當前位置:編程學習大全網 - 源碼下載 - PIC16F877A的4*4矩陣式按鍵程序該怎麽寫

PIC16F877A的4*4矩陣式按鍵程序該怎麽寫

電腦上面只有51的,不行的話明天有877A的,現在的這臺電腦上面沒有源代碼

4×4矩陣式鍵盤識別技術

實驗任務

如圖4.14.2所示,用AT89S51的並行口P1接4×4矩陣鍵盤,以P1.0-P1.3作輸入線,以P1.4-P1.7作輸出線;在數碼管上顯示每個按鍵的“0-F”序號。對應的按鍵的序號排列如圖4.14.1所示

圖4.14.1

硬件電路原理圖

圖4.14.2

系統板上硬件連線

把“單片機系統“區域中的P3.0-P3.7端口用8芯排線連接到“4X4行列式鍵盤”區域中的C1-C4 R1-R4端口上;

把“單片機系統”區域中的P0.0/AD0-P0.7/AD7端口用8芯排線連接到“四路靜態數碼顯示模塊”區域中的任壹個a-h端口上;要求:P0.0/AD0對應著a,P0.1/AD1對應著b,……,P0.7/AD7對應著h。

程序設計內容

4×4矩陣鍵盤識別處理

每個按鍵有它的行值和列值 ,行值和列值的組合就是識別這個按鍵的編碼。矩陣的行線和列線分別通過兩並行接口和CPU通信。每個按鍵的狀態同樣需變成數字量“0”和“1”,開關的壹端(列線)通過電阻接VCC,而接地是通過程序輸出數字“0”實現的。鍵盤處理程序的任務是:確定有無鍵按下,判斷哪壹個鍵按下,鍵的功能是什麽;還要消除按鍵在閉合或斷開時的抖動。兩個並行口中,壹個輸出掃描碼,使按鍵逐行動態接地,另壹個並行口輸入按鍵狀態,由行掃描值和回饋信號***同形成鍵編碼而識別按鍵,通過軟件查表,查出該鍵的功能。

程序框圖

圖4.14.3

匯編源程序

KEYBUF EQU 30H

ORG 00H

START: MOV KEYBUF,#2

WAIT:

MOV P3,#0FFH

CLR P3.4

MOV A,P3

ANL A,#0FH

XRL A,#0FH

JZ NOKEY1

LCALL DELY10MS

MOV A,P3

ANL A,#0FH

XRL A,#0FH

JZ NOKEY1

MOV A,P3

ANL A,#0FH

CJNE A,#0EH,NK1

MOV KEYBUF,#0

LJMP DK1

NK1: CJNE A,#0DH,NK2

MOV KEYBUF,#1

LJMP DK1

NK2: CJNE A,#0BH,NK3

MOV KEYBUF,#2

LJMP DK1

NK3: CJNE A,#07H,NK4

MOV KEYBUF,#3

LJMP DK1

NK4: NOP

DK1:

MOV A,KEYBUF

MOV DPTR,#TABLE

MOVC A,@A+DPTR

MOV P0,A

DK1A: MOV A,P3

ANL A,#0FH

XRL A,#0FH

JNZ DK1A

NOKEY1:

MOV P3,#0FFH

CLR P3.5

MOV A,P3

ANL A,#0FH

XRL A,#0FH

JZ NOKEY2

LCALL DELY10MS

MOV A,P3

ANL A,#0FH

XRL A,#0FH

JZ NOKEY2

MOV A,P3

ANL A,#0FH

CJNE A,#0EH,NK5

MOV KEYBUF,#4

LJMP DK2

NK5: CJNE A,#0DH,NK6

MOV KEYBUF,#5

LJMP DK2

NK6: CJNE A,#0BH,NK7

MOV KEYBUF,#6

LJMP DK2

NK7: CJNE A,#07H,NK8

MOV KEYBUF,#7

LJMP DK2

NK8: NOP

DK2:

MOV A,KEYBUF

MOV DPTR,#TABLE

MOVC A,@A+DPTR

MOV P0,A

DK2A: MOV A,P3

ANL A,#0FH

XRL A,#0FH

JNZ DK2A

NOKEY2:

MOV P3,#0FFH

CLR P3.6

MOV A,P3

ANL A,#0FH

XRL A,#0FH

JZ NOKEY3

LCALL DELY10MS

MOV A,P3

ANL A,#0FH

XRL A,#0FH

JZ NOKEY3

MOV A,P3

ANL A,#0FH

CJNE A,#0EH,NK9

MOV KEYBUF,#8

LJMP DK3

NK9: CJNE A,#0DH,NK10

MOV KEYBUF,#9

LJMP DK3

NK10: CJNE A,#0BH,NK11

MOV KEYBUF,#10

LJMP DK3

NK11: CJNE A,#07H,NK12

MOV KEYBUF,#11

LJMP DK3

NK12: NOP

DK3:

MOV A,KEYBUF

MOV DPTR,#TABLE

MOVC A,@A+DPTR

MOV P0,A

DK3A: MOV A,P3

ANL A,#0FH

XRL A,#0FH

JNZ DK3A

NOKEY3:

MOV P3,#0FFH

CLR P3.7

MOV A,P3

ANL A,#0FH

XRL A,#0FH

JZ NOKEY4

LCALL DELY10MS

MOV A,P3

ANL A,#0FH

XRL A,#0FH

JZ NOKEY4

MOV A,P3

ANL A,#0FH

CJNE A,#0EH,NK13

MOV KEYBUF,#12

LJMP DK4

NK13: CJNE A,#0DH,NK14

MOV KEYBUF,#13

LJMP DK4

NK14: CJNE A,#0BH,NK15

MOV KEYBUF,#14

LJMP DK4

NK15: CJNE A,#07H,NK16

MOV KEYBUF,#15

LJMP DK4

NK16: NOP

DK4:

MOV A,KEYBUF

MOV DPTR,#TABLE

MOVC A,@A+DPTR

MOV P0,A

DK4A: MOV A,P3

ANL A,#0FH

XRL A,#0FH

JNZ DK4A

NOKEY4:

LJMP WAIT

DELY10MS:

MOV R6,#10

D1: MOV R7,#248

DJNZ R7,$

DJNZ R6,D1

RET

TABLE: DB 3FH,06H,5BH,4FH,66H,6DH,7DH,07H

DB 7FH,6FH,77H,7CH,39H,5EH,79H,71H

END

C語言源程序

#include <AT89X51.H>

unsigned char code table[]={0x3f,0x06,0x5b,0x4f,

0x66,0x6d,0x7d,0x07,

0x7f,0x6f,0x77,0x7c,

0x39,0x5e,0x79,0x71};

unsigned char temp;

unsigned char key;

unsigned char i,j;

void main(void)

{

while(1)

{

P3=0xff;

P3_4=0;

temp=P3;

temp=temp & 0x0f;

if (temp!=0x0f)

{

for(i=50;i>0;i--)

for(j=200;j>0;j--);

temp=P3;

temp=temp & 0x0f;

if (temp!=0x0f)

{

temp=P3;

temp=temp & 0x0f;

switch(temp)

{

case 0x0e:

key=7;

break;

case 0x0d:

key=8;

break;

case 0x0b:

key=9;

break;

case 0x07:

key=10;

break;

}

temp=P3;

P1_0=~P1_0;

P0=table[key];

temp=temp & 0x0f;

while(temp!=0x0f)

{

temp=P3;

temp=temp & 0x0f;

}

}

}

P3=0xff;

P3_5=0;

temp=P3;

temp=temp & 0x0f;

if (temp!=0x0f)

{

for(i=50;i>0;i--)

for(j=200;j>0;j--);

temp=P3;

temp=temp & 0x0f;

if (temp!=0x0f)

{

temp=P3;

temp=temp & 0x0f;

switch(temp)

{

case 0x0e:

key=4;

break;

case 0x0d:

key=5;

break;

case 0x0b:

key=6;

break;

case 0x07:

key=11;

break;

}

temp=P3;

P1_0=~P1_0;

P0=table[key];

temp=temp & 0x0f;

while(temp!=0x0f)

{

temp=P3;

temp=temp & 0x0f;

}

}

}

P3=0xff;

P3_6=0;

temp=P3;

temp=temp & 0x0f;

if (temp!=0x0f)

{

for(i=50;i>0;i--)

for(j=200;j>0;j--);

temp=P3;

temp=temp & 0x0f;

if (temp!=0x0f)

{

temp=P3;

temp=temp & 0x0f;

switch(temp)

{

case 0x0e:

key=1;

break;

case 0x0d:

key=2;

break;

case 0x0b:

key=3;

break;

case 0x07:

key=12;

break;

}

temp=P3;

P1_0=~P1_0;

P0=table[key];

temp=temp & 0x0f;

while(temp!=0x0f)

{

temp=P3;

temp=temp & 0x0f;

}

}

}

P3=0xff;

P3_7=0;

temp=P3;

temp=temp & 0x0f;

if (temp!=0x0f)

{

for(i=50;i>0;i--)

for(j=200;j>0;j--);

temp=P3;

temp=temp & 0x0f;

if (temp!=0x0f)

{

temp=P3;

temp=temp & 0x0f;

switch(temp)

{

case 0x0e:

key=0;

break;

case 0x0d:

key=13;

break;

case 0x0b:

key=14;

break;

case 0x07:

key=15;

break;

}

temp=P3;

P1_0=~P1_0;

P0=table[key];

temp=temp & 0x0f;

while(temp!=0x0f)

{

temp=P3;

temp=temp & 0x0f;

}

}

}

}

}

  • 上一篇:散文欣賞:撫今追昔話郵政
  • 下一篇:造夢西遊4白龍馬龍馬技能搭配
  • copyright 2024編程學習大全網