Electronic – Understanding Timers, Counters and Prescaler Registers on the STM32

armcortex-m3interruptsstm32timer

So I'm trying to set up a very simple periodic interrupt on my STM32VL-Discovery board using the libopencm3 libraries on linux. I've looked at the documentation and I just want to verify with somebody that I understand what's going on.

My understanding is that the prescaler register scales the internal clock to another frequency/clock that the timer actually sees. i.e. you set the timer prescaler register to 24Mhz/DesiredHz to get the timer to operate at the desired frequency. Is this correct?

Assuming that's correct, say I want to generate an interrupt on every timer clock cycle. I didn't see anything in the docs that would allow me to do that besides setting the period register to 1. This would generate an overflow interrupt every clock correct?

Best Answer

If you just set the period register to 1, this won't automatically enable interrupt generation. You will need to do 2 additional steps:

Enable the 'update' interrupt for your timer using the TIM_ITConfig() function.
Enable the timer interrupt source in the interrupt controller by calling NVIC_Init() with appropriate parameters.

Of course, you should not forget to enable clocking of your timer peripheral using the APB1PeriphClockCmd().

There's a detailed example on configuring timers on STM32F4-Discovery here: http://visualgdb.com/tutorials/arm/stm32/timers/

STM32VL-Discovery is a bit different, however the main steps are the same.

Related Solutions

Electronic – C18 Timer0 timing calculation using MCC18

The answer is simple. When the overflow bit occurs and the interrupt is raised, we need to first write the counter to TMR0 and reset the TMR0IF flag.

This solution works:

#pragma code 
#pragma interrupt high_priority_interrupt 
void high_priority_interrupt()
{
    if (INTCONbits.TMR0IF)              //TIMER0 overflowed
    {
        //Stuff
        second += 1;
        if (second == 60)
        {
            minute += 1;
            second = 0;
        }

        if (minute == 60)
        {
            measure_and_switch();
            minute = 0;
        }

        WriteTimer0(CLOCK_TICK);        //Write the count to TMR0.
        INTCONbits.TMR0IF = 0;          //Clear TIMER0 overflow bit.
    }
}

I hope this helps anyone. :-)

TIMER0 in PIC18F4550

Unless you have made any changes to the oscillator settings with the OSCCON register it will default to a prescaler of /8 to give 1MHz. If that is true, then the \$F_{CY}\$ will be 256000Hz.

Your prescaler is set to 256, so the clock tick will be at \$\frac{256000}{256}=1000Hz\$

You think you're setting the overflow value to 65535, but you're not. You're actually setting the current timer's high byte to 0xFFFF (which, since it's an 8 bit register is actually 0xFF), and the timer's low byte to 0x00, so the current timer count is 0xFF00 in total. It ticks up, and when it loops from 0xFFFF to 0x0000 it triggers an interrupt. That's 256 ticks away.

So at 1000Hz divided by 256 ticks, that's \$\frac{1000}{256}=3.90625Hz.\$

Which is almost 4 times per second.

So to get exactly one second, with a 1000Hz tick, you need 1000 ticks. So you would set your timer values to 65536 minus 1000, or 64536, which is 0xFC18, so your TIMERH would be 0xFC and your TIMERL would be 0x18.

Best Answer

Related Solutions

Electronic – C18 Timer0 timing calculation using MCC18

TIMER0 in PIC18F4550

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