doan plc

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an mn hoc: iu khin Logic GVHD: Ks.Nguyn Manh Ha

Phn 1: KHI NG NG C IN MT CHIUKCH T C LP

1. Gii thiu v ng c in mt chiu v Ml :

ng c in mt chiu l loi ng c hot ng bng dng in mtchiu. u im ni bt ca ng c in mt chiu so vi cc ng c inxoay chiu l d iu chnh tc v mmen. ng c in mt chiu cng dng kh rng ri nh trong tu in ngm, my khoan, my cn, bngti, cn trc, qut, cnh tay robot,

ng c in mt chiu kch t c lp (Ml) l loi ng c c mchphn ng v mch kch t c cp in c lp nhau. Cc thng s nhmc ca Ml l cng sut Pm (kW), tc nm (v/ph), mmen Mm (Nm),tn s lm vic fm (Hz), in p Um, dng in phn ng I (A) v in tr

phn ng R ( ).

Hnh 1.1

2. Vn khi ng ca Ml : m my ng c in mt chiu c tt, phi thc hin nhng yu

cu sau y:1. M men m my (hay khi ng) Mk phi c tr s cao nht c th

c hon thnh qu trnh m my, ngha l t c tc quy nh trongthi gian ngn nht.

2. Dng in m my (hay khi ng) Ik phi hn ch n mc nh

nht trnh cho dy qun khi b chy hoc nh hng xu n i chiu.Trong khun kh yu cu trn, ngi ta p dng ba phng php m

my sau y:1). M my trc tip.2). M my dng in tr ph.3). M my bng in p thp ( U < Um ).

Khi khi ng trc tip Ml bng cch ng trc tip ng c vo li in,do sut in ng phn ng E = 0 nn dng in phn ng :

f f

U E UI

R R R R

= =

+ +Trong :

SVTH: Nguyn Manh Ha - L p 05SK1 Trang1

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CKT

K1G 2G 3G

Eo

a b

a

b c

d e

fg

1

2

3

I2 I1Ic

xl

1

2

3TN

Rf1 Rf2 Rf3

O


- U l in p mch phn ng.- R l in tr cun dy phn ng.- Rf l in tr ph mch phn ng.Dng I c gi tr rt ln, gp 10 n 20 ln dng in nh mc. Dng

in ny s gy t nng ng c, sinh ra lc in ng ln lm ph hy qu

trnh c ca my v lm cho cc thit b bo v nh cu ch, aptmt tc ngnhm.

m bo an ton cho my, thng thng ngi ta s gii hn dng khing 2,5 ln dng nh mc bng cch a in tr ph vo mch phnng khi khi ng v sau loi b dn in tr ph a tc ng cln xc lp. Phng php ny gi l khi ng bng in tr ph:

Ikb = Inm =ufu

dm

RR

U

+= ( 22,5 ) Im Icp

3. c tnh khi ng ca Ml :

Hnh 1.

Xt Ml khi ng vi 3 cp in tr ph Rf1, Rf2 v Rf3 (xem hnh 1.2.a).Dng khi ng ca ng c c gii hn trong khong 2 1kI I I

(I2=1.1 1.3Ic, I1=2 2.5Im).c tnh khi ng s c dng nh hnh 1.2.b.

Ti a, ng c bt u khi ng vi ton b in tr phRf1+Rf2+Rf3. Tc ng c bt u tng dn ln (on ab). n b, tc ng c l 1, cp in tr ph th nht c ngn mch (Rf1 = 0) ang c ln tc cao hn, in tr ph mch phn ng cn li l Rf2 + Rf3.Do qun tnh nn tc ng c khng tng lin nn c on ngang (bc) sau mi tng tc (on cd). Ti c, tc ng c l 2, cp in tr ph thhai c ngn mch (Rf2=0), ch cn li in tr ph Rf3 trong mch phnng. Do qun tnh nn c on ngang (de) sau ng c tng tc ln tc 3 (on ef). Ti f, cp in tr cui cng c ngn mch (Rf3 = 0). Tng

t s c on ngang (fg) trc khi ng c tng tc ln tc xc lp ( xl),kt thc qu trnh khi ng.


