What is written to the USART_BRR register?

[MMust.5]

What is this value that is written to the USART_BRR register?

What is it for?

I don’t understand why the calculation is done using the formula: fclock / (the divisor, in my formula it is 5 * baudrate)For example, here is an example calculation:

I made the numbers much smaller to simplify the example.

fclock / (divisor, in my formula it is 5*baudrate)

fclock ==Clock frequency==30 Hertz.

Desired baudrate==2

Divisor == 5

30 fclock / 2 baud * 5 divisor==3 this number is written to the USART_BRR register.

3 what is this number?

What is it for?

BRR== Baud Rate Register. But 3 is not the baudrate I needed. I have 2 desired baudrate.

Or does it not matter what my desired data rate is?

Why would I then enter the desired baud rate if that's the case?

Does this formula calculate the speed that will be applied, and not the speed that I need?

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Maybe this will be useful: Shown in the picture

If a marker is set every 3 bits at a clock frequency of 30 Hertz, based on these markers, we get a speed of 10 baud.

But I needed a speed of 2 baud, this formula: fclock / (divisor, in my formula it is 5 * baudrate) does not produce this calculation.

[Tesla DeLorean]

No idea what you're describing

For 16x Oversampling Mode

BRR = APBxClock / Baud

Say USART1, on APB2, with a clock of 24 MHz

24,000,000 / 115,200 = 208.3333 -> 208

[MMust.5]

USARTDIV=fclock / (16*baud)

USARTDIV=24000000/(16*115200)==13

And what is this number 13?

In this case, the USART_BRR register will be written to the number 13 0

HEX:D0

Decimal: 208

What is this number 208? I do not understand. 

This number is similar to the number of samples for each bit at the specified baud rate of 115200

[MMust.5]

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[MMust.5]

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[MMust.5]

It looks like DIV_Mantissa contains the speed required to transfer the entire data packet.

For example: 24000000/(16*115200)==13.02083

The data transfer rate of 13 baud is necessary in order for the entire data packet of 115200 bits * 16 sampling to be transmitted in one bit at this rate. 115200*16==1 843 200

24000000/13.02083==1,843,200

And I still did not understand how to apply the number 208.

What is this number?

If we divide 208/13.02083==16

Then you can get sampling, maybe in order to divide the packet within one bit at a speed of 13 baud, into 115200 bits. 

So 208 is not a baud rate.

Baud rate is 13.02083

And 208 is entered into the USART_BRR register because the last 4 bits of this register are for the fractional part.

Just such a form of recording in this register.

The actual baud rate is 13.02083

[MMust.5]

I think the number 16 in the formula is to combine the integer and fractional parts in one register.

This trick works only with the number 16.

I didn't manage to do it with other numbers.

[gbm]

That's an old problem with STM32 Reference Manuals. Just forget this "division by 16" and "fractional part" and everything becomes simple. For 16x oversampling (default), the value in BRR should be the properly rounded result of dividing the UART input clock by the baud rate. That's it. No "fractions", no "div by 16".

It is almost the same in 8x oversampling mode. The only difference is bit 3 of BRR is not used and the result of division is written to bits 15..4 and 2..0.

Also, see this link:

https://community.st.com/s/question/0D53W00001UxfdySAB/errors-in-uart-baud-rate-sections-of-reference-manuals

[MMust.5]

24,000,000 / 115,200 = 208

Number 208 What is it?

It's not the baud rate.

The baud rate is 13 baud.

24000000/16*115200==13

[MMust.5]

208 This is the number written in the entire register.

In DIV_Mantissa 13 is written. This is HEX==D

In DIV_Fraction 0 is written

All together this is HEX== D0

In decimal it is 208

And the data transfer rate is not 208, but 13