stm8低速时钟校准问题

我大龙哥哥提问时间：2016-8-22 15:28 /

问答
是否解决:

stm8低速时钟校准问题
stm8L10x标准库例程里的这个时钟校准函数一直不理解，求详解啊，一直不理解例程里的低速时钟校准
函数如下：

/**
  * @brief  Update APR register with the measured LSI frequency.
  * @note AWU must be disabled to avoid unwanted interrupts.
  * @note Prescaler calculation:
  *       A is the integer part of LSIFreqkHz/4 and x the decimal part.
  *       x <= A/(1+2A) is equivalent to A >= x(1+2A)
  *       and also to 4A >= 4x(1+2A) [F1]
  *       but we know that A + x = LSIFreqkHz/4 ==> 4x = LSIFreqkHz-4A
  *       so [F1] can be written :
  *       4A >= (LSIFreqkHz-4A)(1+2A)
  * @param  LSIFreqHz Low Speed RC frequency measured by timer (in Hz).
  * @retval None
  */
void AWU_LSICalibrationConfig(uint32_t LSIFreqHz)
{

  uint16_t lsifreqkhz = 0x0;
  uint16_t A = 0x0;

  /* Check parameter */
  assert_param(IS_LSI_FREQUENCY(LSIFreqHz));

  lsifreqkhz = (uint16_t)(LSIFreqHz / 1000); /* Converts value in kHz */

  /* Calculation of AWU calibration value */

  A = (uint16_t)(lsifreqkhz >> 2U); /* Division by 4, keep integer part only */

  if ((4U * A) >= ((lsifreqkhz - (4U * A)) *(1U + (2U * A))))
  {
AWU->APR = (uint8_t)(A - 2U);
  }
  else
  {
AWU->APR = (uint8_t)(A - 1U);
  }

}