baigiang 005 vxl banphim nutnhan

8/3/2019 Baigiang 005 Vxl Banphim Nutnhan

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ai hoc s pha m ky thuat tp HCM khoa ien ien t Nguyen nh Phu

Bo mon ien t cong nghiep bai giang vi ieu khien: ban phm nut nhan 1

VI IEU KHIEN GIAO TIP VI PHM NHNBN PHM

PHM NHN, NT NHN, BN PHM DNG GIAO TIP GIA CON NGI VTHIT BIU KHIN.V D

: BN PHM MY TNH, BN PHM IN THO

I, BN PHM MY BN XNGDU DNG NHP S TIN CN BN, S LT CN BN, MY GIT TNG CBN PHM CHNH CH GIT, CHN MC NC, C 2 DNG GIAO TIP:H THNG T PHM: V D DK NG C BNG 3 PHM: START, STOP, INV, NGHC 3 N 4 PHM CHNH THI GIAN.H THNG NHIU PHM: H THNG BN XNG DU, , MY TNH, IN THOI,

MACH IEN GIAO TIEP HTHNG T PHM

HOAT ONG CUA PHM NHAN:

S DUNG PHM THNG H, NEN CAC NGO VAO THNG MC 1.KHI NHAN TH KN MACH HAY NGAN MACH NOI MASS LAM NGO VAO XUONGMC 0 VA BUONG TAY KHONG NHAN NA TH TR LAI MC 1.CHNG TRNH KIEM TRA XEM CHUNG TA CO NHAN PHM HAY KHONG BANGCACH KIEM TRA MC LOGIC: NEU MC LOGIC LA 1 TH KHONG NHAN, LA 0 THCO NHAN PHMBAI SO 5-1: CHNG TRNH DK ONG C DC QUAY THUAN NGHCH CO 3 PHM

NHAN: START, STOP, INV


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NEU C NGNG TH NHAN START C SE QUAY THUAN NEU C ANG QUAY MA NHAN STOP TH C NGNG NEU C ANG QUAY MA NHAN INV TH C AO CHIEU DUNG 2 NGO RA IEU KHIEN EN XANH O: 2 EN TAT C NGNG XANH SANG TH C QUAY THUAN O SANG TH C QUAY NGHCH LU O IEU KHIEN

2 LED TT

START?

BEGIN

S

LED XANH SNG

STOP?

INV?

S

S

O LED

BEGIN

STOP ?

S

NG C NGNG

START ?

NG C QUAY THUN

INV ?

END

S

S

O CHIU NG C

1. Lenh nhay khi bit = 1 : Cu phap : JB BIT, REL Ma lenh :

0 0 1 0 0 0 0 0

b7 b6 b5 b4 b3 b2 b1 b0

r7 r6 r5 r4 r3 r2 r1 r0

Lenh nay chiem 3 byte va thi gian thc hien lenh la 2 chu ky may Chc nang: neu noi dung cua bit co a ch bit [c xac nh bi byte th 2] bang 1 th vi ieu khien

se nhay en thc hien chng trnh tai a ch rel.

2. Lenh nhay khi bit = 0 :


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Cu phap : JNB BIT, REL Ma lenh :

0 0 1 1 0 0 0 0

b7 b6 b5 b4 b3 b2 b1 b0

r7 r6 r5 r4 r3 r2 r1 r0

Lenh nay chiem 3 byte va thi gian thc hien lenh la 2 chu ky may

Chc nang: neu noi dung cua bit co a ch bit [c xac nh bi byte th 2] bang 0 th vi ieu khiense nhay en thc hien chng trnh tai a ch bang rel.

