[time-nuts] FE-5680A FAQ update: question about frequency synthesizer architecture

Azelio Boriani azelio.boriani at screen.it
Sat Jan 28 09:35:42 EST 2012

And I think it depends on the two frequencies loaded too. The FSELECT
selects between two phase accumulator steps. Maybe the word sent to the Rb
is manipulated to obtain two symmetric values to load.

On Sat, Jan 28, 2012 at 3:21 PM, Javier Herrero <jherrero at hvsistemas.es>wrote:

> El 27/01/2012 19:27, beale escribió:
>> I added a bit to the "electronics" section of the FE-5680A FAQ as below.
>> http://www.ko4bb.com/dokuwiki/doku.php?id=precision_timing:fe5680a_faq#electronic
>> (Note- until today, I had the 8 and 6 digits transposed, calling it the
>> fe5860a. But no one noticed :-)
>> The updated section is below. I measured the 20 MHz input and 5.3 MHz
>> output of the DDS, but I'm puzzled by how the tuning resolution (4.6 mHz)
>> of the DDS output is divided by such a large factor to achieve 0.18 uHz
>> resolution at the final 10 MHz output. Can any frequency synthesizer gurus
>> explain how this is done?
> The frequencies inside the unit are quite similar to those found in the
> FRS-C, 60 and 5.3125MHz. The FRS-C excites the cavity at 60MHz x 114 -
> 5.3125MHz = 6.8346875GHz (a bit over the 6.834682608GHz Rb natural
> resonance - so I suppose that the resonance is driven to 6.8346875GHz using
> the C-Field), so I understand that the FE-5680A operates in the same way.
> Since in the multiplication process the 60MHz frequency is multiplied by
> 114 and the 5.3125MHz only by one, 1Hz offset in the 5.3125MHz frequency
> will need 1/114Hz offset in the 60MHz signal to obtain the same resonant
> frequency.
> I've checked that the DDS is driven by 10MHz, not 20MHz (I've just checked
> it), so the 5.3125MHz is probably an image and not a fundamental DDS
> output. Hence, the minimum DDS step is 2.23mHz. A change of 2.23mHz in the
> 5.3125MHz frequency is compensated by an approximately 20.45uHz change at
> the 60MHz frequency, and so, a 3.41uHz at the 10MHz output, i.e. one part
> in 3.41^-13. This lets to a factor of 19 between the adjustment attainable
> directly by modifiying a 1LSB and the claimed 1.7854^-14 adjustment.
> Probably this is done by modifying the duty cycle of the FSELECT signal. I
> suspect that the 416.6666667 signal at FSELECT is used to produce the
> modulation on the cavity excitation to perform a synchronous detection (the
> same way it is done in the FRS-C at 127Hz), to obtain a null, so the null
> can be slightly "moved" by variying the duty cycle at FSELECT.
> I will try to play a bit more this evening :)
> Best regards,
> Javier
> --
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> Javier Herrero
> Chief Technology Officer                  EMAIL: jherrero at hvsistemas.com
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