Thiet Ke Bo Dieu Khien Dong Co Mot Chieu Kich Tu Doc Lap 20120817041302 31

7/28/2019 Thiet Ke Bo Dieu Khien Dong Co Mot Chieu Kich Tu Doc Lap 20120817041302 31

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2.1.2.1. Phng trnh c tnh cTheo s hnh 2.1a v hnh 2.1b ta c th vit phng trnh cn bng

in p ca mch phn ng nh sau:

U = E + (R + Rf)I. (2-1)

Trong :

U: inp phn ng, V

E: sc in ng phn ng, V

R: in tr mch phn ng,

I: dng in ca mch phn ng, A

Vi: R = r + rcf + rb + rct

r: in tr cun dy phn ng

rcf: in tr cun dy cc t ph

rct: in tr tip xc cun b

Sc in ng E ca phn ng ng c c xc nh theo biu thc

-

PNE . . K. .

2 a= =

(2-2)

Trong :

P: s i cc t chnh

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

a: s i mch nhnh song song ca cun dy phn ng: t thng kch t di mt cc t

: tc gc rad/s

P.NK

2 a=

:h s cu to ca ng c.

T (2-1) v (2-2) ta c

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- - f-

U R R.I

K. k.

+ =

(2-3)

Biu thc (2-3) l phng trnh c tnh c in ca ng c.

Mt khc, m men in t Mt ca ng c c xc nh bi

Mt = K..I (2-4)

t-

MI

K.=

: thay gi tr I vo (2-3) ta c

( )- - f

2 t

U R R.M

K. K.

+ =

(2-5)

Nu b qua tn tht c v tn tht thp th mmen c trn trc ng c

bng m men in t, ta k hiu l M. Ngha l:

Mt = Mc = M

( )- - f

2

U R RM

K. K.

+ =

(2-6)

y l phng tnh c tnh c ca ng c in mt chiu kch t clp.

Gi thit phn ng c b , t thng = const, th cc phng trnh

c tnh c in (2-3) v phng trnh c tnh c (2-6) l tuyn tnh. th

ca chng c biu din trn hnh 2-2 v 2 3 l nhng ng thng.

Theo cc th, khi I = 0 hoc M = 0 ta c:

-0

U

K. = =

(2-7)

0 c gi l tc khng ti l tng ca ng c in mt chiu

kch t c lp.

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dm dm

H2.2. c tnh c in ca ng c

in mt chiu KTL

H2.3. c tnh c ca ng c in

mt chiu KTL

Khi = 0 ta c

-- nm

- f

UI I

R R= =

+ (2-8)

v M = K. .Inm = Mnm (2-9)

Inm v Mnm c gi l dng in ngn mch v m men ngn mch.

Ngoi ra phng trnh c tnh (2-3) v (2-6) cng c th c vit

di dng:

-0

U RIK. K

= =

(2-10)

( )-

2 0

U RM

K. K. = = (2-11)

Trong R = R + Rf,-

0

U

K. =

( )2-

R R.I .M

K. K. = =

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c gi l st tc ng vi gi tr ca M. T phng trnh

c tnh co (2-6) ta thy c 3 tham s nh hng n c tnh c: t thng

ng c, in p phn ng U, in tr phn ng ng c.

2.1.2.2. Xt nh hng ca cc tham s n c tnh ca. nh hng ca in p phn ng.

Gi thit l R = const v = m = const

Khi thay i in p phn ng theo hng gim so vi Um, ta c tc

khng ti.

x

oxm

UU varK.

= =

cng c tnh c( )

2

-

K.

R

= = const

Nh vy, khi thay i in p, t vo phn ng ng c ta c mt

h c tnh c song song vi c tnh c t nhin nh hnh (2-4)

0

H2.4. Cc c tnh ca ng c mt chiu KTL khi gim in p t vo

phn ng ng c (U4 < U3 < U2 < U1 < Um)

Ta thy rng: khi thay i in p (gim in p) th m men ngn

mch, dng in ngn mch ca ng c gim v tc ca ng c cng

gim ng vi mt ph ti nht nh. Do phng php ny cng c sdng iu chnh tc ng c v hn ch dng in khi khi ng.

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b. nh hng ca in tr phn ng.

Gi thit U = Um = const v = m= const.

Mun thay i in tr mch phn ng ta ni thm in tr ph Rf vo

mch phn ng.

Trong trng hp ny tc khng ti l tng.

m0

m

U

K. =

= const

cng c tnh c:( )

2

m

- f

K.M

K R

= =

+= var

Khi Rf cng ln th cng nh, ngha l c tnh c cng dc. ng vi

Rf = 0 ta c c tnh c t nhin.

( )2

m

TN

K.

R

=

TN c gi tr ln nht nn c tnh c t nhin c cng hn tt c

cc ng c tnh c in tr ph.

Nh vy, khi thay i in tr ph Rf ta c mt h c tnh bin tr

c dng nh hnh 2-5.

0

H2.5. Cc c tnh ca ng c mt chiu KTL khi thay i in tr ph

mch phn ng Rf4 > Rf3> Rf2> Rf1.

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ng vi mt ph ti M no , nu Rf cng ln th tc ng c cng

gim, ng thi dng in ngn mch v m men ngn mch cng gim.

Phng php ny hn ch dng in v iu chnh tc ng c

pha di tc c bn.

c. nh hng ca t thng

Gi thit in p phn ng U = Um = const

in tr phn ng R = const. Mun thay i t thng ta thay i dng

in kch t Ikt ng c.

Khi thay i Ikt tc l thay i

th tc khng ti l tng

mox

x

U

K = =

var

cng c tnh c :

( )2

x

-

K

R

= = var

Do cu to ca ng c in, thc t thng iu chnh gim t thng,

nn khi t thng gim th 0x tng, cn s gim. Ta c mt h c tnh c

vi 0x tng dn v cng ca c tnh gim dn khi gim t thng.

02

01

0

2

1

0

01

02

2

1

H2.6. c tnh ca ng c mt chiu KTL khi gim t thng (

2 . Cg sut in t ca my pht, cng sut in t

v cng sut c hc ca ng c l:

PF = EF.I > 0

F = E.I < 0

P = M. > 0

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H 2.11. c tnh c ca h F - D

a. Trong ch ng c; b. Trong ch hm ti sinh

Cc biu thc (2-18) cho ta thy gc phn t th I v th III nng

lng c vn chuyn thun t ngun my pht ng c ti.

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Vng hm ti sinh nm gc t th II v th IV lc ny cho do

o > nn FE E> dng phn ng chy ngc t ng c v my pht

lm cho mmen quay ngc chiu tc quay v th.

PF = EF.I < 0

P = E.I > 0 (2-19)

Pc = M . < 0

Nh vy, dng in ch hm ti sinh ngc chiu vi chiu dng

in ch ng c v nng lng c chuyn vn theo chiu t ti

ng c my pht ngun, my pht F v ng c di chc nng chonhau.

Vng hm ngc ng c trong h F - c gi hn bi c tnh

hm ng nng v trc mmen.

H.2.12. Vng hm ngc ng c trong h F - D

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Sc in ng E ca ng c tr nn cng chiu sut in ng my

pht hoc do rto b ko quay ngc bi ngoi lc ti th nng, hoc do chnh

sut in ng my pht o du. Biu thc tnh cng sut s l.

PF = EP.I > 0

P = E.I > 0

Pc = M.


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Hnh 2.13: S ni dy v s thay th ca chnh lu tia ba pha

t

t

/2

Hnh 2.14. c tnh iu chnh (a) v th thi gian

ca chnh lu hnh tia pha (b)

Khi dng in chnh lu id l lin tc th c th dng c th cc

qu trnh dng in v in p nh trn hnh 2-20 sut in ng chnh lu l

nhng hnh sin ni tip nhau, gi tr trung bnh ca sut in ng chnh lu

tnh nh sau:

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Ed =

2

P

2m

PU sin d d .cos

2

+

=

(2.21)

=

et

: Gc iu khin tnh t thi im chuyn mch t nhin.

P: s xung p p mch trong mt chu k in p xoay chiu.

Edo 2mP

sin Up

U2m l bin in p a vo b chnh lu:

Nu gi gc dn ca van l th c th tnh c thnh phn mt chiu

ca dng in chnh lu chnh l thnh phn sinh momen quay ca ng c.

d 2m 00

P PI i .d U .sin E

2P 2 R 2 2 2

= = +

b. Hin tng chuyn mch

Khi pht xung m mt van Thiristo th in p ant ca pha l

dng hn in p ca pha c van ang dn dng, do m dng in ca

van ang dn gim dn v khng cn dng in ca van k tip s tng dn

ln. Do trong mch c in cm m qu trnh ny xy ra t t, cng ti mt

thi im c hai van u dn dng va chuyn dng cho nahu. Qu trnh ny

gi l chuyn mch gia cc van.

Trong qu trnh chuyn mch v c hai van d dn nen sut in ng

chnh lu bng trung bnh cng ca in p hai pha.

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Hnh 2.15 Hin tng chuyn mch gia cc van Thiristo

: gc iu khin tnh t thi im chuyn mch t nhin khong thigian cn thit van chuyn dng cho nhau (gc trng dn).

Do c chuyn mch nn sut in ng chnh lu b st i (vng gch

cho trn hnh v).

c. Ch dng in gin on.

Hin tng gin on dng in chnh lu xy ra do nng lng in t

tch lu trong mch khi dng in tng khng duy tr tnh cht lin tc

Hnh 2.16: Ch dng in gin on

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Lc ny gc dn van tr ln nh hn 2/p. Dng in qua van tr v

khng trc khi van k tip bt u dn.

