Confused about GPIOTE: pin toggle and interrupt latency

Question

Hi, 
 I have a pretty simple problem but I can't get it right. This is what I want to do: 
 T=000, PIN_x = 0 
T=004, PIN_x = 1 
T=100, call a routine 
T=543, Start over (T=544=0) 
 (T is in cycles, 8MHz) 
 It seems like using PPI and GPIOTE is the correct way for this? 
 
 Will the PIN toggle at EXACT the correct time? (T=0 and T=40) 
 At what time will my first instruction in the routine be called? (is it 100 - latency?) 
 Do the CPU have to do something for resetting the timer? 
 
 Using NRF51822 
 Thanks

Stian Røed Hafskjold · Accepted Answer

Yes, PPI and GPIOTE would be the correct way for this.

You can use three compare registers for this, and set the prescaler to 1 (8MHz):

NRF_TIMER0->PRESCALER = 1;
NRF_TIMER0->CC[0] = 40;
NRF_TIMER0->CC[1] = 100;
NRF_TIMER0->CC[2] = 544;

Use shorts to clear timer on CC[2]. Timer becomes zero right away so it will start to count on 1 next cycle.

NRF_TIMER0->SHORTS = TIMER_SHORTS_COMPARE2_CLEAR_Enabled << TIMER_SHORTS_COMPARE2_CLEAR_Pos;

You can then use CC[2] = 544 to toggle the PIN (as in your T=000, PIN_x = 0), and CC[0] = 40 to toggle the pin again (as in your T=040, PIN_x = 1). Do this with PPI and GPIOTE:

NRF_GPIOTE->CONFIG[0] = (3 << 0) | (PIN1 << 8) | (3 << 16) | (1 << 20); //Task mode, PIN1, Toggle, init: 1
NRF_PPI->CHENSET = 0x3; //enable two channels
NRF_PPI->CH[0].EEP = (uint32_t) &NRF_TIMER0->EVENTS_COMPARE[2];
NRF_PPI->CH[0].TEP = (uint32_t) &NRF_GPIOTE->TASKS_OUT[0];
NRF_PPI->CH[1].EEP = (uint32_t) &NRF_TIMER0->EVENTS_COMPARE[0];
NRF_PPI->CH[1].TEP = (uint32_t) &NRF_GPIOTE->TASKS_OUT[0];

And yes, the time when this happen will be within the 8MHz clock cycle specified in the CC register (T=544/0 and T=40)

Enable interrupt for the CC[1] register:

NVIC_EnableIRQ(TIMER0_IRQn);
NRF_TIMER0->INTENSET = (1 << 17);

Now you can do CPU stuff in the TIMER0_IRQHandler()

void TIMER0_IRQHandler(void){
    if(NRF_TIMER0->EVENTS_COMPARE[1]){
        NRF_TIMER0->EVENTS_COMPARE[1] = 0;
        do_stuff();
    }
}

The CPU would then start executing at T=100 + latency. Since the HF clock and 1.2V peripheral regulator is already running (because of TIMER) the only thing that is started is the 1.7V regulator which takes ~2us to startup (table 32 Power management in the PS). So then T = 100 cycles + 2us.

Clear and start timer after init:

NRF_TIMER0->TASKS_CLEAR = 1;
NRF_TIMER0->TASKS_START = 1;