[SI-LIST] Re: Return Path
- From: steve weir <weirsi@xxxxxxxxxx>
- To: doug@xxxxxxxxxx, darshanmehta2k@xxxxxxxxx, si-list@xxxxxxxxxxxxx
- Date: Fri, 05 Aug 2005 10:45:34 -0700
Doug, I have some real heartburn with some of those representations,
particularly the fluid analogy that speaks of current as the flow of
electrons. When I grew up current was defined as time variation of
electric flux. When an E/M field impinges a chunk of metal the resulting
interaction concentrates the field forming a wave guide. All practical
wave guides leak, be they a microstrip over a plane, a stripline, or
whatever. Some, like a good semirigid coax leak only a little tiny
bit. When they leak too much creating excessive disturbance in nearby wave
guides, we have cross talk problems. I hope that this is what you were
trying to convey.
Regards,
Steve.
At 10:06 AM 8/5/2005 -0700, Doug Brooks wrote:
>Some of the answers you received may have made this sound more complex than
>the issue really is.
>
>First: Current is the flow of electrons. Electrons MUST flow in a closed
>loop. The way they find their way "back" is called the "return path." If
>you have a simple battery with two wires attached to it, one carries the
>outgoing current and the other carries the return current.
>
>Second: Sometimes the return is constrained to a specific wire. Then there
>is no question where it is. Sometimes the return is on a plane. Then there
>CAN be a question where it is, since it can spread around the plane in a
>variety of ways.
>
>Third: WHERE it is on a plane is determined (at least in part) by
>electromagnetic coupling between the signal that generates the return and
>the return itself. The return will follow the path of least IMPEDANCE. For
>high frequency components the lowest impedance path is directly underneath
>the trace carrying the signal. For low frequency components, the return
>will follow a path of least RESISTANCE, which can be a DIFFERENT path than
>for the higher frequency components. Therefore, different parts of the
>return path (different frequency harmonics associated with the return
>current) may flow back to the source on different paths.
>
>Fourth: Since the lowest impedance path (for the higher frequency
>harmonics) is on the plane directly under the trace, this part of the
>return current will "want" to be on the "closest" plane, whether it is a
>power plane or a ground plane. For HIGH frequency harmonics there is no
>difference in these planes. (If that is hard to understand, think of all
>the bypass caps there are connecting the two planes!)
>
>Finally: An issue for signal integrity "can" be (depending on
>circumstances) how the return current actually gets to the plane it wants
>to be on. Sometimes (not always) the way it gets there can cause SI
>problems (especially EMI or crosstalk.) It is not unreasonable to state
>that MOST of the signal integrity issues on boards today are caused by a
>failure to control return currents. Knowing where your return currents are,
>and making proper provision for them, is one of the primary ways of
>controlling signal integrity.
>
>I cover a lot of this in my book "Signal Integrity Issues and Printed
>Circuit Board Design," Prentice Hall.
>
>Doug Brooks
>
>
>
>
>
>
>
>
>
>
>At 10:31 PM 8/4/2005, you wrote:
> >Hello Experts,
> >
> >
> >
> >I have a question on return current. I was going through the book "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 path. Let me
> >describe what I understood.
> >
> >
> >
> >The Signal Path is the active path and the Electric field from Signal will
> >terminate in the return path. The Magnetic field will form a 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 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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