Your 3.1415 cents worth ... is a value-added. At least now, we can use the equation more accurately than before,
right ;)
Thanks and regards.


Douglas L McKean wrote:

> 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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