dieu khien logic3

7/29/2019 Dieu Khien Logic3

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i Hc NngTrng i Hc Bch Khoa

Khoa inB mn Tng - o Lng

GIO TRNH MN HC IUKHIN LOGIC

MN HC DNH CHO CC SINH VIN KHOA INKHO CHNH QUY

Sn v hc trnh: 4 (60 tit)

Ngi bin son:

Lm Tng cNguyn Kim nh

Nng, thng 11 nm 2005


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Chng 0: L thuyt cs B mn Tng o Lng Khoa in

CHNG 0: L THUYT CS(3T)

0.1. Khi nim v logic trng thi:+ Trong cuc sng hng ngy nhng s vt hin tng p vo mt chng ta nh:c/khng; thiu/; cn/ht; trong/c; nhanh/chm...hai trng thi ny i lpnhau hon ton.+ Trong k thut (c bit k thut in - iu khin) khi nim v logic haitrng thi: ng /ct; bt /tt; start /stop+ Trong ton hc lng ho hai trng thi i lp ca s vt hay hin tngngi ta dng hai gi tr 0 &1 gi l hai gi tr logic. Cc nh khoa hc ch xy dng cc hm & bin trn hai gi tr 0 &1 ny. Hm v bin c gi l hm & bin logic. Cs tnh ton cc hm & s gi l i s logic.i s ny c tn l Boole (theo tn nh bc hc Boole).

0.2. Cc hm cbn ca i s logic v cc tnh cht cbn ca chng:

B0.1_ hm logic mt bin:

Tn hm Bng chn l K hiu s Ghi ch

x 0 1

Thut tonlogic kiu rle

kiu khi int

Y0 = 0Hm khng Y0 0 0

Y0 = xx

Hm lun

bng 0

Hm lp Y1 0 1 Y1 =

Hm o Y2 1 0 Y2 = x

Y3 = 1Hm n v Y3 1 1

Y3 = x + x

Hm lun

bng 1B 0.2_ Hm logic hai bin y = f(x1 ,x2 )

Hm hai bin, mi bin nhn hai gi tr 0 &1, nn c 16 gi tr ca hm t y0 y15.

Bng chn l K hiu sx1 0 0 1 1Tn hm

x2 0 1 0 1

Thut tonlogic Kiu rle

Kiu khi int

Ghi ch

Hm

khng Y0 0 0 0 0

Y0 = x1.x 2+

x 1 .x2

Hm lun

bng 0Hm v

Y1 0 0 0 1Y1 = x1.x2

Hm cmx1

Y2 0 0 1 0 Y2 = x1 .x 2

Bin son: Lm Tngc - Nguyn Kim nh 1


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Hm lpx1

Y3 0 0 1 1 Y3 = x1

Hm cmx2

Y4 0 1 0 0 Y4 = x 1. x2

Hm lpx2Y5 0 0 1 1 Y5 = x2

Y6 = x 1. x2+

x1 .x 2Hm hocloi tr

Y6 0 1 1 0

Y6 =x1 x2

Cngmodule

Hm hoc Y7 0 1 1 1 Y7 = x1 + x2

Hm piec Y8 1 0 0 0 Y8 = x 1 .x 2

Hm cng

duY9 0 1 1 1 Y9= 21 xx

Hm ox1

Y10 1 1 0 0 Y10 = x 1

Hm ko

theo x1Y11 1 0 1 1 Y11 = x 2 + x1

Hm ox2

Y12 1 0 1 0 Y12 = x 2

Hm kotheo x2

Y13 1 1 0 1 Y13 = x 1 + x2

Hm

chefferY14 1 1 1 0 Y14 = x 1 +x 2

Hm nv

Y15 1 1 1 1 Y15 = x 1 +x1

x1x2

0 1

0 1 0

1 1 1

Y13 = x 1 + x2

x1x2

0 1

0 1 0

1 1 0

Y12 = x 2

x1x2

0 1

0 1 1

1 1 1

Y15 = 1

x1x2

0 1

0 1 1

1 1 0

Y14 = x 1 +x 2



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x1x2 0 1

0 1 1

1 0 1

Y11 = x 2 + x1

x1x2 0 1

0 1 1

1 0 0

Y10 = x 1

x1x2 0 1

0 1 0

1 0 1

Y9= 21 xx

x1x2 0 1

0 1 1

1 0 1

Y8 = x 1 .x 2

x1x2

0 1

0 0 1

1 1 0

Y6 =x1 x2

x1x2

0 1

0 1 1

1 0 1

Y7 = x1 + x2

x1x2

0 1

0 1 1

1 0 1

Y5 = x2

x1x2

0 1

0 1 1

1 0 1

Y4 = x 1. x2

x1x2

0 1

0 1 1

1 0 1

Y3 = x1

x1x2

0 1

0 1 1

1 0 1

Y2 = x1 .x 2

x1x2

0 1

0 1 1

1 0 1

Y1 = x1.x2

x1x2

0 1

0 0 0

1 0 0

Y0 = 0

* Ta thy rng: cc hm i xng nhau qua trc (y7 v y8 ) ngha l: y0 = y 15, y1 = y 14,y2 = y 13* Hm logic n bin: y = f(x1,x2,x3,..,xn).

1 bin nhn 21 gi tr n bin nhn 2n gi tr; m mt t hp nhn 2 gi trn

Do vy hm c tt c l 2 .2

V d: 1 bin to 4 hm 2

2 bin to 16 hm 23 bin to 256 hm 2

122

232

Kh nng to hm rt ln nu s bin cng nhiu.Tuy nhin tt c kh nng ny u c hin qua cc hm sau:

Tng logicNghch o logicTch logic



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nh l - tnh cht - h s cbn ca i s logic:0.2.1. Quan h gia cc h s:

0 .0 = 0

0 .1 = 0

1 .0 = 0

0 +0 = 00 +1 = 1

1 +0 = 1

1 +1 = 1

0 = 1

1 = 0

y l quan h gia hai hng s (0,1) hm tin ca i s logic. Chng l quy tc php ton cbn ca t duy logic.

0.2.2. Quan h gia cc bin v hng s:A.0 = 0

A .1 =AA+1 = 1A +0 =AA .A = 0A +A = 10.2.3. Cc nh l tng ti s thng:

+ Lut giao hon:A .B =B .A

A +B =B +A

+ Lut kt hp:

( A +B) +C =A +( B +C)( A .B) .C =A .( B .C)

+ Lut phn phi:A ( B +C) =A .B +A .C

0.2.4. Cc nh l c th ch c trong i s logic:A .A =AA +A =A

nh l De Mogan:BA. =A + B

BA+ =A .B

Luthm nguyn:A = A .

0.2.5. Mt sng thc tin dng:A ( B +A) = AA + A .B = AA B +A . B = AA + A .B = A +B



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A(A + B ) = A .B(A+B)(A + B ) = B(A+B)(A + C ) = A +BCAB+A C + BC = AB+A C(A+B)(A + C )(B +C) =(A+B)(A + C )

Cc biu thc ny vn dng tinh gin cc biu thc logic, chngkhng ging nhi s thng.

Cch kim chngn gin v p dng nhtchng minh l thnhlp bng stht.

nhn v t nh

n l

ng0:

0.3. Cc phng php biu din hm logic:0.3.1. Phng php biu din thnh bng:

* Nu hm c n bin th bng c n+1 ct .( n ct cho bin & 1 ct cho hm )* 2

nhng tng ng vi 2n t hp bin.

Bng ny gi l bng s tht hay l bng chn l.V d:

Trong nh c 3 cng tc A,B,C.Ch nh mun n chiu sng khi cng tc A,B, C u hhoc A ng B, C hhoc A hB ng C h.

Vi gi tr ca hm y cho trn ta biu din thnh bng nh sau:

Cng tc n nA B C Y

0 0 0 1 sng

0 0 1 0

0 1 0 1 sng

0 1 1 0

1 0 0 1 sng1 0 1 0

1 1 0 0

1 1 1 0

10 11

0100

x1

x2

* u im ca cch biu din ny l d m ln .* Nhc im: cng knh, c bit khi s bi n.

0.3.2. Phng php biu din hnh hc:a) Hm mt bin biu din trn 1 ng th ng:

b) Hm hai bin biu din trn mt ph



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c) Hm ba bin biu din trong khng gian 3 chiu:

011111

010 110

000 100

001101

X1

X2

X3

d) Hm n bin biu din trong khng gian n chiu0.3.3. Phng php biu din biu thc i s:

Bt k trong mt hm logic n bin no cng c th biu din thnh cc hm ctng chun y v tch chun y .a) Cch vit di dng tng chun y (chun tc tuyn):- Ch quan tm n nhng t hp bin m hm c gi tr bng mt.

- Trong mt t hp (y bin) cc bin c gi tr bng 1 th gi nguyn (xi).- Hm tng chun y s l tng chun y cc tch .

Cng tc n nA B C Y

0 0 0 0 0

1 0 0 1 1

2 0 1 0 x

3 0 1 1 1

4 1 0 0 1

5 1 0 1 x

6 1 1 0 0

7 1 1 1 1

Hm Y tng ng 4 t hp gi tr cc bin ABC = 001, 011, 100, 111Y= A B C +ABC+AB C +ABC

*n gin trong cch trnh by ta vit li:



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f = 1, 3 ,4 ,7Vi N =2 ,5 (cc thtthp bin m khng xc nh )

b) Cch vit di dng tch /chun y ( hi tc tuyn ):- Ch quan tm n t hp bin hm c gi tr ca hm bng 0.- Trong mi tng bin xi = 0 th gi nguyn xi = 1 th o bin ix .- Hm tch chun y s l tch cc tng , t bng trn hm Y tng ng 2 t hpgi tr cc bin: A+B+C = 0 +0 +0, 1 +1 +0

A +B +C, A +B +C Y =( A +B +C)(A +B +C)

*n gin trong cch trnh by ta vit li:f = (0,6)Vi N =2 ,5 (cc thtthp bin m khng xc nh ).

0.3.4. Phng php biu din bng bng Karnaugh:- Bng c dng hnh ch nht, n bin 2n mi tng ng vi gi tr ca 1t hp bin.- Gi tr cc bin c sp xp theo th t theo m vng (nu khng th khngcn l bng Karnaugh na!).*Vi iu slc v m vng:

Gi s cho s nh phn l B1B2B3B4 G3G2G1G0 (m vng)th c th tnh nh sau: Gi = Bi+1 BiV d: G0 = B1 B0 = 1B B0 +B1 0B

G1 = B2 B1 = 2B B1 +B2 1B G2 = B3 B2 = 3B B2 +B3 2B

G3 = B4 B3 = 0 B3 =1.B3 +0. 3B = B3

x2x1

0 1

0

1

x2 x3x1

00 01 11 00

0

1

x3 x4x1x2

00 01 11 10

00

01

11

10



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x3 x4x5x1x2

000 001 011 010 110 111 101 100

00

01

11

10

x4x5x6

x1x2 x3

000 001 011 010 110 111 101 100

000

001

011

010

110

111

101

100

0.4. Phng php ti thiu ho hm logic:Mc ch ca vic ti u ho hm logic thc hin mch: kinh tn gin, vn bom chc nng logic theo yu cu.Tm dng biu din i sn gin nht c cc phng php sau:0.4.1. Phng php ti thiu hm logic bng bin i i s:Da vo cc biu thc phn 0.3 ca chng ny .y =a (b c + a) + (b + c )ab = a b c + a + ba b +c ab = a



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Phng php 1 :y = a ( b c + a) + (b + c )ab = a b c + a + ba b +c ab = a

hoc y = a (b c + a) + (b + c )ab = a b c + a(b+b )(c+c )+ab c = a b c + abc + abc + a b c + ab c +ab c m5 m7 m6 m5 m4 m4

(Phng php 2: dng bng s cp phn sau)

V d 1:

V d 2:



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V d 3:

V d 4:

V d 5:



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0.4.2. Phng php ti thiu ho hm logic bng bng Karnaugh:Tin hnh thnh lp bng cho tt c cc v dphn (1) bng cch bin i biu

thc i s sao cho 1 t hp c mt y cc bin.V d: Cho h thng c s nh sau h thng ny iu khin hai l si L1, L2 vca s S. Cc thng su vo ca l nhit hai mc 10oC & 20oC v m mc

2%.

