From: [email protected]
Date: Mon Jul 31 2000 - 09:44:40 PDT
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"
> 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, non-reflective
lines with greater timing precision (less skew) in Case 1 relative to Case 2.
Michael L. Conn
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