On Wed, Dec 07, 2016 at 08:39:09PM +0100, Richard Cochran wrote:
> > +static s32 gem_ptp_max_adj(unsigned int f_nom)
> > +{
> > + u64 adj;
> > +
> > + /* The 48 bits of seconds for the GEM overflows every:
> > + * 2^48/(365.25 * 24 * 60 *60) =~ 8 925 512 years (~= 9 mil years),
> > + * thus the maximum adjust frequency must not overflow CNS register:
> > + *
> > + * addend = 10^9/nominal_freq
> > + * adj_max = +/- addend*ppb_max/10^9
> > + * max_ppb = (2^8-1)*nominal_freq-10^9
> > + */
> > + adj = f_nom;
> > + adj *= 0xffff;
> > + adj -= 1000000000ULL;
>
> What is this computation, and how does it relate to the comment?
I am not sure what you meant, but it sounds like you are on the wrong
track. Let me explain...
The max_adj has nothing at all to do with the width of the time
register. Rather, it should reflect the maximum possible change in
the tuning word.
For example, with a nominal 8 ns period, the tuning word is 0x80000.
Looking at running the clock more slowly, the slowest possible word is
0x00001, meaning a difference of 0x7FFFF. This implies an adjustment
of 0x7FFFF/0x80000 or 999998092 ppb. Running more quickly, we can
already have 0x100000, twice as fast, or just under 2 billion ppb.
You should consider the extreme cases to determine the most limited
(smallest) max_adj value:
Case 1 - high frequency
~~~~~~~~~~~~~~~~~~~~~~~
With a nominal 1 ns period, we have the nominal tuning word 0x10000.
The smallest is 0x1 for a difference of 0xFFFF. This corresponds to
an adjustment of 0xFFFF/0x10000 = .9999847412109375 or 999984741 ppb.
Case 2 - low frequency
~~~~~~~~~~~~~~~~~~~~~~
With a nominal 255 ns period, the nominal word is 0xFF0000, the
largest 0xFFFFFF, and the difference is 0xFFFF. This corresponds to
and adjustment of 0xFFFF/0xFF0000 = .0039215087890625 or 3921508 ppb.
Since 3921508 ppb is a huge adjustment, you can simply use that as a
safe maximum, ignoring the actual input clock.
Thanks,
Richard
