Bao Cao Plc Nhom 4

7/31/2019 Bao Cao Plc Nhom 4

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i Hc Cng Ngip H Ni

I. Gii thiu v PLC S7-200.1. Cu hnh phn cng.

S7-200 l thit b iu khin kh trnh ca hng Simen,c cu trc theo kiumodul v modul m rng.Cc modul ny c nhiu ng dng lp tnh khc

nhau.Thnh phn c bn ca S7-200 l khi vi x l CPU 212 hoc CPU214. CPU 212 c 8 cng vo v 6 cng ra v c kh nng m rng thm

bng 2 modul m rng. CPU 214 c 14 cng vo v 10 cng ra v c kh nng m rng

thm bng 7 modul m rng.S7200 c nhiu loi modul m rng khc nhau.CPU212 bao gm :- 512 t n, tc 1 KB, lu chng trnh thuc min b nh c ghic v khng b mt d liu nh c giao din vi EEPROM.

- 512 t n nh d liu, trong c 100 t nh c ghi thuc mintrn.- 8 cng vo v 6 cng ra logic.- C th ghp ni thm 2 modul m rng s cng vo ra, bao gm c modultng t.- Tng s logic vo ra cc i l 64 cng vo v 64 cng ra.64 b to thi gian tr, trong c 2 timer c phn gii 1ms, 8 timer c phn gii 10 ms v 54 timer c phn gii 100 ms.- 64 b m( counter), chia lm 2 loi ;loi b m ch m tin, loi va

m tin va m li.368 bit nh c bit, s dng lm cc bit trng thi hoc cc bit t lm ch hot ng.- C ch ngt v x l tn hiu ngt khc nhau bao gm: Ngt truynthng, ngt theo sn ln hay sn xung, ngt theo thi gian v ngt bohiu ca b m tc cao ( 2 KHZ).- B nh khng b mt ngun d liu trong khong 50 gi khi PLC b ngtngun nui.CPU214 bao gm:- 2048 t n thuc min nh c ghi lu chng trnh.

- 2048 t n lu d liu.- 14 cng vo v 10 cng ra logic.- C 7 modul m rng cng vo ra bao gm c u vo alalog.- Tng s cng vo ra cc i l 64 cng vo v 64 cng ra.- 128 timer chia lm 3 loi theo phn gii khc nhau: 4 timer c phngii 1 ms, 16 timer c phn gii 10 ms, 108 timer c phn gii 100 ms.

Lp LTC H in3 K3

Nhm 4

1


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i Hc Cng Ngip H Ni

- 128 b m chia lm 2 loi: loi ch m tin, loi va m tin va mli.- 688 bit nh c bit dung thng bo trng thi v ch lm vic.- Cc ch ngt v x l ngt gm: ngt truyn thng, ngt theo sn lnhoc sn xung, ngt theo thi gian, ngt ca b m tc cao v ngttruyn xung.- 3 b m tc cao vi nhp 2KHz v 7KHz.2 b pht xung nhanh cho dy kiu PTO hoc PMW.- 2 b iu chnh tng t.- Ton b b nh khng b mt d liu trong khong thi gian 190 gi khiPLC b mt ngun nui.Cng truyn thng.S7 200 s dng cng RS 485 phc v vic truyn thng, ghp ni vi ccthit b lp trnh hay cc trm PLC khc.

Ghp ni PLC vi PC thng qua cng RS 232 cn c cp ni PC/PPI vi bchuyn i RS232/RS485.2. Cu trc b nh.B nh ca S7-200 c chia thnh 4 vng v 1 t c nhim v duy tr dliu trong khong thi gian nht nh khi b mt ngun.B nh ca S7-200mang tnh nng ng cao, c v ghi c trong ton vng, ngaoij tr cc

bit nh t bit k hiu l SM ch c truy nhp c.- Vng chng trnh :l min b nh s dng lu tr cc lnh chng

trnh.Vng ny thuc kiu non-volatile c/ghi c.

- Vng tham s: l vng lu gi cc tham s nh: t kha, a ch trmcng nh vng chng trnh, vng tham s thuc kiu non-volatile c/ghic.

- Vng d liu : c s dng ct cc d liu ca chng trnh bao gmcc kt qu ca cc php tnh, hng s c nh ngha trong chng trnh,

b m truyn thng. Mt phn ca vng nh ny thuc kiu non-volatilec/ghi c.

- Vng i tng : Timer, b m, b m tc cao v cc cng vo ratng t c t trong vng nh cui cng.Vng ny khng thuc kiu

non-volatile nhng c ghi c.II. B m tc cao S7-200.B m tc cao dng iu khin v theo di cc qu trnh c tc cao m PLC khng th khng ch c do bi hn ch v thi gian ca vngqut.

