8X8点阵原理及驱动
2010-02-24 00:14阅读：
8X8点阵的实际外观图如下：
一共64个发光点构成，上下两排引脚，每排8个，一共16个
点阵侧面有文字的那面对准自己，
（图 1）
8X8点阵的显示原理：
8X8点阵共由64个发光二极管组成，且每个发光二极管是
放置在行线和列线的交叉点上，当对应的某一行置1电平，某一列置0电平，则相应的
二极管就亮；如要将第一个点点亮，则1脚接高电平 a
脚接低电平，则第一个点就亮了；
如果要将第一行点亮，则第1脚要接高电平，而（a、b、c、d、e、f、g、h
）
这些引脚接低电平，那么第一行就会点亮；如要将第一列点亮，
则第a脚接低电平，而（1、2、3、4、5、6、7、8）接高电平，那么第一列就会点亮.
（图 2）
8*8点阵实际的引脚（图一）的不是一排阳极，一排阴极，（就像键盘一样不是abcd按顺序排的）所以焊接有点麻烦，
8×8LED点阵引脚图如下：
1,2,3,4,5,6,7,8是行（阳极）
a,b,c,d,e,f,g,h是列（阴极）//此处说的都是行阳型8*8点阵，行阴的略
外面的1~16就是实际的管脚，见图一
移位锁存器74HC595原理
74HC595是一个串入并出的芯片，通过一个for(i=0;i<8;i++)来存储数据。
具体来说就是第一个时钟信号来到时低位的数据向高位挪动一位，
在这个程序中是SH_CP 信号，当SH_CP是一个上跳沿时，
传入的形参Data与0x80相与，得到的数为1,则通过SDATA置1,否通过置为0,
并存储在74HC595的相应位置（最低位即Q0那）上，
DS内部也自动左移一位数据
然后dat向左移一位，使次高位变为最高位与0x80相与，
并存储。通过8次后，就可以得到数据，并存储在Q0~Q7中了，
这时ST_CP一个上跳沿，数据即送出去了.
与8*8LED连接如下：
显示0~59
C语言代码：
#include<AT89X52.h>
#define uchar unsigned
char
#define uint unsigned int
void Ser_IN(uchar Data)
{
uchar i;
for(i=0;i<8;i++)
{
SH_CP=0;
//先置为低
DS=Data&0x80;//取数据的最高位
Data<<=1;
//讲数据的次高位移到最高位，为下一次取数据做准备
SH_CP=1; //
再置为高，产生移位时钟上升沿，储存器里的数据移位，数据输入
}
}
void Par_OUT(void)
{
ST_CP=0; //先置为低
ST_CP=1; //再置为高，产生时钟上升沿，上升沿时，数据并行输出
}
uchar code
tab[]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f}; //列
uchar code tabdigit[60][8]={ //字模取模方向 阴码
逆向逐列式
{0x00,0x3E,0x41,0x41,0x3E,0x00,0x00,0x00},
{0x00,0x42,0x7F,0x40,0x00,0x00,0x00,0x00},
{0x00,0x62,0x51,0x49,0x46,0x00,0x00,0x00},
{0x00,0x22,0x49,0x49,0x36,0x00,0x00,0x00},
{0x00,0x38,0x26,0x7F,0x20,0x00,0x00,0x00},
{0x00,0x4F,0x49,0x49,0x31,0x00,0x00,0x00},
{0x00,0x3E,0x49,0x49,0x32,0x00,0x00,0x00},
{0x00,0x03,0x71,0x09,0x07,0x00,0x00,0x00},
{0x00,0x36,0x49,0x49,0x36,0x00,0x00,0x00},
{0x00,0x26,0x49,0x49,0x3E,0x00,0x00,0x00},
{0x02,0xFF,0x00,0x00,0xFF,0x81,0xFF,0x00},
{0x02,0xFF,0x00,0x02,0xFF,0x00,0x00,0x00},
{0x02,0xFF,0x00,0xF3,0x91,0x99,0xCF,0x00},
{0x02,0xFF,0x00,0x42,0x89,0x89,0x89,0x76},
{0x02,0xFF,0x00,0x1C,0x13,0x10,0xFF,0x10},
{0x02,0xFF,0x00,0x4F,0x89,0x89,0x89,0x71},
{0x02,0xFF,0x00,0x7C,0x92,0x91,0x91,0x60},
{0x02,0xFF,0x00,0x01,0xF9,0x05,0x03,0x00},
{0x02,0xFF,0x00,0x76,0x89,0x89,0x89,0x76},
{0x02,0xFF,0x00,0x4E,0x91,0x91,0x91,0x7E},
