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Bai Tap Plc Vocsy
Transcript of Bai Tap Plc Vocsy
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Vocsyit.com
Su tm Page 1
Mc Lc
Bi 1: THIT K H THNG CHUNG BO GI ....................................... 2 Bi 2: N GIAO THNG ............................................................................ 10 Bi 3: THIT K H THNG BM NC DI HM M ...................... 14 Bi 4: IU KHIN MY TRN ................................................................... 19 Bi 5: IU KHIN NG M CA KHO T NG ............................... 23 Bi 6: DY CHUYN CHIT NC VO CHAI ........................................ 27 Bi 7: H THNG M V NG GI SN PHM.................................. 31 Bi 8: IU KHIN BNG CHUYN CN TN ......................................... 34 Bi 9: IU KHIN DAO CT SN PHM ................................................. 40 Bi 10: IU KHIN ROBOT GP HNG ................................................... 44 Bi 11: THIT K CN TRC ....................................................................... 48 Bi 12: X L HA CHT ............................................................................ 52 Bi 13: N GIAO THNG CH NG HP ........................................... 58 Bi 14: BI XE ......................................................................................... 62 Bi 18: MY PHA TR, C PH .................................................................. 66 Bi 19:IU KHIN MY KHOAN .............................................................. 72 BI 1: N GIAO THNG NG HP .......................................................... 76 BI 3: IU KHIN NG M CA KHO T NG .............................. 81 BI 4:DY CHUYN CHIT NC VO CHAI ........................................ 85 BI 5: H THNG M V NG GI SN PHM ................................ 89 BI 6: IU KHIN BNG CHUYN CN TN ........................................ 94 BI 7: IU KHIN MY KHOAN ............................................................ 102 BI 8: IU KHIN DAO CT SN PHM .............................................. 110 BI 9: IU KHIN CNH TAY ROBOT GP HNG ............................. 116 BI 10: THIT K CU TRC ................................................................... 124 BI 11: MY BN V T NG .............................................................. 136 BI 12: TRM TRN VT LIU ................................................................ 155 Bi 1: N GIAO THNG .......................................................................... 164 Bi 2: THIT K H THNG BM NC DI HM M .................... 169 Bi 3: IU KHIN MY TRN ................................................................ 177 Bi 4: IU KHIN NG M CA KHO T NG ............................. 182 Bi 5: DY CHUYN CHIT NC VO CHAI ...................................... 185 Bi 6: H THNG M V NG GI SN PHM................................ 191 Bi 7: IU KHIN BNG CHUYN CN TN ....................................... 196 Bi 8: IU KHIN DAO CT SN PHM ............................................... 203 Bi 9: IU KHIN ROBOT GP HNG ................................................... 211 Bi 10: N GIAO THNG CH NG HP ......................................... 220 Bi 11: BI XE ....................................................................................... 227 Bi 12: IU KHIN MY KHOAN ........................................................... 234 Bi 13: THIT K CHNG TRNH IU KHIN CN TRC............... 237
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BI 14: XE HNG CHY THEO CHNG TRNH .................................. 251
Bi 1: THIT K H THNG CHUNG BO GI
I.Yu cu: Thit k h thng chung bo gi nh sau: - Tit 1: 7h00 xut 1 hi di 7h50 xut 2 hi ngn - Tit 2: 8h00 xut 1 hi di 8h50 xut 2 hi ngn - Tit 3: 9h00 xut 1 hi di 9h50 xut 2 hi ngn - Tit 4: 10h00 xut 1 hi di 10h50 xut 2 hi ngn - Tit 5: 11h00 xut 1 hi di 11h50 xut 3 hi ngn - Tit 6: 12h00 xut 1 hi di 12h50 xut 2 hi ngn - Tit 7: 13h00 xut 1 hi di 13h50 xut 2 hi ngn - Tit 8: 14h00 xut 1 hi di 14h50 xut 2 hi ngn - Tit 9: 15h00 xut 1 hi di 15h50 xut 2 hi ngn - Tit 10: 16h00 xut 1 hi di 16h50 xut 2 hi ngn ng 7h00 sng hm sau chung li reo li
II. Bng phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 Chung
I0.1 Stop Q0.1 n Start
Q0.2 n Stop
Ta c 1 ngy = 24 gi =86400 giy=144 x 600 giy to chu k ny ta dng b Timer T37 vi gi tr t trc l 600 giy v mt b m ln C30 vi gi tr t trc l 144
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III. Lu thut ton:
Bt u 7h00
T37 v C30 bt u m v chung reo 1
hi di 5 giy C30=5 reo 2 hi ngn
C30=6 reo 1
hi di
C30=11 reo
2 hi ngn
C30=12 reo
1 hi di
C30=17 reo
2 hi ngn
C30=23 reo
2 hi ngn
C30=18 reo
1 hi di
C30=24 reo
1 hi di
C30=29 reo
3 hi di C30=30 reo
1 hi di C30=35 reo
2 hi ngn C30=36 reo
1 hi di C30=41 reo
2 hi ngn C30=42 reo
1 hi di C30=47 reo
2 hi ngn
C30=48 reo
1 hi di
C30=53 reo
2 hi ngn
C30=54 reo
1 hi di
C30=59 reo
3 hi di
C30=144 t reset v
bt u m li
MEND
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IV. Chng trnh: Dng LADDER:
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Dng STL:
Network 1 // KHOI DONG
LD I0.0
S M0.0, 1
R M0.1, 1
Network 2 // DEN START SANG
LD M0.0
= Q0.1
Network 3 // DUNG
LD I0.1
S M0.1, 1
R M0.0, 1
Network 4
LD M0.1
= Q0.2
Network 5 // TAO BO DEM THOI GIAN 600S
LD M0.0
AN T37
TON T37, 6000
Network 6 // QUA TRINH LAP LAI SAU MOT NGAY DEM
LD T37
LD C30
CTU C30, 144
Network 7 // CHUONG REO MOT HOI DAI 5S
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LDW= C30, 0
OW= C30, 6
OW= C30, 12
OW= C30, 18
OW= C30, 24
OW= C30, 30
OW= C30, 36
OW= C30, 42
OW= C30, 48
OW= C30, 54
AW
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Network 10 // CHUONG REO
LD M0.2
O M0.3
O M0.4
A M0.0
= Q0.0
Bi 2: N GIAO THNG
I.Yu cu:
Thit k h thng n giao thng ng t vi cc yu cu sau: 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
II. Bng phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 n xanh 1
I0.1 Stop Q0.1 n vng 1
Q0.2 n 1
Q0.3 n xanh 2
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Q0.4 n vng 2
Q0.5 n 2
III. Gin thi gian:
X1
V1
D1
X2
V2
D2
25s
5s
30s
25s
5s
30s
IV. Chng trnh: Dng LADDER:
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Dng STL: NETWORK 1
LD I0.0
O M0.0
AN I0.1
= M0.0
NETWORK 2
LD M0.0
TON T37, 600
S Q0.0, 1
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S Q0.5, 1
NETWORK 3
LDW>= T37, 250
S Q0.1, 1
R Q0.0, 1
NETWORK 4
LDW>= T37, 300
S Q0.3, 1
S Q0.2, 1
R Q0.1, 1
R Q0.5, 1
NETWORK 5
LDW>= T37, 550
S Q0.4, 1
R Q0.3, 1
NETWORK 6
LDW>= T37, 600
R Q0.2, 1
R Q0.4, 1
R T37, 1
Bi 3: THIT K H THNG BM NC DI HM M
I. Yu cu: Thit k h thng bm nc di hm m tha mn yu cu sau: Mc nc c o bi 2 sensor S1,S2 Nhm m bo vn hnh kinh t, cc bm c iu khin hot ng nh sau:
II. Bng phn cng u vo, u ra:
M1 M1 M1
M1 M1
M1
M2
M2
M2
M2
M2
M2
S1
S2
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u vo u ra
I0.0 Start Q0.0 n Start
I0.1 Stop Q0.1 n Stop
I0.2 S1 Q0.2 ng c M1
I0.3 S2 Q0.3 ng c M2
III. Lu thut ton:
IV. Chng trnh: Dng LADDER:
Begin
a gi tr 0 vo mt
vng nh
Khi S1 tc ng th cng 1 vi
gi tr lu trong vng nh
Thc hin php chia nguyn gi tr trong vng nh cho 2
Nu d=1 M2 hot ng
M1 hot ng
S1 thi tc ng
End
S
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Dng STL: Network 1 // KHOI DONG
LD I0.0
S M0.0, 1
R M0.1, 1
Network 2 // DEN START SANG
LD M0.0
= Q0.0
Network 3 // DUNG
LD I0.1
S M0.1, 1
R M0.0, 1
Network 4 // DEN STOP SANG
LD M0.1
= Q0.1
Network 5
LD M0.0
EU
= M0.2
Network 6 // GAN GIA TRI 0 VAO MW1 SAU MOI LAN KHOI DONG
LD M0.2
O C30
MOVW 0, MW1
Network 7 // PHAT HIEN SUON LEN CUA SENSOR_1
