[SI-LIST] Re: Should the signals always return back through GND
- From: "Andrew Ingraham" <a.ingraham@xxxxxxxx>
- To: <nikitanivan@xxxxxxxxxxx>
- Date: Mon, 22 Aug 2005 12:23:53 -0400
> Its always said that GND acts as a return path for a
> signal.
Really? I wasn't aware of this. If everyone says that where you come from,
then maybe you should educate your teachers and/or co-workers that they are
wrong.
What we call GND is often used as a reference point for VOLTAGE, but this
doesn't mean it is where CURRENT flows.
> Do all the signals know that they
> have to return back through GND and not through
> anyother track in the signal layer having
> lower impedance? What are the factors taken into
> consideration to make sure that signals return
> back through GND.
At low frequencies and DC, signal currents return to wherever they came
from, to complete the loop.
If you take a signal source and connect one end of the source to GND and the
other end of it to a wire that goes off somewhere to some load, then the DC
return path will have to get back to GND to complete that loop.
If you take that same source and connect the first end of it to VDD rather
than to GND, then the DC return path will have to end up at VDD and not GND.
It might go by way of GND in order to get back to VDD, but that depends on
the topology, the load, etc. The exact path (GND plane vs. VDD plane, etc.)
that it takes, depends on the relative DC resistances and low frequency
impedances of those paths.
Thus, when the pull-up transistor in a pull-up/pull-down (or "totem-pole")
pair is on, the return path (at DC and low frequencies) will have to get
back to the VDD supply net that connects to that output driver. GND might
not be involved at all.
At high frequencies, signal switching current return paths for a wire or a
trace, are by way of any and every conductor that is nearby, to which field
lines can be drawn. Most of the high frequency current chooses the path(s)
with the lowest impedance. The signal propagates via an electro-magnetic
field in the dielectric, which just happens to cause currents to flow in all
conductors that touch the E-M field.
If a trace runs over a solid plane that connects to, say, some VTT voltage,
then return current will be in that VTT plane. If a trace runs between two
planes, the high frequency return current will be shared between both of
them, regardless of what voltage each one is connected to. If a trace runs
over one plane for several inches, then a different plane for several
inches, and so on, then the return current will have to (or try to) find its
way on each of these planes, in series.
The DC path may take a totally different route. That is, the initial
switching wavefront has a return path associated with it, determined by the
E-M field around and between the conductors; but after a few nanoseconds
the return current may take a different route, eventually determined by DC
resistances and not the E-M field anymore.
Regards,
Andy
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