点阵式LED“0－9”数字显示技术

1． 实验任务

利用8X8点阵显示数字0到9的数字。

2． 电路原理图

图4.25.1

3． 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；

4． 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

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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

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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

			●	●	●
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		●	●	●	●
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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

			●	●	●
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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

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		●	●	●	●	●
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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

		●	●	●	●	●
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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

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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

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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

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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

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00H，00H，32H，49H，49H，49H，3EH，00H

5． 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

6．C语言源程序

#include

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;

        }

    }

}