Trong khong dn van th sut in ng chnh lu bng sut in ng

ca ngun khi dng in bng khng. Khi dng, sut in ng chnh lu

bng sut in ng ca ng c in.

d. Ch nghch lu ph thuc.

Nu trong s hnh (2-19) ta tng gc m ca cc van n gi tr gn

bng

v o chiu sut in ng E bng cch dng ngoi lc bt roto ng

c quay ngc, hoc o chiu dng kch t c th dng in chnh lu vn

theo chiu c nhng sut in ng chnh lu o du cc van dn dng

trong thi gian in p ant m. Cng sut in t ca ng c v ca chnh

lu o du do cc van dn dng trong thi gian in p ant m. Cng

sut in t ca ng c v ca chnh lu l

Pt = EId > 0

Pd = Ed Id


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Hnh 2.17: S thay th chnh lu ng c mt chiu

c tnh c c ng =( )

2

m

k

k

R X

+

Tc khng ti l tng tu thuc vo gc iu khin

doo

m

E cosK

=

Thay i gc iu khin t 0 th sut in ng chnh lu bin

thin t Edo v ta c mt h c tnh tnh song song nhau nm na bn

phi ca mt phng to [, I]

Hnh 2.18: c tnh c h T -

Cc c tnh c bn cua h T - mn hn cc c tnh c ca h F -

bi v c s st p do hin tng chuyn mch gia cc van bn dn gy nn.

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Khi gc iu khin bin thin trong vng 0 2

b bin i lm

vic ch chnh lu, ng c c th lm vic ch ng c nu sut

in ng E cn dng v ch hm ngc nu sut in ng E ichiu (xem hnh 2-24).

Khi tng gc iu khin2

< max v ti c tnh cht th nng

quay ngc chiu ng c th c sut in ng Ed v E u i du. Nu sut

in ng ng c ln hn gi tr trung bnh ca b bin i th dng in

chy qua cc van vn theo chiu c, ng c lm vic ch hm ti sinh.

b. Ch dng in gin on.

Bng phng php ton hc ta c th tnh v v c c tnh c trong

vng dng in gin on l cc don cong nt lin rt dc st trc tung

trn hnh 2-24. H thng khng th p dng cc phng php t ng iu

chnh c bit.

2.3.2.3. Nhn xt chungu im ca h T - l tc ng nhanh cao, khng gy n v d t

ng ho, thun tin cho vic thit lp cc h thng t ng iu chnh vng

nng cao cht lng c tnh tnh v c tnh ng ca h thng.

Nhc im ca h T - l do cc van bn dn c tnh phi tuyn ng

in p chnh lu ra c bin p mch cao, gy tn tht pha trong my

in. H s cos ca h ni chung l thp.H truyn ng iu chnh xung p - ng c mt chiu.

Nguyn l h truyn ng iu chnh xung p - ng c mt chiu.

Trn hnh 2-19a m t s nguyn l h truyn ng iu chnh xung

p - ng c mt chiu.

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Khi kho S thng ta c: U = UN, i = iN, khi kh S ngt th iN = 0, U = 0

v i = i0. Do tc dng duy tr dng ca L. th trn hnh 2-19b m t qu

trnh dng in v in p trong ch dng in lin tc.

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Hnh 2.19: nguyn l lm vic ca h xung p ng c.

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Cn c vo s thay th hnh 2-19 a ta c phng trnh cn bng

in p nh sau:

U = UL + UR + E = 0

diL Ri U 0

dt + = =

Ti thi im t = 0 kho S bt u thng va U = UN

Khi dng in i = Imin. Gii phng trnh

i = ( )t/ t- t/ t-N minU E 1 e I eR +

Vi 0 t < t v T =L

Rl hng s thi gian ca mch phn ng.

Ti thi im t = t kho s bt u ngt khi d i = Imax Khi trong

mch vn tn ti dng in khp mch qua D 0 (do nng lng in t tch lu

trong mch khi dng in tng) v U = D

Nu S thng lin tc t = T th dng in trong mch phng ng s

khngg i v bng

I = Imax = Imin = NU E

R

Nu thi gian thng kho gim n mt gi tr ti hn t = tgh th

dng Imin = 0, v nu t < tgh th h thng s chuyn sang ch dng in ginon.

Ti thi im trn hnh 2-19c l khi dng in gim n gi tr 0. Khi tx

< T th xy ra ch dng in gin on v ging nh h T - th c tnh

c ca ng c trong vng dng in gin on l rt mm.

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CHng III: Phn tch tnh ton mch lc

3.1. Mc ch yu cu

Nh phn tch chng I phn ch yu v truyn ng s dng

ng c in mt chiu c s 600kW, in p ba pha 600V, tc c iu

chnh trn.

Ta s dng h truyn ng T - c phn hi m tc BB cp cho

phn ng kch t ca ng c l kch t c lp. o chiu quay bng cng

tc t chuyn mch o chiu dng in phn ng.

ng c dng phng php hm ng nng kch t c lp. tng

tui th v nng cao hiu qu s dng, khai thc ht cng sut ca ng c v

b bin i, ta s dng qut lm mt c cng sut 0,7 kW qut mt cho

ng c mt chiu.

B i Thiristo cng c lp qut lm mt vi cng sut 7kW.

3.2. La chn ca b chnh luTrong trng hp ny l ng c mt chiu, tc l trn s thy th

c dng R + L + E.

3.2.1. Chnh lu cu mt pha i xng c iu khin.

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Hnh 3.1: Chnh lu cu mt pha i xng:

a) S thay th; b) th thi gian

n gin cho qu trnh phn tch ra xem dng in chnh lu id l

dng lin tc v khng c hin tng chuyn mch gia cc van, tc l ti

mt thi im ch c mt van dn dng.

Ti = ta pht xung cp Thiristo 1 1 v 4. Khi = + ta pht

xung m cp Thiristo 2 v 3, gi l gc m van (hay gc iu khin ). tnh

t thi im chuyn mch t nhin.

U2 l tr s hiu dng. in p pha cun th cp bin p ngun.

Cn c vo bng cc tham s chnh ca cc mch chnh lu trong

quyn ti liu "Hng dn thit k mch in t cng sut" ca thy Phm

Quc Hi ta c cc tham s chnh ca mch chnh lu cu mt pha nh sau:

Gi tr trung bnh ca dng in van Iv = Id/2

Trong Id l tr s trung bnh dng in ra ti. in p ngc ln nht

m van phi chu khi lm vic Ung max = 1,41 U2

Tr s hiu dng dng in cun th cp my bin p ngun Sba = 1,23.Pd

Trong Pd l cng sut mt chiu trn ti.

Pd = Udo .Id.

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H s p mch ca in p chnh lu Km = 0,67

Nhn xt: S ny c th cp dng ti ln ti 100(A) v in p ra ti

t 10 (V) tr ln.

u im ca s ny so vi chnh lu hnh tia l khng nht thit

phi c bin p ngun, khi in p ra ti ph hp vi cp in p ngun xoay

chiu ta c th mc trc tip mch chnh lu v li in. Do st p trn cc

van gp i s hnh tia nn khng thch hp vi ti. Cn dng ln nhng

in p li nh.

3.2.2. Chnh lu hnh tia 3 pha.

t

t

Hnh 32. Chnh lu hnh tia 3 pha: a) s thay th; b) th thi gian

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Cc tham s chnh ca mch chnh lu hnh tia 3 pha nh sau:

Gi tr trung bnh ca in p chnh lu.

2d

3 6UU .cos

2=

Khi = 0 th Udo = 1,17 U2

Tr s trung bnh ca dng in qua van

Iv = dI

3in p ngc ln nht m van phi chu khi lm vic.

Ungmax = 2,45 U2

Tr s hiu dng dng din cun th cp th nht MBA ngun.

I2 = 0,58 Id

Cng sut tnh ton ca my bin p ngun

Sba = 1,.35Pd

Nhn xt: S chnh lu 3 pha hnh tia bt buc phi c bin p

ngun c im trung tnh a ra ti, cng sut my bin p ny ln hn

cng sut mt chiu 1,35 ln.

Tuy nhin st p trong mch van nh nn thch hp vi phm vi in

p lm vic thp. V s dng ngun 3 pha nn khng lm lch in p li v

cho php nng cao cng sut ti ln ln nhiu, mt khc, p mch ca

in p ra sau mch chnh lu gim ng k nn kch thc b lc cng nh

i nhiu so vi s 1 pha.

3.2.3. Chnh lu 3 pha s cu.

Cc tham s chnh ca mch chnh lu 3 pha s cu nh sau:

Gi tr trung bnh ca in p chnh lu.

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Ud = 23 6U cos

cos

Khi = 0 th Udo = 2,34. U2

Tr s trung bnh dng in quan van,

dv

II

3=

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

Ungun max = 26.U

Cng sut tnh ton mch MBA ngun.Sba = 1.05. Pd.

H s p mch ca in p chnh lu Km = 0,057 s chnh lu cu

3 pha (H gn 6 Thiristo chia thnh hai nhm.

Nhm ca tt chung: T1, T3 v T5

Nhm ant chung: T2, T4 v T6

in p cc pha.

U2a = 22.U sin

U2b = 22.U sin(2

3

U2c = 22.U sin(4

3

Gc m c tnh t giao im ca cc na hnh sin.

Hot ng ca s .

Gi thit T5 v T6 ang cho dng chy qua VF = U2c , VG = U2b khi =

1 =6

+ cho xung iu khin m T1. Thiristo ny m v U2a >0. S m ca

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T1 lm cho T5 b kho 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

=

2 =3

6

+ cho xung iu khin m T2. Thiristo ny m v

khi T6 dn dng, n t U2b nn ant T2 m U2b > U2c. S m ca T2 lm cho

T6 b kho li mt cch t nhin v U2b > U2c

Cc xung iu khin lch nhau3

c ln lt a n cc iu khin

cc Thiristo theo th t 1, 2, 3,4, 5, 6, 1

Trong mi nhm khi 1 Thiristo m n s kho ngay Thiristo dn dngtrc n. Tm tt bng.

