Enclosure Aperture Leakage Calculator
Enclosure Aperture Leakage Calculator
What a shielded box actually leaks through its slots, seams, display cut-outs and vents — the aperture’s own shielding effectiveness, the frequency at which it goes to zero, the penalty for many apertures, and the gain a thick wall buys as a waveguide below cutoff.
Leakage through an aperture, seam or vent
a 100 mm seam 0.2 mm wide in a 1.5 mm wall, one of them, at 200 MHz
Leakage is set by the longest dimension, not the area
fresonance = c / 2L the frequency at which that goes to zero
N apertures: −10 log10 N
depth: + K · (t / L) · √(1 − (f/fc)²) K = 8.686 · 2π / λc
round hole: λc = πd / 1.8412, so K = 31.98 slot: λc = 2L, so K = 27.29
- L
- the LONGEST dimension of the opening — a slot’s length, a hole’s diameter. The slot’s width does not appear at all, which is why narrowing a seam without shortening it achieves nothing
- K
- the below-cutoff attenuation constant, derived from the cutoff wavelength rather than quoted. 1.8412 is the first root of the derivative of the Bessel function J1, which sets the TE11 cutoff of a circular guide
- fc
- the opening’s waveguide cutoff. Below it the depth term applies and grows as the frequency falls; at it the term is zero and the opening propagates
- N
- the number of openings close together on the same face. Their leakage POWERS add, so the penalty is 10 log N and not 20 — a hundred holes cost 20 dB
Worked example
a 100 mm seam 0.2 mm wide in a 1.5 mm wall, one of them, at 200 MHz
At 200 MHz the wavelength is 1.499 m. The seam's longest dimension is 100 mm, so λ/2L is 7.495 and the aperture term is 20 log of that = 17.50 dB
The 0.2 mm width appears nowhere in that. Closing the gap to 0.02 mm would change nothing at all; the seam is 100 mm long either way
A slot's waveguide cutoff is c/2L = 1.499 GHz, and 200 MHz is 0.1334 of that, so the depth term is 27.29 × (1.5/100) × √(1 − 0.0178) = 0.41 dB — almost nothing, because the wall is thin compared with the seam's length
One aperture, so no N penalty. The total is 17.9 dB
For comparison: the seam's open area is 20 µm², and a round hole of that area would be 5.046 mm across and worth 43.4 dB — 25.9 dB better for exactly the same hole in the metal
And that penalty is frequency independent: it is 20 log(100 mm ÷ 5.046 mm), which is 25.9 dB at every frequency the model is valid at
The seam stops shielding altogether at 1.499 GHz, where 100 mm is half a wavelength. This page will not print a number above that
A slot against a round hole of the same open area
| Opening | Longest dimension | Open area | SE at 100 MHz |
|---|---|---|---|
| Slot 50 mm × 1 mm | 50 mm | 50 µm² | 29.5 dB |
| Round hole, same area | 7.98 mm | 50 µm² | 45.5 dB |
| Difference | — | none | 15.9 dB |
What depth buys, at frequencies well below cutoff
| Opening | Depth ÷ dimension | Waveguide attenuation |
|---|---|---|
| 3 mm hole in 1 mm sheet | 0.33 | 10.7 dB |
| 3 mm hole in a 6 mm boss | 2.0 | 64.0 dB |
| 2 mm honeycomb cell, 10 mm deep | 5.0 | 159.9 dB |
| 100 mm seam in a 1.5 mm wall | 0.015 | 0.41 dB |
| 10 mm slot in a 10 mm wall | 1.0 | 27.3 dB |
Half-wave resonance: the frequency at which an opening stops shielding
| Longest dimension | f = c / 2L | And the model is only trustworthy below |
|---|---|---|
| 5 mm | 29.98 GHz | 9.993 GHz |
| 10 mm | 14.99 GHz | 4.997 GHz |
| 25 mm | 5.996 GHz | 1.999 GHz |
| 50 mm | 2.998 GHz | 999.3 MHz |
| 100 mm | 1.499 GHz | 499.7 MHz |
| 200 mm | 749.5 MHz | 249.8 MHz |
| 500 mm | 299.8 MHz | 99.93 MHz |
The truth a wall-thickness calculation cannot tell you
A shielded enclosure is not limited by its walls. Half a millimetre of aluminium is worth over a hundred decibels against a plane wave anywhere in the HF band, and no real box comes close to that, because real boxes have lids, seams, connectors, display windows, ventilation and cables. This page is about the openings.
Length, not area. The leakage through a small opening goes as the ratio of the wavelength to twice its longest dimension. The width does not appear. A 100 mm seam 0.1 mm wide leaks exactly as much as a 100 mm seam 2 mm wide, and the instinct to close gaps rather than shorten them is the most common and most expensive mistake in enclosure design. The fix for a long seam is more fastener points, or a conductive gasket that makes the joint continuous. The fix for a display window is a mesh or a conductive coating, which replaces one long opening with many short ones — and that trade is a win, because splitting a slot improves the aperture term by 20 log of the length reduction while costing only 10 log N for the extra openings.
