Here's the specific question- is the isolation provided by a very wide
guard trace the superposition of the isolation of several normal width
guard traces placed side-by-side? Example--- assume 5 mil signal traces,
a 5 mil layer thickness of FR4, and a spacing between them of 20 mils
(between centers). The isolation calculated by formula without any guard
trace is 24.6dB. A stitched-ground guard trace would increase this
isolation to 36.6dB.
NOW- if we instead used a 25 mil wide guard trace, centered such that it
is 25 mils from the other traces, what is the isolation? I calculate
48dB without the guard trace. If we assume superposition, then the 25
mil guard trace is like having five 5 mil guard traces, each providing
12dB of isolation; so we have a total of 60dB isolation. This added to
that of the total spacing is 48dB + 60dB = 108dB of isolation. SO- is
this correct? Can I assume that the wide guard trace is the
superposition of several narrow guard traces? Note that this is a big
improvement on using a normal width guard trace, since the spacing
between the aggressor and victim would have to be 1" to get 104dB of
isolation (with a 5mil guard trace); and the wide guard trace approach
gives the same isolation with a spacing of 80 mils. I appreciate
anyone's thoughts.
-- Regards, Gary L. Sanders, Staff Analog Engineer, [email protected] L3 Communications, Inc. Celerity Systems www.csidaq.com Cupertino, CA dir 408-861-7325 fax 408-873-1397 Ultra Fast Acquisition & Data Generation Systems "The fog is wine. The sun is my gold."**** To unsubscribe from si-list: send e-mail to [email protected]. In the BODY of message put: UNSUBSCRIBE si-list, for more help, put HELP. si-list archives are accessible at http://www.qsl.net/wb6tpu/si-list ****