What is the "minimum" transmit power for uplink cellular transmission
Some GSM systems have tiny on-the-wall cell-sites, thus the "distance" might be just a meter (about 10 wavelengths).

The pathloss will be 22dB + 10db*log10( [distance/wavelength]^2 )

or 22 + 20 = 42dB pathloss. This comes from the geometry of energy spreading out, with spherical behavior around the antenna, and 1/range^2 dropoff in energy density.

Now what receiver signal energy needed? Lets work through a link-budget:

-174dBm/rootHertz thermal noise floor

+60 dB for 1MHz bandwidth (about right for GSM)

+20 dB for SignalNoiseRatio, matching from Antenna-LowNoiseAmplifier, and any upfront diplexor losses



-174 + 60 + 20 = -174 + 80 = -94 dBm (about 5 microVolts RMS, across 50 ohms)

What energy from TX? Simply add the receiver minimum energy (-94) to the PathLoss (42), -94 + 42 = -52dBm.

------    -52 dBm     This is the answer you wanted: much less than 0dBm ------

Given 0dBm is 0.223 volts RMS across 50 ohms, the level of -52 dBm is 12 db below 40 dB below 0 dBm, thus

[0.223 / (100, from -40dB)] / (4, from -12dB)

or 0.223 / 400 = 0.5 milliVolts RMS