Re: [SI-LIST] : Re: approximations for partial self inductance - WHY

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From: Mike Jenkins (jenkins@lsil.com)
Date: Mon Mar 19 2001 - 17:43:51 PST


All,

FWIW, I recall that the IBM Yorktown researchers who developed the
partial inductance concept included on the first page of the research
report the quote from Weber to the effect that inductance makes no
sense unless one considers the entire loop of current. I guess
they were sensitive to not inducing readers to attach any physical
significance to the individual partial inductances.

Regards,
Mike

"Tsuk, Michael" wrote:
>
> Doug McKean wrote:
>
> ------------------------------------------------------------
> Okay, well here goes ...
>
> It's easy to see that if a signal trace had a return trace as a
> wire (shown by a dotted line), the following would cause
> the creation of a loop.
>
> |
> |
> |
> +----------+
> . |
> return . | signal
> trace . Loop | trace
> . |
> +----------+
> |
> |
> |
>
> Obviously from this construction, the inductance of the
> return wire would be less than if the return wire was
> underneath and following the longer path of the signal
> trace. Thus, my questioning the path of less inductance
> rule.
>
> ------------------------------------------------------------
>
> The confusion comes from the fact that you've ignored mutual inductance
> here, which acts to reduce the total inductance of this circuit. The return
> current will choose the path (or combination of paths) that minimize the
> *total* loop inductance, not the partial inductance of the return alone.
>
> In general, you can never ignore mutual inductance. :-( This is
> particularly true if you're dealing with partial inductances, which are only
> useful if all mutuals are included.
>
> Even more interesting to my mind is the sign error you made in calculating
> the direction of the magnetic force on your currents. The magnetic force
> between two parallel currents draws them *together* if the currents are in
> the same direction, and pushes them *apart* if they are in opposite
> directions. Check any electromagnetics text. Your mistake is that you
> assumed a *positive* charge when you equated the current direction with the
> velocity of your particle, but a *negative* charge when you calculated the
> force.
>
> Why the apparent effect of minimizing inductance works in the opposite
> direction is very interesting. I think I have the answer, but I'm not sure.
> I'd appreciate any input people might have.
>
> --
> Michael Tsuk
> Compaq AlphaServer Product Development
> (508) 467-4621
>
> -----Original Message-----
> From: Doug McKean [mailto:dmckean@corp.auspex.com]
> Sent: Monday, March 19, 2001 4:08 PM
> To: si-list@silab.eng.sun.com
> Cc: Doug McKean
> Subject: Re: [SI-LIST] : Re: approximations for partial self inductance
> - WHY
>
> ------------------------------------------------------------
> Okay, well here goes ...
>
> It's easy to see that if a signal trace had a return trace as a
> wire (shown by a dotted line), the following would cause
> the creation of a loop.
>
> |
> |
> |
> +----------+
> . |
> return . | signal
> trace . Loop | trace
> . |
> +----------+
> |
> |
> |
>
> Obviously from this construction, the inductance of the
> return wire would be less than if the return wire was
> underneath and following the longer path of the signal
> trace. Thus, my questioning the path of less inductance
> rule.
>
> As shown above, the signal current path forms the bottom,
> right side, and top parts of a loop. The return current path
> forms the left side of the a loop.
>
> Assume the signal path is bound one-dimensionally by the
> confines of the trace. Assume the return path is bound two
> dimensionally by the confines of the ground plane. In other
> words, the signal path is not free to move at all, but the
> return path is free to move in 2-D (up, down, left, right
> in the above picture).
>
> Now, assume the return path in the ground IS as shown above
> with the signal path and the return path. We have a loop. The
> virtual current loop if you will, circulates causing a soloenoidal
> action creating a magnetic field in the center. As such, using
> the right hand rule for current vs. magnetic fields, we have a
> magnetic field coming out of the monitor. The magnetic field
> lines are normal to the screen of the monitor.
>
> Using the other right hand rule for charges moving in a magnetic
> field by way of the Lorentz force, my thumb points in the direction
> of current flow, my fingers point in the direction of the magnetic
> field, and my palm points in the direction that the a positive
> charge would be pushed. With a negative charge, the push is
> from the back of your hand or toward the signal wire. Since
> the return current is bound only by a plane, it seeks to be
> under the signal trace. And it would continue to balance
> itself there.
>
> Turn the path of the signal current around and the return
> current, everything reverses including the direction of the
> magnetic field, and we still have a Lorentz force pusing the
> return current back to the signal trace.
>
> A DC return current in the ground plane wouldn't cause such
> action. It would follow only the path of least resistance.
>
> This is lots more wordy than if I was face to face and showed
> with the right hand rule for negative charge in a magnetic field.
>
> Regards, Doug McKean
> ------------------------------------------------------------
>
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