Baaif tập PLC

8/3/2019 Baaif tp PLC

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Bi tp mn hc: iu khin logic

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Bi 1: Cnh tay ROBOT bc hng ha.

Cng ngh: loi robot ny hay gp trong cc x nghip cng nghip. Robot nm cht

vt t bng chuyn A b sang bng chuyn B.1. v tr ban u cnh tay pha bng chuyn B2. Khi n START ng cthun khi ng quay theo chiu kim ng h. Khi

gp LS1 n dng li ng thi bng chuyn A c khi ng3. Bng chuyn A a vt tin ti cnh tay. Khi vt chm vo PH1 l cm bin

pht hin vt th cnh tay kp cht ly vt. Khi LS3 tc ng tc l vt ckp cht th ng cngc khi ng quay theo chiu ngc li

4. Khi cnh tay chm vo LS1 th n dng v nh vt xung bng chuyn B5. Bng chuyn B lun hot ng (Khi n START n bt u hot ng)

STARTSTOP

RESET

PH1

LS3 LS1 LS2

Bng chuyn BBng chuyn A


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Lu thut ton:

Bt u

NhnSTART?

N

Y

Khi ng ng c thun

LS1=1?

Khi ng bng chuyn A

PH1=1?

Dng bng chuyn A

LS3=1?

Khi ng ng c ngc

LS2=1?

LS3=0?

Kt thuc

N

Y

N

Y

N

Y

N

Y

Y

N

Nhavt


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Gin thi gian

Start

TG

Stop

PH1

LS1

LS2

Ls3

DCT

DCN

DC Kep

BCA

Phn cng I/O

u vo u ra

I0.0 Start Q0.0 DC ThunI0.1 Stop Q0.1 DC NgcI0.2 PH1 Q0.2 Bng chuyn AI0.3 LS1 Q0.3 DC KpI0.4 LS2I0.5 LS3I0.6 Reset


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Chng trnh vit bng tip im


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Chng trnh vit bng thanh ghi:


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Bi 2: Cu trc

S3

S2

S1S4S5

START

STOP

RESET

ARLAM

TANK

Khi n START trc ko cun trn a mc ln, khi sensor S3 tc ng th trc kongng cun. Sau trc ko di chuyn sang tri, n v tr sensor S4 tc ng th ndng v h mc xung a sn phm vo chng cha (TANK). a sn phmxung khi S2 tc ng th ngng. Sn phm trong thng 20 s. Sau thi gian trnsn phm c nhc ln v di chuyn qua tri cho n khi sensor S5 tc ng thdng li. Mc c hn v tr S2 th chung s reo ln, ngi vn hnh ly smphm v nhn RESET chung ht reo, mc c ko ln v tr S3 ri di chuyn vv tr S1 ri h mc xung v v S2.

Phn cng I/O:

u vo u raI0.0: START Q0.0: NG C SANG TRII0.1: RESET Q0.1: NG C SANG PHI

I0.2: STOP Q0.2: NG C KO LNI1.1: S1 Q0.3: NG C H MC XUNGI1.1: S2 Q0.4: CHUNGI1.3: S3I1.4: S4I1.5: S5


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Gin thi gian

MPhai

MTrai

S5

S4

S3

S2

S1

Stop

PB2

TG

PB1

MLen

MXuong

Coi

T37 20s


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Chng trnh vit bng tip im:


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Chng trnh vit bng thanh ghi


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Bi 3: n giao thng chng hp

Cng ngh: nhng on ng cn sa cha ngi ta lp n xanh ch baogm n xanh v cho mt lung xe i qua. n xanh sng trong khong thi

gian 30s, cn n sang trong khong 60s.

Gin thi gian

Chng trnh vit bng LADER

30 S 15 S

Xanh 1

1

Xanh 2

2


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Bi 4: Bm nc di hm mM1 M1

S2

M1 M2 M1 M2 M1 M2S1

S1: cm bin 1S2: cm bin 2M1: ng c1

M2: ng c2Phn cng I/O:u vo: u raI0.0: Start Q0.0: n StartI0.1: Stop Q0.1: n StopI0.2: S1 Q0.2: ng cM1I0.3: S2 Q0.3: ng cM2


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Lu thut ton

BEGIN

END

a giatr0 vao mt vung nh

Khi S1 tac ng thc hin cng 1 vigiatri trong vung nhvalu lai trong o

Thc hin chia snguyngiatrtrong vung nhcho 2

Nu d= 1?