Hnh 0.1: M t hot ng ca h thng l si

A tc ng khi t0 < 10oC (u o a)B tc ng khi t0 > 20oC (u o b)C tc ng khi m 2% (u o c)(+) tc ng(-) khng tc ngiu kin c thc cho bng sau:

mNhit

W < 2% W 2%

t0 20

oC - + + - - +

20oC

> t0>10

oC+ - + - + -

t0< 10

oC+ + + + - -

Thit b chphnh

L1 L2 S L1 L2 S

L L1 L L2 Ca s L L1 L L2 Ca s

A B C L1 L2 S

0 0 0 1 1 1

0 0 1 1 0 0

0 1 0 x x x

0 1 1 x x x

1 0 0 1 0 1



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1 0 1 0 1 0

1 1 0 0 1 1

1 1 1 1 1 1

Lp bng Karnaugh cho ba hm L1 ,L2 ,SL1 = B .C +A ; L2 = A C +AB C + BC ; S = B + C

0.4.3. Phng php ti thiu hm logic bng thut ton Quire MC.Cluskey:a) Mt snh ngha:+ L tch y ca cc bin.

- nh 1 l hm c gi tr bng 1.- nh 0 l hm c gi tr bng 0.- nh khng xc nh l hm c gi tr khng xc nh x (0 hoc1).+ Tch cc tiu: tch c s bin l cc tiu (t bin tham gia nht) hm c gitr bng 1 hoc l khng xc nh x.+ Tch quan trng: l tch cc tiu hm c gi tr bng 1 tch ny.V d: Cho hm f(x1,x2,x3) c L = 2,3,7 (tch quan trng)

N =1,6 (tch cc tiu)C thnh du theo nh phn hoc thp phn.

b) Cc bc tin hnh:Bc 1: Tm cc tch cc tiu(1) Lp bng biu din cc gi tr hm bng 1 v cc gi tr khng xc nh xng vi m nh phn ca cc bin.(2) Sp xp cc t hp theo th t tng dn (0,1,2,...), t hp gm:

1 ch s 12 ch s 13 ch s 1

(3) So snh t hp th i v i+1 & p dng tnh cht xy +xy = x.Thay bngdu - & nh du v vo hai t hp c.(4) Tin hnh tng t nh (3).



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Bng a Bng b Bng c Bng ds

thpphn

s nhphn

x1x2x3x4

schs 1

sthp

phn

s cs2

x1x2x3x4

Lin

ktx1x2x3x4

2 0010 1 2 0010v 2,3 001-v 2,3,6,7 0-1-

3 0011 3 0011v 2,6 0-10v 2,6,3,7

6 0110 2 6 0110v 3,7 0-11v 6,7,14,15 -11-

12 1100 12 1100v 6,7 011-v 6,14,7,15

7 0111 7 0111v 6,14 -110v 12,14,13,15 11--

13 1101 3 13 1101v 12,13 110-v

14 1110 14 1110v 7,15 -111v

15 1111 4 15 1111v 13,15 11-1v

14,15 111-v

T hp cui cng khng cn kh nng lin kt na, y chnh l cc tch cctiu ca hm f cho & c vit nh sau:0-1- (ph cc nh 2,3,6,7): 1x x3-11- (ph cc nh 6,7,14,15): x2,x3.11-- (ph cc nh 12,13,14,15): x1,x2.V d sau :( v d ny s gii thch cc bc trn ).

Ti thiu ho hm logic bng phng php Quire MC.Cluskey vif(x1,x2,x3,x4), vi cc nh 1 l L = 2,3,7,12,14,15; nh c gi tr khng xcnh l N = 6,13.

Bc 2: Tm tch quan trng tin hnh theo i bc (i =0 n ) cho n khi tmc dng ti thiu.Li : Tp cc nh 1 ang xt bc nh i (khng quan tm n nh khng xcnh x na).Zi: Tp cc tch cc tiu sau khi qua cc bc tm tch cc tiu bc 1Ei : L tp cc tch quan trng.

c thc hin theo tht ton sau:



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*Tip tc v d trn: ( Bc 2)

Kt thc

3. Vit ra cc hm cc tiu

2.Tm cc tch cc tiu ti thiu nh 1

1.Tm cc tch cc tiu

Cho hm vi tp L&N

Bt u

L0 = (2,3,7,12,14,15)

Z0 =( 1x x3,x2x3,x1x2 )

Tm E0 ?Lp bng E0:

L0Z0 2 3 7 12 14 15

1x x3 (x) (x) xx2x3 x x x

x1x2 x x

Ly nhng ct ch c 1 du x v y l tch quan trng.

Tm L1 t L0 sau khi loi nhng nh 1ca L0.Z1 t Z0 sau khi loi nhng tch khng cn thit.

f = 1x x3 +x1x20.5. Bi tp:1) Dng hai phng php ti thiu bng Quire MC.Cluskey & Karnaugh ti thiuho cc hm sau:

1) f (x1x2x3x4) = [2,3,7,(1,6)]



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2) f (x1x2x3x4) = [2,3,7,12,14,15(6,13)]

3) f (x1x2x3x4) = [0,2,3,10,11,14,15]

4) f (x1x2x3x4) = [1,6,(3,5,7,12,13,14,15)]

5) f (x1x2x3x4) = [(3,5,12,13,14,15),6,9,11]

6) f (x1x2x3x4) = [0,2,3,4,6]

(*)n gin biu thc sau dng bng Karnaugh:1) f = 1x 2x x3 +x1x2 3x + x1x2 x3+ x1 2x x32) f = 1x 2x 3x + 1x 2x x3 + 1x x2 x3+ x1 2x x3

3) f = 1x 2x 3x 4x + 1x x2 3x 4x + 1x 2x x3 4x + 1x x2x3 +x1 2x 3x 4x +x1 2x 3x x4

+ x1 2x x3 4x

4) f = ( 43 xx + )+ 1x x3 4x +x1 2x x3 + 1x 2x x3x4 +x1x3 4x

(*)

1) Mch iu khin my photocopy c 4 ng vo & 1 ng ra. Cc ng vo n cccng tc nm dc theo ng di chuyn ca giy. Bnh thng cng tc hv cc ngvo A, B, C, D c gimc cao. Khi giy chy qua mt cng tc th n ng vng vo tng ng xung thp. Hai cng tc ni n A & D khng bao ging cnglc (giy ngn hn khong cch gia hai cng tc ny). Thit k mch c ng ra lncao mi khi c hai hoc ba cng tc ng cng lc, cng bn k v li dng cc thp khng cn quan tm .

Hnh 0.2: M t hot ng ca my in

Cc bi tp ny c trch t bi tp kt thc chng 2.(Mch s _Ng.Hu Phng)

2) Hnh v ch giao im ca trc l chnh vi ng ph. Cc cm bin pht hinc xe c t li C,D (trc l chnh ) & li A ,B (trc ph). Tn hiu ca cm bin



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l thp khi khng c xe v cao khi c xe n giao thng c kim sot theo quy lutsau:

a) n xanh cho trc l chnh mi khi c hai li D & C.b) n xanh cho trc l chnh mi khi li C hoc D c xe nhng c hai li A& B khng c xe.

c) n xanh cho trc l ph mi khi li A hoc B c xe nhng trong khi c haili C & D khng c xe.d) n xanh cho trc l chnh khi cc li u khng c xe. Cc ng ra ca cmbin l cc ng vo ca mch iu khin n giao thng. Mch c ng ra T lm n trc l chnh xanh khi ln cao v ng ra P lm n trc l chnhxanh khi n gin biu thc ti a trc khi thc hin mch.(*) Bi tp dng gin xung:

a 0

y

1 10 0 0 0 1 00 1 1 1 0

0 1 1 0 0 0 0 1 11

000

00 0

11111100 1

0 1 1 1 0 0 0 0 1 1 1 1 1

1

b

b

c

1) y = a b c +ab

2) y = ab+ ac +b c 3) S = a1 + b 2a 3a + b ( 1a a2 + a3)



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Chng 1: Mch thp v mch trnh t B mn Tng - o Lng _ Khoa in

CHNG 1: MCH T HP V MCH TRNH T

1.1. M hnh ton hc ca mch t hp:- Mch t hp l mch m trng thi u ra ca mch ch ph thuc v t hp cc

trng thi u vo cng thi im m khng ph thuc vo thi im trc .- Mch t hp thng c nhiu tn hiu u vo (x1 ,x2 ,x3) v nhiu tn hiu

u ra (y1 ,y2 ,y3 ). Mt cch tng qut c th biu din theo m hnh ton hcnh sau:

Vi: y1 =f(x1 ,x2 ,,xn )y2 =f(x1 ,x2 ,,xn ).

.ym =f(x1 ,x2 ,,xn )

Hnh 1.1: M hnh ton hc ca mch t hp

- Cng c th trnh by di dng vector nh sau:Y =F(X)1.2. Phn tch mch t hp:

- T yu cu nhim v cho ta bin thnh cc vn logic, tm ra bngchc nng ra bng chn l.

- c thc hin theo cc bc sau:

1. Phn tch yu cu:

iu thc logic

ng karnaughBng chn lBng chc nngVn logic thc

Hnh 1.2: Bc phn tch mch t hp

Xc nh no l bin u vo.

Xc nh no l bin u ra. Tm ra mi lin h gia chng vi nhau.iu ny i hi ngi thit k phi nm r yu cu thit k, y l mt vic khkhn nhng rt quan trng trong qu trnh thit k.2. K bng chn l:

- Lit k thnh bng v mi quan h tng ng vi nhau gia trng thi tn hiuu vo vi trng thi hm su ra Bng ny gi l bng chc nng.



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- Tin hnh thay gi tr logic (0 ,1) cho trng thi ta c bng chn l.V d:

Hnh 1.3: Siu khin bng n Y thng qua 2 cng tc A&B

Bng chc nng: Bng chn l:

Kha

A

Kha B Kha C

Ngt Ngt TtNgt ng Ttng Ngt Ttng ng Sng

A B C

0 0 00 1 01 0 01 1 1

1.3. Tng hp mch t hp:Nu s bin tng i t th dng phng php hnh v.Nu s bin tng i nhiu th dng phng php i s.c tin hnh theo s sau:

1.4. Mt s mch t hp thng gp trong h thng:

smch in

slogic

biu thcti thiu

biu thc logic

Bng karnaughhocPP. Mc.cluske

Hnh 1.4: Phng php tng hp mch logic

Cc mch t hp hin nay thng gp l:B m ha (m ha nh phn, m ha BCD) thp phn, u tin.B gii m (gii m nh phn, gii m BCD_ led 7 on) hin th k t.B chn knh.B cng, b so snh.



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B kim tra chn l.ROM , EPROMB dn knh, phn knh.

1.5. Khi nim vmch trnh t(hay mch dy) _ sequential circuits:- u ra ch b kch hot

khi cc u vo ckch hot theo mt trnht no . iu nykhng th thc hin bngmch logic t hp thunty m cn n c tnhnhca FF.

m

2

1

x1x2y1

y2

Z1Z2Y1Y2

cht hp

mchtrnh t

Hnh 1.5: M hnh ton hc ca mch iu khin trnh t

1.6. Mt s phn tnhtrong mch trnh t:1. Rle thi gian:

A

BY

A

Y Y

A

B B

A

Hnh 1.6: Nguyn l lm vic ca cng AND

Y Y

AB

cua FF

R

S

T

S2LS1L

Hnh 1.7: Nguyn l lm vi c ca FF_JKY

>thi gianthiet lap yeu cauA

B

QJ

CLK

K

Y

ln cao trc Aln cao trc A ln cao trc Bln cao trc B

B B

A

S3L


Hnh 1.8: S relay thi gian


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2.Cc mch lt:LoiFF

ng b Khngng b

Bng chnl

Bng kch hnh trng th Gin xung

Qn R S Qn+1 nQn+1R S

0 0 0 0 0 0 x 00 0 1 1 0 1 0 10 1 0 0 1 0 1 00 1 1 x 1 1 0 x1 0 0 11 0 1 11 1 0 01 1 1 xQ'= S+R Q

R-S

Pr

Clr

Q

S

R

Q

CL

Q

R

S

Q

Clr

Pr

RS=0

0 1X0

01

0X

10

Qn D Qn+1 nQn+1D0 0 0 0 0 00 1 1 0 1 11 0 0 1 0 01 1 1 1 1 1

D

Q'n+1=D

1

1

0 10

0

D

Q

Q

CL

Qn J K Qn+1 QnQn+1J K0 0 0 0 0 0 0 x

0 0 1 0 0 1 1 x0 1 0 1 1 0 x 10 1 1 1 1 1 x 01 0 0 11 0 1 01 1 0 11 1 1 0

J-K

Khi J = 1& K =1 thQ lun thayi trngthi nghal mch bdao ngnn JK chlm vic chng b Q'n+1=

X1

0X1X

X0

10

Q

Q

K

J

CL

Qn T Qn+1 nQn+1T

0 0 0 0 0 00 1 1 0 1 11 0 1 1 0 11 1 0 1 1 0

T

Cng

khng cchkhngng b

Q'n+1=TQ

10

1

1

0 0

Q

Q

T

CL

CL

R

S

Q

Q

CLQ

D Q

Clr

Pr

Q

D Q

Clr

Pr

Pr

Clr

Q

K

J

Q

CL

CLQ

T Q

Clr

Pr



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1.7. Phng php m t mch trnh t:Sau y l mt vi phng php nu ra phn tch v tng hp mch trnh t.1.7.1.Phng php bng chuyn trng thi:Sau khi kho st k qu trnh cng ngh, ta tin hnh lp bng. v d ta c bng nhsau:

Trngthi

Tn hiu vo Tn hiu ra

x1 x2 x3 ... Y1 Y2 ...S1 S1 S2 S3 0 1S2 S1 S2 0 0S3 S2 S3 1 1S4S5...

