# Re: [SI-LIST] : Buried Capacitors

Larry Smith ([email protected])
Tue, 29 Jul 1997 20:10:17 -0700

> I am trying to gain a better understanding on the concept of the buried
> capacitors, made of two (2) internal planes on a printed circuit board.
>
> We do not have a general agreement within the design community on the use
> of this technique, so I just wanted to bounce it off of this reflector.
>
> Multi-layer PCB's have power and ground planes as a norm. This technique
> stacks some of these planes very close together (2 mil thin dielectrics)
> to act as a giant capacitor.
>
> The design uses thru-hole vias only.
>
> Are we:
> 1) reducing noise
> 2) using this as an alternative to filter-caps
> 3) using this in conjunction with filter-caps.
> 4) At what frequencies is this practice appropriate.
> 5) anything else you can add.
>
> Thanks
>
> Bob Harrison
> TRW

Bob - we have been doing some work in this area so I will take a crack
at it. The power planes are pure capacitance (very little inductance)
at low frequency. The planes can be considered to be equal-potential-
surfaces with no voltage gradient along them (accept for a small dc
voltage drop). The capacitance is calculated simply as area divided by
thickness times permitivity. For a 10 inch square board and 2 mil FR4
between power and ground plane, this calculates out to 45 nF.

At some frequency, a full wave will stand between the planes along
the length of the board. Assuming velocity of propagation to be
6 inches/nSec, that frequency will be 1/(10/6) = 600 MHz for our 10
inch board. At this frequency, the planes are certainly not
equal potential surfaces as the voltage measured between them can
differ greatly as you move around the board. A reasonable transition
frequency would be 1/10 of 600MHz or 60 MHz. Below that frequency,
consider the planes to be pure capacitance.

Knowing the velocity of propagation and capacitance per area we can
calculate the plane inductance. Vel = 1/sqrt(LC). For our 2 mil thick
dielectric, cap=.45nF/inch2, Vel=6 inch/nSec and inductance=.062 nH/square.
This is a spreading inductance, similar in interpretation to spreading
resistance. It is a very small number and is the reason why the planes
are such pure capacitance.

With inductance and capacitance we can calculate impedance Z0=sqrt(L/C),
which turns out to be .372 Ohm-inch. A plane wave traveling down a
long length of 10 inch wide board will see .372/10 = .0372 Ohms impedance,
again, pretty small.

There is great advantage in having power planes close to each other.
Decoupling capacitance is increased because thickness is in the denominator.
Inductance is decreased because the velocity of propagation must remain
constant. Impedance is also decreased. The power and ground planes are
the means of distributing power. Reducing the dielectric thickness is
effective at increasing high frequency decoupling capacitance and
transporting high frequency power through a lower impedance distribution
system.

regards,
Larry Smith
Sun Microsystems