Khi d1 thng c M1 hoat ng

Khi d0 thng c M2 hoat ng

S1 thi tac ng


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Trng hp 2:


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Lu thut ton

Bt u

a giatr0 vao

mt vung bin

Khi S1 hoc S2 tac ng thchin cng 1 vi giatrtrongvung nhvalu trong o

Thc hin phep chia snguyngiatrtrong vung nhcho 2

Nu d=1Sai

ung

Khi phn d=o hoc M1atac ng + S2 tacng th M2 hoat ng

Khi phn d=1 hoc M2atac ng + S2 tacng th M1 hoat ng

S1 thi tac ng

Bt u

Phn cng I/0

u vo u RaI0.0 Nt Start Q0.0 n StartI0.1 Nt Stop Q0.1 n StopI0.2 S1 Q0.2 ng cM1I0.3 S2 Q0.3 ng cM2


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Bi 5: n giao thng ng t:a) Cng ngh

Way1

Way2

- n xanh X1 sng trong 25 giy- Ri n n vng V1 sng trong 5 giy- n xanh X2 sng trong 25 giy- Ri n n vng V2 sng trong 5 giy

X1V1D1

X2V2D2

StartStop


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b) Gin thi gian

X1

V1

D1

X2

V2

D2

25s

5s

30s

25s

5s

30s

c) Phn cng I/Ou vo u ra

I0.0 Start X1 Q0.0I0.1 Stop V1 Q0.1

1 Q0.2X2 Q0.3V2 Q0.4

2 Q0.5d) Chng trnh bng LADDER


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Bi 6:M HNH H THNG PHA TRN MU SN

1. Yu cu cng nghMu sn c to ra bi 3 mu c bn: Xanh dng, v xanh lc (Q0.1,

Q0.2, Q0.3).- T l cc mu c ci t thng qua thi gian chy ca cc bm.

- Khi bm cc loi sn vo th ng thi ng ctrn (Q0.0) s thc hin trn.- Khi tt c cc bm sn bm th ng ctrn s tip tc hot ng trong

10 giy na ri dng ng ctrn.- X sn trong bnh thng qua van (Q0.4) v ng cx (Q0.5) cho n khi tc

ng ca sensor I0.3 mc thp th dng x.- Nu lng sn trong bnh cha vt mc gii hn cao I0.2 th s ngng bm

v tip trc trn.- Sau khi thc hin xong th tip tc bm v thc hin m khc.


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Motor5 (Q 0.5)Valve (Q 0.4)

Motor4 (Q 0.3)

Green color

Motor3 (Q 0.2)

Blue color

Motor2 (Q 0.1)Red

color

SS

SS

Motor1 (Q 0.0)

Sensor2 (I 0.3)

Sensor1 (I 0.2)

S bnh trn sn.


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2. Phn cng u vo ra

Symbol Address Comment

Start I0.0 Nt nhn startStopped I0.1 Nt nhn stop

Sensor1 I0.2 Sensor xc nh mc caoSensor2 I0.3 Sensor xc nh mc thpMotor1 Q0.0 ng cp trnMotor2 Q0.1 ng cbm sn xanh dngMotor3 Q0.2 ng cbm sn Motor4 Q0.3 ng cbm sn xanh lcValve Q0.4 Van xMotor5 Q0.5 ng cbm x

3. Gin thi gian

Motor1 (Q 0.0)

Motor5 (Q 0.5)

Motor2 (Q 0.1)

Motor3 (Q 0.2)

Motor4 (Q 0.3)

Valve (Q 0.4)

Sensor2 (I 0.3)

Sensor1 (I 0.2)

Stopped (I 0.1)

Start (I 0.0)

T

t

t

t

t

t

t

t

t

t

t

Hnh 2: Gin thi gian h thng trn sn.

