Mesh hole size calculation for RFID signal blocking

[schoor]

TL;DR Summary

Question regarding the hole sizing of a mesh that should block RFID waves from passing through

For a project I am currently facing a problem that I find difficult to solve.

I am developing a shielding portal that uses a mesh to blocks UHF RFID signals. I would like to use aluminium mesh due to it being lightweight and its attenuation properties. The frequency of UHF RFID signals in the ETSI region is 868 MHz and I found that 900 MHz has a wavelength of 33.3cm. Other forums suggest a calculation that divides the wavelenght Lambda by 20 (https://www.researchgate.net/post/Is_steel_mesh_metal_a_good_shielding_to_RF).
That would mean that the max diameter of the mesh holes would be 1,665 CM.

Now I am really unsure if this is a right calculation and if this is the same for all materials. It would be great if someone is able to elaborate more on the principles of hole sizing and the frequency of RF waves.

 

[Baluncore]

At 870 MHz, λ is 345 mm;
The rule of thumb for hole dimension is λ / 20 = 17 mm.
But that does not specify the dB attenuation of transmitted energy, the shape of the hole, slot, or any array factor.

I expect it will require a transmit signal attenuation of about 40 dB, to hide a card. That is going to require two layers of foil separated by a sheet of paper, and folded to envelope the card. The envelope will have a slot on three sides with edge currents, but the sheet of conductive foil will cancel EM fields at the the card antenna.

 

[rude man]

Qualitatively you might think of the mesh aperture as a very short circular waveguide. However, that computes to a mesh hole radius of 3.6 cm for a microwave oven operating at 2.5GHz (all modes). Tyical oven size is more like 0.05cm so quantitatively that doesn't seem to work too well

 

[Baluncore]

Qualitatively you might think of the mesh aperture as a very short circular waveguide.

A waveguide is designed to pass over 99.9% of the energy.
The edges of a hole in a screen must reflect over 99.9% of the energy.
The λ/20 rule of thumb is for a hole in a reflector that will reflect most of the energy, but a proportion of the energy will still pass through.

There are equations that will predict the attenuation due to an aperture or iris in a conductive sheet. A hole in a conductor is a slot antenna element. A mesh is an array of slot antenna elements.

For screening of UHF and microwave, a foil will be lighter than a woven or punched mesh screen. So it is only the skin effect that must be satisfied. Two separated foils, on both sides of a film or paper sheet will satisfy the attenuation requirements.

 

[alan123hk]

Summary:: Question regarding the hole sizing of a mesh that should block RFID waves from passing through

Other forums suggest a calculation that divides the wavelenght Lambda by 20

Those mesh holes are completely useless for blocking RF. Its sole function is to allow air to circulate or let light pass through, so the number and diameter of these holes should of course be as small as possible.

The only way to block RF is to reflect it back or absorb energy.

 

[Baluncore]

The window in the door of a microwave oven has an internal perforated metal plate, usually painted black. That reflects internal microwave energy and so does not allow microwave energy to escape through the window. Light has a shorter wavelength so you can see through the screen if the internal light is on.

 

[Tom.G]

Those mesh holes are completely useless for blocking RF. Its sole function is to allow air to circulate or let light pass through, so the number and diameter of these holes should of course be as small as possible.

The only way to block RF is to reflect it back or absorb energy.

Take a look at:
http://www.ets-lindgren.com/products/shielding/rf-shielded-enclosures/11003/1100303

(above found with:
https://www.google.com/search?&q=screen+room+rf+shielded)

I have worked in a development lab that was about 2 to 3 miles away from a 10kW, AM broadcast, radio station. The field strength in the lab was about 0.5V per meter!

The 'screen room', as it was called, had 3 layers of Copper screen separated by about an inch and all bonded together. It worked really well to keep that radio station out!

The only problem was there was very little air circulation. This caused the temperature to rise rather rapidly when using a dual-beam, vacuum tube, oscilloscope along with the other necessary equipment.

Cheers,
Tom

 

[alan123hk]

The only problem was there was very little air circulation. This caused the temperature to rise rather rapidly when using a dual-beam, vacuum tube, oscilloscope along with the other necessary equipment.

Therefore, several fans are usually installed in the shielded room for air convection, but if there is no air conditioning outside the shielded room, it may not help much.

 

[DaveE]

I'm not sure what you mean by "mesh", but be careful of woven screens. Just because the wires touch, doesn't mean they have good conductivity. Whatever size you end up with, you must have circulating current around the holes.

 

[Baluncore]

Whatever size you end up with, you must have circulating current around the holes.

I believe that whatever polarization the signal has, it can be resolved into two orthogonal components, parallel with the warp and weft of the mesh. The conductive warp and weft will reflect and recombine the two components into the mirror reflected signal. It is not the hole diameter, but the spacing of the lines that decides attenuation.

The same is not true of a slot antenna where the dimension of the slot must be controlled to provide a stable radiation resistance.

 

[hutchphd]

Which is why crossed polaroid films will block all light.

 

[schoor]

Thank you all for the comprehensive replies. This helps a lot and I am also planning on testing this case with a aluminum mesh of around 17mm in diameter.

UHF RFID radiation has a maximum allowed transmitting power of 31dB. As the reading distance drops dramatically when lowering the transmitting power the RFID signal for RFID tags is almost unusable for scanning at distance when smaller then 10db. Therefor the mesh should reduce the signal by 25dB (just to be safe). As @Baluncore stated it the calculation does not say anything regarding attenuation, hole shapes etc. Is there any way to calculate or estimate the max hole size of a aluminum mesh for blocking 870 MHz?

I will also take the conductivity of touching wires into account. To what extent does the thickness of the wires affect the shielding properties?

The signal does not need to be completely blocked but just enough to be unreadable for RFID tags at a distance of around 30 cm behind the shielding.

 

[alan123hk]

This link may be helpful.

http://cdn.lairdtech.com/home/brand...ing Aperture Size Technical Note Download.pdf

 

[schoor]

Perfect, thanks @alan123hk

 

[tech99]

I believe that whatever polarization the signal has, it can be resolved into two orthogonal components, parallel with the warp and weft of the mesh. The conductive warp and weft will reflect and recombine the two components into the mirror reflected signal. It is not the hole diameter, but the spacing of the lines that decides attenuation.

The same is not true of a slot antenna where the dimension of the slot must be controlled to provide a stable radiation resistance.

In his book "Antennas", J D Kraus discusses the transition from rods to slots. But even with the slot, it is the current in the conductor which does the work.