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power transfer using a dipole

I need to calculate the idealized (theoretical) maximum transfer of power into a dipole antenna with various loads.

I thought this would be a simple physics book lookup but all of the answer were more complicated that that and I didn't want to run down a rabbit hole of polarization, impedance matching etc. I'm looking for the textbook optimal/ theoretical best case that you never achieve in real life.

here is the situation: I have a 1 ohm resistor between two wires that can be 1" to 12" long.

I measure an electric field caused by a cell phone or wifi (2.4GHz) and I measure the field strength at 100V/m. If the wires were cut to tune the dipole antenna for maximum sensitivity, how much power could be delivered to the 1 ohm resistor. No matching network is there. just a plain ole 1 ohm resistor. wire lengths should be ~ 1.2" each for this frequency.

would it be the same calculation at different frequencies that tune to the 12" length. I assume this would be 1/10th the freq. so 240MHz.

I also assume that as you move away from the tuned frequency the less energy will be coupled for a given wire length. its been a while. :)

Reply to
mike daniels
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** What need is that?

Exam question ? Maths puzzle ?

** So it's a math puzzle.
** A simple 1/2 wave di-pole at resonance matches to 75 ohms. So using 1 ohm is nuts.

Using a longer ( non resonant) antenna may increase the current flow, hence better with 1 ohm.

Reckon you need some good antenna software for this one.

..... Phil

Reply to
Phil Allison

Razzie accepted Phil. :) This is not a purely academic exercise but a little myth busting. Some in my office believe if a resistor is scratched out with leads between 1" and 12" long and held close to a cellphone or wifi that generates a electric field in the 100V/m range, it can flow 200mA RMS of current through the resistor had heat it up with 40mW of power dissipation.

Another thought was to strap an RTD to the resistor and measure the temperature rise in the resistor since we don't have a way to measure 1+Ghz current. I can benchmark that against the temperature of 200mA DC through the resistor as a reference.

This seems highly unlikely to my mind that that much current could be induced in a 24" max wire with a 1 ohm resistor in the middle but thought I would confer with the folks that deal with antennas and RF on a regular basis. They would know the highly simplified equations

I was hoping there was some simple math to calculate the idealized transfer of energy from a electric field into a dipole antenna into any load. I understand there is a dramatic impedance mismatch for most antennas but i'm looking for the best case.

Reply to
mike daniels

On Tuesday, 3 August 2021 at 15:46:56 UTC+1, snipped-for-privacy@gmail.com wrote:

There are online calculators which will estimate the power that would be received by a well matched receiving antenna from a well matched transmitting antenna at a defined distance and frequency. Here is an example I picked at random:

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that the result is only valid in the "far field" where the separation is more than a few wavelengths (which means around 1m for a cellphone) and where the polarisations match and in the case of your receiving dipole it is at the resonant length. You can assume a gain of maybe -3dB for the cellphone antenna if it is inside the case and +2dB for a resonant dipole in the direction at right angles to the wires of the dipole. Once you have worked out the power that a perfectly matched antenna would receive, you can separately work out the fraction of the power that would be reflected due to the mismatch between the optimum antenna impedance of around 50 to 100 ohms and your 1 ohm load resistor. Also, don't forget that cellphones only transmit for a small fraction of the time and use dynamic power control, so the actual average transmitted power is a small fraction of that in the specifications. If you enter too short a distance the result will suggest that you can receive more power than you have transmitted, proving the unsuitability of such calculators for near-field calculations. Have fun! You are not going to burn your finger on that resistor.

John

Reply to
John Walliker

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