BAUD RATE CALCULATION FOR USART IN STM32

Because some amount of error is allowed in the USART clock rate, and it's the closest achievable value. Typically under 1% is fine. No clock has perfect accuracy.

104.1875 = 0x683 / 16

The BRR uses a fixed-point representation, ie Q12.4

int((104.1875 * 16) + 0.5) = 1667 = 0x683

Personally I've been using BRR = APBCLK / BAUD for 13+ years, it is simpler to explain/compute.

You are basically computing the 16x rate, the USART divides the window into 16 pieces, and can realign the input based on the center time of the bit, basically they have a 16-bit shift register on the input, and they use that to recognize the input/edges.

> Personally I've been using BRR = APBCLK / BAUD for 13+ years, it is simpler to explain/compute.

+1, but the calculation gets more tricky if you want to use the 8x oversampling. OTOH, I'm yet to find a reason for that (except extreme baudrates, which are not a good idea anyway).

JW

No.

Oversampling in context of UART means, how many times one bit is sampled. Normally, it's 16x, so the UART's machine has to be clocked at a frequency which is 16x the UART baudrate.

That's what the integer part of the divider ensures; however, the UART clock generator is not integer-divider but fractional-divider (look up either this or "fractional-N divider"). The resulting clock is not entirely regular, but in average it performs quite well even if the APB clock is not integer-divisible by baudrate*16.

The STM32 UART has an optional mode where oversampling is not 16x but 8x, see OVER8 bit.

Read the UART chapter in RM, this is described there.

JW

Yes.

> And is there any reason for change the oversampling rate?

In the extreme case of very low AHB/APB clock, you can achieve higher baudrates, at the cost of somewhat lower baud mismatch immunity.

JW