怎样用51单片机和led8*8矩阵进行字符汉字显示?

2024-11-20 09:43:49
推荐回答(3个)
回答(1):

8*8也就能显示字符,显示汉字比较吃力。

#include

#include

#define uchar unsigned char

#define uint unsigned int

uchar code Table_of_Digits[]=

{

0x00,0x3e,0x41,0x41,0x41,0x3e,0x00,0x00, //0

0x00,0x00,0x00,0x21,0x7f,0x01,0x00,0x00, //1

0x00,0x27,0x45,0x45,0x45,0x39,0x00,0x00, //2

0x00,0x22,0x49,0x49,0x49,0x36,0x00,0x00, //3

0x00,0x0c,0x14,0x24,0x7f,0x04,0x00,0x00, //4

0x00,0x72,0x51,0x51,0x51,0x4e,0x00,0x00, //5

0x00,0x3e,0x49,0x49,0x49,0x26,0x00,0x00, //6

0x00,0x40,0x40,0x40,0x4f,0x70,0x00,0x00, //7

0x00,0x36,0x49,0x49,0x49,0x36,0x00,0x00, //8

0x00,0x32,0x49,0x49,0x49,0x3e,0x00,0x00, //9

0xff,0x81,0x81,0x81,0x81,0x81,0x81,0xff

};

uchar code xdat[8]={0x80,0x40,0x20,0x10,0x08,0x04,0x02,0x01};

uchar code ydat[8]={0x01,0x02,0x04,0x08,0x10,0x20,0x40,0x80};

uchar i=0,j=0,t=0,Num_Index,key,xi,yi;


sbit we1=P1^1;

sbit we2=P1^3;

//主程序

void main()

{

//P1=0x80;

Num_Index=0; //从0 开始显示

TMOD=0x01; //T0 方式0

TH0=(65536-2000)/256; //2ms 定时

TL0=(65536-2000)%256;

IE=0x82;

key=0;

xi=0;

yi=0;

EX0=1;

IT0=1;

TR0=1; //启动T0

while(1);

}

//T0 中断函数

void ext_int0() interrupt 0

{

key++;

key&=0x03;

}

void LED_Screen_Display() interrupt 1

{

TH0=(65536-2000)/256; //2ms 定时

TL0=(65536-2000)%256;

switch(key)

{

case 0:

P0=0xff;

we1=1;

P0=~Table_of_Digits[Num_Index*8+i];

we1=0;

P0=0xff; //输出位码和段码

we2=1;

P0=xdat[i];

we2=0;

if(++i==8) i=0; //每屏一个数字由8 个字节构成

if(++t==250) //每个数字刷新显示一段时间

{

t=0;

if(++Num_Index==10) Num_Index=0; //显示下一个数字

}

break;

case 1:

we1=1;

P0=~xdat[xi];

we1=0;

we2=1;

P0=ydat[yi];

we2=0;

if(++t==250) //每个数字刷新显示一段时间

{

t=0;

yi++;

if(yi>7){yi=0;xi++;}

if(xi>7)xi=0;

}

break;

case 2:

we1=1;

P0=0x00;

we1=0;

P0=0xff; //输出位码和段码

we2=1;

P0=xdat[i];

we2=0;

if(++t==250) //每个数字刷新显示一段时间

{

if(++i==8) i=0; //每屏一个数字由8 个字节构成

t=0;

}

break;

default:

key=0;

i=0;

j=0;

t=0;

xi=0;

yi=0;

Num_Index=0;

we1=1;

P0=0xff;

we1=0;

we2=1;

P1=0x80;

we2=0;

break;

}

}

回答(2):

首先,刷新的的速度肯定是要提高的,速度不够肯定是闪的。
其次,你没提用什么驱动电路,按理总得用595之类的驱动一下吧,直接用IO去扫每个BIT是不是太累了。

回答(3):