[SI-LIST] Re: Middle (Star) Termination
- From: "George Tang" <gtang@xxxxxxxx>
- To: <weirsp@xxxxxxxxxx>, <dbostan@xxxxxxxxx>
- Date: Thu, 26 Feb 2004 12:41:56 -0800
Routing density is always an issue. Please imagine that you have a 64 bit
bi-directional bus connecting 6 devices together, and each device is a high
pin count BGA, so you can only route ONE trace between the balls of the BGA.
With the star configuration, you will need to route 64 x 6 traces. Please
assume that making a 20 or 30 layer board is out of the question.
The 5pF capacitance is the receiver buffer input capacitance, so you need to
drive that no matter what configuration you use, but the more traces you
connect in parallel, the less current going into each receiver load. I
don't want to give away too much, but you can try this yourself.
The 'STAR' configuration for bus routing is patented. If you use it without
permission, you risk your company being taken to court by the patent holder.
To some, that's a risk that they would like to avoid.
George
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of steve weir
Sent: Wednesday, February 25, 2004 11:31 PM
To: gtang@xxxxxxxx; dbostan@xxxxxxxxx
Cc: si list
Subject: [SI-LIST] Re: Middle (Star) Termination
George,
1) I have never found routing density to be an issue. But with better and
cheaper drivers, the star gets less use these days than it used to.
2) 5pf looking into 50 ohms of a series terminated source will always be
slower than looking into 25 ohms parallel termination. I don't see where
the number of loads that the source drives, assuming it is matched is
visible at the far end of the transmission line.
3) Did the PTO approve a patent on a signal source with multiple loads? I
would like to see someone try and enforce such a patent. I think that is
silly as the Palm space multiplexer patent that is drawing howls of
laughter.
Regards,
Steve.
At 10:17 PM 2/25/2004 -0800, George Tang wrote:
>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.
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