11_LuuDucTuong_DC1201.pdf

1

LI M U

Ngy nay vi s pht trin khng ngng ca nn khoa hc k thut to

ra nhng thnh tu to ln, trong ngnh t ng ha cng gp phn khng

nh vo thnh cng . Mt trong nhng vn quan trng trong cc dy truyn

t ng ho sn xut hin i l vic iu chnh tc ng c. T trc n

nay, ng c mt chiu vn lun l loi ng c c s dng rng ri k c

trong nhng h thng yu cu cao. V vy em c giao ti tt nghip l:

Xy dng h truyn ng in ng c mt chiu s dng b iu khin

PID. Ni dung ti c chia lm 3 chng:

Chng 1. Tng quan v h truyn ng in mt chiu

Chng 2. Xy dng m hnh h truyn ng in mt chiu trn Matlab

v Simulink

Chng 3. Xy dng m hnh vt l b iu khin PID iu khin ng

c in mt chiu

Trong qu trnh lm ti tt nghip, em lun nhn c s hng dn,

ch bo tn tnh v cung cp nhng ti liu cn thit ca thy gio GS TSKH

Thn Ngc Hon. Em xin gi ti hai thy li cm n chn thnh. Tuy nhin, do

thi gian v gii hn ca n cng vi phm vi nghin cu ti liu vi kinh

nghim v kin thc cn hn ch nn bn n ny khng trnh khi nhng

thiu st rt mong s ng gp kin ca thy c bn n ca em c

hon thin hn.

Sinh vin thc hin

Lu c Trng

2

CHNG 1. TNG QUAN V H TRUYN NG IN MT

CHIU

1.1. TNG QUAN V NG C IN MT CHIU

1.1.1. Cu to, phn loi ng c in mt chiu

a. Cu to ca ng c in mt chiu

ng c in mt chiu c th phn thnh hai phn chnh: Phn tnh v

phn ng.

- Phn tnh hay stato hay cn gi l phn kch t ng c, l b phn sinh ra t

trng n gm c:

+) Mch t v dy cun kch t lng ngoi mch t (nu ng c c kch t

bng nam chm in), mch t c lm bng st t (thp c, thp c). Dy

qun kch thch hay cn gi l dy qun kch t c lm bng dy in t, cc

cun dy in t nay c mc ni tip vi nhau.

+) Cc t chnh: L b phn sinh ra t trng gm c li st cc t v dy qun

kch t lng ngoi li st cc t. Li st cc t lm bng nhng l thp k thut

in hay thp cacbon dy 0,5 n 1mm p li v tn cht. Trong ng c in

nh c th dng thp khi. Cc t c gn cht vo v my nh cc bulng.

Dy qun kch t c qun bng dy ng bc cch in v mi cun dy u

c bc cch in k thnh mt khi, tm sn cch in trc khi t trn cc

cc t. Cc cun dy kch t c t trn cc cc t ny c ni tip vi nhau

+) Cc t ph: Cc t ph c t trn cc cc t chnh. Li thp ca cc t

ph thng lm bng thp khi v trn thn cc t ph c t dy qun m cu

to ging nh dy qun cc t chnh. Cc t ph c gn vo v my nh

nhng bulng.

+) Gng t: Gng t dng lm mch t ni lin cc cc t, ng thi lm v

my. Trong ng c in nh v va thng dng thp dy un v hn li, trong

my in ln thng dng thp c. C khi trong ng c in nh dng gang

lm v my.

+) Cc b phn khc:

3

Np my: bo v my khi nhng vt ngoi ri vo lm h hng dy

qun v an ton cho ngi khi chm vo in. Trong my in nh v va np

my cn c tc dng lm gi bi. Trong trng hp ny np my thng

lm bng gang.

C cu chi than: a dng in t phn quay ra ngoi. C cu chi

than bao gm c chi than t trong hp chi than nh mt l xo t cht ln c

gp. Hp chi than c c nh trn gi chi than v cch in vi gi. Gi

chi than c th quay c iu chnh v tr chi than cho ng ch, sau khi

iu chnh xong th dng vt c nh li.

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

+) Phn sinh ra sc in ng gm c:

Mch t c lm bng vt liu st t (l thp k thut) xp li vi nhau.

Trn mch t c cc rnh lng dy qun phn ng.

Cun dy phn ng: Gm nhiu bi dy ni vi nhau theo mt qui lut

nht nh. Mi bi dy gm nhiu vng dy cc u dy ca bi dy c ni

vi cc phin ng gi l phin gp, cc phin gp c ghp cch in vi

nhau v cch in vi trc gi l c gp hay vnh gp.

T trn c gp l cp tri than lm bng than graphit v c ghp st vo

thnh c gp nh l xo.

+) Li st phn ng: Dng dn t, thng dng nhng tm thp k thut in

dy 0,5mm ph cch in mng hai mt ri p cht li gim tn hao do

dng in xoy gy nn. Trn l thp c dp hnh dng rnh sau khi p li th

t dy qun vo. Trong nhng ng c trung bnh tr ln ngi ta cn dp

nhng l thng gi khi p li thnh li st c th to c nhng l thng gi

dc trc. Trong nhng ng c in ln hn th li st thng chia thnh nhng

on nh, gia nhng on y c mt khe h gi l khe h thng gi. Khi

my lm vic gi thi qua cc khe h lm ngui dy qun v li st.

Trong ng c in mt chiu nh, li st phn ng c p trc tip vo

trc. Trong ng c in ln, gia trc v li st c t gi rto. Dng gi rto

c th tit kim thp k thut in v gim nh trng lng rto.

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+) Dy qun phn ng: Dy qun phn ng l phn pht sinh ra sut in ng

v c dng in chy qua, dy qun phn ng thng lm bng dy ng c bc

cch in. Trong my in nh c cng sut di vi Kw thng dng dy c

tit din trn. Trong my in va v ln thng dng dy tit din ch nht,

dy qun c cch in cn thn vi rnh ca li thp.

trnh khi quay b vng ra do lc li tm, ming rnh c dng nm

cht hoc ai cht dy qun. Nm c th lm bng tre, g hay bakelit.

+) C gp: C gp gm nhiu phin ng c c m cch in vi nhau bng

lp mica dy t 0,4 n 1,2mm v hp thnh mt hnh trc trn. Hai u trc

trn dng hai hnh p hnh ch V p cht li. Gia vnh p v tr trn cng cch

in bng mica. ui vnh gp c cao ln mt t hn cc u dy ca cc

phn t dy qun v cc phin gp c d dng.

b. Phn loi, u nhc im ca ng c in mt chiu

- Phn loi ng c in mt chiu

Khi xem xt ng c in mt chiu cng nh my pht in mt chiu

ngi ta phn loi theo cch kch thch t cc ng c. Theo ta c 4 loi ng

c in mt chiu thng s dng:

+) ng c in mt chiu kch t c lp: Phn ng v phn kch t c cung

cp t hai ngun ring r.

+) ng c in mt chiu kch t song song: Cun dy kch t c mc song

song vi phn ng.

+) ng c in mt chiu kch t ni tip: Cun dy kch t c mc ni tp

vi phn ng.

+) ng c in mt chiu kch t hn hp: Gm c 2 cun dy kch t, mt

cun mc song song vi phn ng v mt cun mc ni tip vi phn ng.

- u nhc im ca ng c in mt chiu

Do tnh u vit ca h thng in xoay chiu: sn xut, truyn ti...,

c my pht v ng c in xoay chiu u c cu to n gin v cng sut

ln, d vn hnh... m my in (ng c in) xoay chiu ngy cng c s

dng rng ri v ph bin. Tuy nhin ng c in mt chiu vn gi mt v tr

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nht nh trong cng nghip giao thng vn ti, v ni chung cc thit b cn

iu khin tc quay lin tc trong phm vi rng (nh trong my cn thp,

my cng c ln, u my in...). Mc d so vi ng c khng ng b

ch to ng c in mt chiu cng c th gi thnh t hn do s dng nhiu

kim loi mu hn, ch to bo qun c gp phc tp hn. Nhng do nhng u

im ca n m my in mt chiu vn khng th thiu trong nn sn xut hin

i.

+) u im ca ng c in mt chiu l c th dng lm ng c in hay

my pht in trong nhng iu kin lm vic khc nhau. Song u im ln nht

ca ng c in mt chiu l iu chnh tc v kh nng qu ti. Nu nh

bn thn ng c khng ng b khng th p ng c hoc nu p ng

c th phi chi ph cc thit b bin i i km (nh b bin tn....) rt t tin

th ng c in mt chiu khng nhng c th iu chnh rng v chnh xc m

cu trc mch lc, mch iu khin n gin hn ng thi li t cht lng

cao.

+) Nhc im ch yu ca ng c in mt chiu l c h thng c gp - chi

than nn vn hnh km tin cy v khng an ton trong cc mi trng rung

chn, d chy n.

