STM32F404K8:首次读取后HAL,ADC + DMA被卡住

时间:2019-07-09 09:02:28

标签: c stm32 dma adc nucleo

我正在尝试从三个不同的通道读取ADC值(现在,以后将是6个通道,每个ADC上有3个),我想在经过一段随机时间后访问我的ADC缓冲区,并使用UART稍后。现在它们之间只是通过一个微调电位器连接到+ 3.3V,导致三个引脚上的电压都不同,这意味着我应该能够经常读取ADC吗? 无论如何,要执行此操作,我使用的是STM32f303k8板,其中设置了要使用的ADC1通道2、4和11。我使用HAL库使用STM32CubeMX生成了代码,在其中将其配置为在循环模式下使用DMA,具有连续转换和不同等级的扫描转换。 启动时(使用断点),我可以看到代码的init部分工作正常。 ADC初始化并启动DMA,我什至从ADC1将具有正确值的值(通道1!=通道2!=通道3)从缓冲区中获取值。 问题是继续进行时,进程陷入了无限循环处理程序,并且永远无法到达我的while(1)那里有一些任意语句的地方。 下面的代码。

我一直在遵循不同的指南,以查看是否将其设置错误。其中之一是https://tunizem.blogspot.com/2014/09/using-adc-with-dma-on-stm32.html?showComment=1562660027777#c1229743050555367742。我尝试将采样时间从1.5一直更改为601.5个周期,还尝试将EOCSelection从单次更改为连续。

ADC_HandleTypeDef hadc1;
DMA_HandleTypeDef hdma_adc1;

void MX_DMA_Init(void) 
{
  /* DMA controller clock enable */
  __HAL_RCC_DMA1_CLK_ENABLE();

  /* DMA interrupt init */
  /* DMA1_Channel1_IRQn interrupt configuration */
  HAL_NVIC_SetPriority(DMA1_Channel1_IRQn, 0, 0);
  HAL_NVIC_EnableIRQ(DMA1_Channel1_IRQn);
}


/* ADC1 init function */
void MX_ADC1_Init(void)
{
  ADC_MultiModeTypeDef multimode = {0};
  ADC_ChannelConfTypeDef sConfig = {0};

  /** Common config 
  */
  hadc1.Instance = ADC1;
  hadc1.Init.ClockPrescaler = ADC_CLOCK_SYNC_PCLK_DIV4;
  hadc1.Init.Resolution = ADC_RESOLUTION_12B;
  hadc1.Init.ScanConvMode = ADC_SCAN_ENABLE;
  hadc1.Init.ContinuousConvMode = ENABLE;
  hadc1.Init.DiscontinuousConvMode = DISABLE;
  hadc1.Init.ExternalTrigConvEdge = ADC_EXTERNALTRIGCONVEDGE_NONE;
  hadc1.Init.ExternalTrigConv = ADC_SOFTWARE_START;
  hadc1.Init.DataAlign = ADC_DATAALIGN_RIGHT;
  hadc1.Init.NbrOfConversion = 3;
  hadc1.Init.DMAContinuousRequests = ENABLE;//ENABLE;
  hadc1.Init.EOCSelection = ADC_EOC_SINGLE_CONV;
  hadc1.Init.LowPowerAutoWait = DISABLE;
  hadc1.Init.Overrun = ADC_OVR_DATA_PRESERVED;//OVERWRITTEN;
  if (HAL_ADC_Init(&hadc1) != HAL_OK)
  {
    Error_Handler();
  }
  /** Configure the ADC multi-mode 
  */
  multimode.Mode = ADC_MODE_INDEPENDENT;
  if (HAL_ADCEx_MultiModeConfigChannel(&hadc1, &multimode) != HAL_OK)
  {
     Error_Handler();
  }
  /** Configure Regular Channel 
  */
  sConfig.Channel = ADC_CHANNEL_2;
  sConfig.Rank = ADC_REGULAR_RANK_1;
  sConfig.SingleDiff = ADC_SINGLE_ENDED;
  sConfig.SamplingTime = ADC_SAMPLETIME_61CYCLES_5;
  sConfig.OffsetNumber = ADC_OFFSET_NONE;
  sConfig.Offset = 0;
  if (HAL_ADC_ConfigChannel(&hadc1, &sConfig) != HAL_OK)
  {
     Error_Handler();
  }
  /** Configure Regular Channel 
  */
  sConfig.Channel = ADC_CHANNEL_4;
  sConfig.Rank = ADC_REGULAR_RANK_2;
  if (HAL_ADC_ConfigChannel(&hadc1, &sConfig) != HAL_OK)
  {
     Error_Handler();
  }
  /** Configure Regular Channel 
  */
  sConfig.Channel = ADC_CHANNEL_11;
  sConfig.Rank = ADC_REGULAR_RANK_3;
  if (HAL_ADC_ConfigChannel(&hadc1, &sConfig) != HAL_OK)
  {
    Error_Handler();
  }

