不到100行代碼，51單片機32位數碼管動態掃描顯示

ID:730135 · 發表于 2021-7-10 22:03

51單片機使用573鎖存器，32位數碼管動態顯示，時間仿真顯示
51單片機到底能帶動多少位數碼管呢？

本次演示proteus版本為8.6

單片機源程序如下:

#include "AT89X52.H"
/* */
code unsigned char reg_code[] = {0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90};
code unsigned char reg_code1[] = {0x3f,0x06,0x5b,0x4f,0x66,0x6d,0x7d,0x07,0x7f,0x6f,0x40}; //陰碼（共陰的數碼管的段碼）
#define DELAY_N 1000
unsigned char time[3] = {12,34,50};
void show_time(void);
void get_time(void);
void main(void)
{
unsigned short j;
P2 = 0;
P1_0 = 1;
P1_1 = 1;
P1_2 = 1;
P0 = reg_code1[1];
while(1)
{
show_time();
get_time(); //獲取時間
}
}
void show_time(void)
{
char i;
unsigned short j;
unsigned char buf[8]={10,10,10,10,10,10,10,10};
unsigned char buf1[8]={1,2,3,4,5,6,7,8};
unsigned char buf2[8]={3,2,1,4,5,6,3,2};
buf[0] = time[0]/10;
buf[1] = time[0]%10; //除以10，得到余數
buf[3] = time[1]/10;
buf[4] = time[1]%10;
buf[6] = time[2]/10;
buf[7] = time[2]%10;
for (i=0;i<8;i++)
{
P2 = i<<2;
P1_0 = 1; //讓P0的數據通過573
P0 = reg_code1[buf[i]];
P1_0 = 0; //鎖住這個數據，P0無法通過573
//for(j=0;j<60000;j++);
P1_1 = 1;
P0 = reg_code1[buf1[i]];
P1_1 = 0; //鎖住這個數據
// for(j=0;j<60000;j++);
P1_2 = 1;
P0 = reg_code1[buf2[i]];
P1_2 = 0; //鎖住這個數據
for(j=0;j<200;j++);
P0 = 0; //熄滅所有的段
}
}
//獲得時間
void get_time(void)
{
static unsigned char j;
if(j++>50)
{
j = 0;
time[2]++; //秒
if(time[2] > 59)
{
time[2] = 0;
time[1] ++;
if(time[1] > 59)
{
time[1] = 0;
time[0] ++;
if(time[0] > 23)
{
time[0] = 0;
}
}
}
}
}