bao cao vi dieu khien

8/3/2019 Bao Cao Vi Dieu Khien

1/18

Nhm 29 Lp H C in t -k2

LI NI UK thut vi iu khin hin nay rt pht trin, n c ng dng vo rt nhiu

lnh vc sn xut cng nghip, t ng ha, trong i sng v cn nhiu lnh vckhc na. So vi k thut s th k thut vi iu khin nh gn hn rt nhiu do

n c tch hp li v c kh nng lp trnh c iu khin nn rt tin dngv c ng. Vi tnh u vit ca vi iu khin chng em ng dng nghincu v thc hin ti: Thit k Ampemet in t (vi 3 thang o

0 0,1A, 0 1A, 0 10A), ng thi hin th trn my tnh.

Mc ch ca ti hng n: to ra bc u cho sinh vin th nghimnhng ng dng ca vdk trong thc tin ri t tm ti, pht trin nhiu ngdng khc trong i sng hng ngy cn n.

Trc tin, em xin chn thnh cm n thy gio Nguyn Anh Dng tntnh gip , theo di ng vin, khuyn khch, to mi iu kin em hon

thnh tt ti trn.Em xin gi li cm n n Ban Gim Hiu, ton th thy c v nht l cc thy

c khoa in T -Trng H Cng Nghip H Ni to iu kin thun licng nh ch dn v truyn t kin thc em hon thnh c ti.

Do thi gian lm bi v kinh nghim cn hn ch nn khng trnh khi nhngthiu st . V vy, chng em rt mong s gp ca cc thy c v cc bn.

H ni, thng 5 nm 2010

Nhm thc hin ti

Vng Thnh

Nguyn ng Giang

H Vn Thnh

n mn vi iu khin - 1 - Ampemet in t


2/18


3/18


MC LCLI NI U..

NHN XT CA GIO VIN HNG DN..

CHNG 1: GII THIU LINH KIN V C S L THUYT.. - 4 -

A. KHI X L TRUNG TM (AT89S52)... - 4

KHI QUT VI IU KHIN AT89S52- 4-

TRUYN THNG NI TIP VI AT89S52...-5-

B. KHI CHUYN I ADC (ATMEGA8)....... - 7 -

C. KHI NGUN ...... - 8 -

Gii thiu IC n p 7805... - 8 -D. KHI U O...... - 8 -

E. KHI GIAO TIP VI MY TNH- 8 -

Vi mch MAX232...- 8 -

CHNG 2: NI DUNG THIT K..- 9 -A. S KHI....- 9 -

B. S NGUYN L CA CC KHI....- 9 -

C. S MCH IN....- 12

D. LU GII THUT...- 13 -

E. CHNG TRNH CHNH.....- 14 -

CHNG 3: KT LUN..- 16 -A. NH GI U, NHC IM...- 16 -

B. HNG CI TIN, PHT TRIN..- 16 -



4/18


CHNG 1: GII THIU LINH KIN

V C S L THUYTA. KHI X L TRUNG TM (AT89S52)

KHI QUT VI IU KHIN AT89S52 Cu to v chc nng cc khi ca AT89S52.

CPU( CPU centralprocessing unit)bao gm:

Thanh ghi tch ly A;

Thanh ghi tch ly ph B;

n v logic hc (ALU);

Thanh ghi t trng thi chng trnh;

Bn bng thanh ghi;

Con tr ngn xp. Chc nng cc chn ca AT89C52

Port 0( P0.0=>P0.7)

Port 0 gm 8 chn, ngoi chc nng xut nhp, port 0 cn l bus a hp d liu va ch( AD0-AD7), chc nng ny s c s dng khi 89c52 giao tip vi ccthit b ngoi c kin trc Bus nh cc vi mch nh, mch PIO

Port 1( P1.0=>P1.7)

Chc nng duy nht ca Port 1 l chc nng xut nhp cng nh cc Port khc.Port1 c th xut nhp theo bit v theo byte.

