• From: "Hill, John" <John.Hill@xxxxxxxxxx>
• To: "steve weir" <weirsi@xxxxxxxxxx>, <Sirisha.Godavarthy@xxxxxxxxxxxx>, <Tim.J.Nash@xxxxxxxxxxxxx>, "silist" <si-list@xxxxxxxxxxxxx>
• Date: Mon, 25 Sep 2006 18:03:01 -0400
```Steve,

I am not comfortable with the idea of integrating around a circle. Could
you describe the integration path?

I expect the current will lump up in a line between the source (the
capacitor) and the load (the chip). In the extreme, if the board was a
mile wide, the current would not reach the edge. There is a limit where
further width will not have an effect.=20

The current path will spread out, but I am unsure of how wide it is
going to spread perpendicular to the path of the current. I would also
expect the width of the spreading to be dependant on frequency. But
there is no dependence on frequency or dielectric constant in your
equation. I don't understand how that can be true.

Best regards,

John

-----Original Message-----
From: steve weir [mailto:weirsi@xxxxxxxxxx]=20
Sent: Monday, September 25, 2006 5:38 PM
To: Hill, John; Sirisha.Godavarthy@xxxxxxxxxxxx;
Tim.J.Nash@xxxxxxxxxxxxx; silist
Subject: RE: [SI-LIST] Re: Spreading Inductance

John we solve for inductance between two circular ports by performing=20
integration of the incremental field intensity using the Law of Biot=20
and Savart.  That states that

dB =3D uo*i/4pi * dl sin theta / r^2, magnitude only

alternatively

dB =3D u0i/4pi*dl x r / r^2  to obtain both direction and magnitude

Integrate this formula and you get B.  Since L =3D B/H, and we=20
stipulate H, this gives us L.

So, back to our planes.  When we look at power delivery to an IC die,=20
it is not at the end of a long skinny transmission line, it is inside=20
the extents of the planes.  Current density and hence B both rise as=20
we approach the IC, and diminish as we move away.  This is analagous=20
to what happens with spreading resistance.  In a planar structure we=20
yield resistance per square.

Best Regards,

Steve.

At 02:14 PM 9/25/2006, Hill, John wrote:
>Steve,
>
>I understand that in a transmission line the inductance per unit length
>and capacitance per unit length are related. As you know, the equation
>is:
>
>The relative permeability times the relative dielectric constant
divided
>by the speed of light squared is equal to the inductance per unit
length
>times the capacitance per unit length.
>
>I believe we can model the power planes as a transmission line
>connecting the decoupling capacitor and the load (chip). Given a BC24
>material with a Capacitance of 1 nF per inch (39 nF per meter) with a
>dielectric constant of 4.5. I expect the inductance per unit length to
>be 1.27 nF per meter (32.2 pH per inch)
>
>L=3D(4.5 * 1)/(c^2 * 39 nF per meter)
>
>The Zo would be 179 milliohms. Remember to terminate the transmission
>line or it will show a low impedance resonance at a quarter wavelength
>of electrical length.
>
>Now we still need to deal with the inductance of the connecting vias,
>but I am planning on using an array of laser drilled micro vias.
>
>How did you derive the equation shown below for the L of PLANE_CAVITY?
>It does not contain the things that I would be looking for in such an
>equation. It looks like the equation for the inductance of a via.
>
>So again I ask: Why are you using Ohms per Square?
>
>Best regards,
>
>John
>
>
>
>-----Original Message-----
>From: steve weir [mailto:weirsi@xxxxxxxxxx]
>Sent: Monday, September 25, 2006 11:46 AM
>To: Hill, John; Sirisha.Godavarthy@xxxxxxxxxxxx;
>Tim.J.Nash@xxxxxxxxxxxxx; silist
>Subject: RE: [SI-LIST] Re: Spreading Inductance
>
>John, it is just English / Metric conversion:
>
>1.257E-6 H/m * 0.0254 m/inch =3D 31.92 H/inch
>
>Best Regards,
>
>
>Steve.
>At 08:27 AM 9/25/2006, Hill, John wrote:
> >Steve,
> >
> >I was under the impression that the Permeability of free space was
4PI
> >times 10 to the -7. That is 1.257 X 10 to the -6 henrys per meter.
> >
> >This comes from the square of the speed of light being equal to one
>over
> >the permeability time the dielectric constant. The dielectric
constant
> >of free space being equal to 8.85 X 10 to the -12 farads per meter.