LD I0.2
EU
= M0.3
Network 8 // CONG 1 VAO MW1
LD M0.3
+I 1, MW1
Network 9 // KHOI DONG LAI SAU 32767 LAN
LD M0.3
LD M0.2
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O C30
CTU C30, 32767
Network 10 // THUC HIEN PHEP CHIA NGUYEN
LD M0.3
MOVW MW1, MW5
DIV 2, MD3
Network 11 // DONG CO M1 QUAY
LD I0.2
AW= MW3, 1
LD Q0.3
A I0.3
OLD
= Q0.2
Network 12 // DONG CO M2 QUAY
LD I0.2
AW= MW3, 0
LD Q0.2
A I0.3
OLD
= Q0.3
Bi 4: IU KHIN MY TRN
I. Yu cu :
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S bnh trn sn
S gm c 2 ng ng a 2 loi hai sn mu khc nhau lm c s cho vic to ra mu sn mong mun Hai cm bin bo mc trong bnh: Bo mc cao Sensor 2 Bo mc thp Sensor 1 Mt thit b trn c iu khin bi ng c trn Qu trnh lm vic c thc hin nh sau: Trc tin bm 2 loi sn khc mu nhau vo bnh, loi sn th nht c a vo bnh bng my bm 1, loi sn th hai c a vo bnh bng my bm 2. Sau khi dung dch trong bnh t mc cc i th dng hai my bm v bt u qu trnh trn, qu trnh ny c iu khin bi my trn v thi gian trn cn thit l 5s. Sau khi trn xong, sn phm c a ra rt vo cc hp ng sn qua van x v bm 3. Qu trnh c thc hin lp li 10 ln. Sau dng h thng
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start_M1 Q0.0 Bm M1
I0.1 Start_M2 Q0.1 Bm M2
I0.2 Stop_M1 Q0.2 My trn
I0.3 Stop_M2 Q0.4 Van x
I0.4 Sensor_1 Q0.5 Bm M3
I0.5 Sensor_2
III. Thut ton: Bc 1: Rt loi sn th nht v th hai vo bnh Bc 2: iu hnh ch lm vic khi t mc cao Bc 3: iu khin ng c trn v t thi gian trn Bc 4: a sn phm ra khi bnh trn Bc 5: m s ln trn. Nu 10 ln th dng sn xut Bc 6: Quay li ch lm vic bc 1
IV. Chng trnh: Dng LADDER:
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Dng STL:
Network 1 // BOM SON MAU THU NHAT VAO BINH TRON
LD I0.0
O Q0.0
AN I0.2
AN I0.4
AN C30
= Q0.0
Network 2 // BOM SON MAU THU HAI VAO BINH TRON
LD I0.1
O Q0.1
AN I0.3
AN I0.4
AN C30
= Q0.1
Network 3 // DAT BAO KHI DAT GIA TRI CUC DAI
LD I0.4
S M0.1, 1
Network 4 // TAO TIMER T37 KHONG CHE THOI GIAN TRON
LD M0.1
TON T37, 50
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Network 5 // DONG NGUON CUNG CAP CHO DONG CO TRON
LDN T37
A M0.1
= Q0.2
Network 6 // CHAT LONG SAU KHI TRON DUOC THAO RA
LD T37
AN I0.5
= Q0.4
= Q0.5
Network 7 // DEM SO LAN TRON
LD M0.1
A T37
LD I0.7
CTU C30, 10
Network 8 // DAT BAO KHI MUC TRONG BINH TRON DAT CUC TIEU
LD I0.5
A T37
R M0.1, 1
Bi 5: IU KHIN NG M CA KHO T NG
I. Yu cu:
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Khi xe ang tin v gn ca kho, cm bin Ultrasounic SS1 nhn dng c xe v ca s c m ra n gp gii hn hnh trnh trn LS1 th ca dng li ri xe chy vo. Khi cm bin quang SS2 t pha trong cng ca nhn dng c xe i qua khi ca th ca s c ng li, chm vo gii hn hnh trnh di LS2 th ca dng li. Ch xe ch i mt chiu.
II. Phn cng u vo, u ra:
u vo u ra
I0.0 START Q0.0 Raise_door
I0.1 STOP Q0.1 Down_door
I0.2 SS1
I0.3 SS2
I0.4 LS1
I0.5 LS2
III.Gin thi gian:
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IV. Chng trnh: Dng LADDER:
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Dng STL: Network 1
// KHOI DONG
LD I0.0
O M0.0
AN I0.1
= M0.0
Network 2
// Khi S1 tac dong thi cua keo len
LD M0.0
A I0.2
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S Q0.0, 1
R Q0.1, 1
Network 3
// Khi LS1 tac dong thi cua dung lai tai gioi han tren
LD M0.0
A I0.4
R Q0.0, 1
Network 4
// Phat hien suon xuong cua SS2
LD I0.3
ED
= M0.1
Network 5
// Khi SS2 chuyen tu 1 sang 0 thi cua keo xuong
LD M0.0
A M0.1
S Q0.1, 1
Network 6
// Khi LS2 tac dong thi cua dung lai tai gioi han duoi
LD M0.0
A I0.5
R Q0.1, 1
Bi 6: DY CHUYN CHIT NC VO CHAI
I. Yu cu:
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Chit nc vo chai
Cng on chit nc vo chai theo nguyn tc thi gian, m t hnh 3.13. Nhn nt Start PB, bng ti chy v mang theo chai, cm bin chai pht hin, bng ti dng, bt u m van chit chit nc vo chai. Thi gian chit y l 30s, sau khi chit y bng ti li chy v tip tc chit chai k tip
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 Conveyer motor
I0.1 Stop Q0.1 Fill valve
I0.2 Bottle_sensor
III.Chng trnh: Dng LADDER:
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Dng STL: Network 1 // KHOI DONG DONG CO KEO BANG CHUYEN
LD I0.0
LD M0.0
A M0.1
OLD
S Q0.0, 1
S M0.0, 1
Network 2 // DUNG BANG CHUYEN
LD I0.1
R M0.0, 1
R Q0.0, 1
Network 3 // PHAT HIEN SUON LEN BOTTLE SENSOR
LD I0.2
EU
= M0.2
Network 4 // KHI PHAT HIEN BOTTLE SENSOR CHUYEN TU 0->1 THI
DONG CO BANG CHUYEN DUNG
LD M0.0
A M0.2
R Q0.0, 1
Network 5 // TAO THOI GIAN T37
LD I0.2
A M0.0
TON T37, 300
Network 6 // VALVE CHIET
LD M0.0
AN T37
A I0.2
= Q0.1
Network 7 // PHAT HIEN SUON XUONG VALVE CHIET. KHI VALVE
CHIET CHUYEN TU 1=>0 THI DONG CO CHAY
LD M0.0
A Q0.1
ED
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= M0.1
Bi 7: H THNG M V NG GI SN PHM
I.Yu cu:
Khi nhn cng tc khi ng PB Start bng ti BT1 mang hp ng sn phm di chuyn. Cm bin SS1 nhn dng thng ng tc ng, bng ti BT1 dng li. Bng ti BT2 cha sn phm dch chuyn, sn phm c rt vo hp ng, sn phm c m bi mt cm bin quang hng ngoi SS2 khi m c 10 sn phm ( mi hp cha 10 sn phm) th bng ti BT2 dng, tip tc BT1 dch chuyn ng gi hp mi. ngng qu trnh ta nhn nt PB Stop.
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 Box conveyor
I0.1 Stop Q0.1 Piece conveyor
I0.2 Box sensor
I0.3 Piece sensor
III. Gin thi gian:
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IV. Chng trnh: Dng LADDER:
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Dng STL:
Network 1 // KHOI DONG
LD I0.0
O M0.0
AN I0.1
= M0.0
Network 2 // BOX CONVEYOR
LD M0.0
AN Q0.1
= Q0.0
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Network 3 // PHAT HIEN SUON LEN BOX_SENSOR
LD I0.2
EU
= M0.1
Network 4 // PIECE CONVEYOR
LD M0.1
O Q0.1
A M0.0
AN C30
= Q0.1
Network 5 // DEM SO QUA TAO
LD M0.0
A I0.3
LD C30
CTU C30, 10
Bi 8: IU KHIN BNG CHUYN CN TN
I. Yu cu:
H thng bng chuyn nh hnh v v c hot ng nh sau : Ban u, khi nhn Start th khi ng bng chuyn M1, khi vt chm vo S1 th khi ng bng chuyn M2. Khi S2 tc ng th khi ng bng chuyn M3, sau khi S2
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thi tc ng 10s th dng bng chuyn M2. Qu trnh hot ng tng t cho n bng chuyn M4.
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 M1
I0.1 Stop Q0.1 M2
I0.2 S1 Q0.2 M3
I0.3 S2 Q0.3 M4
I0.4 S3
I0.5 S4
III. Chng trnh: Dng LADDER:
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Dng STL: Network 1 // KHOI DONG
LD I0.0
S M0.0, 1
Network 2 // DUNG
LD I0.1
R M0.0, 1
R Q0.0, 4
Network 3 // KHOI DONG BANG CHUYEN M1
LD M0.0
= Q0.0
Network 4 // KHOI DONG BANG CHUYEN M2
LD M0.0
A I0.2
S Q0.1, 1
Network 5 // KHOI DONG BANG CHUYEN M3
LD M0.0
A I0.3
S Q0.2, 1
Network 6 // KHOI DONG BANG CHUYEN M4
LD M0.0
A I0.4
S Q0.3, 1
Network 7 // PHAT HIEN SUON XUONG S2
LD I0.3
ED
S M0.1, 1
Network 8 // TAO THOI GIAN T37
LD M0.1
TON T37, 100
Network 9 // DUNG BANG CHUYEN M2
LD T37
R M0.1, 1
R Q0.1, 1
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Network 10 // PHAT HIEN SUON XUONG S3
LD I0.4
ED
S M0.2, 1
Network 11 // TAO THOI GIAN T38
LD M0.2
TON T38, 100
Network 12 // DUNG BANG CHUYEN M3
LD T38
R M0.2, 1
R Q0.2, 1
Network 13 // PHAT HIEN SUON XUONG S4
LD I0.5
ED
S M0.3, 1
Network 14 // TAO THOI GIAN T39
LD M0.3
TON T39, 100
Network 15 // DUNG BANG CHUYEN M4
LD T39
R M0.3, 1
R Q0.3, 1
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Bi 9: IU KHIN DAO CT SN PHM
I. Yu cu:
C 3 loi sn phm A, B, C a vo dao ct thnh sn phm nh. Dao ct 1000 sn phm A th thay dao, ct 100 sn phm B th thay dao, ct 10 sn phm C th thay dao. Lc S2 tc ng th dao a xung, khi dao gp S1 th dng v quay ln. Khi dao ct khng c na th ci ku ln, bm reset th ci ht ku.
