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Question about DS92LV2411

ryuji asano
Intellectual 800 points

Hi,Expart.
Please help me.

My Customer want to calculate the below.

[Samling Window = 1UI -trjit]

How do you think about trjit?

Datasheet says as follows.(Page 20&18)

<Page.20>
　tIJT (jitter freq<2MHz) 0.9UI[typ]
　tIJT (jitter freq>6MHz) 0.5UI[typ]

<Page.18(Figure 16. Receiver input Jitter Tolerance)>

tRJIT=RxIN_TOL(Left + Right)
　Samling Window = 1UI -trjit

I have two questions.

１．
I think as follows: tIJT = tRJIT

Is my understanding correct?

２．
2MHz<jitter freq<6MHz

In the above conditions, What UI[typ] is tIJT?


Please teach me.
Best Regards,
Ryuji

  • 0
    TI__Mastermind 29655 points
    Hi Asano-san,

    Determining jitter tolerance can be understood as seeing how much sinusoidal jitter that the deserializer CDR can tolerate and still achieve acceptable performance.

    1. From the description in p.20 and the figure on p. 18, I agree with your understanding that tIJT is conceptually the same as tRJIT. tIJT describes the amount of input jitter (within a certain amplitude) is allowed by the deserializer, and hence, the sampling window is 1 UI minus the specified value in tIJT, just the sampling window is 1 UI minus the specified value tRIJT.

    2. In the situation that the jitter frequency is between 2-6 MHz, the jitter tolerance slopes downward, as is typically seen for a generic jitter tolerance curve. From the characterization data I have found, you can estimate the jitter tolerance by drawing a line from 0.9 UI at 2 MHz to 0.5 UI at 6 MHz with the frequency scaling logarithmically. In other words, you can estimate the tIJT for this intermediate section with the following semi-log line equation:

    tIJT is approximately tIJT (UI) = 1.208 x e ^ (-0.147 x jitter_freq), where 2 MHz < jitter_freq < 6 MHz.

    Thanks,

    Michael