Slot / Aperture Shielding Effectiveness Estimate
Estimate the shielding effectiveness of an aperture from frequency and the longest dimension of the opening, including the reduction for multiple identical openings and the leakage onset frequency.
Worked example
Input: Frequency 1000 MHz, Longest dimension of the opening 10 mm, Number of identical openings 1 pcs
Result: Wavelength λ 299.792 mm, Leakage onset frequency c/(2L) 14989.6 MHz, Estimated shielding effectiveness (SE) 23.5158 dB
Result: Wavelength λ 299.792 mm, Leakage onset frequency c/(2L) 14989.6 MHz, Estimated shielding effectiveness (SE) 23.5158 dB
Formula
SE ≈ 20·log₁₀(λ / 2L) − 10·log₁₀(n) (L < λ/2)
If L ≥ λ/2, SE ≈ 0 dB (the opening leaks like an antenna)
f_max = c / (2L)
How it works
Gaps and openings, not the sheet material, often decide the shielding of a metal enclosure. When the longest side L is shorter than half a wavelength, SE ≈ 20·log₁₀(λ/2L), reduced by 10·log₁₀(n) for n identical openings. At half a wavelength the opening behaves like an antenna and SE drops to about 0 dB. For example a 10 mm opening at 1 GHz gives about 23.5 dB.
Practical tipThis is a rough estimate for a single opening in an ideal thin metal sheet. Real seams can be far worse depending on contact resistance, paint and gasket condition. Break a long seam into short ones with gaskets, and keep ventilation holes small and few.
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Results and summaries are for reference. For certification and test reports use the latest official standard text and calibrated instrument data. Last updated: 2026-10-10