Am 04.07.2013 00:52, schrieb Reinhard Max:
> On Wed, 3 Jul 2013 at 23:35, Frank Schäfer wrote:
>
>> When I wrote the patch in 2009, only the first baud rate programming
>> method was in place.
>> The second method has been added later with commit 8d48fdf6.
>
> Hmm - it would be interesting to know what device and tests that
> change was based on.
Even more interesting would be the Windows driver that has been used for
reverse-engineering. ;)
AFAIK, the standard Windows driver doesn't use the second method (last
checked 2-3 years ago)...
I assume Prolific discloses this extra method/information only to
customers that are asking for what they call "driver customization" in
the datasheet.
>
>> Hmm... the comment says that
>>
>> 12MHz*32 / baud = divisor = (2^buf[1]) * buf[0]
>>
>> But the code below does something different...
>
> According to my tests the comment is true for bits 1..3 of buf[1] and
> the current code seems to work by chance, because it doesn't get used
> for baud rates below 115200.
Yes, I can confirm that. :)
And I'm pretty sure bit 0 of buf[1] belongs to buf[0] !
To summarize our current knowledge:
divisor = 2^A * B
with
A = buf[1] & 0x0E
B = buf[0] + (buf[1] & 0x01) << 8
Example: buf[1] = 0x03, buf[0] = 0x80=128
=> divisor = 2^2 * 0x180 = 4 * 384 = 1536
I've tested this formula and it works from 46 baud to ~1.5MBps.
I'm not sure what happens at higher baud rates - could be a cable problem.
The FTDI-device I'm using as communication partner is also limited to
3.0Mbps.
TODO:
- more validation tests
- B=0 case ? b=value+1 ?
- best method for buf[0], buf[1] determination ? (multiple possibilites
to encode the divisior !)
- ...
>
>> I hope I'll get a FTDI-device for testing tomorrow.
>
> I'll try to hook mine up to a logic analyzer tomorrow, so that I get
> better timing results than from the "estimate the baud rate from the
> time it takes to send and receive a few thousand bytes" method I used
> so far.
Hmmm... would be nice if you could verify the formula with high baud
rates (> 1.5Mbps).
Regards,
Frank
>
>
> cu
> Reinhard
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