## [SI-LIST] Re: Current Carrying Capacity

• From: John Barnes <jrbarnes@xxxxxxxxx>
• To: si-list@xxxxxxxxxxxxx, doug@xxxxxxxxxx, akmishra@xxxxxxxx
• Date: Tue, 09 Jul 2002 09:22:50 -0400

```AK, Doug,
If you would like to be on the conservative side, use
I = 56*diameter
where:
I in Amps.
diameter in inches
will keep the temperature rise in the via below 10C.

Are you concerned just with vias, or also with plated-through holes for
components?  At my previous company we would plate vias solidly into
power/ground planes, because we didn't need to worry about unsoldering
them.  But the plated-through holes into power/ground planes would have
a pad with an 0.006"-wide annular ring, with four 0.015" wide spokes
coming through an 0.012"-wide thermal-isolation ring.  The idea was to
keep the power/ground planes from taking away all the heat when we
manually soldered/unsoldered the component(s).

The ampacity (current-carry capacity, from National Electrical Code
article 310) of a via may be limited by:
*  The cross-sectional area of the barrel.
*  The cross-sectional area between the barrel and an external (on the
top or bottom of the board) pad or power/ground plane.
*  The cross-sectional area between the barrel and an internal pad or
power/ground plane.

For a plated-through hole we also need to consider:
*  The cross-sectional area of the spokes into an external power/ground
plane.
*  The cross-sectional area of the spokes into an external power/ground
plane.

To add to the fun, internal copper layers may be scrubbed and lightly
etched as part of the manufacturing process, so they may be only 74% as
thick as an external copper layer of the same nominal thickness.  For
example, one specification I found gives:
*  Internal 0.5 ounce/ft^2 copper >=      0.00050 inches thick
*  Internal 1 ounce/ft^2 copper >=        0.00100 inches thick
*  Internal 2 ounce/ft^2 copper >=        0.00200 inches thick
*  Internal 3 ounce/ft^2 copper >=        0.00300 inches thick
*  External 0.5 ounce/ft^2 copper >=      0.00068 inches thick
*  External 1 ounce/ft^2 copper >=        0.00135 inches thick
*  External 2 ounce/ft^2 copper >=        0.00270 inches thick
*  External 3 ounce/ft^2 copper >=        0.00405 inches thick

I spent over a year researching the ampacity of printed circuit boards,
from summer 1999 to September 2000.  The vast majority of the published
data, including MIL-STD-275, IPC-D-275, IPC-2221, and copies thereof is
derived from
Jennings, C. W., "Electrical Properties of Printed Wiring Boards",
Sandia Labs SAND75-0663, May 1976.

Another major source, where the authors actually ran the experiments
versus just copying someone else's graphs, was
Friar, Michael E., and McClurg, Roger H., "Printed Circuits and High
Currents," Design News, vol. 23 no. 25, pp. 102-107, December 6,
1968.

The equations that I finally came up with were:
*  external I = 1500*thickness^0.72*width^0.75*deltaT^0.45
*  internal I = 7500*thickness^0.72*width^0.75*deltaT^0.45
where,
*  I in Amps
*  Thickness in inches
*  Width in inches
*  DeltaT in degrees C

These are reasonably close to the equations derived in:
*  McHardy, John, and Gandhi, Mahendra, "Empirical Equation for Sizing
Copper PWB Traces," IPCWorks 1997 technical paper SO6-2-1.
*  Brooks, Douglas, "Temperature Rise in PCB Traces," Proceedings of the
PCB Design Conference West, 1998.

One somewhat surprising result of my study was that the spokes on
plated-through holes tended to limit the ampacity.  For a 10C
temperature rise:
*  Multilayer boards max'ed out at:
-  1.08A going into 0.5 ounce/ft^2 copper, for >= 0.024" holes.
-  1.77A going into 1 ounce/ft^3 copper, for >= 0.026" holes.
-  2.92A going into 2 ounce/ft^2 copper, for >= 0.050" holes.
-  Otherwise I = 56*diameter, where I is in Amps and diameter is
in inches, is conservative.
*  Doublesided cards max'ed out at:
-  2.67A going into 0.5 ounce/ft^2 copper, for >= 0.044" holes.
-  Otherwise I = 56*diameter, where I is in Amps and diameter is
in inches, is conservative.

John Barnes
dBi Corporation
http://www.dbicorporation.com/

Doug Brooks wrote:
>
>  At 04:53 PM 7/8/2002 -0400, AK Mishra wrote:
>
> I am in search for a formula for calculating "Current Carrying capacity" of
> a Via. All the Books talk about Traces only. Help me out..
>
> Thanks!
> UltraCAD has just posted a new article on this subject on its web page
> [1]Title:
> Current Carrying Capacity of Vias
> Some Conceptual Observations
>
> _____________________________________________________________________________
> [       Å
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