VIET CHNG TRNHCHUONG TRINH DIEU KHIEN DONG CO QUAY THUAN NGHICH

START BIT P3.0 ;PHIM START NOI VOI P3.0STOP BIT P3.1INV BIT P3.2

DC1 BIT P1.0 ;XANH DK DONG CODC2 BIT P1.2 ;DO DK DONG COORG 0000HMOV P1,#00 ;TAT HET LED

LB: CLR DC1 ;NGUNG DONG COCLR DC2

JB START,$ ;CHO NHAN PHIM START


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SETB DC1 ;DCO CHAY THUAN - XANH SANG

LB2: JNB STOP,LB ;NEU NHAN THI NHAY VE NGUNG 2 DCJB INV,LB2 ;NHAY VE LB2 KHI KHONG NHAN INVCPL DC1 ;NGHICH DAO COMPLEMENTCPL DC2JMP LB2

$INCLUDE(TV_DELAY.ASM)END

KHUYET IEM LA NHAN PHM B DOI PHM VA TOC O X LYNGUYEN NHAN TH 1 DOI PHM: DO PHM LAM BANG C CAU C KH NEN KHI

NHAN TH LO XO AY PHM NGC LAI LC TAC ONG LAM PHM TIEP XUCNHIEU LAN TRC KHI ON NH VANGUYEN NHAN TH 2: DO TOC O CUA VXL HOAC VK NHANH NEN XEM NH TANHAN NHIEU LAN.


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PHAI CHONG DOI PHM:CHONG DOI PHM: CO 2 CACH:

1. KIEM TRA TRANG THAI CUA PHM NEU CHUYEN TRANG THAI VA TON TAIT NHAT KHOANG 20 MS TH XEM NH CO PHM NHAN.

2. SAU KHI PHAT HIEN PHM NHAN, TH KIEM TRA BUONG PHM HOAC COTHE DELAY 20MS ROI KIEM TRA VA CH CHO EN KHI BUONG PHM

LU O CHONG DOIBEGIN

SPHM ?

DELAY10MS

PHM ?

END

S

PHM ?

DELAY10MS

S

CHUONG TRINH DIEU KHIEN DONG CO QUAY THUAN NGHICH CO C DOI PHIMSTART BIT P3.0STOP BIT P3.1INV BIT P3.2DC1 BIT P1.0 ;XANH

DC2 BIT P1.2 ;DOORG 0000HMOV P1,#00

LB: CLR DC1 ;NGUNG 2 DONG COCLR DC2JB START,$ ;CHO NHAN PHIM STARTSETB DC1 ;DCO CHAY THUAN - XANH SANG

LB2: JNB STOP,LB


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LB1: CALL PHIM_INV ;GOI CHTR KIEM TRA VA CHONG DOIJMP LB2

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXFFFFFFPHIM_INV:JB INV,LBD1 ;KTRA LAN 1, NHAY NEU KHONG AN

CALL DELAY10MSJB INV,LBD1 ;KTRA LAN 2, NHAY NEU KHONG ANCALL DELAY10MSJB INV,LBD1 ;KTRA LAN 3, NHAY NEU KHONG ANCPL DC1 ;DAO CHIEU DONG COCPL DC2JNB INV,$ ;CHO NHA PHIM

LBD1: RET ;THOAT


CHONG DOI CACH 2 LA NHAN PHM THC THI XONG CH BUONG PHM

CHUONG TRINH IEU KHIEN DCO TRONG THC TE CO DELAY CH ONG CDNG

START BIT P3.4STOP BIT P3.5INV BIT P3.6DC1 BIT P1.0 ;XANHDC2 BIT P1.2 ;DODCT1 BIT 00H ;BIT TAMDCT2 BIT 01H

ORG 0000HLB: SETB DCT1 ;NGUNG 2 DONG CO

SETB DCT2CALL OUTDCO

JB START,$ ;CHO NHAN PHIM START

CLR DCT1 ;DCO CHAY THUAN - XANH SANG


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CALL OUTDCO

LB2: JB STOP,LB1JMP LB ;NEU NHAN THI NHAY VE NGUNG 2 DC

LB1: JB INV,LB2CALL DELAY10MSJNB INV,$

SETB DC1SETB DC2CALL DELAY1S

CALL DELAY1SCALL DELAY1SCPL DCT1CPL DCT2CALL OUTDCOJMP LB2

OUTDCO: MOV C,DCT1MOV DC1,CMOV C,DCT2MOV DC2,CRET


HE THONG NHIEU PHM - MA TRAN PHM VI CACH 1 TH MOI PHM S DUNG 1 NGO VAO E KET NOI. 16 PHM SE DUNG 16 NGO VAO TN NHIU NG TN HIU. VI CACH KET NOI DANG MA TRAN TH 16 PHM CH DUNG 8 TN HIEU: 4