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 ng c h
I0.1 Stop Q0.1 ng c nng
I0.2 Reset Q0.2 Ci
I0.3 S1
I0.4 S2
I0.5 SA
I0.6 SB
I0.7 SC
III. Chng trnh: Dng LADDER:
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Dng STL:
Network 1 // KHOI DONG
LD I0.0
S M0.0, 1
Network 2 // DUNG
LD I0.1
R M0.0, 1
Network 3
LD I0.0
O I0.2
MOVW 0, VW1
Network 4
LD M0.0
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A I0.5
+I 1, VW1
Network 5
LD M0.0
A I0.6
+I 10, VW1
Network 6
LD M0.0
A I0.7
+I 100, VW1
Network 7
LD M0.0
A I0.4
AW
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Loi robot ny hay gp trong cc x nghip cng nghip. Robot nm cht vt t bng chuyn B b sang bng chuyn A. v tr ban u cnh tay pha bng chuyn A Khi n START ng c thun khi ng quay theo chiu kim ng h. Khi gp LS2 n dng li ng thi bng chuyn B c khi ng Bng chuyn B a vt tin ti cnh tay. Khi vt chm vo S1 l cm bin pht hin vt th cnh tay kp cht ly vt. Khi S2 tc ng tc l vt c kp cht th ng c ngc khi ng quay theo chiu ngc li Khi cnh tay chm vo LS1 th n dng v nh vt xung bng chuyn A Bng chuyn A lun hot ng (Khi n START n bt u hot ng)
II. Phn cng u vo, u ra :
u vo u ra
I0.0 Start Q0.0 Bng chuyn A
I0.1 Stop Q0.1 Bng chuyn B
I0.2 LS1 Q0.2 ng c thun
I0.3 LS2 Q0.3 ng c ngc
I0.4 S1 Q0.4 ng c kp
I0.5 S2
III. Chng trnh:
Dng LADDER:
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Dng STL:
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NETWORK 1 // KHOI DONG
LD I0.0
S M0.0, 1
NETWORK 2 // DUNG
LD I0.1
R M0.0, 1
NETWORK 3 // BANG CHUYEN A
LD M0.0
= Q0.0
NETWORK 4 // DONG CO THUAN
LD I0.2
O Q0.2
AN I0.3
AN I0.5
AN Q0.3
A M0.0
= Q0.2
NETWORK 5 // BANG CHUYEN B
LD M0.0
AN I0.4
A I0.3
= Q0.1
NETWORK 6 // DONG CO KEP
LD I0.4
O Q0.4
AN I0.2
A M0.0
= Q0.4
NETWORK 7 // DONG CO NGUOC
LD I0.5
AN I0.2
A M0.0
AN Q0.2
= Q0.3
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Bi 11: THIT K CN TRC
I.Yu cu:
Khi n PB1 th cu trc hot ng ko mc ln cho n khi S3 tc ng th dng. Sau trc ko di chuyn ngang sang tri. n v tr S4 tc ng th dng v h mc xung. Khi S2 tc ng th dng v ch y 20s. Ht thi gian 20s, mc c ko ln, n S4 tc ng th dng v di chuyn sang tri. n S5 tc ng, dng v h mc xung. Khi S2 tc ng th dng v chung reo ln. ti y, nhn PB2 th chung ngng reo v mc c ko ln. n S5 tc ng th di chuyn sang phi cho n khi S1 tc ng th dng v h mc xung. n khi S2 tc ng th dng.
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start1 (PB1) Q0.0 ng c sang tri
I0.1 Start2 (PB2) Q0.1 ng c sang phi
I0.2 Stop Q0.2 ng c ko ln
I1.1 S1 Q0.3 ng c h xung
I1.2 S2 Q0.4 Chung
I1.3 S3
I1.4 S4
I1.5 S5
III. Gin thi gian:
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MPhai
MTrai
S5
S4
S3
S2
S1
Stop
PB2
TG
PB1
MLen
MXuong
Coi
T37 20s
IV. Chng trnh: LADDER:
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Bi 12: X L HA CHT
I.Yu cu:
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H thng gm 4 bnh cha c cc bm chuyn cht lng qua h thng. Mi bn c cc cm bin pht hin bn cn hay y. Bn 2 c phn t nung nng vi cm bin nhit. Bn 3 c gn cn khuy trn hai thnh phn cht lng khi chng c bm vo t bn 1 v bn 2. Bn 3 v bn 4 c dung tch gp i bn 1 v bn 2. Hot ng ca h thng nh sau: Bn 1 v bn 2 c y t cc bn cha kim v polime ring bit thng qua bm 1 v bm 2. Bm 1 v bm 2 ngng hot ng khi c tn hiu t cm bin bo bn y. Phn t nung nng trong bn 2 c kch hot nng nhit polime ln 60
0C. Khi cm bin nhit bo tn hiu th tt b nung v kch hot bm 3, bm 4 chuyn hn hp dung dch vo bn 3. Cn khuy c kch hot khi bn ny c dung dch v trong khong ti thiu 60s. Khi bn 3 y, bm 3 v bm 4 dng hot ng. Nu thi gian khuy ln hn 60s, bm 5 s chuyn dung dch vo bn 4 thng qua mt b lc. Bm 5 dng hot ng khi bn 4 y v bn 3 cn. Cui cng, sn phm dung dch c a vo bn cha lu tr nh bm 6. Qu trnh x l n y kt thc v bt u mt chu trnh mi.
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 P1
I0.1 Khi ng P1 Q0.1 P2
I0.2 Khi ng P2 Q0.2 P3
I0.3 S11 Q0.3 P4
I0.4 S12 Q0.4 P5
I0.5 S21 Q0.5 P6
I0.6 S22 Q0.6 Cn khuy
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I0.7 Cm bin nhit Q0.7 B nung
I1.0 S31
I1.1 S32
I1.2 S41
I1.3 S42
I1.4 STOP
III. Lu thut ton:
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IV. Chng trnh: Dng LADDER:
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Dng STL:
NETWORK 1 // KHOI DONG
LD I0.0
O M0.0
AN I1.4
= M0.0
NETWORK 2 // DONG CO P1 HOAT DONG
LD I0.1
O Q0.0
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AN I0.3
A M0.0
= Q0.0
NETWORK 3 // DONG CO P2 HOAT DONG
LD I0.2
O Q0.1
AN I0.5
A M0.0
= Q0.1
NETWORK 4 // BO NUNG HOAT DONG
LD I0.5
O Q0.7
AN I0.7
A M0.0
= Q0.7
NETWORK 5 // DONG CO P3 , P4 HOAT DONG
LD I0.7
O Q0.2
AN I1.0
A M0.0
= Q0.2
= Q0.3
NETWORK 6 // CAN KHUAY HOAT DONG
LD I1.1
AN T37
A M0.0
= Q0.6
NETWORK 7 // TAO THOI GIAN T37
LD I1.1
TON T37, 600
NETWORK 8 // DONG CO P5 HOAT DONG
LD T37
O Q0.4
AN I1.2
A M0.0
= Q0.4
NETWORK 9 // DONG CO P6 HOAT DONG
LD I1.2
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O Q0.5
AN I1.3
A M0.0
= Q0.5
Bi 13: N GIAO THNG CH NG HP
I. Yu cu:
qung ng o, do ma bo nn b st l mt bn. ng ny ch cho php mt xe i qua, nu 2 xe i qua i qua cng mt lc c th gy tai nn. Do , mt h thng n bo c lp t y iu khin cc xe nh theo gin thi gian sau:
II. Phn cng u vo, u ra:
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u vo u ra
I0.0 Start Q0.0 Xanh 1
I0.1 Stop Q0.1 1
Q0.2 Xanh 2
Q0.3 2
III. Chng trnh: Dng LADDER:
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Dng STL:
Network 1 // KHOI DONG
LD I0.0
S M0.0, 1
Network 2
LD I0.1
R M0.0, 1
R Q0.0, 4
Network 3 // TAO THOI GIAN T37
LD M0.0
TON T37, 700
S Q0.0, 1
S Q0.3, 1
Network 4
LD M0.0
AW>= T37, 300
S Q0.1, 1
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R Q0.0, 1
Network 5
LD M0.0
AW>= T37, 350
S Q0.2, 1
R Q0.3, 1
Network 6
LD M0.0
AW>= T37, 650
S Q0.3, 1
R Q0.2, 1
Network 7
LD M0.0
AW>= T37, 700
R T37, 1
R Q0.1, 1
R Q0.3, 1
Bi 14: BI XE
I. Yu cu:
Bi xe c sc cha 100 chic. Nu bi cha y th hin thng bo Xin mi qu khch cho xe vo, nu bi y th thng bo Qu khch thng cm,
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bi y. Khi xe ti, photocell 1 pht hin m barie 1, oto i vo, lc ny barie 2 vn ng. Khi xe vo xong th barie 1 ng li. Khi ti x b 1 la vo thng th barie 2 m ra, xe vo bi, khi photocell 2 thi tc ng th barie 2 ng li. Khi xe ra, photocell 3 tc ng lm barie 3 m ra. Khi xe ra khi th photocell 3 thi tc ng, barie 3 ng li.
