[SI-LIST] Re: Return Path
- From: Darshan Mehta <darshanmehta2k@xxxxxxxxx>
- To: steve weir <weirsi@xxxxxxxxxx>, si-list@xxxxxxxxxxxxx
- Date: Fri, 5 Aug 2005 01:07:38 -0700 (PDT)
Steve,
Now I understood why calculation does not give correct answer and field solvers
are used for Impedance calculation.
Could you please tell me which book I should refer to know more about ?How the
field Solver works?, I am really keen to know about field solvers.
Thanks,
Darshan Mehta
steve weir <weirsi@xxxxxxxxxx> wrote:
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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