Can anyone help me in finding a good reference for minimum transmits power for uplink cellular transmission?

is 0 dBm a good enough power for very low-power transmission in an indoor environment? Will it achieve high enough RSRP values at the receiver (note that in indoor the receiver is very close to the transmitter)

  • \$\begingroup\$ It depends on ambient noise on RSSI and path reflections (Ricean loss) as well as RD sensitivity. Sites have a calculator if you know each of many more parameters \$\endgroup\$ Jun 18, 2018 at 2:45

1 Answer 1


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

  • \$\begingroup\$ :(... Can you make it a little simple please :( \$\endgroup\$
    – SJa
    Jun 18, 2018 at 2:35
  • \$\begingroup\$ The conclusion is yes, if you're close enough, which you easily could be to a micro cell. Presumably that's what's in things like subway stations that used to not have signal, but now often do. \$\endgroup\$ Jun 18, 2018 at 2:57
  • \$\begingroup\$ @Kashan You asked if 0dBm was adequate. My answer showed a much smaller power level, -52dBm, was adequate. And I provided the voltage across an RF-standard resistor, 0.5milliVolts RMS, would be the -52dBm. If you LIKE this answer, then accept it. \$\endgroup\$ Jun 18, 2018 at 5:12
  • \$\begingroup\$ Can you please give me any reference related to this, as I have to quote it in my paper. \$\endgroup\$
    – SJa
    Jun 18, 2018 at 6:54
  • \$\begingroup\$ @Kashan This is link-budget. Google that. And the 0.5 milliVolts is power across a 50_ohm resistor, for -52dBm, or -82dBW. The formula is Power = Voltage_rms^2 / Resistance. \$\endgroup\$ Jun 18, 2018 at 15:56

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