}

 *
 *
 *


uint32_t adcValue1[60];

int main(void)
{
  /* MCU Configuration-----------------------------------------------*/

  /* Reset of all peripherals, Initializes the Flash interface and the 
Systick. */
  HAL_Init();

  /* Configure the system clock */
  SystemClock_Config();

  /* Initialize all configured peripherals */
  MX_GPIO_Init();
  MX_DMA_Init();
  MX_ADC1_Init();
  //MX_ADC2_Init();
  MX_I2C1_Init();
  MX_USART1_UART_Init();
  MX_USART2_UART_Init();

  //Start ADC writing to DMA on scan complete
  if(HAL_ADC_Start_DMA(&hadc1, (uint32_t*)adcValue1, 120) != HAL_OK)
      return 0;


  uint32_t stopwatch = HAL_GetTick();

  /* Infinite loop */
  /* USER CODE BEGIN WHILE */
  while(1)
  {
      if(((uint32_t)HAL_GetTick() - stopwatch) > 49)
      {
          stopwatch = HAL_GetTick();
          HAL_GPIO_TogglePin(GPIOB, GPIO_PIN_3);

          sprintf((char*)buffer, "(%d)\t ADC1, (%d)\t ADC2, (%d)\t ADC3", 
adcValue1[0],
                  adcValue1[1], adcValue1[2]);

      }
  }
  /* USER CODE END 3 */
}
}

我在“ uint32_t秒表= HAL_GetTick();”处有一个断点。在这里我可以看到adcValue1正在读取值。这是正确的,因为所有这三个电压都输入了3.3V,但电位计被设置为在通道2发出2.2V的电压,即低于其他两个电压。但这是我唯一能看到阅读完成的情况,因为它在此之后陷入无限循环。

enter code here

adcValue1   uint32_t [60]   0x20000220 <adcValue1>  
adcValue1[0]    uint32_t    2614    
adcValue1[1]    uint32_t    3638    
adcValue1[2]    uint32_t    3639    
adcValue1[3]    uint32_t    2612    
adcValue1[4]    uint32_t    3639    
adcValue1[5]    uint32_t    3637    
adcValue1[6]    uint32_t    2615    
adcValue1[7]    uint32_t    3642    
adcValue1[8]    uint32_t    3641    
adcValue1[9]    uint32_t    2616    
adcValue1[10]   uint32_t    3642    
adcValue1[11]   uint32_t    3638    
adcValue1[12]   uint32_t    2611    
adcValue1[13]   uint32_t    3637    
adcValue1[14]   uint32_t    3640    
adcValue1[15]   uint32_t    2614    
adcValue1[16]   uint32_t    3639    
adcValue1[17]   uint32_t    3639    
adcValue1[18]   uint32_t    2615    
adcValue1[19]   uint32_t    3643    


 * @brief  This is the code that gets called when the processor receives 
an
 *         unexpected interrupt.  This simply enters an infinite loop, 
preserving
 *         the system state for examination by a debugger.
 *
 * @param  None
 * @retval : None
 */

Default_Handler:
Infinite_Loop:  <--- STUCK HERE
    b   Infinite_Loop
    .size   Default_Handler, .-Default_Handler

1 个答案:

答案 0 :(得分:1)

当存在中断且用户代码中没有处理程序时,将调用

Default_Handler。当您不为该中断提供处理程序时,中断向量表中的所有地址都指向该代码。

您可以检查VECTACTIVE中的SCB->ICSR位(请参阅STM32F4 programming manual)以找出缺少的中断向量,然后在代码中提供适当的处理程序。