When setting up this formula on my Excel spreadsheet, I calculated and also
found in an engineering formula book that:

        One square inch is 0.00064 square meters, not 0.0064 meters (.0254 m
x .0254m).

Jerry A. Hegg
Manager, Package Design
Lockheed Martin Missiles & Fire Control - Dallas
(972) 603-7470

        ----------
        From:  Abd ul-Rahman Lomax [SMTP:[log in to unmask]]
        Sent:  Tuesday, November 23, 1999 1:05 PM
        To:  [log in to unmask]
        Subject:  Re: [DC] Horrible Terrible Mistake ...
        Importance:  High

        At 11:22 AM 11/22/99 -0800, Douglas McKean wrote:
        >[...] the equation for Heat Conducted From Surface to
        >Air.  It's the following ...
        >
        >Q = h * (T1 - T2) * A
        >
        >h  = Heat Transfer Coefficient
        >T1 = temperature in degrees C
        >T2 = temperature in degrees C
        >A  = area in square meters.
        >
        >h is given several values given still or turbulent air.
        >
        >Still Air      h = 23 to 28
        >Turbulent Air, h = 85 to 113
        >
        >One square inch is 0.0064 square meters.
        >
        >Dissipation in Watts for 0.0064 square meters,
        >an ambient of 20C and a raised temp of 40C
        >(a temp differential of 20 degrees C), this
        >works out to be 2.94 Watts. Worst case it
        >and say 3 Watts.

        Now, take that square inch of copper PCB surface and drill plated
holes
        through it, and add copper to the other side. Suppose the holes are
0.050
        inches in diameter and are spaced 0.100 apart. I'll assume there are
100
        holes. The total area of the holes is .2 inches squared, so the
surface is
        reduced to 0.8 square inch on one side, but to this we add the 0.8
square
        inch on the other side, which now has a low thermal resistance to
the first
        side, and the area of the hole walls, which is another 1.0 square
inch. The
        total surface has become 2.6 square inches, which alone would
explain why
        experiment showed greatly improved heat dissipation with holes drill
in the
        board/heat sink. However, the holes will also have an effect in
increasing
        air turbulence, so we would expect the value of h to increase.
Easily, it
        might double.

        In the other direction, cooling by radiation would also increase
because of
        the use of both sides of the board; it is more difficult to quantify
this;
        but my sense is that the increase in pure radiative cooling would be
        greater than a factor of 1.6. But radiation is less significant than
        conduction/convection.

        So I would expect heat dissipation to increase by a factor of five
from
        drilling those holes and using both sides of the board.

        I've asked SMT magazine if they were the ones who printed the
article about
        this technique of improving heat dissipation; I could not find it on
their
        web site index, which was not readily searchable. In any case, the
        information in the article, as I recall, had been somewhat mangled
by
        either poor writing or poor editing and some of the technical
information
        necessary to assess the reported results was missing.

        Next opportunity, I plan to design a test board and verify the
prediction.

        [log in to unmask]
        Abdulrahman Lomax
        P.O. Box 690
        El Verano, CA 95433