[SI-LIST] Re: Spreading Inductance

Jim, for a transmission line over a plane, the image in the plane 
drops off like that.  But for parallel plates of like size the 
distribution is quite different.

Regards,


Steve
At 04:33 AM 9/26/2006, Peterson, James F \(EHCOE\) wrote:
>very interesting discussion. I've always pictured (imagined) current
>density falling off at 1/D^2  away from a direct line between the entry
>and exit points of a ground plane...not a good approximation?
>
>-Jim
>
>-----Original Message-----
>From: si-list-bounce@xxxxxxxxxxxxx [mailto:si-list-bounce@xxxxxxxxxxxxx]
>On Behalf Of steve weir
>Sent: Monday, September 25, 2006 6:35 PM
>To: Hill, John; Sirisha.Godavarthy@xxxxxxxxxxxx; Nash, Tim J (EHCOE);
>silist
>Subject: [SI-LIST] Re: Spreading Inductance
>
>John,
>When we are solving a simple plane spreading inductance problem we
>integrate dB to find total B around the port, in this case a circle.
>Your discomfort with integrating around a circle is no doubt from your
>expectation that yields a zero result.  But for these types of problems
>it does not.  This is not the same as say integrating spring force over
>distance where if we finish where we start we end up with a zero result.
>At each point around the path we have a nonzero increment that adds to
>the final result.
>
>As to the current path spreading out diminishing to zero, well, yes and
>no.  What happens is that the field intensity falls away very quickly
>with initial distance, and then proceeds to fall off slower and slower,
>but it never falls to zero.  So yes, from a practical standpoint for
>digital and basic power distribution, there is a distance beyond which B
>is so small as to be negligible.  If we are dealing with a problem with
>a lot of dynamic range we would not want to use that approximation.
>This is why sometimes moating is a good thing.
>
>I have attached an H field plot that may be helpful.  For anyone who
>cares, this is slide 10 from my 2005 SVCEMC meeting presentation.  As
>you can see, the current distribution concentrates around the ports sure
>enough, but remains pretty intense in all of the area directly between
>the capacitor and the test port, as well as substantially beyond.
>
>Inductance yields frequency dependent impedance, so I am not sure what
>additional effect you are looking for.  If I had superconducting planes
>separated by a loss less diectric, at low frequency I would have
>essentially no impedance, whereas the dimensions of the features yield
>an inductance, and therefore impedance at any higher frequencies.
>
>Best Regards,
>
>
>Steve.
>At 03:03 PM 9/25/2006, Hill, John wrote:
> >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.
> >
> >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]
> >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
> >integration of the incremental field intensity using the Law of Biot
> >and Savart.  That states that
> >
> >dB = uo*i/4pi * dl sin theta / r^2, magnitude only
> >
> >alternatively
> >
> >dB = u0i/4pi*dl x r / r^2  to obtain both direction and magnitude
> >
> >Integrate this formula and you get B.  Since L = B/H, and we stipulate
> >H, this gives us L.
> >
> >So, back to our planes.  When we look at power delivery to an IC die,
> >it is not at the end of a long skinny transmission line, it is inside
> >the extents of the planes.  Current density and hence B both rise as we
>
> >approach the IC, and diminish as we move away.  This is analagous to
> >what happens with spreading resistance.  In a planar structure we 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=(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 = 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 =3D u0 / 2=F0 * H * (( ln( R2/R1 ) * KPERF
> > > > > >)
> >+
> > > > > >ln( R3/R2 ))
> > > > > >- u0 =3D 31.9nH/square
> > > > > >- H =3D cavity height
> > > > > >- R3 =3D radius from die center to bypass cap ring
> > > > > >- R2 =3D radius from die center to via field edge
> > > > > >- R1 =3D 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.
> > > > > >=20
> > > > > >
> > > > > >-----Original Message-----
> > > > > >From: si-list-bounce@xxxxxxxxxxxxx
> > > > >[mailto:si-list-bounce@xxxxxxxxxxxxx] =
> > > > > >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
> > > > > >= dimensionless (i.e. nH/sq) and not a function of the distance
>
> > > > > >between two
> >points,
> > >=
> > > > > >rather a
> > > > > >function of aspect ratio of the plane - L x W.  So, the size of
> >the
> > >=
> > > > > >plane
> > > > > >isn't what affects spreading inductance, but its shape does.
> > > > > >Are
> > > >these
> > > > >=
> > > > > >true
> > > > > >statements?  I want to make sure I understand this correctly.
> > > > > >Best Regards,
> > > > > >Tim
> > > > > >
> > > > > >
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> > > > > >
> > > > >
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