Electronic – Conversion from dBm to Volt/ meter

detectorelectromagneticfield-strengthRFvoltage measurement

I have no experience with electronics but I have a question.
Actually I'm using a circuit for a project that is able to measure RF power at its input in dBm; the circuit provides the integration of a specific component, which is the IC LT5534.
I'm interested in the measurement of the signal strength in Volt/meter, since the law specifies legal limits in units of Volts/meter. I have found (navigating on the web) that in order to convert dBm to Volt/meter, I need to know other parameters ( http://www.qsl.net/pa2ohh/jsvpm.htm ) .
I have also found an RF detector that is able to display in Volt/meter: https://www.amazon.com/Cornet-ED78S-Meter-ElectroMagnetic-Detector/dp/B00P67QLA0/ref=sr_1_3?ie=UTF8&qid=1481739673&sr=8-3&keywords=electrosmog .

How does it work? Is it really possible to convert voltage output (in dBm) to Volt/meter? If the answer is yes, what's the formula?

Best Answer

Please fasten your seatbelts - our flight today will fly over a bit of the math to get from power (dBm) to field strength (V/m).

The most pragmatic way to think about the "strength" of any RF signal is in terms of how much power it can deliver to the terminals of a practical receiving antenna.

Ultimately the laws of physics dictate that for a given receiving antenna, at a given frequency, there is a simple linear relationship between the square root of the power at the receive antenna terminals and the strength (in Volts/meter) of the RF field:

\$Field\ Strength\ = k\cdot\sqrt {P_{rx}}\ \ \ \ \cdots (1)\$

You can think of the factor \$k\$ as the "conversion factor" of that antenna at that specific frequency.

That's it - once you know your antenna conversion factor \$k\$ and the received power (in Watt), you simply calculate the field strength using the above formula.

So, to answer your questions: 1. How does it work? Primarily due to the inverse square law of radiators and aperture. 2. Yes, it's possible. 3. Once you know k for your antenna, use Equation (1) above to calculate RF field strength - see example below.

Example:

The \$k\$ factor for an ideal half-wave dipole, which has a gain of 1.643 (2.14dB) can be found in literature - \$k\$ is essentially the square root of (\$120\pi\ \div \$Antenna Effective Area).

\$k = \sqrt{120\pi} \cdot \sqrt{\frac{4\pi }{1.643\cdot\lambda^2}} \approx\frac{53.7}{\lambda}\$

Let's say we are receiving 0dBm (.001W) at the terminals of our half-wave dipole at 100 MHz. (\$\lambda\$=3m) Using Eq. (1) we see

Field Strength = \$\frac{53.7}{\lambda} \cdot \sqrt{.001\ }\$ = 0.566 V/m

TLDR: Convert dBm to Watt, get square root, multiply by k.