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

Javier Herrero jherrero at hvsistemas.es
Sat Jan 28 14:22:38 EST 2012


Only to note that I measured at the wrong place... the reference 
frequency to the DDS is 20MHz, not 10MHz as I stated :)

Regards,

Javier, EA1CRB

El 28/01/2012 15:21, Javier Herrero escribió:
>
>
> 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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