Lp LTC H in3 K3

Nhm 4

2


3/19

i Hc Cng Ngip H Ni

Trong CPU 212 c 1 b m tc cao v trong CPU 214 c 3 b m tc cao.B m tc cao trong CPU 212 k hiu l HSC0, trong CPU214 khiu c nh s ln lt l HSC0, HSC1, HSC2.

Nguyn tc hot ng ca cc b m tc cao cng tng t nh nguyntc hot ng ca cc b m thng thng trong PLC, tc l cng m theosn ln ca tn hiu u vo.S m s c h thng ghi nh vo mt nh c bit kiu t kp v c gi l gi tr m tc thi ca b m, khiu l CV(current value) .Khi gi tr m tc thi bng gi tr t trc th

b m s pht ra 1 tn hiu bo ngt.Gi tr t trc l 1 s nguyn 32 bitcng c lu trong 1 nh kiu t kp v c k hiu l PV( presetvalue).

Nu ch ngt vo/ra vi b m tc cao c khai bo s dng, cctn hiu ngt sau y s c pht:- Ngt khi PV=CV (i vi HSC0, HSC1, HSC2).

- Ngt khi c tn hiu bo thay i hng m t cng vo (vi HSC1,HSC2).

- Ngt khi c tn hiu bo xa t cng vo( vi HSC1 v HSC2).Mi b m c nhiu ch lm vic khc nhau. Chn ch lm vic cho1 b m bng lnh HDEF.Tng ch lm vic li c nhiu ch hotng khc nhau.Kiu hot ng ca mi b m c xc nh bng nidung ca 1 byte iu khin trong vng nh c bit (special memory) sau c khai bo vi b m nh lnh HSC.Ch c th kch b m khi khai

bo ch lm vic v nh ngha kiu hot ng cho tng ch trongbyte iu khin.1. Nguyn l hot ng ca cc b m tc cao.HSC0( c trong CPU 212 v CPU 214).B m HSC0 c 1 cng vo l I0.0, n ch c mt ch lm vic duy nhtl m tin hoc m li s cc sn ln ca tn hiu u vo ti cng I0.0.Chiu m ( tin hay li ) c qui nh bi trng thi logic ca binSM37.3 nh sau:- SM37.3 = 0 m li theo sn ln ca xung u vo.- SM37.3 =1 m tin theo sn ln ca xung u vo.

Tn s m cc i ca HSC0 l 2KHZ.

HSC0 s dng t kp SMD38(gm 4 byte SMD38,SMD39, SMD40 vSM41) lu gi tr m tc thi CV.Gi tr t trc PV c ghi vo tkp SMD42( bao gm 4 byte SMD42, SMD43, SMD44 v SMD45).C gitr SMD 38 v SMD42 u l nhng s nguyn 32 bit c du( nguyn mhoc nguyn dng).S nguyn l ca HSC0 nh sau.

Lp LTC H in3 K3

Nhm 4

3


4/19

i Hc Cng Ngip H Ni

Hnh 1.1 S nguyn l ca b m HSC0.HSC0 s dng byte SM37 xc nh kiu hot ng nh m tin haym li, cho php hay khng cho php thay i gi tr m tc thi CV,cng nh gi gi tr t trc PV v cho php hay khng cho php kch bm.Kiu hot ng ca HSC0 phi c nh ngha trong SMB37 trc khi

thc hin lnh HDEF.Cu trc byte SMB37 cn gi l byte iu khin ca HSC0 nh sau:

SM37.0 Khng s dngSM37.1 Khng s dng

SM37.2 Khng s dngSM37.3 Chiu m 0-m li, 1-m tin.SM37.4 Cho php i chiu tin: 0- khng cho php, 1- cho phpSM37.5 Cho php sa i gi tr t trc:0-khng cho php, 1- cho

php.

SM37.6 Cho php sa i gi tr m tc thi: 0- khng cho php,1- cho php.SM37.7 1- cho php kch HSC0, 0-khng cho php kch HSC0.

