[SI-LIST] Re: Return Path

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 <weirsi@xxxxxxxxxx> 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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