1.1.2. c tnh c ca ng c in mt chiu

a. Nguyn l lm vic ca ng c in mt chiu

Khi cho in p mt chiu vo, trong dy qun phn ng c in. Cc

thanh dn c dng in nm trong t trng s chu lc tc dng lm rto quay,

chiu ca lc c xc nh bng quy tc bn tay tri.

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

Do c phiu gp chiu dng in d nguyn lm cho chiu lc t tc dng

khng thay i. Khi quay, cc thanh dn ct t trng s cm ng vi sut in

ng E chiu ca sut in ng c xc nh theo quy tc bn tay phi,

ng c chiu s E ngc chiu dng in I nn E c gi l sc phn in

ng. Khi ta c phng trnh: U = E + R.I

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b. c tnh c ca ng c in mt chiu kch t c lp

Khi ngun mt chiu c cng sut khng ln th mch in phn ng

v mch in kch t mc vo hai ngun c lp nhau. Lc ny ng c c

gi l ng c in mt chiu kch t c lp[2].

Hnh 1.1: S ni dy ca ng c in mt chiu kch t c lp

Ta c phng trnh cn bng in p ca mch phn ng nh sau:

U = E + (R + Rf)I (1.1)

Trong :

U: in p 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:

.. . . .

2

P NE K

a (1.2)

Trong :

P: S i cc t chnh

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

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

: T thng kch t di mt cc t

: Tc gc (rad/s)

K = .

2

P N

a: H s cu to ca ng c

T (1.1) v (1.2) ta c:

u fR R .. .

UI

K K (1.3)

Biu thc trn 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 (1.4)

Vi tM

.

IK

: thay gi tr I vo (1.3) ta c

u ft2

R R.M

. ( . )

U

K K (1.5)

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

u f

2

U R R.

K. (K. )

u M

(1.6)

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

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

c tnh c in (1.3) v phng trnh c tnh c (1.6) l tuyn tnh. th ca

chng c biu din trn hnh 1.2 l nhng ng thng.

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

U

K

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

lp.

8

Hnh 1.2: c tnh c in v c tnh c ca ng c in mt chiu

Khi = 0 ta c:

nm

u f

IR R

UI (1.7)

M = K. .Inm = Mnm (1.8)

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

Ngoi ra phng trnh c tnh (1.3) v (1.6) cng c th c vit di dng:

0.. .

U RI

K K (1.9)

02.

. ( . )

U RM

K K (1.10)

Trong :

R = R + Rf,

0.

U

K

2. .

. ( . )

R RI M

K K

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

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.

1.2. CC PHNG PHP IU KHIN TC NG C IN

MT CHIU

- Phng php thay i in tr phn ng

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- Phng php thay i t thng

- Phng php thay i in p phn ng

1.2.1. Phng php thay i in tr phn ng

- y l phng php thng dng iu khin tc ng c in mt chiu

+) Nguyn l iu khin: Trong phng php ny ngi ta gi U = Um, =

m v ni thm in tr ph vo mch phn ng tng in tr phn ng[3].

cng ca ng c tnh c:

2

u f

(k )

R R

M (1.11)

+) Ta thy khi in tr cng ln th cng nh ngha l c tnh c cng dc v

do cng mm hn.

Hnh 1.3: c tnh c ca ng c khi thay i in tr ph

ng vi Rf = 0 ta c cng t nhin TN c gi tr ln nht nn c tnh c t

nhin c cng ln hn tt c cc ng c tnh c c in tr ph. Nh vy,

khi ta thay i Rf ta c mt h c tnh c thp hn c tnh c t nhin.

- c im ca phng php:

+) in tr mch phn ng cng tng th dc c tnh cng ln, c tnh c

cng mm, n nh tc cng km v sai s tc cng ln.

+) Phng php ny ch cho php iu chnh tc trong vng di tc nh

mc ( ch cho php thay i tc v pha gim).

+) Ch p dng cho ng c in c cng sut nh, v tn hao nng lng trn

in tr ph lm gim hiu sut ca ng c v trn thc t thng dng ng

c in trong cn trc.

10

+) nh gi cc ch tiu: Phng php ny khng th iu khin lin tc c

m phi iu khin nhy cp. Di iu chnh ph thuc vo ch s mmen ti, ti

cng nh th di iu chnh D = max / min cng nh. Phng php ny c th

iu chnh trong di D = 3 : 1

+) Gi thnh u t ban u r nhng khng kinh t do tn hao trn in tr ph

ln, cht lng khng cao d iu khin rt n gin.

1.2.2. Phng php thay i t thng

- Nguyn l iu khin:

Gi thit U= Um, R = const. Mun thay i t thng ng c ta thay i

dng in kch t, thay i dng in trong mch kch t bng cch ni ni tip

bin tr vo mch kch t hay thay i in p cp cho mch kch t.

Bnh thng khi ng c lm vic ch nh mc vi kch thch ti a

( = max) m phng php ny ch cho php tng in tr vo mch kch t

nn ch c th iu chnh theo hng gim t thng tc l iu chnh tc

trong vng trn tc nh mc. Nn khi gim th tc khng ti l tng

k

U dmo tng, cn cng c tnh c

uR

k2

gim, ta thu c h c

tnh c nm trn c tnh c t nhin[3].

Hnh 1.4: c tnh c ca ng c khi gim t thng

- Khi tng tc ng c bng cch gim t thng th dng in tng v tng

vt qu mc gi tr cho php nu mmen khng i. V vy mun gi cho

dng in khng vt qu gi tr cho php ng thi vi vic gim t thng th

ta phi gim Mt theo cng t l.

- c im ca phng php:

M

m

2

1 o

o

1

o

2

0 Mc1 Mc2

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+) Phng php ny c th thay i tc v pha tng.

+) Phng php ny ch iu khin vng ti khng qu ln so vi nh mc,

vic thay i t thng khng lm thay i dng in ngn mch.

+) Vic iu chnh tc bng cch thay i t thng l phng php iu

khin vi cng sut khng i.

+) nh gi cc ch tiu iu khin: Sai s tc ln, c tnh iu khin nm

trn v dc hn c tnh t nhin. Di iu khin ph thuc vo phn c ca

my. C th iu khin trn trong di iu chnh D = 3 : 1. V cng sut ca

cun dy kch t b, dng in kch t nh nn ta c th iu khin lin tc vi

1.

+) Phng php ny c p dng tng i ph bin, c th thay i lin tc

v kinh t ( v vic iu chnh tc thc hin mch kch t vi dng kch t

(1 10)%Im ca phn ng nn tn hao iu chnh thp).

y l phng php gn nh l duy nht i vi ng c in mt chiu

khi cn iu chnh tc ln hn tc iu khin.

1.2.3. Phng php thay i in p phn ng

- iu chnh in p phn ng ng c mt chiu cn c thit b ngun nh

my pht in mt chiu kch t c lp, cc b chnh lu iu khin Cc

thit b ngun ny c chc nng bin nng lng in xoay chiu thnh mt

chiu c sc in ng Eb iu chnh nh tn hiu iu khin Uk. V ngun c

cng sut hu hn so vi ng c nn cc b bin i ny c in tr trong Rb

v in cm Lb khc khng. a tc ng c vi hiu sut cao trong gii hn

rng ri 1:10 hoc hn na[3].

Hnh 1.5: S dng b bin i iu khin in p phn ng

ch xc lp c th vit c phng trnh c tnh ca h thng nh sau:

~

BB

LK

Uk E Eb(Uk)

Rb I R

U

12

Eb - E = I(Rb +R) (1.12)

.. .

b b udu

dm dm

E R RI

K K (1.13)

o dk

MU (1.14)

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

khng i, cn tc khng ti l tng th tu thuc vo gi tr in p iu

khin Uk ca h thng, do c th ni phng php iu chnh ny l trit .

xc nh gii iu chnh tc ta rng tc ln nht ca h

thng b chn bi c tnh c c bn, l c tnh ng vi in p phn ng nh

mc v t thng cng c gi gi tr nh mc. Tc nh nht ca di iu

chnh b gii hn bi yu cu v sai s tc v v mmen khi ng. Khi

mmen ti l nh mc th cc gi tr ln nht v nh nht ca tc l:

dm

o

Mmaxmax (1.15)

dm

o

Mminmin (1.16)

tho mn kh nng qu ti th c tnh thp nht ca di iu chnh

phi c mmen ngn mch l: Mnmmin = Mcmax = KM.Mm

Trong KM l h s qu ti v mmen. V h c tnh c l cc ng thng

song song nhau, nn theo nh ngha v cng c tnh c c th vit:

min min

11dmnm dm M

MM M K (1.17)

1

1

11

.maxmax

M

dm

o

dmM

dmo

K

M

MK

M

D (1.18)

13

Hnh 1.6: c tnh c ca ng c khi thay i in p

- Vi mt c cu my c th th cc gi tr 0max, Mm, KM l xc nh, v vy

phm vi iu chnh D ph thuc tuyn tnh vo gi tr ca cng . Khi iu

chnh in p phn ng ng c bng cc thit b ngun iu chnh th in tr

tng mch phn ng gp khong hai ln in tr phn ng ng c. Do c

th tnh s b c: max1

. 10odmM

V th ti c c tnh mmen khng i th gi tr phm vi iu chnh tc

cng khng vt qu 10. i vi cc my c yu cu cao v di iu chnh v

chnh xc duy tr tc lm vic th vic s dng cc h thng h nh trn l

khng tho mn c.

