stm32f103: flash latency and baud rate calculation

Hi all,


I am still relative new to low-level programming microcontrollers, 
currently doing some work on a STM32F103RB using forth.

To get to know the device, I am now trying out some code to change the 
clock from internal8Mhz, to External48Mhz and External72Mhz.

There are two things I do not really understand.

1/ What exactly is the purpose of the Flash Latency? You need to change 
this to 0, 1 or 2 depending on the clock-speed.
As far as I understand it, it defines the speed to access (read) the 
flash memory of the device.

When looking at some C-code I on the internet to set the clock-speed to 
8 to 72 MHz, I noticed that they change the clock FIRST and THEN 
increase the latency from 0 waitstate to 2 afterwards.

I would have expected the code to slow down the access to the flash 
FIRST before stepping up the speed.


2/ Page 791 of the STM32F10x_reference manual say that the speed of a 
USART is Tx/Rx (baud) = f(ck) / (16*USARTDIV).

"USARTDIV is an unsigned fixed point number that is coded on the 
USART_BRR register."

However, the value in the BRR register is a 16 bit value, with the last 
4 bits as the fraction of USARTDIV.
There are some examples on how to calculate the USARTDIV value on page 
791 and 792.


Now, I could be wrong, but it looks to me that, if you want to have a 
certain baudrate, you can just ignore the "*16" part in the formula and 
the "fraction of USARTDIV".

If you just say "Tx/Rx (baud) = f(ck) / USARTDIV" and have USARTDIV 
simply as an 16bit integer value, you get exactly the same result.

Or am I missing something here?
Is there a special reason why this "*16" and "4 last bits = fraction of 
USARTDIV" is needed?




Cheerio! Kr. Bonne.

Hi,



Please ignore question 1. My mistake! :-(


Cheerio! Kr. Bonne.


On 28-10-15 22:52, kristoff Bonne wrote:
> Hi all,
>
>
> I am still relative new to low-level programming microcontrollers,
> currently doing some work on a STM32F103RB using forth.
>
> To get to know the device, I am now trying out some code to change the
> clock from internal8Mhz, to External48Mhz and External72Mhz.
>
> There are two things I do not really understand.
>
> 1/ What exactly is the purpose of the Flash Latency? You need to change
> this to 0, 1 or 2 depending on the clock-speed.
> As far as I understand it, it defines the speed to access (read) the
> flash memory of the device.
>
> When looking at some C-code I on the internet to set the clock-speed to
> 8 to 72 MHz, I noticed that they change the clock FIRST and THEN
> increase the latency from 0 waitstate to 2 afterwards.
>
> I would have expected the code to slow down the access to the flash
> FIRST before stepping up the speed.
>
>
> 2/ Page 791 of the STM32F10x_reference manual say that the speed of a
> USART is Tx/Rx (baud) = f(ck) / (16*USARTDIV).
>
> "USARTDIV is an unsigned fixed point number that is coded on the
> USART_BRR register."
>
> However, the value in the BRR register is a 16 bit value, with the last
> 4 bits as the fraction of USARTDIV.
> There are some examples on how to calculate the USARTDIV value on page
> 791 and 792.
>
>
> Now, I could be wrong, but it looks to me that, if you want to have a
> certain baudrate, you can just ignore the "*16" part in the formula and
> the "fraction of USARTDIV".
>
> If you just say "Tx/Rx (baud) = f(ck) / USARTDIV" and have USARTDIV
> simply as an 16bit integer value, you get exactly the same result.
>
> Or am I missing something here?
> Is there a special reason why this "*16" and "4 last bits = fraction of
> USARTDIV" is needed?
>
>
>
>
> Cheerio! Kr. Bonne.

On Wed, 28 Oct 2015 23:29:48 +0100, kristoff Bonne
<kristoff@skypro.be> wrote:

>> Now, I could be wrong, but it looks to me that, if you want to have a
>> certain baudrate, you can just ignore the "*16" part in the formula and
>> the "fraction of USARTDIV".
>>
>> If you just say "Tx/Rx (baud) = f(ck) / USARTDIV" and have USARTDIV
>> simply as an 16bit integer value, you get exactly the same result.
>>
>> Or am I missing something here?
>> Is there a special reason why this "*16" and "4 last bits = fraction of
>> USARTDIV" is needed?

UART clocks need to be set up to within about 1% of the nominal clock 
speed for synchronisation to occur.

-- 
Stephen Pelc, stephenXXX@mpeforth.com
MicroProcessor Engineering Ltd - More Real, Less Time
133 Hill Lane, Southampton SO15 5AF, England
tel: +44 (0)23 8063 1441, fax: +44 (0)23 8033 9691
web: http://www.mpeforth.com - free VFX Forth downloads

On Wed, 28 Oct 2015 22:52:59 +0100, kristoff Bonne
<kristoff@skypro.be> wrote:

>Now, I could be wrong, but it looks to me that, if you want to have a 
>certain baudrate, you can just ignore the "*16" part in the formula and 
>the "fraction of USARTDIV".
>
>If you just say "Tx/Rx (baud) = f(ck) / USARTDIV" and have USARTDIV 
>simply as an 16bit integer value, you get exactly the same result.
>
>Or am I missing something here?
>Is there a special reason why this "*16" and "4 last bits = fraction of 
>USARTDIV" is needed?

(baud) = f(ck) / (16*USARTDIV)

USARTDIV = f(ck) / (baud * 16)

If the fraction after the division is more than one percent of the
total, e.g. 4.3 vs 4 (approx7.5%) then the UART will not work well.

So take the four bottom bits as a multiply by 16:

USARTDIV = f(ck) / (baud)

and you are good to go. I could be wrong - it's a long time since I
did an STM32F103. If you want to be clever, you can approximate the
last bit of the fraction, but it's rarely useful.

Stephen


-- 
Stephen Pelc, stephenXXX@mpeforth.com
MicroProcessor Engineering Ltd - More Real, Less Time
133 Hill Lane, Southampton SO15 5AF, England
tel: +44 (0)23 8063 1441, fax: +44 (0)23 8033 9691
web: http://www.mpeforth.com - free VFX Forth downloads

Hi Stephen


My appologies for the late reply.



On 28-10-15 23:54, Stephen Pelc wrote:
>> Now, I could be wrong, but it looks to me that, if you want to have a
>> certain baudrate, you can just ignore the "*16" part in the formula and
>> the "fraction of USARTDIV".
>> If you just say "Tx/Rx (baud) = f(ck) / USARTDIV" and have USARTDIV
>> simply as an 16bit integer value, you get exactly the same result.
>> Or am I missing something here?
>> Is there a special reason why this "*16" and "4 last bits = fraction of
>> USARTDIV" is needed?
> (baud) = f(ck) / (16*USARTDIV)
> USARTDIV = f(ck) / (baud * 16)
> If the fraction after the division is more than one percent of the
> total, e.g. 4.3 vs 4 (approx7.5%) then the UART will not work well.
> So take the four bottom bits as a multiply by 16:
> USARTDIV = f(ck) / (baud)
> and you are good to go. I could be wrong - it's a long time since I
> did an STM32F103. If you want to be clever, you can approximate the
> last bit of the fraction, but it's rarely useful.

That's also how I saw it.

I did not really understand why they defined it like that so I thought 
there might have been some "snake below the grass" (as we say in dutch :-) )


Thanks for the reply.



> Stephen
Cheerio! Kr. Bonne.