# Conversion from dBm to Volt/ meter

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?

• That thing on Amazon you linked to is advertised as an electrosmog and paranormal activity detector. What it really is, is a device used to separate the gullible from their money. Do not try to use it as a guide to what you can do. Do not expect it to deliver any kind of accurate measurements.
– JRE
Dec 14 '16 at 20:30
• Thought this sounded familiar. How are you doing with building the circuit for the LT5534? Did you end up buying the development kit, or did you manage to make and assemble your own board? I got my third version of my RF level detector board running this last weekend. Still not as good as I wanted it - still getting too much noise on the level detector output. Might be coming from the power to the detector chip,or power to the LNB, or the LNB itself.
– JRE
Dec 14 '16 at 20:35
• The help of an expert has proved to be fundamental. @JRE I suggest you to buy the development kit for the LT5534; soldering all the components manually requires a lot of handiness. Sometimes it's more convenient to spend a little bit more rather than replicating the board; LT5534s are really small. My version of RF level detector board seems almost perfect Dec 15 '16 at 17:15
• Soldering is not the problem. That is very easy. If I were just interested in measuring ambient RF, I'd have stopped at version 2. I am interested in reliable readings of RF level variations of 0.01 dBm. That means the signal out of the detector has to have less than 50 microvolts ( peak to peak) of noise on it. Right now I am at 100 microvolts of noise. The very first time I used the MAX2015, I got clean measurements repeatable in the 0.1 dBm range. That was on an adapter board I installed inside a satellite detector, not a purpose built PCB like I am using now.
– JRE
Dec 15 '16 at 20:25
• The board I have built includes a controller for the satellite LNB, the MAX2015 RF level detector, an LTC2440 analog digital converter, an LT2050 for filtering and buffering the signal from the MAX2015, a precision reference for the ADC, and various regulators to supply power to the different chips. It also drives a pair of servos to aim the satellite dish and scan it across the target. So, the development kit for the MAX2015 (or the LT5534) won't save me much work in developing my gadget. Besides, the hardware is the fun part.
– JRE
Dec 15 '16 at 20:32

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.

• I think you misunderstood the question. The OP wants to know the relationship between power reported by the circuit and the electric field strength (V/m) in the space surrounding the antenna -- not the voltage at the antenna terminals. Dec 14 '16 at 22:38
• If I am not mistaken, the correct formula for factor conversion 'k' is this: $k = \sqrt{\frac{120\pi}{50}} \cdot \sqrt{\frac{4\pi}{Gr\cdot\lambda^2 }}$ Link (tmgtestequipment.com.au/products/media/bkd0/ref/…). @neonzeon, correct? Dec 20 '16 at 16:35
• @L.Bas my k is not exactly Antenna Factor but you are correct that that $A_{eff}$ should be under the line! Thank you - I edited my response accordingly. Dec 22 '16 at 22:25
• @L.Bas In practice if you use a 73$\Omega$ dipole to deliver power to a 50$\Omega$ coax you lose a small amount of power due to the mismatch - around 0.155 dB. Also account for power loss in the coax from the antenna to your power meter. Dec 22 '16 at 23:06

The transducer that converts an electric field into an electric signal is called an "antenna". Therefore, you would need to know the exact transfer characteristics of the specific antenna in question, at all frequencies of interest, in order to do that conversion.

I'm using a circuit for a project that is able to measure RF power at its input in dBm

If the antenna is receiving a proper electromagnetic wave then, the power per square metre received is E x H i.e. electric field strength in volts per metre x magnetic field strength in amps per metre.

And, for a an EM wave the ratio of E to H is approximately 377 ohms (impedance of free space): -

But, and this is a big "but" you probably don't know the effective aperture of your receiving antenna (measured in square metres usually). You have to find this or you cannot know what the power per square metre is. If you only know power received (and your question says you do) then you could estimate the effective aperture of your antenna. Here are a few examples: -

So take your dBm figure, convert to a real power and using the effective area of your antenna, convert the power to watts per square metre. This then equals E x H and, H is E/377 so power per square metre = E$^2$/377.

• that's a nice summary of effective area for various antennas - can you provide the reference/author? Dec 16 '16 at 17:57
• @neonzeon I found it on google images - that's all I can recall. Dec 16 '16 at 18:31
• @Andy_aka found it on slideplayer uploaded by Joanna Randall but could not figure out who the uploader is. It says "Lecture 17" so would guess it's an Emag course at an engineering school. Dec 16 '16 at 20:40