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Sensor pht hin b tin Q0.0 Barie 1
I0.1 Photocell 1 Q0.1 Barie 2
I0.2 Photocell 2 Q0.2 Barie 3
I0.3 Photocell 3 Q0.3 Thng bo cn ch
I0.4 Start Q0.4 Thng bo ht ch
I0.5 Stop
I0.6 Reset
III. Chng trnh: Dng LADDER:
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Dng STL:
Network 1 // KHOI DONG
LD I0.4
O M0.0
AN I0.5
= M0.0
Network 2 // BARIE 1 HOAT DONG
LD I0.1
AN Q0.4
A M0.0
= Q0.0
Network 3 // BARIE 2 HOAT DONG
LD I0.0
O Q0.1
A I0.2
A M0.0
= Q0.1
Network 4 // TAO XUNG DEM TIEN
LD I0.2
ED
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= M0.1
Network 5 // TAO XUNG DEM LUI
LD I0.3
ED
= M0.2
Network 6 // DEM SO XE TRONG BAI
LD M0.1
LD M0.2
LD I0.6
CTUD C48, 100
Network 7 // THONG BAO CON CHO
LDN C48
A M0.0
= Q0.3
Network 8 // THONG BAO HET CHO
LD C48
A M0.0
= Q0.4
Network 9 // BARIE 3 HOAT DONG
LD I0.3
A M0.0
= Q0.2
Bi 18: MY PHA TR, C PH
I.Yu cu:
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iu khin cc my bn cafe, tr, nc ngt ti cc ni cng cng Cng ngh : Gi s ban u cc lun c t ti ni rt v ch c thc ung c la chn l tr hoc cafe. Trc tin cho ng xu vo, la chn nc ung l tr hoc cafe tng ng n bo chn tr hoc cafe sng ln. Tip theo l chn sa hoc ng. Nu khng chn c th b qua, kt thc vic la chn ta n nt Finish. Lc ny bt van x cc loi thc ung chn, ng v sa c m trong 10s k t khi n nt Finish. Khi sensor bo y cc th ngng van x. Lc ny, ly cc ra ta n nt Cup_remove
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Insert coin Q0.0 Coffee_valve
I0.1 Coffee Q0.1 Tea_valve
I0.2 Tea Q0.2 Sugar_valve
I0.3 Sugar Q0.3 Milk_valve
I0.4 Milk Q0.4 Coffee_light
I0.5 Finish Q0.5 Tea_light
I0.6 Cup_full Q0.6 Remove_cup
I0.7 Cup_remove Q0.7 Cup_full_light
I1.0 Cup_removed
III. Chng trnh:
Dng LADDER:
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Dng STL: Network 1 // KHOI DONG, BO TIEN VAO
LD I0.0
S M0.0, 1
Network 2 // CHON COFFEE
LD I0.1
A M0.0
S M0.1, 1
Network 3 // DEN CHON COFFEE SANG
LD M0.1
= Q0.4
Network 4 // CHON TEA
LD I0.2
A M0.0
S M0.2, 1
Network 5 // DEN CHON TEA SANG
LD M0.2
= Q0.5
Network 6 // CHON SUGAR
LD I0.3
A M0.0
S M0.3, 1
Network 7 // CHON MILK
LD I0.4
A M0.0
S M0.4, 1
Network 8 // KHI CHON XONG, NHAN FINISH
LD I0.5
LD M0.1
O M0.2
ALD
S M0.5, 1
Network 9 // MO VALVE ROT COFFEE
LD M0.5
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A M0.1
= Q0.0
Network 10 // MO VALVE ROT TEA
LD M0.5
A M0.2
= Q0.1
Network 11 // TAO THOI GIAN T37
LD M0.5
TON T37, 100
Network 12 // MO VALVE ROT SUGAR
LD M0.5
AN T37
A M0.3
= Q0.2
Network 13 // MO VALVE ROT MILK
LD M0.5
AN T37
A M0.4
= Q0.3
Network 14 // KHI COC DAY THI DONG TAT CA CAC VALVE, DONG
THOI DEN CUP FULL LIGHT SANG LEN
LD I0.6
R M0.0, 5
= Q0.7
Network 15 // NHAN NUT CUP REMOVE DE LAY COC RA
LD I0.7
O Q0.6
A I0.6
A I1.0
A M0.5
= Q0.6
Network 16 // PHAT HIEN COC DA DUOC LAY RA
LD I1.0
ED
= M1.0
Network 17 // RESET TOAN BO HE THONG
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LD M1.0
R M0.5, 1
Bi 19:IU KHIN MY KHOAN
I.Yu cu:
iu khin my khoan theo yu cu sau: ng c M1 c nhim v tnh tin mi khoan i ln v i xung, ng c M2 c nhim v quay mi khoan.Ban u mi khoan v tr cao nht ( S1 tc ng). Khi n Start th mi khoan i xung khoan vo phi. Khi S3 tc ng th dng li v i ln. n khi S2 tc ng th dng v i xung.Khi S4 tc ng th dng v i ln. Khi S1 tc ng th dng li, kt thc qu trnh khoan.
II.Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 M1_up
I0.1 Stop Q0.1 M1_down
I0.2 S1 Q0.2 M2
I0.3 S2
I0.4 S3
I0.5 S4
III. Chng trnh: Dng LADDER:
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Dng STL: Network 1 // KHOI DONG
LD I0.0
S M0.0, 1
Network 2 // DUNG
LD I0.1
R M0.0, 1
Network 3 // TAO BO DEM C30
LD I0.2
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O I0.3
O I0.4
O I0.5
LD C30
CTU C30, 9
Network 4 // KHOI DONG DONG CO M2
LD M0.0
= Q0.2
Network 5
LD M0.0
AW= C30, 1
S Q0.1, 1
Network 6
LD M0.0
AW= C30, 3
S Q0.0, 1
R Q0.1, 1
Network 7
LD M0.0
AW= C30, 4
S Q0.1, 1
R Q0.0, 1
Network 8
LD M0.0
AW= C30, 6
S Q0.0, 1
R Q0.1, 1
Network 9
LD M0.0
AW= C30, 9
R Q0.0, 1
R M0.0, 1
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BI 1: N GIAO THNG NG HP
I.Yu cu cng ngh qung ng o, do ma bo nn b st l mt bn. ng ny ch cho php mt xe i qua, nu 2 xe i qua i qua cng mt lc c th gay tai nun. Do do, mt h thng n bo c lp t y iu khin cc xe nh theo gin this gain sau:
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II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start_i0_0 Q0.0 Xanh 1 (X1_q0_0)
I0.1 Stop_i0_1 Q0.1 1 (D1_q0_1)
Q0.2 Xanh 2 (X2_q0_2)
Q0.3 2 (D2_q0_3)
III.Chng trnh dng LADER trn S7_200
IV. Dng STL trong S7_200
LD I0.0
S M0.0, 1
LD I0.1
R M0.0, 1
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LD M0.0
AN T37
TON T37, 700
LD M0.0
AW= T37, 300
AW= T37, 350
AW
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VI.Dng STL trong S7_300
A I 0.0
S M 0.0
A I 0.1
R M 0.0
A M 0.0
AN T 0
L S5T#1M10S
SD T 0
NOP 0
L T 0
T MW 10
NOP 0
NOP 0
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A M 0.0
A(
L MW 10
L 700
=I
)
= Q 0.0
A M 0.0
A(
L MW 10
L 400
=I
)
= Q 0.1
A M 0.0
A(
L MW 10
L 350
=I
)
= Q 0.2
A M 0.0
AN Q 0.2
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= Q 0.3
BI 3: IU KHIN NG M CA KHO T NG
I.Yu cu cng ngh Khi xe ang tin v gn ca kho, cm bin Ultrasounic SS1 (cm bin quang) nhn dng c xe v ca s c m ra n gp gii hn hnh trnh trn LS1 th ca dng li ri xe chy vo. Khi cm bin quang SS2 t pha trong cng ca nhn dng c xe i qua khi ca th ca s c ng li, chm vo gii hn hnh trnh di LS2 th ca dng li. Xe ch i mt chiu
II.Bng phn cng u vo ,u ra
u vo u ra
I0.0 START Q0.0 Raise_door
I0.1 STOP Q0.1 Down_door
I0.2 SS1
I0.3 SS2
I0.4 LS1
I0.5 LS2
III.Gin this gain
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IV.Chng trnh LAD trong S7_200
V.Chng trnh STL trong S7_200
LD I0.0
O M0.0
AN I0.1
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= M0.0
LD M0.1
O Q0.0
A M0.0
AN I0.4
AN Q0.1
= Q0.0
LD M0.2
O Q0.1
A M0.0
AN I0.5
AN Q0.0
= Q0.1
LD M0.0
A I0.2
EU
= M0.1
LD M0.0
AN I0.3
ED
= M0.2
VI.Chng trnh LAD vit trong S7_300
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VII.Chng trnh vit STL trong S7_300
A(
O I 0.0
O M 0.0
)
AN I 0.1
= M 0.0
A(
O M 0.1
O Q 0.0
)
A M 0.0
AN I 0.4
AN Q 0.1
= Q 0.0
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A(
O M 0.2
O Q 0.1
)
A M 0.0
AN I 0.5
AN Q 0.0
= Q 0.1
A M 0.0
A I 0.2
FP M 10.0
= M 0.1
A M 0.0
AN I 0.3
FN M 10.1
= M 0.2
BI 4:DY CHUYN CHIT NC VO CHAI
I.Yu cu cng ngh
Cng on chit nc vo chai theo nguyn tc this gain, m t hnh 3.13. Nhn nt Start PB, bng ti chy v mang theo chai, cm bin chai pht hin, bng ti dng, bt u m van chit chit nc vo chai. Thi gain chit y l 30s, sau khi chit y bng ti li chy v tip tc chit chai k tip
II.Bng phn cng u vo, u ra
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u vo u ra
I0.0 Start Q0.0 Conveyer motor
I0.1 Stop Q0.1 Fill valve
I0.2 Bottle_sensor
III.Chng trnh vit LAD trong S7_200
IV.Chng trnh vit STL trong S7_200
LD I0.0
O M0.0
AN I0.1
= M0.0
LD M0.0
AN M0.1
AN Q0.1
= Q0.0
LD M0.1
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O Q0.1
A M0.0
AN T37
= Q0.1
LD M0.0
AN Q0.0
TON T37, 300
LD M0.0
A I0.2
EU
= M0.1
V.Chng trnh vit trong S7_300
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VI.Chng trnh vit STL trong S7_300
A(
O I 0.0
O M 0.0
)
AN I 0.1
= M 0.0
A M 0.0
AN M 0.1
AN Q 0.1
= Q 0.0
A(
O M 0.1
O Q 0.1
)
A M 0.0
AN T 37
= Q 0.1
A M 0.0
AN Q 0.0
L S5T#30S
SD T 0
NOP 0
NOP 0
NOP 0
NOP 0
A M 0.0
A I 0.2
FP M 10.0
= M 0.1
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BI 5: H THNG M V NG GI SN PHM
I.Yu cu cng ngh
Khi nhn cng tc khi ng PB Start bng chuyn hp mang hp ng sn phm di chuyn. Cm bin hp nhn dng thng ng tc ng, bng chuyn hp dng li. Bng chuyn to cha to nhn t mt xilo dch chuyn, sn phm c rt vo hp ng, sn phm c m bi mt cm bin quang hng ngoi (cm bin to) khi m c 10 qu to( mi hp cha 10 qu to) th bng chuyn to dng, bng chuyn hp dch chuyn ng gi hp mi. ngng qu trnh ta nhn nt PB Stop.