Cc bc khai bo khi s dng HSC0:-Np gi tr iu khin ph hp cho SMB37.-Xc nh ch lm vic ph hp cho b m bng lnh HDEF.Do HSC0ch c 1 ch lm vic nn lnh xc nh kiu s l HDEF K0 K0.-Np gi tr m tc thi ban u v gi tr t trc vo SMD 38 v SMD42.-Khai bo ch ngt vo ra v kch tn hiu bo ngt HSC0 bng lnhATCH.-Kch b m bng lnh HSC K0.Sau khi c kch thi b m HSC bt u lm vic v s dng byte SMB36 thng bo trng thi hot ng ca n nh sau:

Lp LTC H in3 K3

Nhm 4

4

HSC0

SMD 38

SMD 42

I0.0HSC0

Gi tr m tcthi CV

Gi tr ttrc PV


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i Hc Cng Ngip H Ni

SM36.0 Khng s dngSM36.1 Khng s dngSM36.2 Khng s dngSM36.3 Khng s dng

SM36.4 Khng s dngSM36.5 chiu ang m, 1-m tin, 0- m li.SM36.6 Kt qu so snh tc thi, 0-nu CV#PV, 1- nu CV=PVSM36.7 Kt qu so snh tc thi 0 nu CVPV

Khi s dng b m HSC0 cng vi ch ngt vo ra, tn hiu bo ngtHSC0 s xut hin khi CV=PV nu tn hiu ngt kiu i2 c khai bo.HSC1( ch c trong CPU 214).HSC1 l b m linh hot, s dng 4 u vo I0.0, I0.1, I0.6, I0.7 vi 12ch lm vic khc nhau.HSC1 s dng t kp SMD 48 ( gm 4 byte SMD48, SMD49, SMD50 vSMD51) lu gi tr m tc thi CV.Gi tr t trc PV c ghi vo tkp SMD52, gm 4 byte(SMD52, SMD53, SMD54 v SMD55).C gi trt trc PV v gi tr tc thi CV u l nhng s nguyn 32 bit c du(snguyn dng v nguyn m).Cc ch lm vic ca b m HSC1 c lit k nh sau:

Ch lmvic

M t

0 m tin hoc m li sn ln ca I0.6. Chiu m c quy

nh bi SM47.3.Hot ng ca HSC1 hon ton c iu khint h thng (khng c tn hiu kch v xa t tn hiu bn ngoi).1 m tin hoc m li sn ln ca I0.6. Chiu m c quy

nh bi SM47.3. HSC1 s dng I1.0 lm tn hiu xa t bnngoi.

2 m tin hoc m li sn ln ca I0.6.Chiu m c quynh bi SM47.3.HSC1 s dng I1.0 lm tn hiu xa t bnngoi v I1.1 lm tn hiu khi pht.

3 m tin hoc m li sn ln ca I0.6, chiu m c quynh bi cng I0.7.Hot ng ca HSC1 hon ton c iukhin bi h thng(khng c tn hiu kch v xa t bn ngoi).

4 m tin hoc m li sn ln ca I0.6.Chiu m c quynh bi cng I0.7.HSC1 s dng I0.1 lm tn hiu xa t bnngoi.

5 m tin hoc m li sn ln ca I0.6.Chiu m c quynh bi I0.7. HSC1 s dng I1.0 lm tn hiu xa t bn ngoi v

Lp LTC H in3 K3

Nhm 4

5


6/19

i Hc Cng Ngip H Ni

I1.1 lm tn hiu khi pht.6 m tin theo sn ln ca I0.6 v li theo sn ln ca

I0.7.HSC1 s dng I1.0 lm tn hiu xa t bn ngoi.7 m tin theo sn ln ca I0.6 v li theo sn ln ca I0.7.

HSC1 s dng I1.0 l tn hiu xa t bn ngoi.8 m tin theo sn ln ca I0.6 v li theo sn ln ca I0.7.HSC1 s dng I1.0 l tn hiu xa t bn ngoi v I1.1 l tn hiukhi pht.

9 m s ln lch trng thi Logic ca 2 cng vo I0.6 v I0.7, tcl kt qu php tnh logic XOR gia I0.6 v I0.7 l 1.Hot ngca HSC1 hon ton c iu khin bi h thng( khng c tnhiu kch v tn hiu xa t bn ngoi).

10 m s ln lch trng thi Logic ca 2 cng vo I0.6 v I0.7, tcl kt qu php tnh logic XOR gia I0.6 v I0.7 l 1.HSC1 s

dng I1.0 lm tn hiu xa t bn ngoi.11 m s ln lch trng thi Logic ca 2 cng vo I0.6 v I0.7, tc

l kt qu php tnh logic XOR gia I0.6 v I0.7 l 1.HSC1 sdng I1.0 lm tn hiu xa t bn ngoi v I1.1 l tn hiu khi

pht.

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Khc vi HSC0, HSC1 c 3 kh nng m :-Tin hoc li theo sn ln ca I0.6( ch 0, 1, 2, 3, 4 v 5).

