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ellie
USA
7 Posts |
 Posted - 05/22/2021 : 08:06:24
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What happens if RF is blocked with faraday material does it just find another direction to dissipate?
We have a wifi home alarm panel that we pass by multiple times a day. The RF readings are extremely high when you walk by it (up to 100RF). On the other side of the wall the unit is mounted to is a bathroom. That room is also receiving high RF.
I have a large faraday phone pouch that out of curiosity I taped to the front of the panel and it decreased the amount of RF when you pass by it by almost half. If I were to create a better cover for it with faraday material what happens to the blocked radiation? Are we just getting it in another direction? I was also thinking of hanging a large framed art piece on the bathroom wall that has the alarm panel a few inches behind it. I would line the art piece in the back with faraday material to protect the bathroom a bit more.
So if the RF is being blocked from the front and the back, what happens to the radiation? I presume it dissipates upward or in any other direction it can get out?
Does the above sound like a decent plan.... as-in would we be getting at least some protection in areas that we previously were not? Or, since I do not know how electrical radiation works, could I be in some way making the situation worse?
To add: We can't move the alarm panel. We actually had two others as well (big house) and one was in the bedroom. I removed the two others but the one in the entry way is the last one so I would like to keep it there. Thanks! |
Edited by - ellie on 05/22/2021 08:07:15
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| Reply #1
Damien68
France
780 Posts |
Posted - 05/22/2021 : 21:16:08
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a metallic surface will cause a reflection of the RF fields, this is calculated by integrating the maxwell equations and by applying a limiting condition which is that the EM field in the metal is zero. depending on the conductivity of the metal the field will enter slightly into the metal which gives what we call a skin thickness but overall the wave will be reflected by being inverted.
During reflection there will be a slight attenuation, a slight loss of power by heat dissipation in the metal but the rest of the energy will be reflected.
if you place an antenna in a closed metal box, what will happen is that the radiation will be reflected a lot of times, and finally all the energy will be dissipated by thermal effect during the reflections. but there will always be a little loss across the shield which will be proportional to the amplitude of the waves inside the shielding, and therefore we will converge towards a steady state which will have a fairly high amplitude of the waves
At this frequency (2.45GHz) the waves are also reflected on walls or any other obstacles (higher is the frequency, more easily the waves are reflected). also the waves can passing by the side of the RF shield and with reflections on the walls or on the ground can bypassing it, but will be well attenuated. because of the currents which will be induced in the shielding, the edge of the RF shielding will also generate a diffraction of the wave (act like a sources or a antenna), all this is complex and is calculated by modelisation and by integrating the 4 maxwell equations.
note: it is not always very good to place shields, especially on cell-phone, because, if it is not completely bogus it disturbs the antenna, so antenna becomes less efficient and to compensate the telephone can then react by emitting louder. and in addition, most often the pellet-shield is not placed on the right side of the phone the solution for a cell-phone is to use a wired hands-free kit and hold your phone more than 30cm away, especially for those who spend time on the phone.
Note2: in this context we should not speak of Faraday. the Faraday cage is an electrostatic phenomenon considering a static electric field.
in the case of RF waves we are no longer in a static context at all and even if there are similarities the observations relating to the Faraday cage no longer apply.
Faraday also worked on dynamic fields but in other contexts (more for engines and others ...). For the present subject the main master is Maxwell and his 4 equations but requires very specific mathematical and physical training to be understood and used.
Note3: for my opinion I think that the heat of the telephone placed for a long time against the ear can be as much or more harmful than the radio waves because it only heats a very localized part of the body and always the same. for Wifi, more than a meter away, the power levels are really very low and don't worry me.
remember that RF measurements made in dBm are on an exponetial scale:
0dBm = 1mW
-10 dBm = 0.1mW
-20dBm = 0.01mW
...
-40dBm= 0.0001mW = 0.1uW = 100 nano Watt
...
and it is by extension the power received by a 'standard' tuned antenna placed at a given location.
Note:in an ideal conductive material the field E is zero because it conducts the current (for the same reasons that there cannot be a voltage at the 2 terminals of a perfect conductor).
when an EM wave arrives on a shield, there is an incident EM wave with a non-zero E field, the E field is zero in the shielding because it is conductive, the current induced by the incident wave in the copper of the shielding will therefore create an inverted E field, and the E field of the incident wave plus the E field of the induced wave will be zero. this is what we call a limit condition |
Mastery is acquired by studying, with it everything becomes simple |
Edited by - Damien68 on 05/27/2021 09:05:13 |
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| Reply #2
ellie
USA
7 Posts |
Posted - 05/29/2021 : 21:15:50
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| Thank you! |
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RF dissipation question
