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In my house, I’ve got a mag mount antenna mounted on a piece of sheet metal on top of rubber and the tip of the antenna goes into my ceiling and possibly contacts or is near to home insulation. I was just wondering if this was a hazard for fire in any way or if I’m OK. The antenna's max is 75 W, but my radio generally never puts that out.

Kevin Reid AG6YO
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Roy Jacobson
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3 Answers3

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The peak voltage at the tip of the antenna is a function of the input power and the field impedance at that tip. The impedance varies based on the geometry of the antenna, but for a typical mag-mount whip it's probably not more than 5000Ω. Voltage $E$, power $P$, and resistance $R$ are related by

$$ P = {E^2 \over R} $$

$$ E = \sqrt{PR} $$

With the worst-case estimate of 5000Ω and 75W, that makes the RMS voltage at the tip of the antenna:

$$ \sqrt{75\:\mathrm W \cdot 5000\:\Omega} = 612\:\mathrm{V_{rms}}$$

If you are concerned about arcing, the peak voltage is probably more relevant:

$$ \sqrt 2 \cdot 612 \:\mathrm{V_{rms}} = 866 \:\mathrm{V_{peak}}$$

So the question is then if 866 V is a hazard. That's high enough that I personally wouldn't take any chances putting it in direct contact with anything.

Even if you never start a fire, the antenna is really designed to have air around it for proper operation, not a ceiling. I'd suggest a different antenna setup.

Phil Frost - W8II
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I think you will be safe as long as there is no metal behind the ceiling. Ideally I would try to have a gap.

darkstar2002
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  • You can try some experiments (outside, and safely) with a piece of paper or something to simulate your situation. At a power levels under 75W into 50 ohms (below microwave frequencies), I'd be shocked (no pun intended) if you could even get a tingle in your fingers much less enough heat/arc to ignite even the driest tinder. – Duston Aug 04 '17 at 21:17
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    @duston 5 W you can just feel is warm, 50 W can cause deep burns if you touch the antenna. Source : myself and several friends at various stages. – tomnexus Aug 04 '17 at 22:10
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    I can verify the 50W comment, but an insulated ceiling shouldn't try to conduct any power away like a 70% saltwater human body does. Unless there is something metal right above it. – SDsolar Aug 05 '17 at 04:04
  • Is the ceiling dry? Is there foil paper on the insulation? Are the structural members wood? Wood conducts a little. Or maybe they are steel? Even worse. Is there any electrical wiring in the ceiling? Are the answers to these questions going to be true forever? Even if the installation isn't hazardous today, it might be difficult to verify that, and it might not always be true. – Phil Frost - W8II Aug 05 '17 at 16:15
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If your antenna is a 1/4 wave vertical, the peak voltage on it will be at the very end of the antenna. So, please don't touch it while you are transmitting.

An antenna that is well-matched has an input impedance of 50 ohms (for most radios) and an output impedance of 377 ohms (this is the impedance of free space, and all an antenna really is, is a matching device from your radio's transmitter to free space).

If you are transmitting 50 watts into a perfectly efficient, perfectly matched antenna, the voltage at the tip is the sqrt(50*377) = 137 volts. You might arc this to a piece of metal if it were within a few thousandths of an inch, and it might conduct a tiny bit of current into high impedance materials.

One concern is the effect on the impedance of the antenna having dielectric materials in the near field. Having things like drywall and fiberglass right around part of the antenna might change its impedance enough to harm the radiation pattern and efficiency.

I would try to remove any materials that are near the antenna.

user2104506
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    Why is the field impedance at the tip of the antenna equal to the impedance of free space? – Phil Frost - W8II Aug 05 '17 at 14:25
  • An antenna is a matching device between a transmitter and free space. Free space has a characteristic impedance of 377 ohms. An antenna is effectively a device with an input impedance of 50 ohms (or whatever your system is) and an output impedance of 377 ohms.

    If you solve the vector Helmholz equations for the fields radiated from surface currents on a PEC, the impedance that resolves a valid solution is the impedance of the medium.

     – user2104506 Aug 06 '17 at 20:04
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    @user2104506 I've never seen a formula in this application that included the Z of free space. Do you have a reference? A related post is here. – Mike Waters Aug 06 '17 at 22:10
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    It appears that this idea of matching to 377 ohms (discussion here) came from the book The Electromagnetic Field by Albert Shadowitz. – Mike Waters Aug 06 '17 at 22:29
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    Sure, but the tip of the antenna isn't free space... – Phil Frost - W8II Aug 06 '17 at 23:40
  • Phil, if you want to pick nits about solving these problems, I could say that the tip of the antenna actually isn't there. Solving an antenna radiation problem is one half of solving an electromagnetic scattering problem, in which the PEC is replaced by the homogeneous (or inhomogeneous) medium and the surface currents resulting from the enforcement of Maxwell's boundary conditions are substituted. The entire problem is solved in free space. I suggest "Computational Methods for Electromagnetics" by A. F. Peterson if you'd like a good elementary primer on solving these sorts of problems. – user2104506 Aug 06 '17 at 23:57
  • @user2104506 Respectfully, we just don't have those books. And while none of us knows everything, Phil certainly doesn't need an elementary primer! :-) Can't you please post just a little more information here to help back up your statements? – Mike Waters Aug 07 '17 at 01:07
  • Asked and answered, but if there's part of my explanation that you would like me to clarify, I am happy to if you tell me what part of it needs clarifying. We are looking at the interface between the antenna (presumed PEC) and free space. The fields at the surface of the wire are determined by the relevant boundary conditions presuming the wire is PEC. But, in any case, when solving these problems for example using a moment method, all of the constitutive parameters used are those of free space (permitivity and permeability), which are the parameters that define the impedance of the medium. – user2104506 Aug 08 '17 at 13:35
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    Assuming a PEC and using boundary conditions, please describe the voltage at various points along the length of the described antenna. – Glenn W9IQ Aug 08 '17 at 23:39
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    Glenn, again, asked and answered. For a 1/4 wave vertical the voltage is at its peak at the end of the radiating element and the current is at its peak at the feed point. Not enough room here to go through all of the math to solve a moment-method radiation problem. – user2104506 Aug 10 '17 at 19:19
  • Nor does it appear anybody is taking into account the insulating properties of the air or the ceiling itself. PLUS, and this is the biggee, do you really think you would want to run full legal power into an antenna in the same room as you? There are legal exposure limits in broadcasting, and I would suggest they should be taken into account with this problem, too. – SDsolar Aug 19 '17 at 07:01