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4. Tnh cc gi tr ca in tr khi ng :Gi s ng c khi ng vi 3 cp in tr ph. c tnh khi ng

u tin v dc nht l ng 1 (xem hnh 1.2.b), sau n ng 2, 3 ngvi cp in tr ph th 2, 3 v cui cng l c tnh c t nhin.

in tr tng ng vi mi c tnh c l:

R1 = R + Rf (1) = R + ( Rf1 + Rf2 + Rf3 )

R2 =R + Rf (2) = R +( Rf2 + Rf3 )

R3 =R + Rf (3) = R +(Rf3 )

- Ti b, ta c 121

mU EIR

=

- Ti c, ta c1

1

2

mU EIR

=

Trong qu trnh khi ng, 1

2

Iconst

I= =

1 1 2

2 2 3

I R R

I R R= = =

3

2

2 3

3

1 2

R R

R R R

R R R

=

= =

= =

13 31.

m

R U

R R I= =

Trong 11

mURI

=

Cc gi tr in tr ph ca tng cp l:

3 3

2 2 3

21 1 2

( 1)

( 1)

( 1)

f

f

f

R R R R

R R R R

R R R R

= =

= =

= =

5. Qu trnh qu c ca Ml :


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Xt QTQ khi khi ng Ml vi 3 cp in tr ph (xem hnh 1.2a), s c4 giai on QTQ khi ng :

1

M

2

3

b

a

c

d e

f g

1

2

3

TN

Mc M2 M1

xl

O

I

O t

xl

I2

Ic

I1

t1 t2 t3 t4

a) b)Hnh 1.3

a). Giai on 1: on ab, c tnh 1 :

f f1 f2 f3 1 f1 f2 f3R =R +R +R R =R +R +R +R

Theo c tnh 1, cng c tnh c:2

1 21

1 1

( )k M M

R

= =

- Hng s thi gian c:1 2 3

1 2 2

1

1

( )( )

( ) ( )

f f f

c

J R R R RJ JT s

k k

R

+ + += = =

- iu kin ban u, im a: b1 = 0, Mb1 = M1

- iu kin xc lp: xl1 (xc nh theo c tnh c), Mxl1 = M2

- Phng trnh QTQ trong giai on 1 l:

1

1

1

1

(1 )

( )

c

c

tT

xl

tT

c c

e

M M M M e

=

= +

- Khi = 1, QTQ s chuyn sang giai on 2 .

b). Giai on 2: on bcd, c tnh 2 :

f f2 f3 2 f2 f3R =R +R R =R +R +R


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- Theo c tnh 2, cng c tnh c:2

1 22

2 1 2

( )k M M

R

= =

- Hng s thi gian c:

2 3

2 2 2

2

2

( )( )

( ) ( )

f f

c

J R R RJ JT s

k kR

+ += = =

- iu kin ban u, im c: b2 = 1, Mb2=M1


- Phng trnh QTQ trong giai on 2 l :

2

2

2 1 2

1

( ) )

( )

c

c

tT

xl xl

tT

c c

e

M M M M e

= +

= +

- Khi = 2, QTQ s chuyn sang giai on 3

c). Giai on 3: on def, c tnh 3 :

f f3 3 f3R =R R =R +R

- Theo c tnh 3, cng c tnh c:2

1 23

3 2 3

( )k M MR

= =

- Hng s thi gian c:3

3 2 2

3

3

( )( )

( ) ( )

f

c

J R RJ JT s

k k

R

+= = =

- iu kin ban u, im c: b3 = 2, Mb3 = M1



3

3

3 2 3

1

( ) )

( )

c

c

tT

xl xl

tT

c c

e

M M M M e

= +

= +

- Khi = 3, QTQ s chuyn sang giai on 4

d). Giai on 4: on fg, c tnh TN:


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f 4 R =0 R =R

- Theo c tnh TN:

2

4 3 22

( )

( )( )( )

TN c TN

k J J J R

T sR kk

R

= = = = =

- iu kin ban u, im g: b4 = 3, Mb4 = M1

- iu kin xc lp: xl4 = xl, Mxl4 = Mc


4

4

3

1

( )

( )

c

c

tT

xl xl

tT

c c

e

M M M M e

= +

= +

- Khi xl, M Mc xem nh kt thc QTQ khi ng.