- Trong phm vi ph ti cho php c th coi c tnh c tnh ca h truyn ng

mt chiu kch t c lp l tuyn tnh. Khi iu chnh in p phn ng th

cng c c tnh c trong ton di l nh nhau, do st tc tng i s t

gi tr ln nht ti c tnh thp nht ca di iu chnh. Hay ni cch khc, nu

ti c tnh c thp nht ca di iu chnh m sai s tc khng vt qu gi

tr sai s cho php, th h truyn ng s lm vic vi sai s lun nh hn sai s

cho php trong ton b di iu chnh. Sai s tng i ca tc c tnh c

thp nht l:

max0

max

U k2

0min

Mnm Mm

M,I

U k1

min

14

min min

min min

o

o o

s (1.19)

min.

dmcp

o

Ms s (1.20)

V cc gi tr Mm, 0min, scp l xc nh nn c th tnh c gi tr ti

thiu ca cng c tnh c sao cho sai s khng vt qu gi tr cho php.

lm vic ny, trong a s cc trng hp cn xy dng cc h truyn ng in

kiu vng kn.

- Nhn xt: C 3 phng php trn u iu chnh c tc ng c in mt

chiu nhng ch c phng php iu chnh tc ng c in mt chiu bng

cch thay i in p U t vo phn ng ca ng c l tt nht v hay c

s dng nht v n thu c c tnh c c cng khng i, iu chnh tc

bng phng v khng b tn hao.

1.3. GII THIU MT S H TRUYN NG IN MT CHIU

- H truyn ng my pht - ng c mt chiu (F - )

- H truyn ng xung p - ng c (XA - C

- H truyn ng chnh lu - ng c (CL - C)

1.3.1. H truyn ng my pht - ng c in mt chiu (F - )

- Cu trc h F - v c tnh c bn:

H thng my pht - ng c (F - ) l h truyn ng in m b bin i in

l my pht in mt chiu kch t c lp. My pht ny thng do ng c s

cp khng ng b ba pha ko quay[3].

Tnh cht ca my pht in c xc nh bi hai c tnh: c tnh t

ho l s ph thuc gia sc in ng my pht vo dng in kch t v c

tnh ti l s ph thuc ca in p trn hai cc ca my pht vo dng in ti.

Cc c tnh ny ni chung l phi tuyn do tnh cht ca li st, do cc phn ng

ca dng in phn ng trong tnh ton gn ng c th tuyn tnh ho cc

c tnh ny:

EF =KF. F. F =KF. F.C.iKF (1.21)

Trong :

15

KF: l h s kt cu ca my pht

C = F/ iKF l h s gc ca c tnh t ho.

Nu dy qun kch thch ca my pht c cp bi ngun p l tng

UKF th: IKF =UKF/rKF

Sc in ng ca my pht trong trng hp ny s t l vi in p kch thch

bi h s hng KF nh vy c th coi gn ng my pht in mt chiu kch t

c lp l mt b khuych i tuyn tnh:

EF = KF.UKF

Hnh 1.7: S nguyn l h truyn ng my pht ng c

Nu t R = RF + RD th c th vit c phng trnh cc c tnh ca

h F - nh sau:

.K K

FKF

K RIU

(1.22)

2K K

FKF

K RU M

(1.23)

,o KF KDKD

MU U

U (1.24)

Cc biu thc trn chng t rng, khi iu chnh dng in kch thch ca

my pht th iu chnh c tc khng ti ca h thng cn cng c

tnh c th gi nguyn. Cng c th iu chnh kch t ca ng c c di

iu chnh tc rng hn.

Uk

iKF

UKF

~

K

F UF=U

F

MS

M

~

Uk UK

iK

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- Cc ch lm vic ca h F-

Hnh 1.8: Cc trng thi lm vic ca h F -

Trong h F - khng c phn t phi tuyn no nn h c nhng c tnh

ng rt tt, rt linh hot khi chuyn cc trng thi lm vic. Vi s c bn

nh hnh 1.7 ng c chp hnh c th lm vic ch iu chnh c c

hai pha: Kch thch my pht F v kch thch ng c , o chiu quay bng

cch o chiu dng kch thch my pht, hm ng nng khi dng kch thch

my pht bng khng, hm ti sinh khi gim tc hoc khi o chiu dng

kch t, hm ngc cui giai on hm ti sinh khi o chiu hoc khi lm

vic n nh vi mmen ti c tnh cht th nng h F - c c tnh c c

bn gc phn t ca mt phng to [ ,M].

+) gc phn t th I v th III tc quay v mmen quay ca ng c lun

cng chiu nhau, sc in ng my pht v ng c c chiu i nhau v

FE E , c . Cng sut in t ca my pht v ng c l:

PF = EF.I > 0

P = E.I < 0

Pc = M. > 0

Cc biu thc ny ni ln rng nng lng c vn chuyn thun chiu t

ngun my pht ng c ti.

+) Vng hm ti sinh nm gc phn t th II v th IV, lc ny do o

nn FEE , mc d E, EF mc ngc nhau, nhng dng in phn ng li chy

ngc t ng c v my pht lm cho mmen quay ngc chiu tc quay.

Uktf = 0 (III)

(IV)

(I) (II)

o

M

17

Cng sut in t ca my pht, cng sut in t v cng sut c hc ca ng

c l:

PF = EF.I < 0

P = E.I > 0

Pc = M. < 0

Ch do dng in i chiu m cc bt ng thc trn cho ta thy nng

lng c chuyn vn theo chiu t ti ng c my pht ngun, my

pht F v ng c i chc nng cho nhau. Hm ti sinh trong h F - c

khai thc trit khi gim tc , khi hm o chiu quay v khi lm vic n

nh vi ti c tnh cht th nng.

- c im ca h F - :

+) Cc ch tiu cht lng ca h F - v c bn tng t cc ch tiu ca h

iu p dng b bin i ni chung. u im ni bt ca h F - l s chuyn

i trng thi lm vic rt linh hot, kh nng chu qu ti ln, do vy thng s

dng h truyn ng F - cc my khai thc trong cng ngip m.

+) Nhc im quan trng nht ca h F - l dng nhiu my in quay, trong

t nht l hai my in mt chiu, gy n ln, cng sut lp t my t nht

gp ba ln cng sut ng c chp hnh. Ngoi ra do cc my pht mt chiu c

t d, c tnh t ho c tr nn kh iu chnh su tc .

1.3.2. H truyn ng xung p ng c (XA - C)

B bin i xung p l mt ngun in p dng iu chnh tc ng c

in mt chiu[3].

Hnh 1.9: S nguyn l v gin xung

18

ci thin dng sng ca dng in phn ng ta thm vo mch mt

van m V0. C th s dng thyristor hoc transistor cng sut thay cho kha

K trn. Khi ng ct kha K, trn phn ng ng c s c in p bin i

theo dng xung vung. Khi trng thi dng lin tc th gi tr trung bnh ca

in p ra s l:

1

0

1 .1

t

CKCK

d UUT

tUdt

TU (1.25)

Trong :

t1 : L thi gian kha trng thi ng

t2 : L thi gian kha trng thi m

Tck : Thi gian thc hin mt chu k ng m kha

CK

1

T

t: L rng ca xung p

Vy ta c th coi b bin i xung ng tr vi ngun lin tc c in p ra Ud

v Ud c th thay i c bng cch thay i rng xung . Mt khc, thi

gian mt chu k ng ct ca kha K rt nh so vi hng s thi gian c hc ca

h truyn ng, nn ta coi tc v sc in ng phn ng ng c khng thay

i trong thi gian Tck.

- c tnh iu chnh ca h XA - C

IK

RR

K

U.

..

.

m

bb

m

(1.26)

MK

RR

K

U.

).(.

.2

m

b

m

(1.27)

Khi thay i ta c h ng thng song song c cng = const v

tc khng ti l tng o thay i theo . Nu ngun v cng ln th ta c th

b qua Rb, khi cng ca c tnh c ca h c cng l:

constR

KTN

b

2

m ).( (1.28)

19

Tc khng ti l tng o ph thuc vo ch l gi tr gi nh. N c

th tn ti nu nh dng trong h l lin tc k c khi gi tr dng tin n 0. V

vy hai biu thc trn ch ng vi trng thi dng lin tc.

