[time-nuts] Updated Divider Jitter Results - 74HC390

Bruce Griffiths bruce.griffiths at xtra.co.nz
Sun Apr 5 00:42:28 UTC 2009


John

The parameters for a simple model for the Wavecrest input jitter can be
derived from your measurements as

For each channel:
Jitter = SQRT[8E-24 + 2.53E-7/(S*S)]
Where S is the input signal slew rate at the trigger threshold

Input noise ~ 503 uV rms. (2.53E-7 = square of input rms noise)
8E-24 = square of Wavecrest input channel intrinsic jitter (2.8ps rms).

Bruce

John Ackermann N8UR wrote:
> Bruce Griffiths said the following on 04/04/2009 07:30 PM:
>   
>> John
>>
>> I can't find a spec for the Wavecrest 2075 input amplifier/trigger
>> circuit noise but it could be as high as 1mV rms given its 800MHz+ input
>> bandwidth.
>>
>> If the noise is 1mV rms:
>> Then an input signal slew rate of 1V/ns is required to keep the jitter
>> contribution of the amplifier input noise below 1ps rms.
>> A 3 stage limiter cascade with an overall slope gain of about 12x can be
>> used to increase the slew rate of a 10MHz 2V pp input signal to 1V/ns.
>> With an appropriate distribution of limiter stage gain and bandwidth,
>> the jitter contribution due to limiter noise and Wavecrest input noise
>> can be held below1.2ps rms.
>> The jitter contribution due to amplifier input noise with such an input
>> signal connected directly to the Wavecrest input would be about 16ps rms.
>>     
>
> Bruce --
>
> A simple experiment just verified your hunch that input amp noise is a
> limiting factor.  Using the Wavecrest in time interval ("total
> propagation delay") mode with signal going into channel 1, then through
> a 4 foot cable into channel 2 to generate about 6 ns of delay (with a
> tee at channel 2 providing a 50 ohm load):
>
> A 10 MHz sine wave at 1.0V P-P shows a 100K sample jitter of 23 ps.
>
> A 10 MHz pulse train at 1.0V P-P from a 5359A Time Synthesizer (with < 5
> ns transition time) shows a 100K sample jitter of 4 ps.
>
> I think I missed the point of your suggestion a couple of messages ago,
> but I get it now -- to measure the input circuit performance, I probably
> need to use two input circuits, one fed to start and the other to stop,
> so the Wavecrest can get a better slew rate to deal with, then divide by
> sqrt(2).  That'll be part of tomorrow's experiments...
>
> John
>
>
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