- Cc ct ca bng ghi: bin u vo (tn hiu vo): x1, x2, x3 ; hm u ra y1,y2, y3

- S hng ca bng ghi r s trng thi trong cn c ca h (S1 ,S2 ,S3).- giao gia ct tn hiu vo xi vi hng trng thi Sj ghi trng thi ca mch.

Nu trng thi mch trng vi trng thi hng l trng thi n nh.- giao gia ct tn hiu ra Yi v hng trng thi Sj chnh l tn hiu ra tng

ng.* iu quan trng l ghi y v ng cc trng thi trong cc ca bng, chai cch:Cch 1:

Nm r d liu vo, nm su v quy trnh cng ngh ghi trng thi nnh hin nhin.

Ghi cc trng thi chuyn r rng (cc trng thi n nh 2 d dng nhnra).

Cc trng thi khng bit chc chn th trng v s b sung sau.Cch 2:

Phn tch xem tng in trng thi. Vic ny l logic, cht ch, rrng.

Tuy nhin rt kh khn, nhiu khi khng phn bit c cc trng thitng t nh sau.



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V d ta c bng sau:

Bin(x)Trng

thi(S)

S1 S2/1 S4/1 S3/0S2 S4/1 S2/0 S4/1S3 S1/1 S1/1 S1/1S4 S3/1 S4/0 S2/0S5 S5/0 S3/0 S4/0

1.7.2. Phng php hnh trng thi:M t cc trng thi chuyn ca mt mch logic tng t. hnh gm: cc

nh, cung nh hng, trn cung ny ghi tn hiu vo/ra & kt qu. Phng php ny

thng dng cho hm ch mt u ra.a. hnh Mealy: hnh Mealy chnh l s chuyn trng thi thnh hnh.ta thc hin chuyn t bng trng thi sang hnh: Bng c 5 trng thi; l nm nh ca hnh. Cc cung nh hng trn ghi hai thng s: bin tc ng, kt qu hm khi

chu s tc ng ca bin.

1 2/

3 4

5

1

(+)/100

/1(++)/1

0

0

0

/0

0

Hnh 1.10: hnh Mealy

b. hnh Moore: hnh Moore cng thc hin chuyn bng trng thi thnh hnh. T bng

trng thi hay t hnh Moore ta chuyn sang hnh nh sau:Vi nh l cc gi tr trng thi: cung nh hng; bin ghi tc ng.



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Bc 1: T cc bng trng thi ta tm ra cc trng thi & gi tr tng ng.V d: bng bn c 5 trng thi t S1 S5 nhng ch c: S1 c gi tr S1/1; S5 c gi trS5/0

Cn cc trng thi: S2 , S3 , S4 c 2 gi tr 0 & 1 nn ta c 6 nh.Vy tng cng, hnh Moore c 8 nh. nh ny gn tng ng vi cc Q, t Q1

n Q8 .Q1 = S2/0 ; Q2 = S3/0 ; Q3 = S4/0 ;Q4 = S5/0 ; Q5 = S1/1 ; Q6 = S2/1 ; Q7 = S3/1

Q8 = S4/1Bc 2:Tin hnh thnh lp bng nh sau:(T bng trng thi ta tin hnh in nh Qi vo v d gc u bn tri, ging vi S2

bn bng trng thi ta c S4 /1 Q8 in Q8 vo ny, tng t nh vycho n ht). ct tn hiu ra l kt qu ca tng nh Q tng ng.Bc 3: Tin hnh v th Moore tng t hnh Mealy.

* th Moore c nhiu nh hn hnh Mealy. Nhng bin u ra n gin hnMealy.

Q1/0 Q2/0 Q4/0Q3/0

Q7/1 Q8/1Q6/1Q5/1

(++)

(++)

(+)

(+)

Hnh 1.11: hnh Moore

1.7.3. Phng php lu :Phng php ny m t h thng mt cch trc quan, bao gm cc khi cbn sau:

1) Khi ny biu th gi tr ban u chun b sn sng hoc cho h thnghot ng.2) Thc hin cng vic (x l, tnh ton ...).3) Khi kim tra iu kin v a ra mt trong hai quyt nh.4) Kt thc cng vic.

V d ta c s thut ton sau:Chuyn a) sang hnh Moore; hnh c su nh, nm nh l trng thi ca z, mtnh cn li l trng thi bt u v kt thc.



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1.8 Grafcet_Cng c m t mch trnh ttrong cng nghip:1.8.1. Hotng theo logic trnh tca thit btrong cng nghip:

Trong dy chuyn sn xut cng nghip my mc thng hot ng theo trnht logic cht ch nhm m bo cht lng sn phm, an ton cho ngi v thit b.Cu trc hot ng trnh t ca dy chuyn a ra yu cu cho iu khin ng thi

cng gi cho ta s phn nhm logic ca hot ng trnh t bi cc tp hp con camy mc v cc thut ton iu khin bng chng trnh con. S khi ca hiukhin qu trnh c th hin theo s sau:

Tn hiuvo

Qu trnhCu trciu khin

trnh t

Hnh 1.12: S khi ca hiu khin qu trnh

Mt qu trnh cng ngh bao gm ba hnh thc hot ng sau:+ Hon ton tng+ Bn tng+ Bng tay

Trong qu trnh h thng lm vic, m bo an ton v linh hot, hiu khin cnphi c s chuyn i d dng t tng bn tng hoc bng tay v

ngc li nh vy h mi p ng c yu cu thc t.Trong qu trnh lm vic, s khng bnh thng (s c) ca h thng c rtnhiu loi; v vy trong qu trnh phn tch h thng c gng m t chng mt cch y nht, ngha l cc s kin v li a s phi c nh ngha trc. Trong vn vs c ngi ta thng phn ra lm 3 nhm sau:

+ H hng mt b phn trong cu trc iu khin.+ H hng cu trc trnh t iu khin.+ H hng b phn chp hnh.Khi thit k h thng phi tnh n cc phng n khc nhau nh: vic dng

my khn cp, x l tc ngn vt liu v nhiu hin tng nguy him khc ng thi

cho php ngi vn hnh can thip ngay im xy ra s c hoc c lp vng xy ra sc.Grafcet l cng c rt hu ch thit kv thc hin y cc yu cu ca

h thng tng ho cc qu trnh cng ngh.



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1.8.2.nh ngha Grafcet:Grafcet l t vit tt ca ting Php Graphe fontionnel de commande tape

transition, l hnh chc nng cho php m t cc trng thi hot ng ca h thngv biu din qu trnh iu khin vi cc trng thi chuyn bin t trng thi ny sangtrng thi khc, l mt graphe nh hng v xc nh bi cc phn t sau:

G := {E, T, A, M}Trong :+ E = {E1, E2, E3, ... , Em} l mt tp hu hn cc trng thi(giai on ) ca h thng,c k hiu bng hnh vung. ng vi mi trng thi sao cho hnh vi iu khin lkhng thay i, hnh vi c th hot ng hoc l khng hot ng. iu khinchnh l thc hin cc mnh logic cha cc bin vo/ra h thng c c trng

thi xc nh trong h v y cng chnh l mt trong cctrng thi ca Grafcet.(P, M: a.Ek)Ej Trng thi Ej hnh 1.13 l s phi hp gia bin ra P vM, vi M = a.Ek, trong Ekl bin c trng cho hot

ng ca trng thi Ekcn a l bin u vo ca h.Hnh 1.13

+ T = {t1, t2, t3, ... ti} l tp hu hn cc chuyn trng thi, biu din bng du gchngang. Gia hai trng thi lun tn ti mt chuyn trng thi, chuyn trng thi nyc dng hm Bool gn vi mt chuyn trng thi mt tip nhn .

Vic thc hin chuyn trng thi tj hnh 1.14 c thchin bi tch Ev.a. c , trong Ev l bin c trng cho shot ng trng thi Ev, cn a, c l cc bin vo. iu kin chuyn trng thi tj l tj = Ev.a.c .

ctj Ev.a.

Hnh 1.14

tjVic chuyn trng thi tj hnh 1.15 c thc hin biiu kin logic: Ev.(a), trong Ekl bin c trng chos hot ng trng thi Ek, cn a biu din s thay i t0 ln 1ca bin u vo a.

Ev.(a)

Hnh 1.15

+ A = {a1, a2, a3, ... ai} l tp cc cungnh hngni gia 1 trng thi vi 1 chuyntrng thi hoc 1 chuyn trng thi vi mt trng thi.+ M = {m1, m2, m3, ... mi} l tp cc gi tr(0,1). Nu mi = 1 th trng thi i l hotng, ngc li trng thi i khng hot ng.1.8.3. Mt sk hiu dng trong Grafcet:

a) Hnh vung c nh s nh hnh 1.16 a), b) biu th trng thi; hnh ch nhtbn phi dng m t hot ng ca trng thi .

b) Hai hnh chnht lng vo nhau c nh s, biu th trng thi khi u.c) Hnh vung nh s c km theo du chm .biu th trng thi hot ng.



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2.3 3Khi ng

ut ht

dng bng ti

h ia

1

a) b)c) d)

Hnh 1.16a, b k hiu trng thi ; c trng thi khi u; d trng thi hot ng


32

1

t12 t13t79 t89

b) OR9

7 8

9

d) AND

7 8

t789

Hnh 1.17

t/q/2s

d)

10

9

d

c)

8

7

c

b)

6

5

b

a)

4

3

a) OR

c) AND

1

2

t123

3

Hnh 1.18


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d) Du gch ngang biu th cho vic chuyn trng thi. Trng thi c chuynkhi iu kin chuyn c tho mn. Xem hnh 1.17

e) Cc k hiu phn nhnhhnh 1.18:

Hnh 1.18 a) khi TT1 ang hot ng nu t12 tho mn th TT2 hot ng; nut13 tho mn th TT3 hot ng; nu t12 v t13 cng tho mn th TT2 v TT3cng hot ng gi l TT OR. Tng t cho hnh 1.18 b).Hnh 1.18 c) TT1 ang hot ng nu t123 tho mn th c hai TT2 v TT3 hotng gi l trng thi AND.Hnh 1.18 d) TT7 v TT8 ang hot ng nu t789 tho mn th TT9 hot ngtrng thi ny gi l TTAND.

f) Hnh 1.19 a) cho php thc hin bc nhy, nu ang hot ng TT2, iukin a tho mn th h thng s chuyn hot ng t TT2 sang TT5 b qua TT3v TT4; ngc li nu a khng tho mn th cc trng thi 3, 4, 5 ln lt s

c thc hin.

a

2 6

a d

3 7

eb

4 8

c f5 9

a) b)

Hnh 1.19

Hnh 1.19 b) nu iu kin f cha tho mn th TT8 s quay v li TT7, nu f tho mnth TT8 mi chuyn sang TT9.1.8.4. ng dng Grafcet:

Ta xt mt v d c th m t hot ng ca h thng tng iu khin qutrnh. H thng trn c s cng ngh hnh 1.20. Thng X dng cha ncchun b cho h thng trn. Trc khi ng cM ko cnh khuy trn yu cuthng Y phi c nc; cn 1 v 2 cn vt liu; lc ng cM khi ng cnhkhuy cng l lc hai bng ti C1, C2 c khi ng a hai vt liu A, B vothng trng Y.



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YSn phm raEv

PV1

M Nmin

NlimNma

x

2

1

C1

C2

V5

V3V4

V2

B

A

Hnh 1.20: S cng ngh ca h thng trn

Trnh t khuy trn nh sau:- Nu mc vt liu thng trn l min (Nmin) th h thng lm vic ch t

ng (AUT) Cp tn hiu cho mcc van V1, V2, V3.- Bm P c khi ng bm nc t thng X vo thng Y.- Khi khi lng cn trn cc cn 1, 2 th van V2, V3ng li.- Nc trong thng Y tng dn cho n khi t mc max (Nmax) th bm P dng

v van V1ng li.- Khi vic chun b nguyn vt liu trn xong, ng ckhuy M bt u hot

ng ng thi cc van V4, V5 m, bng ti C1, C2 hot ng a liu vothng Y.

- Qu trnh trn c tnh bng thi gian t2, sau thi gian t2 th c tn hiu Ft2 xuthin v ct ng ckhuy M kt thc qu trnh trn.

- Nlim l tn hiu cc hn trn cm hot ng khi thng trn Y qu y.- Trc khi ng cM hot ng th van Ev m tho ht vt liu trong thng Y

ra ngoi n mc min (Nminng), ng thi vt liu trn cn 1, 2 ht th vanV4, V5 tng ng li nhng bng ti C1, C2 cn phi quay thm mt on naa ht vt liu trn bng ti xung thng Y.

- V l do an ton, h thng cn c nt dng khn cp (AU) khi h thng c s cbt thng, ng thi trc khi h thng hot ng li cn c tn hiu t li choh thng (REP).



