnh lu o chiu s dng s cu ba pha. Do hai b chnh

lu cng u vo mt ti nn gi tr trung bnh ca chng phi bng nhau.



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2.2.3 H iu truyn ng iu chnh xung p ng c

+ S nguyn l

+ Nguyn l lm vic:

in p v dng in ca ng c u, i ch c gi tr dng khi kha S thong ta c u

= uN, khi kha S ngt iN = 0; u = 0 v i = iD0 do tc dng duy tr dng ca in cm L. Cc

gi tr trung bnh ca in p v dng in phn ng U, I v do sc in ng E ca

ng c khi ng v ngt lien tc kha S, s c xc nh nu bit lut ng ngt kha v

cc thong s ca mch. Nu ng ngt kha S vi tn s khng i th hot ng ca mch

tng t nh chnh lu mt pha, mt na chu k. Qu trnh dng in v in p trong ch

dng in lin tc:

- U + UL + UR + E = 0

L

EUi

L

R

dt

di

Ti thi im t = 0+ kha S bt u thng v U = UN, i = Imin, nu coi s E khng i

trong mt chu k ng ngt ca kha S .

Ti thi im t = t kha S bt u ngt:

i = Imax = )1( /TutN e

R

EUImin.e

-t/Tu

Lc ny U = 0 do van D0 dn nn

L

Ei

L

R

dt

di

Ti t = 0+, i = Imax

i =R

E (1- e-t/Tu ) + Imax.e

-t/Tu; t t < T

S

ik

iiN

UN D0

iD0

U

L R

E

UR UL



22/86



Ti t = T t, tc l ti t = T, i = Imin ta c

i = Imin =R

E (1- e-(T-t)/Tu ) + Imax.e

-(T-t)/Tu

Nu S thng lin tc t = T th dng in trong mch phn ng s khng i v bng

I = Imax = Imin =R

EUN

+ Phng trnh c tnh c

IK

R

K

UUN

.

T phng trnh c tnh c ca h thng iu chnh xung p ng c ta c c tnh

iu chnh v c tnh c ca h thng

+ iu chnh xung p o chiu

h truyn ng c th lm vic ch hm ti sinh, c th dung s iu chnh

xung p loi B.



23/86



Trong dng in phn ng c th o chiu, song sc in ng ng c ch c

chiu dng. Khi kha S1 v van D1 vn hnh, dng in phn ng lun dng, cng sut

in t ca ng c l Pt = I.E < 0, my in lm vic ch ng c.

o chiu dng in ta a kha S2 v van D2 vo vn hnh cn kha S1 th b ngt.

Nu E > 0 th dng in chy ngc li chiu ban u do mch ch c ngun duy nht l sE, cng sut in t ca ng c Pt = I.E > 0, cng sut ny c tch vo in cm L. Khi

S2 ngt, trn in cm L sinh ra s t cm UL > 0, cng chiu vi s quay E, tng hai

s ny tr nn ln hn in p ngun UN lm van D2 dn dng ngc v ngun v tr li

ngun phn nng lng tch ly trong in cm L trc .

2.2.4 H thng bin tn ng c

a. Bin tn ngun p

S nguyn l

S gm 4 khi chnh: ngun in mt chiu, mch lc, nghch lu c lp ngun

p v ng c.Ngun mt chiu v mch lc to ra in p mt chiu c gi tr iu chnh c,

nghch lu gm 6 kha bn dn S1.S6 v cn 6 van khng iu khin D1D6, cc kha

nghch lu c ng ct theo th t nht nh to thnh in p xoay chiu 3 pha t l n

ng c chp hnh, gc dn ca cc kha l 1800, thi im cc kha S1, S3, S5v S2, S4, S6

bt u dn lch nhau 1200, do in p ra ca nghch lu cng lch nhau v thi gian l



24/86



1200. in p dy ca nghch lu c dng xung ch nht vi rng l 1200 v tha mn

iu kin phn tch thnh chui iu ha

uab =

32ud.

1 6cos

1

k

k

k

.sin(kct +

6

)

k = 1 + 6C ; C = 0, 1, 2

Thnh phn iu ha c bn c bin

U1

abm =

32ud = 1,103.ud

V c gi tr hiu dng l U1ab =

6ud = 0,78.ud

Gi tr hiu dng ca chui

Uab =3

2ud = 0,816.ud

Bin tng song hi bc k

Uk

abm =.

32

kud

Gi tri in p ng c c iu chnh hoc bi iu chnh bin in p mt chiu

bng chnh lu iu chnh hc ng b bm xung p. in p cng c th iu chnh bngiu chnh thi gian ng cc ca kha S hoc l iu ch rng xung p bng chnh

nghch lu.

Phng php sau c s dng rng ri nht l cc truyn ng cng sut nh, do

c u im l va iu chnh c in p, va lm sin ha in p t vo ng c. Vi s

lng cc xung c rng thch hp, phng php iu ch rng xung c th l m trit

tiu cc song hi bc cao.

b. Bin tn ngun dngBin tn ngun dng c u im l tng c cng sut n v my, mch lc n

gin m vn thc hin hm ti sinh ng c.

Ngun in mt chiu cp cho nghch lu phi l ngun dng in, tc l dng in

khng ph thuc vo ti m ch ph thuc vo tn hiu iu khin. to ngun dng in

mt chiu thng dng chnh lu iu khin hoc bm xung p mt chiu c b iu dng

in c cu trc t l tch phn (PI), mch lc l in khng tuyn tnh c tr s in cm



25/86



ln. Do c ngun dng in mt chiu nn vic chuyn mch gia cc van bn dn c th

thc hn bng in p trn cc t chuyn mch.

+ S nguyn l ca b bin tn dng in

+ Nguyn l lm vic

Th t ng ct cc van T1 T6 v dng in cc pha ca ng c. Trong khong thi

gian van T1 v T6 th dng in cc pha ia = -ib, cc t chuyn mch c np in v c cc

tnh nh trn s nguyn l. Khi bt u t xung m ln T2 th T2 dn v ni cc dng

ca t in C62 ln catt ca T6 lm cho T6 kha. Do ti c tnh cht in cm m dong ib

khng ct ngay, dng ny khp mch qua D6 t C62 song song vi mch ni tip C46 C42 T2, np cho t C62, in p trn C62 tng tuyn tnh cho n khi xut hin dng in ic qua

pha C ca ti, bt u s chuyn dng cho D6 cho D2, cng l s chuyn dng t pha B sang

pha C. Kt thc qu trnh chuyn mch khi ib = 0 v ic = Id t in C62 c phn cc ngc



26/86



li, chun b cho ln phn mch sau. Do mch stato ni hnh sao nu dng in cc pha l

xung hnh ch nht c rng l 2/3 v chiu cao ng bng Id.

2.3 Chn phng n truyn ng:

Qua qu trnh phn tch hai h thng F - v T- ta thy chng c nhng u im

nhc im nht nh.C hai h thng u p ng c yu cu cng ngh t ra.

Nhng xt v ch tiu kinh t, k thut th mi h thng t c nhng c im khc

nhau. C th ta thy h F - d iu chnh tc , chuyn i trng thi hot ng linh hot

v c tnh h thng nm u bn gc phn t.Vi h thng F - khi lp t chim din

tch ln, cng knh nhng hiu sut li khng cao.Khi lm vic li gy n o, rung ng

mnh, cng lp t ln, vn u t cao.