CHO HANG VA 4 CHO COT GOI LA MA TRAN 44 = 16 PHM V DU MA TRAN 88 = 64 PHM


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U IEM TIET KIEM NG IEU KHIEN KHUYET IEM: CHNG TRNH PHC TAP

H0=P3.4

C3=P3.3C2=P3.2C1=P3.1C0=P3.0

0 4 8 C123

5

67

VCC=1

0 1 1 1

H1=P3.5H2=P3.6H3=P3.7

TRONG MA TRAN 44 TH CO 4 HANG VA 4 COTHANG C CHON LA TN HIEU VAO COT C CHON LA TN HIEU RA

H0=P3.4

C3=P3.3C2=P3.2C1=P3.1C0=P3.0

0 4 8 C12

3

5

6

7

VCC=1

0 1 1 1

H1=P3.5H2=P3.6

H3=P3.7

HANG TH TREO IEN TR LEN NGUON VCC NEN MC LOGIC LA 1

CAC PHM NHAN THNG H NEN HANG LUON MC 1. H3H2H1H0 =1111COT LA TN HIEU RA CHUNG TA IEU KHIEN CHO 1 COT MC 0, 3 COT CON LAI MC 1.V DU COT C0 = 0, C1C2C3=111


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H0=P3.4

C3=P3.3C2=P3.2C1=P3.1C0=P3.0

0 4 8 C123

5

67

VCC=1

0 1 1 1

H1=P3.5H2=P3.6H3=P3.7

NEU NHAN PHM SO 0 TH NOI HANG H0=1(5V) VI C0=0 (0V): KHAC AP NGANMACH LAM CHO H0 = 0 DO CO IEN TR NEN KHONG SAO, VY:

NHAN PHM SO 0 TH LAM H0=0 HAY H3H2H1H0 = 1110 NHAN PHM SO 1 TH LAM H1=0 HAY H3H2H1H0 = 1101 NHAN PHM SO 2 TH LAM H2=0 HAY H3H2H1H0 = 1011 NHAN PHM SO 3 TH LAM H3=0 HAY H3H2H1H0 = 0111 NEU NHAN PHM SO 4 TH NOI HANG H0=1(5V) VI C1=1 (5V): CUNG AP 5V:

KHONG CO G THAY OICOT C0=0 CH PHAT HIEN 4 PHM 0, 1, 2, 3NEU KHONG NHAN 4 PHM 0, 1, 2, 3 MA NHAN CAC PHM KHAC TH TA PHAI

LAM C1 = 0, C0C2C3=111

H0=P3.4

C3=P3.3

C2=P3.2C1=P3.1C0=P3.0

0 4 8 C123

5

67

VCC=1

01 1 1

H1=P3.5H2=P3.6H3=P3.7

NHAN PHM SO 4 TH LAM H0=0 HAY H3H2H1H0 = 1110 NHAN PHM SO 5 TH LAM H1=0 HAY H3H2H1H0 = 1101 NHAN PHM SO 6 TH LAM H2=0 HAY H3H2H1H0 = 1011 NHAN PHM SO 7 TH LAM H3=0 HAY H3H2H1H0 = 0111

LAM C2 = 0, C0C1C3=111


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H0=P3.4

C3=P3.3C2=P3.2C1=P3.1C0=P3.0

0 4 8 C123

5

67

VCC=1

11 0 1

H1=P3.5H2=P3.6H3=P3.7

NHAN PHM SO 8 TH LAM H0=0 HAY H3H2H1H0 = 1110 NHAN PHM SO 9 TH LAM H1=0 HAY H3H2H1H0 = 1101 NHAN PHM SO A TH LAM H2=0 HAY H3H2H1H0 = 1011 NHAN PHM SO B TH LAM H3=0 HAY H3H2H1H0 = 0111

