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

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From: Brian Young (brian.young@motorola.com)
Date: Wed Mar 14 2001 - 11:29:43 PST


There is an interesting phenomenon here that I would
like to point out because I don't think it is
widely appreciated. Transmission lines include the
current return path and are characterized by a per-unit-
length inductance. The inductance equations (7.9) in
Johnson's book and (5.53) in my book for the round wire
have a d*ln(d) dependence on length, d, which is
definitely not linear. Once the return path has been
established, the linear dependence on length is recovered.

Dr. Johnson provides perfect examples with his equations
[1]-[4], where the inductance is linear with respect to
length while varying in magnitude depending on the
assumption regarding the return path.

The round wire formula we have been discussing does not
specify the return path, so it is a partial inductance.
A partial inductance is not a complete description of the
problem-you have to complete the loop with the partial
inductances of the other sections of the loop plus the
partial mutual inductance between all the sections. Once
you do this, the loop inductance is real and physical.
If the loop looks like a transmission line (i.e. long
compared to its uniform cross section), then the inductance
will be linearly dependent on length.

From Dr. Johnson's comments, I'm confused about his formula.
Dr. Johnson wrote:

> My earlier formula was a gross approximation which
> ignored the position of the returning current path, .... It made the
> crude assumption that the return path was approximately
> coaxial and located at a distance S=5.43*H.

Assuming that the discussion is still about (7.9) from his book,
then a coaxial return path assumption should have yielded
an inductance formula that is linear with respect to length.
However, (7.9) contains the partial inductance dependence
of length*ln(length). What am I missing?