Thi im M Kho

1 = /6 + T1 T5

2 = 3/6 + T2 T6

3 = 5/6 + T3 T1

4 = 7/6 + T4 T2

5 = 9/6 + T5 T3

6 = 11/6 + T6 T4

Gi tr trung bnh ca in p trn ti

ng bao gi trn biu din in th cau F v VF. ng bao gi di

biu din in th cau G v VG.

in p trn ti l Ud = VF -VG l khong cch thng ng gia hai

ng bao.

Ud =

5

62

6

3 6U62Usin d .cos

2

+

=

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Ud =

7

62

3

6

3 6U62Usin d .cos

2 2

+

2 + = 3

a)

Hnh 3.3 Chnh lu 3 pha s cu: a) s thay th; b) th thi gian

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Nhn xt: Chnh lu 3 pha s cu l loi c s dng nhiu nht

trong thc 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 in t hn cc loi trn ng

thi cng sut MBA cng ch xp x cng sut ti. Cng sut mch chnh lu

ny c th rt ln n hng trm kW.

Nhc im ca n l mt sp p trong mch van gp i s hnh

tia nn khng ph hp vi cp in p ra ti di 10V.

3.2.4. Kt lun

Sau khi a ra cc s trn ta c nhn xt:

Cng sut ti ca ta ln nn nu ta dng s c 1 pha s gy ra mo

in p li.

Nu dng chnh lu s hnh tia th ta cn phi c bin th. Mt

khc, do s ny c h s p mch ca in p chnh lu tng i ln

nn phi dng thm b lc.Nu dng chnh lu 3 pha s cu th s m bo c yu cu v

cng sut ca ti ln. Mt khc in p ca ng c trong trng hp nyl

600V v vy ta c th mc trc tip chnh lu 3 pha s cu vo li in

m khng cn phi qua bin th. H s p mch ca in p chnh lu cau s

ny rt thp (0,057). V th c thun li l ta khng phi dng b lc.

Vy sau khi cn c vo cng sut v in p m ti yu cu ,v ichiu vi u v nhc im ca tng s l chn chnh lu cu 3 pha c

iu khin l hp l nht.

3.3 Tnh ton mch ng lc

m bo a ra n ti in p 600V cng sut ng c 600 kW cn

phi b cc st p do in tr, in cm ca dy qun MBA v st p trn van

bn dn. V vy in p trn van thc t l Ud.

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Ud = Ut + Ur + Ux + Uvr

Uv l st p trn van Uv = 2A.

St p trn tr dy qun bin p. Do in p Ud to thnh nh in p 2

pha ca ngun nn in tr gy st p l hai in tr pha. V vy c h s 2

trong nhiu biu thc sau:

Ur = 2Id rba = 2Id = 2Id =22

ba

U

S /3r

Ur = Id2

d

d

U 6.

2,34 1,05Ud.I

= 1,044.rUd

St p do trng dn gy ra bi in cm dy qun pha bin p

Ux =22

d ba d xba

UI X I

3 3 S /3

=

= x d29

.U2,34 .1,05

= 0,497xUd

Ud =( ) ( )

dt v

r x

U U 600 2,01

1 1,044 0,497 1 1,044.0,02 0,497.0,07

+ +=

+ +

Ud = 638

U2 = dU 638

2,34 2,34= = 272 (V)

r = 2%, x = 7% l lng st p % do in tr v in cm dy qun

bin p.

3.3.1. Tnh chn van cho mch ng lc.

Mch ng lc gm c 6 van Thiristo lp theo s cu.

in p ngc m mi van phi chu l v van c mc theo s cu

3 pha nn in p ngc t ln van c tnh theo cng thc.

Ungmax = 2,45U2 = 2,45.272 = 666,4.

36


37/91

in p ca van c chn phi tho mn iu kin.

Uv > Ku.Ungmax

y Ku l h s d tr in p cho van.

Do thc t in p li khng n nh v c php dao ng, mt

khc c nhiu yu t nh hng ngu nhin trn mng in nn h s d tr

in p ly trong khong 1,7 n 2,2

Ta chn Ku = 2, v th van c chn phi c

UV = 2. 666,4 = 1332,8 (V)

Dng in ma mi van phi chu lIV > KV .Itbv

Trong :

Itbv l gi tr trung bnh dng in thc t qua an

Iv l dng trung bnh ca van c chn

Kv l h s d tr v dng in cho van.

i vi s sau ta c dng in trung bnh thc t qua van.

Itbv = dI 100

3 3= = 333,3 (A)

Trong : Id l dng nh m ca ng c.

Id = Im = 1000 (A)

Itbv = Im = 333,3H s d tr dng in ca van tu thuc vo c im lm vic ca

van, chng hn cng dng in qua van thuc loi ln bhau trung bnh,

c hay lm vic iu kin qu ti hau khng , iu kin mi trng ch d

lm mt. Vi iu kin c th ca nh my thp ta thy.

Kv = 2

Vy van m ta chn phi c: 2. 333,3 = 666,6 (A)

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Iv > Kv .Itbv = 2.333,3 = 666,6 (A)

Nh vy ta phi chn van tho mn hai ch tiu.

Ch tiu in p Uv > 1332,8 (V)

Ch tiu dng in Iv > 666,6 (A)

Cn c vo bng tham s Thiristo do Lin X sn xut (ti liu TCS

ca thy Phm Quc Hi) ta chn Thiristo k hiu T -800, c cc thng s

sau.

Icp = 800 (A) Id = 14.000 (A) cp du/dt 4-6

Tx = 3600(A) Id = 30 (mA) cp tph = 1Cp di /dt = 1-4

U = 2,1 (V)

Uk = 7 (V)

Ik = 400 (mA) Cp in p (10 - 24) 100 (V)

3.3.2. Tnh ton thit k cun khng bo v hn ch tc tng dng

di/dt.

V mch chnh lu mc trc tip vo li in khng c my bin p

nn ta phi c cun khng ny.

Tc tng dng di/dt s ln nht khi in p trc khi van dn l cao

nht, tng ng vi in p li tng thm 5%, do cun khng m ta cn

thit k gii c h s t cm tho mn:

L >( )

vmax6

cp

U 21,05.600

di /dt 50.10= = 18.10-6H = 18H

Ta chn in cm hn ch L = 20H.

Da vo ti liu "Thit k - thit b in in t cng sut" ca tc gi

Trn Vn Thnh ta thit k tnh ton cun khng sau:

I = 2f.L

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39/91

I. Tng tr cun khng.

+ St p trn cun khng : Uck = I.Z

I l dng in xoay chiu i qua cun khng vo b chnh lu cu pha

3pha: I = 0,816.Id.

Pck = Uck.I = I2.Z = (0,816.Id)2

Pck = (0,816.100)2.2.3,14.50.20.19-6 = 4181 (VAR)

+ Tnh tit din li thp cun khng.

SFe = K ckP

f

vi K =5 6 ta thy k = 5,5

SFe = 5,5818150

= 50 cm2 .

Vy ta chn S = 55cm2

+ Tnh s vng dy cn c ca cun khng.

W =

ck

dq

U

4,44.K .f.B.S

Trong : Kdq : l h s dy qun Kdq = 1,1 1,3 ta chn Kdq = 1,1

B - l mt t cm cau li thep

B = 1,1 1,8 (T) ta chn B = 1,2

Vy : W =6

4

0,816.100.2.3,14.50.200.10

4,44.1,1.50.1,2.55.10

+Chn tit din dy qun cun khng.

dng in i qua cun khng l:

I = 0,816.100 = 816 (A)

Ta chn dy qun khng l dy ng. cun khng pht nhit ta chn

mt dng in qua cun khng l .

J = 3A /mm2

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SCU =I 816

J 3= = (mm2).

Vy cun khng hn ch tc tng dng di/di m ta va thit k c

cc thng s chnh sau.

SFe = 55cm - tit din li t

SCu = 272mm2 - tit din dy ng.

W = 3,1 vng - s vng qun.

3.3.3. Tnh mch bo v qu p RC mc song song vi van.

Hnh 3.4: Bo v Thiristo

+ H s qu p ca van c tnh theo cng thc sau:

K =ngcp

ngtt

U

U

Trong : Ungcp - l in p ngc ln nht m c th chu.

Ungtt -l in p ln nht t ln van tong ng in p li tng ln

5%. V vy:

K =2000

2.1,05.600 = 2,25

Cn c vo ti liu hng dn thit k in t cng sut ca thy Phm

Quc Hi ta c phng php tnh tr s R v C. Ta c th tra * *maz minR ,R ,

v C* theo k sau:

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41/91

Hnh 3.5: Cc h s C*, R*max, R*min theo h s K

tra th 3 - 5 vi K =2,2 c C* = 0,25 ; *minR = 2; R*max

Tr s dng in nh qua van.

IVmax = V maxU

Rt

Trong .

