smlogan
Community Member
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Posts: 51
Boston, MA
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Hi Peter,
All good questions! I'll provide my two cents on the issue.
First, I believe the following approximations may only be used if the LVDS buffers have sufficient bandwidth relative to the random and deterministic jitter components. Specifically, as the total jitter causes the eye to be closed, the waveform presented to an LVDS buffer may not adequately achieve the required logic high and low level amplitude. In this case, one must examine the noise margin of the series of buffers. Clearly, at some level of total jitter, the series connection of buffers will not produce the input waveform accurately - i.e., a logic high or low at the input will not produce a logic high or low at the output of the chain. I think by examining the noise margin of the series connected buffers, one may estimate what level of total jitter must be examined in more detail.
If, however, the total jitter is not close to a level that impinges on the noise margin, I think the following approximations might be in order.
In the case of random jitter, which should be a minor contributor, I would expect that over the frequency range of interest, for N LVDS buffers the output random jitter may be obtained by the sum of the squares.
sigma_RJtotal = power(N*power(sigma_rj,2),0.50)
This is an approximate relationship as long as the same integration period is used for the random jitter for all buffers - as well as the output jitter.
For deterministic jitter, indeed, each LVDS buffer will contribute jitter and the total peak-to-peak jitter may be obtained as the sum of the deterministic jitter added by each individual buffer - if each buffer contributes the same amount DJpp_per_buffer.
DJpp_total = N * DJpp_per_buffer
where: DJpp_per_buffer = Peak-to-peak DJ for each buffer in the chain
However, as the capacitive load on the buffers may vary, it is highly likely you will need to consider the DJ contribution of each buffer separately and sum the peak-to-peak of each buffer. The DJ will be a function of buffer load.
I realize this is different than the method suggested in the prior reply. If that hypothesis were correct, one could simply place an infinite number of buffers in series and only see an output deterministic jitter representative of a single buffer. However, in reality, despite the fact that a buffer may improve the transient response of a waveform, its output edges at the threshold are modulated. Intuitively, as an example, if the deterministic jitter is caused by a modulation of the power supply, in effect the propagation delay of the buffer is modulated. Each buffer has its own propagation delay that is modulated and hence the total modulation is the sum of the modulation of each buffer - not the modulation of just the last buffer in the chain.
Feel free to comment - and I hope I've managed to interpret your question correctly!
Shawn
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