Da vo cc phng trnh QTQ ca (t), M(t) trong 4 giai on, tac th v c c tnh (t) v M(t) khi khi ng vi m = 3 cp in tr

ph nh hnh 1.3a

6. Tnh thi gian khi ng :

tk = tq + t1 + t2 + t3 + t4 (ng vi m = 3 cp)T phng trnh M(t), ta c th tnh c cc thi gian:

1

2

ln ci cic

M Mt T

M M

=

vi i=1 4

4 4

1

1 1 2ln

c

k q i cii i c

M M

t t t T M M= =

= = =

Phn 2 : TNG QUAN V PLC S7-200

I. Gii thiu v PLC :1. PLC :

PLC (Programmable Logic Controller) l thit b logic kh trnh cdng trong cc ng dng cng nghip iu khin my mc v cc dychuyn sn xut t ng. N theo di cc u vo, a ra cc quyt nh, v

iu khin cc u ra theo t ng ha my mc cc dy chuyn sn xut.


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PLC hin nay c ng dng rt ph bin, chng hn nh trong ccdy chuyn ng chai, ng gi sn phm t ng, iu khin thang my,

bm, my trn, iu khin khi ng, dng v o chiu ng c in,

2. Nhng thun li ca PLC :

- Kch thc gn, d bo qun.- D lp trnh v thay i chng trnh.- Dung lng b nh ln, thch hp cho cc chng trnh phc tp.- Chng nhiu tt, lm vic tin cy trong mi trng cng nghip.

- Giao tip c vi cc thit b thng minh khc nh: my tnh, nimng, cc mun m rng,

- Gi c hp l.3. So snh PLC vi cc h thng iu khin khc :

Ch tiuso snh

RLE MCH S MY TNH PLC

Gi thnh kh thp thp cao thpKch thc ln rt gn kh gn rt gn

Tc x l chm rt nhanh kh nhanh nhanhKh nng chng

nhiuxut sc tt kh tt tt

Lp tmt thi gianthit k v lp

t

mt thi gianthit k

mt nhiuthi gian lp

trnh

lp trnh vlp t n

ginKh nng iu

khin tc v phctp

rt kh kh kh n gin rt n gin

Thay i iukhin

rt kh kh kh n gin rt n gin


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Cng tc bo trkm do c rtnhiu cng tc

km nu ICc hn

km do cnhiu mch

in tchuyn dng

tt do ccmun ctiu chun

ha

II. Cu trc v hot ng ca PLC :1. Cu trc ca PLC :

Cc thnh phn c bn ca mt PLC l cc mun u vo, n v x ltrung tm - CPU (Central Processing Unit) v cc mun u ra (xem hnh2.2).

Cac

mounau vao

CPU

n vxltrung tam

Cac

mounau ra

Thiet blap trnh

Moungiam sat

Hnh 2.2

Cc mun u vo: nhn cc tn hiu tng t v s t cm bin vchuyn i thnh cc tn hiu logic cho CPU x l.

CPU: a ra cc quyt nh v thc thi cc cu lnh ca chng trnhiu khin c lu sn trong trong b nh.

Cc mun u ra: nhn ch th t CPU thay i tn hiu u raphc v cho vic iu khin thit b.

Thit b lp trnh: c dng np chng trnh vo PLC, cho bitPLC s lm g khi nhn c cc tn hiu vo.

Mun gim st: dng hin th thng tin ca tin trnh v cho phpnhp vo cc thng s iu khin mi.CPU gm 3 phn: b x l, h thng b nh v h thng ngun cung

cp


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CPU(n vxltrung tam)

Bonh

Nguoncung cap

Hnh 2.3

C 3 loi b nh chnh l ROM, RAM v EPROM:

ROM (Read Only Memory) l b nh ch c, dng lu tr ccchng trnh c nh, khng th thay i.

RAM (Random Access Memory) l b nh truy xut ngu nhin, dng lu tr cc d liu tm thi.

EPROM (Erasable Programmable Read-Only Memory) l ROM c thlp trnh v xa c.

2. Hot ng ca mt PLC :

u tin, cc mun vo nhn cc tn hin bn ngoi t cc cm bin,cng tc, chuyn i sang dng s v a n cho CPU x l. Khi CPUnhn c tn hiu ng vo, n s x l v a cc tn hiu iu khinqua ng ra n cc thit b cn iu khin.