30/214


S cu trc ca h thng:

BA

NminNmax

Nlim

Ev

V5

V4

V3

V2

V1

C2

C1

Cc thit bca qu trnh

Cu trc cahiu khin

Hnh 1.21: S cu trc chung ca h thng trn

y:M, P, C1, C2, V1, V2, V3, V4, V5, Ev l bin iu khin qu trnh: AUT, AU, REP.

A, B, Nmin, Nmax, Nlim l tn hiu qu trnh a viu khin trng thi.Vi v d c th ny chng ta cn lu n hai phng thc iu khin sau:1. Phng thc lm vic tng theo chu k. Chu ky l chu k trn, ngha

l h thng thc hin xong mi m trn. Mt m trn c bt u bng tnhiu iu khin AUT (iu kin bt u l P, M, V1, V2... trng thi cha lmvic).

2. Phng thc kho khi c s c, khi c s c ngu nhin th h thng phi cdng khn cp bng lnh AU. Lc ny phi cht li ngay kt quang x l,n khi no s cc khc phc xong th c hot ng theo trnh tt li

bng lnh REP vi vic tnh n hoc khng tnh n iu kin khi ng ban

u.Ban u chng ta bt u i vo thit k h thng cha c lnh AU v REP tham gia, l Grafcet hnh 1.22. Trng thi khi u trong trng hp ny l TT1. Gi s cciu kin u l thng mc min, c cu chp, hnh trng thi tt (sn sng lmvic)th trng thi 2, 3, 4 c thc hin (van V1 m, bm P quay, van V2, V3 ma vtliu xung cn 1, 2). Khi nc trong thng dng ln n mc max (Nmax) th h thngchuyn sang trng thi 5. Khi khi lng trn cn 1 (tn hiu bo A), khi lng



31/214


trn cn 2 (tn hiu bo B) th h thng chun sang trng thi 6, 7. Trng thi 5, 6, 7biu hin cho nguyn liu trong mt m trn chun b xong. Khi cc iu kin Nmax,A, B tha mn th h thng s chuyn sang trng thi 8, tng ng ng ctrn Mhot ng, thi gian t2c tnh, van V4, V5 m, bng ti C1, C2 hot ng. Khi x htliu trn hai cn 1, 2 th van V1, V2 dng v thi gian t1 c tnh hai bng ti chy

thm 1 thi gian na (t1). Sau thi gian ny bng ti dng v tn hiu F t1 xut hin vh thng chuyn sang trng thi 9, ti y M vn cn hot ng n khi thi gian t2 ktthc h thng s chuyn v trng thi ngh chun b cho chu k tip theo.

AU AUREP_

10

11

P, V1

Mmin

Ft2

M, t2

Ft1

M, V4, V5, C1, C2, t1

A BNmax

V3V2

9

8

5

2

6

3

7

4

1

AU_Nmin

Hnh 1.22

Hnh 1.22 b) xt cho trng hp s c v khc phc xong s c tin hnh chy

li h thng.



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Hnh 1.23 v 1.24 tnh n cc trng hp s c v t li.

AU

AU

AU

Ft1.M10

REP.M10

C1, C2, t1:M10E .M10

Nmin.M10

1615

13 14

12

AU

AU

M10

REP.

10

11

P, V1

Mmin.M10

Ft2.

M, t2

Ft1.

M, V4, V5, C1, C2, t1

B.M10Nmax.M10 A.M10

V3.M10V2.M10

9

8

5

2

6

3

7

4

1

AU_Nmin.M10

AU

Hnh 1.23



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AU.REP

AU

AU

V2:M14

F2.

M,t2

F1.

M,V4,V5,C1,C2,t1

AU

AU

Nmax.M14 A.M14 B.M14

V3:M14P,V1,Nlim:M14

AUT.M14.Nlim

M14.Nmin Ft1.M14

E :M14 C1,C2,t1:M14

REP.M14

15

15

15

15

15

15

15

15

15

15

15

15

14

M14

14

15

Hnh 1.24



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Chng 2: Biu khin lp trnh PLC B mn Tng - o Lng _ Khoa in

CHNG 2: BIU KHIN LP TRNH PLC

2.1. c im biu khin logic kh trnh (PLC):Programmable Control Systems

Programmable Logic Controller (PLC)

Sra i ca biu khin PLC:- Nm 1642, Pascal pht minh ra my tnh ckh dng bnh rng.n nm 1834 Babbage hon thin my tnh ckh "vi sai" c kh nng tnh tonvi chnh xc ti 6 con s thp phn.- Nm 1808, Joseph M.Jaquard dng cc l trn tm ba th kim loi mng, sp xpchng trn my dt theo nhiu chiu khc nhau iu khin my dt tng thchin cc mu hng phc tp.- Trc nm 1904, Hoa K v c s dng mch rle trin khai chic my tnhin tu tin trn th gii.- Nm 1943, Mauhly v Ackert ch to "ci my tnh" u tin gi l "my tnh v tch

phn sin t" vit tt l ENIAC. My c: 18.000 n in t chn khng. 500.000 mi hn th cng. Chim din tch 1613 ft2. Cng sut tiu thin 174 kW. 6000 nt bm. Khong vi trm phch cm.

Chic my tnh ny phc tp n ni ch mi thao tc c vi pht li v h hng xut hin. Vic sa cha lp t li n in t chy li phi mt n c tun.Ch ti khi p dng k thut bn dn vo nm 1948, a vo sn xut cng nghip vo

nm 1956 th nhng my tnh in t lp trnh li mi c sn xut v thng miho.

S pht trin ca my tnh cng km theo k thut iu khin tng. Mch tch hp in t - IC - nm 1959. Mch tch hp gam rng - LSI - nm 1965. B vi x l - nm 1974. D liu chng trnh - iu khin. K thut lu gi...Nhng pht minh ny nh du mt bc rt quan trng v quyt nh trong

vic pht trin t k thut my tnh v cc ng dng ca n nh PLC, CNC,... lc ny

khi nim iu khin bng ckh v bng in t mi c phn bit.n cui thp k 20, ngi ta dng nhiu ch tiu phn bit cc loi k thut iukhin, bi v trong thc t sn xut i hi iu khin tng th nhng h thng mytnh ch khng iu khin n l tng my.


S pht trin ca PLC em li nhiu thun li v lm cho n cc thao tc my trnn nhanh, nhy, d dng v tin cy hn. N c kh nng thay th hon ton cho ccphng php iu khin truyn thng dng rle (loi thit b phc tp v cng knh);


35/214


kh nng iu khin thit b d dng v linh hot da trn vic lp trnh trn cc lnhlogic cbn; kh nng nh thi, m; gii quyt cc vn ton hc v cng ngh;kh nng to lp, gi i, tip nhn nhng tn hiu nhm mc ch kim sot s kchhot hoc nh ch nhng chc nng ca my hoc mt dy chuyn cng ngh.

Nh vy nhng c im lm cho PLC c tnh nng u vit v thch hp trong

mi trng cng nghip: Kh nng khng nhiu rt tt. Cu trc dng module rt thun tin cho vic thit k, m rng, ci to nngcp... C nhng modul chuyn dng thc hin nhng chc nng c bit haynhng modul truyn thng kt ni PLC vi mng cng nghip hoc mngInternet...

Kh nng lp trnh c, lp trnh d dng cng l c im quan trng xphng mt h thng iu khin tng . Yu cu ca ngi lp trnh khng cn gii v kin thc in t m ch cnnm vng cng ngh sn xut v bit chn thit b thch hp l c th lp trnhc. Thuc vo h sn xut linh hot do tnh thay i c chng trnh hoc thayi trc tip cc thng s m khng cn thay i li chng trnh.

2.2. Cc khi nim cbn v PLC:Cc thnh phn ca mt PLC thng c cc modul phn cng sau:1. Modul ngun.2. Modul n v x l trung tm.3. Modul b nhchng trnh v d liu.

4. Modul u vo.5. Modul u ra.6. Modul phi ghp ( h trcho vn truyn thng ni b).7. Modul chc nng ( h trcho vn truyn thng mng).

n vx l

trung tm

Panel lptrnh, vn

hnh,

gim st.

Khi ng vo

Qunl

vicphighp

Ngun

B nhd liu

Khi ng ra

B nhchngtrnh

Hnh 2.1: M hnh tng qut ca mt PLC



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2.2.1. PLC hay PC: thc hin mt chng trnh iu khin s th yu cu PLC phi c tnh nng

nh mt my tnh (PC). CPU (n v x l trung tm). B nhchnh (RAM, EEPROM, EPROM...), b nhmrng. Hiu hnh. Port vo/ra (giao tip trc tip vi thit biu khin). Port truyn thng (trao i thng tin vi mi trng xung quanh). Cc khi chc nng c bit nh: T, C, cc khi chuyn dng khc.

2.2.2. So snh vi h thngiu khin khc:

h m

PLC x l mt bit.PLC x l t ng

Rle, linh kin in t, mchin t, c- thu kh

Lin kt cng Lin ktch c

RAM -EEPROM

ROM -EPROM

B nhthayi c

Kh lp trnht do

Thay i c

Khng thay

i

Quy trnh cng Quy trnh mm

B nhkh lp trnhTp xc vt l

Vi chc nng c lu tr bng :

iu khin

Hnh 2.2: Nhng c trng lp trnh ca cc loi iu khin

PLC c u im vt tri so vi cc h thng iu khin cin nh rle, mch t hp

in t, IC s. Thit b cho php thc hin linh hot cc thut ton iu khin s thng quangn ng lp trnh. Biu khin s nh gn. D dng trao i thng tin vi mi trng xung quanh nh: TD (text display),OP (operation), PC, PG hay mng truyn thng cng nghip, k c mnginternet.



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Thc hin chng trnh lin tc theo vng qut.2.3. Cu trc phn cng ca PLC:2.3.1.n v xl trung tm (CPU Central Procesing Unit):

Thng trong mi PLC c mt n v x l trung tm, ngoi ra cn c mt sloi ln c ti hai n v x l trung tm dng thc hin nhng chc nng iu

khin phc tp v quan trng gi l hot standby hay redundant.a) n v xl "mt -bit": Thch hp cho nhng ng dng nh, chn thun l logicON/OFF, thi gian x l di, nhng kt cu n gin nn gi thnh h vn c thtrng chp nhn.b)n v xl "t- ng":

X l nhanh cc thng tin s, vn bn, php tnh, o lng, nh gi, kim tra. Cu trc phn cng phc tp hn nhiu. Gi thnh cao.

* Nguyn l hot ng:- Thng tin lu tr trong b nhchng trnh gi tun t (do c iu khin vkim sot bi bm chng trnh do n v x l trung tm khng ch).- B x l lin kt cc tn hiu (d liu) n l (theo mt quy nh no - do thutton iu khin) rt ra kt qu l cc lnh cho u ra.- S thao tc tun t ca chng trnh i qua mt chu trnh y ri sau li bt uli tu thi gian gi l "thi gian qut".-o thi gian m b x l x l 1 Kbyte chng trnh lm ch tiu nh gi gia cc PLC. Nh vy b vi x l quyt nh kh nng v chc nng ca PLC.

Bng 2.1: So snh b vi x l 1 bt v b vi x l t ng


B xl mt - bit B xl t- ngX l trc tip cc tn hiu u vo(a chn).

Cc tn hiu vo/ra ch c thc

a ch ho thng qua t ng.Cung cp lnh nh hn, thngthng ch l mt quyt nhc/ khng.

Cung cp tp lnh ln hn, i hiphi c nhng kin thc v vi tnh.

Ngn ngu vo n gin, khngcn kin thc tnh ton.

Ngn ng u vo phc tp dngcho vic cung cp lnh ln.

Kh nng hn ch trong vic x ld liu s (khng c chc nngton hc v logic).

Thu thp v x l d liu s.

Chng trnh thc hin lin tip,

khng b gin on, thi gian cachu trnh tng i di.

Cc qu trnh thi gian ti hn

c a ch ho qua cc lnh ginon hoc chuyn i iu khinkhn cp.

Ch phi c vi my tnh ngin.

Phi ghp vi my tnh hoc hthng cc my tnh.

Kh nng x l cc tn hiu tngt b hn ch.

X l tn hiu tng t c uvo v u ra.


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2.2.3. B nh: Bao gm c RAM, ROM, EEPROM.Mt ngun in d phng l cn thit cho RAM duy tr d liu ngay c khi

mt ngun in chnh.B nhc thit k thnh dng modul cho php d dng thch nghi vi cc

chc nng iu khin vi cc kch ckhc nhau. Mun rng b nhch cn cm thnhvo rnh cm chsn trn modul CPU.2.3.4. Khi vo/ra:

Hot ng x l tn hiu bn trong PLC: 5VDC, 15VDC (in p cho h TTL &CMOS). Trong khi tn hiu iu khin bn ngoi c th ln hn. khong 24VDVn 240VDC hay 110VAC n 220VAC vi dng ln.

