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STM32L4 DMA Normal Mode number of bytes transferred check

Question asked by botha.antonie on Feb 1, 2018
Latest reply on Apr 11, 2018 by Kurt Alber

Good Day


If I call:

HAL_ADC_Start_DMA(&hadc1, (uint32_t*)getADCBuffer, adcBufferLength); //uint16_t getADCBuffer[adcBufferLength] and adcBufferLenght is defined as 1000

and then call: 


before the "adcBufferLength" was reached, how can I check how many new bytes I have in my getADCBuffer for every iteration of the code?

Can I do this:

adcNumSamplesCaptured = (adcBufferLength - (uint16_t)(DMA1_Channel1->CNDTR));

where obviously DMA1_Channel1 is associated with the ADC Configuration. 

In normal mode, will CNDTR return to "adcBufferLength" automatically every time? The Reference Manual for my MCU says: 


"If the channel is configured in non-circular mode, no DMA request is served after the last
transfer (that is once the number of data items to be transferred has reached zero). In order
to reload a new number of data items to be transferred into the DMA_CNDTRx register, the
DMA channel must be disabled.


I should never reach "0" items left to transfer, is there something that I must do? 

Should I call "CLEAR_BIT(DMA1_Channel1->CCR,DMA_CCR_EN);" before calling

"HAL_ADC_Start_DMA(&hadc1, (uint32_t*)getADCBuffer, adcBufferLength);" ?


I find that sometimes "adcNumSamplesCaptured is larger than the number of non-zero bytes in "getADCBuffer", and can't quite figure out why.


My ADC DMA Configuration:

 /* ADC1 DMA Init */
    /* ADC1 Init */
    hdma_adc1.Instance = DMA1_Channel1;
    hdma_adc1.Init.Request = DMA_REQUEST_0;
    hdma_adc1.Init.Direction = DMA_PERIPH_TO_MEMORY;
    hdma_adc1.Init.PeriphInc = DMA_PINC_DISABLE;
    hdma_adc1.Init.MemInc = DMA_MINC_ENABLE;
    hdma_adc1.Init.PeriphDataAlignment = DMA_PDATAALIGN_HALFWORD;
    hdma_adc1.Init.MemDataAlignment = DMA_MDATAALIGN_HALFWORD;
    hdma_adc1.Init.Mode = DMA_NORMAL;
    hdma_adc1.Init.Priority = DMA_PRIORITY_HIGH;
    if (HAL_DMA_Init(&hdma_adc1) != HAL_OK)
      _Error_Handler(__FILE__, __LINE__);



Kind Regards