LAM C3 = 0, C0C1C2=111

H0=P3.4

C3=P3.3

C2=P3.2C1=P3.1C0=P3.0

0 4 8 C123

5

67

VCC=1

11 1 0

H1=P3.5H2=P3.6H3=P3.7

NHAN PHM SO C TH LAM H0=0 HAY H3H2H1H0 = 1110 NHAN PHM SO D TH LAM H1=0 HAY H3H2H1H0 = 1101 NHAN PHM SO E TH LAM H2=0 HAY H3H2H1H0 = 1011 NHAN PHM SO F TH LAM H3=0 HAY H3H2H1H0 = 0111

12CAU HOI LAM SAO E PHAN BIET CAC PHM ?CAU TRA LI LA MOI PHM CO MOT MA SO HAY MA PHM (MP) KHAC NHAUCHUNG TA XAY DNG: PHM 0 CO MA LA 00H, PHM 1 CO MA LA 01, TNG TCHO CAC PHM CON LAICO THE TEN PHM VA MA KHONG TRUNG NHAU.VAY KHI QUET PHM TH PHAI TAO RA MA PHM NHAN


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H0=P3.4

C3=P3.3C2=P3.2C1=P3.1C0=P3.0

0 4 8 C123

5

67

VCC=1

0 1 1 1

H1=P3.5H2=P3.6H3=P3.7

QUET PHM XUAT MC 0 LAN LT RA CAC COT BAT AU T COT C0, SAU OKIEM TRA CAC HANG H3H2H1H0=1111.

NEU H3H2H1H0=1111(UNG) TH KHONG NHAN PHM HOAC NHAN PHMKHAC, PHAI QUET COT TIEP THEO CHO EN KHI HET TAT CA CAC COT.

NEU H3H2H1H0 1111(SAI) TH CO NHAN PHM PHAI XEM HANG NAO BANG0.