-Tin theo sn ln ca I0.6 v li theo sn ln ca I0.7( ch 6, 7 v 8).-Tin hoc li theo s ln lch logic gia I0.6 v I0.7, tc l s ln php tnhXOR gia I0.6 v I0.7 l 1( ch 9, 10, 11).Chiu m tin hoc m li trong ch 0, 1,2 c quy nh bi trngthi logic ca bin SM47.3 nh sau:

SM47.3=0 m li theo sn ln ca I0.6.SM47.3=1 m tin theo sn ln ca I0.6.

Lp LTC H in3 K3

Nhm 4

6

HSC1

SMD48

SMD52

I0.6

I0.7

I1.0

I1.1

HSC1

Gi tr m tc thi

Gi tr t trc

Hnh 1.2: s nguyn l ca bm HSC1


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i Hc Cng Ngip H Ni

V trong ch 3, 4, 5 bi u vo I0.7 nh sau:I0.7 = 0 m li theo sn ln ca I0.6.I0.7 =1 m tin theo sn ln ca I0.6.

Tn s bin i trng thi cc i cho php tn hiu u vo ti cng I0.6v I0.7 l 7KHz.HSC1 c 2 tn s m. Trong nhng ch t 0 n 8, tn s m caHSC1 bng tn s thay i trng thi tn hiu u vo, do tn s m cc ica HSC1 trong nhng ch ny l 7 KHz.Ring trong ch 9, 10, 11ty theo khai bo s dng th tn s m c th bng hay ln hn 4 ln tns bin thin trng thi kt qu ca php tnh XOR gia I0.6 v I0.7( mi lntrng thi ca I0.6 v I0.7 lch nhau th gi tr m tc thi CV cng tngln 4 n v ).Do trong ch 9, 10, 11 tn s cho php ca HSC1 l 28KHz.H thng s dng bit SM47.2 nh ngha tn s m ca HSC1trong ch 9, 10, 11 nh sau:

- SM47.2 =0 tn s m gp 4 ln tn s thay i trang thi gia I0.6XOR I0.7.- SM47.2 =1 tn s m bng tn s thay i trng thi I0.6 XOR I0.7.Trong cc ch 1, 2, 4, 5, 7, 8, 10 v 11 HSC1 s dng tn hiu reset bnngoi l I1.0.Bit SM47.0 c h thng s dng nhm nh ngha kiu resetcho HSC1 cho cc ch nh sau:

SM47.0 = 0 reset khi I1.0 c gi tr logic 1.1 reset khi I1.0 c gi tr logic 0.

Cng tng t nh vy cho cc ch 2, 5, 8 v 11 khi HSC1 s dng tnhiu kch t bn ngoi qua cng I1.1, h thng s dng bit SM37.1 nhngha kiu kch nh sau:

SM47.1 = 0 HSC1 lm vic khi I1.1 c gi tr logic 1.1 HSC1 lm vic khi I1.1 c gi tr logic 0.

HSC1 s dng byte SMB47 xc nh kiu hot ng nh : m tin haym li, cho php hay khng cho php thay i gi tr tc thi CV, cng nhgi tr t trc PV, cho php hay khng cho php kch b mKiu hotng ca HSC1 phi c nh ngha trong SMB47 trc khi thc hinlnh HDEF.Cu trc byte SMB47 hay l byte iu khin ca HSC1 nh sau:

SM47.0 Kiu reset cho tn hiu xa ti cng I1.0( ch 1, 2, 4,5, 8, 10 v 11).

SM47.1 Kiu kch cho tn hiu khi ng ti cng I1.1( ch 2, 5, 8 v 11),

SM47.2 Tn s m ca HSC1 ( ch 9, 10 v 11).SM47.3 Chiu m 0-m li, 1-m tin.SM47.4 Cho php i chiu m 0- khng cho php, 1- cho

Lp LTC H in3 K3

Nhm 4

7


8/19

i Hc Cng Ngip H Ni

php.SM47.5 Cho php sa i gi tr t trc, 0- khng cho php,

1- cho php.SM47.6 Cho php sa i gi tr m tc thi, 0- khng cho

php, 1- cho php.SM47.7 1- cho php kch HSC1, 0- Hy HSC1.

