At 01:54 PM 12/30/98 -0500, Eisermann, Phil wrote...
>
>        Then multiply Irms*500 to get the number of circular mils.
>
>        For example, assume Irms = 40mA, you'd need
>        0.04 * 500 = 20 circular mils. so you need a wire
>        with a diameter of sqrt(20) = 4.5 circular mils.
>        That's between #37AWG and #36AWG. use #36AWG.
>
>        (#36AWG has a diameter of 5 mils. That's an
>        area of 25 circular mils, or 19.63E-6 sq. in)

Because there is a potential error here of > 10%, I thought I would point it out
:

Circular area is PI*r^2, so #36 is 3.1416 * 2.5 mils ^2 = 19.63 mils^2.

For 20 mils^2, the diameter would be 2 * sqrt(20 mils^2 /PI) = 5.05 mils

Somehow, you got the in^2 correct, but the mils^2 wrong!?

Mike