Actually, I'm glad someone brought this up.
I've considered this slightly incorrect
with all due respect to Clyde Coombs.

The better way IMHO is to consider the inner diameter
of the material as best case since you have slightly
more material proceeding out to the limits of the
hole, and the outer diameter of the material as
worst case since it assumes all the material is
out at this point.  Middle of the road would be
to do average  A = (OD + ID)/2.

Just my 3.1415 cents worth ...

Doug

On Sat, 5 Sep 1998, Lum Wee Mei wrote:

> I hope the following information can be of some help to you.
>
> (1)    With reference to "Printed Circuit Handbook by Clyde F. Coombs Jr" :
> Consider a 0.04" diameter hole plated to a 2-oz internal power plane with 0.001" minimum copper in the
> hole. The copper hole wall is equivalent to:
>     0.04" x 0.001" x PI = 125 x 10 exp -6 sq.in
> At the interconnection to the power plane, there will be :
>     0.04" x 0.0028" x PI = 352 x 10 exp -6 sq. in (Note : 0.0028" is the thickness of 2-oz copper)
> With reference to the current-carrying capacity chart of MIL-STD-275 or now IPC-D-275, a 125 x 10 exp -6
> sq. in conductor will carry approximately 2A with a 10 deg. C rise above ambient.
>
> (2)    With reference to " Handbook of Printed Circuit Design, Manufacture, Components & Assembly by
> Giovanni Leonida" :
> I = (W/K) exp0.538 where the value of K =
> 3.60 for 1-oz/10 deg C above ambient        1.97 for 2-oz/10 deg C above ambient
> 1.97 for 1-oz/20 deg C above ambient        1.10 for 2-oz/20 deg C above ambient
> 1.45 for 1-oz/30 deg C above ambient        0.80 for 2-oz/30 deg C above ambient
> 1.20 for 1-oz/40 deg C above ambient        0.65 for 2-oz/40 deg C above ambient
>
> 1.30 for 3-oz/10 deg C above ambient
> 0.80 for 3-oz/20 deg C above ambient
> 0.53 for 3-oz/30 deg C above ambient
> 0.45 for 3-oz/40 deg C above ambient
> W is the width of the conductor in mils.
>
> Example : Current which can flow through a conductor with W = 0.15in, 3-oz/sq. ft at temperature rise of 40
> deg C is
> I = (150/0.45)exp0.538 = 333.3exp0.538 = 22.8A. Since the conductor width is more than 0.04", the value
> obtained should be decreased by 50%.
>
>
> Regards.
>
>
> Scott Holthausen wrote:
>
> > Hey TechNeter's
> >
> > Does anybody have an equation that could help a person determine the
> > current handling capability of through hole vias?  Our practice has
> > normally been to use multiple vias to connect components with high current
> > paths to plane layers.  However, that isn't good enough for one of our
> > engineers who wants more than an approximation, and wishes to find out
> > whether we are putting more vias on the board than we need to.
> >
> > Thanks,
> > Scott Holthausen
> >
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