II.Bng phn cng vo/ra
u vo u ra
I0.0 PB Start Q0.0 Bng chuyn hp
I0.1 PB Stop Q0.1 Bng chuyn to
I0.2 CB hp
I0.3 CB to
III.Gin thi gian
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IV.Chng trnh vit trong S7_200 dng LAD
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V.Chng trnh vit trong S7_200 dng STL
LD I0.0
O M0.0
AN I0.1
= M0.0
LD M0.0
AN M0.1
= Q0.0
LD M0.0
AN I0.2
ED
S M0.1, 1
LD M0.0
AN Q0.0
AN C10
= Q0.1
LD M0.0
A I0.3
EU
LD C10
O I0.1
CTU C10, 10
LD M0.0
AW= C10, 10
R M0.1, 1
VI.Chng trnh vit trong S7_300 dng LAD
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VII.Chng trnh STL vit trong S7_300
A(
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O I 0.0
O M 0.0
)
AN I 0.1
= M 0.0
A M 0.0
AN M 0.1
= Q 0.0
AN I 0.2
FN M 10.0
S M 0.1
A M 0.0
AN Q0.0
AN M 0.3
= Q0.1
A M 0.0
A I0.3
FP M 10.1
= M 0.2
A M 0.2
CU C 10
BLD 101
A I 1.3
L C#10
S C 10
A(
O I 0.1
O M 0.3
)
R C 10
L C 10
T MW 2
NOP 0
A C 10
= Q 1.0
A M 0.0
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A(
L MW 2
L 10
==I
)
R M 0.1
= M 0.3
BI 6: IU KHIN BNG CHUYN CN TN
I.Yu cu cng ngh
H thng bng chuyn nh hnh v v c hot ng nh sau : Ban u, khi nhn Start th khi ng bng chuyn M1, khi vt chm vo S1 th khi ng bng chuyn M2. Khi S2 tc ng th khi ng bng chuyn M3, sau khi S2 thi tc ng 10s th dng bng chuyn M2. Qu trnh hot ng tng t cho n bng chuyn M4. II.Phn cng u vo/u ra
u vo u ra
I0.0 Start Q0.0 M1
I0.1 Stop Q0.1 M2
I0.2 S1 Q0.2 M3
I0.3 S2 Q0.3 M4
I0.4 S3
I0.5 S4
III.Chng trnh dng LAD trong S7_200
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IV.Chng trnh STL vit trong S7_200
LD I0.0
O M0.0
AN I0.1
= M0.0
LD M0.0
A I0.2
= M0.1
LD M0.1
EU
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= M0.2
LD M0.2
MOVW 1, MW2
LD M0.3
SLW MW2, 1
LD I0.1
O T39
MOVW 0, MW2
LD M3.0
A I0.3
LD M3.1
A I0.4
OLD
LD M3.2
A I0.5
OLD
AN M0.3
= M0.3
LD M0.0
= Q0.0
LD M3.0
O M3.1
AN T37
= Q0.1
LD Q0.2
TON T37, 100
LD M3.1
O M3.2
AN T38
= Q0.3
LD Q0.3
TON T38, 100
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LD M3.2
O M3.3
AN T39
= Q0.3
LD M3.3
TON T39, 100
V.Chng trnh dng LAD vit trong S7_300
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Su tm Page 99
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Su tm Page 100
VI.Chng trnh dng STL vit trong S7_300
A(
O I 0.0
O M 0.0
)
AN I 0.1
= M 0.0
A M 0.0
A I 0.2
= M 0.1
A M 0.1
FP M 10.0
= M 0.2
A M 0.2
JNB _001
L W#16#1
T MW 2
_001: NOP 0
A M 0.3
JNB _002
L W#16#1
L MW 2
SLW
T MW 2
_002: NOP 0
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A(
O I 0.1
O T 2
)
JNB _003
L W#16#0
T MW 2
_003: NOP 0
A(
A M 3.0
A I 0.3
O
A M 3.1
A I 0.4
O
A M 3.2
A I 0.5
)
AN M 0.3
= M 0.3
A M 0.0
= Q 0.0
A(
O M 3.0
O M 3.1
)
AN T 0
= Q 0.1
A Q 0.2
L S5T#10S
SD T 0
NOP 0
NOP 0
NOP 0
NOP 0
A(
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O M 3.1
O M 3.2
)
AN T 1
= Q 0.2
A Q 0.3
L S5T#10S
SD T 1
NOP 0
NOP 0
NOP 0
NOP 0
A(
O M 3.2
O M 3.3
)
AN T 2
= Q0.3
A M 3.3
L S5T#10S
SD T 2
NOP 0
NOP 0
NOP 0
NOP 0
BI 7: IU KHIN MY KHOAN
I.Yu cu:
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Su tm Page 103
iu khin my khoan theo yu cu sau: ng c M1 c nhim v tnh tin mi khoan i ln v i xung, ng c M2 c nhim v quay mi khoan.Ban u mi khoan v tr cao nht ( S1 tc ng). Khi n Start th mi khoan i xung khoan vo phi. Khi S3 tc ng th dng li v i ln. n khi S2 tc ng th dng v i xung.Khi S4 tc ng th dng v i ln. Khi S1 tc ng th dng li, kt thc qu trnh khoan.
II.Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 M1_up
I0.1 Stop Q0.1 M1_down
I0.2 S1 Q0.2 M2
I0.3 S2
I0.4 S3
I0.5 S4
III.Chng dng LAD vit trong S7_200
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Su tm Page 104
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Su tm Page 105
IV.Chng trnh dng STL vit trong S7_200
LD I0.0
S M0.0, 1
R M0.1, 1
LD I0.1
S M0.1, 1
R M0.0, 1
LD M0.0
= Q0.2
LD M0.0
A I0.0
A I0.2
= M0.2
LD M0.2
EU
= M0.3
LD M0.3
MOVW 1, MW2
LD M0.4
SLW MW2, 1
LD M0.1
O M1.0
MOVW 0, MW2
LD M3.0
A I0.4
LD M3.1
A I0.3
OLD
LD M3.2
A I0.5
OLD
LD M3.3
A I0.2
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OLD
AN M0.4
= M0.4
LD M3.1
O M3.3
= M1_up_q0_0
LD M3.0
O M3.2
= Q0.1
LD M3.4
R M0.0, 1
S M1.0, 1
V.Chng trnh dng LAD vit trong S7_300
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Su tm Page 108
VI.Chng trnh dng STL vit trong S7_300
A I0.0
S M 1.0
R M 1.1
A I 0.1
S M 1.1
R M 1.0
A M 1.0
= Q 0.0
A I 0.0
A M 1.0
A I 0.2
= M 1.2
A M 1.2
FP M 10.0
= M 1.3
A M 1.3
JNB _001
L W#16#1
T MW 2
_001: NOP 0
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A M 1.4
JNB _002
L W#16#1
L MW 2
SLW
T MW 2
_002: NOP 0
A(
O M 1.1
O M 1.4
)
JNB _003
L W#16#0
T MW 2
_003: NOP 0
A(
A M 3.0
A I 0.4
O
A M 3.1
A I 0.3
O
A M 3.2
A I0.5
O
A M 3.3
A I 0.2
)
AN M 1.4
= M 1.4
O M 3.1
O M 3.3
= Q 0.1
O M 3.0
O M 3.2
= Q 0.2
A M 3.4
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R M 1.0
S M 1.4
BI 8: IU KHIN DAO CT SN PHM
I.Yu cu cng ngh
C 3 loi sn phm A, B, C a vo dao ct thnh sn phm nh. Dao ct 1000 sn phm A th thay dao, ct 100 sn phm B th thay dao, ct 10 sn phm C th thay dao. Lc S1 tc ng th dao a i ra ct sn phm, khi dao gp S2 th dng v thu dao v v tr ban u chun b ct sn phm mi. Khi dao ct khng c na th ci ku ln, bm reset th ci ht ku.
