AT89C518个led灯实现定时器流水灯,1秒亮灭
时间: 2024-02-13 14:06:12 浏览: 79
流水灯_流水灯_pine1oe_
好的,我可以帮你解决这个问题。首先,AT89C518是一种单片机,我们可以通过编写程序来控制它的IO口,从而实现LED灯的控制。以下是一个简单的程序示例,实现了定时器流水灯,每个LED灯亮灭时间为1秒:
```C
#include <REG52.H>
sbit LED1 = P1^0;
sbit LED2 = P1^1;
sbit LED3 = P1^2;
sbit LED4 = P1^3;
sbit LED5 = P1^4;
sbit LED6 = P1^5;
sbit LED7 = P1^6;
sbit LED8 = P1^7;
void init_timer0(void);
void main()
{
init_timer0();
while(1)
{
LED1 = 1;
LED2 = 0;
LED3 = 0;
LED4 = 0;
LED5 = 0;
LED6 = 0;
LED7 = 0;
LED8 = 0;
delay(1000);
LED1 = 0;
LED2 = 1;
LED3 = 0;
LED4 = 0;
LED5 = 0;
LED6 = 0;
LED7 = 0;
LED8 = 0;
delay(1000);
LED1 = 0;
LED2 = 0;
LED3 = 1;
LED4 = 0;
LED5 = 0;
LED6 = 0;
LED7 = 0;
LED8 = 0;
delay(1000);
LED1 = 0;
LED2 = 0;
LED3 = 0;
LED4 = 1;
LED5 = 0;
LED6 = 0;
LED7 = 0;
LED8 = 0;
delay(1000);
LED1 = 0;
LED2 = 0;
LED3 = 0;
LED4 = 0;
LED5 = 1;
LED6 = 0;
LED7 = 0;
LED8 = 0;
delay(1000);
LED1 = 0;
LED2 = 0;
LED3 = 0;
LED4 = 0;
LED5 = 0;
LED6 = 1;
LED7 = 0;
LED8 = 0;
delay(1000);
LED1 = 0;
LED2 = 0;
LED3 = 0;
LED4 = 0;
LED5 = 0;
LED6 = 0;
LED7 = 1;
LED8 = 0;
delay(1000);
LED1 = 0;
LED2 = 0;
LED3 = 0;
LED4 = 0;
LED5 = 0;
LED6 = 0;
LED7 = 0;
LED8 = 1;
delay(1000);
}
}
void init_timer0(void)
{
TMOD &= 0xF0;
TMOD |= 0x01;
TH0 = (65536-50000)/256;
TL0 = (65536-50000)%256;
ET0 = 1;
EA = 1;
TR0 = 1;
}
void timer0_isr(void) interrupt 1
{
TH0 = (65536-50000)/256;
TL0 = (65536-50000)%256;
}
void delay(unsigned int ms)
{
unsigned int i,j;
for(i=0;i<ms;i++)
for(j=0;j<1000;j++);
}
```
在以上代码中,我们使用了定时器0来实现1秒的定时,同时使用了延时函数进行LED灯的亮灭控制。具体实现中,我们将8个LED灯分别接在P1口的8个引脚上,并依次点亮和熄灭。我们在定时器0的中断服务程序中,不断地更新定时器的初值,从而实现定时器的连续工作。
需要注意的是,以上代码只是一个简单的例子,如果您需要更精确的定时器控制,可能需要根据具体情况进行调整。希望对您有所帮助!
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