Converting dBm to v/m?

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**Tom Blahovici**:

Hello,

Is it possible to convert a signal as displayed on a panadapter to V/m? In other words if I have a signal of 20 dBm what would the signal strength be in V/m?

Thanks

**Donald L. Collinson**:

Tom,

Congrats on the new license, and welcome to the hobby. The answer to your question is no, you cannot convert directly from one to the other because you're mixing units of measure.

dBm is a measure of power, expressed as dB with respect to a milliwatt. V/m is a measure of field strength, and can only be converted to and from a power density expressed as W/m^2 or dBm/m^2. That conversion is fairly straightforward.

Exactly what is it you're trying to do? If we knew that, we might be able to provide more help.

73,

Don, K2DC

**DAVE CUTHBERT**:

Yes you can do this conversion. You need to know the "Antenna Factor" or AF or your antenna. A rule-of-thumb is to say a 30 MHz dipole has an AF of 0 dB. A 60 MHz dipole an AF of 6 dB, a 15 MHz dipole an AF of -6 dB.

Dipole AF = 20LOG(F/30)

The AF tells us the loaded (50 ohm) terminal voltage of a dipole immersed in an E-field. The 30 MHz dipole in an E-field of 1 V/m will output 1 volt. The 60 MHz dipole 1/2 volt and the 15 MHz dipole 2 volts.

Now convert your 20 dBm to dbuV. In a 50 ohms system add 107 dB and you have 127 dBuV. Add the AF. If you AF is -19 dB (an 80 meter dipole at 3.5 MHz) the E-field is 127 - 19 = 108 dBuV.

Converting 108 dBuV to volts we get 0.25 volts.

**Peter Chadwick**:

But bear in mind that the uncertainty is quite large. Ground reflections can change things by 6dB or more. For qualification laboratories, the measurement of radiated power or field strength is usually required to be to better than plus or minus 6 dB to a 95% certainty. And that is on a known calibrated test site or in an anechoic chamber (where it's usually much less than 6 dB, but often plus or minus 2 dB)

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