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BEGIN

0, 1, 2, 3

MP:=00C3C2C1C0=1110

END

S

H3H2H1H0=1111

MP:=04C3C2C1C0=1101

4, 5, 6, 7

S

H3H2H1H0=1111

MP:=08C3C2C1C0=1011

8, 9, 10, 11

S

H3H2H1H0=1111

MP:=12C3C2C1C0=0111

12, 13, 14, 15

S

H3H2H1H0=1111

MP:=FFH

H3H2H1H0=1110L 0H3H2H1H0=1101 L 1H3H2H1H0=1011 L 2H3H2H1H0=0111 L 3

H3H2H1H0=1110 L 4H3H2H1H0=1101 L 5H3H2H1H0=1011 L 6H3H2H1H0=0111 L 7

H3H2H1H0=1110 L 8H3H2H1H0=1101 L 9H3H2H1H0=1011 L 10

H3H2H1H0=0111 L 11

H3H2H1H0=1110 L 12H3H2H1H0=1101 L 13H3H2H1H0=1011 L 14H3H2H1H0=0111 L 15

MP=00

MP=04

MP=08

MP=12

XOAY PHI H3H2H1H0

C = 0

END

S

MP:= MP +1

S

NEU KHONG CO NHAN PHM TH THOAT VI MA PHM = FFH

NU NHN PHM S 0

1 0

H3 H2 H1 H0

11 X

C

TRC KHI THC HIN

SAU KHI THC HIN

1 1

H3 H2 H1 H0

1x 0

C

NU NHN PHM S 0 TH THOT VI MP = 00H

NU NHN PHM S 1, M PHM XUT PHT: MP=00


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0 1

H3 H2 H1 H0

11 X

C

TRC KHI THC HIN

XOAY LN 1, MP:=MP+1

1 0

H3 H2 H1 H0

1x 1

C

XOAY LN 2

1 1

H3 H2 H1 H0

xx 0

C

NU NHN PHM S 1 TH THOT VI MP = 01H

NU NHN PHM S 7, M PHM XUT PHT: MP=04

1 1

H3 H2 H1 H0

10 X

C

TRC KHI THC HIN

XOAY LN 1, MP:=MP+1 = 05

1 1

H3 H2 H1 H0

0x 1

C

XOAY LN 2, MP:=MP+1 = 06

0 1

H3 H2 H1 H0

xx 1

C

XOAY LN 3, MP:=MP+1=07

x 0

H3 H2 H1 H0

xx 1

C

XOAY LN 4

x x

H3 H2 H1 H0

xx 0

C

NU NHN PHM S 7 TH THOT VI MP = 07HLU O GON HN


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S

KEYPAD

DEM:=4MP:=0

C3C2C1C0 = 1110

H3H2H1H0 = 1111

MP:=MP+4XOAY TRI C3C2C1C0

DEM:= DEM -1

MP:=FF

END

XOAY PHI

C=0

DEM=0

MP:=MP+1

S

S

BAI SO 6-1: CTR QUET PHM MA TRAN VA HIEN TH TREN 1 LED 7 OAN VIET CHNG TRNH

H0=P3.4

C3=P3.3C2=P3.2C1=P3.1C0=P3.0

0 4 8 C12

3

5

6

7

VCC=1

0 1 1 1

H1=P3.5H2=P3.6

H3=P3.7

;CHUONG TRINH CON QUET MA TRAN PHIM 44;PORT 3 NOI VOI MA TRAN 4x4

DEM EQU 40H ;SO LAN QUET COT


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MP EQU 41H ;MAPHIMC3C0 EQU 42H ;C3C2C1C0

KEYPRESS16: MOV MP,#00MOV DEM,#4MOV C3C0,#11111110B ;C0=0, H3H2H1H0=1111

KEYP2: MOV P3,C3C0 ;P3.0= 0=C0 (OUT)MOV A,P3 ;P3=H3H2H1H0C3C2C1C0 (IN)ANL A,#0F0H ;(A)=H3H2H1H0-0000SWAP A ;(A)= 0000 -H3H2H1H0CJNE A,#00001111B,KEYP1 ;NHAY NEU CO NHAN PHIM

MOV A,MP ;MP:=MP+4ADD A,#4MOV MP,AMOV A,C3C0 ;ROTATE LEFT C3C0RL AMOV C3C0,ADJNZ DEM,KEYP2

MOV MP,#0FFH ;THOAT KHONG NHAN PHIMRET

KEYP3: INC MP ;TANG MA PHIMKEYP1: RRC A ;XU LY PHIM NHAN

JC KEYP3 ;NHAY KHI C=1RET

BAI SO 6-2: CTR TEST QUET PHM MA TRAN VA HIEN TH TREN 1 LED 7 OANCHNG TRNH TEST QUT MA TRN PHM XEM NG HAY SAI BNG CCHGI CHNG TRNH QUT PHM V GII M HIN TH M PHM TRN 1 LED7 ON.

ORG 0HMOV DPTR,#CODE7SEG


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MAIN1: CALL KEYPRESS16MOV A,MPCJNE A,#0FFH,MAIN2JMP MAIN1

MAIN2: MOV P1,MP ;CHO HIEN THI O P1MOVC A,@A+DPTRMOV P0,AJMP MAIN1

CODE7SEG: DB 0C0H,0F9H,0A4H,0B0H,099H,092H,082H,0F8HDB 080H,090H,088H,083H,0C6H,0A1H,086H,08EH

BAI SO 6-3: CTR CON QUET PHM MA TRAN 88NU L MA TRN 88 TH LU THAY I CI G?

S

KEYPAD

DEM:=8MP:=0

C7C0 = 11111110

H7H0 = 11111111

MP:=MP+8XOAY TRI C7C0

DEM:= DEM -1

MP:=FF

END

XOAY PHI

C=0

DEM=0

MP:=MP+1

S

S

;CHUONG TRINH CON QUET MA TRAN PHIM 88;HANG LA P3, COT LA P1


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DEM EQU 40HMP EQU 41HC7C0 EQU 42H ;C7=>C0

KEYPRESS64: MOV MP,#00MOV DEM,#8MOV C7C0,# 11111110B ;C0=0

KEYP2: MOV P1,C7C0 ;P1=C7C6C5C4 C3C2C1C0MOV A,P3 ;(A)=H7H6H5H4 H3H2H1H0CJNE A,#11111111B,KEYP1 ;NHAY NEU CO NHANMOV A,C7C0 ;ROTATE LEFT

