ATMega2560 timer not increasing again after overflow

cmicrocontrollertimer

I want to generate waveform with f = 196 Hz using only normal mode.
Here is the code I use. Whenever the timer overflows TOV1 is set and PIND0 will toggle. TOV1 is cleared after that and I wait for another overflown to happen so PIND0 will toggle again.

I tried my code, with PIND0 connected to the LED. On power up, the LED turned on and off a short time after that but it didn't turn on again. Because it had turned off means that the overflown happened so PIND is toggled but I don't know why it didn't turn on again. I even tried to not clear TOV1 but get the same result as TOV1 is cleared.

#include <avr/io.h>
#define F_CPU 25690112UL
int main(void)
{
    TIFR1 |= (1<<TOV1); // clear timer 1 overflown flag
    DDRD = 0xFF; // set port D as output
    TCCR1B = 0b00000001; // using internal I/O clock with prescaler 1
    while(1){
        while(TIFR1 & (1<<TOV1)){
            PIND ^= (1<<PD0); // toggle bit 0 on port D
            TIFR1 |= (1<<TOV1); // clear TOV1 by adding 1 to TOV1
        }
    }
}

Can anyone see an error in my code?

Best Answer

The PIND ^= (1 << PD0) line is off.

The PIN register holds the actual physical state pins, and would not normally be written. The PORT register sets the state of the pin drivers.

However, as a common read-only register is a waste of I/O address space, Atmel added a neat feature to the pin register: Writing a bit mask to PIN will toggle the corresponding bits in PORT. The line 'PIND ^= mask' is then approximately the same as PORTD XOR PIND XOR mask. Since PIND reads the pins PORTD drives, PIND is usually = PORTD, which results in this being equivalent to 'PORTD = mask'.

Try substituting 'PIND' with 'PORTD', -or- '^=' with just '='

Related Solutions

Electronic – Timer overflow on MSP430

Not sure exactly which series MSP you're using so you'll have to look up the specifics. But the basic premise should apply across the board I think.

The following example is from the file msp430g2xx1_ta_03.c in the code snippets that can be downloaded from the TI Wiki:

#include <msp430.h>

int main(void)
{
  WDTCTL = WDTPW + WDTHOLD;                 // Stop WDT
  P1DIR |= 0x01;                            // P1.0 output
  TACTL = TASSEL_2 + MC_2 + TAIE;           // SMCLK, contmode, interrupt

  _BIS_SR(LPM0_bits + GIE);                 // Enter LPM0 w/ interrupt
}

// Timer_A3 Interrupt Vector (TAIV) handler
#pragma vector=TIMERA1_VECTOR
__interrupt void Timer_A(void)
{
 switch( TAIV )
 {
   case  2: break;                          // CCR1 not used
   case  4: break;                          // CCR2 not used
   case 10: P1OUT ^= 0x01;                  // overflow
            break;
 }
}

The key is checking the TAIV register to see what caused the interrupt. Take a look at page 416 of the CC430 Family User's Guide:

TAIV Register Description

Now the code example is looking for a value of 10 in the TAIV register and the User Guide indicates that 0x0E is the overflow. So, again, check your specific chip's documentation. But the TAIV register is what will tell you what the cause of the interrupt is.

Electronic – STM8L Timer 1 unexpectedly low overflow frequency

Ok, so the CPU is running at 16MHz. You have a "count to" value of 1. The counter starts at 0. After 1 tick the counter will be at 1, so the interrupt occurs and the counter resets.

Second tick you enter the ISR and start pushing registers to the stack. The counter increases to 1, so the interrupt occurs and the counter resets.

Third tick, you're still in the ISR from before, still pushing registers on to the stack, and you will be for the foreseeable future. But you want it to enter the ISR again, and you're already there. So the interrupt is delayed.

Fourth tick. You're still pushing to the stack, the counter increases to 1, so the interrupt triggers. The interrupt is already queued, so nothing at all happens. The counter resets to 0.

Eventually, after many many many ticks, and missed interrupts, you finish the ISR and return to normal program flow. Immediately the ISR runs again because of the delayed interrupt from earlier, and the whole cycle starts again.

What you are trying to do - run an interrupt every other CPU clock tick, is not physically possible. Interrupts take time to execute.

You cannot trigger an interrupt faster than the interrupt takes to process.

Best Answer

Related Solutions

Electronic – Timer overflow on MSP430

Electronic – STM8L Timer 1 unexpectedly low overflow frequency

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