[time-nuts] cheap 5V OCXO in 14DIP has about 1E-9 drift per day

Bruce Griffiths bruce.griffiths at xtra.co.nz
Sat Apr 9 20:16:26 UTC 2011


There is no subharmonic energy in the input signal to a standard 
regenerative divide by 2 yet, correctly adjusted, a regenerative divide 
by 2 circuit output has very low phase noise over and above that of the 
input signal.
The loop starts when the input signal exceeds the threshold required for 
the loop gain to exceed unity.

The 16MHz is necessary for the loop to function:
The mixer mixes down the 26MHz to a pair of conjugate frequencies, 10MHz 
and 16MHz.
Thermal and device noise is sufficient to start the process.

10MHz = 26MHz - 16MHz
16MHz = 26MHz - 10MHz

The tuned circuits in the feedback loop are required to suppress the 
loop gain at other frequencies to avoid unwanted outputs such as 13MHz 
(fin/2) and 39MHz (3 fin/2).

Bruce

Greg Broburg wrote:
> In looking at the CRD idea, it is not obvious to me how either
> of the 16M0 Hz or the 10M0 Hz signal would contain any
> relevant energy that would give the precision necessary for
> the desired ideal result.
>
> The only harmonic relationship that I can see is to square up
> the clean analog and divide by 13 down to 2M0 Hz, thence use
> the 5th and 8th harmonics at 10M0 and 16M0 but then why
> use the 16M0 at all?
>
> I expect that I am missing something obvious here
> a little nudge may help.
>
> Regards;
>
> Greg
>
>
> On 4/7/2011 1:23 PM, Chris Albertson wrote:
>> On Thu, Apr 7, 2011 at 12:01 PM, Bruce Griffiths
>> <bruce.griffiths at xtra.co.nz>  wrote:
>>> A conjugate regenerative divider with 2 parallel (16MHz&  10MHz) low Q
>>> bandpass filters should suffice.
>> Never having used one of those, I'm thinking I'd need two mixers, a
>> 10MHz bandpass filter and a 16 MHz bandpass filter.  I'm guessing
>> there would need to be some gain in the system too.
>>
>> How stable are these?  Seems to me If the bandpass filters were
>> temperature sensitive the 10MHz output would drift around.    What
>> types of filters and mixers are typically used?
>>
>> I wonder if "conjugate regenerative divider" qualifies as an
>> inexpensive converter
>>
>>
>>
>
>
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