Richard, The caveat in Henry Ott's statement about 20 dB, are the words, "properly designed." For equipment that is not "properly designed," 20 dB may be a little on the lean side. Boards that are not "properly designed" are very common.
Since it is difficult to predict shielding performance (or shielding need) with accuracy (see my previous posting), a cautious designer will either over-design, or provide a contingency to add additional shielding if the need arises. I have done the latter and been very glad about it. Jim ------------------------------------------------------------------------ --------------------------- Dr. Jim Knighten NCR 17095 Via del Campo San Diego, CA 92127 Telephone: 619-485-2537 Fax: 619-485-3788 e-mail: [email protected] ---------- From: WOODS, RICHARD [SMTP:[email protected]] Sent: Tuesday, July 21, 1998 6:33 AM To: 'emc-pstc' Subject: Designing Openings for EMC Compliance 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.