Cc bit SM47.0, SM47.1 v SM 47.2 c gn cho b m bng lnhHDEF, cc bit cn li c gn cho b m sau lnh HSC.Cc bc khai bo s dng HSC1 ( nn thc hin ti vng qut u tin).- Np gi tr iu khin ph hp cho SM47.- Xc nh ch lm vic cho b m bng lnh HDEF.- Np gi tr m tc thi ban u SMD48 v gi tr t trc vo SMD52.- Khai bo s dng ch ngt vo/ra v kch tn hiu bo ngt HSC0 bng

lnh ATCH.- Kch b m bng lnh HSC.Khi s dng lnh HSC1 cng vi ch ngt vo ra, cc tn hiu bo ngtsau y s c pht:- Bo ngt khi CV=PV nu tn hiu bo ngt kiu 13 c khai bo.- Bo ngt khi c tn hiu bo thay i chiu m t I0.7, nu tn hiu khai

bo ngt kiu 14 c khai bo.- Bo ngt khi HSC1 b reset bi I1.0, nu tn hiu kiu 15 c khai bo.

Sauk hi c kch hot bng lnh HSC b m HSC1 s dng byte

SMB46 thng bo trng thi hot ng ca n nh sau:SM46.0 Khng s dng.SM46.1 Khng s dng.SM46.2 Khng s dng.SM46.3 Khng s dng.SM46.4 Khng s dng.SM46.5 Chiu ang m.1- m tin, 0- m li.SM46.6 Kt qu so snh tc thi.0- nu CV#PV, 1- nu CV=PV.SM46.7 Kt qu so snh tc thi. 0- nu CVPV

Mc d c kch bng lnh HSC nhng trong cc ch 2, 5, 8 b mHSC1 ch thc s lm vic khi c tn hiu start t cng vo I1.1.HSC2( ch c trong CPU 214).B m tc cao HSC2 c nguyn l lm vic ging nh HSC1.C 2 bm HSC1 v HSC2 lm vic c lp, khng nh hng n nhau.Cc uvo I0.6 v I0.7, I0.1 v I0.0 c thay th bi cc u vo I0.2, I0.3, I0.4,I0.5 trong HSC2.

Lp LTC H in3 K3

Nhm 4

8


9/19

i Hc Cng Ngip H Ni

Cc ch lm vic ca HSC2:Ch

lm vicM t

0 m tin hoc m li sn ln ca I1.2. Chiu m c quy

nh bi SM57.3.Hot ng ca HSC2 hon ton c iu khint h thng (khng c tn hiu kch v xa t tn hiu bn ngoi).1 m tin hoc m li sn ln ca I1.2. Chiu m c quy

nh bi SM57.3. HSC1 s dng I1.4 lm tn hiu xa t bnngoi.

2 m tin hoc m li sn ln ca I1.2.Chiu m c quynh bi SM57.3.HSC1 s dng I1.4 lm tn hiu xa t bnngoi v I1.5 lm tn hiu khi pht.

3 m tin hoc m li sn ln ca I1.2, chiu m c quynh bi cng I1.3.Hot ng ca HSC2 hon ton c iu

khin bi h thng(khng c tn hiu kch v xa t bn ngoi).4 m tin hoc m li sn ln ca I1.2.Chiu m c quy

nh bi cng I1.3.HSC2 s dng I0.1 lm tn hiu xa t bnngoi.

5 m tin hoc m li sn ln ca I1.2.Chiu m c quynh bi I1.3. HSC2 s dng I1.4 lm tn hiu xa t bn ngoi vI1.5 lm tn hiu khi pht.

6 m tin theo sn ln ca I1.2 v li theo sn ln ca I1.3.Hotng HSC2 hon ton c iu khin bi h thng ( khng c

tn hiu xa v ngt t bn ngoi).7 m tin theo sn ln ca I1.2 v li theo sn ln ca I1.3.HSC1 s dng I1.4 l tn hiu xa t bn ngoi.

8 m tin theo sn ln ca I1.2 v li theo sn ln ca I1.3.HSC2 s dng I1.4 l tn hiu xa t bn ngoi v I1.5 l tn hiukhi pht.

9 m s ln lch trng thi Logic ca 2 cng vo I1.2 v I1.3, tcl kt qu php tnh logic XOR gia I1.2 v I1.3 l 1.Hot ngca HSC1 hon ton c iu khin bi h thng( khng c tnhiu kch v tn hiu xa t bn ngoi).

10 m s ln lch trng thi Logic ca 2 cng vo I1.1 v I1.1, tcl kt qu php tnh logic XOR gia I1.2 v I1.3 l 1.HSC2 sdng I1.4 lm tn hiu xa t bn ngoi.

11 m s ln lch trng thi Logic ca 2 cng vo I1.2 v I1.3, tcl kt qu php tnh logic XOR gia I1.2 v I1.3 l 1.HSC sdng I1.0 lm tn hiu xa t bn ngoi v I1.4 l tn hiu khi

Lp LTC H in3 K3

Nhm 4

9


10/19

i Hc Cng Ngip H Ni

pht.HSC2 s dng t kp SMD58 lm thanh ghi cho gi tr m tc thi v tkp SMD62 cho gi tr t trc.