Khi dng in nh th h s chuyn trang thi t dng lin tc sang

trng thi dng gin on. Khi cc phng trnh c tnh iu chnh ni trn

khng cn ng na m lc ny c tnh ca h l nhng ng cong rt dc.

Hnh 1.10: c tnh c ca h

- Nhn xt:

+) Tt c c tnh iu chnh ca h XA C khi dng in gin on u c

chung mt gi tr khng ti l tng, ch ngoi tr trng hp = 0.

+) B ngun xung p cn t van dn nn vn u t t, h n gin chc chn.

+) cng ca c tnh c ln.

+) in p dng xung nn gy ra tn tht ph kh ln trong ng c. Khi lm

vic trng thi dng in gin on th c tnh lm vic km n nh v tn

tht nng lng nhiu.

1.3.3. H truyn ng chnh lu - ng c in mt chiu (CL - C)

- S nguyn l:

Hnh 1.11: S nguyn l ca h chnh lu - ng c in mt chiu

+

-

CL L

K

C

KT

20

H truyn ng chnh lu c iu khin - ng c in mt chiu (CL - C)

c b bin i l cc mch chnh lu c iu khin, c sc in ng Ed ph

thuc vo gi tr ca xung iu khin ( tc l ph thuc vo gc iu khin hay

gc m Tiristor )[3].

in p chnh lu Ud ( hay Ed ) l in p khng ti u ra, c dng p mch

vi s ln p mch l n trong mt chu k 2 ca in p th cp my bin p.

+) Vi s chnh lu hnh tia: n = m, trong m l s pha

+) Vi s hnh cu: n = 2.m, trong m l s pha

Gi s in p th cp ca my bin p c dng hnh sin vi biu thc l:

u2 = U2m.sin t = U2m.sin , ( vi = t ) (1.29)

Trong khong = ( 0 2 ) th dng in p v dng in lp li nh chu k ban

u nn ta ch cn xt trong mt chu k T = 2 .

- S thay th ca h CL C.

Hnh 1.12: S thay th ca h chnh lu - ng c in mt chiu

Khi van dn th ta c phng trnh cn bng in p nh sau:

dt

di.LR.IEu

dd2 (1.30)

Suy ra: dt

di.LR.iEsin.U

ddm2 (1.31)

Trong :

R = Rba + R + Rk

L = Lba + L + Lk

Vi: 2

1

212ba )

W

W.(RRR (1.32)

E

R

Tiristor

L

Ud

~

21

2

1

212ba )

W

W.(LLL (1.33)

- Trng thi dng lin tc

trng thi dng lin tc, khi van ny cha kha th van k tip m, vic m

van k tip l iu kin cn kha van ang dn. Do vy, in p ca chnh

lu s c dng ng bao ca in p th cp my bin p.

Gi tr trung bnh ca in p chnh lu:

n

2

m2

n

2

2d d.sin.U.2

ndt.u.

2

nU (1.34)

cos.Ucos.U.n

sin.n

dom2

Trong :

t.

)n2

(o : L gc m ca van

nsin.U.

nU m2do : L in p mt chiu ln nht u ra chnh

lu ng vi = 0

U2m: L tr bin ca in p th cp my bin p

n: L s ln p mch trong mt chu k

+) B qua st p trn van, ta c phng trnh c tnh c nh sau :

MK

R

K

U do2

mm ).(.

cos. (1.35)

Trong :

.2

u kh ba ba v

nR R R R X R

R: L in tr ca phn ng ng c

Rkh: L in tr ca cun khng lc

Rba: L in tr ca my bin p, vi 2

1

212ba )W

W.(RRR

22

Xba: L in khng my bin p, vi 2

1

212ba )W

W.(XXX

Rv: L in tr ca cc van ( Rv rt nh c th b qua )

baX.2

n: L in tr ng tr do qu trnh chuyn mch

+) cng ca c tnh c:

R

KM

d

dM2

m ).( (1.36)

Hnh 1.13: c tnh c ca h chnh lu - ng c mt chiu khi dng lin tc

- Trng thi dng gin an

Khi in khng trong mch khng ln, nu sc in ng ca ng c

ln th dng in ti s tr thnh gin on. trng thi ny th dng qua van

bt k s bng 0 trc khi van k tip m. Do vy trong mt khong dn ca van

th sc in ng ca chnh lu bng sc in ng ngun: ed = U2 , vi 0

, trong l khong dn.

Khi dng in bng 0 th sc in ng ca chnh lu bng sc in ng

ca ng c: ed = E , vi < n

2

Vy ta c in p trung bnh ca chnh lu l :

0

n

2

m2

0

n

2

2d d.Ed.sin.U.2

nd.Ed.u.

2

nU (1.37)

)n

2.(E)cos1.(U.

2

nm2

Udo

Ud1

Ud2

Ud3

o

o1

o2

o3

M( I )

23

Vy : )n

2.(E)cos1.(U.

2

nU m2d (1.38)

c tnh c ca h CL - C khi dng in gin an:

Hnh 1.14: c tnh c ca h chnh lu - ng c khi dng gin on

- Nhn xt:

+) u im: H truyn ng chnh lu - ng c c tc ng nhanh cao,

khng gy n v d t ng ha, do cc van bn dn c h s khuch i cng

sut rt cao, v vy rt thun tin cho vic thit lp h thng t ng iu chnh

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

khc, vic dng h chnh lu - ng c c kch thc v trng lng nh gn.

+) Nhc im: H truyn ng chnh lu - ng c c cc van bn dn l cc

phn t phi tuyn tnh, do dng in p chnh lu ra c bin p mch

cao, gy nn tn tht ph trong my in mt chiu.

24

CHNG 2. XY DNG M HNH H TRUYN NG IN

MT CHIU TRN MATLAB V SIMULINK

2.1. M HNH TON CA NG C IN MT CHIU

Khi t ln dy qun kch t mt in p no th trong dy qun kch

t s c dng in v mch t ca my s c t thng . Tip t mt gi tr

in p U ln mch phn ng th trong dy qun phn ng s c dng in I

chy qua, tng tc gia dng in phn ng v t thng kch t to thnh

mmen in t. Vy ta c cc phng trnh c bn ca ng c mt chiu.

- Phng trnh cn bng in p phn ng:

U = E + I(R + Rf) (2.1)

- Sc in ng phn ng E c tnh theo biu thc:

E = k.. (2.2)

- Mmen in t ca ng c c xc nh:

Mdt = k..I (2.3)

- Phng trnh cn bng m men ca ng c:

M(t) MC(t) = J

dt

d (2.4)

Trong :

R: L in tr cun dy phn ng

E: L sc in ng phn ng ng c

Rf: L in tr ph

I: L dng phn ng

K: L h s cu to ca my in

M: L m men ng c

U: L in p t vo phn ng ng c

: L tc gc ng c

: L t thng ng c

Chuyn cc phng trnh trn sang dng ton t Laplace:

U(p) = R.I(p) + L.I(p).p + E(p) (2.5)

M(p) - MC(p) = J(p). (p).p (2.6)

25

E(p) = K. (p) (2.7)

M(p) = k..I(p) (2.8)

- Ta thnh lp c phng trnh c tnh c nh sau:

= Mk

RR

k

IpLU puu .).(

..2 (2.9)

- Hm truyn ca ng c nh sau:

)(.1

/1)(

.

1EU

pT

REU

pLRI

U

U

UU

(2.10)

T cc phng trnh trn ta c s cu trc ca ng c in mt chiu nh

sau:

Hnh 2.1: S cu trc ng c in mt chiu

- La chn thng s m phng

ng c s dng l ng c mt chiu B1T20E ca hng YASKAWA.

