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

[Mike Brown <bmgman@xxxxxxxxxx>]

Samuel,

One driver to two receivers is not at all unusual.  What may be unusual 
is their positions on the bus topology.  It would help if you could post 
a description of the topology of the net (approximate lengths between 
terminators, drivers, and stubs), and the particulars of the board 
stackup and trace cross section.  What frequency are you trying to 
drive?  Is the signal crossing any cuts in the plane(s) above/below it, 
assuming that such exist?

Without knowing more, I'd venture that the 50 ohm termination is 
probably close unless the board is of unusual construction. I would also 
note that there needs to be a short-lead bypass capacitor very near 
 each terminator.  Anything in the range of 1 nF to 0.1 uF ought to be a 
good start.  Perhaps not optimum, but probably "close enough".  Until I 
can see the bus topology, I can't suggest whether or not a single 
terminator would work.

I'm not about to rehash the recent thread(s) on bypassing.  Study them 
at your leisure.

Samuel Dadel wrote:

> Hi Mike,
>
>   Thanks much for your answer - it helps a great deal.
>
>   We have picked up 50ohm parallel termination resistors.
>
>   Though one unique thing about our bus configuration is that
>
>   one GTL driver drives 2 GTL receivers. Do you think this would
>
>   bring down the effective impedance?
>
>   We are constrained for time, and if we have leeway to only experiment
>
>   with one GTL bus termination, what should we terminate the resistor 
> values to be?
>
>   45 ohm - 40 ohms? 
>
> Thanks and greatly appreciate your answer,
>
> Samuel
>
>  Mike Brown wrote:
>
>     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
>     effective
>     im! pedance 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 sw! ing 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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