Port 2( P2.0=>P2.7)

Port 2 ngoi chc nng l cng vo/ra nh Port 0 v 1 cn l byte cao ca bus ach khi s dng b nh ngoi.

Port 3

Mi chn trn Port 3 ngoi chc nng xut nhp cn c mt chc nng ring, cth nh sau:

Bit Tn Chc nng

P3.0 RXD D liu nhn cho Port ni tip

P3.1 TXD D liu truyn cho Port ni tip

P3.2 INT0 Ngt bn ngoi 0

P3.3 INT1 Ngt ngoi 1

P3.4 TO Ng vo ca Timer/counter0

P3.5 T1 Ng vo ca Timer/counter1



5/18


P3.6 /WR Xung ghi b nh d liu ngoi.

P3.7 /RD Xung c b nh d liu ngoi.

Chn /PSEN :

/PSENl chn iu khin c chng trnh b nh ngoi.

Chn ALE.

ALE l tn hiu iu khin cht a ch c tn s bng 1/6 tn s dao ng ca viiu khin. Tn hiu ALE c dng cho php vi mch cht bn ngoi nh7473.

Chn /EA.

Tn hiu /EA cho php chn b nh chng trnh l b nh trong hay ngoi.EA=1 th thc hin chng trnh trong RAM ni. EA=0 thc hin RAM ngoi.

RST( reset)

Ng vo reset trn chn s 9. khi RST=1 th b vi iu khin s c khi ngli thit lp ban u.

XTAL1, XTAL2

2 chn ny c ni song song vi thch anh tn s max=33 Mhz. to daong cho b vi iu khin.

VCC, GND:Cung cp ngun nui cho b vi iu khin cp qua chn 20 v 40.

TRUYN THNG NI TIP VI AT89S52

8051 c hai chn c dng chuyn cho truyn v nhn d liu ni tip. Hai

chn ny c gi l TxD v RxD v l mt phn ca cng P3 ( l P3.0 vP3.1). chn 11 ca 8051 l P3.1 c gn cho TxD v chn 10 (P3.0) c dngcho RxD. Cc chn ny tng thch vi mc l-gch TTL. Do vy chng ihi mt b iu khin ng truyn chng tng thch vi RS232. Mt biu khin nh vy l chp MAX232 s c ni r hn trong khi giao tip mytnh. cho php truyn d liu gia my tn PC v h thng 8051 m khng cbt k li no th chng ta phi bit chc rng tc baud ca h 8051 phi phhp vi tc baud ca cng COM my tnh PC.

1) Tc Baud

8051 truyn v nhn d liu ni tip theo nhiu tc khc nhau. Tc truyn ca n c th lp trnh c. iu y thc hin nh s tr gip cab nh thi Timer1.

Cc gi tr ca thanh ghi TH1 trong Timer1 cho cc tc baud khc nhau.



6/18


Tc baud thng dng l 9600

2) Thanh ghi SBUF

SBUF l thanh ghi 8 bt c dng ring cho truyn thng ni tip trong 8051.

i vi mt byte d liu cn phi c truyn qua ng TxD th n phi ct trong thanh ghi SBUF. Tng t nh vy SBUF gi mt byte d liu khi nc nhn bi ng RxD ca 8051. Khi mt byte c ghi vo thanh ghiSBUF n c ng khung vi cc bt Start v Stop v ng truyn ni tipquan chn TxD. Tng t nh vy, khi cc bt c nhn ni tip t RxD th8051 m khung n loi tr cc bt Start v Stop ly ra mt byte t d liunhn c v t n vo thanh ghi SBUF.

3) Thanh ghi iu khin ni tip SCON

Thanh ghi SCON l thanh ghi 8 bt c dng lp trnh vic ng khung bt

bt u Start, bt dng Stop v cc bt d liu cng vi vic khc.