`

Kiu hot ng cho tng ch hot ng ca HSC2 c quy nh trongbyte iu khin SMB57. Trng thi lm vic ca HSC2 thng bo quaSMB56.Cu trc byte SMB57, cn c gi l byte iu khin ca HSC2, nh sau:SM57.0 Kiu reset cho tn hiu xa ti cng I1.4( ch 1, 2,

4, 5, 8, 10 v 11).SM57.1 Kiu kch cho tn hiu khi ng ti cng I1.5( ch

2, 5, 8 v 11),SM57.2 Tn s m ca HSC2 ( ch 9, 10 v 11).SM57.3 Chiu m 0-m li, 1-m tin.SM57.4 Cho php i chiu m 0- khng cho php, 1- cho

php.SM57.5 Cho php sa i gi tr t trc, 0- khng cho

php, 1- cho php.SM57.6 Cho php sa i gi tr m tc thi, 0- khng cho

php, 1- cho php.SM57.7 1- cho php kch HSC2, 0- Hy HSC2.

Cc bit SM57.0, SM57.1, SM57.2 c gn cho b m bng lnhHDEF.Cc bt cn li c gn sau lnh HSC.Sau khi c kch bng lnhHSC b m HSC2 s dng byte SMB56 thng bo trng thi hot ngca n nh sau:

Lp LTC H in3 K3

Nhm 4

10

HSC2

SMD58

SMD62

I1.2

I0.3

I1.4

I1.5

HSC2

Gi tr m tc thi

Gi tr t trc

Hnh 1.3: s nguyn l ca b

m HSC2


11/19

i Hc Cng Ngip H Ni

SM56.0 Khng s dng.SM56.1 Khng s dng.SM56.2 Khng s dng.SM56.3 Khng s dng.

SM56.4 Khng s dng.SM56.5 Chiu ang m.1- m tin, 0- m li.SM56.6 Kt qu so snh tc thi.0- nu CV#PV, 1- nu CV=PV.SM56.7 Kt qu so snh tc thi. 0- nu CVPV

Ging nh HSC1, HSC2 c 3 kh nng m.- Tin hoc li theo sn ln ca I1.2 ( ch 0, 1 ,2 ,3 ,4 v 5).- Tin theo sn ln ca I1.2 li theo sn ln ca I1.3 ( ch 6, 7 v 8).- Tin hoc li s ln lc gi tr logic gia 2 cng vo I1.2 v I1.3, tc l sln php tnh XOR gia I1.2 v I1.3 c kt qu 1 ( ch 9, 10 v 11).Chiu m tin hay li trong ch 0,1 v 2 c quy nh bi trng thi logicca bit SM57.3 v trong ch 3, 4 v 5 bi u vo I1.5 nh sau:

SM57.3, I1.3 =0 m li theo sn ln ca I1.2=1 m tin theo sn ln ca I1.2

Tn s bin i cc i cho php ca tn hiu u vo ti cng I1.2 ( v I1.3)l 7KHz.Trong nhng ch t 0 n 8 tn s ca HSC2 cng l7KHz.Trong ch 9,10 v 11 , tn s m c th bng hoc gp 4 ln tns bin thin trng thi kt qu php tnh XOR gia I1.2 v I1.3.

SM57.2 = 0 Tn s m gp 4 ln tn s bin thin I1.2XORI1.3.=1 Tn s m bng tn s bin thin I1.2XORI1.3.

Trong cc ch 1, 2, 4, 5, 7, 8, 10 v 11 bit SM 57.0 c h thng sdng nhm nh ngha kiu reset ca HCS2 cho cc ch ny nh sau:SM57.0 = 0 HSC2 lm vic khi I1.5 c gi tr logic 1.

=1 HSC2 lm vic khi I1.5 c gi tr logic 0.Khi s dng HSC2 cng vi ch ngt vo ra , cc tn hieuj bo ngt sauy s c pht:- Bo ngt khi CV=PV nu tn hiu bo ngt kiu 16 c khai bo.- Bo ngt khi tn hiu bo thay i chiu m t I1.3 nu tn hiu bo ngt

kiu 17 c khai bo .- Bo ngt khi HSC2 b reset bi I1.4 nu tn hiu bo ngt kiu 18 c

khai bo.

