单片机点阵式LED显示技术数字“0－9”

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”

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

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

00H，00H，27H，45H，45H，45H，39H，00H
数字“3”代码建立如下图所示

00H，00H，22H，49H，49H，49H，36H，00H
数字“4”代码建立如下图所示

00H，00H，0CH，14H，24H，7FH，04H，00H
数字“5”代码建立如下图所示

00H，00H，72H，51H，51H，51H，4EH，00H
数字“6”代码建立如下图所示

00H，00H，3EH，49H，49H，49H，26H，00H
数字“7”代码建立如下图所示

00H，00H，40H，40H，40H，4FH，70H，00H
数字“8”代码建立如下图所示

00H，00H，36H，49H，49H，49H，36H，00H
数字“9”代码建立如下图所示

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)%6;
TR0=1;
ET0=1;
EA=1;
while(1)
{;
}
}
void t0(void) interrupt 1 using 0
{
TH0=(65536-3000)/256;
TL0=(65536-3000)%6;
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;
}
}
}