Trong sut qu trnh hot ng, CPU qut d liu hoc trng thi cacc thit b ngoi vi thng qua ng vo, sau thc hin cc chng trnhtrong b nh. Chng trnh c vit dng STL (lit k lnh) hoc LAD(hnh thang) c dch ra ngn ng my trc khi lu vo b nh. Khithc hin xong chng trnh, qu trnh truyn thng ni b v kim tra lic thc hin. Cui cng CPU s gi hoc cp nhp tn hiu ti cc thit

b iu khin thng qua ng ra. Nh vy mt vng qut s gm 4 khu:c tn hiu ng vo, thc hin chng trnh, truyn thng ni b, tkim tra li v gi cp nhp tn hiu ng ra.


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Vong quetPLC

Hnh 2.4

Thc t, khi thc hin chng trnh, PLC khng cp nhp tc thi ccng ra m qua 2 bc: trong sut qu trnh thc hin chng trnh, tn hiua ra ng ra s c lu tm thi trong chng trnh ni vi cc mclgc l ON v OFF. Khi chng trnh c thc hin xong, CPU schuyn cc tn hiu lu ra cc ng ra thc i iu khin ccthit b.

Thng thng, thi gian thc hin mt vng qut rt ngn, khong t 1ms n 100 ms, ph thuc ch yu vo di chng trnh PLC v tc

giao tip gia PLC vi cc thit b ngoi vi. B vi x l cng ch c th cc tn hiu ng vo khi tn hiu ny tc ng trong thi gian ln hnmt chu k qut. Nu thi gian tc ng ng vo nh hn mt chu kqut th b vi x l xem nh khng c tn hiu ny. Tuy nhin trong thc tsn xut, thng thng cc h thng chp hnh l cc h thng c kh nntc qut nh trn hon ton c th p ng c cc chc nng ca dychuyn sn xut.

III. PLC S7-200 :

1. Gii thiu v S7-200 :

S7-200 l mt PLC c kch thc nh do hng Siemens ca c snxut. Cc chc nng cp ngun v cng I/O c thit k on-board.

S7-200 rt thch hp cho cc ng dng n nh iu khin my trn,thang my, khi ng v o chiu ng c, N cng c th c dngtrong cc ng dng cng nghip phc tp nh cc dy chuyn ng chaiv ng gi sn phm,

2. Cu trc phn cng ca S7-200 (CPU 214) :


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Hnh 2.5

S7-200 (CPU 214) c 14 cng u vo v 10 cng u ra, mt cng lp

trnh giao tip vi my tnh np chng trnh vo PLC. Cc tnh nngca CPU 214 nh sau:

B nh:

Chng trnh: 6 KB

D liu ngi dng: 5 KB

Loi b nh: EEPROM

Loi b nh chp nh: EEPROM

Thi gian lu d liu khi PLC mt in: 50 gi

I/O:

S lng I/O c sn: 14 vo/10 ra

S mun m rng ti a: 7

S cng I/O s m rng ti a: 94 vo/14 ra

S cng I/O tng t m rng ti a: 28 vo/7 ra hoc 0 vo/14 ra

Tp lnh:

Tc x l lgc: 0,37 s/lnh

S lng rle ni: 256

S lng b m: 256

S lng timer: 256


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S lng rle iu khin tun t: 256

Vng lp for/next: c

Tnh ton trn s nguyn (+-*/): c

Tnh ton trn s thc (+-*/): c

3. Cu trc chng trnh ca S7-200 :

Chng trnh cho S7-200 c th c vit bng STEP7 Micro/WINhoc STEP7 Micro/DOS, c cu trc gm chng trnh chnh, ccchng trnh con v cc chng trnh x l ngt:

- Chng trnh chnh kt thc bng lnh MEND.

- Chng trnh con l mt b phn ca chng trnh, kt thc bng lnhRET. Cc chng trnh con phi vit sau lnh kt thc chng trnhchnh (MEND).

- Chng trnh ngt l mt b phn ca chng trnh, c dng khi cn,kt thc bng lnh RETI v cng phi c vit sau lnh MEND.


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Main Program...MEND

Chng trnh chnh

SBR 0...RET

Chng trnh con 1

SBR n...RET

Chng trnh con n+1

INT 0...RETI

Chng trnh xl ngat 1

Chng trnh xl ngat n+1

INT n.