Khi giao tip vo ra c vai tr giao tip gia mch vi in t ca PLC vimch cng sut bn ngoi.Thc hin chuyn mc in p tn hiu v cch ly bngmch cch ly quang (Opto-isolator) trn cc khi vo ra. Cho php tn hiu nhi quav ghim cc tn hiu c mc cao xung mc tn hiu chun. Tc dng chng nhiu tt

khi chuyn cng tc bo v qu p t ngun cung cp in ln n in p 1500V.Ng vo: nhn trc tip tn hiu t cm bin.Ng ra: l cc transistor, rle hay triac vt l.

2.3.5. Thit b lp trnh: C 2 loi thit b c th lp trnh c l Cc thit b chuyn dng i vi tng nhm PLC ca hng tng ng.My tnh c ci t phn mm l cng c l tng nht.

2.3.6. Rle: Rle l b nh1 bt, c tc dng nh rle ph trvt l nh trong mchiu khin dng rle truyn thng gi l cc r le logic. Theo thut ng my tnh thrle cn c gi l c, k hiu l M. C rt nhiu loi rle chng ta s kho st k hni vi loi cc PLC ca hng.

2.3.7. Modul qun l vic phi ghp: Dng phi ghp b PLC vi cc thit b bnngoi nh my tnh, thit b lp trnh, bng vn hnh v mng truyn thng cngnghip.2.3.8. Thanh ghi (Register): l b nh16 bit hay 32 bit lu tr tm thi khi PLCthc hin qu trnh tnh ton.- Thanh ghi cht (Latch register) duy tr ni dung cho n khi n c chng ln bngni dung mi.- Thanh ghi chuyn dng (Special register).

- Thanh ghi tp tin hay thanh ghi b nhchng trnh (Program memory registers).- Thanh ghi iu chnh gi trc t bin trbn ngoi (External adjusting register).

- Thanh ghi ch mc (Index register).2.3.9. Bm (Counter): k hiu l C.a) Phn loi theo tn hiu u vo:- Bm ln.- Bm xung.- Bm ln - xung, bm ny c cchuyn dng chn chiu m.- Bm pha ph thuc vo s lch pha gia hai tn hiu xung kch.



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- Bm tc cao (high speed counter), xung kch c tn s cao khong vi kHz nvi chc kHz.b) Phn loi theo kch thc ca thanh ghi v chc nng ca bm:- Bm 16 bit: thng l bm chun, c gi trm trong khong -32768 32767.- Bm 32 bit: cng c th l bm chun nhng thng l bm tc cao.

- Bm cht: duy tr ni dung m ngay c khi PLC b mt in.2.3.10. Bnh th (timer): k hiu l T, c dng nh cc s kin c quan tmn vn thi gian, bnh th trn PLC c gi l bnh th logic. Vic t chcnh th thc cht l mt bm xung vi chu k c th thay i c. Chu k caxung tnh bng n vms gi l phn gii. Tham s ca bnh th l khong thigian nh th, tham s ny c th l bin hoc l hng nhp vo l s nguyn.2.4. Gii thiu mt s nhm PLC ph bin hin nay trn th gii:1. Siemens: c ba nhm

CPU S7 200:

CPU 21x: 210; 212; 214; 215-2DP; 216.

CPU 22x: 221; 222; 224; 224XP; 226; 226XM. CPU S7300: 312IFM; 312C; 313; 313C; 313C-2DP+P; 313C-2DP; 314;

314IFM; 314C-2DP+P; 314C-2DP; 315; 315-2DP; 315E-2DP; 316-2DP; 318-2

CPU S7400: Lin h cataloge Siemens.2. Mitsubishi: H FX3. Omron: H CMQ4. Controtechnique: H Compact TWD LCAA 10DRP; TWD LCAA 16DRP; TWDLCAA 24DRP...

5. ABB: Ba nhm AC 100M

AC 400M AC 800M, y l loi c 2 module CPU lm vic song song theo ch dphng nng.

2.5. Tng quan v h PLC S7-200 ca hng Siemens:C hai series: 21x (loi c khng cn sn xut na) v 22x (loi mi). V mt

tnh nng th loi mi c u im hn nhiu. Bao gm cc loi CPU sau: 221, 222,224, 224XP, 226, 226XM trong CPU 224XP c h tranalog 2I/1O onboard v 2port truyn thng.



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Bng 2.2: Cc loi CPU S7-200

Bng 2.3: So snh cc thng s v c im k thut ca series 22x



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Bng 2.4: M s v cc thng s vin p ngun v I/O

Bng 2.5: Cc thng s v cng sut tiu th v dng in I/O

2.6. Cu trc phn cng ca S7-200:2.6.1. Hnh dng bn ngoi:1. Cc n trng thi:

n RUN-mu xanh: Chnh PLC ch lm vic v thc hin chngtrnh c np vo b nhchng trnh. n STOP-mu vng: Chnh PLC ch STOP, dng chng trnh angthc hin li (cc u ra u ch off).

n SF-mu , n bo hiu h thng b hng c ngha l li phn cng hochiu hnh. y cn phn bit r li h thng vi li chng trnh ngidng, khi li chng trnh ngi dng th CPU khng th nhn bit c vtrc khi download xung CPU, phn mm lp trnh lm nhim v kim tratrc khi dch sang m my.



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Hnh 2.3: CPU S7-200 module

n Ix.x-mu xanh: Chnh trng thi On/Off ca u vo s. n Qx.x-mu xanh: Chnh trng thi On/Off ca u vo s. Port truyn thng ni tip: RS 485 protocol, 9 chn s dng cho vicphi ghp vi PC, PG, TD200, TD200C, OP, mng bin tn, mng cngnghip.

Tc truyn - nhn d liu theo kiu PPI tc chun l 9600 baud.Tc truyn - nhn d liu theo kiu Freeport l 300 38400 baud.

Hnh 2.4: Cu trc ca port RS 485



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Bng 2.6: M t chc nng ca cc chn ca port RS 485

2. Cng tc chn ch: Cng tc chn ch RUN: Cho php PLC thc hin chng trnh, khi

chng trnh gp li hoc gp lnh STOP th PLC s tng chuyn sang ch STOP mc d cng tc vn ch RUN (nn quan st n trng thi). Cng tc chn ch STOP: Khi chuyn sang ch STOP, dng cng bcchng trnh ang chy, cc tn hiu ra lc ny u v off. Cng tc chn ch TERM: cho php ngi vn hnh chn mt trong haich RUN/STOP t xa, ngoi ra ch ny c dng download chngtrnh ngi dng.

3. Vt chnh nh tng t: Mi CPU c t 1 n 2 vt chnh nh tng t, c th xoayc mt gc 270, dng thay i gi tr ca bin s dng trong chng trnh.4. Pin v ngun nui b nh: S dng t vn nng v pin. Khi nng lng ca t b cn

kit PLC s tng chuyn sang s dng nng lng t pin.2.6.2. Giao tip vi thit b ngoi vi:a) Thit b lp trnh loi PGxx c trang b sn phn mm lp trnh, ch lp trnh cvi ngn ng STL.b) My tnh PC: Hiu hnh Win 95/98/ME/2000/NT4.x.Trn c ci t phn mm Step7 Micro/Win 32 v Step7 Micro/Dos. Hin nay huht s dng Step7 Mcro/Win 32 version 3.0, 3.2, 4.0. V4.0 cho php ngi lp trnh cth xem c gi tr, trng thi cng nh th ca cc bin. Nhng ch s dng ctrn my tnh c ci t hiu hnh Window 2000/ WinNT v PLC loi version minht hin nay. Sau y l cch ci t v giao tip gia PC-PLC:



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Hnh 2.6: Ci t phn mm STEP7 trn Window 95/98/ME/2000/NT

Sau khi thc hin xong, trn mn hnh s xut hin:



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Hnh 2.7: Giao tip gia PC/PG vi PLC thng qua PC/PPI cable


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Sau khi tin hnh lp t phn cng xong, ta tin hnh thit lp truyn thng giaPC/PG vi PLC. l thit lp tc , s bit d liu truyn/nhn, bit chn l, cngCOM, a ch PLC, thi gian Time out... c tin hnh nh sau:

Hnh 2.8: Thit lp kt ni gia PC/PG ti PLC

1. Kch chut vo biu tng Communications trn Group bar.2. Kim tra vic thit lp truyn thng.3. Kch double vo biu tng Refresh d tm a ch v cc thng sca PLC.4. Nu khng nhn c phn hi t PLC hoc Window vn khng thit lpc truyn thng th kch vo Set PG/PC interface sau kch double voPC/PPI cable.

5. nh du vo PC/PPI cable v chn properties...


6. Vo PPI/Addres t a ch 2 (a ch mc nh) v tc truyn l 9.6kbps. Vo Local connection/ connnection to chn port kt ni(COM1/COM2/USB), chp nhn vic la chon ny bng nt OK.


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7. Vo kch double biu tng Refresf ln na xem s kt ni gia PCv PLC.

c) Giao tip vi mng cng nghip:Nu l mng PPI th ch cn u ni v ni trc tip vo Port truyn thng caCPU.

Nu l mng Profibus - DP phi c thm modul EM 277.Nu l mng Ethernet hoc internet phi c thm modul CP 243-1/ CP 243-1IT.

Nu l mng AS-I phi c thm modul CP 243-2.Ngoi ra cn c thm TD200 (Text Display) dng hin th v thng bobng text, c thiu chnh trc tip gi tr ca bin trong chng trnh ngidng, ng vai tr nh mt panel vn hnh.

TP070 loi ny l Touch panel, c thit kc bit cho S7-200, c chc nng nhHMI (Human Mechanical Interface).

Tip!: Gi phn mm STEP 7 Micro/Win32 V3.x cng c chia ra nhiu modul.Modul chnh dng thc hin nhnh chc nng cbn, mt s modul chuyn dngnh: USS hay Modbus, S7-200 Toolbox: TP_Desinger cho OP 070 ( cu hnh choTO 070), Microcomputing limited, ActiveX components h trvic truyn thnggia PC vi PLC qua cc ngn lp trnh khc. S7-200 OPC server for random OPCclients cng s dng cho vic truy xut d liu vi S7-200.2.6.3. Giao tip gia sensor v ccu chp hnh:S7-200 c hai loi cbn:AC/DC/RLY_loi ny in p ngun cung cp t 85 264VAC, tn s 47 63 Hz;in p vo: c ngun cung cp in p chun cho sensor l 24VDC.in p ra: loi ny s dng ngun in ngoi, c th l DC hoc AC nhng khng

vt qu 220V. Nu s dng i vi nhng thit b tiu th c cng sut b khongchng vi Woat th c th ly trc tip ngun ca cm bin.Sau y l th d v mch in giao tip gia PLC vi cm bin v ccu chp hnh lng c1 chiu c o chiu quay.



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Hnh 2.9: S mch in giao tip gia CPU 221loi AC/DC/RLY v ccu chp hnh



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Hnh 2.10: S mch giao tip gia CPU 224 AC/DC/RLYvi sensor v ccu chp hnh

Hnh 2.11: S mch giao tip gia CPU 224 DC/DC/DCvi sensor v ccu chp hnh

DC/DC/DC_Ngun nui 24VDC.Ngun nui cm bin 24VDC.u ra Transitor hcolector ngun cung cp 24VDC.

2.7. Cu trc b nhS7-200:2.7.1. Phn chia b nh: B nhc chia lm 4 vng cbn, hu ht cc vng nhu c kh nng c/ghi ch tr vng nhc bit SM (Special Memory) l vng nhc s chc, s cn li c thc/ghi c.

Vng nh chng trnh: L min b nhc dng lu gi cc lnh.chng trnh. Vng ny thuc kiu non-valatie c/ghi c. Vng nhtham s: L min lu gi cc tham s nh t kho, a ch trm...cng ging nh vng chng trnh, vng ny thuc kiu (non-valatile) c/ghic.



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Vng dliu:c s dng ct cc d liu ca chng trnh bao gm ktqu ca cc php tnh, hng sc nh ngha trong chng trnh, bmtruyn thng... Vng i tng: Timer, bm, bm tc cao v cc cng vo/ra tngtc t trong vng nhcui cng. Vng ny khng thuc kiu non-valatilenhng c/ghi c.

Hai vng nhcui cng c ngha quan trng trong vic thc hin mt chng trnh.Do vy sc trnh by chi tit mc tip theo.

Hnh 2.12: B nhtrong v ngoi ca S7200

T

Min nhngoiEEPROMi tng

Tham s

D liu

Chng trnh

D liu

Tham s

Chng trnh

D liu

Tham s

Chng trnh

2.7.2. Vng nhdliu v i tng v cch truy cp:Vng nhd liu l vng nhng, n c th truy cp theo tng bit, byte, tn(worrd), t kp (double word) v cng c th truy nhp c vi mng d liu. cs dng lm min lu tr d liu cho cc thut ton, cc hm truyn thng, lp bng,cc hm dch chuyn, xoay vng thanh ghi, con tra ch...Vng i tngc s dng lu gi d liu cho cc i tng lp trnh nh cc

gi tr tc thi, gi trt trc ca Counter hay Timer. D liu kiu i tng baogm cc thanh ghi ca counter, Timer, cc bm tc cao, bm vo/ra tng tv cc thanh ghi AC (Accumulator).

