FreeLists / si-list / [SI-LIST] Re: Return Path

[steve weir <weirsi@xxxxxxxxxx>]

Darshan, please read Dr. Bogatin's book thoroughly.  He demonstrates:
1. That all closed form formulas for wave guide impedance are 
approximations with limited accuracy over finite ranges.
2. Line to line coupling for several configurations.
3. That to get a really accurate answer you need to use a field solver.

Steve.
At 12:24 AM 8/5/2005 -0700, Darshan Mehta wrote:

>Steve and Abhijit,<?xml:namespace prefix = o ns = 
>"urn:schemas-microsoft-com:office:office" />
>
>
>
>Once again Thanks for clarifying my doubts.
>
>
>
>The formula for Capacitance calculation for Stripline configuration is as 
>follows (pF/inch);
>
>
>
>C = 1.4 * Er / ln [ 2.4 * (b / w)]
>
>
>
>Where b = total height of Dielectric between the Plane Layer and w is the 
>width of Trace.
>
>
>Generally we take the height between the Plane layers (return current 
>path) as the b in the capacitance calculation. But as per your mail if the 
>nearby conductors also act as a return current, what dielectric height I 
>should consider while calculating the Capacitance.
>
>Thanks,
>
>Darshan Mehta
>
>
>
>
>steve weir <weirsi@xxxxxxxxxx> wrote:
>Darshan, Abhijit, I find Dr. Bogatin's book "Signal Integrity Simplified"
>is an excellent text that presents the material clearly without losing
>technical accuracy. It belongs right up there with other excellent books
>such as those by Dr. Johnson, Dally & Poulton, Young, and Hall. And I am
>not just saying that because it features Teraspeed's development of "Final
>Inch" for Samtec.
>
>Abhijit is correct. Any conductor is a better wave guide than
>dielectric. Either provide conductors to supply the return path by
>intention, or your signals will find other conductors on their own.
>
>Steve.
>At 11:53 PM 8/4/2005 -0700, Abhijit Mahajan wrote:
> >Darshan
> >
> >Nearby signals traces can also act as return path. One possible way to
> >improve return path situation (on two layer boards) is to use coplanar
> >waveguide configuration. It can be useful for two layer boards having =
> >high
> >speed traces, and also when the board thickness is on the higher side.=20
> >
> >BTW, I haven't read this book but it certainly sounds like a good one =
> >since
> >it got you thinking so much on return currents!
> >
> >Abhijit.
> >=20
> >
> >-----Original Message-----
> >From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx] =
> >On
> >Behalf Of Darshan Mehta
> >Sent: Friday, August 05, 2005 11:35 AM
> >To: steve weir; si-list@xxxxxxxxxxxxx
> >Subject: [SI-LIST] Re: Return Path
> >
> >Steve,
> >
> >=20
> >
> >Thanks for the reply.=20
> >
> >=20
> >
> >So you mean to say that if we don't have Plane Layer, the Electric and
> >Magnetic field will be coupled to the nearby conductor. If we have a =
> >Plane
> >layer also present on the board, does the nearby conductor act as a =
> >Return
> >path? My basic question is, on what basis we should think that this =
> >conductor
> >will act as a return path for given signal. If the Plane shape is =
> >present on
> >the board, does the Plane as well as the nearby conductors will act as a
> >return current or only the Plane shape will act like return current =
> >path.=20
> >
> >=20
> >
> >Thanks,
> >
> >Darshan Mehta
> >
> >
> >steve weir wrote:
> >Darshan, if you have a board with no dedicated plane, then the fields =
> >will
> >just spread out, coupling the signal into multiple conductors. A field =
> >solver
> >can determine the coupling on each line for a given configuration.
> >
> >Steve.
> >At 10:31 PM 8/4/2005 -0700, Darshan Mehta wrote:
> > >Hello Experts,
> > >
> > >
> > >
> > >I have a question on return current. I was going through the book=20
> > >"Signal Integrity Simplified" and came to know that return current path =
> >
> > >can be a Power Plane or Ground Plane. I am still confused about return=20
> > >path. Let me describe what I understood.
> > >
> > >
> > >
> > >The Signal Path is the active path and the Electric field from Signal=20
> > >will terminate in the return path. The Magnetic field will form a=20
> > >circular loop around the signal and it will be coupled with return path =
> >
> > >so that equal and opposite current will flow in return path. Normally=20
> > >if we have Power Plane or Ground place, the return path will be easy to =
> >
> > >find out. Let's assume, if we have a 2 layer board with no copper shape =
> >
> > >drawn on it, how to find the return path for the signal? Please help me
> >understanding this.
> > >
> > >
> > >
> > >Thanks,
> > >
> > >Darshan Mehta
> > >
> > >
> > >
> > >
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