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Will antenna insulation protect from RF burns?

I understand that antennas should normally be built with their insulation on. Will that insulation protect from RF burns? Up to what power level?

If not: What prevents the feedline from causing RF burns, assuming you don't have a balun?

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  • $\begingroup$ "I understand that antennas should normally be built with their insulation on." what is the source of that understanding? $\endgroup$
    – webmarc
    Commented Jun 18 at 1:08

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Good question. The answer is no.

You have to separate shock from RF burn because they are two different things that are not directly related. Voltages on wire antennas can reach lethal voltages measured in kilovolts. Wire insulation can offer protection, but you do not want to count on it. Dipole antennas operating at legal limits will have very high voltages at the ends, measuring a few kVs. End-fed antennas have high voltages above a kV at the feed point. The wire insulation you would likely use is rated 600 volts; flashovers do happen, and if you are in the path, you will have a bad hair day. That covers shock.

RF burn is a different animal. It does not matter if the wire or element is insulated or not. It does not require physical or electrical contact but rather proximity and frequency. If you made a 70 cm dipole, wrap one element around your hand, key up 100 W, and you will learn a nasty lesson and, at a legal limit, get a severe burn. If you did the same thing with a 160 to 20 m dipole with one wrap around your hand, you will not likely notice anything because the wavelength is too long.

The thing to remember about RF burn is the frequency dominates the potential for RF burn. The higher the frequency, the further away you must be. For example, say a 2.4 GHz dish antenna operating at 1000 W. You can inflict severe RF burns on someone from a great distance. Take a 300 ft vertical tower operating at 2 MHz with 10,000 W and walk right up to the tower without ill effects. You might be shocked if you touched the tower, but there is no RF burn.

Follow the electrical codes, and you will not have to worry about it because wire antennas must be placed out of reach with proper clearances and materials.

Insulation can keep the copper pure and corrosion-free on the surface. RF runs on the outside of conductors; corrosion increases resistance.

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  • $\begingroup$ "The higher the frequency, the further away you must be" - I assume you mean "the further way you must be to avoid RF burns". Can you elaborate on that? $\endgroup$ Commented Jun 18 at 3:12
  • $\begingroup$ @SRobertJames it's not completely correct, but Dereck's claim is practical for ham frequencies: Conductive materials absorb RF energy. The higher the frequency, the higher is the absorbtion rate. A 20 cm thick block of gelatin will absorb almost 100% of the energy of a wave front aimed at it at 2.4 GHz, but at 30 MHz, only a tine percentage will be absorbed by the 20 cm of material. (You're pretty close to a block of gelatin.) There's two different physical aspects here at play: first, the general property that slightly conductive objects much smaller than the wavelength can only convert … $\endgroup$ Commented Jun 18 at 10:53
  • $\begingroup$ … a very limited amount of the incident power to heat, because they necessarily must emit the wave at the other end of the block, the skin depth being larger than their thickness. The other effect, and that's one that does not generally increase with frequency, is that human materials interact in non-ohmic ways with RF at different frequencies, differently. Your microwave oven is at 2.4 GHz for a specific reason – water molecules resonate especially well with that frequency. Go to 3 GHz, and the effect is less pronounced – your oven would work less well to burn your flesh, although the $\endgroup$ Commented Jun 18 at 11:05
  • $\begingroup$ frequency is higher. But, in general, Dereck is of course right: The higher the frequency, the more likely is it that a body of fixed thickness will absorb all the energy. And absorbing, wild example, 1% of 100 W, distributed quite evenly through your whole body, will not hurt you. But absorbing 10% of 100W, in the first 0.2 mm of skin that it hits, will leave a lesion. Now, worse is of course if your body is cooked from the inside … and that's why we don't stand in front of the radar dishes, kids. $\endgroup$ Commented Jun 18 at 11:05
  • $\begingroup$ Or, if you are Percy Spencer, you notice your snack has melted in your pocket and get an idea... $\endgroup$
    – clvrmnky
    Commented Jun 18 at 19:47
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RF burning is what heats your food in the microwave oven. It would do the same, whether or not the inner sheet metal surfaces are isolated or not.
It heats organic material by converting RF energy to heat, primarily by absorption through the forced movement of polar molecules in the high-speed electric field, especially water. (This effect is frequency-dependent, and low frequencies in general do not interact with small bodies very much; but 2.4 GHz, for example, is much more effective at heating water than, say, 3 GHz. The permittivity of water has a higher loss tangent at 2.4 GHz than at 3 GHz. In essence, if you make plate capacitor and fill it with watery stuff rather than air, it will get very low-Q when used at 2.4 GHz, and slightly higher-Q at 3 GHz.)

Direct, low frequency current flow is what burns you when you touch a high voltage power line; it flows as surface current in lines of high current density, and heats things by means of ohmic resistance, mostly.

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Insulated wire used for whatever purpose can protect from shocks and burns from DC, AC & AC at RF. The question is, what power are you putting into that wire, and where on the standing voltage wave of that wire, are you presenting a better return path for the voltage to its source? If you have wire with a 600V insulation rating, that insulation is good for 10s of thousands of Watts at a voltage node, but if you are at the maximum voltage point of the standing voltage wave, then maybe only 120 Watts.

Here is the relationship of voltage and current distributed along a wire, that is a half wavelength, and this distribution repeats indefinitely, with only the current direction changing ever halfwave. Current vs. Voltage Distribution along a halfwave antenna

What prevents RF burns on coax is the insulation, but also that the return path is copper braid or aluminum foil, and you are not a better path for that voltage.

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