UVmax = 2 .1,05.600 = 890 (V)

Rt =d

d

U 638 0,64I 100

= =

IVmax =890

0,64= 1390 (A)

Trong mch chnh lu tn s chuyn mch ca van bng tn s li v vy.

di

dt

(max) = 2f.Imax = 2.3,14.50.1390

di

dt= 436460A/S = 0,436 (A)/s

+ Xc nh din tch Q tn ti trong van trc khi kho Q theo h c

tnh trong s tra cu [Q=f(di/dt ; IVmax]

Vi loi T-800 tra th Q = f (di/dt; Imax; I)

Trong s tay vi di/dt v IVmax va tnh c Q = 50 As = 50.10-6

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T ta c

6' *

min

ngtt

2Q 2.50.10C C

U 890

= = .0,25 = 0,02.10-6(F)

Ta chn t C = 0,1F

+ Tnh R, vi s cu c hai pha tham gia dn din cho ti nn gi tr

L' = 2L nen

6ngtt

6

U 890.2.18.10

2Q 2.50.10

= = 18

V vy in tr nm trong phm vi

R*min. 18 RR*max.18

36 R 72

Chn in tr R = 60 ()

Lc ny dng phng ln nht ca t bng.

ICmax = Ungtt/R = 890/60 = 15 (A)3.3.4. Chn cu dao ng ct cho mch lc

Dng in i qua cu dao l

I = 0,816 .Id = 0,816.1000 = 816 (A)

Cu dao ny t gi li xoay chiu, v th ta chn cu dao c k

hiu Po - 3 do Lin X sn xut c cc thng s:

Um = 1200 (A)

Im = 1000 (A)

3.3.5. Chn cu ch bo v cho mch lc.

Cu ch c sau cu dao bo v cho mch lc.

Ta chn cu ch loi H do Lin Xo sn xut c cc thng s sau ;

42


43/91

Idc =KI 1,7.816

1,5=

i vi ng c mt chiu c phn ng c cp in bi b bin i

ta ly K = 1,7

l h s ph thuc vo iu kin khi ng ta ly = 1,5

Idc =1,7.816

1,5=925

Ta chn dy chy c I = 1500 (A) .

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H 3.6. S mch lc

44


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Chng IV

Phn tch tnh ton mch iu khin

4.1. Yu cu chung i vi mch iu khin- m bo pht xung vi cc yu cu m van nh:

bin , rng, sn xung ngn.

- m bo phm vi iu chnh gc iu khin minmax tng ng vi

in p ra ti khi thay i ca mch lc.

- m bo tnh i xng vi c knh iu khin, khng vt qu 1 0

30

in, tc l gc iu khin i vi mi van khng c lch qu gi trtrn,

tc ng ca mch iu khin nhanh.

- m bo mch hot ng n nh v tin cy khi li in xoay chiu

dao ng c v gi tr trn in p v tn s.

- m bo cch ly gia mch iu khin v mch lc.

- Khng gy nhiu i vi cc h thng iu khin in t khc

xuang quanh.

- C kh nng bo v qu p, qu dng, mt pha v bo hiu khi c

s c.

C hai loi iu khin c bn l h ng b v h khng ng b.

Trong bn n ny ta s dng h ng b v nguyn tc iu khin dc.

h ny gc iu khin m van lun c xc nh xut pht t mt thi

im c nh ca in p mch lc. V vy trong mch iu khin phi c mtkhu thc hin nhim v ny gi l khu ng b pha m bo mch iu

khin hot ng theo nhp ca in p lc. hot ng theo nguyn tc iu

khin dc th trong mch phi c mt khu to ra in p ta U T c dng c

nh (thng l xung rng ca) theo chu k do nhp ng b ca khu ng

pha. Khu so snh xc nh im cn bng ca hai in p ta UT v in p

iu khin Uk pht khu to ra xung m van hay gc iu khin thay i

45


46/91

do s thay i tr s ca Uk. Trn th l s di chuyn theo chiu dc

ca trc bin . Trong thc t ngi ta ch yu s dng nguyn tc ny.

4.2. Tnh chn bin p cho ngun iu khin.

4.2.1. Mt s c im ca bin p ngun iu khin.

Ngun iu khin thng c cng sut khng qu 1KVA, nn bin p

cp ngun cho n l loi cng sut nh. Do , vic tnh ton khc nhiu so

vi loi bin p lc th hin cc im sau: Mt trong nhng nhim v ch

yu ca thit k bin p lc l tnh ton dng kch thc mch t. Trong khi

bin p cng sut nh thng c sn cc loi li chun ho v c ch to

sn. St p trong bin p lc ch yu l do cun cm, cn bin p iu

khin l do in tr cun dy.

*Kt lun: Mc ch tnh ton bin p iu khin l da vo cng sut

ca bin p cn thit, ta chn trn cc cun th cp in p khi mang ti v

khng pht nng qu mc cho php.

4.2.2. Tnh ton bin p 3 pha.

Khi ngun iu khin l 3 pha ta c hai phng php thit k. Mt

l dng 3 bin p loi 1 pha u thnh 3 pha hoc dng 1 bin p 3 pha.

bin p 1 pha ta ch cn xc nh cc tham s ca mt bin p theo bng 16

(ti liu hng dn thit k mch in t cng sut) ca thy Phm Qu Hi.

cng sut ca bin p 1 pha = 1 phn 3 cng sut 3 pha, cn pha B, C tnh

ton tng t.

Bin p ngun iu khin pha A c hai cun dy th cp. Cun th cp

th th nht cho in p ng pha v cun th cp th hai cho cc mch

46


47/91

iu khin pha AUfa

Ufab

Ufac

A

B

C

D3 D5 D1

D2D4D0

C3

C4

C5

C6

7815

78915

C1

C2

+24V

+15V

15V

Ngun iu khinMBA ng pha

H4.0. Mch cp ngun iu khin

Khuch i thut ton v bin p xung.

+ Vi mch to bin p ng pha, U21 = 12 (V)

+ Vi mch to ngun n nh cho I21 = 100(mA)

Khuch i thut ton v bin p xung

U22 = 20 (V); I22 = 1(A)

Do cng sut ca my bin p ngun l

P21 = 12. 0,1 = 1,2 (W)

P21 = 20. 1 = 20 (W)

P2 = P21 + P22 = 1,2 + 20 = 21,2 (W)

Sba = Ssd. P2

Trong :

Ksd l h s s tra trong bng 1.1 [TL-HDTCS] ca thy Phm

Quc Hi theo mch chnh lu Ksd = 1,05.

Sba = 1,05. 21,2 = 22,26 (VA)

P2 - cng sut tc dng ca cun th cp

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48/91

Sba - cng sut ton phn ca my bin p ngun xc nh c li thp

da vo cng sut MBA c.

2ba

th csmax th cs

S .10

S .S 2,22.f.B .S.K .K=

T kinh nghim thc t, tit din li thp Sth c tnh theo cng thc

sau:

thS 1,2 P 1,2 21,2= = = 5,5 cm2.

Cc tham s trong biu thc tra trong bng 15[TL-HD TCS].

Kth = 0,9, S = 2,4 (A/mm2

); Bmax = 1,3.Kcs = 0,25.

2

cs

22,26.105,5.S

2,22.50.1,3.2,4.0,9.0,25=

cs

28,56S

5,5= = 5,19 cm4

Tra bng cng thc my bin p cng sut nh, chn li st c tit dinS = 7,28cm2, c ghp bng l thp loi II 20 x 40 c b dy 0,35mm c cc

thng s sau:

Hnh 4.1: Kch thc li thp cun lc mt chiu

Tnh s vng cc cun dy s cp v th cp l:

- S vng dy s cp l:

48


49/91

4

max th

E.10W

4,44.f.B .S=

Trong :

E= U0. (1 - 0,01. U0%).

U0 l st p trn cun s cp U% = 13 (V)

E0 = 220 (1 - 0,01. 13) =191,4 (V)

S vng dy cun th cp l:

41

21max th

E .10W

4,44.f.B .S=

Trong :

E1 = 12 (1 - 0,01. 18) = 9,84 (V)

4

21

9,84.10W

4,44.50.1,3.5,5= = 62 (vng).

S vng dy cun th cp l

4

222

max thE .10W 4,44.f.B .S=

Trong :

E1 = 20 (1 - 0,01. 18) = 16,4 (V)

4

22

16,4.10W

4,44.50.1,3.5,5= = 103 (vng).

Dng in chy trong cun s cp l

11

P 21,2I

U 220= = = 0,096 (A)

Ta ly mt dng in l

J = 2,5 (A/mm2).

Nn tit din ca cun s cp l

11 I 0,096S J 2,5

= = = 0,038 (mm2)

49


50/91

Do ta chn dy c ng knh l d1 = 0,3mm

Tng t S21 = 0,04 mm2. Chn d21 = 0,23 mm

S22 = 0,4 mm2. Chn d22 = 0,7 mm

Ta chn D0 D5 loi 1N5420 c I = 2A, U = 70 (V)

Chn C1, C2 l t lc ngun IC n p loi 2200F - 35 (V)

Chn t C3, C4 l t lc sau n p loi 470(F)-25(V)

Chn t C5, C6 l t lc hi nhiu cao tn chn 10(F).

4.3. Khu ng pha

a)

psin10

b)

50


51/91

H4.2. Khu ng pha: a. S nguyn l; b. Dng in p u ra

Khu ng pha gm c khuch i thut ton OA1, cc in tr R1, R2,

VR1 v bin p ng pha. mch lm vic ch chnh lu, trn th

ng in p. U0 phi giao nhau vi ng in p ng pha i qua im 0,

v th.

00 pU 2U .sin10= = 1,41.12. sin100 = 2,93 (V)

Vi in p U0 l gi tr hiu dng in p ng pha in p cp cho

VR1 l 12(V) ta chn bin tr VR1 = 10 (k), lc dng i qua bin tr l:

12I10.000

= = 1,2 (mA)

in tr m ta chnh nh trn VR1 l:

0c

U 2,93R

I 1,2= = = 2,44 (k).

Vy ta chn R1 = 10 (k); R2 = 15 (k).

4.4. Khu to in p ta.

a)

51


52/91

b.

