[Andre Jute]

HEATSINKING

No 57 in Andre's Handy Shortcuts
also
No 113 in Andre's Rules of Thumb for the Mathematically Ten-Thumbed

Let's say you have a component (resistor, tube, whatever) which needs
to be kept from overheating. You can make a heatsink by bolting
aliminium angles to a backplate to make fins.

The permissible temperature rise is the permissible temperature less
the room temperature which we assume is 30 degrees.

The heatsink rating is usually given in degrees Celsius of rise per
Watt of power dissipated by the device, assuming a perfect junction.
Thus, divide the permissible temperature rise by the power of the
device.

The minimum area of heatsinking you want to achieve this rating is
appromimately 50 divided by the square root of the area in square
centimetres.

You count all area exposed to free-flowing air: both sides of the
outside fins plus the outside of the backplate but not its inside if
enclosed.

So we have

50/(sqrtA in cm) = (permissable rise in degrees C)/(power in watts)

which is good for checking if an existing heatsink will do the job, or
we can calculate how big a heatsink must be to do the job with this
more convenient formulation

A = (C/(W*50))*(C/(W*50))

and the answer is in square centimetres.

HTH.

Andre Jute
Visit Jute on Amps at http://members.lycos.co.uk/fiultra/
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