Nope. The area is a measure of the current charging this
capacitance, whether lumped or distributed.
Say, for example, you had a section of line lower than Zo due
to extra capacitance (from higher dielectric constant, for
instance). If you drew the bounce diagram (or used a TDR)
and calculated the AREA (area above the line is negative),
then AREA=Zo*C/2 would give the correct extra capacitance.
> I still thought of a via as a distributed LC circuit which often "appeared"
> more capacitive than inductive, however, if I happened to build a via with
> an impedance which matched my line, then I would get no reflection and
> according to your formula, no capacitance, when in fact the via still has
> capacitance associated with it (only the "ratio" of L to C matches that of
> the line). Comments? Thanks in advance for your feedback.
You're right. The formula gives "extra" capacitance above
and beyond what's needed for Zo.
> > _____________. . . . . . . . ._____________
> > / \ _/
> > / \ Zo*C/2 __/
> > / \________/
> > /
> > ____/
ps: This also works for (extra) series inductance, using AREA above
the line and the time constant L/(2*Zo).
pps: The formula needs correction for incorrectly terminated lines,
unless one cuts off the area calculation before the TDR "sees"
the termination (or lack of it).
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Mike Jenkins Phone: 408.433.7901 _____ LSI Logic Corp, ms/G715 Fax: 408.433.7461 LSI|LOGIC| (R) 1525 McCarthy Blvd. mailto:Jenkins@LSIL.com | | Milpitas, CA 95035 http://www.lsilogic.com |_____| ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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