Re: [SI-LIST] : A simple topology

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From: Silvio ORSI (Silvio.Orsi@cern.ch)
Date: Mon Jul 31 2000 - 09:25:11 PDT


Hi Kent,

My two cents: case 1 is always better than case 2, at least regarding
the
reflections coming from the T-point.
  Case 1, L0 = 5.00", L1 = 1.00", L= 6.00"
  Case 2, L1 = 1.00", L2 = 5.00", L= 6.00"
Let's consider the outgoing signal at RX1. In both cases 1 and 2 the
reflections from RX2 and RX1 to the T-point and then back to RX1 will
cancel each other (one is positive, from RX2, the other negative), while
the reflection from T-point back to TX and again to RX1 will arrive much
before in case 2 (8", less than half ns, depending on epsilon) compared
to
case 1 (16", exactly the double in your example). Of course for
not-so-high frequency this difference may not be that important, but for
high frequencies rising and falling edges are affected even by this tiny
time delay. (a negative reflection on the rising edge is never ok)

              / RX1
             /
            /
     TX ___/ T
        L0 \
            \
          L1 \
              \ RX2

Regards,
        Silvio

On Mon, 31 Jul 2000 Kent.Zhang@smartm.com wrote:

>
>
> Let's take a net with one driver and two
> identical receivers and analyze a simple topology.
>
> Driver to T-point = L0
> T-point to both receivers = L1
> Total length from Driver to receiver L = L0 + L1
>
> Case 1, L0 = 5.00", L1 = 1.00", L= 6.00"
> Case 2, L1 = 1.00", L2 = 5.00", L= 6.00"
>
> Question:
>
> Is Case 1 ALWAYS better than Case 2 ? If not,
> under what conditions? Variables can be types of
> drivers and receivers, frequency, trace impedance,
> stackup etc.
>
>
> Kent Zhang
> SMART Modular Technologies
>
>
>
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