Vng nhdliu v i tngc chia ra nhiu min nhnh vi nhng ng dngkhc nhau. Chng c k hiu bng ch ci u ca tn ting Anh. Thng s, chcnng, gii hn ca cc vng nh tng ng vi tng CPU c m t qua cc bngsau:



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Bng 2.7: c im v gii hn vng nhca CPU S7 22x.

a ch truy nhp c quy c vi cng thc: Truy nhp theo bit:- Vit: tn min (+) a ch byte (+).(+) ch s bit (t 07).- c: ngc li, v d: V12.7_bit 7 ca byte 12 trong vng nhV.

M8.2_bit 2 ca byte 8 trong vng nhM.

Truy nhp theo byte:- Vit: tn min (+) B (+) a ch ca byte trong min.- c: ngc li, v d: VB32_byte 32 trong vng nhV. truy nhp theo Word (tn):- Vit: tn min (+) W (+) a ch byte cao ca t trong min.- c: ngc li, v d: VW180_Word 180 trong vng nhV, t ny gm c2 byte 180 v 181.



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15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

VW180VB180 b te cao VB181 b te th

Truy nhp theo double Word (tkp):- Vit: tn min (+) D (+)a ch byte cao ca t cao trong min.- c: ngc li, v d: VD8_double Word 8 trong vng nhV, t kp nybao gm 4 byte 8, 9, 10, 11.

31 24 23 16 15 8 7 0

VD8 B te 8 B te 9 B te 10 B te 11

Tt c cc byte thuc vng d liu u c th truy nhp bng con tr. Con trquy nh trong vng nhV, L hoc cc thanh ghi AC1, AC2, AC3. Mi con tr gm4 byte, dng lnh MOVD. Quy c s dng con tr truy nhp nh sau:

Truy nhp con tr

a ch:&a ch byte (cao) l ton hng ly a ch ca byte, t hoc t kp m con trang

ch vo. V d:- AC1=&VB10, thanh ghi AC1 cha i ch ca byte 10 thuc vng nhV.- VD100=&VW110, t kp VD100 cha a ch byte cao (VB110) ca tnVW110.

- AC2=&VD150, thanh ghi AC2 cha a ch ca byte cao (VB150) ca tkp VD150.

Truy nhp con tr dliu:*con tr d liu l ton hng ly ni dung ca byte, t hoc t kp m con trang ch vo. V d nhi php gn a ch trn th:- *AC1 = VB10, ly ni dung ca byte VB10.- *VD100 = VW110, ly ni dung ca tn VW110.- *AC1 = VD150, ly ni dung ca t kp VD150.Php gn a ch v s dng con tr nh trn cng c tc dng vi nhng

thanh ghi 16 bit ca Timer, bm thuc vng i tng hay cc vng nhI, Q, V, M,AI, AQ, SM.



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AC1

VB109VB110

VB111

VB112VB113 AC0

5 6 7 8

a ch VW112

1 2

3 4

5 6

7 8

AC1

VB109VB110

VB111

VB112VB113

AC0

1 2 3 4

a ch VW110

1 2

3 45 6

7 8

MOVD &VW110, AC1 to con tra ch bng ccha a ch ca byte cao VB110 vo thanh ghi AC1.

MOVD *AC1, AC0, a gi tr trong word VW110vo trong thanh ghi AC0.

+D +2, AC1 cng 2 vo gi tra ch ca con trVW110 ri lu gi trong thanh ghi AC1.

MOVD *AC1, AC0, a gi tr trong word VW112vo trong thanh ghi AC0.

Hnh 2.13: Cch to v s dng con tra ch

2.7.3 Mrng cng vo ra:S module m rng tu thuc vo tng loi CPU, s module tng ng vi

tng loi CPU c trnh by theo bng 2.3. Cch mc ni cc module mrng cmc ni tip (theo mt mc xch) v pha bn phi ca module CPU.Cc module s hoc tng tu chim ch trn bm vo/ra tng ng

vi u vo/ra ca module. V d v cch khai bo a ch trn cc module mrng:

Hnh 2.14: Ghp ni CPU 224XP vi module mrng



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Hnh 2.15: Ghp ni CPU 212 vi module mrng

Hnh 2.16: Ghp ni CPU 214 hoc 215 vi module mrng



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Chng 3: Ngn nglp trnh v ng dng B mn Tng o Lng Khoa in

CHNG 3: NGN NGLP TRNH V NG DNG

3.1.Gii thiu cc ngn nglp trnh:Lp trnh cho S7 200 v cc PLC khc ca hng Siemens da trn 3 phng

php cbn:

Phng php hnh thang (Ladder logic _ LAD).Phng php khi hm (Function Block Diagram _ FBD).Phng php lit k cu lnh (Statement List _ STL).

Chng ny s gii thiu cc thnh phn cbn ca ba phng php v cch s dngchng trong lp trnh.Nu chng trnh c vit theo ngn ng LAD (hoc FBD) th c th chyn sangngn ng STL hay FBD (hoc LAD) tng ng. Nhng khng phi bt c chngtrnh vit theo STL no cng chuyn sang ngn ng LAD hay FBD c. B tp lnhSTL c trnh by trong gio n ny u c mt chc nng nh cc tip im, cundy, cc hp (trong LAD) hay IC s trong FBD.

Nhng lnh ny phi phi hp c trng thi cc tip im quyt nh vgi tr trng thi u ra hoc gi tr logic cho php hoc khng cho php thc chcnng ca mt (hay nhiu) cun dy hoc hp. Trong lp trnh lgic thng hay s dnghai ngn ng LAD v STL v n gn gi hn i vi chuyn ngnh in. Sau y lnhng nh ngha cn phi nm khi bt tay vo thit k mt chng trnh:3.1.1.nh ngha v LAD: LAD l ngn ng lp trnh bng ha. Nhnh thnh phnc bn dng trong LAD tng ng vi nhng thnh phn c bn dng trong bngmch rle.

+ Tip im c hai loi: Thng ng

Thng h+ Cun dy (coil): ( )+ Hp (box): M t cc hm khc nhau, n lm vic khi c tn hiu a nhp. C cc nhm hp sau: hp cc bnh thi, hp cc bm, hp di chuynd liu, hp cc hm ton hc, hp trong truyn thng mng...+ Mng LAD: L mch ni cc phn t thnh mt mng hon thin, cc phn tnh cun dy hoc cc hp phi c mc ng chiu. Ngun in c haing chnh, mt ng bn tri th hin dy nng, mt ng bn phi l dytrung tnh (neutral) nhng khng c th hin trn giao din lp trnh. Mtmch lm vic c khi cc phn tc mc ng chiu v kn mch.

3.1.2. nh ngha v STL: L phng php th hin chng trnh di dng tp hpcc cu lnh. to ra mt chng trnh bng STL, ngi lp trnh cn phi hiu rphng thc s dng 9 bit trong ngn xp (stack) logic ca S7 200.

Ngn xp l mt khi 9 bit chng ln nhau t S0S8, nhng tt c cc thut tonlin quan n ngn xp u lm vic vi bit u tin v bit th hai (S0 v S1) cangn xp. gi tr logic mi c thc gi hoc ni thm vo ngn xp. Hai bit S0 vS1 phi hp vi nhau th ngn xp c ko ln mt bit.Ngn xp ca S7 200 (logic stack):



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S0

S1

S2

S3

S4S5

S6

S7

S8

3.2.Vng qut (thc hin chng trnh) v cu trc ca mt chng trnh:

Stack0 bit u tin ca ngn xp.Stack1 bit th hai ca ngn xp.Stack2 bit th ba ca ngn xp.Stack3 bit th t ca ngn xp.Stack4 bit th nm ca ngn xp.Stack5 bit th su ca ngn xp.Stack6 bit th by ca ngn xp.Stack7 bit th tm ca ngn xp.Stack8 bit th chn ca ngn xp.

PLC thc hin chng trnh theo vng lp. Mi vng lp c gi l vng qut (scan).Cc giai on ca vng qut:

Khi gp lnh vo/ra tc thi ngay lp tc h thng dng tt c mi cng vickhc, ngay c chng trnh x l ngt thc hin chng trnh ny trc tip vi cngvo/ra.

Nu sdng cc ch ngt, chng trnh con tngng vi tng tn hiu

ngtc son tho v cit nhmt b phn ca chng trnh. Chng trnhxl ngt chc thc hin trong vng qut khi xut hin tn hiu bo ngt v cthxy ra bt cthiim ntrong vng qut.



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3.3.Tp lnh S7-200:Tp lnh ca S7-200 c chia lm 3 nhm:

1. Cc lnh m khi thc hin th lm vic c lp khng ph thuc vo gi trlogic ca bit u tin trong ngn xp (gi l nhm lnh khng iu kin).2. Cc lnh ch thc hin khi bit u tin trong ngn xp c gi tr bng 1 (gil nhm lnh c iu kin).3. Cc nhn lnh nh du v tr trong tp lnh (gi l nhm lnh iu khinchng trnh).

! Cc ngn ngsdng chI (Immediately) ch ngha tc thi.

Cy lnhTp lnh BitTp lnh can thip vo thi gian h thngTp lnh truyn thngTp lnh so snhTp lnh bin iTp cc bmTp lnh ton hcTp lnh ton hcTp lnh iu khin ngt

Tp lnh cc php tnh logic bin iTp lnh di chuyn d liuTp lnh iu khin chng trnhTp lnh thao tc vi thanh ghi (dch/quay vng thanh ghi)Tp lnh lm vic vi chuiTp lnh lm vic vi bng d liuTp cc bnh thiTp lnh gi chng trnh con v chng trnh ngt


Hnh 3.3: M t cy lnh vi SIMATIC S7-200


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1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

11

1

1

1

1

1

1

2

2

2

22

3

Hnh 3.4: M t cy lnh bit

2

2

Hnh 3.5:M t cy lnh can thip vo thi gian h thng

2

22

2

2

2

Hnh 3.6:M t cy lnh truyn thng



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11

1

1

11

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

11

1

1

1

1



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1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

11

1

1

1

1

1

1

1

1

11

1

1

1

Hnh 3.7:M t cy lnh so snh



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22

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

Hnh 3.8:M t cy lnh bin i

2

2

2

2

2

2

Hnh 3.9:M t cy lnh cc bm

2

2

2

Hnh 3.10:M t cy lnh cc bnh thi

2

2

2

2

2


Hnh 3.11:M t cy lnh iu khin ngt


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2

2

2

2

2

2

Hnh 3.12:M t cy lnh hc kiu Floating-Point

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

22

2

2

Hnh 3.13:M t cy lnh ton hc kiu Integer

2

2

2

2

22

2

2

2

2

2


Hnh 3.14:M t cy lnhphp tnh logic bin i


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2

2

2

2

2

2

2

2

2

Hnh 3.15:M t cy lnhdi chuyn d liu

3

3

3

3

3

3

3

3

3

3

3

Hnh 3.16:M t cy lnhiu khin chng trnh

2

2

2

2

2

2

2

2

2

2

2

Hnh 3.17:M t cy lnhiu khin chng trnh



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2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

Hnh 3.18:M t cy lnhlm vic vi chui

2

2

2

2

2

2

2

2

!1_Cc lnh khngiu kin.Hnh 3.19:M t cy lnhlm vic vi bng d liu

2_Cc lnh c iu kin.3_Cc lnh iu khin chng trnh.

3.4. C php v cch ng dng SIMATIC struction S7-200:3.4.1. Ton hng v gii hn cho php:



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Bng 3.1: Gii hn ton hng ca CPU S7-200 series CPU 22x

3.4.2. SIMATIC instructions:1. SIMATIC Bit Logic Instructions:

STL LADM t

(Description)Ton hng(Operands)

Kiu dliu

(Data

Types)



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LD

A

O

Tip im thng mscng khi bit = 1

bit: I, Q, M, V, SM,T, C, S, L

Bool

LDN

AN

ON

Tip im thng ng scmkhi bit = 1 bit: I, Q, M, V, SM,T, C, S, L Bool

LDI

AI

OI

Tip im thng msngtc thi (khng ph thuc vo

chu k vng qut)bit: I Bool

LDNI

AIN

OIN

Tip im thng ng s mtc thi (khng ph thuc vo

chu k vng qut)bit: I Bool

NOT o gi tr logic ca bit u tintrong ngn xp

Khng Khng

EU

Bit u tin trong ngn xp cgi tr bng 1 (trong khong thigian ng bng 1 chu k vngqut) khi pht hin sn ln catn hiu u vo.


Bool

ED

Bit u tin trong ngn xp cgi tr bng 1 (trong khong thigian ng bng 1 chu k vngqut) khi pht hin sn xung

ca tn hiu u vo.


