Re: [SI-LIST] : Spreading Inductance?

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From: Larry Smith (ldsmith@lisboa.eng.sun.com)
Date: Thu Aug 17 2000 - 09:29:34 PDT


Bob - spreading inductance is something like spreading resistance for a
sheet of conductive material. Most of us are familiar with resistance
in ohms/square. If you have a strip (sheet) of conductive material
with a length and width, the resistance in the length direction is
simply L/W * Rs. L/W gives you the number of squares. Multiply that
by sheet resistance in ohms per square and you get the resistance in
Ohms. If we have a point source and a point sink with some diameter in
a sheet of material, the resistance between the two points can be found
by counting the number of curvilinear squares in series and parallel
and calculating the resistance. This is often called a spreading
resistance problem. The physics has a lot to do with current traveling
through a conductive material. Well, that is the concept.

It turns out that we can do exactly the same thing for inductance of a
power plane pair and call it spreading inductance. The physics is no
longer related to conductivity but rather permeability. As current
spreads out on the voltage plane (and an opposite current "spreads in" on
the ground or return plane), it creates a magnetic B field between the
planes. Inductance is essentially a measurement of the energy stored
in this B field.

Think of the power plane pair as a bunch of transmission lines with a
width and thickness, divided up into long narrow strips. It is easy to
think of capacitance per inch of these transmission line strips. There
is a velocity associated with the transmission lines strips that is
sqrt(L/C) where L is inductance per inch and C is capacitance per
inch. So if you know the velocity and capacitance, you know the
inductance. Extend these 1 dimensional transmission lines concepts to
two dimensional power planes and you get capacitance per square inch
and inductance per square (a lot like ohms per square). Inductance per
square is the spreading inductance. Attached is another email from
last march on the same subject.

regards,
Larry Smith
Sun Microsystems

> From: "Bob Perlman" <bobperl@best.com>
> To: si-list@silab.eng.sun.com
> Date: Wed, 16 Aug 2000 16:40:53 -0700
>
> Hi -
>
> I've seen the term "spreading inductance" used repeatedly here.
> Would someone be so kind as to define it?
>
> Thanks,
> Bob Perlman


attached mail follows:


This is an excellent place to apply the "basic question" principles
from today's other thread on SI-list. Given a pair of power planes
in FR4 material (eR=4), we know that the velocity of propagation
is light_speed/sqrt(4) = 1.5x10**8 M/sec = vel.

The capacitance is simply Ca = eR/thk where Ca is capacitance per
area (Farads/M**2) and thk is the thickness of the dielectric.

The inductance the power planes is calculated from vel = sqrt(La/Ca)
where La is the spreading inductance (Henries per square of
material, just like spreading resistance in a sheet of material).

The impedance is Zl = sqrt(La/Ca). The units Zl for power planes
come out in Ohm-M. This is interpreted as the impedance for a
plane wave propagating down a strip of this material that is 1
meter wide. Divide by the actual width to get the actual impedance
in Ohms.

Note that we have a homogenous environment between the two power
planes. Things get a little more complicated if we have an
non homogenous environment (ie both FR4 and air dielectrics involved).

So, if we compare the two power plane pairs below with 4 and 40 mil
separations:

        The velocities are constant.
        
        The capacitance of the 4 mil planes is 10X the 40 mil plane.
        
        The inductance of the 40 mil plane is 10X the 4 mil plane.
        
        The impedance of the 40 mil plane is 10X the 4 mil plane.
        
To directly answer Mark's questions, with the same amount of noise
stimulation, there will be much more noise between the 40 mil power
planes than between the 4 mil planes, in fact 10X. The reason is
because of the increased impedance.

It turns out that one of the most important consequences of thin power
planes is the 10X decrease in inductance. This greatly increases the
effectiveness of the discrete decoupling capacitors that are mounted on
the power planes.

I vote for the s-s-G-P-s-s stackup. Not only does it have better
capacitance properties, but it also has better inductance and impedance
properties. SI and EMI noise are greatly reduced. Microstrip
transmission lines work just fine.

regards,
Larry Smith,
Sun Microsystems

> Date: Fri, 17 Mar 2000 10:29:33 -0600 (CST)
> From: mjs <mjs@enteract.com>
> To: "'si-list@silab.eng.sun.com'" <si-list@silab.eng.sun.com>
> Subject: [SI-LIST] : Fast edges with limited plane capacitance
> MIME-Version: 1.0
>
>
> Let's assume that a power subsystem has a low, flat impedance up to a few
> hundred MHz, and has a pair of realtivly unbroken planes. Only problem is
> that the stackup is s-G-s-s-P-s since the engineer has insisted that EMI
> will be a problem unless all noisy digital signals are 'sandwiched'
> between the planes. This board also has parts with 1-2ns edge rates.
>
> I am arguing that s-s-G-P-s-s is the preferred stackup, as this would
> allow the planes to be 4-5mils apart instead of 40mils on an .062" card,
> yielding much greater plane capacitance.
>
> My questions is this: How does a lack of planar capacitance contribute
> to increasing EMI? It seems that not having the proper plane capacitance
> would tend to slow edge rates and possibly be one of the lesser SI sins.
>
> Also: Is there any validity to the s-G-s-s-P-s 'copper sandwich'
> decreasing EMI?
>
> Regards,
> Matt Stanik
> PCB Design Engineer
>
>
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