Di y l m t cc bt khc nhau ca SCON:

Bt SM0, SM1y l cc bt D7 v D6 ca thanh ghi SCON. Chng c dng xc nh

ch ng khung d liu bng cch xc nh s bt ca mt k t v cc btStart v Stop. Cc t hp ca chng l:

SM0 SM1

0 0 Ch ni tip 0



7/18


0 1 Ch ni tip 1, 8 bt d liu, Start, Stop

1 0 Ch ni tip 2

1 1 Ch ni tip 3

Trong bn ch ta ch quan tm n ch 1. Trong thanh ghi SCON

khi ch 1 c chn th d liu c ng khung gm 8 bt d liu, 1 btStart, 1 bt Stop tng thch vi cng COM ca IBM PC v cc PC tngthch khc. Quan trng hn l ch ni tip 1 cho php tc baud thay iv c thit lp bi Timer1 ca 8051. Trong ch ni tip 1 th mi k tgm c 10 bt c truyn trong c bt u l bt Start, sau l 8 bt d liuv cui cng l bt Stop.

Bt SM2

Bt SM2 l bt D5 ca thanh ghi SCON. Bt ny cho php kh nng a x l ca

8051. i vi cc ng dng ca chng ta t SM2 = 0 v ta khng s dng 8051

trong mi trng a x l. Bt REN

y l bt cho php thu (Receive Enable), bt D4 ca thanh ghi SCON. Khibt REN cao th n cho php 8051 thu d liu trn chn RxD ca n. V kt qu lnu ta mun 8051 va truyn v nhn d liu th bt REN phi c t ln 1. Khit REN th b thu b cm.. Bt ny c th c dng khng ch mi vicnhn d liu ni tip v n l bt cc k quan trng trong thanh ghi SCON.

Bt TB8 v RB8

Bt TB8 l bt SCON.3 hay l bt D3 ca thanh ghi SCON. N c dng

cho ch ni tip 2 v 3.

Bt TI v RI

Cc bt ngt truyn TI v ngt thu RI l cc bt D1 v D0 ca thanh ghi SCON.

Cc bt ny l cc k quan trng ca thanh ghi SCON. Khi 8051 kt thc truynmt k t 8 bt th n bt TI bo rng n sn sng truyn mt byte khc. Bt TIc bt ln trc bt Stop. Cn khi 8051 nhn c d liu ni tip qua chnRxD v n tch cc bt Start v Stop ly ra 8 bt d liu t vo SBUF, saukhi hon tt n bt c RI bo rng n nhn xong mt byte v cn phi ly i

ko n b mt c RI c bt khi ang tch bt Stop.

B. KHI CHUYN I ADC (ATMEGA8)

Chuyn i ADC trn ATMEGA8

Chip AVR ATmega8 ca Atmel c tch hp sn cc b chuyn i ADC vi phn gii 10 bit. C tt c 6 knh n (cc chn ADC0 n ADC5). ADC trn



8/18


AVR cn c nui bng ngun in p ring chn AVCC, gi tr in p cpcho AVCC khng c khc ngun nui chip (VCC) qu +/-0.3V. ADC trongAVR l loi chuyn i xp x ln lt (successive approximation ADC).

in p tham chiu cho ADC trn AVR c th c to bi 3 ngun: dng in

p tham chiu ni 2.56V (c nh), dng in p AVCC hoc in p ngoi ttrn chn VREF. Cc chn trn PORTC ca chip ATmega8 c dng cho bADC, chn PC0 tng ng knh ADC0 v chn PC5 tng ng vi knh ADC5.

H thng ca ta dng chn ADC5/

C. KHI NGUN

Gii thiu IC n p 7805

in p ng vo : 5V 18V

in p ng ra : 4.8V -5.2V

Nhit hot ng : 00C 1250C

Nhit bo qun : -650C 1500C

D. KHI U O

S dng 3 in tr cng sut cho 3 thang o: 0.22 Om, 2.2 Om, 22 Om (5W)

IC LM358 hn dng v n nh dng.

Mi thang o c 1 cng tc ring chuyn mch.

