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Design >> Mixed-Signal Design >> Stability analysis of a Bang-Bang CDR loop
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Message started by Sumilak1994 on Mar 18th, 2019, 10:05am

Title: Stability analysis of a Bang-Bang CDR loop
Post by Sumilak1994 on Mar 18th, 2019, 10:05am

Hi,
I am simulating the stability model of a Half Rate Bang Bang CDR, but am confused about the value for Kpd.

In most places, I found Kpd defined as a linear function of σ (sigma) as follows:

For Gaussian distribution: (1/σ)√(2/pi)

For Uniform distribution: 1/(σ2√3)

But, I am confused how this "sigma" relates to jitter - whether it's only the random jitter component or the deterministic jitter component or both.

Also, why is only the jitter component of the data and not the clock taken into consideration while determining the Kpd ?

I am in a bit of a fix.   :-/





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