H4.3. Khu to in p t: a. S nguyn l; b. Dng in p

Khi in p u ra ca OA1 tc l im B m th in p ra ca OA2

ti im C t C7 R3 D8 OA1 mt OA2. it n p ta chn

loi c ngng l UDZ = 9V. Thi gian ca t l t1 tng ng vi 100. nh vy

trong khong thi gian t1 in p trn t phi tng t 0 n 9V.

Khi in p u ra l dng th u ra ca OA2 l m, do t C7

phng t C7 OA2 mt ngun R4 VR2 t C7. Dng in qua

t khi ny ngc chiu vi khi np, in p trn t C7 gim dn.

c 4

1U U iR .dt

C=

V R4 = const nn

c d

4 2

1 UU U . .t

C R VR

=

+(4-1)

52


53/91

Trong : U = 9V - l in p nh rng ca.

Sau khong thi gian t = t2 - t1 tng ng vi 900 - 100 = 800 th in p

trn t s gim ti 0 ta c.

( )0

30

1 80t . 4,44.10 s

50 360= =

Chn t C7 c in dung C7 = 220 nF = 220.10-9(F)

Thay cc gi tr vo cng thc (4-1) ta c;

39

4 2

1 120 9 .4,44.10

R VR220.10

= =

+

( )3

4 2 9

12.4,44.10R VR 2609

220.10

+ = =

Chn R4 = 20 (k). Chit p VR2 chn 15 (k)

Chnh nh chn VR2 = 10 (k).

Trong qu trnh np in p trn t C7 c tnh theo cng thc sau:

c c c1 1U i dt .i .tC C

= = (4-2)

Trng :

bhc 3 4

3 4 2

U 0,7 Ui iR iR

R R VR

= =

+

Nh phn tch trn th in p trn t C7 s tng t 0 n 9V

trong thi gian t1. Nh vy thi gian np l.0

31 0

1 10t t . 0,55.10

50 360= = = (s)

Vi khuch i thut ton c ngun l 12V c Ubh= 10V, thay vo cng

thc (4-2) ta c.

bh

c 3 4 2

U 0,71 UU .t

C R R VR

=

+

53


54/91

Khi t = t1 th Uc = 9V

Uc =3

9

3

1 10 0,7 12,.0,55.10

R 26909220.10

R3 = 2319 (). Chn R3 = 2,4 (k)

4.4. Khu so snh to xung.

UT

R5

+OA3

R6

Uk

D9Uss

UUT

UkUTU

t

t

a)

b)

0

Uk

H4.3. Khu so snh to xung a. S nguyn l; b.Dng in p

Phn t chnh l OA3 c cp ngun +12V loi LM 324 c cc thng

s k thut sau:

Ung = 3V 22V

Nhit lm vic: t = - 550 1250C

Cng sut tiu th: P = 650 (mW) = 0,65 (W)

54


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Tng tr u vo: Zv = 70 ()

Tng tr u ra: Zr = 1 (M) = 1000.000 ()

H s khuch i: K0 = 5.104

Dng in u ra: Iv = 25 (mA).

Tc tng p:du

dt= 0,5 V/s

Nhim v so snh 2 in p ta v in p iu khin kt qu u ra

l cc xung ch nht sau khi qua it D9 loi b phn m, ta c dng

in p nh trn s . in tr R5 v R6 c nhim v cch ly gia Uk v UTng thi hn ch dng khng gy qu ti cho tng trc. Ta chn R5 = R6 =

15 k.

4.5. Khu to xung chm.

55

UxsD10

OA4

R7

C8

+

R8R9

t

Uxs

a)

b)

E


56/91

H4.5. Khu to xung chm

a. S mch in nguyn l; b. Dng in p

y l mch dao ng a hi to xung chm. Chu k ca in p tnh

theo cng thc.

97 8

8

2RT 2.R .C .ln 1

R

= +

(4-3)

gim chnh lch in p tnh gia 2 ca vo ca OA4 ta chn R9

< R8, chn R9 = 8 (k), R8 = 24 (k).

Tn s ca mch dao ng ta chn f = 1,5 KHz, chn t C 8 l loi tgm c k hiu 2A2245J. C:

C8 = 22.104pF = 220.10-9 (F)

Thay cc gi tr vo cng thc (4-3) ta c:

973

1 2.82.R .220.10 ln 1

241,5.10

= +

R7 = 2966 () chn R5 = 3,3 (k).

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4.6. Khi khuch i xung chm

H4.6. S nguyn l khi khuch i xung chm

Theo nh loi thiristor chn chng III l loi T-800 c in p

iu khin l: Uk = 7(V); Ik = 0,4 (A) khu dao ng a hi to xung chm

c f = 1,5KHz. Vy tham s dng in cun s cp l.

U1 = U2. K = 7.3 = 21 (V)

I1 = ( )kI 0,4

0,13 AK 3

= =

Phn t chnh ca khu khuch i xung l cc transitor Q42 v Q41

c ghp darlington. it D43 v in tr R41 nhm tiu tan nng lng

tch lu ca cun dy trong giai oan kho ca cc bng bn n. Ngun cng

sut phi c tr s ln hn U1 b s st p trn in tr. V vy ta chn

Ec=24V. cn c vo tr s ca Ecs v I1 ta chn bng Q42 loi 2N3705 c

Icmax=0,5(A), Ucc = 30 (V), min = 50.

in tr R41 thng chn t kh nng dn dng ln nht cho php ca Q42.

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cs41

cmax

E 24R 48

I 0,5> = = ().

Vy chn R41 = 60 ().

Bng Q41 chn loi BC107 c U = 45; Icmax = 0,1 (A); min = 110.

Theo (ti liu HDTKMDTCS), ta tnh R43 t iu kin m bo ho

tt cho Q41 v Q42 sau.

1 2 cs43

1

. .E 50.110.24R 846153

S.I 1,2.0,13

< = = ().

Trong :S- H s d tr m bo cc bng m bo ho. Mt khc R43 phi

tho mn iu kin l hn ch dng khng gy qu ti cho cng AND. T

ta chn R43 = 15 (k). Cng AND loi CMO4084.

Theo (ti liu- HDTKMDTCS) in tr tch xung R44 phi tho mn

f D

44 43T X

2.U .sin U

R .RU

< (4-4)

Trong : Uf gi tr hiu dng ca in p ng pha Uf = 10V.

min gc iu khin nh nht min = 300

UD st p trn i t D45 = 0,7 (V)

UTX - in p u ra CMOS c ngun cp l 12V nn in p

u ra mc cao 11(V).

Thay cc gi tr vo cng thc (4-4) ta c:

0

44

2.10.sin30 0,7R .15

11

= 87 (k)

Vy chn R44 = 5,6 (k).

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4.7. Bin p xung

Bin p xung l loi bin p c bit, trong in p t ln pha s

cp c dng xung ch nht m khng phi in p hnh sin, iu ny dn n

ch lm vic v phng php tnh rt khc so vi bin p thng thng.

Tn s in p lm vic ca bin p xung c th vi trm Hz n hng trm

KHz. Nhim v ca bin p xung l cch ly mch lc v mch iu khin,

phi hp tr khng tng khuch i xung v cc iu khin van lc.

Bin p xung lm vic tn s cao nn li dn t trng l li ferit. Do

tn tht trong bin p tng mch theo tn s nn cng t cm cng phi

gim ng k so vi tn s 50Hz.

B

H

0,6

0,2

0 80 160 240

H4.6. c tnh t ho li ferit

Do dng in qua cc cun dy BHX khng lin tc m c dng xung

nn tr s hiu dng ca dng in nh. V vy, tit din dy qun BAX

thng khng tnh t tr s dng in nh thng thng m hay c chn t

iu kin m bo bn c hc vi ng knh dy qun trong khong

(0,20,4)mm.

Do ch lm vic ca bin p xung l t ho 1 phn nn theo th ta

chn.

B = 0,2 T v H = 30 (mA).

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Do xung ca b dao ng a hi to ra l i xng nn thi gian c

xung bng thi gian ngh (tx = tn).

Tn s xung chm l 1,5 KHz tng ng chu k 1 xung l.

( )3XC 3XC

1 1T 0,66.10 s

f 1,5.10= = =

Thi gian c xung l:

Tx = 0,5. TXC = 0,5. 0,66. 10-3 = 0,03.10-3 (s)

Th tch ca li t l:

ba 2 2 x xK .U .I .t . UV B. H =

I2 l dng in cun th cp do tn = tx nn I2 = 0,5.Ik = 0,5. 0,4 =

Ux l st p xung cho php thng ly bng (0,2 0,2)

y ta ly Ux = 0,15 (V)

T s bin p: Kba = 3

in p th cp U2 = Uk= 7 (V).

Thay cc gi tr va tnh vo cng thc (4-3) ta c:

33.7.0,5.0,4.0,33.10 .0,5V

0,2.30

= =

Ta chn li ferit 3019

ng knh ngoi: 30 (mm)

ng knh trong: 19 (mm)

V = 6,19 (cm3)

F = 1,28 (cm3)

Vy s vng dy cun s cp:

31 2

1 4ba

U .t 9.0,33.10

B.S 0,2.1,38.10

= =

= 107 (vng)

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Vy s vng dy cun th cp:

12

107

K 3

= = = 36 (vng)

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f

Hnh 4.9. th thi gian ca mch to xung

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* Nguyn l lm vic ca mch to xung.