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 ng c tin
I0.1 Stop Q0.1 ng c li
I0.2 Reset Q0.2 Ci
I0.3 S1
I0.4 S2
I0.5 SA
I0.6 SB
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I0.7 SC
III.Chng trnh dang LAD trong S7_200
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Su tm Page 112
IV.Chng trnh dang STL trong S7_200
LD I0.0
O M0.0
AN I0.1
= M0.0
LD I0.0
O I0.2
MOVW 1000, VW1
LD M0.0
A I0.5
-I 1, VW1
LD M0.0
A I0.6
-I 10, VW1
LD M0.0
A I0.7
-I 100, VW1
LD M0.0
A I0.3
AW>= VW1, 0
S Q0.0, 1
R Q0.1, 1
LD M0.0
A I0.4
AW>= VW1, 0
S Q0.1, 1
R Q0.0, 1
LD M0.0
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AW= VW1, 0
S Q0.2, 1
LD I0.2
R Q0.2, 1
V.Chng trnh dang LAD trong S7_300
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VI.Chng trnh dang STL trong S7_300
A(
O I 0.0
O M 0.0
)
AN I 0.1
= M 0.0
A(
O I 0.0
O I 0.2
)
JNB _001
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L 1000
T MW 1
_001: NOP 0
A M 0.0
A I 0.5
JNB _002
L MW 1
L 1
-I
T MW 1
_002: NOP 0
A M 0.0
A I 0.6
JNB _003
L MW 1
L 10
-I
T MW 1
_003: NOP 0
A M 0.0
A I 0.7
JNB _004
L MW 1
L 100
-I
T MW 1
_004: NOP 0
A M 0.0
A I 0.3
A(
L MW 1
L 0
>=I
)
S Q 0.0
R Q 0.1
A M 0.0
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A I 0.4
A(
L MW 1
L 0
>=I
)
S Q 0.1
R Q 0.0
A M 0.0
A(
L MW 1
L 0
==I
)
S Q 0.2
A I 0.2
R Q 0.2
BI 9: IU KHIN CNH TAY ROBOT GP HNG
I.Yu cu cng ngh
Loi robot ny hay gp trong cc x nghip cng nghip. Robot nm cht vt t bng chuyn B b sang bng chuyn A. v tr ban u cnh tay pha bng chuyn A
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Khi n START ng c thun khi ng quay theo chiu kim ng h. Khi gp LS2 n dng li ng this bng chuyn B c khi ng Bng chuyn B a vt tin ti cnh tay. Khi vt chm vo S1 l cm bin pht hin vt th cnh tay kp cht ly vt. Khi S2 tc ng tc l vt c kp cht th ng c ngc khi ng quay theo chiu ngc li Khi cnh tay chm vo LS1 th n dng v nh vt xung bng chuyn A Bng chuyn A lun hot ng (Khi n START n bt u hot ng)
II. Phn cng u vo, u ra
u vo u ra
I0.0 Start Q0.0 Bng chuyn A
I0.1 Stop Q0.1 Bng chuyn B
I0.2 LS1 Q0.2 ng c thun
I0.3 LS2 Q0.3 ng c ngc
I0.4 S1 Q0.4 ng c kp
I0.5 S2
III. Chng trnh dng LAD trong S7_200
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Su tm Page 119
IV. Chng trnh dng STL trong S7_200
LD I0.0
O M0.0
AN I0.1
= M0.0
LD M0.0
A I0.0
A I0.2
= M0.1
LD M0.1
EU
= M0.2
LD M0.2
MOVW 1, MW1
LD M0.3
SLW MW1, 1
LD I0.1
O M0.4
MOVW 0, MW1
LD M2.0
A I0.3
LD M2.1
A I0.4
OLD
LD M2.2
A I0.5
OLD
LD M2.3
A I0.2
OLD
AN M0.3
= M0.3
LD M0.0
= Q0.0
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LD M2.1
= BC_B_q0_1
LD M2.0
= Q0.2
LD M2.3
= Q0.3
LD M2.2
S Q0.4, 1
LD M2.4
R Q0.4, 1
= M0.4
V. Chng trnh dng LAD trong S7_300
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Su tm Page 122
VI. Chng trnh dng STL trong S7_300
A(
O I 0.0
O M 0.0
)
AN I 0.1
= M 0.0
A M 0.0
A I 0.0
A I 0.2
= M 0.1
A M 0.1
FP M 10.0
= M 0.2
A M 0.2
JNB _001
L W#16#1
T MW 1
_001: NOP 0
A M 0.3
JNB _002
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L W#16#1
L MW 1
SLW
T MW 1
_002: NOP 0
A(
O I 0.1
O M 0.4
)
JNB _003
L W#16#0
T MW 1
_003: NOP 0
A(
A M 2.0
A I 0.3
O
A M 2.1
A I 0.4
O
A M 2.2
A I 0.5
O
A M 2.3
A I 0.2
)
AN M 0.3
= M 0.3
A M 0.0
= Q 0.0
A M 2.1
= Q 0.1
A M 2.0
= Q0.2
A M 2.3
= Q 0.3
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A M 2.2
S Q 0.4
A M 2.4
R Q 0.4
= M 0.4
BI 10: THIT K CU TRC
I.Yu cu cng ngh
Khi n PB1 th cu trc hot ng ko mc ln cho n khi S3 tc ng th dng. Sau trc ko di chuyn ngang sang tri. n v tr S4 tc ng th dng v h mc xung. Khi S2 tc ng th dng v ch y 20s. Ht thi gian 20s, mc c ko ln, n S4 tc ng th dng v di chuyn sang tri. n S5 tc ng, dng v h mc xung. Khi S2 tc ng th dng v chung reo ln. ti y, nhn PB2 th chung ngng reo v mc c ko ln. n S5 tc ng th di chuyn sang phi cho n khi S1 tc ng th dng v h mc xung. n khi S2 tc ng th dng. II. Phn cng u vo, u ra
u vo u ra
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III. Gin thi gian
I0.0 Start1 (PB1) Q0.0 ng c sang tri
I0.1 Start2 (PB2) Q0.1 ng c sang phi
I0.2 Stop Q0.2 ng c ko ln
I1.1 S1 Q0.3 ng c h xung
I1.2 S2 Q0.4 Chung
I1.3 S3
I1.4 S4
I1.5 S5
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MPhai
MTrai
S5
S4
S3
S2
S1
Stop
PB2
TG
PB1
MLen
MXuong
Coi
T37 20s
IV.Chng trnh dng LAD trong S7_200
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Su tm Page 127
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Su tm Page 128
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Su tm Page 129
V.Chng trnh dng STL trong S7_200
LD I0.0
S M0.0, 1
R M0.1, 1
LD I0.2
R M0.0, 1
S M0.1, 1
LD M0.0
A I1.1
A I1.2
= M0.2
LD M0.2
EU
= M0.3
LD M0.3
MOVW 1, MW1
LD M0.4
SLW MW1, 1
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LD M1.1
O M0.1
MOVW 0, MW1
LD M2.0
A I1.1
A I1.3
LD M2.1
A I1.4
OLD
LD M2.2
A I1.2
A T37
OLD
LD M2.3
A I1.4
OLD
LD M2.4
A I1.5
OLD
LD M2.5
A I0.1
OLD
LD M2.6
A I1.3
OLD
LD M2.7
A I1.1
OLD
LD M1.0
A I1.2
OLD
AN M0.4
= M0.4
LD M2.0
O M2.3
O M2.6
= Q0.2
LD M2.1
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O M2.4
= Q0.0
LD M2.2
O M2.5
O M1.0
= Q0.3
LD M2.7
= Q0.1
LD M2.2
A I1.2
TON T37, 200
LD M2.5
= Q0.4
VI.Chng trnh dng LAD trong S7_300
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Su tm Page 132
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Su tm Page 133
VII.Chng trnh dng STL trong S7_300
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Su tm Page 134
A(
O I 0.0
O M 0.0
)
AN I 0.2
= M 0.0
A M 0.0
A I 1.1
A I 1.2
= M 0.1
A M 0.1
FP M 10.0
= M 0.2
A M 0.2
JNB _001
L W#16#1
T MW 1
_001: NOP 0
A M 0.3
JNB _002
L W#16#1
L MW 1
SLW
T MW 1
_002: NOP 0
A(
O I 0.2
O M 1.0
)
JNB _003
L W#16#0
T MW 1
_003: NOP 0
A(
A M 2.0
-
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Su tm Page 135
A I 1.1
A I 1.3
O
A M 2.1
A I 1.4
O
A M 2.2
A I 1.2
A T 0
O
A M 2.3
A I 1.4
O
A M 2.4
A I 1.5
O
A M 2.5
A I 0.1
O
A M 2.6
A I 1.3
O
A M 2.7
A I 1.1
O
A M 1.0
A I 1.2
)
AN M 0.3
= M 0.3
O M 2.0
O M 2.3
O M 2.6
= Q 0.2
O M 2.1
O M 2.4
= Q 0.0
O M 2.2
O M 2.5
O M 1.0
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= Q 0.3
A M 2.7
= Q 0.1
A M 2.2
A I 1.2
L S5T#20S
SD T 0
NOP 0
NOP 0
NOP 0
NOP 0
A M 2.5
= Q 0.4
BI 11: MY BN V T NG
I.Yu cu cng ngh
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Su tm Page 137
C 3 loi xu :xu 5, xu 10 ,xu 20,c 3 nt n mua v:N1 mua mt v,N2 mua 2 v,N3 mua 3 v.C 3 ca b xu:1 ca b xu 5,1 ca b xu 10,1 ca b xu 20.ng vi loi xu no th b vo ca xu y,cn b vo ca xu khc th khng va.C 1 ca thi tin vi loi xu 5.Mi v bn 15 xu,khi mua 2 v b 40 xu thi li 2 xu 5
II.Lu thut ton
III.Xc nh s u vo v u ra
u vo u ra
I0.0 Sensor pht hin b xu 5 Q0.0 Ra 1 v
Bt u
n N1,N2 hoc N3
Sau 2 pht m
MW4M
W2
Xut s v tng ng vi s v ngi mua n
Thi li tin d
Xa vng nh b cng (MW4)
Ly MW4 chia s
nguyn cho 15
Thng=2
Thi li s tin bng vi s phn d ca php chia
Xut ra 2 v Xut ra 1
v
Lu tng s tin ca v cn mua vo trong vng bin
Thng=1
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Su tm Page 138
I0.1 Sensor pht hin b xu 10 Q0.1 Ra 2 v
I0.2 Sensor pht hin b xu 20 Q0.2 Ra 3 v
I1.1 N1 mua 1 v Q1.1 Thi 1 ng xu 5
I1.2 N2 mua 2 v Q1.2 Thi 2 ng xu 5
I1.3 N3 mua 3 v Q1.3 Thi 3 ng xu 5
Q1.4 Thi 4 ng xu 5
IV.Chng trnh LAD vit trong S7_200
