【学习笔记】单片机的40个经典实验之30:点阵式 LED“0-9”数字显示技术

  一. 实验任务

  利用 8X8 点阵显示数字 0 到 9 的数字。

   . 电路原理图


  三. 硬件系统连线

  (1). 把“单片机系统”区域中的 P1 端口用 8 芯排芯连接到“点阵模块” 区域中的“DR1-DR8”端口上;

  (2). 把“单片机系统”区域中的 P3 端口用 8 芯排芯连接到“点阵模块” 区域中的“DC1-DC8”端口上;

  四. 程序设计内容

  (1). 数字 0-9 点阵显示代码的形成

  如下图所示,假设显示数字“0”

  1 2 3 4 5 6 7 8

  00 00 3E 41 41 41 3E 00

  因此,形成的列代码为 00H,00H,3EH,41H,41H,3EH,00H,00H;只要把这 些代码分别送到相应的列线上面,即可实现“0”的数字显示。

  送显示代码过程如下所示

  送第一列线代码到 P3 端口,同时置第一行线为“0”,其它行线为“1”,延时 2ms 左右,送第二列线代码到 P3 端口,同时置第二行线为“0”,其它行线为 “1”,延时 2ms 左右,如此下去,直到送完最后一列代码,又从头开始送。

  数字“1”代码建立如下图所示

1 2 3 4 5 6 7 8

  其显示代码为 00H,00H,00H,00H,21H,7FH,01H,00H

  数字“2”代码建立如下图所示

  1 2 3 4 5 6 7 8

  00H,00H,27H,45H,45H,45H,39H,00H

  数字“3”代码建立如下图所示

  1 2 3 4 5 6 7 8

  00H,00H,22H,49H,49H,49H,36H,00H

  数字“4”代码建立如下图所示

  1 2 3 4 5 6 7 8

  00H,00H,0CH,14H,24H,7FH,04H,00H

  数字“5”代码建立如下图所示

  1 2 3 4 5 6 7 8

  00H,00H,72H,51H,51H,51H,4EH,00H

  数字“6”代码建立如下图所示

  1 2 3 4 5 6 7 8

  00H,00H,3EH,49H,49H,49H,26H,00H

  数字“7”代码建立如下图所示

  1 2 3 4 5 6 7 8

  00H,00H,40H,40H,40H,4FH,70H,00H

  数字“8”代码建立如下图所示

  1 2 3 4 5 6 7 8

  00H,00H,36H,49H,49H,49H,36H,00H

  数字“9”代码建立如下图所示

  1 2 3 4 5 6 7 8

  00H,00H,32H,49H,49H,49H,3EH,00H

  五. 汇编源程序

  TIM EQU 30H

  CNTA EQU 31H

  CNTB EQU 32H

  ORG 00H

  LJMP START

  ORG 0BH

  LJMP T0X

  ORG 30H

  START: MOV TIM,#00H

  MOV CNTA,#00H

  MOV CNTB,#00H

  MOV TMOD,#01H

  MOV TH0,#(65536-4000)/256

  MOV TL0,#(65536-4000) MOD 256

  SETB TR0

  SETB ET0

  SETB EA

  SJMP $

  T0X:

  MOV TH0,#(65536-4000)/256

  MOV TL0,#(65536-4000) MOD 256

  MOV DPTR,#TAB

  MOV A,CNTA

  MOVC A,@A+DPTR

  MOV P3,A

  MOV DPTR,#DIGIT

  MOV A,CNTB

  MOV B,#8

  MUL AB

  ADD A,CNTA

  MOVC A,@A+DPTR

  MOV P1,A

  INC CNTA

  MOV A,CNTA

  CJNE A,#8,NEXT

  MOV CNTA,#00H

  NEXT: INC TIM

  MOV A,TIM

  CJNE A,#250,NEX

  MOV TIM,#00H

  INC CNTB

  MOV A,CNTB

  CJNE A,#10,NEX

  MOV CNTB,#00H

  NEX: RETI

  TAB: DB 0FEH,0FDH,0FBH,0F7H,0EFH,0DFH,0BFH,07FH

  DIGIT: DB 00H,00H,3EH,41H,41H,41H,3EH,00H

  DB 00H,00H,00H,00H,21H,7FH,01H,00H

  DB 00H,00H,27H,45H,45H,45H,39H,00H

  DB 00H,00H,22H,49H,49H,49H,36H,00H

  DB 00H,00H,0CH,14H,24H,7FH,04H,00H

  DB 00H,00H,72H,51H,51H,51H,4EH,00H

  DB 00H,00H,3EH,49H,49H,49H,26H,00H

  DB 00H,00H,40H,40H,40H,4FH,70H,00H

  DB 00H,00H,36H,49H,49H,49H,36H,00H

  DB 00H,00H,32H,49H,49H,49H,3EH,00H

  END

  六. C 语言源程序

  #include <AT89X52.H>

  unsigned char code tab[]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f};

  unsigned char code digittab[10][8]={

  {0x00,0x00,0x3e,0x41,0x41,0x41,0x3e,0x00}, //0

  {0x00,0x00,0x00,0x00,0x21,0x7f,0x01,0x00}, //1

  {0x00,0x00,0x27,0x45,0x45,0x45,0x39,0x00}, //2

  {0x00,0x00,0x22,0x49,0x49,0x49,0x36,0x00}, //3

  {0x00,0x00,0x0c,0x14,0x24,0x7f,0x04,0x00}, //4

  {0x00,0x00,0x72,0x51,0x51,0x51,0x4e,0x00}, //5

  {0x00,0x00,0x3e,0x49,0x49,0x49,0x26,0x00}, //6

  {0x00,0x00,0x40,0x40,0x40,0x4f,0x70,0x00}, //7

  {0x00,0x00,0x36,0x49,0x49,0x49,0x36,0x00}, //8

  {0x00,0x00,0x32,0x49,0x49,0x49,0x3e,0x00} //9

  };

  unsigned int timecount;

  unsigned char cnta;

  unsigned char cntb;

  void main(void)

  {

  TMOD=0x01;

  TH0=(65536-3000)/256;

  TL0=(65536-3000)%256;

  TR0=1;

  ET0=1;

  EA=1;

  while(1)

  {;

  }

  }

  void t0(void) interrupt 1 using 0

  {

  TH0=(65536-3000)/256;

  TL0=(65536-3000)%256;

  P3=tab[cnta];

  P1=digittab[cntb][cnta];

  cnta++;

  if(cnta==8)

  {

  cnta=0;

  }

  timecount++;

  if(timecount==333)

  {

  timecount=0;

  cntb++;

  if(cntb==10)

  {

  cntb=0;

  }

  }

  }

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