[time-nuts] DIY TimePod

John Swenson johnswenson1 at comcast.net
Mon Jun 13 16:54:05 UTC 2016


The sampling will be done by a set of ECL flops. The FPGA is reading the 
already sampled ECL outputs. (with ECL to CMOS converters) I'm using a 
hex ECL register with one clock input for all six flops,

The ECL input circuit is a differential amplifier, I will be feeding the 
CMOS level input to one side and the other side will be a very low noise 
reference voltage set to half the CMOS voltage (1.65V for 3.3V square 
wave).

I really don't know the input noise of that differential amp, but it is 
probably much better than a normal CMOS input. I guess I will find out!

The particular chip says it has a max of 100fs additive jitter on the 
output from the input clock. But that is output jitter, I don't really 
care about output jitter, it is sampling jitter that is important here, 
I'm not sure how those two correlate.

John S.

On 6/13/2016 6:28 AM, Chris Caudle wrote:
> On Sun, June 12, 2016 10:50 pm, John Swenson wrote:
>> I'm just doing phase noise measurements of digital clocks (square waves)
>> so it seems to me I don't need some of the circuitry in the TimePod, in
>> particular the digitally controlled RF attenuators and the ADCs
>> themselves. My idea is to use LVPECL flip-flops to sample the DUT and
>> reference clocks, convert the differential outputs to CMOS and feed the
>> FPGA inputs from that. Yes you loose AM noise riding on top of the
>> square wave, but is that really necessary for just square wave phase
>> noise measurements?
>
> Are the FPGA inputs low enough noise for that?  With an ADC the time
> resolution is a combination of clock noise and input noise, for most high
> quality ADC the effective time resolution you can achieve by analyzing the
> output data stream is much higher than the resolution of the clock period.
> Can you achieve similar with just a single bit quantizer based on the FPGA
> CMOS inputs?
>
>




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