Trong giai on CNH - HH ngy nay vi xu th chung hng ti mc tiu yu cu

ti u nht m bo tnh khoa hc, gn nh khng gy n, t nh hng n mi trngxung quanh.Vi h truyn ng F - mc d c nhiu u im nhng cn nhiu hn ch

cha p ng c yu cu CNH - HH hin nay.

Ngy nay vi nn cng nghip hin i ngi ta ang dn tin hnh thay th h thng

truyn ng F - bng cc h truyn ng khc.Vi h truyn ng T - c h s khuych

i ln, d t ng ho do tc ng nhanh chnh xc, cng sut tn hao nh.Kch thc nh

v gn nh.

Vi s pht trin mnh m ca khoa hc cng ngh xu hng t ng ho cc hthng t ng, gia cng chnh xc, nn iu khin h thng c thc hin bng cch lp

ghp h thng vi cc b iu khin t ng nh PLC, vi x l.

Nhn chung h thng T - p ng c yu cu t ra.Vi nhng u im v nhng

c im ph hp cch truyn ng. Vy em quyt nh chn phng n truyn ng T - .

Bi v h T- c ch tc ng nhanh v d t ng ho, ph hp vi yu cu ca s

pht trin khoa hc k thut l ti u ho, t ng ho gia cng chi tit chnh xc, tin

cy cao gim c sc lao ng v tng nng sut, kch thc c kh gn nn phn c khca my gn to nn tnh thm m ca h thng.V th kinh t vn u t v chi ph vn

hnh thp hn nhiu so vi h truyn ng F-.

2.4 Thit k mch ng lc:

Mun xy dng s mch ng lc phi chn c s chnh lu lm b bin

i cho mch, nh bit chng 3, h T- c tc ng nhanh, cao, khng gy n o v

d t ng ho do cc van bn dn c h s cng sut cao, iu thun tin cho cc h



27/86



thit k cc h t ng iu khin nng cao cht lng cc c tnh tnh v ng ca

h.Mt khc cc mch chnh lu iu khin c sc in ng Ed ph thuc vo gc iu

khin nn c th dng lm ngun iu khin in p phn ng hoc dng kch thch ng

c.

Tc truyn ng in trong h thng ny c iu chnh bng cch bin i gcthng chm ca van, tc l iu chnh s, , , E

ca b bin i. Khi bin i t 0

n2

, tr s E bin thin t Em n 0. R rng l cc c tnh c v c tnh tc ca

truyn ng in l mt h cc ng thng song song vi nhau.Cc ng thng ct

trc tung ti nhng im tng ng vi cc tc khng ti l tng

0 =m

vm

K

UE

cos.

c tnh c v c tnh tc ca truyn ng in khi b bin i l vic trng thi

ng in gin on khng th biu din bng gii tch c

Trng hp b bin i l vic trng thi dng in lin tc, nu bin i gc thng

van th tc khng ti l tng gi nh 0' bin i trong mt phm vi rng. Nu

20

th 0' =0 v ng c lm vic trng thi hm ng nng.

Khi 2

, 0'0 , h thng truyn ng in s lm vic trng thi hm ti sinh. Khi

, nng lng do my in mt chiu sinh ra s c b bin i van bin thnh nng

lng in xoay chiu truyn vo li cung cp. Kh i ny, b bin i van lm vic

trng thi nghch lu vi c im sau: dng in trong mch ca ngun xoay chiu chy

di tc dng ca s... E ca ng c. Trong phn ln khong dn ca van, dng in ny

chy ngc chiu s, d, ca cc cun dy my bin p.

khng xy ra hin tng t bin nghch lu vi dng in trong mch ln gp

nhiu ln so vi lc lm vic bnh thng, d gy hng trong b bin i, m trc ht lgy nguy him cho cc van, ta phi hn ch gc thng van:

)(max

Trong : l khong thi gian phc hi tnh cht ngt ca van.

Khi tn s li l 50Hz, gc phc hi tnh cht ngt ca van ion bng khong 120. i

vi Tiaristo, thi gian phc hi tnh cht ngt khng qu 150 s , tng ng vi 30



28/86



Thng thng, khi phn tch hot ng ca b bin i trng thi nghch lu, ngi

ta s dng khi nim gc thng trc ca van

Tng ng loi tr hin tngt bin nghch luta c iu kin hn ch

sau: min .

Trong thc t cc h thng truyn ng, van, ngi ta p dng ba phng php bin

i chiu mmen ng c sau:

a) Bin i chiu t thng ng c khi chiu dng in phn ng khng i (hnh a).

(a) (b) (c)

b) Bin i cc tnh in p phn ng nh cc cng tc chuyn i (b o chiu) (hnh b).

c)Bin i cc tnh in p phn ng nh b bin i van hai nhm (hnh c).

Khi p dng hai phng php u, ng c c xung cp t b iu khin n (b

bin i mt nhm). Tuy nhin, khi rt kh thc hin chuyn i t trng thi ng c

sang trng thi hm vi chiu quay khng i. S u l r nht v n gin nht, song

c nhc im l thi hn o chiu ln, bng khong (0,5 - 2,5)s ,( do hng s thi gian

ca cun dy kch t ng c khng ln). S th hai tuy c thi hn o chiu nh hn

nhng van khng th di 0,1s v trong qu trnh o chiu, phi m bo th t tc ng

nht nh trong h thng iu khin truyn ng in.

i vi cc h thng truyn ng yu cu o chiu nhanh v cn c trng thi ng

c hay trng thi hm trong cng mt chiu quay ca ng c, ngi ta s dng cc s

c hai nhm van (b bin i kp). Mi nhm dn dng in theo mt chiu nn b bin i

c kh nng dn in theo c hai chiu. B bin i nh vy c th c ni theo nhiu s

khc nhau.C 2 b chnh lu iu khin l s u cho v s song song ngc.V

mt nguyn l th s u cho hoc s song song ngc hot ng tng t nh nhau.



29/86



Khi BB ny lm vic th BB kia ngh, khi i ch ca BB th dng in qua ti -

c i chiu. Thc t ngi ta hay s dng s u song song ngc vi cc phng

php iu khin khc nhau. Trong s song song ngc, c hai nhm van u c cung

cp t mt nhm dy cun th cp ca my bin p.

Khi h thng truyn ng in lm vic trng thi ng c, mt nhm van, v d 1V,lm vic trng thi chnh lu cn nhm kia 2V b kho hoc chun b lm vic trng thi

nghch lu. Trong trng hp th hai, loi tr hin tng truyn nng lng do 1V bin

i vo li qua 2V ta phi bo m E2> E1. Nu 2V lm vic trng thi chnh lu, th

1V phi c kho hoc chun b lm vic ch nghch lu. Khi , tng ng vi

trng hp trn, ta c Ed1 Ed2: Nh vy ni chung Edn1 Edc2, trong Edn1 , Edc2 - s..

ca cc nhm van b bin i lm vic trng thi nghch lu v chnh lu.