Thng s ng c:

+) Cng sut nh mc: Pm = 20 (W)

+) in p nh mc phn ng: Um = 21,3 (V)

+) Tc quay nh mc: n m = 2200 (vng/pht)

+) Dng in nh mc: Im = 0,59 (A)

+) in cm phn ng: Lu = 9,1 (mH)

+) in tr phn ng: Ru = 15,7 ( )

+) Mmen qun tnh ca ng c: J = 1,18.10-6 (kg.m2)

+) Hng s momen Km = 0,037 (N.m/A)

+) Hng s thi gian iu khin Tk = 0,0001 (s)

+) Hng s thi gian chuyn mch chnh lu: Tv = 0,001 (s)

26

+) Hng s thi gian ca my bin dng: Ti = 0,001 (s)

+) Hng s thi gian ca my pht tc: T = 0,01 (s)

Ta c:

Hng s thi gian phn ng:

u

uu

R

LT =

9,1.10-3

15,7 = 5,796.10

-4 (s)

Tc gc ca rto:

55,9

n =

2200

9,55 = 230,37 (rad/s)

H s khuch i ca b bin i:

Chn Uk = 5 (V)

Vy Kbd = UmUk

= 21,4

5 = 4,28

Hm truyn ca b bin i:

Wb = (1 )(1 )

bd

dk v

K

T P T P (2.11)

3 7 24,28 4,28

(1 0,0001. )(1 0,001. ) 1 1,1.10 . 10 .s s s s

- M hnh ng c in mt chiu

Hnh 2.2: M hnh ng c in mt chiu

27

2.2. TNG HP MCH VNG DNG IN

Khi b qua sc in ng E ta c s sau:

Hnh 2.3: S cu trc ca mch vng dng in

Trong :

(1 )(1 )

bd

dk v

K

T P T P: Hm truyn ca b bin i

bdK : H s khuch i ca b bin i

dkT : Hng s thi gian mch iu khin

vT : Hng s thi gian ca s chuyn mch ca van bn dn

pT

K

i

i

1:Hm truyn ca cm bin dng in

iK : H s khuch i ca cm bin dng in

iT : Hng s thi gian ca cm bin dng in

Thu gn ta c s nh hnh v:

Trong iS 0 l hm truyn ca i tng

Hnh 2.4: S thu gn ca mch vng in

iS0.1/

(1 )(1 )(1 )(1 )

bd i u

dk v i u

K K R

T p T p T p T p (2.12)

V ivdk TTT ;; s thi gian rt nh nn b qua thnh phn bc cao l cc hng s

iR iS 0 idU I

-

28

iS0 .1/

1 ( ) (1 )

bd i u

dk v i u

K K R

T T T p T p (2.13)

t: vdkisi TTTT

iS0.1/

(1 )(1 )

bd i u

si u

K K R

T p T p (2.14)

p dng tiu chun mdul ti u ta c hm truyn ca h thng:

MCF 22221

1

pp (2.15)

M MCK FFii

ii

SR

SR

0

0

1 iR

)1(

11

0 Mii FS

Thay vo ta c:

iR(1 )(1 )

.2 (1 )

u si u

bd i

R pT pT

K K p p (2.16)

Chn = siT ta c b iu chnh dng:

iR1

12

u u

bd i si u

R T

K K T pT (2.17)

Ri l khu t l tch phn PI

- H s khuch i dng:

Chn Uid = 5(V)

Ki = UidIkm

= 5

8,470,59

Tsi = Ti + Tk + Tv = 0,001 + 0,0001 + 0,001 = 2,1.10-3 (s)

- Hm truyn ca b iu chnh dng in l:

iR1

12

u u

bd i si u

R T

K K T pT =

4

3 4

15,7.5,796.10 11

2.4,28.8,47.2,1.10 5,796.10

= 0,06. 103,52s

s

- Khu phn hi dng in: pT

K

i

i

1 =

8,47

1 0,001.s (2.18)

29

2.3. TNG HP MCH VNG TC

Ta c s mch vng tc :

Hnh 2.5: S mch vng tc ng c in mt chiu

Trong :

. .

u

m c

R

K T p : Hm truyn ca i tng iu khin

2.

( )

uc

m

R JT

K: Hng s thi gian c hc ca ng c

pT

K

1 : Hm truyn ca my pht tc

K : H s khuch i ca my pht tc

T : Hng s thi gian ca my pht tc

S thu gn:

Hnh 2.6: S thu gn ca mch vng tc

Trong :

0S =.

. .(1 2 )(1 )i m c si

K Ru

K K T p T p T p (2.19)

V siT v T rt nh nn ta b qua cc thnh phn bc cao:

0

.

. . 1 (2 ) )i m c si

K RuS

K K T p T T p (2.20)

30

t: sT =2 siT + T v .

.

u

i m

R KK

K K

Ta c: So = K

Tc.p(1+Ts.p) (2.21)

p dng tiu chun mdul i xng ta c hm truyn ca h thng:

FMC = 3322 8841

41

ppp

p (2.22)

M MCK FF0

0

1 SR

SR R

)1(

11

0 MCFS (2.23)

Vy R = Tc

2.K.Ts.1+4.Ts.p

4.Ts.p (2.24)

- H s khuch i ca my pht tc:

Chn dU = 5 (V)

K = dU

=

5

230 = 0,022

Ts = 2.Tsi+ T = 2.2,1.10-3 + 0,01 = 0,0142 (s)

- Hm truyn ca b iu chnh tc :

R = Tc

2.K.Ts.1+4.Ts.p

4.Ts.p (2.25)

Vi : . 15,7.0,022

1,1. 8,47.0,037

u

i m

R KK

K K

6

2 2

. 15,7.1,18.100,021

0,037m

uc

R JT

K

Vy : 0,021 1 0,672. 11,83

12.1,1.0,0142 4.0,0142.

sR

s s

- Khu phn hi tc :

pT

K

1 =

0,022

0,01. 1s (2.26)

31

2.3.1. M hnh mch vng tc khi c mch vng dng in

Hnh 2.7: M hnh mch vng tc khi c mch vng dng in

+) p ng dng in:

Hnh 2.8: p ng dng in ca ng c khi c 2 b iu khin

32

+) p ng tc

Hnh 2.9: p ng tc ca ng c khi c 2 b iu khin

Nhn xt: Khi c 2 b iu khin l mch vng dng in v mch vng

tc th sau khi khi ng khong 0,6s th dng in v tc ng c dn n

inh. Sau khi ng c nhn ti th khong 0,3s tc ca ng c v dng in

ca ng c cng t gi tr nh mc.

2.3.2. M hnh mch vng tc khi b qua mch vng dng in

S thu gn ca m hnh mch vng tc khi b qua mch vng dng in:

Hnh 2.10: S thu gn ca mch vng tc khi b qua mch vng dng

Khi b qua mch vng dng in th hm truyn ca i tng:

0

. .

. . .(1 . )(1 . )(1 . )(1 . )

bd m

u u dk v

K K KS

R J s pT T s T p T p (2.27)

V ivdk TTT ;; s thi gian rt nh nn b qua thnh phn bc cao l cc hng s

Vy: 0. .

. . .(1 . ) 1 ( ).

bd m

u u dk v

K K KS

R J s p T T T T p (2.28)

33

t: dk vT T T T

p dng tiu chun ti u modul ta c hm truyn ca h thng:

MCF 2 21

1 2 2p p (2.29)

M MCK FF0

0

.

1 .

R S

R S 1

0

1

( 1)MCR

S F (2.30)

Chn = dk vT T T T = 0,0001 + 0,001 + 0,01 = 0,0111 (s)

- Vy hm truyn ca b iu khin tc :

. . . 11

2. . . . .

u u

bd m u

R T J sR

K K K T p T (2.31)

4 6

4

15,9.5,796.10 .1,18.10 . 11

2.4,28.0,037.0,022.0,011 .5,796.10

s

s

39,354 9,12.10 s

- H s khuch i ca my pht tc:

Chn dU = 5 (V)

K = dU

=

5

230 = 0,022

- Khu phn hi tc :

0,022

1 . 1 0,01

K

T p s (2.32)

Hnh 2.11: M hnh mch vng tc khi b qua mch vng dng in

34

+) p ng dng in:

Hnh 2.12: p ng dng in ca ng c khi c mch vng tc

+) p ng tc :

Hnh 2.13: p ng tc ca ng c khi c mch vng tc

Nhn xt: Lc u gi tr dng in v tc ca ng c tng ln v dn n

nh, sau khi nhn ti khong 0,6s th th tc v dng in ca ng c gn

nh khng cn dao ng na v t gi tr nh mc. Khi c b iu khin tc

th thi gian tc ng c n nh c rt ngn i rt nhiu so vi

khi cha c b iu khin.

35

CHNG 3. XY DNG M HNH B IU KHIN PID IU

KHIN NG C IN MT CHIU

3.1. NGUYN L XY DNG B IU KHIN PID

- Lut iu khin t l P

Tn hiu iu khin U(t) t l vi tn hiu vo e(t).

Phng trnh vi phn m t ng hc:

U(t) = Km.e(t) (3.1)

Trong :

U(t): Tn hiu ra ca b iu khin

e(t): Tn hiu vo

Km: H s khuch i ca b iu khin

Xy dng bng s thut ton:

Hnh 3.1: S khuch i thut ton biu din lut iu khin t l

+) u im: B iu khin c tnh tc ng nhanh khi u vo c tn hiu sai

lch th tc ng ngay tn hiu u ra.

+) Nhc im: H thng lun tn ti sai lch d, khi tn hiu sai lch u vo

ca b iu khin b th khng gy tn hiu tc ng iu khin, mun khc

phc nhc im ny th ta phi tng h s khuch i Km. Nh vy h thng s

km n nh.

