STM32的内部温度传感器读数仅在复位后更新。

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英文:

Internal Temperature Sensor Reading of STM32 only updating after Reset

问题

我们正在读取STM32-F410RB - NUCLEO 64板的内部温度传感器读数。

问题:

按照程序设置,每隔1秒获取温度传感器的读数,但获取到的是相同的读数。
只有在进行复位后,温度读数才会更新。

STM32的内部温度传感器读数仅在复位后更新。

当前设置

  1. Windows 10 64位
  2. STM32 Nucleo 64 - F410RB
  3. 使用CubeMX IDE
  4. 用于串行监控的PuTTy

已完成的STM32 CUBE MX IDE配置

  1. 系统核心 - 串行线
  2. 模拟 - ADC1 - 启用温度传感器通道
    参数设置 - ADC设置 - 启用扫描转换和连续转换
    排名 - 采样时间 - 480周期
  3. USART2 - 9600波特率

主函数的代码片段

int main(void)
{
  
  HAL_Init();
  SystemClock_Config();
  MX_GPIO_Init();
  MX_USART2_UART_Init();
  MX_ADC1_Init();
  HAL_ADC_Start(&hadc1);

  while (1)
  {
	  uint16_t readValue;
	  float tCelsius;
	  readValue = HAL_ADC_GetValue(&hadc1);
	  tCelsius = ((float)readValue) * 3300.0 / 4095.0; // 转为mV
	  tCelsius = (tCelsius - 500.0) / 10.0; // 转为摄氏度
	  UART_SEND_TXT(&huart2, "Temperature = ", 0);
	  UART_SEND_FLT(&huart2, tCelsius, 1);
	  HAL_Delay(1000);
  }
}

请帮忙解决。提前致谢。

英文:

We are reading Internal Temperature Sensor reading of STM32-F410RB - NUCLEO 64 Board.

Problem:

Getting the reading of Temperature sensor every 1 second as programmed, but the same readings are got.
Only when Reset is done, the Temperature Reading get updated

STM32的内部温度传感器读数仅在复位后更新。

Current Setup

  1. Windows 10 64bit
  2. STM32 Nucleo 64 - F410RB
  3. Using CubeMX IDE
  4. PuTTy for Serial Monitoring

STM32 CUBE MX IDE Configurations Done

  1. System Core - Serial Wire
  2. Analog - ADC1 - Enabled Temperature Sensor Channel
    Parameter Settings - ADC Setting - Scan Conversion & Continues Conversion Enabled
    Rank - Sample Time - 480 Cycles
  3. USART2 - 9600 Baud Rate

CODE Snippet of Main

int main(void)
{
  
  HAL_Init();
  SystemClock_Config();
  MX_GPIO_Init();
  MX_USART2_UART_Init();
  MX_ADC1_Init();
  HAL_ADC_Start(&hadc1);

  while (1)
  {
	  uint16_t readValue;
	  float tCelsius;
	  readValue = HAL_ADC_GetValue(&hadc1);
	  tCelsius = ((float)readValue) * 3300.0 / 4095.0; // To mV
	  tCelsius = (tCelsius - 500.0) / 10.0; // To degrees C
	  UART_SEND_TXT(&huart2, "Temperature = ", 0);
	  UART_SEND_FLT(&huart2, tCelsius, 1);
	  HAL_Delay(1000);
  }
}

Please kindly help. Thanks in Advance.

答案1

得分: 2

只有一个HAL_ADC_Start在循环外部。因此,转换只启动一次。

在循环内部,您只是通过HAL_ADC_GetValue重复接收上次转换的结果。

您需要每次调用HAL_ADC_Start来启动常规组转换,然后通过HAL_ADC_PollForConversion等待转换结束,最后通过HAL_ADC_GetValue获取结果。

英文:

There is only one HAL_ADC_Start outside the loop. Therefore conversion starts only once.

Inside the loop you just repeatedly receive the last result of the conversion by HAL_ADC_GetValue.

You need to start regular group conversion by calling HAL_ADC_Start each time, then wait conversion to end by HAL_ADC_PollForConversion and, at last, get the result by HAL_ADC_GetValue

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  • 本文由 发表于 2023年5月29日 14:26:21
  • 转载请务必保留本文链接:https://go.coder-hub.com/76355092.html
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