I think you should be careful trying to calculate current carrying capacity this 
way. For one thing, although 1oz. copper is theoretically 1.4mils, with plating 
and tolerances and processing it can be more like 1.8-2.0 for external layers 
and .98 mil for internal when all is said and done (see Table  3-11 and 3-12 of 
IPC-6012)
To calculate the trace width itself you are going to have to factor in the 
acceptable temperature rise and other factors like proximity to heat sinks 
(planes?), and as long as your pad is as wide as your trace you should be 
okay, right? If your pad is wider than your trace, I can't imagine what the 
worry would be. This article might help:
http://www.frontdoor.biz/HowToPCB/HowToPCB-extra/HowTo2152.pdf

Unless what you were really leading up to was the current through the solder 
joint itself, and I have never heard anyone discuss that (solder much poorer 
conductor than copper)

Jack

.
On Tue, 20 Jul 2010 11:34:24 -0400, Forrester, Michael (H USA) 
<[log in to unmask]> wrote:

>I'm a bit confused on how to calculate the current capability of a SMT
>pad.  We have a 2220 size chip bead that has 8A flowing through it.
>Our pad geometry is built per IPC specifications but the manufacturers
>datasheet calls out for a PAD twice the width (part max is 10A). 
>So the question came up, with our current pad geometry (to IPC), what is
>the current capability?  I understand that it is based on the 
>cross sectional area of the copper.  The current density of copper is
>4.02 A/sq-mm.  For a 1 oz copper trace the area is 
>1.4 mils x trace width.  But what do I use for a PAD?  I have an width,
>length, and thickness.  For instance I have a SMT pad that 
>is 1.5mm X 6mm using 1 oz copper with multiple traces coming off of the
>pad.  What is the current carring capability of that pad?   
>Or am I thinking about this wrong?  Thank you.
> 
>Best Regards,
> 
>Michael Forrester
>Sr. Product Engineer
> 
>Siemens Healthcare Diagnostics
>101 Silvermine Rd.
>Brookfield, CT  06804
>PH: (203) 740-6452

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