- Lut iu khin tch phn I

Tn hiu iu khin U(t) t l vi tch phn ca tn hiu u vo e(t).

Phng trnh vi phn m t ng hc:

U(t) = 0 0

1( ). ( ).

t t

i

e t dt e t dtT

(3.2)

36

Trong :

U(t): Tn hiu iu khin

e(t): Tn hiu vo ca b iu khin

Ti: Hng s thi gian tch phn

+) Xy dng s mch khuch i thut ton

Hnh 3.2: S mch khuch i thut ton biu din lut iu khin tch phn

Ta c: 1

. .

r

v

U

U R C p (3.3)

+) u im: B iu khin tch phn loi b c si lch d ca h thng, t

chu nh hng tc ng ca nhiu cao tn.

+) Nhc im: B iu khin tc ng chm nn tnh n nh ca h thng.

- Lut iu khin vi phn D

Tn hiu ra ca b iu khin t l vi vi phn tn hiu vo.

Phng trnh vi phn m t ton hc:

( )( ) d

de tU t T

dt (3.4)

Trong :

e(t): Tn hiu vo ca b iu khin

U(t): Tn hiu iu khin

Td: Hng s thi gian vi phn

+) Xy dng s bng khuch i thut ton

37

Hnh 3.3: S khuch i thut ton biu din lut iu khin vi phn

Ta c:

( ).

.

r

r

v

dU tU RC

dt

URC p

U

(3.5)

+) u im: Lut iu khin vi phn p tnh tc ng nhanh y l mt c

tnh m trong iu khin t ng thng rt mong mun.

+) Nhc im: Khi trong h thng dng b iu khin c lut vi phn th h

thng d b tc ng bi nhiu cao tn. y l loi nhiu thng tn ti trong

cng nghip.

Cc lut t l, vi phn, tch phn thng tn ti nhng nhc im ring.

Do vy khc phc cc nhc im trn ngi ta thng kt hp cc lut

li c b iu khin loi b cc nhc im , p ng cc yu cu k thut

ca cc h thng trong cng nghip.

- B iu khin t l tch phn PI

Phng trnh vi phn m t quan h tn hiu vo, ra ca b iu khin.

1 2

0

0

( ) . ( ) . ( )

1( ) ( ) ( )

t

t

m

i

U t K e t K e t dt

U t K e t e t dtT

(3.6)

Trong :

e(t): L tn hiu vo ca b iu khin

U(t): L tn hiu ra ca b iu khin

Km = K1: L h s khuch i

38

1

2

i

KT

K: L hng s thi gian tch phn

+) Xy dng bng s khuch i thut ton

Hnh 3.4: S khuch i thut ton biu din b iu khin PI

Ta c:

1

2 3 3 0

1 2

2 3 3 1

1. ( )

.

(1 ). . .

t

r v v

r

v

RU U U t dt

R R C

U R R

U R R C R p

(3.7)

- B iu khin t l vi phn PD

Phng trnh vi phn m t quan h vo ra ca b iu khin

1 2

( )( ) . ( ) .

( )( ) ( )m d

de tU t K e t K

dt

de tU t K e t T

dt

(3.8)

Trong :



Km = K1: L h s khuch i

39

2

1

d

KT

K: L hng s thi gian vi phn

+) Xy dng bng s khuch i thut ton

Hnh 3.5: S khuch i thut ton biu din b iu khin PD

Trong :

1

2

21

2 1

. . .

. .1 .

vr v d d

d dr

v

dURU U R C

R dt

R R CU Rp

U R R

(3.9)

Tn hiu ra ca b iu khin lch pha so vi tn hiu vo mt gc , y

l c im tc ng nhanh ca h thng. Khi h thng s dng b iu khin t

l vi phn d b tc ng bi nhiu cao tn, tn ti sai lch d, nhng li p ng

c tnh tc ng nhanh. Nn b iu khin ny thng c s dng trong h

thng t c nhiu cao tn v cn tnh tc ng nhanh.

Bng thc nghim hoc l thuyt ta xc nh cc tham s Td, Km b iu

khin p ng c tnh h thng.

- B iu khin t l vi tch phn PID

ci thin cht lung ca cc b iu khin PI, PD ngui ta kt hp ba

lut iu khin t l, vi phn, tch phn tng hp thnh b iu khin t l vi

40

tch phn (PID). C c tnh mm do ph hp cho hu ht cc i tng trong

cng nghip[5].

Phng trnh vi phn m t quan h tn hiu vo ra ca b iu khin:

1 2 3

0

0

( )( ) . ( ) . ( ) .

1 ( )( ) ( ) ( )

t

t

m d

i

de tU t K e t K e t dt K

dt

de tU t K e t e t dt T

T dt

(3.10)

Trong :



Km = K1: H s khuch i

3

1

d

KT

K : Hng s thi gian vi phn

1

2

i

KT

K : Hng s thi gian tch phn

+) Xy dng bng khuch i thut ton

Hnh 3.6: S khuch i thut ton b iu khin PID

41

Ta c:

1

2 3 3 0

21 2

2 1 1 3 3

1. . . ( ).

.

. .1 .

. . .

t

vr v d d v

d dr

v

dURU U R C U t dt

R dt R C

R R CU R Rp

U R R R R C p

(3.11)

+) c tnh lm vic ca b iu khin PID rt linh hot, mm do. gii tn s

thp th b iu khin lm vic theo quy lut t l tch phn. gii tn s cao th

b iu khin lm vic theo quy lut t l vi phn khi 1

.i dT T b iu khin

lm vic theo quy lut t l.

+) B iu khin c ba tham s Km, Ti v Td.

Khi ta cho Ti = , Td = 0 th b iu khin lm vic theo lut t l

Khi Ti = b iu khin lm vic theo lut t l - vi phn

Khi Td = 0 b iu khin lm vic theo lut t l - tch phn

Tn hiu ra ca b lch pha so vi tn hiu vo mt gc , y l c tnh mm

do ca b iu khin. Nu ta chn c b tham s ph hp cho b iu khin

PID th h thng cho ta c tnh nh mong mun, p ng cho cc h thng

trong cng nghip. c bit nu ta chn b tham s tt b iu khin s p ng

c tnh tc ng nhanh, y l c im ni bt ca b iu khin.

Trong b iu khin c thnh phn tch phn nn h thng trit tiu c sai

lch d. Bng thc nghim hoc l thuyt ta xc nh cc tham s Km, Ti ,Td

b iu khin p ng c tnh h thng.

3.2. XY DNG M HNH VT L

3.2.1. S khi ca h thng

Hnh 3.7: S khi ca b iu khin ng c mt chiu

42

Mch iu khin tip nhn gi tr in p t v gi tr in p phn hi t

my pht tc, sau x l tn hiu v cp tn hiu xung PWM v tn hiu nhn

bit chiu ca ng c vo mch cng sut iu khin ng c.

3.2.2. Xy dng mch cng sut

- Gii thiu IC MC33883

MC33883 l mt IC kch FET chuyn dng iu khin cu H.

+) in p ngun VCC2 cung cp cho IC t 5,5V n 28V

+) in p ngun VCC t 5,5V n 55V

+) Hot ng nhit t -40o C n 125o C

+) C th p ng tn s bm xung PWM ln n 100Khz

+) S chn ca MC33883

Hnh 3.8: S chn ca MC33883

Bng 3.1: M t cc chn chc nng ca MC33883

Chn K hiu Chc nng

1, 11 VCC, VCC2 Chn cp ngun cho thit b

2, 13 C2, C1 Hai chn ny ni vi mt t in

4, 19 SRC_HS1, SRC_HS2 Ngun u ra ca MOSFET

5, 18 GATE_HS, GATE_HS2 Hai chn ny ni vi chn iu khin

ca MOSFET

6, 17 IN_HS1, IN_HS2 u vo tn hiu logic, tc ng ln

chn iu khin ca MOSFET

7, 16 IN_LS1, IN_LS2 Hai chn u vo tn hiu logic, dng

iu khin cc cng ca MOSFET

43

8, 15 GATE_LS1, GATE_LS2 u vo chn iu khin ca

MOSFET

9, 12, 14 GND1, GND_A, GND2 Chn ni t ca thit b

20 G_EN Chn chung cho php

Hnh 3.9: S dng MC33883 iu khin 4 MOSFET

- MOSFET IRF540

IRF540 l MOSFET knh dn loi N c thng s nh sau:

+) in p nh mc 100 V

+) Dng in nh mc 25o C l 33 A

+) Dng in nh l 100 A

+) in p iu khin VGS l 20 V

+) in p ngng iu khin VGS t 2 n 4 V

+) in p ri trn van khi dn hon ton l 1,2 V

+) Dng r trng thi kha ID look l 25 A

+) Din tr khi dn RDS on l 44 m

+) Dng r cc G trng thi kha IG look l 100 nA

+) Tc tng trng in p du

dt l 7 V/ns

+) Nhit lm vic 175o C

44

3.2.3. Khu phn hi tc

- Hi tip tc l mt phn khng th thiu trong h thng iu khin tc

ng c. iu khin c tc ca ng c ngi ta phi bit c gi tr

u ra ca h thng c nhng iu chnh thch hp u vo t c tc

mong mun. phn hi tc ngi ta thng hay dng my pht tc v

cc b o tc xung s.