2.Khai bo s dng b m tc cao.Khai bo s dng cc b m HSC0, HSC1, HSC2 nn c thc hin tivng qut u tin khi bit SM0.1 c gi tr logic l 1.Vic khai bo thng

Lp LTC H in3 K3

Nhm 4

11


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i Hc Cng Ngip H Ni

c t chc thnh 1 chng trnh con v chng trnh con ny c goibng lnh CAAL trong vng qut u.Cng vic khai bo s dng chng trnh con bao gm:- Np gi tr v kiu hot ng ph hp cho byte iu khin.- Xc nh ch lm vic cho b m bng lnh HDEF.- Np gi tr ban u v gi tr t trc .- Khai bo ch ngt vo ra v kch tn hiu bo ngt.- Kch b m vi kiu lm vic c ghi trong byte iu khin bng

lnh HSC.HDEF: lnh xc nh ch lm vic ca b m tc cao.Tn ca bm c ch nh bng ton hng HSC.Ch lm vic c chn l nidung ca ton hng MODE.C php s dng lnh HDEF trong LAD v STS nh sau:

LAD STL Ton hng

HDEF HSC MODE

HSC CPU212:0(byte) CPU 214 02MODE CPU212:0(byte) CPU214:0(HSC0)

011(HSC12)

HSC : Lnh t kiu lm vic cho b m tc cao.Tn ca b m cch nh bng ton hng N. Kiu lm vic c t l ni dung ca byte iukhin b m.C php s dng lnh HSC trong LAD v STL nh sau:

LAD STL Ton hng

HSC N

N CPU212:0(t n) CPU 214 02

3. S dng hm pht xung tc cao.CPU214 s dng hai cng ra l Q0.0 v Q0.1 pht dy xung tn s caohoc tn hiu iu xung theo rng.

Lp LTC H in3 K3

Nhm 4

12

EN

HSC

MODE

HDEF

EN

N

HSC


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i Hc Cng Ngip H Ni

Dy xung c k hiu bng PTO( pulse train output) cn tind hiu iu ng xung k hiu l PMW(pulse width modulation).PTO l 1 dy xung vung tun hon, c chu k l 1 s nguyn nm trongkhong t 250s 65535s.PMW cng l dy xung vung tun hon, c chu k l 1 s nguyn nmtrong khong t 250s 65535s.Khc vi PTO, rng xung mi chu kc th quy nh c v l 1 s nguyn nm trong khong 0s 65535s.

Nu rng xung c quy nh ln hn chu k ca PMW th dy xung sl 1 tn hiu u c gi tr logic l 1.Ngc li khi quy nh rng xung

bng 0 th dy xung s u c tn h iu logic 0.PTO v PMW c h thng pht ra theo 2 cng Q0.0 v Q0.1. Cc ngun

pht m PTO v PMW s dng l:- 1 byte iu khin 8 bit.- 1 t n ghi chu k chung.

- 1 t kp ghi s xung ca dy.a ch nhng nh ny nh sau:

Cng ra(bit)

Byte iukhin( byte)

Chu k(t n)

rngxung(t n)

S xung(t kp)

Q0.0 SMB67 SMW68 SMW70 SMD72

Q0.1 SMB77 SMW78 SMW80 SMD82

Cc nh ny phi c np cc gi tr thch hp trc khi thc hin lnh

to xung PLS .Lnh PLS s to ra 1 dy xung c kiu c quy nh trongbyte iu khin vi chu k , rng xung v s xung t trc.Mi khi thay i ni dung cc nh trn, bt buc phi thc hin li lnhPLS.PLS : Lnh pht dy xung ti cng Q0.0 hoc Q0.1 theo cu trc c nhngha trong byte iu khin v cc nh v chu k, rng.Cng xung phtra c ch nh trong ton hng x( 0 cho Q0.0 v 1 cho Q0.1) ca lnh.C php s dng lnh PLS nh sau:

LAD STL Ton hng

Lp LTC H in3 K3

Nhm 4

13


14/19

i Hc Cng Ngip H Ni

PLS x X CPU214 0,1

(t n)

Lnh PLS s dng cc byte iu khin SMB67 v SMB77 theo cu trcsau:

Q0.0 Q0.1 Mc chSM67.0 SM77.0 i chu k: 1- cho php , 0- khng cho php.SM67.1 SM77.1 i rng xung: 1- cho php, 0- khng cho php.

SM67.2 SM77.2 i s m xung cho PTO: 1- cho php, 0- khng chophp.SM67.3 SM77.3 n v thi gian: 1-ms, 0- s.SM67.4 SM77.4 Khng s dngSM67.5 SM77.5 Khng s dngSM67.6 SM77.6 Ch kiu xung:0-PTO, 1-PMWSM67.7 SM77.7 Khai bo: 1- kch, 0 -hy.Mun to nhng xung c nhiu dng khc nhau th ta s dng chng trnhx l ngt tng ng khai bo dng xung mi.