RL AMOV C7C0,AMOV A,MPADD A,#8MOV MP,ADJNZ DEM,KEYP2MOV MP,#0FFH ;THOAT KHONG NHAN PHIM

RETKEYP3: INC MPKEYP1: RRC A

JC KEYP3RET


BAI SO 6-4: CTR CON QUET PHM MA TRAN 54NU 20 PHM MA TRN 54 TH LU THAY I CAI G? 5 HANG + 4 CT = 9NG TN HIEU NEN KHONG DUNG C 1 PORT MA PHAI DUNG 2 PORT


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F

KEYPAD

DEM:=4 (S CT)MP:=0

C3C2C1C0 = 1110

H4-H0 = 11111

TNO PRESS

PRESSED

MP:=MP+ 5(S HNG)XOAY TRI C3C2C1C0

DEM:= DEM -1

TF

MP:=FF

END

X PHI H4-H0

C=0

TF

DEM=0

MP:=MP+1

;CHUONG TRINH CON QUET MA TRAN PHIM 54

;HANG LA P3, COT LA P1 (5 HANG + 4 CT = 9)DEM EQU 40HMP EQU 41HC3C0 EQU 42H ;C3=>C0

KEYPRESS16: MOV MP,#00MOV DEM,#4

MOV C3C0,# 1110B ;C0=0

KEYP2: MOV P1,C3C0 ;P1=xxxx C3C2C1C0MOV A,P3 ;(A)=XXX H4 H3H2H1H0ANL A,#00011111B ;XOA CAC BIT DUCJNE A,#00011111B,KEYP1 ;NHAY NEU CO NHANMOV A,C3C0 ;ROTATE LEFT

RL A


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MOV C3C0,AMOV A,MPADD A,#5MOV MP,ADJNZ DEM,KEYP2MOV MP,#0FFH ;THOAT KHONG NHAN PHIMCLR C ;BAO HIEU KHONG NHANRET

KEYP3: INC MPKEYP1: RRC A

JC KEYP3

SETB C ;BAO HIEU CO NHAN PHIMRET


BAI SO 6-5: CTR CON QUET PHM MA TRAN 44 B DOI PHM;CHUONG TRINH CON QUET MA TRAN PHIM 44 B DOI

; CHTR QUET PHIM HIEN THI TREN 8 LED DE THAY HIEN TUONG DOI PHIMORG 0HMOV DPTR,#CODE7SEGCALL XOAVNHTHI

MAIN1: CALL KEYPRESS16MOV A,MP

CJNE A,#0FFH,MAIN2 ;SS VOI FFH, KHAC FFH THI CO NPCALL HIENTHI ;NEU KNP THI HIEN THIJMP MAIN1 ;QUAY LAI LAM TIEP

MAIN2: CALL DICHDATAMOVC A,@A+DPTRMOV 37H,A ;(37H) SE LUU SO MOI NHATJMP MAIN1

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx


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;CT CON DICH CHUYEN VUNG NHO HIEN THI TU 37H DEN 30H SANG TRAI;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxDICHDATA:

MOV 30H,31HMOV 31H,32HMOV 32H,33HMOV 33H,34HMOV 34H,35HMOV 35H,36HMOV 36H,37HRET

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

;CT CON XOA VUNG NHO HIEN THI TU 37H DEN 30H;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXOAVNHTHI:

MOV 37H,#07FH ;HIEN THI DAU CHAMMOV 36H,#0FFHMOV 35H,#0FFHMOV 34H,#0FFH

MOV 33H,#0FFHMOV 32H,#0FFHMOV 31H,#0FFHMOV 30H,#0FFHRET

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx;CT CON HIEN THI

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxHIENTHI: MOV P0,37H

MOV P2,#11111110BCALL DELAY1MSMOV P2,#0FFH

MOV P0,36H

MOV P2,#11111101B


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CALL DELAY1MSMOV P2,#0FFH

MOV P0,35HMOV P2,#11111011BCALL DELAY1MSMOV P2,#0FFH



MOV P0,32H


MOV P0,31HMOV P2,#10111111BCALL DELAY1MS

MOV P2,#0FFH

MOV P0,30HMOV P2,#01111111BCALL DELAY1MSMOV P2,#0FFHRET

$INCLUDE(TV_DELAY.ASM)