Khi h thng truyn ng in lm vic trng thi hm ti sinh, mt nhm van lm

vic trng thi nghch lu, cn nhm van kia b kho hoc chun b lm vic ch

chnh lu. Trng hp ny cng phi m bo quan h Edn1 Edc2

Trng thi lm vic ca b bin i van o chiu ph thuc rt nhiu vo phng

thc iu khin c hai nhm van. Khi iu khin chung, tn hiu iu khin c a vo c

hai nhm van sao cho m bo c Edn1 Edc2. Trng hp ny cn hn ch dng in

cn bng chy gia hai nhm van di tc dng ca cc tr s tc thi ca s.. cc nhm

van. Do trong mch ca b bin i ngi ta ni cc cun khng cn bng CB1 - CB4,

nh hnh v trn.

loi tr dng in cn bng ngi ta s dng phng php iu khin ring cc

nhm van ca b bin i. Khi cc tn hiu iu khin (xung) ch c a vo nhm

van ang lm vic, cn nhm van kia (nhm khng lm vic ti thi im dang xt)

khng c xung iu khin nn n b ngt. thay i trng thi lm vic ca b bin i

ngi ta dng mt thit b bin i c bit ban u lm mt xung iu khin trn nhm

van ang lm vic, ri sau mt khong thi gian ngn (5 - 10 s) a xung iu khin

ln nhm van th hai. Vi th t nh vy th trng ti dng in gin on ca b bin i

s tng ng vi qu trnh chuyn i h thng truyn ng in t trng ti dng c sang

trng ti hm v ngc li

Cc c tnh tc v c tnh c ca truyn ng van o chiu c iu khin ri ng

cc nhm van ph thuc vo cch phi hp cc gc iu khin u im ca phng thc

iu khin ring so vi iu khin chung l:

- Khng cn ni cun khng cn bng



30/86



- Tn dng c kh nng my bin p (Ed.d.cmax = Em) gim bt c xc sut t

bin nghch lu v thi gian bin i lm vic trng thi nghch lu t hn.

- Gim c tn tht nng lng v nng cao hiu sut truyn ng in do khng

c dng in cn bng. tuy nhin, khi iu chnh ring, yu cu v tin cy ca thit b

chuyn i ( chuyn xung iu khin t nhm van ny sang nhm van kia) cao hn; cmt khong h khi chuyn i b bin i t trng thi chnh lu sang trng thi nghch

lu, nn thi hn ca qu trnh qu tng ln. Cng cn ch rng, phng tc iu khin

ring khng th s dng cho cc h truyn ng c kh nng lm vic trng ti khng ti

l tng hoc trong vng gn (nh thang my chng hn) v trng thi dng in gin

on ca b bin i tng ng vi cc ph ti nh ca ng c.

hn ch dng in ngn mch, thng ngi ta chon my bin p c in p ngn

mch ln Unm = 8 - 10%, do xBAcng c tr s ln. Kt qu l thnh phn in tr ca b

bin i do hin tng gim p trong qu tr nh chuyn mch cc van2

.mxR BA

cm

kh

ln. l nguyn nhn chnh lm mm c tnh c ca truyn ng in.Nu b bin i

c ni vo li xoay chiu khng qua my bin p th ngi ta phi ni cun khng hn

ch dng in vo mch xoay chiu ni tip vi b bin i.Cm khng xBA vRBA ca my

bin p nn hin tng trn vn tn ti.

m rng phm vi iu chnh tc ca h truyn ng ny, c th s dng cc hi

tip ging nh trong h thng F - , nh hi tip dng theo dng in phn ng, hi tipm theo tc v theo in p. hn ch dng in trong b bin i v trong phn ng

ng c, s dng hi tip m dng in c ngt; khi truyn ng in c c tnh my

xc. V b bin i van c h s khuych i ln, nn trong nhiu trng hp c th thc

hin cc hi tip k trn khng qua khuych i trung gian

Trong s gm c:

-My bin p BA: Lm nhim v cung cp ngun cho mch

-CK l cun khng dng lc ngun 1 chiu gi l cun khng san bng-BD l cc my bin dng c s dng ly tn hiu m dng in, a tr li

khng ch u vo mch iu khin.

-Cc b R-C c mc song song vi cc Tiristor trong cc qu trnh chuyn mch v

bin thin du/dt ; di/dt

- l ng c n 1 chiu kch t c lp, dng truyn ng cho h thng.



31/86



Hnh 2 5: Mch ng lc truyn ng cho my tin



32/86



CHNG III

TNH CHN CC THIT B MCH NG LC

3.1 ng c:

Kiu

Pm(kW)

n(vg/ph)

Im(A)

n

max

U(V)

10 1090 55 1

900

220

Im =mm

m

U

P

.=

85,0.220

1000053,47 (A)

R ng c:+ R ng c c tnh gn ng nh sau:

R = 0.5.(1 - ). udm

udm

I

U0,5.(1 - 0,85).

47,53

220= 0,308 ( )

L L in cm phn ng ng c c tnh theo cng thc Umanxki Lindvit:

Ta c: L = 0,5. mm

m

Inpn

U

....2

60.0,5.

47,53.1090.2.2

60.220= 0,0283 ( H ) = 28,3 ( mH )

Trong : 5,0 l h s ly cho ng c khng c cun b

H s khuch i ng c c tnh:

K =

308,0.47,53220

1090

. uuu

m

RIU

n5,35 (v/ph.V)

3.2 Tnh chn Tiristor:

Tnh chn da vo cc yu t c bn dng ti, iu kin to nhit lm vic v cc

thng s c bn ca van c tnh nh sau:

* in p ngc ln nht m Tirisitor phi chu

u

d

nvnvnK

UKUKU 02max ..

Do ta tnh in p ngc ln nht nn Ud = Ud0



33/86



Unmax = Knv. u

d

K

U

3.220

6.3.1

220.6

= 230,3 (V)

Vi KnV = 6 ; Ku =

6.3

* in p ngc m van cn chn

UnV = KdtU.Unmax = 2.230,3 = 460,6(V)

Ly bng 461 (V)

Vi KdtU H s d tr in p thng (1,6 2 ). Chn KdtU = 2

* Chn theo iu kin dng in

ITbV KI .ITtbmaxTrong : KI l h s d tr dng in, ta ly KI = 3

ITtbmax = Im/ 3

Nh vy ta c: ITbV 95,27 A

T cc thng s UnV, ImV ta chn loi TiristorFT100DY10 c cc thng s

in p ngc van UnV = 500 (V)

Dng in ngc van InV = 100 (A)nh xung dng in Ix = 2000 (A)

Dng in xung iu khin Ik = 0,1 (A)

in p xung iu khin Uk= 2 (V)

Dng in r Ir = 0,015 (A)

St p trn Tiristor trng thi dn U = 2 (V)

Tc bin thin in p dU/dt = 100 ( V/s)Thi gian chuyn mch tcm = 15 ( s )

Nhit lm vic cc i Tmax = 1250C

3.3 My bin p mch ng lc

a) Chn my bin p: c t ni dy / lm mt bng khng kh t nhin

b) Tnh cc thng s c bn:



34/86



- Cng sut biu kin ca my bin p:

SBA = Ks.P = Ks. m

mP

1,05. 310.

85,0

1000012,35 (kVA)

- in p s cp MBA:

Uf= 380 (V)

- in p pha th cp MBA:

Phng trnh cn bng in p khi c ti:

Ud0.cos min = Ud + 2 UV + Udn + UBA

Trong :

min = 150

- Gc st p khi c s suy gim in p li

UV = 2(V) - St p trn Tiristor.

Udn 0 - St p trn dy ni.

UBA = U1 + UX - St p trn in tr v in tr khng MBA.

Chn s b:

UBA = 6%.Ud = 6%.220 = 13,2 (V)

T phng trnh in p khi c ti ta c:

Ud0 =

015cos

2,1302.2220245,57 (V)

- in p th cp MBA:

U2 = 63

57,2450

V

d

K

U104,98 105 (V)

- Dng in hiu dng th cp MBA

I2 =3

2.Id =

3

2.95,27 = 77,78 78 (A)

- Dng in s cp MBA:

I1 = KBA.I2 =1

2

U

U.I2 =

380

105.78 = 21,55 (A)

Tnh ton s b mch t



35/86



- Tit din s b tr

QFe = kQ.fm

SBA

.