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Su tm Page 139
-
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Su tm Page 140
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Su tm Page 141
IV.Chng trnh STL vit trong S7_200
LD I1.1
O I1.2
O I1.3
S M0.0, 1
LD M0.0
EU
= M0.1
LD M0.1
O M0.7
MOVW 0, MW2
LD I1.1
MOVW 15, MW4
LD I1.2
MOVW 30, MW4
LD I1.3
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Su tm Page 142
MOVW 45, MW4
LD M0.0
TON T37, 1200
LD I0.0
+I 5, MW2
LD I0.1
+I 10, MW2
LD I0.2
+I 20, MW2
LDW>= MW2, MW4
S M0.2, 1
LD M0.2
EU
= M0.3
LD M0.3
MOVW MW2, MW6
-I MW4, MW6
LD M0.3
LPS
AW= MW4, 15
= Q0.0
LRD
AW= MW4, 30
= Q0.1
LPP
AW= MW4, 45
= Q0.2
LD M0.3
LPS
AW= MW6, 5
= Q1.1
LRD
AW= MW6, 10
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= Q1.2
LRD
AW= MW6, 15
= Q1.3
LPP
AW= MW6, 20
= Q1.4
LDW>= MW2, MW4
NOT
S M0.4, 1
LD M0.4
A T37
S M0.5, 1
LD M0.5
EU
= M0.6
LD M0.6
MOVW MW4, MW10
DIV 15, MD8
LD M0.6
LPS
AW= MW10, 1
= Q0.0
LPP
AW= MW10, 2
= Q0.1
LD M0.6
LPS
AW= MW8, 5
= Q1.1
LPP
AW= MW8, 10
= Q1.2
LD Q0.0
O Q0.1
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O Q0.2
LD Q1.1
O Q1.2
O Q1.3
O Q1.4
ALD
= M0.7
LDW= T37, 1200
R M0.0, 1
R M0.2, 1
R M0.4, 1
R M0.5, 1
V.Chng trnh LAD vit trong S7_300
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Su tm Page 145
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Su tm Page 146
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Su tm Page 147
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Su tm Page 148
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Su tm Page 149
VI.Chng trnh STL vit trong S7_300
O I 1.1
O I 1.2
ON I 1.3
S M 0.0
A M 0.0
FP M 30.0
= M 0.1
A(
O M 0.1
O M 0.7
)
JNB _001
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Su tm Page 150
L W#16#0
T MW 2
_001: NOP 0
A I 1.1
JNB _002
L 15
T MW 4
_002: NOP 0
A I 1.2
JNB _003
L 30
T MW 4
_003: NOP 0
A I 1.3
JNB _004
L 45
T MW 4
_004: NOP 0
A M 0.0
L S5T#2M
SD T 0
NOP 0
L T 0
T MW 20
NOP 0
NOP 0
A I 0.0
JNB _005
L 5
L MW 2
+I
T MW 2
_005: NOP 0
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A I 0.1
JNB _006
L 10
L MW 2
+I
T MW 2
_006: NOP 0
A I 0.2
JNB _007
L 20
L MW 2
+I
T MW 2
_007: NOP 0
L MW 2
L MW 4
>=I
S M 0.2
A M 0.2
FP M 30.1
S M 0.3
A M 0.3
JNB _008
L MW 2
L MW 4
-I
T MW 6
_008: NOP 0
A M 0.3
= L 20.0
A L 20.0
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A(
L MW 4
L 15
==I
)
= Q 0.0
A L 20.0
A(
L MW 4
L 30
==I
)
= Q 0.1
A L 20.0
A(
L MW 4
L 45
==I
)
= Q 0.2
A M 0.3
= L 20.0
A L 20.0
A(
L MW 6
L 5
==I
)
= Q 1.1
A L 20.0
A(
L MW 6
L 10
==I
)
= Q 1.2
A L 20.0
A(
L MW 6
L 15
==I
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Su tm Page 153
)
= Q 1.3
A L 20.0
A(
L MW 6
L 20
==I
)
= Q 1.4
A(
L MW 2
L MW 4
>=I
)
NOT
S M 0.4
A M 0.4
A T 0
S M 0.5
A M 0.5
FP M 30.2
= M 0.
A M 0.6
JNB _009
L MW 4
L 15
/I
T DBW 8
_009: NOP 0
A M 0.6
= L 20.0
A L 20.0
A(
L MW 10
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L 1
==I
)
= Q 0.0
A L 20.0
A(
L MW 10
L 2
==I
)
= Q 0.1
A M 0.6
= L 20.0
A L 20.0
A(
L MW 8
L 5
==I
)
= Q 1.1
A L 20.0
A(
L MW 8
L 10
==I
)
= Q 1.2
A(
O Q 0.0
O Q 0.1
O Q 0.2
)
A(
O Q 1.1
O Q 1.2
O Q 1.3
O Q 1.4
)
= M 0.7
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L MW 20
L 1200
==I
R M 0.0
R M 0.2
R M 0.4
R M 0.5
BI 12: TRM TRN VT LIU
I.Yu cu cng ngh
Mt trm trn vt liu c yu cu cng ngh nh sau:Ban u khi trm mi bt u hot ng chng ta nhn nt Start th van V1 v V2 bt u x cc vt liu t hai Xilo xung cc cn s c gn cc cm bin o lng S1 v S2.Khi S1 v S2 tc ng th khi ng hai bng chuyn A v B,sau 5s th van x V3 v V4 c m x cc vt liu xung bn trn,van V3 v V4 x trong vng 20s th ngng x. + Sau khi van V3 v V4 ngng x 5s th bng chuyn A v B ngng hot ng
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+Sau khi van V3 v V4 ngng x th V1 v V2 x vt liu ln cn.S1 v S2 tc ng th ng van xa V1 v V2. +Van V3 v V4 x th m ng thi van V5 sau 5s th khi ng ng c M1 +n khi mc nc n S6 tc ng th : -ng van V5 sau 5s th dng ng c M1 -Khi ng ng c khuy sau 50s th ng ng c khuy +ng c khuy ngng th m van x V6 sau 2s th khi ng c bm x hn hp ra II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 V1
I0.1 Stop Q0.1 V2
I1.0 Sensor1 Q0.2 V3
I1.1 Sensor2 Q0.3 V4
I1.2 S5 Q0.4 V5
I1.3 S6 Q0.5 V6
Q0.6 Bng chuyn A
Q0.7 Bng chuyn B
Q1.0 Bm M1
Q1.1 Bm M2
Q1.2 ng c khuy
III. Chng trnh LAD trong S7_200:
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IV. Chng trnh STL trong S7_200:
LD SM0.1
MOVB 2#0, MB0
LD I0.0
S M0.0, 1
LD I0.1
MOVB 2#0, MB0
LD M0.0
LPS
EU
MOVB 2#1, VB0
LPP
A Q0.6
MOVB 2#0, VB0
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LD M0.0
LPS
AN I1.0
AN Q0.2
= Q0.0
LPP
AN I1.1
AN Q0.3
= Q0.1
LD M0.0
LD V0.0
A I1.0
A I1.1
LD V1.0
A I1.2
OLD
O Q0.6
ALD
AN T38
= Q0.6
= Q0.7
LD Q0.6
O Q0.7
TON T37, 50
A T37
LPS
TON T38, 200
AW
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A M0.1
= Q0.4
TON T39, 50
A T39
AN M0.2
= Q1.0
LD M0.2
TON T40, 50
AW= T40, 50
R M0.1, 1
LD I1.3
EU
S M0.2, 1
LD M0.0
A M0.2
= Q1.2
TON T41, 500
A T41
S M0.4, 1
R M0.2, 1
LD M0.0
A M0.4
= Q0.5
TON T42, 50
A T42
AN M0.5
= Q1.1
LD I1.2
EU
S M0.5, 1
MOVB 2#1, VB1
LD M0.5
TON T43, 50
A T43
R M0.4, 1
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Bi 1: N GIAO THNG
I.Yu cu:
Thit k h thng n giao thng ng t vi cc yu cu sau: - 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
II. Bng phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 n xanh 1
I0.1 Stop Q0.1 n vng 1
Q0.2 n 1
Q0.3 n xanh 2
Q0.4 n vng 2
Q0.5 n 2
III. Gin thi gian:
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X1
V1
D1
X2
V2
D2
25s
5s
30s
25s
5s
30s
IV. Chng trnh: A. Dng S7-200
Dng LADDER:
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Dng STL: NETWORK 1 : TO ROLE TRUNG GIAN LD START
O M0.0
AN STOP_
= M0.0
NETWORK 2: TO THI GIAN LD M0.0
AN T37
TON T37, 600
NETWORK 3: KHOI DONG
LD M0.0
LPS
AW
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AW>= T37, 300
AW= T37, 300
AW= T37, 550
AW
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Bi 2: THIT K H THNG BM NC DI HM M
I. Yu cu: Thit k h thng bm nc di hm m tha mn yu cu sau:
- Mc nc c o bi 2 sensor S1,S2 - Nhm m bo vn hnh kinh t, cc bm c iu khin hot ng nh
sau:
II. Bng phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 n Start
I0.1 Stop Q0.1 n Stop
I0.2 S1 Q0.2 ng c M1
I0.3 S2 Q0.3 ng c M2
III. Lu thut ton:
M1 M1 M1
M1 M1
M1
M2
M2
M2
M2
M2
M2
S1
S2
Begin
a gi tr 0 vo mt vng nh
Khi S1 tc ng th cng 1 vi End
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IV. Chng trnh: A.Dng S7- 200
1.Dng LADDER
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2. Dng STL Network 1 // Khi ng LD I0.0
S M0.0, 1
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R M0.1, 1
Network2// n START sng LD M0.0
= DEN_START
Network 3// Dng LD STOP_
S M0.1, 1
R M0.0, 1
Network4// n Stop sng LD M0.1
= DEN_STOP
Network5// To xung trong mt vng qut LD M0.0
EU
= M0.2
Network6// Gn 0 vo vng nh MW1 LD M0.2
MOVW 0, MW1
Network7// To xung trong mt vng qut LD S1
EU
LD S2
EU
OLD
= M0.3
Network8// Cng gi tr trong vng nh LD M0.3
+I 1, MW1
Network9// Gn 0 cho MW1 nu MW1>2 LDW> MW1, 2
MOVW 1, MW1
Network10// Thc hin php chia nguyn LD M0.3
MOVW MW1, MW5
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DIV 2, MD3
