[time-nuts] FE-5680A FAQ update: question about frequency synthesizer architecture
jherrero at hvsistemas.es
Sat Jan 28 13:38:53 EST 2012
It can be... but the resolution of the center frequency of two DDS
frequencies would be limited to half DDS LSB... so not enough. I will
try to monitor FSELECT and also to extract some SPI data if I found some
free time today or tomorrow :)
El 28/01/2012 15:35, Azelio Boriani escribió:
> 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.
>>> (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
>> 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 Herrero
>> Chief Technology Officer EMAIL: jherrero at hvsistemas.com
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