Your equation does not tell you what the correct termination value is - it tells you what the minimum permissible value is while remaining within the DC spec of the driver. That is solely a function of the driver, and not a function of the interconnection Zo. The greater-than-specified swing is not atypical. It merely means that the driver has a better-than-spec current sink capability. I would expect that of most parts at nominal Vdd. You don't mention whether this is a steady-state condition, or whether it is a transient value. You need to look at the waveforms at the driver and at the terminated ends of the bus to figure out what the right impedance is. As you try to analyze the waveforms, imagine the driver to be a TDR with a very low source impedance (6-8 ohms). You can probably make a pretty good guess at the effectiveimpedance of the line by using the loaded line correction factor (1/sqrt(1+Cl/Co))* Zo. A 50 ohm board can easily look like 40 ohms or less (much less,in some configurations!) when loaded with distributed C. Hopefully, your loads have minimum stub length off of the main run. If the stubs are short, they may be treatable as a capacitive load for purposes of this estimation. If they are not, you probably need to resort to simulation to find the best termination value. There's not enough info for a specific diagnosis. Regards Mike Samuel Dadel wrote: >Hi! All, I am new to transmission lines: so I would greatly appreciate of someone could give me some leads on the following. I am employing a dual-ended parallel termination scheme for a GTL+ bus. The driver has the capacity to sink 50mA. The Vol is .5 and the Voh is 1.5, and the Vref is 1.0 [Nominal]. When employing a 50ohm termination scheme I see a voltage swing of 1.15 Volts at the driver, with Vol gone down to 0.3 Volts. The nominal values > > are 1.5 and 0.5. I get bit errors, that go away when Vref > > is lowered to 0.8 Volts. > > Could this be a case of unmatched termination? In addition: these is this additional equation that I cannot relate to the characteristic impedance of the trace. Iol[max] = .050 Amps. Iol = Vtt - Vol / R[eff]. R[eff] = 1.5 - 0.5 / .050 = 20 ohms. For a dual ended termination scheme Rtt = 20 * 2 = 40 ohms.Thanks,Daniel > > > >--------------------------------- >Do You Yahoo!? >Yahoo! Finance - Get real-time stock quotes > >------------------------------------------------------------------ >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 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 > > > ------------------------------------------------------------------ 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 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