Th tc khai bo s dng hm pht xung tc cao tt nht nn c thchin trong vng qut u tin v gm nhng bc sau:- Ghi gi tr cho byte iu khin.- Np cc gi tr v chu k ( rng xung ) v s xung ca dy vo nhng

tng ng.- Khai bo s dng ch ngt.- Thc hin lnh PLS.

Xung s c pht ngay bng sn ln ca xung u tin sau khi thchin lnh PLS.

III. Thc hin bi ton o tc o tc ng c .

Lp LTC H in3 K3

Nhm 4

14

EN

Q0.x

PLS


15/19

i Hc Cng Ngip H Ni

1. Cu to v nguyn l c bn ca Encoder.Cu to.

Hnh 1.4: Cu to b Encoder.

Nhn trn hnh ta thy encoder gm: 1 tm trn c khc l, 1 H thng LEDpht v thu.b. Nguyn l hot ng cn bn.

Nuyn l c bn ca encoder, l mt a trn xoay, quay quanh trc. Trna c cc l (rnh). Ngi ta dng mt n led chiu ln mt a. Khi aquay, ch khng c l (rnh), n led khng chiu xuyn qua c, ch cl (rnh), n led s chiu xuyn qua. Khi , pha mt bn kia ca a,

ngi ta t mt con mt thu. Vi cc tn hiu c, hoc khng c nh sngchiu qua, ngi ta ghi nhn c n led c chiu qua l hay khng.Sxung m c v tng ln n tnh bng s ln nh sng b ct.

Nh vy l encoder s to ra cc tn hiu xung vung v cc tn hiu xungvung ny c ct t nh sng xuyn qua l. Nn tn s ca xung u ra s

ph thuc vo tc quay ca tm trn . i vi encoder mnh ang dngth n c 2 tn hiu ra lch pha nhau 90. Hai tn hiu ny c th xc nhc chiu quay ca ng c.

Lp LTC H in3 K3

Nhm 4

15


16/19

i Hc Cng Ngip H Ni

Hnh 1.5 : Cu to bn trong c bn ca Encoder.Trn mch nguyn l encoder c tch hp sn v c gn trc tip vo

ng c nn ta ch cn cp in vo cho b encoder v ly 1 dy tn hiu rakhi ng c quay l c xung ra ti chn ca encoder v in p cp choencoder l 5V.Vi loi Encoder gn trc tip trn ng c gm 4 dy (Khng k hai dyngun cho in p 12V vo ng c quay). Trong dy l dy 5V, dyxanh l dy GND, Dy vng v dy trng l hai dy tn hiu lch pha nhau90.Gi s ta s dng loi a m ha bao gm 8 l c khc u nhau, nhvy khi ng c quay ti u ra tn hiu Encoder ta thu c tm xung

vung. xc nh chiu quay ca ng c (thun, ngc) cn s dng 2 cm binkp gm 2 led quang, 2 transitor quang v 2 a m ha.Khi a quay ta nhnc 2 xung hnh ch nht lch nhau 1 gc 90 .Chiu quay c xc nh

bng v tr tng i gia 2 tn hiu u ra.Tc bng 0 tc l xung tiptheo khng bao gi ti.Bit c s xung trong mt chu k ta s tnh c tc quay ca ng c:

n [vng/pht]=60N/4N0Tn.Trong : - Tn : chu k iu chnh tc , y l chu k m xung tnh

bng giy.

- N0 : s xung trong 1 vng, cn gi l phn gii ca b cmbin tc .

- N : s xung trong thi gian Tn.

Lp LTC H in3 K3

Nhm 4

16


17/19

i Hc Cng Ngip H Ni

2. Lu thut ton o tc ng c.

Lp LTC H in3 K3

Nhm 4

RET

PV=CV

Thc hin ngt.

Khai bo kiu mv lm vic mi.

RET

S

17

I 1.1 khi phtChng trnh

chnh

Trong vng qut u tin:- gi chng trnh con

khi to gi tr ban u.- Gi chng trnh khai

bo ch ngt.

MEND

SBR:0

- Np gi tr cho byte SMB47.-Xc nh ch lm vic cho

b m bng lnh HDEF.-Np gi tr PV v CV.-Khai bo ch ngt vo/ra.-Kch b m.

ITR:0

m s ln lch trngthi logic gia I.6 vI0.7


18/19

i Hc Cng Ngip H Ni

3. Chng trnh iu khin.

Lp LTC H in3 K3

Nhm 4

18


19/19

i Hc Cng Ngip H Ni

Lp LTC H in3 K3

Nhm 4

19

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