Trong : - kQ l h s ph thuc phng thc lm mt, ly kQ = 6.

- m: s tr MBA, ly m = 3

- f: Tn s ngun xoay chiu, ly f = 50 Hz

Thay s vo ta c

QFe = 6. 50.3

1235054,44 (cm

2)

- ng knh tr

d =

44,54.4.4 FeQ 8,32 (cm)

Chun on ng knh tr theo tiu chun d = 8 (cm)

- Chn loi thp 330 cc l thp c dy 0,5 mm

Chn mt t cm trong tr BT = 1 (T)

- Chn t s:

h

m2,3 h = 2,3.d = 2,3.8 = 18,4(cm)

Thng thng m = (2 2,5 )

Ta chn chiu cao ca tr l 19 (cm)

c) Tnh ton dy qun

S vng dy mi pha s cp MBA

W1 = 1.10.44,54.50.44,4380

...44,4 41

TFe BQfU 314,42 (vng)

Ly W1 = 315 vng

- S vng dy mi pha th cp MBA

W2 =1

2

U

U.W1 = 315.

380

10587,03 (vng)



36/86



Ly W2 = 87 vng

- Chn s b mt dng in trong MBA.

Vi dy dn bng ng, MBA kh, chn J1 = J2 = 2,75 (A/mm2)

- Tit in dy dn s cp MBA

S1 = 1

1

J

I

75,2

55,217,83 (mm

2)

Chn dy dn tit din ch nht cp cch in B

Chun ho tit din theo tiu chun S1 = 7,829 mm2

Kch thc dy dn c k c cch in Sd.d = a1 . b1 = 2,36 3,55 ( mm mm)

- Tit din dy dn th cp MBA

S2 = 75,2

78

2

2

J

I28,36 (mm

2)

Chn dy dn tit din ch nht cp cch in B

Chun ho tit din theo tiu chun S1 = 29,6 mm2

Kch thc dy dn c k c cch in Sd.d = a2 . b2 = 3,35 9,0 ( mm mm)

* Kt cu dy dn s cp

Thc hin kiu dy qun ng tm b tr theo chiu dc trc:

- Tnh s b s vng dy trn 1 lp ca cun s cp:

W11 =

95,0.67,0

5,1.219.

.2

1

c

gk

b

hh22,6 (vng)

Ly W 11 = 23 vng

Vi: h l chiu cao tr

kc = 0,95 Hng s p cht

hg l khong cch t gng n cun dy s cp

Chn s b khong cch in gng l 1,5 cm

- Tnh s b lp dy qun cun s cp:

n11 = 11

1

W

W

23

31513,69 (lp)



37/86



Chn s lp l 14 lp

Nh vy c 315 (vng) chia thnh 14 lp mi lp c 23 (vng)

- Chiu cao thc t ca cun s cp:

11 1

. 23.0,6716,220,95c

W bh

k (cm)

- Chn ng dy qun lm bng vt liu cch in c b dy S01 = 0,1 cm

- Khong cch t tr n cun dy s cp

a01 = 1,0 (cm)

- ng knh trong ca ng cch in

Dt = dFe + 2.a01 - 2.S01 = 8 + 2.1 - 2.0,1 = 9,8 (cm)

- ng knh trong ca cun s cp

Dt1 = Dt + 2.s01 =9,8+ 2.0,1 = 10 (cm)

- Chn b dy gia hai lp dy cun s cp cd11 = 0,1 (cm)

- B dy cun s cp

Bd1 = (a1 + cd11 ).n11 = ( 0,236 + 0,1).14 = 4,7 ( cm)

- ng knh ngoi ca cun s cp

Dn1 = Dt1 + 2.Bd1 = 10 + 2.4,7 = 19,4 (cm)

- ng knh trung bnh cun s cp

DTB1 =

2

4,1910

2

11 ntDD

14,7 (cm)

- Chiu di dy qun s cp

L1 = W1. .DTB1 = 315. .14,7 = 14,54 (m)

- Chn b dy gia cun s cp v th cp

a12 = 1,0 (cm)

- Chn s b chiu cao cun th cp

h1 = h2 =16,22 (cm)

- Tnh s b s vng dy trn mt lp



38/86



212

2

16,22. .0,95 7,9

1,95c

hW k

b (vng)

- Tnh s b s lp dy qun th cp

n12 = 12

2

W

W

9,787

11,01 (lp)

Chn s lp dy qun th cp n12 = 11 ( lp), 11 lp mi lp c 7 vng.

- Chiu cao thc t ca cun th cp

12 2. 7,9.1,95 16,20,95c

W bh

k (cm)

- Chn ng dy qun lm bng vt liu cch in c b dy S01 = 0,1cm

- Khong cch t tr n cun dy s cpa12 = 1,0 (cm)

- ng knh trong ca cun th cp

DT2 = Dn1 + 2. a12 = 19,4 + 2.1 = 21,4 (cm)

- Chn b dy gia hai lp dy cun s cp cd22 = 0,1 (cm)

- B dy cun th cp

Bd2 = (a2 + cd22 ).n22 = ( 0,335 + 0,1).11 = 4,785 ( cm)

- ng knh ngoi ca cun th cp:

Dn2 = DT2 + 2.Bd2 = 21,4 + 2.4,785 = 30,97 (cm)

- ng knh trung bnh cun th cp:

DTB2 =

2

97,304,21

2

22 nTDD

26,19 (cm)

- Chiu di dy qun th cp:

L2 = W2. .DTB2 = 87. .26,19 = 7,15 (m)

* Tnh cc thng s MBA

- in tr ca cun s cp MBA 750C

R1 =1

1.S

L = 0,02133.

829,7

54,1439,6.10

-3( )



39/86



Trong : 02133,0 ( ) in tr sut

- in tr cun th cp MBA nhit 750C

R2 =2

2.S

L = 0,02133.

6,29

5,75,4.10

-3( )

- in tr ca MBA quy i v pha th cp

RBA = R2 + R1.

1

2

W

W= 39,6.10

-3+ 5,4.10

-3.

315

8741,1.10

-3( )

- St p trn in tr MBA

UR = RBA.Id = 41,1.10-3

.95,27 = 3,92 (V)

- in khng MBA quy i v pha th cp

72112

2210..

3...8

dd

qd

d

BA

BBa

h

rWX

XBA = 8.2 .87

2.

72 10.314.10.3

785,47,401,0.

2,16

26,110,54 ( )

- in cm MBA quy i v pha th cp:

LBA

= 314

54,0

314

BAX

1,72.10-3

(H) LBA

= 1,72 (mH)

- St p trn in khng MBA

UX =

3.XBA.Id =

3.0,54.95,27 = 49,1 (V)

Rt =

3.XBA =

3.0,54 = 0,51 ( )

- St p trn MBA

UBA = 2222 1,4992,3XR UU 49,25 (V)

in p trn ng c khi gc m 0min 15

BAVd UUUU 2cos. min0

U = 245,57.cos 150

2.2 49,25 = 183,95 (V)



40/86



3.4 Tnh chn cun khng lc

Xc nh gc m max

Chn gc m cc tiumin = 15

0.Gc m cc tiu l d tr c th b tr c s

gim in p li.

- Khi gc m nh nht min th in p trn ti l ln nht

Ud max = Ud0.cos min = Udm

V tng ng vi tc ng c ln nht

- Khi gc m ln nht min th in p trn ti l nh nht

Ud min = Ud0.cos max

V tng ng vi tc ng c nh nht

Ta c:max = arccos

0

min

d

d

U

U= arccos

2

min

.34,2 U

Ud (1)

Trong : Udmin c xc nh nh sau

D =

uudmd

uudmm

RIU

RIU

n

n

.