Bin p ng pha l bin p th cp c im gia ni mt v th 2

u cun th cp s cho ra 2 in p ngc pha nhau 1800. Hai in p ny

sau khi qua D6 v D7 c nn na chu k nn in p ti im A chnh l

in p ng pha UA(Uf) c dng nh (hnh 4.2). in p ng pha ny i

qua R2 c a vo u o ca OA1, in p ny c so snh vi U0 c

a vo u khng o ca OA1. V b bin i ch lm vic ch chnh

lu nn gc iu khin = 0 n 900 tnh t thi im chuyn mch t nhin.

m rng phm vi iu chnh gi tr U0 ta ly nh 00 fU 2.U .sin10= (100

tng ng vi khong thi gian ca t np to sn rng ca i ln mch

to in p ta).

in p u ra ca OA1 tc l in p ti im B c dng nh (hnh

4.3) l dy xung hnh ch nht, ng vi khong thi gian in p ti im

B m th in p ny thng qua i t D8 v R3 c a ti u o ca OA2

lm cho in p u ra OA2 ti im C7 dng t C2 c np t im C

t C7 R3 D8 im B OA1 mt OA2 im C. Thi gian

np ca t C7 ph thuc vo in dung ca C7 v tr s R3. Thng thng

mch c thit k vi iu kin R3 0; Uk


65/91

ng c tn s t 6 n 10KHz. Sau khi i qua D ct b hn m ta c

in p ti im E c dng nh (hnh 4.5). Xung chm v in p ca khu

to xung ti im D c chia ra lm 2 knh v c a ti cc cgn AND.

kt hp vi mch tch xung D45, R44 kt qu l in p ti cc baz ca

transtor Q41 v Q42 c dng nh (hnh 4.9). Cc in p ny c khuch i

bng cc cp transitor u kiu darlington sau c truyn qua bin p

xung a n cc iu khin thyristor.

4.8. Tng hp h thng truyn ng in

4.8.1. ng c in mt chiu KTL

Khi dng in kch t ng c khng i, hoc khi ng c c kch

thch bng nam chm vnh cu th t thng kch t l hng s.

Phng trnh ng hc ca ng c c vit dng ton t laplace l:

K = const = CU

U(P) = R . I(P). (1 + T) + CU. (P)

Vi ---

LT R= l hng s thi gian ca mch phn ng.

R, L l in tr v in cm ca mch phn ng.

Phng trnh chuyn ng ca h thng

M(P) = MC(P) = J. P. (P)

Trong :

J l m men qun tnh ca cc phn t chuyn ng quy v trc ng

c. T cc cng thc trn ta xy dng c s cu trc m t ng c in

mt chiu.

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Hnh 4.10: S cu trc m t ng c in mt chiu

Thng s ca ng c

Um = 600V R = 0,4Im = 1000A IKT = 25,5/6,6A

Pm = 600 kW UKT = 262/68VNm = 1300 v/p J = 15,5 Kg/m2

S i cc Zp= 2Trong L l in cm ca mch phn ng c tnh theo cng thc

nm- L

-m p m

UL K .

I .Z .n= (1)

Nu mch phn ng c cun b th h s KL = 1,4 1,9

Nu mch phn ng khng c cun b th h s KL = 5,5 5,7.ng c s dng c cun b l ly R2 = 1,9 thay cc gi tr va tnh

vo cng ghc (1) ta c.

-

600L 1,9.

1000.2.1300= = 0,00044 H

( )---

L 0,00044T 0,0011 s

R 0,4= = =

H s i cc tnh theo cng thc.

um um - um um -

m m

U I .R U I .RCu K .9,55

n

= = =

600 1000.0,4Cu .9,55 1,47

1300

= = V/rad/s

T cc thng s va tnh c ta c cu trc ca ng c nh sau:

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Hnh 4.11: Quan h gia Uk v gc 4.8.2. B iu chnh c k

Hm truyn t ca b chnh lu thyristor tng qut l

BBBB

BB

KU

1 P.T=

+

Trong KBB l h s khuch i ca b bin i, TBB l hng s thi

gian ca b bin i c tnh theo cng thc.

BB

1T

2mf=

Vi m = 6 l s xung p u ra trong mt chu k in p li.

F = 50 l tn s ca in p li.

V th

BB

1T 0,00178

2.6.50= =

tm KBB ta phi lp bng tm quan h gi Uk , , E.

in p chnh lu cu ba pha l:

d

600E 2,34. .cos

3=

Nh tnh mch to xung th nh rng ca l 9V, thi gian np

tng ng 100, thi gian phng tng ng 800.

67

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Hnh 4.11: Quan h gia Uk v gc Quan h gia Uk v gc

Cn c vo tam gic ABC ta c quan h

rm k k0

rm rm

U U U1

U U90

= =

k

rm

U90 1

U

=

Ta c bng

Uk 0 1 2 3 4 5 6 7 8 9

0 90 81,1 72,2 63,2 54,3 45,4 36,5 2,76 18,7 9,8

E(V) 0 125,

4

247,

8

365,

5

473 569,

2

651,

6

718,

3

767,

8

798,7

Cn c vo quan h gia Uk , , E ta dng c th th hin quan

h gia Uk v Ed nh sau:

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k

Hnh 4.12: Quan h gia Uk v Ed

V ng c c Um = 600V ta d d phng in p li dao ng st

p, v st p trn cc van khi dn dng, st p trn cun khng nn ng vi

Um ta chn Ed = 652 V.

Khi Ed = 652 V th Uk= 6 v = 36,50.

Khi h s ca BB l. BB652

K 108,76

= =

Vy BBBBBB

K 108,7U 1 P.T 1 0,0017.P= =+ +

4.8.3. Khu phn hi dng in

Phn hi dng in l khu qun tnh c hm truyn.

( )( )i i

i i

U P KW P

I 1 T P= =

+

Trong : I: dng in i vo phn ng ng c

Ti: hng s thi gian ca khu truyn ph thuc vo t lc C, in

tr R, h s in cm v ni tr ca bin dng ta ly Ti = 0,005s.

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Hnh 4.13: Khu phn hi dng in

Ki: h s truyn ca khu phn hi dng ini

i

UK

I=

Ta ly in p phn hi Ui = 10V

Khi ng c chy ch nh mc

C I = 1000A

V th i10

K 0,011000

= =

T ta c:

( ) iii

K 0,01W P

1 T P 1 0,005P= =

+ +

4.8.4. Khu phn hi tc .k

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Hnh 4.14: Hm truyn ca khu phn hi tc c dng.

( )( )

( )ft f

U P KW P

P 1 T P

= =

+

Trong : Tf l hng s thi gian ca b lc ly Tf = 0,005s.

UK

W

= : khi ng c chy vi tc nh mc th ta c U= Umax= 1(V)

mm

n 1300136rad/s

9,55 9,55 = = =

U 10K 0,096v.S/radW 136

= = =

Vy ta c: ( )ftf

K 0,096W P

1 T P 1 0,005P

= =+ +

5.8.5. Tng hp mch vng dng in.

Hnh 4.15: Cu trc s mch vng dng in

Ri l b iu chnh dng in

TBB, Ti, T l hng s thi gian ca BB, trong trng hp h thng

truyn ng in c hng s thi gian c hc rt ln, hn thi gian in t

ca mch phn ng th ta c th coi s ca ng c khng nh hng n

qu trnh iu chnh ca mch vng dng in (coi E = 0).

Hm truyn t ca mch vng dng in.

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( )( ) ( ) ( )

iBB

-0

i BB -

KK .

RS P

1+P.T 1 P.T 1 PT=

+ +

V Ti v TBB l nh ta dt Ts = Ti + TBB lc ny biu thc S0(P) c thvit li nh sau.

( )( ) ( )

iBB

-0

s -

KK .

RS P

1+P.T 1 PT=

+

p dng tiu chun ti u m dun ta tm c hm truyn ca b iu

chnh dng in c dng khu PI.

( ) -iBB i

s-

1 P.TR P

K .K.a.T .P

R

+=

Ta ly hng s a = 0.

k

Hnh 4.16: S nguyn l b iu chnh Ri

BB i1 s-

K .KR C .2TR=

Trong Is = Ti + TBB = 0,005 + 0,0017 = 0,0067

KBB = 108,7; Ki = 0,01; R = 0,4.

Ta chn C = 5,7F, t ta c.

BB i1 s

6-

K .K 108,7.0,01.2.0,0067R .2T 7748

C.R 4,7.10 .0,4= = = ()

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Ta chn R1 = 15 k

Tnh R3: V

u3 - 3

TR .C T R C= =

3 6

0,0011R 2,34

4,7.10= =

Chn R3 = 1 k

Hm truyn ca mch vng dng in l.

( ) uiBB

s-

1 PT 1 0,0011PR P K 108,7.0,1.2.0,0067.P.a.T .P

0,4R

+ += =

( )i27,5 0,03P

R PP

+=

108,727,5+0,03P

4.8.6. Tng hp mch vng tc .

mc trn ta tng hp c mch vng dng in trong phn ny ta

s s dng biu thc kt qu trong b qua nh hng ca sc in ng

ca ng c.

( )

( ) ( )3 2i i i s ss s s

I P 1 1 1 1. .

U P R K 1 2T P 1 T P1 2T .P 2T .P 2T P= =

+ ++ + +

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V 1 + Ts. P 1 v th n gin ta c th dng biu thc gn ng

tnh ton hm truyn t ca mch vng dng in.

( )

( )di i s

U P 1 1

.U P K 1 2.T .P= +

S khi cu trc ca h iu chnh tc .

P

Hnh 4.17: S khi ca h iu chnh tc

Trong Sw l sensor tc c hm truyn l khu qun tnh vi h s

truyn l Kw v hng s thi gian (lc) Tfw.Thng Tfw c gi tr nh khi ta t.

2T's = 2. Ts + Tfw i tng iu chnh cc hm truyn.

( )-

0i u c s

R .K 1S .

K .C .T P 2T .P 1=

+

Theo tiu chun ti u mdun c th xc nh c hm truyn ca b

iu chnh tc l b t l.

