FreeLists / si-list / [SI-LIST] Re: Termination resistance value for GTL+ bus

[Mike Brown <bmgman@xxxxxxxxxx>]

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
 >
 >
 >
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