S dng 2 que o dng o

E. KHI GIAO TIP VI MY TNH Vi mch MAX232

Vi mch ny nhn mc RS232 c gi t my tnh v bin i tn hiu ny thnhtn hiu TTL cho tng thch vi vi iu khin v n cng thc hin ngc li bini tn hiu TTL t vi iu khin thnh mc +12, -12 cho ph hp vi hot ngca my tnh.



9/18


CHNG 2: NI DUNG THIT K

A. S KHI

B. S NGUYN L CC KHI


KHI X L

TRUNG TM

KHI NGUN

KHIGIAO TIPMY TNH

KHIU O

KHICHUYN I

ADC


10/18


KHI S L TRUNG TM

V C C

D B 3

D B 7

D 5L E D

D B 8

F

3

M O S I

D 3

L E D

F 1

D B 1

R 5 74 7 0V C C

R E S E T

S E LD 4

L E

D B 2

R

5

5

4

7

0

Y 11 2 M H z

D B 9

V C C

T X D

V C C

S C K

D B 4

R X D

D B 0

R 11 0 K S W 1

R E S E T

S C K

D B 5

M I S O

S W 2

V C C

+ C 5

1 0 u F

F

1

S E L

U 1 3

4 7 0

23456789

1

R 1

R 2R 3R 4R 5R 6R 7R 8

1

V C C

R

5

4

4

7

0

D B 6

M I S O

J 8

C O N 1 0 A

1 23 45 67 89 1 0

C 2 33 3 p F

R 2 54 . 7 K

D 2

L E D

R

5

6

4

7

0

F 2

U 1

A T 8 9 S 5 2

2 93 0

4

0

2

0

3

1

1 9

1 8

9

3 93 83 73 63 53 43 33 2

12345

678

2 12 22 32 42 52 62 72 81 0

1 11 21 31 41 51 61 7

P S E NA L E

V

C

C

G

N

D

E

A

X 1

X 2

R S T

P 0 . 0 / A D 0P 0 . 1 / A D 1P 0 . 2 / A D 2P 0 . 3 / A D 3P 0 . 4 / A D 4P 0 . 5 / A D 5P 0 . 6 / A D 6

P 0 . 7 / A D 7

P 1 . 0P 1 . 1P 1 . 2P 1 . 3P 1 . 4

P 1 . 5 / M O S IP 1 . 6 / M I S OP 1 . 7 / S C K

P 2 . 0 / A 8P 2 . 1 / A 9

P 2 . 2 / A 1 0P 2 . 3 / A 1 1P 2 . 4 / A 1 2P 2 . 5 / A 1 3P 2 . 6 / A 1 4P 2 . 7 / A 1 5P 3 . 0 / R X D

P 3 . 1 / T X DP 3 . 2 / I N T 0P 3 . 3 / I N T 1P 3 . 4 / T 0P 3 . 5 / T 1P 3 . 6 / W RP 3 . 7 / R D

C 1 6

1 0 4 p F

R E S E T

F 3

C 2 23 3 p F

M O S I

F

2

KHI CHUYN I ADC

D B 9

D B 1

D B 4D B 3

U 1 2

A T M E G A 8

1

23456

7

891 0

1 11 21 3

1 41 51 61 7

1 81 9

2 02 1

2 2

2 32 42 52 62 72 8

P C 6 ( R E S E T )

P D 0 ( R X D )P D 1 ( R X D )P D 2 ( I N T 0 )P D 3 ( I N T 1 )P D 4 ( X C K / T 0 )

V C C

G N DP B 6 ( X T A L 1 / O S C 1 )P B 7 ( X T A L 2 / O S C 2 )

P D 5 ( T 1 )P D 6 ( A I N 0 )P D 7 ( A I N 1 )

P B 0 ( I C P 1 )P B 1 ( O C 1 A )P B 2 ( S S / O C 1 B )

P B 3 ( M O S I / O C 2 )P B 4 ( M I S O )P B 5 ( S C K )

A V C C

A R E F

G N D

P C 0 ( A D C 0 )P C 1 ( A D C 1 )P C 2 ( A D C 2 )P C 3 ( A D C 3 )

P C 4 ( A D C 4 / S D A )P C 5 ( A D C 5 / S C L )

