topic Re: FRDM-K20D50M clock system in Kinetis Microcontrollers

FRDM-K20D50M clock system

danielegener — Tue, 03 Mar 2015 10:34:51 GMT

Hi,

i am new in ARMs, new in Kinetis and new in Processor Expert.

I try to setup the clock system of this Evaluation system. To test the setup i want to toggle a pin in the main function. But i cannot see a clean toggle signal. An other idea was to toggle the pin in an timer interrupt. If i choose Period of 100 Khz, i can see a clean toggle signal on the oscilloscope. If i change the period to 50 Khz or 1 Mhz i cannot see a clean signal.

Does anybody an idea what's going wrong?

Thanks in advance!

Cheers,

Daniel

Re: FRDM-K20D50M clock system

mjbcswitzerland — Mon, 02 Nov 2020 13:12:30 GMT

Hi Daniel

1MHz is too fast for the K20D50 - the maximum interrupt rate that it can handle is about 500kHz (interrupts take about 2us at 50MHz).

However 250kHz and 50kHz interrupts on two flex timers is not a problem:

Here I have measured two FlexTimer interrupts toggling the LEDs on the board at 50kHz and 250kHz rates. The 50kHz toggle is very stable but the faster one has some jitter.

The binary is attached.

The third LED is togged at 2.5Hz by a software timer and the UART on the OpenSDA virtual COM connection is fully operational at the same time.

I don't know if or why the PE configuration may have problems with specific speeds but you can compare the register setups to maybe see something.

I set up the FlexTimers in the uTasker framework as follows:

TIMER_INTERRUPT_SETUP timer_setup; // interrupt configuration parameters
_CONFIG_PORT_OUTPUT_FAST_LOW(A, PORTA_BIT2, (PORT_SRE_FAST | PORT_DSE_LOW)); // configure port outputs (LEDs)
_CONFIG_PORT_OUTPUT_FAST_LOW(C, PORTC_BIT3, (PORT_SRE_FAST | PORT_DSE_LOW));
timer_setup.int_type = TIMER_INTERRUPT;
timer_setup.int_priority = PRIORITY_TIMERS;
timer_setup.int_handler = timer_int; // interrupt handler
timer_setup.timer_reference = 0; // FlexTimer 0
timer_setup.timer_mode = (TIMER_PERIODIC); // period timer interrupt
timer_setup.timer_value = TIMER_US_DELAY(20); // 20us periodic interrupt
fnConfigureInterrupt((void *)&timer_setup); // configure for first timer

timer_setup.timer_reference = 1; // FlexTimer 1
timer_setup.int_handler = timer_int_1M; // interrupt handler
timer_setup.timer_value = TIMER_US_DELAY(4); // 4us periodic interrupt
fnConfigureInterrupt((void *)&timer_setup);  // configure for secondtimer


The interrupt handlers do:

static void timer_int(void)
{
    _TOGGLE_PORT(C, PORTC_BIT3);
}


static void timer_int_1M(void)
{
    _TOGGLE_PORT(A, PORTA_BIT2);
}

Note however that generally a timer will not be used to perform such a function because it is much easier and more accurate to configure a PWM output on the FlexTimer to generate a high precision output without any interrupt overhead.

Regards

Mark

Kinetis: µTasker Kinetis support

K20: µTasker Kinetis TWR-K20D72M support / µTasker Kinetis FRDM-K20D50M support / µTasker Kinetis TWR-K20D50M support / µTasker Kinetis support

Hardware Timers: http://www.utasker.com/docs/uTasker/uTaskerHWTimers.PDF

For the complete "out-of-the-box" Kinetis experience and faster time to market

Re: FRDM-K20D50M clock system

david_diaz — Wed, 04 Mar 2015 21:16:18 GMT

Hello Daniel,

Indeed, as Mark said, the maximum interrupt rate that the device can handle is about 500kHz (interrupts take about 2us at 50MHz).

This is because the Interrupt Latency.

The term interrupt latency refers to the number of clock cycles required for a processor to respond to an interrupt request, this is typically a measure based on the number of clock cycles between the assertion of the interrupt request up to the cycle where the first instruction of the interrupt handler exected.

Ideally, the interrupt latency should be low.

The interrupt latency of the ARM Cortex M4 is, in the best case, 12.

In addition, as in any program code, the ISRs take time to execute.

In the case of your application and using Processor Expert, the ISR takes about 80 clock cycles to execute it and I takes about 10 to Return Latency.

In conclusion, if we add all the clock cycles that requires to execute an interrupt, we have 102 clock cycles that means:

1 clock cycle -> 20 ns

102 clock cycle -> 2.040 us

Hope this information will be useful for you.

Have a great day. :smileyhappy:

Best Regards,

David Diaz.

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