.

minmin

max

CTFtuudmd RRIDUDU .1cos..34,21 min2min

Udmin = 035,0.27,95).1100(15cos.105.34,2.100

1 0 5,67 (V)

Thay vo phng trnh (1) ta c

max = arccos0

min

d

d

U

U= arccos

105.34,2

67,588

0

Ta c cng thc tnh in cm ca cun khng lcTa thy rng khi gc m cng tng th bin thnh phn sng hi bc cao cng ln,

c ngha l p mch ca in p, dng in cng tng ln.S p mch ny lm xu ch

chuyn mch ca vnh gp ng thi gy ra tn hao ph di dng nhit trong ng c.

hn ch s p mch ny ta phi mc ni tip vi ng c mt cun khng lc ln

udmm II .1,0



41/86



Ngoi tc dng hn ch thnh phn sng hi bc cao, cun khng lc cn c tc dng

hn ch vng dng in gin on

in khng lc cn c tnh khi gc m max

Ta c: dt

di

LiRIREuU udud ~~ ..

Cn bng hai v:

dt

diLiRU u ~. v

dt

diLiRu ~. nn : dt

diLU

Trong cc thnh phn bc cao, th thnh phn sng bc k =1 c mc ln nht gn

ng ta c: U~ = U1m.sin 6 nn

I = 11

~6cos.

...2.1

LfUdtU

L

m Im.cos 16

Vy udmm

m ILf

UI .1,0

...2.6

1

Suy ra Lm

f

If

U

.1,0...6.2

1

LCKL = L - L - LBA

Trong : LCKL - in cm cun khnglc cn phi mc thm

L - in cm cn thit lc thnh phn sng hi dng in vi I*1% < 10%

L - in cm ng c

LBA - in cm bin p

6 l s xung p mch trong mt chu k in p

Vi: U1m = max2

2

max0

61.16

cos..2

tg

Ud

U1m =

02

2

0

8861.16

88cos.105.34,2.2tg 15,74 (V)

L 97,53.1,0.50..6.2

74,15

1,55.10

-3( H ) = 1,55 (mH)

in cm phn ng c: L = 3,68 (mH)



42/86



LCKL = L - L - LBA

LCKL = L - (L + LBA ) = 1,55 (3,68 + 1,72) = - 3,85 (mH)

V in cm ca phn ng ln nn ta khng cn t cun khng lc dng in p

mch

3.5 Bo v qu in p cho van

- Bo v qu in p: Do qu trnh ng ct cc tiristor c thc hin nhanh, khi c

chuyn mch cc in tch tch t trong cc lp bn dn phng ra to ra d ng in ngc

trong khong thi gian ngn, s bin thin nhanh ny to ra sc in ng cm ng rt ln

trong cc in cm lam cho qu in p gia katt v ant ca tiristor. Ta thc hin mc

song song R,C vi tiristor s to ra mch vng phng in tch trong qu trnh chuyn mch

nn tiritor khng b qu in p.

Theo kinh nghim chn

R1 = ( 30 50 ) R1 = 50

C1 = ( 0,22 0,3 ) F C1 = 0,25 FT

R1 C1



43/86



CHNG IV

THIT K H THNG IU KHIN

cc van ca b chnh lu c th m ti thi im mong mun th ngoi iu kin ti

thi im trn van c in p thun th trn cc iu khin G v K ca van phi c in

p iu khin (thng gi l tn hiu iu khin). c h thng tn hiu iu khin xut

hin ng theo yu cu m van ngi ta s dng mch in to ra cc tn hiu gi l

mch iu khin.

in p iu khin cc Tiristor phi p ng c cc yu cu cn thit v cng sut,

bin cng nh thi gian tn ti. Do c im ca Tiristor l khi van m th vic tn ti

tn hiu iu khin na hay khng cng khng nh hng n dng qua van. V th hn ch

cng sut ca mch pht tn hiu iu khin v gim tn tht trn vng cc iu khin to ra

cc tn hiu iu khin Tiristor c dng xung.Trong h thng truyn ng ta dng cc h thng pht xung iu khin ng b, khng

ch theo nguyn tc pha ng vi s khi nh sau:

- Khi 1: Khi ng b ha v pht xung rng ca khi ny c nhim v ly tn hiu

ng b ha v pht ra in p hnh rng ca a n khi so snh.

- Khi 2 : Khi so snh c nhim v sa snh hai tn hiu in p hnh rng ca URC v

in p iu khin Uk pht ra xung in p a ti mnh to xung.

- Khi 3: Khi to xung c nhim v to ra cc xung iu khin a ti

U1: in p li xoay chiu cung cp cho b chnh lu

urc: in p ta hnh rng ca ly t u ra ca khi BH - FXRC

uk: in p iu khin mt chiu dng iu khin gi tr gc m cc iu khin

ca Tiristor.



44/86



4.1Khu ng b ha v pht xung rng ca

4.1.1 S dng hai transistor

a. Thit b ca mch gm

- BA l my bin p ng b xoay chiu mt pha gm mt cun dy pha s cp v

hai cun dy pha th cp c cc tnh ngc nhau. ly tn hiu ng b v hai cun dy

pha th cp cn li c lp vi hai cun dy trn dng cung cp in p ngun nui cho

mch iu khin.

- Trn mch ra ca cun dy th cp ly tn hiu ng b c cc phn t l mch to

in p rng ca, trong :

+ Mch gm Tr2, Z, R4, WR l mch n nh dng np t.

+ URC l in p rng ca u ra ca s .+ U0 l in p n nh trn it n p DZ; ic1, ic2 l dng in cc gp Tr1 v Tr2

b. Nguyn l lm vic ca khu ng b ha v pht xung rng ca:

in p Ucb2 gia cc pht ra v cc Tr2 l Ucb2 = U0 - ic2.RWR, vi RWR l tr s in

tr ca bin tr WR. Do st p gia cc pht va cc gc ca mt Tranzitor hu nh khng

i nn ta xem Ucb2 = A = const, vy ta c: ic2 = (U0- Ucb2)/RWR = 1 const mt khc ta li c

dng in qua cc gp Tr2 l khng i

Ta gi thit rng: ti t = 0 th Ub = 0 v bt u chuyn sang chu k dng, ti t =0 th in p trn t C = 0

Vy sau thi im t = 0 th Ub > 0 nn it D c t in p thun, D s m dn

n c dng in t cun th cp BA i qua R2 v D, nu b qua st p rt nh trn cun

dy my bin p ng b ha v trn it D th trn R2 c t in p bng ton b sc

in ng th cp BA tc l Ub. in p st trn R2 lc ny c th dng t vo cc

pht Tr2 cn th m dt vo cc gc Tr1, do vy mch gc pht Tranzitor b t in p ng-

o

UB

a

R1

D ic2

Tr2

R2

BA

uc URC

Dz

C*

+UccWR

Ul

*

ic1

R3

Tr1

R4

U0



45/86



c v Tr1 kha v t c np in bi dng cc gp Tr2 c gi tr n nh. in p trn

t tng dn theo quy lut UC = I.t/c y l quy lut tuyn tnh.

