C51 Programming 16-Interrupts (Timer Interrupts 3)

To use the timer/counter interrupt, the following conditions must be met.

    1) Interrupts are always enabled EA = 1;

    2) The interrupt source is allowed to be turned on ET0 = 1 or ET1 = 1;

    3) Set the working mode of the timer/counter (set TMOD)

    4) Load the timer and set the initial calculation value of the counter.

    5) Start the timer (TR0 = 1 or TR1 = 1 in TCON)

    6) Interrupt service function

Through the study of some previous interrupt articles, I believe that except for the 4th point above, using timers/counters is not difficult.

Load the timer. As we know from the previous timer chapter, there are 4 modes for timer/counter. They are 13-bit manual load (mode 0), 16-bit manual load (mode 1), 8-bit auto-reload (mode 2), and 8-bit manual load (mode 3). According to the number of bits of the timer/counter, TH0 (1) and TL0 (1) are required to form the number of bits of the timer/counter.

The input clock pulse is obtained by dividing the output of the crystal oscillator by 12, so the timer is a counter of the machine cycle, each machine cycle +1, that is, the value of the number of bits of TH0 (1) and TL0 (1) that make up the timer/counter is +1 in each machine cycle. When the value overflows, a timer/counter interrupt is generated.

The following uses method 1 and method 2 to perform timing and light up the LED. 

Method 1

/****************************************

Header:

File Name: main.c

Author: adam

Date: September 19, 2020

**********************************************/

#include "reg52.h"

#include "stdio.h"

typedef unsigned char uchar;

typedef unsigned int uint;

uint countms = 0;

void DelayMs(uint ms);

void main(){

   EA = 1; //General interrupt enabled

   ET0 = 1; //Timer T0 interrupt enabled

   TMOD = 0x01; //Set to mode 1, timer mode: C/T=0; do not start external INT0 control to start timing: GATE=0,

   //Timing 1ms, machine cycle 1us, 1000 machine cycles 1ms

   //16-bit timer overflow condition: full value 65535 (#FFFF) + 1 overflows, so timing 1ms requires (65535+1-1000);

   // Divide the value into two registers, each of which can hold 256 numbers.

   TH0 = (65536-1000)/256; //Load initial value TH     

   TL0 = (65536-1000)%256; //Load initial value TL

   TR0 =1; //Start timer TR0

   while(1)

   {

        DelayMs(1000);

        P0=~P0; 

   }

}

void DelayMs(uint ms)

{

        while(countms        countms=0; 

}

void Timer0(void) interrupt 1

{

    //Reinstall the initial value

    TH0 = (65535-1000)/256;

    TL0 = (65535-1000)%256;  

    countms++; //Every ms, countms++

}

Method 2

/****************************************

Header:

File Name: main.c

Author: adam

Date: September 19, 2020

**********************************************/

#include "reg52.h"

#include "stdio.h"

typedef unsigned char uchar;

typedef unsigned int uint;

uint count = 0;

uint countms = 0;

void DelayMs(uint ms);

void main(){

   EA = 1; //General interrupt enabled

   ET0 = 1; //Timer T0 interrupt enabled

   TMOD = 0x02; //Set to mode 2, timer mode: C/T=0; do not start external INT0 control to start timing: GATE=0,

   //Since the mode is only 8 bits, the maximum timing can only be 0.256us, so first time 100us, and then time ms through 100us.

   TH0 = 256-100; //Load value TH, automatic reload     

   TL0 = 256-100; //Load value TL, overflow generates interrupt.

   TR0 =1; //Start timer TR0

   while(1)

   {

        DelayMs(100);

        P0=~P0;

   }

}

void DelayUs(uint us){  

      while(countus      count = 0;

}

void DelayMs(uint ms)

{

           while(countms            DelayUs(10);   

            countms++;  

        }      

        countms=0;

}

void Timer0(void) interrupt 1

{  

    countus++; //every ms, countms++   

}

Note: In the above two examples, the interrupts are 1ms and 100us respectively. It is not recommended to use the timer to set the interrupt to a smaller time.

Although the timer can be set to generate an interrupt in 1us, 1us is only one machine cycle. After the interrupt is responded, the interrupt service function is entered. It may take four machine cycles to execute a simple statement, plus the protection and restoration of the scene, which will take up a long machine cycle and make the timing inaccurate.

When GATE = 1, the timer/counter is affected by the INT0/INT1 level. INT0/INT1 = 1 starts counting, and INT0/INT1 = 0 stops counting. When GATE = 1, after the microcontroller is powered on, the initial values ​​of P0~P3 are all high levels, so there is no need to connect an external high level to start the timer/counter. Generally, INT0/INT1 uses the following circuit. When the button is pressed, it is low level and the timer/counter stops counting.