Date: Mon Jul 31 2000 - 14:54:09 PDT
Don't forget that you specified this as a uni-directional topology. Three
bi-directional buffers may drive a different solution.
> -----Original Message-----
> From: MikonCons@aol.com [mailto:MikonCons@aol.com]
> Sent: Monday, July 31, 2000 11:45 AM
> To: Kent.Zhang@smartm.com; firstname.lastname@example.org
> Cc: email@example.com
> Subject: Re: [SI-LIST] : A simple topology
> Per your questions below,
> > Driver to T-point = L 0
> > 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.
> Case 1 certainly has the edge on Case 2 for the following reasons.
> 1. Use a series-terminated driver that will match the L0
> segment of line and
> absorb the load reflection(s). This segment is the longest
> and routing a
> single trace is more space-efficient than routing two (as in Case 2).
> 2. Size the line impedances so that the receiver segments
> (L1) are twice
> that of the driver segment (L0). This construction assures no line
> mismatches at the T-junction.
> 3. Matching the lengths of the two L1 segments is much
> easier over the
> shorter distance of Case 1 versus Case 2.
> 4. The reflections from the receivers will (because of the short L1
> segments) accurately coincide in phase at the T-junction
> (again a matched,
> non-reflective interface) and the combined signal will
> propagate to the
> driver and be series terminated (i.e., no reflection).
> In other words, less routing space is used to achieve clean,
> lines with greater timing precision (less skew) in Case 1
> relative to Case 2.
> Mike Conn
> Michael L. Conn
> Owner/Principal Consultant
> Mikon Consulting
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