I have read on the following page that the maximum permissible exposure MPE standard FCC have is 1 mW/cm2 for RF wave above 1500 MHz for general population. http://www.rfcafe.com/references/electrical/fcc-maximum-permissible-exposure.htm

But I remember that power density of such waves is measured only at a particular distance from the transmitter but there is no mention of any distance from which the measurements are taken. Let me know if I got anything wrong.

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    \$\begingroup\$ The point of the regulation isn't to give you a safe distance. It is telling you how much RF power is allowed to reach the surface of your body. That is given as mW per square centimeter - that's power per unit of area. You don't give darn about the distance from the transmitter. You use a measuring device that measures RF power in the units given in the regulation. This independent if the distance from you to some transmitter - and that is intentional. The regulation makes perfect sense. \$\endgroup\$ – JRE Sep 14 '18 at 10:43
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    \$\begingroup\$ If you need to find a distance from a transmitter such that exposure is under the limit, then you need to know the power of the transmitter, the efficiency of the antenna, and its directionality. From all of that, you can estimate a boundary line around the antenna such that anyone outside the line can be assumed to be under the exposure limit. \$\endgroup\$ – JRE Sep 14 '18 at 10:47
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    \$\begingroup\$ Stay outside the fence that the operator puts up. Keeping you outside of the exposure limit is the responsibility of the operator, and the reason why the FCC defined the limit as it did. If you suspect that an antenna is placed such that you can be over exposed outside of the operator's fence then you need to have the level measured using the proper equipment by someone who knows how to operate it. \$\endgroup\$ – JRE Sep 14 '18 at 11:13
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    \$\begingroup\$ You get a qualified person with calibrated equipment to measure the exposure in and around your house. If it is too high, you sue the operator(s) and/or the landowner. \$\endgroup\$ – JRE Sep 14 '18 at 11:26
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    \$\begingroup\$ The FCC limits are not vague in any way. They set a limit, and the operator has to see to it that ordinary people aren't exposed to anything above that limit. Engineers use programs to plan things, then can measure to be sure the limits are met. \$\endgroup\$ – JRE Sep 14 '18 at 12:50

Ok, if I have a 100W transmitter (for example), feeding an isotropic radiator (a theoretical construct that radiates equal power in all directions, in practise you need to take antenna pattern into account), and I am going for a 5mW/cm^2 power limit, lets see what the distance is?

100,000mW = 5mW * area(cm^2), so the area of the sphere is 100000/5 cm^2 = 20,000cm^2 = 4*Pi *r^2, r = sqrt (5,000/Pi) = ~40cm, looks like 100W into an isotropic transmitter is pretty much an non issue.

Now if that was a 10,000W set into an antenna having say 20dB of gain (Large TV transmitter for example), then we have an EIRP of 1MW in the pattern (which will be a flatish disk, but lets see how far we would have to be if we were standing in line with the antennas:

1000,000,000mW / 5mw/cm^2, turning the handle give about 40m, seems about right, 100W = 40cm, 1MW = 40M, factor of 10,000 in power gives factor of 100 in distance, which works with the area of a sphere going as r^2.

  • \$\begingroup\$ Is using surface area of sphere really giving you the value? \$\endgroup\$ – ObsessionWithElectricity Sep 14 '18 at 16:45
  • \$\begingroup\$ Of course, emit energy at the centre, have it radiate outwards in a sphere, the ratio of 1 square cm to the surface area gives you the fraction of the power incident on that square cm. Antenna gain complicates the thing only slightly. The sums are not hard. Note that this breaks down very close to the antenna as it can no longer be viewed as a point source, but the pattern also breaks down in the near field so meh, near field math is harder, avoid going there. \$\endgroup\$ – Dan Mills Sep 14 '18 at 16:54

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