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I've seen people refer to avoiding touching an antenna or bare transmission line that might be transmitting. What voltage potentials and currents are commonly found on typical ham radio antennas? Do both the voltage and current increase with increased transmitter power, or mainly one more than the other?

What are the general safety measures one should take concerning antennas, specifically from the power going into the antenna purposefully? Antenna safety around powerlines and other power sources is a different question.

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    $\begingroup$ Maybe a duplicate of ham.stackexchange.com/questions/558/… ? $\endgroup$ – Dan KD2EE Nov 26 '13 at 12:25
  • $\begingroup$ @DanKD2EE I don't think it's necessarily a duplicate, but they are related. The question you linked appears to be about indirect exposure (being exposed to radiated RF) whereas this question is about direct conduction (conductor to skin) of the voltages and currents on an antenna that is connected to a live transmitter. They are both very much valid IMO. $\endgroup$ – a CVn Nov 26 '13 at 19:11
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Most antennas have a standing wave along there length and are therefore effective impedance transformers. The feed impedance of a dipole might be of the order of 73Ω, at its ends, the impedance will be at least 2kΩ, if not higher.

Solving for voltage at the antenna ends, we will have:

$P=\frac{V^2}{R}\Rightarrow V_{rms}=\sqrt{P\cdot R}$

Peak voltage is indeed $\sqrt{2}\cdot V_{rms}$, but the voltages at the ends of a dipole antenna are balanced with respect to ground so: $V_{peak}=\frac{\sqrt{2}}{2}V_{rms}$, only half that value.

Assuming a transmitter power of 1kW: $V_{peak}=\frac{\sqrt{2}}{2}\sqrt{P\cdot R}=\sqrt{\frac{P\cdot R}{2}}=\sqrt{\frac{10^3 \cdot \not 2 \cdot 10^3}{\not 2}}=10^\frac{6}{2}=10^3=1kV$

If the antenna is loaded with a coil, the impedance will be transformed up to an even higher value. This is how a Tesla coil works. The resulting corona effect might be quite dramatic as shown in the picture below (1kW on 80m in short W4JRW dual-band dipole @ HB9DWU).

1kW on 80m

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Touching a "live" (transmitting) antenna could impart a serious RF burn if the power level was high enough. This is different from the classic "electric shock" obtained by putting a finger into a live wall socket. Radio frequencies heat tissue (non-ionizing radiation), proximity and contact with high RF fields will cause a RF burn.

There is a risk of electrocution if the antenna is in contact with an overhead power line, although a well placed antenna should NOT be near any such hazards.

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  • $\begingroup$ So why you don't get a shock while standing near the antenna? Also suppose you create a 60hz antenna, can you power wirelessly an electric device by it? $\endgroup$ – Ronen Festinger Jun 8 '16 at 22:45
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RF burns aside (these often occur without physical contact), the voltage on the antenna can reach very high levels. There is a good discussion of this at http://forums.qrz.com/showthread.php?243998-Voltage-at-antenna.

Paraphrasing the linked discussion, Ohm's law applies, so if you're dumping 50W into a 35 ohm load:

$P = I^2R$, so $I = \sqrt{\frac{50}{35}} = 1.195$

$V = IR = 1.195 \cdot 35 = 41.83 V_{RMS} \cdot \sqrt{2} = 59.16 V_{PEAK}$

Toward the tip of the antenna, the impedence changes, so the voltage can be orders of magnitude higher (and the current lower).

(Yes, I know V should be E, but old habits are hard to break. I also know that I'm using the RMS for a sine wave, but it's a reasonable approximation.)

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    $\begingroup$ Note, however, that this assumes a purely resistive load (i.e., that the antenna is resonant at the transmitting frequency) and a perfect impedance match between the transmission line and the antenna. In most real-world situations there is a reactive component in the load and the voltage can be much higher. For example, the ARRL Handbook has a chart (Table 20.1) for a 100 ft. center-fed dipole at 1500 watts. The calculation above would give about 300 volts; at 1.8MHz, with an RG-213 coaxial cable, the actual voltage is around 1500; with a 450-ohm ladder line it's nearly 11,000. $\endgroup$ – Pete NU9W Nov 27 '13 at 18:57
  • $\begingroup$ @PeteNU9W, very good point. The calculations above are a drastic simplification, and caution should always be used when simplified numbers are applied to the real world. $\endgroup$ – KD8TGR Nov 27 '13 at 20:00

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