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From The Basic Principles of Shielding:

Apertures, or holes, have SE. The SE of an aperture and ultimately the entire electronic enclosure is determined by the size, shape and number of the apertures. The formula is:

1006_F1_eq1

Where: λ = Wavelength k = 20 for a slit or 40 for a round hole L = Longest dimension of the aperture If there is more than one hole, we subtract from the original formula: the total number of holes within half a wavelength.

Likewise, Analyse on shielding effectiveness of board level shielding with apertures says:

$$\mathrm{SE\ (dB)} = K \log_{10} \left (\frac{λ}{2 L} \right ) - 20 \log_{10}(n)$$

Where: n = numbers of apertures within a half wavelength

and Rule-of-Thumb for Calculating Aperture Size says "n=number of apertures within λ/2"

But what does this mean? Number of holes in a circle with half-wavelength radius? (When I try to calculate this way, the first part of the expression is 275 dB, while the second half is 277 dB, producing a negative SE.) Number of holes along a line half a wavelength long?

Can you work through an example calculation for, say, a microwave oven door with triangularly-packed holes?

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    \$\begingroup\$ I have seen similar equation in a book by Henry Ott. The book says that this equation is applicable to a linear array of closely spaced apertures. But this same expression is used for calculating SE of two dimensional array of apertures also. In that case \$n\$ is the maximum number of holes lined up in a straight line, horizontal, vertical, or diagonal. \$\endgroup\$ – nidhin Jun 2 '14 at 18:27
  • \$\begingroup\$ @nidhin He also uses 10 log10(n) instead of 20 log10(n). hmmm \$\endgroup\$ – endolith Jun 2 '14 at 19:48
  • \$\begingroup\$ yes. He says that "For a linear array of closely spaced apertures, the reduction in shielding effectiveness is proportional to the square root of the number of apertures (n)". Hence 10log instead of 20log. \$\endgroup\$ – nidhin Jun 3 '14 at 3:09

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