We have been doing chip carrier and printed circuit board analyses with
our newly developed software tool SPEED97 so I did some simulation on your
"10 inch square board and 2 mil FR4 between power and ground plane".
Just as what you said, we found package resonance at about 590 MHz,
830MHz, and so on. The power and ground plane structure is capacitive at
the low frequency and its capacitance is about 45 nF in the low frequency
limit.
However, the power and ground planes in a PCB are somewhere connected to a
power supply that has a very small internal impedance. In that case, the
power and ground plane structure is inductive instead of capacitive at the
low frequency. The value of this inductance depends on any factors, such
as where the power supply is connected to the power and ground planes,
how many connections are made between the power supply and the power and
ground planes, and the radius of via(s) connected to the power and ground
planes, etc. The inductance value does not seem to be easy to estimate.
Moreover, we found package resonance at frequencies much lower than 600
MHz.
Some examples that I tested have been placed in SIGRITY's web site
(www.sigrity.com). These examples show the input impedances of the power
and ground plane structure from DC to 1 GHz, and the effective capacitance
and inductance from DC to 50MHz. You can also download these examples
and a demo SPEED97 program from the web site so you can run these examples
on your own machine.
Regards,
Raymond Chen
SIGRITY
On 7/29, Larry Smith wrote:
> 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
>