0
$\begingroup$

According to this sales listing for a 2.4Ghz antenna, it can handle up to 20 watts of power but has an impedance of 50 ohms.

That means that (at 20 watts), the input would need to be at more than 50 volts to keep the heat generated to be less than 8 watts

20 watts / 50 volts = 0.4 amps

0.4 amps ^ 2 * 50 ohms = 8 watts of heat

50 volts is enough to shock someone (and that's probably a low estimate)

Am I missing something here?

$\endgroup$
3
$\begingroup$

The answer is "it depends".

It depends on the Antenna design, but lets look at an example:

A typical (license free / home / consumer) installation of WiFi is 100mW (not a 1000mW as your post indicates)

A typical antenna has a 50 Ohm impedance, and lets take an example:

For a 1/4 wave vertical antenna the voltage distribution has a voltage minimum at feed point, and a voltage maximum at the end. And lets take that this antenna has a perfect Resistance part of the complex Impedance = 50 + 0j Ohm (in practice this may vary a lot)

Ohm's Law will show you E = SQRT(P*R) = SQRT(0.1*50) = 2.23 V (at the antenna tip)

This is only true for the standard quarter-wave vertical, other antenna designs may have different voltage distribution.

An antenna may have a rating of "20 W" but that only indicates the maximum power input possible with that antenna before "things go bad".

You will have to check local legislation on what the maximum power is you may run, but as indicated above in many regions of the world this is 100mW (YMMV)

Often this is expressed in ERP, so it will take antenna gain into account. Example: If your antenna has a unity gain of 3dB, then you need to ensure that the transmitter (the WiFi Router) is throttled back to 50mW in order to be compliant with a 100mW-ERP legislation. This assumes no cable loss, so the actual calculation will be more complex taking cable loss / connector loss into account.

$\endgroup$
1
$\begingroup$

This doesn't actually answer your question per se, but I think you have an important misconception: An antenna is not a resistor.

Of course you know it's not, but in particular, an antenna having a 50 Ω input impedance does not dissipate as much heat as a 50 Ω resistor would for the same input. The majority of the energy is radiated, not converted to heat — that's what makes it an antenna!

An antenna having some particular resistive input impedance means that the electrical characteristics at its input (that is, how a connected transmitter will be affected by it) are equivalent to a resistor of the same value, not that it behaves the same way otherwise.

Suppose you indeed put 20 W into your 50 Ω antenna. The only thing the 50 Ω number tells you is the ratio between current and voltage on the feed line — if we work it out then it is 32 V and 632 mA.

That same ratio of voltage and current exists at one place on the antenna: the feed point, where the feed line connects to the antenna element(s). Everywhere else, it will be different, and in no case is it primarily due to the resistance of the metal in the antenna.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.