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標題:
4個74HC595+單片機的點陣顯示Proteus仿真圖和程序
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作者:
zcy8520
時間:
2019-6-3 14:58
標題:
4個74HC595+單片機的點陣顯示Proteus仿真圖和程序
0.png
(14.18 KB, 下載次數: 97)
下載附件
2019-6-3 20:42 上傳
單片機源程序如下:
#include<reg51.h>
#define uchar unsigned char
#define uint unsigned int
sbit SH_CP=P1^0;
sbit DS=P1^1;
sbit ST_CP=P1^2;
uchar code tab[]=
{
/*-- 文字: 東 --*/
/*-- 楷體_GB231212; 此字體下對應的點陣為:寬x高=16x16 --*/
0x02,0x00,0x02,0x00,0x02,0x00,0x7F,0xFC,0x04,0x00,0x09,0x00,0x11,0x00,0x21,0x00,
0x3F,0xF8,0x01,0x00,0x09,0x20,0x11,0x10,0x21,0x08,0x41,0x04,0x05,0x00,0x02,0x00,
/*-- 文字: 方 --*/
/*-- 楷體_GB231212; 此字體下對應的點陣為:寬x高=16x16 --*/
0x02,0x00,0x01,0x00,0x01,0x00,0xFF,0xFE,0x04,0x00,0x04,0x00,0x04,0x00,0x07,0xF0,
0x04,0x10,0x04,0x10,0x04,0x10,0x08,0x10,0x08,0x10,0x10,0x10,0x20,0xA0,0x40,0x40,
/*-- 文字: 紅 --*/
/*-- 楷體_GB231212; 此字體下對應的點陣為:寬x高=16x16 --*/
0x10,0x00,0x10,0x00,0x21,0xFC,0x24,0x20,0x44,0x20,0xF8,0x20,0x10,0x20,0x20,0x20,
0x40,0x20,0xFC,0x20,0x40,0x20,0x00,0x20,0x1C,0x20,0xE0,0x20,0x43,0xFE,0x00,0x00,
/*-- 文字: 太 --*/
/*-- 楷體_GB231212; 此字體下對應的點陣為:寬x高=16x16 --*/
0x01,0x00,0x01,0x00,0x01,0x00,0x01,0x00,0x01,0x00,0xFF,0xFE,0x01,0x00,0x01,0x00,
0x02,0x80,0x02,0x80,0x04,0x40,0x04,0x40,0x0A,0x20,0x11,0x10,0x21,0x08,0xC0,0x06,
/*-- 文字: 陽 --*/
/*-- 楷體_GB231212; 此字體下對應的點陣為:寬x高=16x16 --*/
0x00,0x00,0x7C,0x00,0x45,0xFC,0x49,0x04,0x49,0x04,0x51,0x04,0x49,0x04,0x49,0x04,
0x45,0xFC,0x45,0x04,0x45,0x04,0x69,0x04,0x51,0x04,0x41,0x04,0x41,0xFC,0x41,0x04,
/*-- 文字: 升 --*/
/*-- 楷體_GB231212; 此字體下對應的點陣為:寬x高=16x16 --*/
0x01,0x20,0x07,0xA0,0x3C,0x20,0x04,0x20,0x04,0x20,0x04,0x20,0x04,0x20,0xFF,0xFE,
0x04,0x20,0x04,0x20,0x04,0x20,0x08,0x20,0x08,0x20,0x10,0x20,0x20,0x20,0x40,0x20,
};
uchar a[]={0x7f,0xff,0xbf,0xff,0xdf,0xff,0xef,0xff,0xf7,0xff,0xfb,0xff,0xfd,0xff,0xfe,0xff,
0xff,0x7f,0xff,0xbf,0xff,0xdf,0xff,0xef,0xff,0xf7,0xff,0xfb,0xff,0xfd,0xff,0xfe};
void delayms(unsigned int ms)
{
uchar j;
while(ms--)
for(j=0;j<123;j++);
}
void writbyte(uchar dat)
{
uchar k;
for(k=0;k<8;k++)
{
dat=dat<<1;
DS=CY;
SH_CP=0;
SH_CP=1;
}
}
void main()
{
uchar num,temp,n,m,count;
while(1)
{
// for(n=0;n<6;n++)
for(m=0;m<20;m++)//m移動速度
{
temp=0;
for(num=0;num<16;num++)
{
writbyte(tab[2*(num+count)+1]);//行顯示數據高8位
writbyte(tab[2*(num+count)+0]);//低8位
writbyte(a[2*temp+1]);//選擇行高8位
writbyte(a[2*temp]);//選擇
temp++;
if(temp==16)
{temp=0;}
ST_CP=0;
ST_CP=1;
delayms(1);
}
}
count++;
if(count==80)
{
count=0;
}
}
}
復制代碼
仿真原理圖如下(proteus仿真工程文件可到本帖附件中下載)
新建文件夾.rar
(15.84 KB, 下載次數: 91)
2019-6-3 14:58 上傳
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硬件
下載積分: 黑幣 -5
作者:
guodengshan
時間:
2019-6-10 19:15
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