Hi,

I know your question have been more than 2 months old and you may already have the answer.
However, I would like to share what I have discovered in the "Printed Circuits Handbook" that may
be useful to you. The information is as follows:

Consider a 0.04" diameter hole plated to a 2-oz internal power plane with 0.001" minimum copper
in the hole. The copper hoel wall is equivalent to 0.04 x 0.001 x PI = 125x10exp-6 sq.in of
copper.

At the interconnection to the power plane, there is 0.04 x 0.0028 x PI = 352x10exp-6 sq.in
Note : 2-oz copper thickness = 0.0028in

The copper plating of the hole wall is the smallest conductor in the path and for purposes of
calculation in current-carrying capacity, consider it an internal conductor of a multilayer.
Referring to the current-carrying capacity curves of MIL-STD-275 or now IPC-D-275, a
125x10exp10-6 sq,in conductor will carry approximately 2A with a 10 degree C rise above ambient.

Regards.


Rolf Buchholz wrote:

> Hello everyone,
>
> i search some tables or calculation rules to ascertain the current value for
> vias or Pads.
>
> Rolf
>
> [log in to unmask]
>
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