Thanks to all of you who replied to my request. I received at least four different formulas ranging from the very complex to the very simple. Some equations were based upon the wave guide beyond cutoff principle, but some people said that the wave guide formula does not apply to thin panels. Most people seem to rely upon a simple equation which can be found in Henry Ott's book and other sources. The equation is based upon the fact/assumption (your choice) that a slot dimension of 1/2 wavelength has zero loss. The equation then becomes A = 20 log (lambda/2L). For multiple apertures, the reduction in shielding is approximately proportional to the square root of the number of openings: A = - 10 log n.
Several people indicated that I should avoid using equations with dB when addressing Mechanical Engineers and Designers, so I made the equation even simpler. Henry Ott indicates that enclosures of properly designed commercial equipment only needs 20 dB of shielding. I then added my own assumptions. I have assumed that the clock frequency is no higher than 33 MHz which, based upon experience, can present harmonics up to about 600 MHz. Therefore, I will design for 20 dB of shielding at 600 MHz. Using these assumptions and combining the two equations above, one is left with the equation L(cm) = 2.5/sqr(N). Now this is an equation the Mechanical Engineers and Designers can live with. Comments? Richard Woods Sensormatic Electronics [email protected] Views expressed by the author do not necessarily represent those of Sensormatic.
