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

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From: Howard Johnson (howiej@sigcon.com)
Date: Fri Mar 16 2001 - 12:15:46 PST


Hi Brian,

I guess I left out a step or two in my hurried explanation.

Let's start with my message of 3/14/01:

>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]
>
>

Compare this with the crude approximation in my book (7.9),

L = 5.08*H*(ln(4*H/D)+1) [7.9]

Now all we need do is ask ourselves, "At what distance S
must a coaxial return path be located in order to generated
the approximation [7.9]?"

The answer is, S must equal 5.43*H. This is found by
equating the two approximations,

  5.08*H*(ln(2*S/D)) = 5.08*H*(ln(4*H/D)+1)

Divide out the common 5.08*H terms on both sides,

  (ln(2*S/D)) = (ln(4*H/D)+1)

Now recognize that ln(x)+1 = ln(x*e), where e=2.718...

  (ln(2*S/D)) = ln(4*e*H/D)

And now equate the arguments of the logarithm:

  2*S/D = 4*e*H/D

Divide both sides by two, and multiply both by D:

  S = 2*e*H

And recognize the constant 2*e = 5.436563656...

The approximation [7.9] is good for the inductance
of a via where you have a number of return paths
all located within 5.43*H of the via in question.

If you are working with a 0.063" thick board with
eight-to-ten layers, the outer two solid planes are
probably about 0.050" apart. This height times 5.436
would be .2718". If you've got a bunch of return
paths about this far away, the approximation
works OK.

If you hold the position of the return paths
constant and vary the inter-plane separation,
inductance changes linearly with interplane height
(which equates roughly to via length).

If you scale the distance to the return paths
at the same time you vary the inter-plane
height, always keeping S=5.43*H, you get formula
[7.9].

I hope this clears up any ambiguity about
what I meant.

Note that all the approximations I have provided
so far have the property that they give you
the inductance of a via and its associated
return path only covering that part of the via
embedded between the planes. The part of
the via that sticks up above the planes
to touch your capacitor pads in not included.
If you want to model that part (and I think that
at today's speeds you *need* that part), check
out the models in my article:
http://www.signalintegrity.com/articles/edn/pinductbypass.htm
   
Best regards,
Dr. Howard Johnson

At 02:48 PM 3/14/01 -0800, you wrote:
>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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