LED 8X8的显示屏怎么产生0~9

只看该作者 · 2010-4-20 16:53

我现在学LED显示屏，不知道怎么操作，很郁闷：
1.不知道用什么器件，还需要锁存器吗？
2.哪位大哥能传个C的程序和电路图看看

谢谢！

只看该作者 · 2010-4-20 17:02

本帖最后由 liao_fangxing 于 2010-4-20 17:11 编辑

1、直接用I/O驱动或用HT1623等驱动IC
2、上网荡吧

显示屏相当于控制大量（如8*8=64个）发光管亮与不亮来组合显示一个内容如下图

只看该作者 · 2010-4-20 17:18

谢谢！可是我试了，显示不是稳定，不知道怎么办

只看该作者 · 2010-4-20 19:18

去看看振南的视频吧！

只看该作者 · 2010-4-20 19:50

你要的是不是这么效果呢？
http://v.youku.com/v_playlist/f4321694o1p1.html

只看该作者 · 2010-4-20 20:21

#include<reg52.h>
#include<intrins.h>
#define uchar unsigned char
#define uint unsigned int

uchar flag,num;
uchar code dula[]={0xfd,0xfb,0xf7,0xef,0xdf};
uchar code wela[]={0x3c,0x24,0x24,0x24,0x3c,
   0x18,0x18,0x18,0x18,0x18,
   0x3c,0x20,0x3c,0x04,0x3c,
   0x3c,0x20,0x3c,0x20,0x3c,
   0x1c,0x1c,0x3c,0x18,0x18,
         0x3c,0x04,0x3c,0x20,0x3c,
   0x3c,0x04,0x3c,0x24,0x3c,
   0x3c,0x20,0x20,0x20,0x20,
   0x3c,0x24,0x3c,0x24,0x3c,
   0x3c,0x24,0x3c,0x20,0x3c,};

void delayms(uint x)
{
uint i,j;
for(i=x;i>0;i--)
for(j=110;j>0;j--);
}

void init()
{
TMOD=0X01;
TH0=(65536-50000)/256;
TL0=(65536-50000)%256;
EA=1;
TR0=1;
ET0=1;
}
void main()
{
uchar i;
init();
while(1)
{
for(i=0;i<5;i++)
{
P3=dula[i];
P1=wela[flag+i];
delayms(10);
}
}
}

void t0_time() interrupt 1
{
TH0=(65536-50000)/256;
TL0=(65536-50000)%256;
num++;
if(num==20)
{
num=0;
flag=flag+5;
if(flag>45)
flag=0;
}
}

只看该作者 · 2010-4-20 20:22

我自己写出来，谢谢大家！突然之间明白过来怎么显示！

只看该作者 · 2010-4-21 10:20

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
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;
}
}
} 还有汇编的希望
能帮助你