( ) i u c P'- s 2

K C T 1R P . K

R .K 2.T .a = =

Trong thng ly a2 = 2.

Hng s thi gian c hc ng c.

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( )uc 2 2R .J 0,4.15,5

T 2,8 sCu 1,47

= = =

Trong mc 4.4.1 ta tnh c.

Ki = 0,01, Kw = 0,096

2T's = 2Ts + Tfw = 2. 0,067 + 0,005 = 0,0184 (s)

Thay cc gi tr vo cng thc trn ta c.

( )0,01.1,47.2,8 1

R P .0,4.0,096 2.0,0184

= = 29.

y l b iu chnh t l Kp = 29 v th.

6p

5

RK 29

R= = ta chn R5 = R7 = 10 k.

T ta c R6 = KPR5 38,3. 10 =383 k

Hnh 4.18: Cu trc b iu chnh

Cu

JP1

Cu

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Hnh 4.19: S cu trc ca h thng

4.9. o chiu quay ca ng c.

4.9.1. o chiu t thng

Chiu t lc tc dng vo dng in c xc nh theo quy tc bn tay

tri. Khi o chiu t thng hay o chiu dng in th lc t c chiu ngc

li. Vy mun o chiu quay ca ng c in mt chiu c th thc hin

mt trong hai cch sau.

Hoc o chiu t thng (qua o chiu dng in kch t).

Hoc o chiu dng in phn ng.Trong dy truyn cn ng c s dng h truyn ng vi cng sut

ln (v cng sut kch t ch khong 18% cng sut truyn ng) v hu nh

khng o chiu nn ta s dng phng php o chiu bng t thng.

Hnh 4.20. S o chiu ng c bng kch t

Phng php o chiu t thng thc hin nh nhng v mch t thng

c cng sut nh hn mch phn ng. Tuy vy cun kch t c s vng dy

ln, h s t cm ln do thi gian o chiu tng ln (thi gian qu ln)

nn phng php o chiu t thng tc tng qa, khng tt.

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4.9.2. S mch kch t

Hnh 4.21. S mch kch t

Hot ng ca s : Hai mch chnh lu hot ng ring bit. Mch

ny hot ng (c pht xung iu khin) th mch li hon ton ngh (b

ngt xung iu khin). V vy loi tr c hin tng dng qun v khng

cn cun khng cn bng. Song qu trnh o chiu cn c thi gian cht (nh

nht l vi ms) cho van ca mch phi ngng hot ng kp phc hi tnh

cht kho ri mi bt u pht xung cho mch kia hot ng. V vy cn c

mt khi logic iu khin o chiu tin cy.

4.9.3. Qu trnh o chiu.

Khi ta nhn nt dng ct ngun ng c ngng hot ng. Nhng theo

qun tnh ng c cn chy mt thi gian sau mi ngng hn. V vy n sinh

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ra mt nng lng trong mch phn ng. Mun lm mt nng lng ta

phi mc mt in tr R trit tiu nng lng pht sinh .

Khi bt u ng in cho ng c, tc ng c bng khng nn

dng in ng c Im=m

-

U

R= (10 - 20) Im rt ln to ra m men ngn mch.

mnm m m nm

-

RM M K . K I

R= = =

Cng rt ln v c th gy ra hu qu xu. M men m my qu ln s

to ra cc xung lc ng lm h truyn ng b git, khng tt v mt c hc,hi my, c th gy ra nguy him gy trc, v bnh rng.

Mmm = Mnm = K. dm. Imm.

Vy, m bo an ton cho ng c v cc c cu truyn ng cng

nh trnh nh hng xu ti li in, phi hn ch dng in khi m my,

khng cho vt mc gi tr Imm =(1,5 - 2,5)Im.

Ngha l phi thm in tr ph vo mch phn ng Rp sao cho.

( )m mmm m- p -

U UI 1,5 2,5 I

R R R = =

+

P ln th chn Imm nh

Trong qu trnh m my, tc ng c tng, sc in ng ca

ng c E = K.

cng tng dn v dng in ng c b gim.

- P

U EI

R R

=

+

Do m men ng c gim.

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= +

a. th c tnh c lc m my b. S m my ng c in mt chiu KTLHnh 4.22

Do m men ng c ny cng gim. ng c m my theo mi ten

trn ng c tnh c 1 (Hnh 4.22).

Nu c gi nguyn Rp trong mch phn ng th khi ng c gim t

Mmm xung bng MC, ng c s quay n nh vi tc thp B.

Do vy, khi m mn gim i 1 mc no th phi ct in tr Rp

trong mch phn ng nh ng tip im K ng c tr v lm vic (hy

tip tc m my) trn c tnh t nhin ti im E. Lc ny m men ng c

ME li ln hn nhiu m men ti MC nn ng c tip tc tng tc nhanh ti

im A th M = Mc v ng c s chy n nh vi tc A .

Khi ng tip im K ct in tr ph Rp ra khi mch rto th ngay

lp tc ng c chuyn t im lm vic D trn c tnh c nhn to 1 sang

lm vic trn c tnh c t nhin 2, do qun tnh c khi chuyn c tnh, tc

ng c khng kp thay i trong mt khong thi gian qu ngn nn on

chuyn i DE l nm ngang.

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Chng 5: M phng h thng bng simulink

5.1. Gii thiu phn mm simulink

Matlab l chng trnh phn mm tr gip cho vic tnh ton v hin

th. N c th chy trn hu ht cc loai my tnh v c iu khin bi s

lng lon cc lnh, cc tp lnh. Cc lnh ny ngy cng c m rng nh

cc phn Toolbox (th vin tr gip) khc nhau hay thng qua cc hm ng

dng c to lp bi ngi s dng.

Simulink l mt Toolbox h tr c lc cho vic m hnh ho, m

phng v phn tch mt h thng ng. Simulink cho php m t h thng

tuyn tnh, h phi tuyn, cc m hnh trong thi gian lin tc, gin on hay

mt h kt hp c lin tc v gin on. H thng cng c th c nhiu tc

khc nhau c ngha l cc phn khc nhau ly mu v cp nht s liu tc

khc nhau.

m hnh ho Simulink s dng cc i tng ho gi l Graphic

Programming Unit - GPU. N c xy dng trn c s cc ngn ng lp

trnh OOP, to iu kin ht sc thun li cho vic thay i gi tr cc thuc

tnh trong nhng khi thnh phn. Loi hnh lp trnh ny c xu hng c

s dng nhiu trong k thut bi u im ln nht can l tnh trc quan, d

vit v hnh dung i vi ngi lp trnh khng chuyn nghip x nh nhng

ngi mun b nhiu thi gian cho vic nghin cu thm mt ngn ng lp

tnh mi.

Th vin ca Simulink cng bao gm tan b th vic cc khi nh

khi nhn tn hiu, cc ngun tn hiu, cc phn t tuyn tnh v phi tuyn,

cc u ni chun. iu c bit chng trnh ny l ngi s dng c th

to ra cc khi th vin ring ca mnh. Cc m hnh bi ton trong Simulink

c xy dng c th bc hay cn gi l xy dng theo m hnh phn cp,

iu cho php ngi s dng c th xy dng m hnh theo hng POP-

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DOWN hoc POP-UP. Ngi dng va c th quan st h thng mc tng

quan, va c th t c mc c th bng cch nhy kp vo tng khi xc

nh xem chi tit mo hnh ca tng khi. Vi cch xy dng kiu ny,

ngi dng c th hiu c su sc t chc ca mt m hnh v nhng tc

ng qua li ca cc phn trong m hnh nh th no.

Sau khi xy dng m hnh h thng, ngi s dng c th m phng n

trong Simulink bng cch nhp lnh trong ca s ca Matlab hoc s dngc

c menu c sn. Vic s dng cc menu c bit thch hip cho cc cng vic

c s tc ng qua la ln nhau, cn s dng dng lnhthwng hay c s

dng khi chy mt lot cc m hnh. Cc b Scope v khi hin th khc chophp nhiu s dng c th theo di kt qu trong khi ang chy m phng.

Hn na, ngi s dng c th thay i thng s h thng mt cch trc tip

v nhn bit c cc nh hng n m hnh. Kt qu m phng c th t

vo Matlab x l a ra my in hay hin th.

Cn nhn mnh iu quan trng nht trong vic m phng mt h

thng hay mt qu trnh l vic xy dng c m hnh h thng qu trnh . s dng tt chng trnh ny, ta cn nm chc cc kin thc c bn v

iu khin t ng, xy dng m hnh ton hc theo quan im l thuyt iu

khin t ng v t xy dng m hnh bi ton cn gii quyt.

5.2. M phng h thng bng simulink.

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th hnh 5.2 th hin ba c tnh ca ng c.

Hnh 5.2a th hin c tnh ca in p.

Hnh 2.5b th hin c tnh ca dng in.

Hnh 5.2.c th hin c tnh ca tc khi khi ng th ng c mang

ti ngay t u, sau 1s qu trnh khi ng kt thc v ng c lm vic

ch xc lp.

Khi khi ng ban u c mang ti.

in p tng dn xp x in p nh mc sau 1,5S.

Dng in m my tng ln vt dng nh mc nhng do tc ngca mch phn hi dng in nn n gim dn v t ti gi tr nh mc sau

1,55.

Tc tng ln vt gi tr nh mc sau 3,5s ta ng ti cho h thng,

ng c lm vic c ti.

in p n nh v t gi tr nh mc 600V.

Dng in n nh v t gi tr nh mc 1000A.

Tc n nh 1300V/ph

5.3. ngha ca qu trnh m phng.

