[time-nuts] Why are 1PPS signals so skinny?

Attila Kinali attila at kinali.ch
Tue May 15 20:19:34 UTC 2012


On Tue, 15 May 2012 15:47:46 -0400
Mike S <mikes at flatsurface.com> wrote:

> On 5/15/2012 2:45 PM, shalimr9 at gmail.com wrote:
> > The narrow pulses are easily filtered by the power supply because the
> > frequency distribution of the power consumption has a much smaller
> > component at 1Hz.
> 
> But, since PPS is the leading edge, if the power draw for a longer pulse 
> width causes timing problems, then a short pulse would too, unless the 
> edge can somehow see into the future to know how long the pulse will last.

Not really. Keep in mind that you have capacitors everywhere.
One or two small ones at the output driver directly. Maybe
a few mid sized ones in the power distribution, if the device
is a bit bigger. Bigger capacitors at the devices power supply,
resp. power input. Even the mains socket can be regarded as a very
very big capacitor.

Each of these capacitors stores energy, proportional to its size
(in Farad, i leave the Voltage dependence out for this). If the
PPS pulse is short, it contains very little energy, which means
the energy can be supplied by the small capacitors at the output
driver. The longer the pulse gets, the more energy it needs.
And the more energy it needs, the more capacitors further back
in the power chain get involved. Which means that more subsystems
of the device see the power spike of the output driver (and its assciated
voltage drop). Which might have a negative effect on their performance.

Now, why not place very big capacitors at the output driver?
Because big capacitors are expensive, even more so fast and big capacitors.
And you want to have very fast capacitors to have a sharp rising
edge. As a rule of thumb, the frequency limit of a capacitor is
inversly proportional to its size, if all other factors are kept the same.
It is possible to use different materials or different build-ups which
have better "high speed" properties, but these get very expensive very fast.

So, the rising edge of the pulse, the one we usually look at, is determined
by the output drivers slew rate and the fast capacitors that provide its
energy. But the form of the rest of the pulse is determined by the power needs
of the pulse and the size of the surrounding capacitors (and the impedance
that leads to them). And keep in mind, that it's no use of having a fast
rising edge, if the pulse colapses a couple ns later.

Hope that explains it

			Attila Kinali

-- 
Why does it take years to find the answers to
the questions one should have asked long ago?




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