8051 MCU (STC89C52) two countdown timers count synchronously

Here, timer 0 is required to provide 5ms precise timing, so each timing cycle requires (5*10^-3)/(1*10^-6)=5000 times of adding 1 count, so the initial value of timer 0 is set to 65536 - 5000 = 60536 = EA84H.

Here we first implement two countdown timers working simultaneously, and the next step is to consider how to implement two countdown timers working asynchronously.

#include <STC89C5xRC.H>

void delay() //Use timer 0 to achieve 5ms precise timing

{

TMOD = 0x01;

TH0 = 0xEA;

TL0 = 0x84; //65536 - 5000 = 60036

TF0 = 0; //Set the overflow flag to 0

TR0 = 1; //Start timer 0

while(TF0 == 0);

TR0 = 0; // Pause timer 0

}

void disp_digit(int d, int r)

{

unsigned char code DIG_CODE[10] = {0x3f, 0x06, 0x5b, 0x4f, 0x66, 0x6d, 0x7d, 0x07, 0x7f, 0x6f};

//First countdown

//Display the single digit

P2 = 0; // P2 = 0 -> (P24, P23, P22) = (0, 0, 0) -> the first number from the right lights up

P0 = DIG_CODE[d % 10];

delay(); //5ms precise timing

//Display the tens digit

P2 = 1 << 2; //P2 = 0000 0100 -> (P24, P23, P22) = (0, 0, 1) -> the second digit from the right lights up

P0 = DIG_CODE[d / 10];

delay(); //5ms precise timing

//Second countdown

//Display the single digit

P2 = 6 << 2; //P2 = 0001 1000 -> (P24, P23, P22) = (1, 1, 0) -> The seventh number from the right lights up

P0 = DIG_CODE[r % 10];

delay(); //5ms precise timing

//Display the tens digit

P2 = 7 << 2; //P2 = 0001 1100 -> (P24, P23, P22) = (1, 1, 1) -> The eighth number from the right lights up

P0 = DIG_CODE[r / 10];

delay(); //5ms precise timing

}

int main()

{

int i;

int sec1, sec2;

while(1)

{

sec1 = 15;

sec2 = 15;

while(sec1 >= 0)

{

for(i = 0; i < 50; i++)

{

disp_digit(sec1, sec2); // takes about 20ms

}//20ms*50=1000ms=1s

sec1 --;

sec2 --;

}

}

return 0;

}