- My pht tc l my in nh, lm vic ch my pht v thc hin chc

nng bin i chuyn ng ca trc thnh tn hiu in p ra.

Phng trnh c tnh ca my pht tc:

UF = K.n = K1. d

dt (3.12)

Trong :

UF: L in p ra ca my pht tc

K, K1: L h s khuch i

n: L vn tc quay ca roto (vng/pht)

: L gc quay

Cc yu cu i vi my pht tc l: tuyn tnh ca c tnh cao, h s

khuch i K = UFn

ln, in p ra phi i xng.

My pht tc mt chiu c cu trc v nguyn l hot ng nh my in

mt chiu cng sut nh, kch thch bng nam chm vnh cu hoc kch thch

c lp. Yu cu i vi my pht tc mt chiu l in p mt chiu c cha t

thnh phn xoay chiu tn s cao v t l vi tc ng c, khng b tr nhiu

v gi tr v du so vi bin i i lng o. in p mt chiu pht ra khng

ph thuc vo ti v nhit , m bo yu cu trn my pht tc mt chiu

phi c t thng khng i trong ton vng iu chnh tc . V vy phi hn

ch tn tht mch t bng vic s dng vt liu t c t tr hp v s dng l

thp k thut in mng (hn ch tn tht dng in xoay).

Nhc im ca my pht tc mt chiu l chnh xc ph thuc vo

ph ti. Mt khc nhit cun dy thay i nh hng ti in tr phn ng

45

my pht lm in p ra ca my pht thay i (do in p ri mch phn ng

thay i). in p u ra ca my pht cn b thay i do in tr ca chi than.

nh hng ca phn ng phn ng ti h s t l (nht l khi vng tc cao).

3.2.4. Xy dng mch iu khin

Hnh 3.11: Cu trc mch iu khin ng c mt chiu

Tn hiu t v tn hiu phn hi t my pht tc c a qua mt mch

tr, in p u ra ca mch tr s c a vo b iu khin. in p iu

khin Udk s c a n khu so snh, khu so snh s so snh in p iu

khin v in p rng ca to ra tn hiu xung, tc ng vo van cng sut.

- Mch iu khin s dng khuch i thut ton (operational amplifier), thng

c gi tt l op-amp l mt mch khuch i "DC-coupled" (tn hiu u vo

bao gm c tn hiu BIAS) vi h s khuch i rt cao, c u vo vi sai, v

thng thng c u ra n. Trong nhng ng dng thng thng, u ra c

iu khin bng mt mch hi tip m sao cho c th xc nh li u ra,

tng tr u vo v tng tr u ra. Cc mch khuch i thut ton c nhng

ng dng tri rng trong rt nhiu cc thit b in t thi nay t cc thit b

in t dn dng, cng nghip v khoa hc.

- Vi mch khuch i thut ton LM358 xy dng b iu khin PID

Hnh 3.12: S chn ca vi mch LM358

46

LM358 l mt vi mch tch hp sn 2 khuch i thut ton. Ngun cung

cp cho LM358 tm t 3V~32V, p ti a ng vo t 0~32V i vi ngun n

v cng tr 16V i vi ngun i. y l mch khuch i c hi tip v c

in tr rt cao, cho nn khng lm nh hng xu n tn hiu cm bin, c kh

nng chng nhiu cao.

+) li khuch i in p DC ca LM324 ti a khong 100 dB.

+) Tn s hot ng ca LM324 l 1MHz.

- Khu in p t

Hnh 3.13: S nguyn l khi iu chnh in p t

Ta c th thay i gi tr in p t thng qua bin tr R4. in p t

qua mt mch lc RC, nh vy ta s c ng c tnh in p t mt hn.

Ta chn: Gi in p t U = 5V, R1 = 50 (K)

R4 = 50 (K), C9 = 1(F), R2 = R9 = 10 (K)

- Khu mch tr

47

Hnh 3.14: S nguyn l mch tr

Ta c: UphVnR10

+ U-Vn

R12 = 0

Chn R10 = R12 ta c: U = 2.Vn - Uph

M Vp = Uset.R13R13+R11

, chn R13 = R11 ta c: Vp = Uset2

M Vn = Vp vy: U = Uset - Uph

Ta chn: R10 = R11 = R12 = R13 = 10 (K)

- Mch iu khin PID

48

Hnh 3.15. S nguyn l mch PID

Khi PID gm 3 khu: T l, tch phn v vi phn. Gi tr in p ra t

mch tr c a vo khi ny, u ra ca khi PID qua mt mch cng th ta

s c in p iu khin.

+) Khu t l: Ta s dng 2 bin tr R3 v R8 thay i h s P ca b iu

khin, s dng kha sw1 ta c th ngt c b ny ra khi mch iu khin

Theo tnh cht ca khuch i thut ton h s khuch i ca khu t l:

K = UpU

= - R3R8

in p ra ngc pha so vi in p vo, bin tr R3 gy ra hi tip m song

song theo in p lm cho h s khuch i gim xung.

49

Vy Up = - U. R3R8

(3.14)

Ta chn R3 = 100 (K) v R8 = 10 (K)

+) Khu tch phn: Ta s dng bin tr R5 thay i h s I ca b iu khin,

s dng kha sw2 ta c th ngt c b ny ra khi mch iu khin.

Theo tnh cht ca khuch i thut ton ta c:

UI = 1

R5.C7

U.dt (3.15)

in p ra t l vi tch phn in p vo

R5.C7 gi l hng s tch phn

Ta chn R5 = 100 (K) v C7 = 10 (F)

+) Khu vi phn: Ta s dng bin tr R6 thay i h s D ca b iu khin,

s dng kha sw3 ta c th ngt c b ny ra khi mch iu khin.

Theo tnh cht ca khuch i thut ton ta c:

UD = C8.R6.dU

dt (3.16)

in p ra t l vi tch phn in p vo

C8.R6 gi l hng s vi phn

Ta chn R6 = 50 (K) v C8 = 100 (nF)

- Mch cng in p

Hnh 3.16: S nguyn l mch cng in p

50

Mch cng in p thc hin nhim v cng gi tr in p Up, UI, UD li

Nu R15 = R16 = R17 = R18, theo tnh cht ca khuch i thut ton ta c:

Uk = - (UP + UI + UD) (3.17)

Ta chn: R15 = R16 = R17 = R18 = 10 (K)

- Khu nhn bit chiu ca tn hiu iu khin

Hnh 3.17: S nguyn l khu nhn bit chiu ca tn hiu iu khin

Ta chn R23 = 4,7 (K) v D1 l diode zener loi DZ5V1.

- Mch tch tn hiu chiu v ln tn hiu iu khin s dng IC CD4052

CD4052B l mt b dn knh - phn knh 4 knh tng t. C hai ng

chn u vo nh phn l A v B, v mt hn ch u vo. Hai tn hiu u vo

la chn 1 trong 4 cp knh phi c bt v kt ni cc yu t u vo tng

t v ra s c u ra.

+) S chn ca CD4052

Hnh 3.18: S chn ca CD4052

51

+) Khi tch knh d liu vo chn COM OUT/IN, ra 4 knh CHANNEL I/O.

Ngc li, khi dn knh th d liu song song vo cc chn CHANNEL

OUT/IN v ra chn COM OUT/IN.

+) 2 ng chn l A, B

+) Chn INH (inhibit) cho php d liu c php truyn ra

Hnh 3.19. Cu trc mch ca CD4052

Bng 3.2. Hot ng ca CD4052

52

Hnh 3.20: S nguyn l mch tch tn hiu iu khin dng CD4052

in p iu khin c a vo chn 13 ca CD4052, thc hin tch

knh d liu. in p iu khin c tch xang 2 knh X0 v X1. Nu khng

c tn hiu bt iu khin, chn X0 s c ni vi Y0 v ta c u ra Y. Nu c

tn hiu bit iu khin th chn X1 s c ni vi Y1, in p iu khin c th

m ln t chn X1 in p iu khin s c qua mt mch khuch i thut

ton o, nh vy ta s c u ra Y.