n thi im t = th Ub = 0 v bt u chuyn sang na chu k m.Van D b t

in p ngc v kha li do vy in p ng b khng tc ng n mch gc pht ca

Tr1 na lc ny di tc ng ca ngun cung cp mt chiu qua in tr nh thi n R1

trong mch nh thin trong kiu phn p gm R1 v R2 m Tr1 m. Khi Tr1 m th t ngng

np v bt u phng in qua mch gp pht ca Tr1v in tr bo v Tranzitor R3.Ngita tnh chn cc in tr R1, R2 v Tr2 sao cho Tr1 m bo ha vi dng cc gp l 1

Vy t C s ngng phng in khi in p trn t gim xung bng st p bo ha ca

Tr1 cng vi st p trn R3 gy nn bi dng m bo ho ca Tr1; UR3 = iR3, st p bo ha

trn mt Tranzitor rt nh nn ta c th b qua, mt khc R3 v I cng c gi tr rt nh (1 -

5 mA) nn ta c th b qua st p trn R3. Nh vy th t C phng n in p bng khng

ti t = vt v do Tr1 vn m nn t vn gi nguyn gi tr in p bng khng cho n thi

im t =2 . Ti thi im ny th Ub = 0 v li bt u chuyn sang dng, it D li

c t in p thun v li m v Tr1 li b kha, do vy t C li c np tng t nh

khi t = 0 v s lm vic ca s lp li nh chu k va xt. in p rng ca trn u ra

cng chnh l in p trn t C v dng in p ra URC c cho trn th in p. Vi s

ny th bin in p rng ca khng ph thuc vo bin in p ng b, dng

in p ra gn ging hnh rng ca v di sn trc(giai on np t) cng t n

1800

ta s s dng sn ny ca URC.

Trong s th R3 l in tr hn ch dng phng ca t C qua Tr1 m bo v Tr1,

cn WR iu chnh tin in p rng ca cho ph hp vi yu cu.

Nhn xt:S ny cho dng in p rng ca chnh xc nhng do c in tr bo v

R3 m in p trn t khng gim v khng (0V) c. Mt khc, in tr ti nh s nh

hng n dng in p uRC



46/86



4.1.2 S dng IC khuch i thut ton

Nguyn l hot ng:

na chu k dng Tr kho, in p m qua R3,R

4dn ti u vo o ca IC khin

in p ra ca IC c gi tr dng v t C c np bi in p u ra ny. Dng np cho t

c xc nh l: ic = iv - iI nu IC l l tng th iv = 0 nn ic = - iI

constRR

ui ccI

43

Nn ic = const v in p trn t tuyn tnh.

na chu k m, D kho, Tr m nh cp in tr nh thin R1, R2; t C phng in

qua Tr. in p trn t gim v 0V

Gin in p nh hnh v:

Nhn xt: S ny c u im l dng in p ta rt chnh xc, dung lng ca t Ccn rt nh nn khng cn in tr bo v Tr. Mt khc, do in tr u ra ca IC nh nn

dng in p ra hu nh khng ph thuc vo in tr ti mc u ra ca IC. in p ra

c dng gn l tng

4.2 Khu so snh

to ra mt h thng xung xut hin mt cch chu k vi chu k bng chu k in

p rng ca (cng l chu k ngun cung cp cho b chnh lu) v iu khin c thi im

o

a

ic1

Urc

*

D

Ul

R1+Ucc

-Ucc

iv-i1

iv+

Tr

-

+

IC

R2

R3

uc

*

Udb

BA

C

WR

uv



47/86



xut hin cc xung ta s dng mch so snh. C th thc hin khu so snh theo nhiu mng

khc nhau, y ta dng IC khuych i thut ton v cch ni hai tn hiu URC v Uk theo

cch tng hp song song s nh sau.

0

0

-udk

udk

t

t

uRCuRC

a. Thit b ca mch gm

- IC1 l IC khuych i thut ton c nhim v khuych i v so snh tn hiu URC v

Uk. URC l in p rng ca c chu k theo in p thun t ln cc van mch ng lc,

cn Uk l in p iu khin.

- it D2 bo v u ra ca mch so snh

b. Nguyn l lm vic

Cc in p rng ca URC v in p iu khin Rk c a vo mch so snh vi

cc tnh khc nhau.

C th trn s ta c URC > 0 cn Uk < 0, IC thut ton lm nhim v so snh v tithi im dkRC UU th u ra khi so snh Ura s thay i trng thi c th:

Khi dkRC UU : Ura < 0 < 900

dkRC UU : Ura > 0 > 900

dkRC UU : Ura i chiu

Qu trnh ny c m t trn gin in p ca mch iu khin

4.3 Khu to xung:

m bo yu cu v chnh xc ca thi im xut hin xung,s i xng ca

xung cc knh khc nhau.Ta thit k cho khu so snh lm vic vi cng rut ra nh, do

xung ra ca khu so snh cha p ng cc thng s yu cu ca cc iu khin

Tiristor.V vy phi thit k thm mch khuych i xung, sa xunggi l mch to xung

+Udk

+Ucd

R3IC+

_R2

R1

Ur

+Ucc

-Ucc

Urc



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Thit b ca mch bao gm

- R1 C1, D2, Tr2, Tr3c nhim v sa xung.

- Tr3, Tr4, D3, D4, BAX c nhim v khuych i v truyn xung cung cp cho cc G

ca Tiristor.

Nguyn l lm vic ca mch to xung

- Xung truyn n cc iu khin Tiristor dng my bin p xung BAX. My bin p

xung ghp gia u ra ca tng khuych i cng sut xung vi cc iu khin G v K ca

Tiristor

- Khuych i xung: dng tng khuych i alinhtn mch khuych i c h s

khuych i l: 21 trong d 21 , l h s khuych i ca Tr3,Tr4)

- Sa xung: Khi in p u ra ca khu so snh c gi tr dng, t C1 s np (D2kho Tr3, Tr4 m bi xung dng theo ng + USS R7 - C1 Tr3 Tr4 - USS

nn UC1 : UCC(+) in p u ra ca khu so snh c gi tr m, it D2 phn cc thun, Tr3

v Tr4 kha, t C1 phng in ( + C1 ) R7 USS D2 (-C1) t C1 phng nhanh v 0

v np li vi in p c cc tnh ngc li vi hng s thi gian

= R7.C1. Do Tr3 v Tr4 khng kha li ngay m dn kha li ty thuc , qu

trnh gi l qu trnh sa xung.

Xut pht t nguyn l hot ng ca khu so snh ta thy: Khi thay i tr s in p

iu khin Uk thay i gc m th di ca cc xung ra ca khu so snh thay i.

* Mch sa xung

Nh vy s xut hin tnh trng mt s trng hp di xung qua ngn khng

m cc Tiritor hoc di xung qu ln, gy tn tht ln cho mch pht xung.

Mch sa xung nhm khc phc cc vn nu trn. Mch lm vic theo nguyn tc

khi c xung vo vi di khc nhau nhng mch vn cho xung ra c di bng nhau

theo yu cu v gi nguyn thi im xut hin ca mi xung. S nguyn l ca mtmch sa xung nh hnh v:



49/86



Trong s : Uv l in p vo ca mch, chnh l in p ra ca khu so snh c

mc bo ha dng v m. Cc phn t R11 v C2 s quyt nh di ca xung ra.

Nguyn l lm vic:

- Khi in p vo UV mc bo ha dng cng vi in tr nh thin R12, Tr6 m

bo ha, t C2 c np vi cc tnh nh pha trn ( qua C2 R11 Tr6 ). Tr6 m bo ha

lm im F c mc logic 0. Mc logic ny tn ti trong sut qu trnh UV bo ha dng.

