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Issue with uart USART_ISR_EOBF bit

Forum|Forum|2 years ago
January 24, 2024
10 replies
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Hi

I have a stm32L433, and setup Uart1 for standard TX and RX with interrupt for both send and receive (9600N81 setup). I don't use any hardware handshake . Within the IRQ routine occasionally the USART_ISR_EOBF bit in the ISR is set , and I am not using smartcard mode , and the EOBIE bit is not set in the CR2 register.. I have the below , and even after its cleared it then keeps getting set and cant get any data.

void USART1_IRQHandler(void)

{

unsigned char data;

if(USART1->ISR & USART_ISR_FE)

USART1->ICR |= USART_ICR_FECF;

if(USART1->ISR & USART_ISR_EOBF)

USART1->ICR |= USART_ICR_EOBCF;

if(USART1->ISR & USART_ISR_RXNE) //RX interupt

{

data=USART1->RDR;

...

Many thanks

Scott

STM32L4 series

Best answer by Tesla DeLorean

Is the RMW to ICR the appropriate approach?

What else in ISR is set. Show configuration of the USART and the registers when it fails.

The Noise, Overrun or Framing Errors are more probable.

Andrew Neil

Super User

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Bob S

Super User

I would not expect that bit getting set to prevent receiving incoming bytes. But I have zero experience using smartcard mode. And, yes, I understand (or presume) that you are not using smartcode mode, just normal UART. From the code you posted, it looks like it should flow down to the RXNE block whether or not EOBF is set. Is that happening?

Are you using the HAL libraries? It doesn't look like it from your code above, or at least you are not using HAL code to respond to the interrupt.

For STM32 code it is good practice when checking interrupt flags to check BOTH the ISR bit AND the IE bit, and only respond if both are set. Look at the HAL UART IRQ handler for examples (even if you aren't using HAL), in stm32l4xx_hal_uart.c HAL_UART_IRQHandler().

SSmit.13Author

Senior

Hi

Most times it works fine, its the other times the EOBF is set. When its set, it doesnt run down to the RXNE routine.

You mention "check BOTH the ISR bit AND the IE bit". I check the ISR register as below, whats the IE bit?

Thanks

Scott


void SetupUart1Port(void)
{
	//////////////////////////
	//setup serial port...
	//UART 1
	//PB6-TX PB7-RX
	GPIOB->MODER &= ~0b1111000000000000; //clear pins PB6 , PB7
	GPIOB->MODER |= 0b1010000000000000; //alternate function for both pins

	GPIOB->AFR[0] &= ~0xff000000; //mask out
	GPIOB->AFR[0] |= 0x77000000; //AF7 for both PB6 and PB7

	GPIOB->OSPEEDR &= ~0xf000; //mask out
	GPIOB->OSPEEDR |= 0xf000; //speed high for PB7 and PB6

	GPIOB->PUPDR &= ~0xf000; //mask out pullup
	GPIOB->PUPDR |= 0x5000; //pullup

	RCC->APB2ENR |= RCC_APB2ENR_USART1EN; //enable uart 1 clock

	//??RCC->CCIPR &= ~0xc0;//mask out clock
	RCC->CCIPR &= ~0xf;//mask out clock
	RCC->CCIPR |= 0x5;//select SYSCLCK

	USART_Init(USART1,BAUD9600);
	NVIC_SetPriority(USART1_IRQn,1);
	NVIC_EnableIRQ(USART1_IRQn);
	USART1->CR1 &= ~USART_CR1_TXEIE; //clear tx interupt com 1
	USART1->CR1 |= USART_CR1_RXNEIE; //rx interupt com 1
}


void USART_Init(USART_TypeDef *USARTx,int Baud)
{
	USARTx->CR1 &= ~USART_CR1_UE; //disable uart
	USARTx->CR1 &= ~USART_CR1_M0;// 8 bits
	USARTx->CR1 &= ~USART_CR1_M1;// 8 bits
	USARTx->CR2 &= ~USART_CR2_STOP;// 1 stop bit
	USARTx->CR1 &= ~USART_CR1_PCE;// no parity
	USARTx->CR1 &= ~USART_CR1_OVER8; //oversample
	USARTx->BRR = Baud; //set baud
	USARTx->CR1 |= (USART_CR1_TE | USART_CR1_RE); //enable tx and rx
	USARTx->CR1 |= USART_CR1_UE; //enable uart

	while((USARTx->ISR & USART_ISR_TEACK)==0); //verify uart is ready for transmission
	while((USARTx->ISR & USART_ISR_REACK)==0); //verify uart is ready for reception
}

void USART1_IRQHandler(void)
{
	unsigned char data;

	if(USART1->ISR & USART_ISR_FE)
	 USART1->ICR |= USART_ICR_FECF;

	if(USART1->ISR & USART_ISR_EOBF)
 USART1->ICR |= USART_ICR_EOBCF;

	if(USART1->ISR & USART_ISR_RXNE) //RX interupt
	{
 data=USART1->RDR;
 ......
 }
}

Bob S

Super User

> When its set, it doesnt run down to the RXNE routine.

The code you posted SHOULD still get to the RXNE test even if EOBF is set. Are you saying that when EOBF is set RXNE is not?

> whats the IE bit

The "interrupt enable" bit in the CR1 register, for example RXNEIE. You code should only respond to an interrupt if the interrupt/ISR flag is set AND the corresponding IE bit is set. Normally this is as issue only when enabling and disabling interrupts, as it is possible to enter the IRQ handler after your code has cleared the IE bit (depends on timing or IRQ vs timing of code and instruction pipelining). This is mostly "good programming practice" but is likely not the issue here.

