I am trying to derive the conversion from dBm to dB\$\mu\$V. So, please correct me If I am wrong.
\begin{align}
\textrm{dBm}
&= 10 \log_{10} \left( \frac{P}{1\textrm{mW}} \right) = 10 \log(P) + 30 \\
&= 10 \log_{10} \left( \frac{ V^2}{Z} \right) + 30\\
&= 10 \log_{10} \left( V^2 \right) – 10 \log_{10}(Z) + 30\\
&= 20 \log_{10} \left( V \right) – 10 \log_{10}(Z) + 30\\
&= 20 \log_{10} \left( \frac{V}{1\mu V} 10^{-6} \right) – 10 \log_{10}(Z) + 30\\
&= \underbrace{20 \log_{10} \left( \frac{V}{1\mu V} \right)}_{\textrm{dB}\mu V} + \underbrace{20\log_{10}\left( 10^{-6} \right)}_{-120} – 10 \log_{10}(Z) + 30\\
&= \textrm{dB}\mu V -120 – 10 \log_{10}(Z) + 30\\
&= \textrm{dB}\mu V -90 – 10 \log_{10}(Z) \\
\Leftrightarrow \\
\textrm{dB}\mu V &= \textrm{dBm} + 90 + 10 \log_{10}(Z)
\end{align}
Best Answer
dBm is short for dBmW/m^2. The measurement is milli-watts.
dBuV is short for dbuV/m^2. The mesurement is in micro-volts.
In order to convert from volts to watts you have to assume some sort of impedance.
For a free space propagation of an RF signal we assume the impecance of free space (which is 376.73 ohms).
Voltage and power are related as follows...
W = V^2 / Z
Or
V = sqrt(W * Z)
Assuming a 1mW signal and Z = the impedance of free space we get...
V = sqrt(0.001 * 376.73) = 0.61378
in db-volts this is...
dBV = 20 * log10(0.61378) = -4.2dBV
To convert dBV to dBuV we just multiply by 1 million (so add 120dB)
-4.24dBV = (-4.2 + 120)dBuV = 115.8dBuV
Therefore 1dBmW = 115.8dBuV when Z = 376.73 ohms
This matches the conversion chart you linked to.
Your derivation above comes up with the same answer if you put in 376.73 ohms for Z.
dBuV = dBm + 90 + 10*log(376.73) = dBm +90 + 25.8 = dBm + 115.8