[SI-LIST] Re: Middle (Star) Termination

There are a couple of different concepts being mixed together here, I =
think.

A driver driving multiple series terminated lines isn't quite the same =
thing as the star termination.

A low-impedance driver can launch a half-amplitude wave into the load =
t-lines with proper selection of series termination Rt:
                                       =20
                       Rt
Vs --/\/\/\----+-----/\/\/\------ Zo             =20
         Zs    |        .                =20
               +        .             =20
               |        .                   =20
               +-----/\/\/\------ Zo  =20
                                                      =20
The condition to launch Vs/2 is (Zs + (Rt/n)) =3D Zo/n,    =20
  with the limit on n such that Zo/n >=3D Zs.


The superficially similar topology for the star termination is:


                Zo        Rt               Rt                    =20
Vs --/\/\/\---/\/\/\----/\/\/\----+-----/\/\/\------ Zo            =20
      Zs              ^           |        .                   =20
                      A           +        .                   =20
                                  |        .                      =20
                                  +-----/\/\/\------ Zo               =20
                                                                =20
A match at node A requires (Rt + Rt/(n-1)) + Zo/(n-1) =3D Zo =20

As given earlier in the thread, Rt =3D Zo*(n-2)/n.       =20

This leads to at most a half-amplitude signal at node A, which is =
further attenuated by the star such that the wave launched into the =
rightmost Zo's is considerably smaller than 1/2 amplitude: in fact, it =
is 1/4 for n=3D3.    =20

Rt can be reduced to increase the launch amplitude, at the expense of a =
negative reflection coefficient at node A.  There's proably an optimum =
reflection coefficient at that node which minimizes settling time to =
some specified voltage at the far end of the load lines.  I haven't read =
Grebenkemper's paper; I would not be surprised if he has analyzed it.

The slow risetimes discussed in this thread are due more to the low =
launch amplitude into the destination nodes than to any ZoC time =
constant at the far end of the lines.

Regards

Mike



-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]On Behalf Of steve weir
Sent: Thursday, February 26, 2004 1:31 AM
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=20
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=20
the number of loads that the source drives, assuming it is matched is=20
visible at the far end of the transmission line.

3) Did the PTO approve a patent on a signal source with multiple loads?  =
I=20
would like to see someone try and enforce such a patent.  I think that =
is=20
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 =3D
> > termination works, at least at a high level.=3D20
> > If anyone is interested, here is a summary:=3D20
> > Resistors with identical values are connected at the
> > junction of the =3D
> > star. The value is calculated such that the resistor
> > of the driven leg =3D
> > plus the parallel combination of the resistors in
> > the undriven legs plus =3D
> > the parallel combination of the impedances of the
> > undriven leg traces =3D
> > EQUAL the impedance of the driven leg trace. The
> > resistor value is =3D
> > calculated as (N-2/N)Zo where N is the number of
> > legs in the star and Zo =3D
> > is the characteristic trace impedance. The
> > equivalent circuit yields a =3D
> > matched impedance (almost) on the driven leg to the
> > parallel =3D
> > combinations of all of the undriven legs/resistors.
> > A half wave is =3D
> > propagated at the driven resistor (side closest to
> > the driver). When the =3D
> > wave hits the receivers, it doubles as it reflects
> > (off of infinite =3D
> > impedance juncture). The reflected wave propagates
> > back to the driver =3D
> > leg which looks almost like an infinite length
> > matched impedance, so =3D
> > very little reflection occurs. There are of course
> > details such as the =3D
> > output impedance of the driver, etc that are not
> > covered in this =3D
> > 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 =3D
> > 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 =3D
> > a middle termination scheme?
> > > In the book" High Speed Digital Design" by
> > Johnson&Graham the topic =3D
> > seems to be glossed over. It essentially says that
> > providing resistors =3D
> > connected in the middle of a star network that have
> > impedance of 1/3 Zo =3D
> > work by halving the voltage at the node. I've used
> > Hyperlynx to simulate =3D
> > this arrangement and it works well. I've searched
> > the web and haven't =3D
> > found a satisfactory explanation.
> > >=3D20
> > > Thanks in advance.
> > >
> >
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> > Principal Consultant
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