Regards,
Brian Young

Howard Johnson wrote:
>
> Dear Itzhak Hirshtal and Brian Young,
>
> The difficulties with approximating the inductance
> of a via are even worse than you
> may have suspected. Both approximations are flawed whether
> you use +1 or -3/4, (or, as I have also seen, -1).
>
> The issue of the exact constant (1, -3/4, or something
> else) depends critically on your assumption about
> the path of returning signal current. (Current always
> makes a loop; when signal current traverses the via,
> a returning signal current flows SOMEWHERE in
> the opposite direction.). It is a principle
> of Maxwell's equations that high-speed returning signal
> current will flow in whatever path produces the
> least overall inductance.
>
> Let's do an example involving a signal via that
> dives down through a thick, multi-layer board.
> If the signal in question changes reference
> planes as it traverses the via,
> then the returning signal current will also have to
> change planes, meaning that the returning signal
> current will flow through one or more vias (often
> leading to bypass capacitors) as it moves from
> plane to plane. For example, if the signal starts
> out on the top layer, the returning signal current
> is flowing on the nearest reference plane (call it
> layer 2). If the via conducts the signal current
> down to the bottom layer (16), then the returning
> signal current at that point must be flowing on
> the nearest (bottom-most) reference plane, call it 15.
> Somehow the returning signal current has to hop from
> reference plane 2 to reference plane 15 in the
> vicinity of the via.
>
> If you examine the space between the planes, the
> magnetic fields within are created partly by
> the signal current, and in equal measure (but in
> differnt locations) by the returniing signal
> current, which flows on different vias. The
> total magnetic flux between the outgoing and
> returning vias defines the inductance.
> Specifically, to calculate the effective
> inductance of via (A), you must first specify the
> location of the return path, via (B), and then
> calculate the total magnetic flux in the area
> between the two vias. The total magnetic flux
> generated by a signal current of one amp, in units
> of webers, equals the inductance.
> In the case of more complex return-path
> configurations, other considerations apply.
> I think at this point that the following
> formulii for the effective series inductance
> of a via are pretty good:
>
> For a signal which pops from one side of the
> plane, through a via, to the opposite side
> of the same plane (i.e., the return current
> doesn't have to jump planes), the via
> inductance is very, very low. This is a best-case
> scenario. I don't know a good way to make this
> calculation except with a true 3-D E&M field solver.
>
> For a signal which first uses reference-plane A,
> and then changes (through a via) to use
> reference-plane B, I'll do several examples. In
> all cases the separation between reference planes
> is H. (It doesn't matter if there are other
> unused reference planes in the way, only the
> spacing between the two reference planes A and B
> matter).
>
> If the return current is carried mainly on one nearby
> via, where the spacing from signal via to return via
> is S and the via diameter is D:
>
> L = 5.08*H*(2*ln(2*S/D)) [1]
>
> If the return current is carried mainly on two vias
> equally spaced on either side of the signal via,
> where the spacing from signal via to either return via
> is S and the via diameter is D:
>
> L = 5.08*H*(1.5*ln(2*S/D) + 0.5*ln(2)) [2]
>
> If the return current is carried mainly on four vias
> equally spaced in a square pattern on four sides
> of the signal via, where the spacing from signal via
> to any return via is S and the via diameter is D:
>
> L = 5.08*H*(1.25*ln(2*S/D) + 0.25*ln(2)) [3]
>
> If the return current is carried mainly on a
> coaxial return path completely encircling the signal
> via, where the spacing from signal via
> to the return path is S and the via diameter is D:
>
> L = 5.08*H*(ln(2*S/D)) [4]
>
> The last formula I hope you will recognize as the
> inductance of a short section of coaxial cable with
> length H and outer diameter 2*S. I hope this
> recognition will lend credence to the idea that
> the position of the returning current path is
> an important variable in the problem.
>
> My earlier formula was a gross approximation which
> ignored the position of the returning current path,
> and omission which I greatly regret. It made the
> crude assumption that the return path was approximately
> coaxial and located at a distance S=5.43*H. As you
> note, when the inductance really matters a
> more accurate approximation is needed.
>
> To obtain a result as low as 5.08*H*(ln(2*S/D)-1)
> you would have to assume the return path were coaxial
> and located at a ridiculously small separation of
> S=.735*H, or that the return path were a single via
> located at some even closer distance.
>
> On my web site http://signalintegrity.com under "articles"
> there is a write-up about calculating the inductance of
> a bypass capacitor that includes the above formulas for
> vias, as well as some handy ways to estimate the
> inductance of the capacitor body.
>
> By the way, if you find a flaw in THIS write-up,
> please let me know.
>
> Best regards,
> Dr. Howard Johnson
>
> >>On the two versions of the equation, it looks to me like the version
> >>in Johnson's book has a typo. When d>>r, the external partial
> >>self-inductance of a straight round wire is
> >>
> >>L=5.08d*{ln(2d/r)-1}nH,
> >>
> >>where d is the length in inches, and r is the radius in inches.
> >>The external inductance is a good approximation at high frequencies
> >>where the skin effect shields the internal metal of the wire. At
> >>low frequencies, the internal self-inductance needs to be
> >>added to the external partial self-inductance to obtain
> >>
> >>L=5.08d*{ln(2d/r)-3/4}nH,
> >>
> >>which is the formula from Gover, as Eric pointed out.
> >>
> >>It seems that Johnson's book has the first (high-frequency) version
> >>with a sign error on the 1 because he has
> >>
> >>L=5.08h*{ln(4h/d)+1}nH,
> >>
> >>where h is the length in inches, and d is the diameter in inches.
> >>
> >>
> >>This formula should not be used for vias because it assumes that
> >>the length is much greater than the diameter. To compute partial
> >>self-inductance for vias, you should use the more complex formula
> >>that does not have this assumption built in. The correct formula
> >>is (5.49) from my book. This is the external partial self-inductance,
> >>so if you want the low frequency inductance, you need to add the
> >>internal inductance from (5.45).

-- 
***************************************************************
* Brian Young                           phone: (512) 996-6099 *
* Somerset Design Center                  fax: (512) 996-7434 *
* Motorola, Austin, TX               brian.young@motorola.com *
***************************************************************

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