{0xE2,0x91,0x89,0x86,0x00,0xFF,0x81,0xFF},
{0xE2,0x91,0x89,0x86,0x00,0x02,0xFF,0x00},
{0x79,0x49,0x4F,0x00,0x7A,0x4A,0x4E,0x00},
{0x79,0x49,0x4F,0x00,0x49,0x49,0x49,0x36},
{0x79,0x49,0x4F,0x00,0x1E,0x10,0xFF,0x10},
{0x79,0x49,0x4F,0x00,0x4E,0x4A,0x7A,0x00},
{0x79,0x49,0x4F,0x00,0xFE,0x92,0x92,0x60},
{0x79,0x49,0x4F,0x00,0x79,0x05,0x03,0x00},
{0x79,0x49,0x4F,0x00,0x76,0x89,0x89,0x76},
{0x79,0x49,0x4F,0x00,0x4E,0x91,0x91,0x7E},
{0x42,0x89,0x89,0x76,0x00,0xFE,0x82,0xFE},
{0x42,0x89,0x89,0x76,0x00,0x02,0xFF,0x00},
{0x42,0x89,0x89,0x76,0x00,0xF9,0x89,0x8F},
{0x42,0x89,0x89,0x76,0x00,0x89,0x89,0xFF},
{0x42,0x89,0x89,0x76,0x00,0x0F,0x08,0xFF},
{0x42,0x89,0x89,0x76,0x00,0x9E,0x92,0xF2},
{0x42,0x89,0x89,0x76,0x00,0xFF,0x89,0xF9},
{0x42,0x89,0x89,0x76,0x01,0x01,0xFD,0x03},
{0x42,0x89,0x89,0x76,0x00,0xFF,0x89,0xFF},
{0x42,0x89,0x89,0x76,0x00,0xCF,0x89,0xFF},
{0x1F,0x10,0xFF,0x00,0x7E,0x81,0x81,0x7E},
{0x1F,0x10,0xFF,0x00,0x02,0xFF,0x00,0x00},
{0x1F,0x10,0xFF,0x00,0xE2,0x91,0x89,0xC6},
{0x1F,0x10,0xFF,0x00,0x42,0x89,0x89,0x76},
{0x1F,0x10,0xFF,0x00,0x1E,0x10,0xFF,0x10},
{0x1F,0x10,0xFF,0x00,0x8F,0x89,0x89,0xF9},
{0x1F,0x10,0xFF,0x00,0xFF,0x89,0x89,0xF9},
{0x1F,0x10,0xFF,0x00,0x03,0xF9,0x05,0x03},
{0x0F,0x08,0xFF,0x00,0x76,0x89,0x89,0x76},
{0x1F,0x10,0xFF,0x00,0x4E,0x91,0x91,0x7E},
{0x4F,0x49,0x79,0x00,0x3E,0x41,0x41,0x3E},
{0x4F,0x49,0x79,0x00,0x00,0x02,0x7F,0x00},
{0x4F,0x49,0x79,0x00,0x62,0x51,0x49,0x46},
{0x4F,0x49,0x79,0x00,0x42,0x89,0x89,0x76},
{0x4F,0x49,0x79,0x00,0x1E,0x10,0x7E,0x10},
{0x4F,0x49,0x79,0x00,0x4F,0x49,0x79,0x00},
{0x4F,0x49,0x79,0x00,0x7F,0x49,0x79,0x00},
{0x4F,0x49,0x79,0x00,0x01,0x7D,0x03,0x01},
{0x4F,0x49,0x79,0x00,0x76,0x89,0x89,0x76},
{0x4F,0x49,0x79,0x00,0x0E,0x91,0x91,0x7E},
};
uint timecount;
uchar a;
uchar b;
void main(void)
{
TR0=0;
TMOD=0x01;
TH0=(65536-1000)/256;
TL0=(65536-1000)%256;
EA=1;
ET0=1;
TR0=1;
while(1);
}
void t0(void) interrupt 1
{
TH0=(65536-1000)/256;
TL0=(65536-1000)%256;
Ser_IN(tab[a]); //8X8点阵列扫描
Ser_IN(tabdigit[b][a]);//送行扫描数据
Par_OUT(); //显示
a++;
if(a==8)
{
a=0;
}
timecount++;
if(timecount==1000)
{
timecount=0;
b++;
if(b==60)
{
b=0;
}
}
}