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$INCLUDE(TV_KEY16.ASM)CODE7SEG: DB 0C0H,0F9H,0A4H,0B0H,099H,092H,082H,0F8H

DB 080H,090H,088H,083H,0C6H,0A1H,086H,08EHEND

BAI SO 6-6: CTR CON QUET PHM MA TRAN 44 CHONG DOI BANG DELAY;CHUONG TRINH CON QUET MA TRAN PHIM 44 B DOI; CHTR QUET PHIM HIEN THI TREN 8 LED DE THAY HIEN TUONG DOI PHIM

ORG 0HMOV DPTR,#CODE7SEGCALL XOAVNHTHI

MAIN1: CALL KEYPRESS16MOV A,MPCJNE A,#0FFH,MAIN2 ;SS VOI FFH, KHAC FFH THI CO NPCALL HIENTHI ;NEU KNP THI HIEN THIJMP MAIN1 ;QUAY LAI LAM TIEP

MAIN2: CALL DICHDATAMOVC A,@A+DPTRMOV 37H,A ;(37H) SE LUU SO MOI NHATCALL DELAY100MS ;LAM CHAM LAIJMP MAIN1

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx;CT CON DICH CHUYEN VUNG NHO HIEN THI TU 37H DEN 30H SANG TRAI

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxDICHDATA:MOV 30H,31HMOV 31H,32HMOV 32H,33HMOV 33H,34HMOV 34H,35HMOV 35H,36HMOV 36H,37H


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RET;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx;CT CON XOA VUNG NHO HIEN THI TU 37H DEN 30H;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxXOAVNHTHI:

MOV 37H,#07FH ;HIEN THI DAU CHAMMOV 36H,#0FFHMOV 35H,#0FFHMOV 34H,#0FFHMOV 33H,#0FFHMOV 32H,#0FFHMOV 31H,#0FFH

MOV 30H,#0FFHRET

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx;CT CON HIEN THI;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxHIENTHI: MOV P0,37H

MOV P2,#11111110B




MOV P0,34H

MOV P2,#11110111B


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MOV P0,30H

MOV P2,#01111111BCALL DELAY1MSMOV P2,#0FFHRET

$INCLUDE(TV_DELAY.ASM)$INCLUDE(TV_KEY16.ASM)CODE7SEG: DB 0C0H,0F9H,0A4H,0B0H,099H,092H,082H,0F8H

DB 080H,090H,088H,083H,0C6H,0A1H,086H,08EHEND

BAI SO 6-7: CTR CON QUET PHM MA TRAN 44 CHONG DOI;CHUONG TRINH CON QUET MA TRAN PHIM 44 CO CHONG DOI; CHTR QUET PHIM HIEN THI TREN 8 LED DE THAY HIEN TUONG DOI PHIM

ORG 0HMOV DPTR,#CODE7SEGCALL XOAVNHTHI


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MAIN1: CALL KEYPRESSMOV A,MPCJNE A,#0FFH,MAIN2 ;SS VOI FFH, KHAC FFH THI CO NPCALL HIENTHI ;NEU KNP THI HIEN THIJMP MAIN1 ;QUAY LAI LAM TIEP

MAIN2: CALL DICHDATAMOVC A,@A+DPTRMOV 37H,A ;(37H) SE LUU SO MOI NHAT

JMP MAIN1

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx;CT CON DICH CHUYEN VUNG NHO HIEN THI TU 37H DEN 30H SANG TRAI;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxDICHDATA: MOV 30H,31H

MOV 31H,32HMOV 32H,33HMOV 33H,34H

MOV 34H,35HMOV 35H,36HMOV 36H,37HRET

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx;CT CON XOA VUNG NHO HIEN THI TU 37H DEN 30H;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

XOAVNHTHI:MOV 37H,#07FH ;HIEN THI DAU CHAMMOV 36H,#0FFHMOV 35H,#0FFHMOV 34H,#0FFHMOV 33H,#0FFHMOV 32H,#0FFH

MOV 31H,#0FFH


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MOV 30H,#0FFHRET

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx;CT CON HIEN THI;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxHIENTHI: MOV P0,37H