4. Chng trnhTITLE=CHUONG TRINH DIEU KHIEN HE THONG TRON SONBEGIN

Network 1 // Nhap thoi gian bom cho cac loai son, ta co the nhap tren cac man hinhgiao dien hoac WinCC.LD SM0.1MOVW +50, VW10AENOMOVW +60, VW12AENOMOVW +70, VW14Network 2 // khoi dong he thong tron

// Network CommentLD I0.0O M0.0


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AN I0.1= M0.0Network 3 // khoi dong pump 1LD M0.0TON T37, VW10

AN T37AN I0.2= Q0.1Network 4 // khoi dong pump 2LD M0.0TON T38, VW12AN T38AN I0.2= Q0.2

Network 5 // khoi dong pump 3LD M0.0TON T39, VW14AN T39AN I0.2= Q0.3Network 6 // Tron them khoang thoi gian 10 sLD M0.0LPSA T37A T38A T39TON T40, +10LPPAN T40= Q0.0Network 7 // xa son sau khi tronLD M0.0

A T40AN I0.3= Q0.4= Q0.5Network 8 // reset va hoat dong laiLD I0.3EUR T37, 3


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Bi 7: H THNG NG THNG CC HP SN

1. Yu cu cng nghMc ch: ng gi cc hp sn sau khi c pha trn.- Bng chuyn th nht c iu khin vi motor1 (q0.0) chuyn cc hp sn

sau khi c pha trn.- Khi hp sn i n gp sensor1 pht hin th bng ti dng li v robot (Q0.3)hot ng a hp sn vo thng. Bng chuyn Q0.0 tip tc hot ng.

- Khi bn hp sn trong thng th bng ti Q0.1 hot ng cho n khisensor2 (I0.3) tc ng thi dng li.

- V h thng li tip tc hot ng b vo thng tip theo...

Sensor2 (I 0.3)

Robot (Q 0.2)

Motor2 (Q 0.1)

Motor1 (Q 0.0)

Sensor1 (I 0.2)

Hnh 5: H thng ng thng cc hp sn.

2. Phn cng u vo, u ra


Start I0.0 Nt nhn startStops I0.1 Nt nhn stopSensor1 I0.2 Sensor 1, bo v tr hp sn

Sensor2 I0.3 Sensor 2, bo v tr thng ng hp snMotor1 Q0.0 ng cko bng chuyn 1Motor2 Q0.1 ng cko bng chuyn 2Robot Q0.2 Xung kch vo rbt


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3. Gin thi gian

.

Bi 8: H THNG IU KHIN TNG CA RA VO CA BN XE

1. Yu cu cng ngh- H thng s tng mca khi c xe vo ra thng qua cm bin quang (I 0.5).

Ca sng sau khi chi 10s khng c xe.- Khi c xe n sensor cm bin nhn dng (I0.5) s tc ng, ng cQ0.0 s

ko ca ln cho n khi tc ng cng tc hnh trnh I0.4 th dng li.- Nu sensor cm bin nhn dng khng tc ng th lc tnh thi gian tr sau

10 giy th Q0.1 tc ng ng ca li.

Robot (Q 0.2)

Motor2 (Q 0.1)

Motor1 (Q 0.0)

Sensor2 (I 0.3)

Sensor1 (I 0.2)

Stop (I 0.1)

Start (I 0.0)

1 2 3 4 1 2 3 4

t

t

t

t

t

t

t


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Motor

Motor_up (Q0.1)

Motor_down (Q0.0)

Sensor (I 0.4)

Position_close(I 0.2)

Position_open (I 0.3)

2. Phn cng u vo, u ra


Start I0.0 Nt nhn startStopped I0.1 Nt nhn stopPosition_close I0.2 Cng tc hnh trnh bo ca ng

Position_open I0.3 Cng tc hnh trnh bo ca mSensor I0.4 Sensor siu m xc nh c xe cn ra vo caMotor_up Q0.0 ng cko ca lnmotor_down Q0.1 ng cko ca xung

3. Gin thi gian

Position_close (I 0.2)

10s

Motor_down (Q 0.0)

Motor_up (Q 0.1)

Position_open (I 0.3)

Stopped (I 0.1)

Start (I 0.0)

T

Sensor sieu am (I 0.4)

t

t

t

t

t

t

t

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Baaif tập PLC