..RETI

Thc hien trong mot vong quet

Thc hien khi c chng trnhchnh goi

Thc hien khi cotn hieu bao ngat

Hnh 2.6

4. Phng php lp trnh LAD v STL :


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Network 1

LD I0.2O I0.3= Q0.1

Q0.1I0.2

I0.3

a) dang LAD

b) dang STL

Hnh 2.7

a) Phng php LAD (hnh thang) :

LAD l phng php s dng cc k hiu ha biu din cc phnt v cc lnh lgc ca chng trnh PLC:

- Tip im: thng m -| |- hoc thng ng -|/|-

- Cun dy: -( )-

- Hp: thay th cho cc timer, b m v cc hm ton hc,

b) Phng php STL (lit k lnh):

STL l phng php th hin chng trnh PLC di dng tp hp cacc cu lnh n.

5. Mt s lnh thng dng ca S7-200 :


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LAD STL M t

n LD n Np gi tr logic ca tipim n vo bt u tin trnngn xp.

n LDN n Np gi tr lgc o catip im n vo bt u tintrn ngn xp.

n = n Sao chp gi tr ca nhngn xp ra tip im n.

I0.0 I0.1 Q0.0 LD I0.0

A I0.1

= Q0.0

Thc hin php AND lgc.

I0.2 Q0.1

I0.3

LD I0.2

O I0.3

= Q0.1

Thc hin php OR lgc.

MENDMEND Kt thc khng iu kin

chng trnh hin hnh.

Txx

IN

PT

TON

TON Txx n Khai bo timer s hiu xxkiu TON to thi giantr tnh t khi u vo INc kch. Nu gi tr mtc thi ln hn hoc bnggi tr t trc PT th T-bitc gi tr lgc bng 1. Cth reset timer kiu TON

bng lnh R hoc gi trlgc 0 ti u vo IN.

Txx

IN

PT

TONR

TONR Txxn

Khai bo timer s hiu xxkiu TONR to thi giantr tnh t khi u vo INc kch. Nu gi tr mtc thi ln hn hoc bnggi tr t trc PT th T-bit

c gi tr lgc bng 1. Chc th reset timer kiuTONR bng lnh R cho T-


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bit.

Cxx

CTUCU

PV

R

CTU Cxxn

Khai bo b m tin theosn ln ca CU. Khi gi trm tc thi C-word Cxx

ln hn hoc bng gi trt trc PV, C-bit (Cxx)c gi tr lgc bng 1. Bm c reset khi u voR c gi tr lgc bng 1.B m ngng m khi C-word Cxx t gi tr cc i32767.

Cxx

CTUDCU

PV

R

CTUD Cxx

n

khai bo b m tin/li,

m tin theo sn ln caCU v m li theo snln ca CD. Khi gi tr mtc thi C-word Cxx lnhn hoc bng gi tr ttrc PV, C-bit Cxx c gitr lgc bng 1. B mngng m tin khi C-wordt gi tr cc i 32767 v

ngng m li khi t gitr cc tiu -32767. CTUDreset khi u vo R c gitr lgc bng 1.

Phn 3 : LP TRNH PLC IU KHIN QU TRNHKHI NG Ml BA CP IN TR PH

1. Yu cu bi ton :Lp trnh PLC iu khin khi ng ng c1GH5 132-OJG6-6VV1

ca hng Siemens vi 3 cp in tr ph. S liu ca ng c nh sau:- Pm = 25 KW- nm = 2000 vng/pht- U = 220V- Im = 120 A- R = 0,08

- J = 0,15 kgm2

.


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2. Gii thiu v cng ngh :

Hnh 3.1Giao tip

gia ngi dngvi ng c rt

n gin, chgm mt nt nkhi ng vmt nt ndng. y l ccnt n khng ckh nng t gi.

Khi ngi dng n nt khi ng, b iu khin PLC s iu khin atn hiu ra cc CTT ngn mch dn cc in tr ph trn mch phn ng v

a tc ng ng c nhanh chng t ti gi tr xc lp, kt thc qu trnhkhi ng. dng my, ngi dng ch vic n nt dng. PLC s iu khina tn hiu ra CTT hm ng in tr hm vo trong thi gian c nhsn nhm gim tc ng c v 0.

