RE: [SI-LIST] : Thin Power Plane Dielectrics

Ray Anderson (raymonda@radium.eng.sun.com)
Tue, 19 Oct 1999 15:06:55 -0700 (PDT)

(apologies if you receive multiple copies of this message,
we were experiencing network difficulties earlier today and
an earlier send may or may not have been trashed)

Andrew.Ingraham@compaq.com wrote:

> I have seen warnings about "buried capacitance" layers that use high-Er
> materials. While this is OK at lower frequencies, the advantage falls apart
> at higher frequencies because the propagation delay through the buried
> capacitance layer is more. Even though the capacitance per unit area is
> large, the capacitance per nanosecond (risetime) may not be.
>
> Andy

A good point! The main argument usually made regarding the use of of
high Er dielectric materials is that they enable the inclusion of very
high amounts of capacitance/area in a PCB. While this is true, there
is more to be considered.

While a high Er constant enables a higher capacitance, it also has another
effect: that of slowing down the time of flight of signal associated with
the high Er dielectric layer.

Consider the scenario where a planar stackup is modified only by
changing the Er of the substrate material from say 4.0 to 40 .
Let's analyze what happens:

First the capacitance per unit area within the layer increases by the
ratio of the dielectric constants since:

Eo*Er*A
C = -------
D

(changing Er from 4 to 40 causes capacitance/unit area to increase 10X)


Second, the time of flight increases by the square root of the ratio of
the dielectric constants since:

Dist
Tf = ---------
C/sqrt(Er)

(changing Er from 4 to 40 causes Tof to increase 3.16 times)


Since the time of flight has increased, it takes longer for charge stored in
the dielectric to reach a "point of consumption" or sink point on the plane.
So if a circuit needs current in say 1ns and charge is traveling 1/3.16
slower then the radius of the "effective capacitance" is 1/3.16 the size
of the original radius.

We know that the area of a circle is: A= (pi * r^2) . So if the radius
is 1/3.16 the original then the area is 1/10 the original area.

Since the capacitance / unit area increased 10 X then the
"available capacitance" is 1/10 * 10 = 1 X the original !

Therefore even though the capacitance per unit area has increased
10X, the available capacitance to supply current within a specified
time hasn't increased a bit. At lower frequencies the slowing of the
time of flight isn't an issue, but then again we usually aren't depending
on the charge storage capailities of the planes at those frequencies anyway.

Decreasing the separation between the planes WILL improve matters. Making
the interplanar spacing 1/10 of the original spacing increases the
capacitance by a factor of 10 and doesn't affect the speed of propagation.
In addition, the close plane spacing reduces the planar spreading inductance
giving you a big gain there too.


Ray Anderson
Sun Microsystems Inc.


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