51单片机使用573锁存器,32位数码管动态显示,时间仿真显示
51单片机到底能带动多少位数码管呢?
本次演示proteus版本为8.6
单片机源程序如下:
- #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}; //阴码 (共阴的数码管的段码)
- #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(); //获取时间
- }
-
- }
- 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,得到余数
- 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; //让P0的数据通过573
- P0 = reg_code1[buf[i]];
- P1_0 = 0; //锁住这个数据,P0无法通过573
- //for(j=0;j<60000;j++);
-
-
- P1_1 = 1;
- P0 = reg_code1[buf1[i]];
- P1_1 = 0; //锁住这个数据
- // for(j=0;j<60000;j++);
-
- P1_2 = 1;
- P0 = reg_code1[buf2[i]];
- P1_2 = 0; //锁住这个数据
-
- for(j=0;j<200;j++);
- P0 = 0; //熄灭所有的段
-
- }
- }
- //获得时间
- 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;
- }
- }
- }
-
- }
-
- }
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