Depth is the one thing that helps a hole. Below its cutoff frequency a hole is a waveguide that does not propagate, and the field decays exponentially down it. The attenuation is about 32 decibels for every unit of depth over diameter in a round hole, and about 27 for a slot, which is why a honeycomb vent panel with 2 mm cells 10 mm deep gives over 150 dB while a 2 mm hole in foil gives nothing. Both constants come straight from the cutoff wavelength — πd/1.8412 for a circular guide, 2L for a rectangular one — and both collapse as the frequency approaches cutoff.
Where this model stops. These are small-aperture approximations. They assume the opening is much smaller than a wavelength, that the surrounding sheet is a good conductor bonded all round, that nothing is coupled to the opening from inside, and that the enclosure has no cavity resonance nearby. Every one of those assumptions fails somewhere. The page prints the frequency at which the longest dimension reaches half a wavelength — where the model says zero and the reality is an efficient slot antenna — and a frequency a third of that, above which it is already flattering. It will not print numbers above resonance, because extrapolating a small-aperture formula into the microwave region is how a leaky enclosure comes to look sealed on paper.
What is still missing after all that. Coupling. The formulas above describe how well an opening blocks a field that happens to arrive at it; they say nothing about how strongly whatever is inside the box is driving it. A switching node routed along the inside of a seam is a completely different problem from the same seam with a ground plane between it and everything else, and the difference is tens of decibels. Enclosure shielding is as much about internal layout as it is about the metalwork.
And cables beat all of it. An enclosure with perfect seams still has cables leaving it, and a cable shield terminated with a pigtail carries interference straight through the wall as if it were not there. That is the transfer impedance page, and it is usually the first thing to fix. The solid-wall page gives the upper bound this page erodes, and the RE102 margin page is where the requirement lives.
Frequently asked questions
Does making a seam narrower help?
No — not in this model and not much in practice. Leakage is set by the longest dimension of the opening, which for a seam is its length between bond points. Closing a 1 mm gap to 0.1 mm leaves a 100 mm slot either way. What helps is shortening it: more fasteners, a conductive gasket, a continuous weld. Halving the length between bonds buys 6 dB.
Why is a round hole so much better than a slot of the same area?
Because the answer depends on the longest dimension and not on the area. A 50 × 1 mm slot and a 8.0 mm round hole pass exactly the same open area, and the slot is 16 dB worse at every frequency — 20 log of the ratio of the two dimensions. This is why ventilation panels are drilled with many small holes rather than cut as a few long louvres.
How much does wall thickness help?
Through the waveguide-below-cutoff effect, and only in proportion to depth over the longest dimension. A 3 mm hole in 1 mm sheet gets about 11 dB from depth; the same hole through a 6 mm boss gets about 64 dB. For a long seam the ratio is tiny and the effect is negligible. And it all disappears as the frequency approaches the opening’s cutoff.
What happens above the half-wave frequency?
The opening stops being an aperture and becomes an antenna. At λ = 2L the model gives exactly zero, and above it the real structure has resonances that a small-aperture approximation cannot describe at all. This page clamps the aperture term at zero there and says so, rather than printing a negative number that would look like a calculation.
Do a hundred small holes really only cost 20 dB?
10 log N, yes — provided they are on the same face and close together compared with a wavelength, so their leakage powers add. That is why a perforated panel works: a hundred 3 mm holes give up 20 dB against one 3 mm hole, but each of them is worth far more than one large opening of the same total area. Holes spread over different faces of the enclosure do not combine like this and should be assessed separately.
Can I add this to the wall’s shielding effectiveness?
No. They are not in series in that sense — the enclosure leaks by whichever route is worse, so the aperture number effectively replaces the wall number wherever it is lower. In almost every real enclosure the aperture is lower by a wide margin, which is the point of this page.
Related calculators
References
- Ott HW. Electromagnetic Compatibility Engineering. Wiley, 2009. Chapter 6 Shielding, sections 6.10 Apertures, 6.10.1 Multiple Apertures, 6.10.2 Seams and 6.11 Waveguide Below Cutoff. Section numbers and titles verified against the author’s own published detailed contents listing.
- Schelkunoff SA. Electromagnetic Waves. D. Van Nostrand Company, New York, 1943. The waveguide cutoff wavelengths from which the depth-attenuation constants on this page are derived — πd/1.8412 for the circular TE11 mode and 2L for the rectangular TE10 mode.
- IEEE Std 299-2006, IEEE Standard Method for Measuring the Effectiveness of Electromagnetic Shielding Enclosures. The measurement these estimates would be compared against. Copyrighted; named for the method only.
- MIL-STD-461G, paragraph 5.18 RE102, radiated emissions, electric field and paragraph 5.21 RS103, radiated susceptibility, electric field. Aperture leakage works in both directions and the same opening limits both requirements.