- Khi in p UV mc bo ha m, t C2 phng in qua D1 t th m ln mch

pht gc ca Tr6 lm Tr6 kha dn n im F c mc logic 1, ngha l u ra nhn c

xung ra. Do in tr ngc ca Tr6 rt ln nn Ura > UCC. Khi C2 phng ht in tch,

n s c np theo chiu ngc li. Nh c R12 m th (+) li t ln pht gc ca Tr6

lm u ra li c mc logic 0. Mc d cn xung m u vo nhng nh R12 m Tr6 mbo ha. Thi gian tn ti xung c xc nh bi biu thc: tx = R11.C1.ln2.

di xung ch ph thuc vo gi tr ca R11 v C2 do cc xung ra lun c gi tr

khng i.

Thit b u ra ( Mch truyn xung ): Thng thng c hai cch truyn xung t u h

thng iu khin mch G K ca Tiritor l truyn xung trc tip v truyn xung qua my

bin p xung.

Truyn xung qua BAX c u im:

- m bo s cch ly tt v in gia mch ng lc v mach iu khin b chnh lu.

- D dng thc hin vic truyn ng thi cc xung n cc Tiritor mc ni tip nhau

hoc song song bng cch dung BAX nhiu cun th cp.

- D dng phi hp gia in p ngun cung cp cho tng khuch i cng sut xung

v bin xung cn thit trn cc iu khin ca Tiritor nh vic chn t s BAX hp l.



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- BAX v c bn kt cu ging nh bin p bnh thng cng sut nh. Hot ng

ca BAX tng t nh bin p thng vi dng in khng sin hoc c th xc nh nh l

phi tuyn v s bng khng khi mch t bo ha. BAX c mch t rt chng bo ha, n ch

hot ng trong thi gian ngn.

* Mch khuch i xung: khuch i cng sut ca xung iu khin, hin nay ph binnht l cc s Tiritor v Transistor. Hnh bn l s mch khuch i xung dung

Transistor kh ph bin hin nay.

Tn hiu vo UV ca mch khuch i xung s dng hai Transistor mc ni tip ( cn

gi l kiu Darlinhtn ). Tt7 v Tr8 mc ni tip tng ng mt Transistor c h s

khuch i dng in: 21.

Chc nng ca cc phn t trong s :

D2 l it c chc nng gim dng in qua cun dy s cp ca BAX khi Transistor

kha, ng thi hn ch qu in p trn Transistor. D3 bo v cun dy th cp BAX

nh i vi D2 ca mch s cp. D4 ngn xung m c th ti cc iu khin ca Tiritor

nh cc Transistor khc. Cc in tr hn ch xung p u vo v dng in cc gp ca

transistor.

Nguyn l lm vic ca s :

Tn hiu vo ca mch iu khin l tn hiu ra ca mch gi xung l tn hiu logic.

Gi txv l thi gian tn ti ca mt xung in p vo

tbh l thi gian tnh t lc c dng in mt chiu qua cun dy s cp ca BAX (

khi Tr7 v Tr8 m bo ha ) n lc li thp bo ha t

txr thi gian tn ti ca xung ra



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* Xt trng hp tbh > txv:

- Trong khong thi gian t = 0 t1, cha c xung vo, khng c dng qua BAX nn

th cp ca my khng c tn hiu.

- Khi t = t1 xut hin xung vo, Tr7,Tr8 m bo ha nn cun W1 c dng in chy

qua, lm cm ng sang pha th cp xung in p, to dng in qua D4 n mch G K

ca Tiritor.

- Khi t = t2 ( lc ny mch cha bo ha ) mt xung vo. Tr7,Tr8 ng dng in s

cp gim v khng qua D2. Bn th cp c s cm ng ( ngc chiu vi ban u do t

cm ) nhng nh D4 m xung m khng truyn ti Tiritor. Xung dng m kp qua R17 v

D3 tiu tn trn in tr.

Nh c D2 v D3 m khng xut hin in p t cm rt ln tren dy qun s cp ca

BAX.

* Khi tbh < txv :

- Khi t < t1 cha c u vo, Tr7 v Tr8 kha, khng c xung iu khin

- Khi t = t1 xut hin xung lm Tr7, Tr8 mbo ha lm xut hin xung iu khin.

- Khi t = t1 + tbh mch t BAX b bo ha, t thng li thp bng const nn mt xung

cm ng trn W2.

- Khi t = t2 dng in s cp v khng lm xut hin xung m trn dy qun th cpnhng khng a n mch G K nh ni trn.

Nh vy thi gian lm vic ca mch t BAX c nh hng rt ln n di ca

xung iu khin. Khi tbh > txv th di xung iu khin bng di xung vo. Cn trong

trng hp ngc li, di xung iu khin chnh bng thi gian bo ha mch t ca

BAX. Do cn cho BAX c thi gian bo ha t ln.

4.4 Mt s mch khc

4.4.1 Mch to ngun niDo trong mch c s dng cc vi mch khuych i thut ton, ta cn phi s dng

hai ngun nui ngc du ni tip nhau v c im chung l im ni mt. Ta thit k

mch ny nh sau:



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Ngun cung cp cho ton mch iu khin c ly trn cun th cp ca my bin p

ng b xoay chiu mt pha v c chnh lu qua b chnh lu cu 3 pha.

Trong :

+ T C3, C5, IC7815 dng n nh in p v cung cp ngun nui cho cc knh

iu khin.

+ T C4, C6, IC7915 dng n nh in p v cung cp ngun nui cho cc knh

iu khin.

4.4.2 Khu phn hi tc

i vi h truyn ng ngoi yu cu v phm vi iu chnh tc th n nh tc

khi lm vic cng rt quan trng. Trong h truyn ng ny ta thit k mch phn hi m

tc nng cao c tnh c

Tc ng c c truyn n my pht tc. My pht tc l mt my pht in mtchiu c in p ra t l vi tc ng c. Tn hiu phn hi ly trn WR4 v a vo khu

tng hp tn hiu (KTG) x l

7915

+15V

- 15V

7815



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4.4.3 Khi phn hi m dng in

trnh dng in trong ng c tng qu mc cho php khi khi ng, hm, o

chiu hay gp qu ti ta s dng mch in hn ch dng in phn ng. y ta s

dng mch phn hi m dng in. S mch in nh hnh v:

My bin dng TI cch ly gia mch ng lc v mch iu khin. in p ra ca T I

c chnh lu nh cu ba pha ( m bo cho dng in trong cun s cp ca T I l xoay

chiu ).Tn hiu phn hi dng in c ly mt phn trn bin tr R ri c a vo lc

v khuch i bi IC1; IC2. in p m trn in tr R4 c tc dng nh mt ngng; in

p u ra IC2 c tnh nh sau:

Ta chn R5 = R6

15.

2 16

7

rICr

uR

Ru

IurIC .2

ngr

IIR

Ru

R

R .

2 4

6

41

42

15.ngI

l h s phn p

Khi IU < Ing, in p u ra IC2 c du dng nn cc it kho, mch phn hi cha

c tc dng



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Khi IU > Ing, in p ra c gi tr m, lc ny mch phn hi dng in tham gia

khng ch gc m lm gim dng phn ng

4.5 Tnh ton mch iu khin

Vic tnh ton mch iu khin thng c tin hnh t tng khuch i ngc tr ln

Mch iu khin c tnh xut pht t yu cu v xung m Tiristor.Cc thng s c

bn tnh mch iu khin

Tiristor FT100DY10

in p ngc van UnV = 500 (V)

Dng in ngc van InV = 100 (A)

nh xung dng in Ix = 2000 (A)

Dng in xung iu khin Ik = 0,1 (A)

in p xung iu khin Uk= 2 (V)

Dng in r Ir = 0,015 (A)

St p trn Tiristor trng thi dn U = 2 (V)

Tc bin thin in p du/dt = 100 (V/s)

Thi gian chuyn mch tcm = 15 (s )

Nhit lm vic cc i Tmax = 1250C

rng xung iu khin tx = 1,67 (ms) tng ng vi 30

in.