D B 8

D B 2

D B 0

D B 6

C 1 5

1 0 4 p

V CL 1

1 0 0 u H

D B 5A D C

D B 7



11/18


KHI NGUN

U 1 0 7 8 0 5

1 3

2

I N O U TGN

DV C C

J 4

7 . 5 - 1 5 V

12

+

C 8

1 0 0 0 u F

C 1 1

1 0 4 p F

D 1

L E D+

C 9

1 0 0 u F

1 2 V

C 1 0

1 0 4 p FR 2 8

4 7 0

KHI U O

R 4 3

2 . 2

F 2

F 1

J 7

1 2

S W 3

-

+

U 5 A

L M 3 5 8

3

21

8

4

F 3

S W 5

R 3 32 2

R 5 30 . 2 2

A D C

1 2 V

S W 4

KHI GIAO TIP MY TNH



12/18


R 2 I N

R 2 I N +C 21 0 u F

T X DT 2 O U T

+

C 3

1 0 u F

P 1

C O M 9

59483

7261

+

C 1

1 0 u F

R X

U 3M A X 2 3 2

1

3

4

5

1

6

1

5

26

1 29

1 1

1 0

1 3

81 4

7

C 1 +

C 1 -

C 2 +

C 2 -

V

C

C

G

N

D

V +V -

R 1 O U TR 2 O U T

T 1 I N

T 2 I N

R 1 I N

R 2 I NT 1 O U T

T 2 O U T

V C C

T 2 O U T

+

C 4

1 0 u F

C. S MCH IN



13/18




14/18


D. LU THUT GII

NO

YES

NO NO

YES YES YES

NO

YES


Phmcnhn

Begin

Gi gi tr dng in ln

my tnh

End

c ADCtnh U1

c ADCtnh U2

c ADCtnh U3

Tnhdng in

I = U1 / R1

Tnhdng in

I = U2 / R2

Tnhdng in

I = U3 / R3

Hin thng bo

Kim tra thang o

ThangoI

ThangoII

ThangoIII

I


15/18


E. CHNG TRNH CHNH

#include #include

#define phim1 P2_1

#define phim2 P2_0

#define phim3 P2_2

#define phim4 P1_4

unsigned int i,adch, adcl;

float dongdien,adc;

void delay_ms(unsigned int t)

{

while(t--)

for(i=0;i>3)&0x03; adcl=P0;

adc=((float)adch*256+adcl)*5/1023;

}

void thang_do() // Tinh dong dien

{

//============== Thang do 0-0.1A =====================//

if(phim1==0&&phim2==1&&phim3==1)

{

dongdien=(adc/10);

if(dongdien


16/18


P1_3=1;

}

//=============== Thang do 0-1A =====================//

else if(phim1==1&&phim2==0&&phim3==1)

{dongdien=adc;

if(dongdien


17/18


{

SCON = 0x52;

TMOD = 0x20;

TH1=TL1=-3;

TR1=1;while(1)

{

while(phim4==1); //Bam phim de bat dau do

while(1)

{

dong_dien();

thang_do();

}}

}

CHNG 3: KT LUN



18/18


1. U IM

H thng o c nhiu thang o v dng in ln c th l:

Thang o 1: 0-0,1A

Thang o 2: 0-1A

Thang o 3: 0-10A

Kt qu ca h thng c hin th trn my tnh nn kt qu rt trc quan: cc con s v thng bo cho tng thang o.

2. NHC IM

Do h thng phi o di dng in ln 0-10A nn sai s ln.

3. HNG CI TIN V PHT TRINHn ch sai s cho cc thang o.

Ti Liu Tham Kho

1)o lng iu khin bng my tnh - Ng Din Tp

2) Gio trnh ghp ni my tnh (l thuyt +thc hnh) V TrungKin Khoa in T - H Cng Nghip H Ni

3)Gio trnh VB6.0

4)Lp trnh giao tip bng VB - H Thanh Lm,L Vn Bn HNng Lm TP HCM


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