Shielding Effectiveness Calculator
Calculate electromagnetic shielding effectiveness, absorption loss, reflection loss, and skin depth. Evaluate enclosure materials per MIL-STD-285.
Formula
Reference: MIL-STD-285, Schulz et al.
How It Works
The Shielding Effectiveness Calculator computes electromagnetic attenuation for conductive enclosures — essential for EMC compliance (CISPR 32, FCC Part 15), medical device immunity (IEC 60601-1-2), and military specifications (MIL-STD-461G). EMC engineers use this to achieve 40-80 dB shielding required for sensitive electronics protection.
Per Henry Ott's 'EMC Engineering' and MIL-HDBK-419A, shielding effectiveness SE = A + R + B, where A is absorption loss, R is reflection loss, and B is re-reflection correction (negligible when A > 10 dB). Absorption loss A = 8.686 x t/delta, where t is thickness and delta = sqrt(2/(omega x mu x sigma)) is skin depth. At 1 GHz, copper skin depth is 2.1 um; a 1mm copper sheet provides A > 400 dB.
Reflection loss R = 20 x log10(Z0/4Zs), where Z0 = 377 ohm (free space) and Zs = sqrt(omega x mu/sigma) is shield impedance. Copper at 1 GHz has Zs = 0.026 ohm, giving R = 20 x log10(377/(4 x 0.026)) = 67 dB. Total SE for copper exceeds 100 dB — but real enclosures have apertures.
Per Ott, apertures dominate shielding failure. A single slot of length L reduces SE to approximately 20 x log10(lambda/(2L)) at frequencies where L > lambda/2. A 10cm slot (f_cutoff = 1.5 GHz) provides only 0 dB shielding at 1.5 GHz and negative SE (resonant amplification) above. CISPR 32 Class B requires 40 dBuV/m limit — enclosure apertures must be sized to provide 20+ dB margin.
Worked Example
Problem: Design aluminum enclosure (sigma = 3.77e7 S/m, mu_r = 1) with 2mm wall thickness for 40 dB shielding at 1 GHz. Maximum ventilation slot length?
Solution per Ott:
- Skin depth at 1 GHz: delta = sqrt(2/(2 x pi x 1e9 x 4 x pi x 1e-7 x 3.77e7)) = 2.6 um
- Absorption loss: A = 8.686 x 0.002/2.6e-6 = 6680 dB (wall is not limiting factor)
- Reflection loss: Zs = sqrt(2 x pi x 1e9 x 4 x pi x 1e-7/3.77e7) = 0.032 ohm; R = 20 x log10(377/(4 x 0.032)) = 66 dB
- Enclosure SE without apertures: >100 dB
- For 40 dB at 1 GHz with apertures: SE_aperture = 20 x log10(lambda/(2L)); lambda = 0.3m at 1 GHz
- 40 = 20 x log10(0.3/(2L)); 100 = 0.3/(2L); L = 1.5mm maximum slot length
- For 20 ventilation slots: use honeycomb waveguide-beyond-cutoff filter (5mm cells provide >60 dB at 1 GHz)
Result: 2mm aluminum provides >100 dB material SE, but slots must be <1.5mm for 40 dB. Use honeycomb filter for ventilation.
Practical Tips
- ✓Size apertures to lambda/20 maximum — per Ott, this provides 26 dB margin versus lambda/2 resonance. At 1 GHz (lambda=30cm), max aperture = 15mm; at 3 GHz, max = 5mm.
- ✓Use conductive gaskets at all seams — EMI gaskets (BeCu fingerstock, conductive foam) maintain <10 mohm contact resistance required for 40+ dB SE per MIL-HDBK-419A.
- ✓Place EMI filters at cable entry points — feedthrough capacitors provide 40-60 dB; Pi-filters provide 60-80 dB. Filter must be bonded to enclosure for proper ground reference.
Common Mistakes
- ✗Assuming material SE equals enclosure SE — material provides 60-100+ dB but apertures (seams, ventilation, displays) typically limit real enclosures to 20-60 dB. Per Ott, a single untreated seam can reduce SE to <10 dB.
- ✗Using DC conductivity for high-frequency calculations — skin effect confines current to surface; surface finish (oxidation, paint) can add 10-20 dB loss. Use measured surface resistance or specify conductive finish.
- ✗Ignoring cable penetrations — unfiltered cables act as slot antennas inside shielded enclosures. Per MIL-STD-461G, all cables must be filtered at point of entry or use shielded/filtered connectors.
Frequently Asked Questions
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