# Shielding effectiveness calculation

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:

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.

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

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?

• 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. – nidhin Jun 2 '14 at 18:27
• @nidhin He also uses 10 log10(n) instead of 20 log10(n). hmmm – endolith Jun 2 '14 at 19:48
• 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. – nidhin Jun 3 '14 at 3:09