Tn s xung iu khin fx = 3 (kHz)

mt i xng cho php 04

in p ngun nui mch iu khin U = 12 (V)

Mc st bin xung sx = 0,15

4.5.1 Tnh chn khu to in p ch o

Chn bin tr : RWR3 = 4,7 K cng sut tiu tn trn bin tr l

wPR 0478,0

4700

152

3

Chn R30 = 4,7 K , 1w



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4.5.2 Tnh chn my bin p xung

- Chn vt liu lm li st l Ferit HM. Li c dng h nh xuyn, lm vic trn mt

phn ca c tnh t ho c 3,0B (T), 30H (A/m), khng c khe h khng kh.

- T s MBA xung thng l m = 2 3, ta chn m =3.

- in p cun th cp MBA xung: U2 = Uk = 2 (V)

- in p t ln cun s cp MBA xung:

U1 = m.U2 = 3.2 = 6 (V)

- Dng in th cp MBA xung:

I2 = Ik = 0,1 (A)

- Dng in s cp MBA xung:

I1 = I2 /m = 0,1/3 = 0,03 (A)

4.5.3 Tnh tng khuch i cui cng

Chn transistor cng sut loi 2SC9111 lm vic ch xung c cc thng s:

Transistor loi p-n-p vt liu bn dn Si

* Tnh chn khu tng hp tn hiu

Khu tng hp tn hiu ch c nhim v tng hp tn hiu m khng yu cu c h s

khuch i ln. S b chn h s khuch i ca khu ny bng 2; KTH = 2

* Tnh chn khu hn ch gc m ca b bin i

T hnh v ta thy c quan h:

612

dku

phn trc ta chn0

min 15 do gc phi 150

. Tc l

1261

2

dku uk 5 (V)

Vi ukmax = 5 (V), uTGmax = 13 V

rcm

dkrcm

U

uU 2

2



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4,0TG

kHC

u

uK

4.5.4 Tnh khu so snh

Khuch i thut ton A2 ca khu so snh chn IC loi TL084 c cc thng s

- in p ngun nui: Vccmax = 18 V. Chn Vcc = 12 V

- Hiu in th gia hai u vo: 30 V

- Nhit lm vic: Tmax = 125oC

- Cng sut tiu th: P = 180 mW = 0,18 W

- Tng tr u vo: Rin = 106

( M )

- Dng in u ra: Ira = 30 (pA)

- Tc bin thin in p cho php: du/dt = 100 (V/s)

*Chn: Trong nu ngun nui Vcc = 12 V th in p vo A2 l Uv 12V. Dng invo c hn ch Iv < 1mA. Chn R4 = R5 = 15K. Khi dng vo A2 l:

Iv = 310.1512

0,8 (mA)

4.5.5 Chn cng and cho khu khu chia xung

Ton b mch iu khin phi dng 12 cng AND nn ta chn hai IC loi 4081 hCMOS. Mi IC c cng AND c cc thng s sau :

- Ngun nui IC Vcc = 318 V.

- Chn Vcc = 12 V .

- Nhit lm vic: - 40o 80o .

- in p ng vi mc logic 1: 2 4,5 V

- Dng in nh hn 1 mA .Cng sut tiu th P = 2,5 (mW / 1 cng )

4.5.6 Tnh chn khu phn hi tc

Cn c vo tc nh mc ca ng c v sai lch tnh ca h thng ta chn my

pht tc c cc thng s sau:

in p nh mc: Um = 230 (V)



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Dng in nh mc: Im = 0,5 (A)

Tc nh mc: nm = 1300 (v/ph)

in tr nh mc: R = 0,8 ( )

H s ca my pht tc:

1300

8,0.5,0230.

m

FTFTFT

m

FTFT

n

UIU

n

E 0,18

Do in p ra ca khuch i thut ton Ur Ecc= 12V nn ta iu chnh bin tr Rs1

sao cho U 12 V. vy =dmn

U

1300

12= 0,009

Vy = 0,009 (Vphut/vong)

4.5.7 Khu phn hi m dng c ngt

Do khi s dng phn hi m tc n nh tc i vi chnh lu tiristo th vn

khi tc ng c bin thin ngay lp tc gy nn s tng gim qu mc ca dng in

phn ng v ca tiristor khng th chp nhn c. Do cn phn hn ch dng in

mt cch t ng, ta dng khu phn hi m dng c ngt

Trong s : A6 l cc b khuych i thut ton. tn hiu phn hi dng c ly trn in

tr iu chnh R thng qua b bin dng v b chnh lu cu 1 pha .Sau vo u vo A6

so snh vi Udk bi iot D10

Khu ngt c tc dng khi c qu dng phn ng tng qu dng ngt khu ngt tc

dng hn ch dng in

Chn loi bin dng loi 150:5 , cp chnh xc 0,5

H s ca bin dng: PI =5

150= 0,03

Chn ngun to in p ngt dng:

Dng in cn hn ch ca ng c c chn theo yu cu khi ng vi ti c th.

Vy ta chn dng in ngt ca c tnh vi ng c mt chiu l:

Ing = 1,5.Im = 1,5.53,97 = 81(A)

in p ngt: Ung = 0,1.Ing = 0,1.81 = 8,1 (V).

Mun gii hn c dng in Igh = 1,5.Im ta cn chn ngun cp -E sao cho ta khng

ch Ung c in p l 9 V.



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in p phn hi dng in a vo A3 l u ra A4 sau khi khuch i, trong qu

trnh khi ng th u ra A4 ri vo vng bo ha nn c in p xp x bng in p

ngun cp cho A4 l Ubh Ucc = 12V.

Vy ta c:kd

bh

I

U =

97,53.2

12= 0,11 (V/A)



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CHNG V

XY DNG S CU TRC CA H TRUYN NG

5.1 Tnh cc tham s c bn.

*Cc tham s c bn:

: H s phn hi m tc = 0,009

R : Tng in tr mch phn ng R= 0,308 ().

I: Tr s dng in trn ti, tnh theo dng nh mc ng cIm = 53,97 (A)

K: H s khuch i ca ng c.

Tnh

97,53.308.0220

1090

..

1

mm

m

dm

IRU

n

CeK

5,35

Tnh Kb:

tnh h s khuch i ca b bin i (K) ta xy dng c tnh biu din quan h

Ud = f(Uk) sau tuyn tnh ho c tnh ny ra c tnh h s gc ca on c tnh . H

s ca on c tnh c l h s khuch i ca b bin i

Kb = tg =dk

d

U

U

Quan h Ud = f(Uk) xut pht t hai quan h: Ud = f() v = f(Uk)

* Xy dng quan h Ud = f():

Coi h thng lm vic ch dng in lin tc: Ud = Ud0.cos

Trong : Ud0 =230 (V)l in p chnh lu khng ti ca b bin i

l gc iu khin. Cho bin thin t = (0 /2) ta c cc tr s Ud lp

thnh bng sau:

0 /12 /

6

/4

/3 /2

U

d (V)

230 215 1

93

157,

7

1

11,5



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* Xy dng quan h = f(Uk): Khi thay i gi tr in p iu khin (Uk) th gi tr

gc iu khin cng thay i theo.ng vi mi (Uk) khc nhau ta

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