[time-nuts] Question for my new GPSDO

[Attila Kinali]

On Tue, 15 Oct 2019 18:43:29 -0700
Hal Murray <hmurray at megapathdsl.net> wrote:

> You can get a few bits by reducing the tuning range.  Suppose you follow the 
> DAC with an OpAmp with a gain of 1/8 while the reference is adjustable.  That 
> adds 3 bits of resolution by decreasing the output range.  The downside is 
> that you have to manually set the adjustment.

Unless you use fixed, metal foil resistors, then the temperature
coefficient and the 1/f noise of the resistors will dominate the
system. Using a pot is even worse, as these have quite high 1/f noise.
If you go through the math, it turns out that using two DACs with
metal foil resistors is both going to be lower in noise/tempco
and also giving you the ability to tune the "midrange point" in software.

 
> That adds the op-amp to the temperature consideration list.

The opamp should be on this list anyways. The input offset tempco of
most opamps is high enough that it beats those of the DAC.
Hence, either precision bipolar or auto-zero/chopper opamps
have to be used.
 

> > While temperature coefficient is important, it is not as important as you
> > think. Unless you want to operate your GPSDO outside or where you expect
> > large temperature swings. Instead, what you do is give your whole GPSDO
> > enough thermal mass, such that you don't see a significant temperature change
> > at time scales shorter than the loop time constant.  ...
> 
> How big a box of whatever do I need to get the thermal time constant large 
> relative to 100 or 1000 seconds?

Let us do a quick back-of-the envelope calculation:
Let us assume we have to keep the temperature within 0.1°C within 1000s
Let us further assume we have a 1°C temperature jump.
First thing we need is an estimate of the thermal resistance of the
system. Something in the order of 10K/W to 100K/W are probably sane
values. Let's go with 10K/W for a worst-case scenario.
Now we go from a temperature difference of 1K to 0.9K over 1000s.
This is low enough that we can assume the temperature changes linearly,
which gives us an average temperature difference of 0.95K. Over 1000s
with 10K/W that means we are transfering 0.95/10*1000=95W·s. Ie we need
to provide 95J over a temperature difference of 0.1K, which means we need
a heat capacity of 950J/K. Going to your physics book, you see that
aluminium has 900J/(K·kg) of heat capacity. Which means we need about 1kg
of aluminium.


				Attila Kinali 

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