> >
> >Where does the per square come from?
> >
> >Best regards,
> >
> >John
> >
> >-----Original Message-----
> >From: si-list-bounce@xxxxxxxxxxxxx
>[mailto:si-list-bounce@xxxxxxxxxxxxx]
> >On Behalf Of steve weir
> >Sent: Monday, September 25, 2006 10:59 AM
> >To: Hill, John; Sirisha.Godavarthy@xxxxxxxxxxxx;
> >Tim.J.Nash@xxxxxxxxxxxxx; silist
> >Subject: [SI-LIST] Re: Spreading Inductance
> >
> >John, this is the permeability of free space.  It is an observed
> >constant behavior.
> >
> >Best Regards,
> >
> >
> >Steve.
> >At 07:30 AM 9/25/2006, Hill, John wrote:
> > >Steve,
> > >
> > >Where does the 31.9nH/Square come from?
> > >
> > >Best regards,
> > >
> > >John
> > >
> > >
> > >
> > >
> > >---------------------------------------
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> > >-----Original Message-----
> > >
> > >From: si-list-bounce@xxxxxxxxxxxxx
> >[mailto:si-list-bounce@xxxxxxxxxxxxx]
> > >On Behalf Of steve weir
> > >Sent: Tuesday, September 19, 2006 12:28 AM
> > >To: Sirisha.Godavarthy@xxxxxxxxxxxx; Tim.J.Nash@xxxxxxxxxxxxx;
silist
> > >Subject: [SI-LIST] Re: Spreading Inductance
> > >
> > >Tim, those equations come from a presentation I made to the SCVEMC
> > >January 2005.  You can find the presentation in whole on both the
> > >X2Y, and Teraspeed web
> > >sites:  http://www.x2y.com/bypass/method/does_position_matter.pdf,
> > >and http://www.teraspeed.com/publications.html  Note that they
apply
> > >for specific circumstances of calculating spreading inductance
> > >between a power ring, and a surrounding bypass capacitor ring.
This
> > >would also work for a single Vdd / Vss pair.  However for more
> > >complex pin arrays, you really need to solve the plane.  Tools from
> > >IBM, KAW, Ansoft, or Sigrity are all capable of that task.
> > >
> > >Steve.
> > >At 05:57 AM 9/18/2006, Sirisha Godavarthy wrote:
> > > >Hi Tim,
> > > >I got some info regarding spreading inductance.
> > > >Hope it helps.
> > > >
> > > >Plane spreading inductance:
> > > >
> > > >- L of PLANE_CAVITY =3D3D u0 / 2=3DF0 * H * (( ln( R2/R1 ) * =
KPERF )
+
> > > >ln( R3/R2 ))
> > > >- u0 =3D3D 31.9nH/square
> > > >- H =3D3D cavity height
> > > >- R3 =3D3D radius from die center to bypass cap ring
> > > >- R2 =3D3D radius from die center to via field edge
> > > >- R1 =3D3D radius from die center to pkg power / ground
attachments.
> > > >- KPERF scaling factor for perforation in via field
> > > >* For perforations small compared to wavelength area ratio
> > > >
> > > >Thanks&Regards,
> > > >Sirisha.
> > > >
> > > >Thanks&Regards,
> > > >Sirisha.
> > > >=3D20
> > > >
> > > >-----Original Message-----
> > > >From: si-list-bounce@xxxxxxxxxxxxx
> > >[mailto:si-list-bounce@xxxxxxxxxxxxx] =3D
> > > >On Behalf Of Nash, Tim J (EHCOE)
> > > >Sent: Friday, September 15, 2006 7:56 PM
> > > >To: silist
> > > >Subject: [SI-LIST] Spreading Inductance
> > > >
> > > > From what I have gathered, spreading inductance of a plane is =
=3D
> > > >dimensionless
> > > >(i.e. nH/sq) and not a function of the distance between two
points,
>=3D
> > > >rather a
> > > >function of aspect ratio of the plane - L x W.  So, the size of
the
>=3D
> > > >plane
> > > >isn't what affects spreading inductance, but its shape does.  Are
> >these
> > >=3D
> > > >true
> > > >statements?  I want to make sure I understand this correctly.
> > > >Best Regards,
> > > >Tim
> > > >
> > > >
> > > >-- Binary/unsupported file stripped by Ecartis --
> > > >-- Type: application/x-pkcs7-signature
> > > >-- File: smime.p7s
> > > >
> > > >
> > >
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