[SI-LIST] Re: Middle (Star) Termination
- From: "George Tang" <gtang@xxxxxxxx>
- To: <dbostan@xxxxxxxxx>
- Date: Wed, 25 Feb 2004 22:17:38 -0800
The down side to the star termination is shown below:
1. The trace routing density increases drastically, since you need to route
a separate trace from the (electrical) center point to each load.
2. This works well for the ideal case when the loads behave as open
circuits. But if the load capacitance is 4 or 5pF and the driver puts out a
sub-nanosecond rise time, the wave form at the receiver will have a
significantly slower rise-time.
3. The star termination is patented, so you may need to pay a fee to use it.
With some clever engineering, you might be able to come up with a custom bus
design that eliminates the above problems.
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of Dan Bostan
Sent: Wednesday, February 25, 2004 1:50 PM
To: Scott.Newton@xxxxxxxxxxxxxxxx; Robert Haller
Cc: si list
Subject: [SI-LIST] Re: Middle (Star) Termination
I think this termination technique is well explained
in the Motorola data sheet for clock drivers driving
multiple loads.
/dan
--- "Newton, Scott" <Scott.Newton@xxxxxxxxxxxxxxxx>
wrote:
> 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.
> >
>
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>
> --
> 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
>
>
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