MOV P0,36HMOV P2,#11111101B




MOV P0,33H


MOV P0,32HMOV P2,#11011111BCALL DELAY1MS

MOV P2,#0FFH


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MOV P0,30HMOV P2,#01111111BCALL DELAY1MSMOV P2,#0FFHRET

CODE7SEG: DB 0C0H,0F9H,0A4H,0B0H,099H,092H,082H,0F8HDB 080H,090H,088H,083H,0C6H,0A1H,086H,08EH

;CHUONG TRINH CON QUET MA TRAN PHIM 44;PORT 3 NOI VOI MA TRAN 4X4

DEM EQU 40HMP EQU 41H

C3C0 EQU 42H ;C3C2C1C0LAP EQU 43HMP1 EQU 44H

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxKEYPRESS: MOV LAP,#10 ;nhap so dem 10 lanKEYPRESS1 : CALL KEYPRESS16 ;Neu co phim an thi co c=1

JC KEYPRESS2 ;kiem tra tiep neu c = 1

RET ;Neu khong co phim nhan thi co c=0

KEYPRESS2: DJNZ LAP,KEYPRESS1 ;Quay ve lap lai chong nayMOV MP1,MP ;Cat noi dung ma phim

KEYPRESS3: MOV LAP,#10 ;Nhap so dem 10 lan cho nha phimKEYPRESS4: CALL KEYPRESS16 ;Co phim nhan hay khong

; CALL HIENTHI


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JC KEYPRESS3 ;Co thi kiem tra lai

DJNZ LAP,KEYPRESS4 ;Khong thi lap lai 50 lan va dambao

MOV MP,MP1RET ;ket thuc mot chuong trinh con

;xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxKEYPRESS16: MOV MP,#00

MOV DEM,#4MOV C3C0,#11111110B ;C0=0

KEYP2: MOV P3,C3C0

MOV A,P3 ;P3=H3H2H1H0C3C2C1C0ANL A,#0F0H ;(A)=H3H2H1H0-0000SWAP A ;(A)= 0000 - H3H2H1H0CJNE A,#00001111B,KEYP1 ;NHAY NEU CO NHANMOV A,MPADD A,#4MOV MP,A

MOV A,C3C0 ;ROTATE LEFTRL AMOV C3C0,ADJNZ DEM,KEYP2MOV MP,#0FFH ;THOAT KHONG NHAN PHIMCLR CRET

KEYP3: INC MPKEYP1: RRC A

JC KEYP3SETB CRET

$INCLUDE(TV_DELAY.ASM)


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END

BAI SO 6-2: CHNG TRNH IEU KHIEN ONG C DC QUAY THUAN NGHCH COPHM IEU KHIEN DUNG BAN PHM MA TRAN

START LA PHM SO 0, STOP LA PHM SO 1, INV LA PHM SO 2PHM SO 0 CO MA LA 00H, PHM SO 1 CO MA LA 01H, PHM SO 2 CO MA LA 02H,

LU O IEU KHIENBEGIN

STOP ?

S

NG C NGNG

START ?

NG C QUAY THUN

INV ?

END

S

S

O CHIU NG C

VIET CHNG TRNH;CHUONG TRINH DIEU KHIEN DONG CO DC DUNG BAN PHIM MA TRAN

START EQU 00H ;PHIM 0STOP EQU 01H ;PHIM 1INV EQU 02H ;PHIM 2DC1 BIT P1.0DC2 BIT P1.2

ORG 0000H


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MOV P1,#00LB: CLR DC1

CLR DC2

LB1: CALL KEYPRESS ;GOI QUET PHIM TRONG THU VIENMOV A,MPCJNE A,#START,LB1

SETB DC1 ;CHO DONG CO CHAYLB2: CALL KEYPRESS ;GOI QUET PHIM TRONG THU VIEN

MOV A,MPCJNE A,#STOP,LB3

JMP LB ;NHAY VE DUNG DONG CO

LB3: CJNE A,#INV,LB2CPL DC1CPL DC2JMP LB2

$INCLUDE(TV_KEY16.ASM)

END

baigiang 005 vxl banphim nutnhan

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