All, I haven't found the paper yet, but I've realized how the star = termination works, at least at a high level.=20 If anyone is interested, here is a summary:=20 Resistors with identical values are connected at the junction of the = star. The value is calculated such that the resistor of the driven leg = plus the parallel combination of the resistors in the undriven legs plus = the parallel combination of the impedances of the undriven leg traces = EQUAL the impedance of the driven leg trace. The resistor value is = calculated as (N-2/N)Zo where N is the number of legs in the star and Zo = is the characteristic trace impedance. The equivalent circuit yields a = matched impedance (almost) on the driven leg to the parallel = combinations of all of the undriven legs/resistors. A half wave is = propagated at the driven resistor (side closest to the driver). When the = wave hits the receivers, it doubles as it reflects (off of infinite = impedance juncture). The reflected wave propagates back to the driver = leg which looks almost like an infinite length matched impedance, so = very little reflection occurs. There are of course details such as the = output impedance of the driver, etc that are not covered in this = summary. Scott -----Original Message----- From: Robert Haller [mailto:rhaller@xxxxxxxxxx] Sent: Wednesday, February 25, 2004 12:54 PM To: Newton, Scott Cc: si list Subject: Re: [SI-LIST] Middle (Star) Termination Scott, John Grebenkemper wrote an excellent paper that is available in = the Designcon98 precedings "Network Topology Analysis Using The Reflection Coefficient" that does and excellent job of not only explaining the theory of middle termination but also presents the math. Regards Bob Newton, Scott wrote: > Can anyone please provide a brief explanation regarding the theory of = a middle termination scheme? > In the book" High Speed Digital Design" by Johnson&Graham the topic = seems to be glossed over. It essentially says that providing resistors = connected in the middle of a star network that have impedance of 1/3 Zo = work by halving the voltage at the node. I've used Hyperlynx to simulate = this arrangement and it works well. I've searched the web and haven't = found a satisfactory explanation. >=20 > Thanks in advance. > ------------------------------------------------------------------ > To unsubscribe from si-list: > si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field > > or to administer your membership from a web page, go to: > //www.freelists.org/webpage/si-list > > For help: > si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field > > List technical documents are available at: > http://www.si-list.org > > List archives are viewable at: =20 > //www.freelists.org/archives/si-list > or at our remote archives: > http://groups.yahoo.com/group/si-list/messages > Old (prior to June 6, 2001) list archives are viewable at: > http://www.qsl.net/wb6tpu > =20 > -- Robert J. Haller (rhaller@xxxxxxxxxx) Principal Consultant Signal Integrity Software Inc. 6 Clock Tower Place, Suite 250 Maynard, MA 01754 Phone: (978) 461-0449, ext 15 ------------------------------------------------------------------ To unsubscribe from si-list: si-list-request@xxxxxxxxxxxxx with 'unsubscribe' in the Subject field or to administer your membership from a web page, go to: //www.freelists.org/webpage/si-list For help: si-list-request@xxxxxxxxxxxxx with 'help' in the Subject field List technical documents are available at: http://www.si-list.org List archives are viewable at: //www.freelists.org/archives/si-list or at our remote archives: http://groups.yahoo.com/group/si-list/messages Old (prior to June 6, 2001) list archives are viewable at: http://www.qsl.net/wb6tpu