Network11// ng c M1 hot ng LD S1
AD= MD3, 1
O S2
= DONG_CO_M2
Network12// ng c M2 hot ng LD S1
AD= MD3, 0
O S2
= DONG_CO_M1
B. Dng S7 -300
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Bi 3: IU KHIN MY TRN
I. Yu cu :
S bnh trn sn
S gm c 2 ng ng a 2 loi hai sn mu khc nhau lm c s cho vic to ra mu sn mong mun Hai cm bin bo mc trong bnh:
- Bo mc cao Sensor 2 - Bo mc thp Sensor 1
Mt thit b trn c iu khin bi ng c trn Qu trnh lm vic c thc hin nh sau: Trc tin bm 2 loi sn khc mu nhau vo bnh, loi sn th nht c a vo bnh bng my bm 1, loi sn th hai c a vo bnh bng my bm 2. Sau khi dung dch trong bnh t mc cc i th dng hai my bm v bt u qu trnh trn, qu trnh ny c iu khin bi my trn v thi gian trn cn thit l 5s. Sau khi trn xong, sn phm c a ra rt vo cc hp ng sn qua van x v bm 3. Qu trnh c thc hin lp li 10 ln. Sau dng h thng
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 Bm M1
I0.1 Stop_ Q0.1 Bm M2
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I0.2 Sensor_1 Q0.2 My trn
I0.3 Sensor_2 Q0.3 Van x
Q0.4 Bm M3
II. Thut ton: - Bc 1: Rt loi sn th nht v th hai vo bnh - Bc 2: iu hnh ch lm vic khi t mc cao - Bc 3: iu khin ng c trn v t thi gian trn - Bc 4: a sn phm ra khi bnh trn - Bc 5: m s ln trn. Nu 10 ln th dng sn xut - Bc 6: Quay li ch lm vic bc 1
IV. Chng trnh: A.Dng S7- 200
Dng LADDER:
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Dng STL:
Network 1 // TAO ROLE TRUNG GIAN
LD START
O M0.0
AN STOP_
AN C30
= M0.0
Network 2 // BOM 1,2 HOAT DONG
LD START
O M0.0
AN STOP_
AN C30
= M0.0
Network 3 // KHOI DONG MAY TRON
LD M0.0
A S2
AN BOM_3
AN T37
= MAY_TRON
TON T37, 50
Network 4 // BOM 3 VAN XA LAM VIEC
LD T37
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O VAN_XA
A M0.0
A S1
= VAN_XA
= BOM_3
Network 5 // TAO BO DEM SO LAN TRON
LD M0.0
A T37
LD C30
O STOP_
CTU C30, 10
B. Dng S7- 300
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Bi 4: IU KHIN NG M CA KHO T NG
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I. Yu cu:
Khi xe ang tin v gn ca kho, cm bin Ultrasounic SS1 nhn dng c xe v ca s c m ra n gp gii hn hnh trnh trn LS1 th ca dng li ri xe chy vo. Khi cm bin quang SS2 t pha trong cng ca nhn dng c xe i qua khi ca th ca s c ng li, chm vo gii hn hnh trnh di LS2 th ca dng li. Ch xe ch i mt chiu.
II. Phn cng u vo, u ra:
u vo u ra
I0.0 START Q0.0 Raise_door
I0.1 STOP Q0.1 Down_door
I0.2 SS1
I0.3 SS2
I0.4 LS1
I0.5 LS2
III.Gin thi gian:
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IV. Chng trnh: Dng LADDER:
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Dng STL: Network 1
// TAO ROLE TRUNG GIAN
LD START
O M0.0
AN STOP_
= M0.0
Network 2
// CUA MO KHI PHAT HIEN XE
LD SS1
O Raise_door
A M0.0
AN LS1
= Raise_door
Network 3
// CUA DONG KHI XE DA VAO TRONG
LD SS2
ED
O Down_door
A M0.0
AN LS2
= Down_door
Bi 5: DY CHUYN CHIT NC VO CHAI
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I. Yu cu:
Chit nc vo chai
Cng on chit nc vo chai theo nguyn tc thi gian, m t hnh 3.13. Nhn nt Start PB, bng ti chy v mang theo chai, cm bin chai pht hin, bng ti dng, bt u m van chit chit nc vo chai. Thi gian chit y l 30s, sau khi chit y bng ti li chy v tip tc chit chai k tip
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 Conveyer motor
I0.1 Stop Q0.1 Fill valve
I0.2 Bottle_sensor
III.Chng trnh: A.Dng S7_200
Dng LADDER:
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Dng STL: Network 1 // TAO ROLE TRUNG GIAN
LD Start
S M0.0, 1
Network 2 // BANG TAI HOAT DONG
LD M0.0
AN Bottle_sensor
= Conveyer_motor
Network 3 // CHIET NUOC VAO CHAI
LD Bottle_sensor
S Fill_valve, 1
R Conveyer_motor, 1
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TON T37, 300
A T37
S Conveyer_motor, 1
R Fill_valve, 1
R T37, 1
Network 4 // DUNG HE THONG
LD Stop
R Conveyer_motor, 2
B. Dng S7- 300
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Bi 6: H THNG M V NG GI SN PHM
I.Yu cu:
Khi nhn cng tc khi ng PB Start bng ti BT1 mang hp ng sn
phm di chuyn. Cm bin SS1 nhn dng thng ng tc ng, bng ti BT1 dng li. Bng ti BT2 cha sn phm dch chuyn, sn phm c rt vo hp ng, sn phm c m bi mt cm bin quang hng ngoi SS2 khi m c 10 sn phm ( mi hp cha 10 sn phm) th bng ti BT2 dng, tip tc BT1 dch chuyn ng gi hp mi. ngng qu trnh ta nhn nt PB Stop.
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II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 Box conveyor
I0.1 Stop Q0.1 Piece conveyor
I0.2 Box sensor
I0.3 Piece sensor
III. Gin thi gian:
IV. Chng trnh: A.Dng S7_200
Dng LADD
Dng lnh S, R
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B. Dng S7_ 300
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Bi 7: IU KHIN BNG CHUYN CN TN
I. Yu cu:
H thng bng chuyn nh hnh v v c hot ng nh sau : Ban u, khi nhn Start th khi ng bng chuyn M1, khi vt chm vo S1 th khi ng bng chuyn M2. Khi S2 tc ng th khi ng bng chuyn M3, sau khi S2 thi tc ng 10s th dng bng chuyn M2. Qu trnh hot ng tng t cho n bng chuyn M4.
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 M1
I0.1 Stop Q0.1 M2
I0.2 S1 Q0.2 M3
I0.3 S2 Q0.3 M4
I0.4 S3
I0.5 S4
III. Chng trnh: A. Dng S7_200
Dng LADDER:
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Dng STL: Network 1 // KHOI DONG
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LD I0.0
S M0.0, 1
Network 2 // DUNG
LD I0.1
R M0.0, 1
R Q0.0, 4
Network 3 // KHOI DONG BANG CHUYEN M1
LD M0.0
= Q0.0
Network 4 // KHOI DONG BANG CHUYEN M2
LD M0.0
A I0.2
S Q0.1, 1
Network 5 // KHOI DONG BANG CHUYEN M3
LD M0.0
A I0.3
S Q0.2, 1
Network 6 // KHOI DONG BANG CHUYEN M4
LD M0.0
A I0.4
S Q0.3, 1
Network 7 // PHAT HIEN SUON XUONG S2
LD I0.3
ED
S M0.1, 1
Network 8 // TAO THOI GIAN T37
LD M0.1
TON T37, 100
Network 9 // DUNG BANG CHUYEN M2
LD T37
R M0.1, 1
R Q0.1, 1
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Network 10 // PHAT HIEN SUON XUONG S3
LD I0.4
ED
S M0.2, 1
Network 11 // TAO THOI GIAN T38
LD M0.2
TON T38, 100
Network 12 // DUNG BANG CHUYEN M3
LD T38
R M0.2, 1
R Q0.2, 1
Network 13 // PHAT HIEN SUON XUONG S4
LD I0.5
ED
S M0.3, 1
Network 14 // TAO THOI GIAN T39
LD M0.3
TON T39, 100
Network 15 // DUNG BANG CHUYEN M4
LD T39
R M0.3, 1
R Q0.3, 1
B.Dng S7_300
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Bi 8: IU KHIN DAO CT SN PHM
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I. Yu cu:
C 3 loi sn phm A, B, C a vo dao ct thnh sn phm nh. Dao ct 1000 sn phm A th thay dao, ct 100 sn phm B th thay dao, ct 10 sn phm C th thay dao. Lc S2 tc ng th dao a xung, khi dao gp S1 th dng v quay ln. Khi dao ct khng c na th ci ku ln, bm reset th ci ht ku.
II. Phn cng u vo, u ra:
u vo u ra
I0.0 Start Q0.0 ng c h
I0.1 Stop Q0.1 ng c nng
I0.2 Reset Q0.2 Ci
I0.3 S1
I0.4 S2
I0.5 SA
I0.6 SB
I0.7 SC
Thut ton: Ta quy cc sn phm B v C theo A. Do dao ct 1 sn phm B tng ng vi dao ct 10 sn phm A Dao ct 1 sn phm C tng ng vi dao ct 100 sn phm A
III. Chng trnh: A. Dng S7_200
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Dng LADDER:
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Dng STL:
Network 1 // Khoi dong chuong trinh
// Network Comment
LD I0.0
S M0.0, 1
Network 2 // Dung chong trinh
LD I0.1
R M0.0, 1
R Q0.1, 1
R Q0.0, 1
R Q0.2, 1
Network 3 // Tao bien de cong
LD I0.0
MOVW 0, VW1
R Q0.2, 1
EU
S Q0.0, 1
R Q0.1, 1
Network 4 // Cat san pham A
LD M0.0
