UART interrupt data loss

[KMale.1]

I used USART interrupt for RS485 encoder communication. I send 8 bit data and encoder reply with a 16 bit data. The first 8 bits out of 16 bits are received correctly. But second 8 bits didn't get correctly. I observed receiving data using oscilloscope. By comparing received data and oscilloscope signal, I found that start bit of the second byte always missed. Therefor RDR register take the first 0 bit of the second byte as the start bit.

eg: first byte(0,1,1,0,0,0,0,1,1,1) second byte(0,1,1,0,0,0,1,0,1,1)

Bolded bits are start and stop bits. In second byte, underlined bit is taken as start bit.

Here is my code.

 HAL_UART_IRQHandler(&huart2);

 /* USER CODE BEGIN USART2_IRQn 1 */

 UART_HandleTypeDef *huartx =&huart2;

  uint32_t isr  = READ_REG(huartx->Instance->ISR);

  uint32_t cr1   = READ_REG(huartx->Instance->CR1);

  if (__HAL_UART_GET_IT(&huart2, UART_IT_RXNE)==SET){

   USART2->ISR;

   RX[count] = (USART2->RDR);

   count++;

   if(count==2){

   __HAL_UART_ENABLE_IT(&huart2, UART_IT_TXE);

   count=0;

   }

  }

  if(HAL_RS485Ex_Init(&huart2, UART_DE_POLARITY_HIGH, 1/16,1/16)==HAL_OK){

   if (((isr & USART_ISR_TXE) != RESET) && ((cr1 & USART_CR1_TXEIE) != RESET)){

   HAL_GPIO_WritePin(GPIOD, GPIO_PIN_4, 1); //DE ON

   USART2->ISR;

   USART2->TDR = TX[0];

   __HAL_UART_ENABLE_IT(&huart2, UART_IT_TC);

   __HAL_UART_DISABLE_IT(&huart2, UART_IT_TXE);

   i=1;

   }

  }

  if (((isr & USART_ISR_TC) != RESET) && (i==1)){

    HAL_GPIO_WritePin(GPIOD, GPIO_PIN_4, 0); //DE OFF

    __HAL_UART_DISABLE_IT(&huart2, UART_IT_TC);

    __HAL_UART_CLEAR_FLAG(&huart2, UART_FLAG_TC);

    i=0;

    }

 /* USER CODE END USART2_IRQn 1 */

Could anyone please help me to solve this issue

Thank you.

[KnarfB]

The other day I implemented RS485 using DMA like

        HAL_UART_Receive_DMA( &huart1, (uint8_t*)&absa_response, sizeof(absa_response) );
	// send request
	HAL_GPIO_WritePin( re_port, re_pin, GPIO_PIN_SET ); // hi inactive
	HAL_GPIO_WritePin( de_port, de_pin, GPIO_PIN_SET ); // hi active
	static const uint8_t request = 0b00000010; // data readout absa
	HAL_UART_Transmit( &huart1, (uint8_t*)&request, 1, 0 ); // no timeout, this makes function return faster
	HAL_GPIO_WritePin( de_port,de_pin, GPIO_PIN_RESET );
	HAL_GPIO_WritePin( re_port, re_pin, GPIO_PIN_RESET ); // lo active

and processed the response in HAL_UART_RxCpltCallback.

Works as a charme with 2500000 baud on a F0 device @ 48 MHz.

[waclawek.jan]

Which STM32? What is the primary clock?

Sounds like baudrate mismatch. Verify that the incoming data baudrate matches the expected baudrate, and verify your STM32's clock.

JW

[KMale.1]

Thank you for the reply. I'll try this.

[KMale.1]

I'm using STM32 F767zi. My SYSCLK is 180MHz and USART2 clk is 45MHz. Encoder baudrate is 38400bps and I set USART2 baudrate equals to that.

[waclawek.jan]

I mean, what's the source of clock, HSI or any other internal RC oscillator - which are imprecise and drift significantly with temperature - or crystal oscillator (HSE)?

And by verification, I mean measuring using oscilloscope or logic analyzer.

JW

[KMale.1]

I used HSE crystal oscillator(25MHz). I didn't verify the baud rate using oscilloscope and I'll do it.