Bool

bit

bit

bit

bit

NOT

Hnh 3.20: V d minh ho lnh LD, NOT, ED trong chng trnh LAD v STL



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S

bit

n

SIbit

n

STL LADM t

DescriptionTon hngOperands

Kiu dliu

Data Types

= bitCun dy u ra trng thi ONkhi c dng in iu khin iqua.

bit: I, Q, M, V, SM,T, C, S, L Bool

=I bit

Cun dy u ra trng thi ONtc thi (khng ph thuc vochu k vng qut) khi c dngin iu khin i qua.

bit: QBool

S bit, n

Set 1 mng gm n tip im,tnh t tip im "bit" (n


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2. SIMATIC Compare Byte Instructions:Hnh 3.21: V d minh ho lnh =, S, R trong chng trnh LAD v STL

STL LADM t

(Description)

Ton hng(Operands)

Kiu d liu(Data Types)

COMPARE BYTE

LDB=

AB=

OB=

Lnh so snh gi tr ca hai byteIN1 v IN2.Trng thi tip im l ng khilnh so snh IN1= IN2 l ng.

IB, QB, MB,

VB, SMB, SB,LB, AC,

Constant,

VD, AC,

LD

Byte

LDB

AB

OB

Lnh so snh gi tr ca hai byteIN1 v IN2.Trng thi tip im l ng khilnh so snh IN1 IN2 l ng.

IB, QB, MB,

VB, SMB, SB,LB, AC,

Constant,

VD, AC,

LD

Byte

LDB=

OB>=

Lnh so snh gi tr ca hai byteIN1 v IN2.Trng thi tip im l ng khilnh so snh IN1>= IN2 l ng.

IB, QB, MB,

VB, SMB, SB,LB, AC,

Constant,

VD, AC,

LD

Byte

COMPARE WORD (COPARE INTEGER)LDW=

AW=

OW=

Lnh so snh gi tr ca haiWord IN1 v IN2.

Trng thi tip im l ng khilnh so snh IN1= IN2 l ng.

IW, QW, MW, VW,SMW, SW, LW,

AC, Constant, VD,

AC, LD

Word

LDW

AW

OW


Trng thi tip im l ng khilnh so snh IN1 IN2 l ng.

IW, QW, MW, VW,SMW, SW, LW,

AC, Constant, VD,

AC, LDWord

LDW>

AW>

OW>


Trng thi tip im l ng khilnh so snh IN1 > IN2 l ng.

IW, QW, MW, VW,

SMW, SW, LW,

AC, Constant, VD,

AC, LD

Word

LDW>=

AW>=

OW>=

Lnh so snh gi tr ca haiWord IN1 v IN2.Trng thi tip im l ng khilnh so snh IN1 >= IN2 l ng.

IW, QW, MW, VW,

SMW, SW, LW,

AC, Constant, VD,

AC, LD

Word

LDW

Lnh so snh gi tr ca haiDoubleWord IN1 v IN2.

Trng thi tip im l ng khilnh so snh IN1 > IN2 l ng.

ID, QD, MD, VD,

SMD, SD, LD, AC,

Constant, VD,

AC, LD

Double

Word

LDDW>=

ADW>=

ODW>=

Lnh so snh gi tr ca haiDoubleWord IN1 v IN2.

Trng thi tip im l ng khilnh so snh IN1 >= IN2 l ng.

ID, QD, MD, VD,SMD, SD, LD, AC,

Constant, VD,

AC,

LD

DoubleWord

LDDW=

Lnh so snh gi tr ca hai sthc IN1 v IN2.Trng thi tip im l ng khilnh so snh IN1 >= IN2 l ng

ID, QD, MD, VD,

SMD, SD, LD, AC,

Constant, VD,

AC, LD

Real

LDRDIN2

IN1

>=DIN2

IN1

=RIN2

IN1


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AR (ln hn).

STL LADM t

DescriptionTon hngOperands

Kiu dliuData

Types

Table Fine

TBL: IW, QW,VW, LW, SW,

MW, SMW, AIW,

T, C, VD, AC,

LD

Word

PTN: IW, QW,VW, LW, SW,

MW, SMW, AIW,

T, C, AC,

Constant, VD,

AC, LD

INT

INDX: LW, SW,MW, SMW, AIW,

T, C, VD, AC,

LDWord

FND= TBL,

PARNT, INDX

FND TBL,

PARNT, INDX

FND< TBL,

PARNT, INDX

FND> TBL,

PARNT, INDX

Thc hin vic tm kimtrong bng xc nh biTBL , bt u t v trd liu INDX nhchd liu PARNT. Lut

tm kim c quy nhbi CMD c gi tr t 1n 4 tng ng =, ,.

Khi tm thy , INDX sch vo d liu utin tm c trong bngv lnh c kt thc.Do tm kim dliu tip theo, INDXphi c tng gi tr l

v gi li lnh ny. Nunh khng tm thyINDX c gi trngbng gi tr ca bmEC.

CMD: Constant

Byte

D_T_TBLEN ENO

TBL

PTN

INDX

CMD



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Bng 3.5: S khc nhau gia bng d liu nh ngha bng lnh ATT, FIFO, LIFO vlnh FIN

Hnh 3.35: V d v cch s dng lnh tm kim FND



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STL LADM t

(Description)

Ton hng(Operands)

Kiu dliu

(Data

Types)

Fisrt - In - Fisrf - Out


MW, T, C, VD,

AC, LD INT

FIFO TABLE,

DATA

Lnh ly d liu utin ca bng ra khibng. Nu bng trngc ngha l d liu trong c ly ra ht, hayEC=0, bit SM1.4=1. Dliu ly ra c ghi voDATA (kiu t). Cc dliu cn li c dn lnv tr trn lp chtrng va mi b ly i.

Khi lnh thc hin xongni dung ca EC gimi mt n v.

DATA: IW, QW,VW, LW, SW,

MW, SMW, AIW,

T, C, AQW, VD,AC, LD

Word

FIFOEN ENO

TBL DATA

Hnh 3.36: V d v cch s dng lnh FIFO.



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STL LADM t

Description

Ton hngOperands

Kiu dliuData

Types

Last - In - Fisrf - Out


MW, T, C, VD,

AC, LD INT

LIFO TABLE,DATA

Lnh ly d liu cuicng ca bng ra khibng tc l d liu cnhp sau cng. Nubng trng c nghal d liu trong cly ra ht, hay EC=0, bitSM1.4=1. D liu ly raghi vo DATA(kiu t). Cc d liucn li c dn ln v

tr trn lp ch trngva mi b ly i. Khilnh thc hin xong nidung ca EC gim imt n v.

DATA: IW, QW,VW, LW, SW,

MW, SMW, AIW,T, C, AQW, VD,

AC, LD

Word

LIFOEN ENO

TBL DATA

Hnh 3.37: V d v cch s dng lnh LIFO



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STL LADM t

Description

Ton hngOperands

Kiu dliu

Data Types

Memory Fill

IN: IW, QW, VW,

LW, SW, MW, SMW,AIW, T, C, AC,

Constant, VD, AC,

LD

Word

N: IB, QB, MB, VB,

SMB, SB, LB, AC,

Constant, VD, AC,

LD

ByteFILL IN, OUT,

N

Lnh in gi trcha trong Word INvo mng bt u ta ch Word OUT.N l s tn camng, 1


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ORB IN1, OUT

Lnh thc hin OR gia ccbit tng ng ca hai ByteIN1 v IN2, kt qu ghi voByte OUT.

a hai Byte

IN1 v IN2, kt qu ghi voByte OUT.

XORB IN1,

OUT

Lnh thc hin XOR gia ccbit tng ng ca hai ByteIN1 v IN2, kt qu ghi voByte OUT.

OUT:IB, QB, MB,

VB, SMB,

SB, LB, AC,

VD, AC,

LD

Byte

And Word, Or Word, Exclusive Or Word

ANDW IN1,OUT

Lnh thc hin AND gia ccbit tng ng ca hai WordIN1 v IN2, kt qu ghi voWord OUT.

ORW IN1, OUT

Lnh thc hin OR gia ccbit tng ng ca hai WordIN1 v IN2, kt qu ghi voWord OUT.

IN1, IN2:

IW, QW, VW,LW, SW, MW,

SMW, AIW, T,

C, AC,Constant, VD,

AC, LD

XORW IN1,

OUT

Lnh thc hin XOR gia ccbit tng ng ca hai WordIN1 v IN2, kt qu ghi voWord OUT.

OUT:IW, QW, VW,LW, SW, MW,

SMW, T, C,

AC, VD,

AC, LD

Word

And DWord, Or DWord, Exclusive Or DWord

ANDD IN1,

OUT

Lnh thc hin AND gia ccbit tng ng ca hai t kpIN1 v IN2, kt qu ghi vo tkp OUT.

ORD 1, OUT

Lnh thc hin OR gia ccbit tng ng ca hai t kpIN1 v IN2, kt qu ghi vo tkp OUT.

IN1, IN2:ID, QD, VD,LD, SD, MD,

SMD, HD, AC,

Constant, VD,

AC, LD

XORD IN1,

OUT

Lnh thc hin XOR gia ccbit tng ng ca hai t kpIN1 v IN2, kt qu ghi vo tkp OUT.

OUT:ID, QD, VD,

LD, MD, SMD,

AC, VD,

AC, LD

Double

Word

WOR_BEN ENO

IN1 OUTIN2

WXOR_BEN ENO

IN1 OUTIN2

WAND_WEN ENO

IN1 OUTIN2

WOR_WEN ENO

IN1 OUTIN2

WXOR_WEN ENO

IN1 OUTIN2

WAND_DWEN ENO

IN1 OUTIN2

WOR_DWEN ENO

IN1 OUTIN2

WXOR_DWEN ENO

IN1 OUTIN2



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Hnh 3.39: V d v cch s dng lnh AND, OR, XOR

STL LADM t

Description

Ton hngOperands

Kiu dliuData

Types

Invert Byte, Invert Word, Invert DWord

INVB OUT

Lnh o tng bit cabyte u vo IN, kt qaa ra u ra OUT.Thng th u vo vra cng a ch.

IN:IB, QB, MB,VB, SMB, SB, LB,

AC, Constant,

VD, AC, LDOUT:IB, QB, MB,

VB, SMB, SB, LB,AC, Constant,

VD, AC, LD

Byte

INV_BEN ENO

IN1 OUT



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INVW OUT

Lnh o tng bit ca tn u vo IN, kt qaa ra u ra OUT.u ra OUT.Thng th u vo vra cng a ch.Thng th u vo vra cng a ch.

IN: IW, QW, VW,LW, SW, MW,

SMW, AC, AIW , T,

C, Constant, VD,

AC, LD

IN: IW, QW, VW,LW, SW, MW,

SMW, AC, AIW , T,

C, Constant, VD,

AC, LDOUT: IW, QW,

VW, LW, SW, MW,

SMW, AC, T, C,

VD, AC, LD

OUT: IW, QW,VW, LW, SW, MW,

SMW, AC, T, C,

VD, AC, LD

WordWord

INVD OUT

Lnh o tng bit ca tkp u vo IN, kt qaa ra u ra OUT.Thng th u vo vra cng a ch.

IN: ID, QD, VD,LD, SD, MD, SMD,

HD, AC, Constant,

VD, AC, LDOUT: ID, QD, VD,LD, SD, MD, SMD,

AC, VD, AC, LD

DWord

INV_EN ENO

IN1 OUT

INV_DWEN ENO

IN1 OUT

Hnh 3.40: V d v cch s dng lnh INVB, INVW, INVD10. SIMATIC Stack Logic Instructions:

Cc lnh tip im trong i s Boolean cho php to lp c cc mch logic(khng c nh). Trong LAD cc mch ny biu din thng qua cu trc mch, mc nitip hay song song cc mch tip im thng ng v cc tip im thng m. STL cth s dng cc lnh A (And) v O (Or) cho cc tip im mc ni tip v song song lthng h hoc cc lnh AN (And Not) v ON (Or Not) cho cc tip im mc ni tipv song song l thng ng. Gi tr ca cc bit trong ngn xp thay i tu thuc votng lnh. Trong phn ny chng ta s i su hn v s lm vic ca cc bit trong ngnxp, vic hiu v nm bt v ngn xp l iu rt cn thit trong vn lp trnh dngngn ng STL.

Ngoi nhng lnh lm vic trc tip vi tip im, S7-200 cn c 5 lnh c bitbiu din cc php tnh ca i s Boolean cho cc bit trong ngn xp, c gi l cclnh stack logic. Trong LAD khng dng nhng lnh ny. STL s dng cc lnh ny



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thc hin nhng php ton ca phng trnh c nhiu biu thc con. Sau y l bng tmtt c php v hng dn cch s dng lnh.