Tesla DeLoreanAnswer

Guru

Is the RMW to ICR the appropriate approach?

What else in ISR is set. Show configuration of the USART and the registers when it fails.

The Noise, Overrun or Framing Errors are more probable.

Tips, Buy me a coffee, or three.. PayPal VenmoUp vote any posts that you find helpful, it shows what's working..

SSmit.13Author

Senior

What I have noticed is if I clear the flags before checking the RXNE flag, then the RXNE gets cleared. If I put at the end then works as expected. I am thinking I dont even have to check this flag as I dont use this feature.

void USART1_IRQHandler(void)
{
	unsigned char data;
	if(USART1->ISR & USART_ISR_FE)
	 USART1->ICR |= USART_ICR_FECF;

	if(USART1->ISR & USART_ISR_EOBF)
	 USART1->ICR |= USART_ICR_EOBCF; <<this clears the RXNE bit

	if(USART1->ISR & USART_ISR_RXNE) //RX interupt
	{
		data=USART1->RDR;
 ...
 }
}

Below is the snap shot prior to moving the above to the end..

Best Regards

Scott

Bob S

Super User

As @Tesla DeLorean said (or strongly hinted at) - do not do a read-modify-write (RMW) to the ICR (oops, I missed that). And the "|=" operation is a RMW. Just write the (single) bit for the flag you want to clear.

SSmit.13Author

Senior

I thought the only way is to either do a |= or read in the register then | the bit and write the result back in. Either way its read in. If not how would I set the bit without doing the just mentioned?

Tesla DeLorean

Guru

The important thing to understand is that these aren't "memories", but bits into a state machine or other combinational logic. The Reference Manual will likely explain or indicate expected behaviour. Doing register level coding requires a lot higher level of awareness to what the peripheral designer implemented, and why.

Writing SET bits can be done with a singular write instruction.

Using |= or &= can have other side effects if it reads high or low bits. It's also very inefficient.

The one that most often caused issues was TIM->SR &= ~1; You'd occasionally end up clearing bits you didn't mean too if an interrupt was latched between the read and write operations.

Doing this is also very inefficient, the optimizer can't fold actions on a "volatile" memory interactions. Your doing a pair of LOAD/STORE operations when you could do it in half of that.

	GPIOB->AFR[0] &= ~0xff000000; //mask out
	GPIOB->AFR[0] |= 0x77000000; //AF7 for both PB6 and PB7

	GPIOB->OSPEEDR &= ~0xf000; //mask out
	GPIOB->OSPEEDR |= 0xf000; //speed high for PB7 and PB6

Do this

GPIOB->AFR[0] = (GPIOB->AFR[0] & ~0xff000000) | 0x77000000; //AF7 for both PB6 and PB7

Tips, Buy me a coffee, or three.. PayPal VenmoUp vote any posts that you find helpful, it shows what's working..

Bob S

Super User

Just write to it. Only bits that are "1" do anything.

if(USART1->ISR & USART_ISR_EOBF) {
 USART1->ICR = USART_ICR_EOBCF; // Clear EOBCF and only EOBCF
}

SSmit.13Author

Senior

I never knew this about only 1 bits do anything, I must have missed that.

Thanks for letting me know

:grinning_face:

Bob S

Super User

Show your current code.

SSmit.13Author

Senior

Below is the UART handler. When I placed a beakpoint on the USART1->ICR = USART_ICR_EOBCF; line ,the RXNE =1. As soon as I step onto the next instruction the RXNE bit is cleared.

void USART1_IRQHandler(void)
{
	unsigned char data;

	if(USART1->ISR & USART_ISR_EOBF)
	 USART1->ICR = USART_ICR_EOBCF; //after stepping pass this the RXNE is cleared

	if(USART1->ISR & USART_ISR_RXNE) //RX interupt
	{
		data=USART1->RDR;
	 ....
 }

Scott

Pavel A.

Super User

So change it as following:

void USART1_IRQHandler(void)
{
	unsigned char data;
	if(USART1->ISR & USART_ISR_RXNE) //RX interupt
	{
		data=USART1->RDR;
	 ....
 }
	if(USART1->ISR & USART_ISR_EOBF)
	 USART1->ICR = USART_ICR_EOBCF; // why this at all, in plain UART mode?
}

SSmit.13Author

Senior

I previously done this, but was wanting to check what you previously said about , only a 1 bit effects the bits in the register. I removed the above line , but was curious why the RXNE bit was cleared. Still dont know what single write without a read would work.

Pavel A.

Super User

Bits in the ICR register are write-only, so there's no sense in reading it anyway. It can read as all 0's or return some random value.

Bob S

Super User

> As soon as I step onto the next instruction the RXNE bit is cleared.

One possible explanation - if you have the SFR view open to show the UART registers, it re-reads those registers after every step. If the IDE reads/diplays the RDR register, that will clear the RXNE bit. The IDE should **NOT** automatically read that register for exactly this reason, and I don't think it does. But I don't recall for sure. Or - do you by chance have a "watch" set on the RDR?

> Still dont know what single write without a read would work.

Look in the ref manual. The bits in the ICR register are of type "rc_w1", which it says means "Software can read as well as clear this bit by writing 1. Writing 0 has no effect on the bit value." Not the most clear explanation to me, but it means you can read the register if you want., but you don't have to. To clear a bit, write a "1" to that bit position. Writing a "0" to a bit has no effect. THAT is why writing a single bit works to clear that bit, you are writing 0's to all the other bits.

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