Chn R21 = R22 = 10 (K )

- Mch to xung dao ng dng IC NE555

+) 555 l mt loi linh kin kh l ph bin by gi vi vic d dng to c

xung vung v c th thay i tn s ty thch, vi s mch n gin, iu

ch c rng xung. N c ng dng hu ht vo cc mch to xung ng

ct hay l nhng mch dao ng khc. y l linh kin ca hng CMOS sn

xut[1].

in p u vo: 2 - 18V (Ty tng loi LM555, NE555, NE7555)

53

Dng in cung cp: 6mA - 15mA

in p logic mc cao: 0,5 - 15V

in p logic mc thp: 0,03 - 0,06V

Cng sut ln nht l: 600mW

+) S chn ca NE555

Hnh 3.21: S chn ca NE555

IC NE 555 gm c 8 chn

+) Chn s 1 (GND): Cho ni GND ly dng cp cho IC hay cn gi l chn

chung.

+) Chn s 2(TRIGGER): ng vo ca 1 tn so p. Mch so p dng cc

transistor PNP, mc p chun l 2.Vcc/3.

+) Chn s 3(OUTPUT): Chn ny l chn dng ly tn hiu ra logic. Trng

thi ca tn hiu ra c xc nh theo mc 0 v 1, 1 y l mc cao n tng

ng vi gn bng Vcc nu (PWM=100%) v mc 0 tng ng vi 0V nhng

m trong thc t mc 0 ny ko c 0V m n trong khong t (0.35 ->0.75V).

+) Chn s 4(RESET): Dng lp nh mc trng thi ra. Khi chn s 4 ni

masse th ng ra mc thp. Cn khi chn 4 ni vo mc p cao th trng thi

ng ra ty theo mc p trn chn 2 v 6. Nhng m trong mch to c dao

ng thng hay ni chn ny ln VCC.

+) Chn s 5 (CONTROL VOLTAGE): Dng lm thay i mc p chun trong

IC555 theo cc mc in p ngoi hay dng cc in tr ngoi cho ni GND.

Chn ny c th khng ni cng c nhng m gim tr nhiu ngi ta

thng ni chn s 5 xung GND thng qua t in t 0.01uF n 0.1uF cc t

ny lc nhiu v gi cho in p chun c n nh.

54

+) Chn s 6(THRESHOLD) : l mt trong nhng chn u vo so snh in p

khc v cng c dng nh 1 chn cht.

+) Chn s 7(DISCHAGER) : c th xem chn ny nh 1 kha in t v chu

iu khin bi tng logic ca chn 3. Khi chn 3 mc p thp th kha ny

ng li.ngc li th n m ra. Chn 7 t np x in cho 1 mch RC lc IC555

dng nh mt mch dao ng.

+) Chn s 8 (Vcc): l chn cung cp p v dng cho IC hot ng. Khng

c chn ny coi nh IC cht. N c cp in p t 2V ->18V (Ty tng loi

555 thp nht l NE7555).

Hnh. 3.22: Mch to dao ng dng NE555

Khi t C4 np in ta c: T1 = 0,693.C4.(R7 + R27)

Khi t C4 phng in ta c: T2 = 0,693.C4.R27

Vy chu l xung l: T = T1 + T2 = 0,693.C4.(R7 + 2.R27)

to dao ng c tn s 10Khz, tc l chu k dao ng T = 1

f =

1

104 = 10

-4 s

Ta chn: R7 = 50 (K ), R27 = 100 (K ), C4 = 1 (nF)

55

Vy: T1 = 0,639.10-9 .(50000 + 100000) = 1,04.10

-4 (s)

T2 = 6,93.10-9 .100000 = 6,93.10

-5 (s)

- Mch to xung rng ca dng kha Transistor

Hnh 3.23: Mch to xung rng ca dng Transistor

Khi transistor m, t C3 phng in qua transistor, Uc = 0. Khi transistor

kha t C3 np in t +12V qua R29, in p trn t thay i theo quy lut hm

m vi hng s thi gian = R29.C3 [1].

Uc = 12.(1 - e

t

) (3.18)

ly on tuyn tnh ca in p trn t c th chn T = 1

3.

+) Chn transistor l loi A1015 c cc thng s sau:

Ic = 150 mA = 0,15 (A)

VCB0 = -50 (V)

VCE0 = -50 (V)

Pcmax = 400 (mW)

Tn s hot ng 1 kHz

+) Dng in cc i qua Baz l IB = IC

HFE =

0,15

90.1,2 = 2 (mA)

56

M IB = 12-0,7

R28+R29 Vy R28 + R29 = 5650 ( )

Ta c Un = 12V, T = 10-4 s, vy R29.C3 = 3.10

-4

Chn R 28 = 3 (K ), R29 = 3 (K ), t C3 = 0,1 (F)

- Mch so snh

Hnh 3.24: Mch so snh in p

y l mch so snh hai in p vo l: in p rng ca v in p

iu khin Uk (ly t bn ngoi vo)[1]

Ti thi im bng nhau v gi tr tuyt i ca 2 in p ny, trong phn

sn s dng ca in rng ca th mch pht ra mt xung in p, xung ny

c a qua khi to xung n c th thay i c di cng sut, dc

sn trc. C ngha l khi so snh l ni quyt nh gi tr gc iu khin

th so snh in p:

57

Hnh 3.25: th so snh in p

Mun xc nh c thi im m van cng sut ( gc m ) th ta tin

hnh so snh hai tn hiu Uk v Urc. in p rng ca c a vo ca o ca

khu khuch i thut ton qua R25 so snh vi in p iu khin c a

vo ca khng o, in p iu khin c a vo ca khng o ca khuch

i thut ton qua R24.

+) Nu Uc < Uk th tn hiu ra l dng Ur > 0.

+) Nu Uc > Uk th tn hiu ra l m Ur < 0.

+) Nu Uc = Uk th l thi im pht xung m van cng sut. Vy u

ra ca khuch i thut ton l mt chui xung m dng lin tip. Mun thay

i gc m ca van cng sut th ta thay i gi tr ln ca in p iu

khin Uk.

+) it D2 dng loi b phn xung m. V vy in p ra ch cn phn xung

dng.

+) Tnh ton khu so snh

Chn in tr R24 = R25 = R26 = 4,7 (k )

it D2 dng gii hn in p u ra chn loi DZ5V1

58

- Xy dng mch iu khin

Hnh 3.26: S nguyn l khu in p t v mch tr

59

Hnh 3.27: S nguyn l b iu khin PID

60

Hnh 3.28: S nguyn l khu nhn bit chiu v tch in p iu khin

Hnh 3.29: S mch to xung rng ca v khu so snh in p

61

3.3. KT QU THC NGHIM

Sau hn ba thng nghin cu em hon thnh n tt nghip: Nghin

cu tng quan v h truyn ng in mt chiu, i su xy dng b iu khin

PID cho ng c in mt chiu vi cc kt qu t c nh sau:

- Tm hiu tng quan v ng c mt chiu

- Cc phng php iu khin tc ng c mt chiu

- Xy dng m hnh h truyn ng in mt chiu trn Matlab & Simulink

- Xy dng b iu khin PID ng dng cho ng c mt chiu.

Hnh 3.31: M vt l iu khin tc ng c in mt chiu

62

KT LUN

ti iu khin ng c mt chiu s dng b iu khin PID tuy khng

phi l mt ti mi, nhng qua phn nh c tnh nghim tc trong

vic hc hi v vn dng cc kin thc vo vic thc hin ti.

Sau thi gian ba thng nghin cu em hon thnh ti vi cc kt qu

t c nh sau: Tm hiu tng quan v ng c mt chiu, cc phng php

iu khin tc ng c mt chiu, xy dng m hnh h truyn ng in

mt chiu trn Matlab & Simulink v l thuyt iu khin t ng t lm c

s cho vic xy dng b iu khin PID ng dng cho ng c mt chiu.

Tuy nhin bn n vn cn mt s vn tn ti, hn ch cn gii

quyt:

+) Vic kim sot cc tham s ca b iu khin PID l kh kh khn.

+) Cha quan st c mt cch trc quan tc ng c trn my tnh.

Do vy, hng pht trin tip theo ca ti s l:

+) ng dng cm bin o dng in ACS712 xy dng h thng iu khin

gm 2 mch vng tc v dng in cho ng c in mt chiu.

+) Thit k giao din trn my tnh cho php quan st c p ng tc .

+) Xy dng b iu khin PID cho h thng iu khin v tr.

63

TI LIU THAM KHO

1. Nguyn Bnh (1996), in t cng sut. NXB Khoa Hc K Thut

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

(2008), iu chnh t ng truyn ng in. NXB Khoa hc v k thut

3. Bi Quc Khnh, Nguyn Vn Lin (2005), C S Truyn ng in. NXB

Khoa hc v k thut

4. Nguyn Phng Quang (2006), Matlab & Simulink dnh cho k s iu

khin t ng. NXB Khoa hc v k thut

5. Design PID circuit - http://www.ecircuitcenter.com/Circuits/op_pid.htm

11_LuuDucTuong_DC1201.pdf

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