L cng c quan trng cho php kho st cc i tng, h thng hay

qu trnh k thut vt l, m khng nht thit phi c i tng thc. c

trang b mt cng c m phng mch v c hiu bit v cc phng php m

hnh ho, ngi k s s c kh nng rt ngn thi gian v gim chi ph

nghin cu pht trin sn phm mt cch ng k, iu ny c bit c

ngha khi sn phm l cc h thng thit b k thut phc tp vi gi tr kinh

t ln.

5.4. M t qu trnh m hnh, m phng.

y ta s dng m hnh, m phng b bin i thc t tnh ti cc

yu t phi tuyn.

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+ Ta dng ng c in mt chiu kch t c lp c cc thng s sau:

Um = 600V n =1300 V/ph

Im = 1000A J = 15,5 kg/m2.

Pm = 600 kW R = 0,4

L = 0,00044 (H) T = 0,0011 (S)

B bin i dng s cu 3FA c iu khin (Thyristor converter).

c nhn tn hiu iu khin t b pht xung (Synchronized).

B pht xung iu khin c cp ngun t (Synchronization

Voltages).+ Hm truyn ca b iu chnh dng (PI Curent Regulator) u vo

nhn tn hiu t khu phn hi tc u ra gi ti khi iu khin lpha-

deg iu khin gc m.

+ Hm truyn ca b iu chnh tc (Speed Controller) u vo

nhn tn hiu t my pht tc v khu to gia tc.

u ra gi tn hiu ti u vo ca PI.

+ Khu Scope nhn cc dng tn hiu: in p, dng in, tc .

* Nguyn l hot ng ca m hnh, m phng.

Khi ng c lm vic tc ca ng c c n nh nh b phn hi

tc v b phn hi dng in tc thc t ca ng c c so snh vi

tc t do vy trn Scope ta nhn thy tc ca ng c bm st ng

t ().

Dng in thc t ca ng c s thay i khi ng c mang y ti v

khng ti c hin th trn Scope v dng in thc t ta thy bm theo

ng theo dng in t.

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Kt lun

Qua thi gian thc hin n tt nghip, vi ti "Thit k b iu

khin ca ng c in mt chiu kch t c lp" gip em hiu nhng

vn l thuyt v thc t lin quan n ti nhm cng c thm nhng

kin thc hc trng.

c s hng dn nhit tnh ca thy gio Trn Trng Minh v cc

thy c gio trong b mn T ng ho v cc bn cng vi s n lc ca bn

thn, em hon thnh ti thit k tt nghip ny.

ti thit k tt nghip cp n c bn ca phng php iu

khin ng c in mt chiu kch t c lp. ti tt nghip cng cp

n phng php xy dng m hnh quan st in p, dng in, tc thng

qua vic m phng bng chng trnh MATLAB SIMULINK trn my tnh.

Cc kt qu m hnh ho trn my tnh cho thy c s khng

nh h thng truyn ng in xut trong n hon ton c th pdng c trong thc t.

Do thi gian c hn nn n c nhiu thiu st, mong thy c v cc

bn gp cho em.

Em xin chn thnh cm n thy c gio v cc bn!

Hi Phng, ngy 15 thng 11 nm 2004

Sinh vin

Phm Hu Qun

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Ti liu tham kho

1.Nguyn Mnh Tin, V Quang Hi

Trang B in - in t my gia cng kim loi

2. Phm Quc Hi

Hng dn thit k mch in t cng sut

3. Bi Quc Khnh, Nguyn Vn Lin, Nguyn Th Hin

Truyn ng in - NXB KHKT - 2001

4. Bi Quc Khnh, Phm Quc Hi, Nguyn Vn Lin, Dng Vn Nghi.

iu chnh t ng truyn ng in - NXB KHKT - 1999.

5.Nguyn Bnh

in t cng sut - NXB KHKT - 2000.

6. Dng Minh Tr

S tay linh kin - NXB KHKT - 1998.7. Phm Cng Ng

L thuyt iu khin t ng - NXB KHKT - 1998.

8. Xun Th

K thut in t- NXB Gio dc - 1997.

9. V Quang Hi

Trang b in - in t cng nghip - NXB Gio dc - 2003.

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Mc lc

Trang

2.1.2.1. Phng trnh c tnh c..........................................1

2.1.2.2. Xt nh hng ca cc tham s n c tnh c . .. .42.2. iu chnh tc ng c in mt chiu..........7

2.2.1. Nguyn l iu chnh in p phn ng..................72.2.2. Nguyn l iu chnh t thng ng c.................11

2.3. H thng bin i - ng c (BB - ) .............122.3.1. H thng truyn ng my pht ng c mtchiu (F ) ...................................................................13

2.3.1.1. Cu trc h F v cc c tnh c bn................132.3.1.2. Cc ch lm vic ca h F D...........................15

2.3.2. H thng chnh lu ng c mt chiu ..............182.3.2.1. Chnh lu bn dn lm vic vi ng c in. .......182.3.2.2. c tnh ca h truyn ng chnh lu Thiristo ng c mt chiu...............................................................222.3.2.3. Nhn xt chung......................................................24

CHng III: Phn tch tnh ton mch lc..................................28

3.1. Mc ch yu cu ..........................................283.2. La chn ca b chnh lu...............................28

3.2.1. Chnh lu cu mt pha i xng c iu khin.....28

3.2.2. Chnh lu hnh tia 3 pha........................................303.2.3. Chnh lu 3 pha s cu.....................................313.2.4. Kt lun................................................................35

3.3 Tnh ton mch ng lc..................................353.3.1. Tnh chn van cho mch ng lc.........................363.3.2. Tnh ton thit k cun khng bo v hn ch tc tng dng di/dt..........................................................383.3.3. Tnh mch bo v qu p RC mc song song vi van.................................................................................40

3.3.4. Chn cu dao ng ct cho mch lc ..................423.3.5. Chn cu ch bo v cho mch lc........................42H 3.6. S mch lc.............................................................44

Chng IV.............................................................................45

Phn tch tnh ton mch iu khin........................................45

4.1. Yu cu chung i vi mch iu khin ..........454.2. Tnh chn bin p cho ngun iu khin..........46

4.2.1. Mt s c im ca bin p ngun iu khin....464.2.2. Tnh ton bin p 3 pha........................................46

4.3. Khu ng pha...............................................50

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4.4. Khu to in p ta......................................514.4. Khu so snh to xung...................................544.5. Khu to xung chm.......................................554.6. Khi khuch i xung chm............................57

4.7. Bin p xung..................................................594.8. Tng hp h thng truyn ng in...............654.8.1. ng c in mt chiu KTL...............................654.8.2. B iu chnh c k..............................................674.8.3. Khu phn hi dng in......................................694.8.4. Khu phn hi tc ............................................705.8.5. Tng hp mch vng dng in............................714.8.6. Tng hp mch vng tc ..................................73

4.9. o chiu quay ca ng c............................76

4.9.1. o chiu t thng...............................................764.9.2. S mch kch t...............................................774.9.3. Qu trnh o chiu..............................................77

Chng 5: M phng h thng bng simulink............................80

5.1. Gii thiu phn mm simulink.........................805.2. M phng h thng bng simulink...................815.3. ngha ca qu trnh m phng......................825.4. M t qu trnh m hnh, m phng.................82

Kt lun...............................................................................84

Ti liu tham kho.................................................................85------------ ------------..............................................................88Li cam oan.........................................................................91

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Trng i hc Bch Khoa H NiKhoa in

B mn t ng ho xncn

------------ ------------

n tt nghip ti: thit k b iu khin ng c mt chiu kch t c lp

Gio vin hng dn : TS. Nguyn mnh tinGio vin hng dn : Trn trng minh

Sinh vin thc hin : Phm hu qunLp : TH k1 - HPM s sinh vin : 4063

Hi Phng - 11/2004B Gio dc v o to

Trng i hc Bch Khoa H Ni---------------

Cng ho x hi ch ngha Vit Namc lp T do Hnh phc

---------***---------

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Nhim v n tt nghip

H tn sinh vin : Phm Hu Qun S hiu sinh vin: 4063

Kho : 1 Hi Phng Khoa: in Ngnh: TH XNCN

1. u thit k.

Thit k b iu khin ca ng c mt chiu kch t c lp

2. Cc s liu ban u..............................................................................................................................

.............................................................................................................................

.............................................................................................................................

3. Ni dung cc phn thuyt minh v tnh ton.

Chng I : Khi nim v cng nghip cn.

Chng II : iu khin ng c mt chiu.

Chng III : Phn tch tnh ton mch lc.

Chng IV : Phn tch tnh ton mch iu khin.

Chng V : M phng h thng bng Simulink.

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

4. Cc bn v, th (Ghi r cc loi bn v v kch thc cc bn v):

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

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5. H tn cn b hng dn.

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

6. Ngy giao nhim v n:...........................................................................

7. Ngy hon thnh n: ...............................................................................

Ch nhim b mn(K, ghi r h tn)

Ngy thng nm 2004Cn b hng dn

(K, ghi r h tn)

Hc sinh hon thnh v np n tt nghip ngy thng nm 2004

Ngi duyt

(K, ghi r h tn)

Sinh vin

(K, ghi r h tn)

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Li cam oan

Em xin cam oan ti tt nghip ny l do em t thit

k di s hng dn ca thy gio Trn Trng Minh. Cc sliu v kt qu trong ti l hon thnh trung thc.

hon thnh bn n ny, em ch s dng nhng ti

liu tham kho c ghi trong bng cc ti liu tham kho,

khng s dng ti liu no khc m khng c lit k phn

ti liu tham kho.

Sinh vin

Phm Hu Qun

Thiet Ke Bo Dieu Khien Dong Co Mot Chieu Kich Tu Doc Lap 20120817041302 31

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