STL LADM t

Description

Ton hngOperands

Kiu dliuData

TypesAnd Load

ALD none

Lnh t hp gi tru tin v gi tr ca bitth hai trong ngn xp bng php tnh .Kt quc ghi li vo bit u tin cangn xp. Gi tr cn li c ko ln 1 bit.

none none

Or LoaD

OLD none

Lnh t hp gi tru tin v gi tr ca bitth hai trong ngn xp bng php tnh .Kt quc ghi li vo bit u tin cangn xp. Gi tr cn li c ko ln 1 bit.

none none

Logic PuSh

LPS none

Sao chp gi tr ca bit u tin vo bit thhai trong ngn xp. Ga tr cn li byxung 1 bit. Bit cui cng by ra ngoi.

none none

Logic ReaD

LRD noneLnh sao chp gi tr ca bit th hai vo bitu tin ca ngn xp, cc gi tr cn lica ngn xp vn gi nguyn.

none none

Logic PoP

LPP noneLnh ko ngn xp ln 1 bit theo nguyntc bit sao ln bit trc.

none none

LoaD Stack

LDS n none

Lnh sao chp gi tr ca bit th n (ngnxp c 9 bit th bit th nh c tnh l1...n bit cui cng l 8) ca ngn xp lnbit u tin. Cc gi tr cn li ca ngn xpby li xung 1 bit, bit cui cng byra khi ngn xp.

n: 18 Byte

Hnh 3.41: M t hot ng ca lnh LDS



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Hnh 3.42: M t hot ng ca lnh ALD v OLD

Hnh 3.43: M t hot ng ca lnh LPS, LRD, LPP

Hnh 3.44: V d v cch s dng lnh ALD, OLD, LPP, LPS, LRD



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11. SIMATIC Conversion Instructions:Cc hm i kiu d liu cho php thc hin vic i kiu d liu t kiu ny sang kiukhc. Sau y l cc lnh bin i kiu d liu trong STL v LAD:

STL LADM t

Description

Ton hng

Operands

Kiu dliu

DataTypes

BCD to Integer and Integer to BCD

BCDI OUT

Lnh chuyn i mt snh_thp phn IN sang snguyn v lu kt quvo OUT.

Gii hn ca IN:09999.

IN: IW, QW, VW,

LW, MW, SMW,

AIW ,AC, T, C,

Constant, VD, AC,

LD, SW.OUT: IW, QW, VW,

LW, MW, SMW, AC,

T, C, VD, AC,

LD, SW.

Word

IBCD OUT

Lnh chuyn i mt snguyn IN sang snh_thp phn v lu ktqu vo OUT.Gii hn ca IN:09999.

IN: IW, QW, VW,LW, MW, SMW,

AIW ,AC, T, C,

Constant, VD, AC,

LD.OUT: IW, QW, VW,

LW, MW, SMW, AC,

T, C, VD, AC,

LD.

Word

Double Integer to Real

DTR IN, OUT

Lnh chuyn i snguyn 32 bit IN sang sthc (32 bit) v lu ktqu vo OUT.

IN: ID, QD, VD, LD,MD, SMD, AC, HD,

Constant, VD, AC,

LD, SD.OUT:ID, QD, VD,LD, MD, SMD, AC,

HD, VD, AC, LD,SD.

DWord

Round

ROUND IN,

Lnh chuyn i s thc

IN thnh s nguyndouble Integer (lm trns) v kt qa lu vo

IN: ID, QD, VD, LD,

MD, SMD, AC,Constant, VD, AC,

LD, SD.

Real

BCD_IEN ENO

IN OUT

I_BCDEN ENO

IN OUT

DI_REN ENO

IN OUT

RONUDEN ENO

IN OUT



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OUT OUT.Nu phn l >= 0.5 thc lm trn v pha lnhn 1 n v.

OUT:ID, QD, VD,LD, MD, SMD, AC,

HD, VD, AC, LD,SD.

DINT

Truncate

IN: ID, QD, VD, LD,

MD, SMD, AC,

Constant, VD, AC,

LD, SD.

Real

TRUNC IN,

OUT

Hm chuyn i s thc32 bit c du sang snguyn 32 bit c du. OUT:ID, QD, VD,

LD, MD, SMD, AC,

HD, VD, AC, LD,

SD.

DINT

Double Integer to Integer and Integer to Double IntegerIN: IW, QW, VW,

LW, MW,SW, SMW,AIW ,AC, T, C,

Constant, VD, AC,

LD.

INT

ITD IN, OUT

Lnh chuyn i snguyn 16 bit sang s

nguyn 32 bit.OUT: ID, QD,VD,

LD, MD,SD, SMD,

AC, VD, AC, LD.

DINT

IN: ID, QD,VD, LD,MD,SD, SMD,

AC,Constant, VD,

AC, LD.

DINT

DTI IN, OUT

Lnh chuyn i snguyn 32 bit sang s

nguyn 16 bit.OUT: IW, QW, VW,LW, MW,SW, SMW,

AC, T, C, VD, AC,

LD.

INT

Integer to Real, Byte to Integer and Integer to Byte

(Integer toReal) none

Khng c lnh chuyni trc tip ny. Ta cth thc hin c bng

cch dng lnh ITD(chuyn s nguyn 16 bitthnh s nguyn 32 bit)sau dng tip lnhDTR (chuyn s nguyn32 bit sang s thc ).

none none

TRUNCEN ENO

IN OUT

I_DIEN ENO

IN OUT

DI_IEN ENO

IN OUT



102/214


IN: IB, QB, MB,SMB, VB, SB, LB,

AC, Constant, VD,

AC, LD.

Byte

BTI IN, OUT

Lnh chuyn i gi trca Byte IN thnh gi trInteger 16 bit v lu vo

OUT.OUT: IW, QW, VW,

LW, MW,SW, SMW,AC, T, C, VD, AC,

LD.INT

IN: IW, QW, VW,LW, MW,SW, SMW,

AC, T, C, AIW,

Constant, VD, AC,

LD.

INT

IBT IN, OUT

Lnh chuyn i gi trtrong Word IN thnh gi

Byte v lu gi tr nyvo OUT.

OUT: IB, QB, MB,

SMB, VB, SB, LB,AC, VD, AC, LD.

Byte

I_BEN ENO

IN OUT



103/214


Hnh 3.45: V d minh ho cch s dng cc lnh chuyn i

STL LADM t

Description

Ton hngOperands

Kiu dliuData

Types

Decode

IN: IB, QB, MB,SMB, VB, SB,

LB, AC, Constant,

VD, AC, LD

Byte

DECO IN, OUT

Lnh t gi tr logic 1 vobit ca tn OUT c chs (trng s ca bit thuc

Word) bng s nguyn

nm trong nibble (4 bit)thp ca byte u vo IN.Cc bit cn li ca tn

c gi tr logic bng 0.

OUT: IW, QW,

VW, LW,MW,SW, SMW,

AC, T, C, AIW,

VD, AC, LD.

Word

IN: IW, QW, VW,

LW, MW,SW,

SMW, AC, T, C,

AIW, VD, AC,

LD.

Word

ENCO IN, OUT

Lnh xc nh ch s cabit thp nht trong tnIN c gi tr logic 1v ghi

kt qu ny vo nibblethp nht ca byte u ra

OUT.

OUT: IB, QB,

MB, SMB, VB,

SB, LB, AC, VD,

AC, LD.

Byte

ENCOEN ENO

IN OUT

DECOEN ENO

IN OUT



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Hnh 3.46: V d v cch s dng lnh DECO

Hnh 3.47: V d v cch s dng lnh ENCO

STL LAD M t(Description)

Ton hng(Operands)

Kiu dliu

(Data

Types)

Segment



105/214


IN: IB, QB, MB,SMB, LB, VB, AC,

Constant, VD, AC,

SB, LD.SEG IN, OUT

Lnh xut cc bit chothanh ghi 7 on tngng vi ni dung ca 4bit thp nht ca byteu vo IN. Kt qu

c chi vo byte ura.

OUT: IB, QB, MB,

SMB, LB, VB, AC,VD, AC, SB, LD.

Byte

ASCII to Hexa and Hexa to ASCII

IN, OUT:IB, QB, MB, SMB,

LB, VB, VD, AC,

SB, LD.

Byte

ATH IN,OUT, LEN

Thc hin php bin imt chui k t c di c ch th trongton hng LEN, bt ubng k t chnh trongton hng IN, sang snguyn h cs 16 vghi vo vng nhk t

byte c chnh biOUT. di cc ica chui k t l 255.Nhng k t hp l lnhng k t c mASCII t 3039 v4146 (cs 16, ngvi cc k t t 09,AF ). Nu m ho mtk t b sai th qu trnhm ho b dng li v bit

SM1.7 c gi tr logicbng 1.

LEN:IB, QB, MB, SMB,

LB, VB, AC,

Constant, VD, AC,SB, LD.

Byte

IN, OUT:IB, QB, MB, SMB,

LB, VB, VD, AC,

SB, LD.

HTA IN,OUT, LEN

Thc hin i mt dych vit trong h cs16 thnh chui k t mASCII. Dy su voc lu trong mng btu bng IN v c di l LEN. di cci ca dy s l 255.Chui k tu ra c

ghi vo mng c byteu l OUT.

LEN:IB, QB, MB, SMB,

LB, VB, AC,

Constant, VD, AC,

SB, LD.

Byte

SEGEN ENO

IN OUT

ATH

ENIN OUTLEN

ATHEN

IN OUTLEN



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Hnh 3.48: V d v cch s dng lnh ATH, HTA

Hnh 3.49: V d v cch s dng lnh SEG

Hnh 3.50: M hin th thanh ghi 7 on12. SIMATIC Clock Instrutions:

Tuyt i khng s dng lnh c/ghi (TODR/TODW) thi gian thc cng mtlc trong chng trnh chnh v chng trnh x l ngt. Khi mt lnh TODR hocTODW thc hin th khi gi chng trnh x l ngt, cc lnh lm vic vi ng hthi gian thc trong chng trnh x l ngt s khng c thc hin na. Bit SM4.5 sc mc logic 1 trong nhng trng hp nh vy.



107/214


ng h thi gian thc ch c i vi CPU214 tr ln. c th lm vic ving h thi gian thc th CPU s cung cp 2 lnh c/ghi gi tr cho ng h. Nhng gitrc c hoc ghi c vi ng h thi gian thc l cc gi tr v ngy, thng, nmv cc gi tr v gi, pht, giy.

Cc d liu c/ghi vi ng h thi gian thc trong LAD, STL c di 1 bytev phi c m ho theo kiu s nh thp phn BCD (Ex: 16#95 cho nm 95).Chng nm trong bm gm 8 byte lin nhau theo th t nh sau:

Hnh 3.51: Bm 8 byte ca lnh ng h thi gian thcCc gi trca cc thng sphi nm trong gii hn:

CPU S7-200 khng thc hin kim tra li ngy thng, ngy ca tun iu chnhli ngy thng. Gi tr v ngy thng nh l February 30 c thc chp nhn. Do bn s phi chc chn rng ngy thng ca bn a vo l ng.

STL LAD

M tDescription

Ton hngOperands

Kiu dliuData

Types

Read Real-Time Clock and Set Real-Time Clock

TODR T

Lnh c ni dungca ng h thigian thc vo bm 8 byte c chnh trong lnh bngton hng T.

TODW T

Lnh ghi ni dungca bm 8 byte

c chnh tronglnh bng ton hngT vo ng h thigian thc.

T: VB,IB, QB, MB,

SMB, SB, LB, VD,

AC, LD.Byte

READ_RTCEN ENO

T

SET_RTCEN ENO

T



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13. SIMATIC Program Control Instrutions:Cc lnh ca chng trnh, nu khng c nhng lnh iu khin ring, sc

thc hin tun t t trn xung di trong mt vng qut. Lnh iu khin chng trnhcho php thay i th t thc hin lnh. Chng cho php chuyn th t nh: ng l ral lnh tip theo, ti mt lnh bt c no khc ca chng trnh; trong ni iu khinchuyn n phi c nh du trc bng nhn chch. Nhm lnh iu khin chngtrnh gm: lnh nhy, lnh gi chng trnh con, nhn ch ch (hay gi n gin lnhn), phi c nh du trc khi thc hin lnh nhy hay lnh gi chng trnh con.

Vic t nhn cho lnh nhy phi nm trong chng trnh. Nhn ca chng trnhcon hay nhn ca chng trnh x l ngt phi c khai bo u chng trnh. Khngth dng lnh JMP chuyn iu khin t chng trnh chnh vo nhn bt k trongchng trnh con hoc chng trnh x l ngt. Ngc li cng khng c php t mtchng trnh con hay chng trnh x l ngt nhy ra ngoi chng trnh chnh .

Lnh gi chng trnh con l lnh chuyn quyn iu khin n chng trnh con.Sau khi chng trnh con thc hin xong th quyn iu khin li c chuyn v lnh

tip theo trong chng trnh chnh ngay sau lnh gi chng trnh con. T mt chngtrnh con c th gi mt chng trnh con khc trong n, c th g

dieu khien logic3

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