Electronic – Timer/counter overflow and TIFR register

avr

I recently planned to use AVR timer/counter for my projects. I have studied a number of documents about how to set it up. I use ATTiny2313A to test it out and learn about its features. I came across a problem when tried to toggle a pin for every few seconds.

According to the datasheet the TIFR has the following bit arrangement:

enter image description here

As you see the first bit is assigned to TOV0. When a counter overflow occurs this bit is set to 1. I used this fact and developed a code as following:

#include <avr/io.h>

int main(void)
{
uint16_t count=0, M=500;
uint8_t n=0x02;
DDRB=0xff;
TCNT0=0x00;
TCCR0B=(1<<CS00) | (1<<CS02); //use prescaler 1024


   while (1)
   {
       while ((TIFR & n) == 0); // control passes this loop after TIFR sets to n,
                                   // which is when counter overflows

       TCNT0 = 0x00;
       TIFR=n; // clear TIFR
       count++;

       if (count>=M)
       {
           PORTB ^= (1 << PB0);
           count=0;
       }

return (0);
   }

The code works as expected. I expect that when an overflow occurs TIFR changes to 0x02 as TOV0 sets to one. But that seems not to be the only scenario. If I change the variable n to something like n=0x04 and build the program I see the same behavior, meaning that the LED connected to PB0 still toggles with the same frequency as it did previously. The second bit in TIFR corresponds to OCF0B. So if the code wants to work as before this bit must change to 1 when an overflow happens. But does it? I don't think the output compare flag is enabled so OCF0B can't change.

Best Answer

The OCF0B is a compare match flag, which will get set regardless of whether the interrupt is enabled or not. It will get set exactly once per overflow cycle, just like a broken clock will show the correct time twice a day. This is why you are getting the exact same period time with this flag or with the overflow flag.

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