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Undocumented behaviour STM32C011F4U6TR

ah2
Associate III

Hi,

We notice the startup time from power up can be affected by setting one GPIOB pin to input. The startup time here refers to the time from SystemInit() until main(). Basically it contains scatterload and __rt_entry to my understanding.

Specifically, the sample code below, built with ARM Compiler 6.16 Tool, would demonstrate that

#define MEASURE_PIN 5

int main(void)
{
  IO_SET_PIN(GPIOA, MEASURE_PIN, 1); 

  // Enable IWDG
  IWDG->KR = 0xCCCC;
  // Enable write access
  IWDG->KR = 0x5555;
  // Set prescaler
  IWDG->PR = 0;
  // Set reload
  IWDG->RLR = 1;

  // Wait for the registers reload
  while (IWDG->SR)
  {
  }
    
  // refresh the watchdog.
  IWDG->KR = 0xAAAA;

  while(1);
  return 0;
}
void SystemInit(void)
{
  // enable GPIOA clock
  RCC->IOPENR |= RCC_IOPENR_GPIOAEN | RCC_IOPENR_GPIOBEN;

  // set GPIOA pin5 to output
  IO_SET_PIN_DIR(GPIOA, MEASURE_PIN, 1);

  // set GPIOA pin5 to low
  IO_SET_PIN(GPIOA, MEASURE_PIN, 0);
  
  // Set GPIOB pin7 to input
  // GPIOB->MODER &= 0xffff3fff;  
} 

Basically GPIOA pin 5 is toggled to low and high for the time between SystemInit() and main() to be measured.

main() also has IWDG setup and ends with a while loop so that IWDG reset happens periodically since power up.

On the oscilloscope it looks like the screenshots below.

powerup_time_without_gpiob_input.png2nd_iwdg_reset_without_gpio.png

Respectively from the screenshots, it can be observed that the 1st startup(power up) takes 254 μs while the 2nd startup (IWDG reset) takes 132 μs.

Now if uncommenting this line GPIOB->MODER &= 0xffff3fff; and redo the measurement, we get the results below.

powerup_time_with_gpiob_input.png

We can see that having the GPIOB pin 7 set to input reduces the startup time from 254 μs to 132 μs and the rest startup remains taking the same time.

Could you please confirm the following:

  1. What causes the differences between 1st and 2nd startup?
  2. Why setting GPIOB ping 7 to input would affect the startup timing?

Just in case, this is how Reset_Handler looks like in our startup file. It is basically without any customization.

; Reset handler routine
Reset_Handler    PROC
                 EXPORT  Reset_Handler                 [WEAK]
        IMPORT  __main
        IMPORT  SystemInit  
                 LDR     R0, =SystemInit
                 BLX     R0
                 LDR     R0, =__main
                 BX      R0
                 ENDP

  Anta

21 REPLIES 21

@waclawek.jan wrote:

> And initalizing the ram-variables (in __rt_entry) shouldn't take up any time as I don't see any global or static variable in your code. I suspect the internal clock is not fully stabilized yet.

0x400 bytes starting at 0x20000020 are set to zero - there are 256 word writes.

If the loop takes 6 cycles to execute, which IMO it can, that's around 130us at 12MHz clock.

JW


That's quite a lot of zero-initialized variables. Or does it always clear the entire block reserved in the linker file?

Kudo posts if you have the same problem and kudo replies if the solution works.
Click "Accept as Solution" if a reply solved your problem. If no solution was posted please answer with your own.

I'm just describing what I see in the disasm. I am not ARM/Keil so can't explain their decisions... ;)

I'm not sure the described phenomenon can be explained without having deep access into the 'C0's innards. I was asking for the chip revision because of the "first SRAM access may fail" erratum, although the erratum is not clear about what constitutes the circumstances of failure and what exactly are its consequences. But regardless of that erratum, one of the mechanisms I can envisage is, that after poweron reset a hardware process runs, which performs some operation across the whole SRAM array, and thus the first SRAM access is delayed by some time - well say it precalculates the parity, one word per cycle, 6kBytes at 12MHz would take around 130us...

The difference between the "with gpio" and "without gpio" is, that in the former the first access to RAM is the push which happens *before* the observed pin is pulled low, i.e. the lengthy RAM operation happens before the pin is pulled low and thus the pulse has the expected length; whereas in the latter (without gpio) the first access to RAM is the first zero write for the initialization, which is *after* the pin is pulled low, thus prolongs the observed pulse.

This is just a theory and could be proven or disproven by a carefully crafted test (possibly in asm), where one pin edge would be generated before the first access to SRAM, and another edge (either on the same or other pin) after that access. I don't have a 'C0 to experiment.

In any case, the total time between the moment when POR releases the processor and the moment when the observed pin is pulled high at beginning of main() should be roughly the same in both cases. This could be proven or disproven by powering the circuit from a source which can go from 0V to 3V quickly enough (in a few us).

JW