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”

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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

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00H，00H，27H，45H，45H，45H，39H，00H

数字“3”代码建立如下图所示

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00H，00H，22H，49H，49H，49H，36H，00H

数字“4”代码建立如下图所示

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00H，00H，0CH，14H，24H，7FH，04H，00H

数字“5”代码建立如下图所示

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00H，00H，72H，51H，51H，51H，4EH，00H

数字“6”代码建立如下图所示

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00H，00H，3EH，49H，49H，49H，26H，00H

数字“7”代码建立如下图所示

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00H，00H，40H，40H，40H，4FH，70H，00H

数字“8”代码建立如下图所示

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00H，00H，36H，49H，49H，49H，36H，00H

数字“9”代码建立如下图所示

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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 <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;

}