3. Tnh thi gian khi ng :

m =55,9

dmn

=55,9

2000= 209,4 (rad/s)

Mm =dm

dmP

1000.=

4,209

1000.25= 119,39 Nm

- chn M1=2,5Mm=2,5 . 119,39 = 298,5 NmMmen cn : Mc Mm= 119,39 NmM2=1,3Mm=1,3 . 119,39 = 155,2 NmI1= 2,5 Im=2,5 . 120 = 300 A

-

- k =m

U

= 4,209

220= 1,1025

- (k )2 = 7,84- Khi ng vi 3 cp in tr ph (m=3) bi s dng in khi khi

ng = 31.IR

U

u

= 3300.08,0

220= 2,092


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- Cc cp in tr ph l:

Rf1 = 2( -1).R = 2,0932 (2,093 - 1).0,08 = 0,383 Rf2 = ( -1).R = 2,092(2,093 - 1).0,08 = 0,183 Rf3 = ( -1).R = (2,093 - 1).0,08 = 0,08744

- Thi gian t1, t2, t3:

Tc1 = 2321

)(k

)(

+++

ufufufuRRRRJ

=

=1025,1

)08744,0183,0383,008,0.(15,0 +++= 0,1

t1 = Tc1lnc

c

MM

MM

2

1 = 0,1.ln39,1192,155

39,1195,298

= 0,161(s)

Tc2 = 232

)(k

)(

++

ufufuRRRJ

=1025,1

)08744,0183,008,0.(15,0 ++= 0,0477

t2 = Tc2lnc

c

MM

MM

2

1 = 0,0477.ln39,1192,155

39,1195,298

= 0,077 (s)

Tc3 = 23

)(k

)(

+

ufu RRJ=

1025,1

)08744,008,0.(15,0 += 0,0228

t3 = Tc3ln cc

MM

MM

2

1

=0,0228.ln 39,1192,15539,1195,298

= 0,0367(s)

4. Tnh thi gian hm ng nng :

thm = Tc . lnc

c

M

MM +1

Tc = J.M

= J.

1M

b

M1= 2,5.Mm = 298,5 Nmb = m = 119,39(rad/s)

Tc = 0,15. 5,2984,209 = 0,105 (s)

thm = 0,105.ln 39,11939,1195,298 +

= 0,132 (s)


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5. Phn cng vo/ra cho cc thit b :

CKT

H

Rh

K

1G 2G 3G

Eo

Hnh 3.2

a ch Chc nngI0.0 Nt dngI0.1 Nt khi ng

Q0.0 Tn hiu ng K Q0.1 Tn hiu ng 1GQ0.2 Tn hiu ng 2GQ0.3 Tn hiu ng 3GQ0.4 Tn hiu ng HT33 Timer 1 (16x10ms=160ms)T34 Timer 2 (8x10ms=80ms)T35 Timer 3 (4x10ms=40ms)T36 Timer 4 (13x10ms=130ms)


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6. S Thut ton:


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Hm:


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7. Gin thi gian

Q 0 . 4

T 3 6

T 3 5

Q 0 . 3

T 3 4

Q 0 . 1

Q 0 . 2

T 3 3

Q 0 . 0

I 0 . 1

I 0 . 0

1 6 0 m s

8 0 m s

4 0 m s

1 3 0 m s


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8. Chng trnh LAD:


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I 0 . 0 I 0 . 1

Q 0 . 0

Q 0 . 0

N E T W O R K 5

N E T W O R K 4

N E T W O R K 3

N E T W O R K 2

N E T W O R K 1

Q 0 . 1Q 0 . 0 T 3 3

I N T O N

P T 1 0 m s

Q 0 . 1

Q 0 . 1

T 3 3Q 0 . 0

1 0 m sP T

T O NI N

T 3 4Q 0 . 1 Q 0 . 2Q 0 . 0

Q 0 . 0 T 3 4

Q 0 . 2

Q 0 . 2

1 6

8


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N E T W O R K 7

N E T W O R K 8

Q 0 . 3

Q 0 . 3

T 3 5Q 0 . 0

I 0 . 0 T 3 6

Q 0 . 4

Q 0 . 4

N E T W O R K 9

Q 0 . 4 T 3 6

I N

P T

T O N

1 0 m s

N E T W O R K 6

Q 0 . 1 Q 0 . 3Q 0 . 2 T 3 5

I N T O N

P T 1 0 m s

Q 0 . 0

4

1 3

doan plc

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