In a cellular connection, How much is power density (in W/m2 or mW/cm2) of the RF energy from the phone to the tower? If not possible in above units then can I know in dbm ? My phone has:

Head SAR: 0.75 W/kg at 10g tissue

Body SAR : unknown

Connection : LTE

Is it true that the power emitted increases when we get low reception?


closed as too broad by Chris Stratton, winny, pipe, Dmitry Grigoryev, R Drast Sep 4 '18 at 10:59

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    \$\begingroup\$ Please be careful with your edits. You have deleted some of your original question so you make the answers previous to your edit look irrelevant. This doesn't do any favours for the folks who gave of their time answering your question. Use 2 x <Enter> for paragraph breaks. \$\endgroup\$ – Transistor Sep 4 '18 at 17:36
  • \$\begingroup\$ ObsessionWithElectricity - Hi, I've reversed your edit to add "[Answer found]" to the title. That is not the procedure here, unlike some forums. If one of the answers given does answer your question, you accept it (which adds the "green tick" to that answer) to indicate that your question has been answered. Please read the tour and help center for more information about how the site works. Thanks. \$\endgroup\$ – SamGibson Sep 6 '18 at 4:59

Yes the phone also transmits back to the tower.

In GSM the phone transmitter maximum power is 33 dBm. In LTE the phone transmitter maximum power is 23 dBm. The transmitter power is smaller, when the phone is closer to the base station, as it smaller amount of power is sufficient for reliable communication.

EDIT: Below is a simplification, but I believe it matches your current knowledge on the subject.

Imagine you had an isotropical antenna (which actually is physically impossible, but bear with me). An isotropical antenna transmits the RF energy spherically. Assume you are transmitting 1 W of power. The 1 W power now divides evenly to the surface area of the sphere. However, the surface are of the sphere depends on the distance.

Assume we are 10 meters from the transmitting isotropical antenna. The surface area of the sphere will be: \begin{equation} \ A = 4*pi*r^2 = 1257m^2, \end{equation}

The transmitted power is divided evenly to the sphere area. Thus, the power density is \begin{equation} \ Power density = 1 W / 1257 m^2 = 795 uW/m^2, \end{equation}

If we are a hundred meters away, the area of the sphere is 125714 m^2. So the power density is

\begin{equation} \ Power density = 1 W / 125714 m^2 = 7.95 uW/m^2, \end{equation}

So you see, the further you are from the transmitting antenna, the smaller the power density (or field strength).

EDIT 2: Yes, LTE (and GSM) transmits with a greater power when the connection gets worse.

  • \$\begingroup\$ How much is it for LTE? And what does the 2 W really means? I heard the unit of power density of RF waves was in W/m2 or mW/cm2. How much will it be in W/m2? \$\endgroup\$ – ObsessionWithElectricity Sep 3 '18 at 3:48
  • \$\begingroup\$ @ObsessionWithElectricity use this table to do some conversions: rapidtables.com/electric/dBm.html -23 dBm = 0.2 Watt this does not tell you about field strength, just the amount of power transmitted. \$\endgroup\$ – Bimpelrekkie Sep 3 '18 at 6:30

Does the phones also transmit RF waves back to the tower

Of course it does, how else would you otherwise be able to talk to someone over the phone?

Cell phones transmit at a power of up to a couple of Watt.

Basestations can transmit a bit more power but still only up to a few Watt per antenna.

The amount of power received in a phone or basestation varies a lot depending on distance and other circumstances.

The lowest usable level is generally about -60 dBm (just my guess, 30 dB above a -90 dBm sensitivity level) which is about 1 uW or 0.000001 W. Yeah, that's very little power! It's a miracle that it works at all but it does.

This isn't much different for GSM, UMTS or LTE as the physical limits of what can be done don't change.

  • \$\begingroup\$ The sensitivity for GSM is roughly between -110 dBm ... - 100 dBm, and thats for mobile devices. I assume its better for base stations. \$\endgroup\$ – user94729 Sep 3 '18 at 7:35
  • \$\begingroup\$ It’s a miracle? That’s because you didn’t see how much power a satellite signal is on the earth’s surface \$\endgroup\$ – PDuarte Sep 4 '18 at 17:50

The radios, whether in the handset (smartphone) or in the basestation, must obey the same laws of physics.

Assume thermal noise floor is 4 e-21 watts per Hertz, with bandwidth of 1MHz. The noise power increases linearly with bandwidth, thus noise power is 4 e-21 Watts * 1MHz = 4 e-15 watts.

Add a factor of 100 for multipathing, for PI-network matching losses, for noise figure in the very first transistor of the Receiver, etc. We have 4e -15 * 100, or

4e-13 watts.

In systems, the antenna (assume a vertical whip quarterwave, or dipole halfwave) collects energy from approximately one square wavelength.

At 300MHz, with wavelength of 1 meter, the needed energy density will be

4e-13 watts/square meter.

Now it gets interesting. At 3,000MHz, the radio physics has not changed, thus the same 4e-13 watts must be collected, but the simple antennas will be 10X smaller, the collection area will be 100X smaller, thus the energy density

must be 100X higher, or 4e-11 watts/square meter.

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    \$\begingroup\$ So how does this in any way answer the question? \$\endgroup\$ – Bimpelrekkie Sep 3 '18 at 6:20
  • \$\begingroup\$ The radios are symmetric. The math shows the needed power density. \$\endgroup\$ – analogsystemsrf Sep 3 '18 at 21:53
  • \$\begingroup\$ If you say that phones emit 2 W then phone battery should last very short. Why? Let's say a GSM phone has 1 Ah battery at 3.7 V then it means it only has 3.7 W of power. If only the GSM uses the power that means it should last less than 2 hrs. But that is not the case. \$\endgroup\$ – ObsessionWithElectricity Sep 4 '18 at 14:52
  • \$\begingroup\$ GSM phone batteries last for more than 3-4 days. Even on call they last 6-8 hours \$\endgroup\$ – ObsessionWithElectricity Sep 4 '18 at 14:55
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    \$\begingroup\$ I like your visceral understanding and rules of thumb, rather than running straight to equations. I fear you've still pitched it at too high a level for this question though. \$\endgroup\$ – tomnexus Sep 5 '18 at 19:09

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