From: Michael Nudelman ([email protected])
Date: Wed Jan 03 2001 - 16:18:08 PST
This depends how you routed your pairs.
If you have un-coupled pair (two independent transmission lines) - the ret.
currents will flow into the ref. plane(s).
If the pair is tightly coupled - almost all current will flow in the opposite
If it is loosely coupled, the current will be shared between the plane and the
I simulated some of the broadside-coupled pairs I use; when I removed planes,
the impedance changed about 15%, which means also, that the pair was tightly
coupled and most current would flow in the opposite conductor versus the ref.
So, if you have 2 100Ohms lines, the resulting impedance is <= 50Ohm. The
tighter the coupling, the lower the impedance will be.
And - the fast changing current always tries the least inductance path. Which
alleviates understanding of where and why it will flow. For example, all other
things being equal, the current will go in the closest parallel conductor
(plane or wire) to minimize the loop area (the inductance is proportional to
the loop area).
"Dill, Franz @ Celerity" wrote:
> Please excuse my 'newbie-like' questions, my inclusion in this mailing list
> is more for curiosity and personal advancement/understanding than as a
> First, is this statement valid?
> - In a differential pair, one 'leg' of a signal's return current path is
> through the complementary 'leg' of a differential pair and not through the
> ground or power planes (Assuming equal trace lengths, Zo=50 single-ended,
> Zo=100 diff. impedance - using ECL logic as an example).
> Now, assuming the above statement is true:
> If the differential impedance is NOT 100 Ohms (Differential traces NOT
> routed differentially) how does this effect the return current path? Does
> the return current begin to flow through the ground and power planes rather
> than through the differential pair?
> Thanks SI gurus!
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