# Help With Cell Signals

1. Oct 6, 2012

### apmsj

I'm trying to learn more about how cell phones operate. Right now I don't understand the power involved in the signal between the cell site and mobile phone. I've read that the average cell phone has a signal strength that ranges in between -80dBm and -120dBm. I've also read that cell phones output between 27 to 33 dBm. What exactly does this mean? Is this related to how much power the cell phone needs to use to effectively communicate with the cell site?
Why is -80dBm considered a better signal than -120dBm? Is this the ampltiude of the signal being transmitted by the cell site? How is this calculated? Thanks - Adam

2. Oct 6, 2012

### vk6kro

The first figures refer to the power available at the cell phone's antenna terminals.

The second figures are the audio power at the speaker in the cell phone. This depends mainly on the gain control setting and the maximum audio power of the device. This is interesting, but not really related to the received signal strength.

There is an article in Wikipedia here:
http://en.wikipedia.org/wiki/DBm
which explains the way these are calculated.

dBm refers to the ratio of the power being quoted and 1 mW. 1 mW is 0.001 watts.

For example a 5 watt signal would be calculated like this:
power (dBm) = 10 log ( 5 watts / 0.001 watts) = 36.98 dBm

Here is a table of conversions.
http://www.minicircuits.com/pages/pdfs/dg03-110.pdf

It is very common to quote a figure in microvolts for the signal reaching a receiver or the minimum signal a receiver can receive.

3. Oct 7, 2012

### apmsj

So when the cell phone reports that it has -90dbm, it means that

P = 10^(1/10(-90-30))= 10^-12 W

is available at the antennae? I thought that negative numbers in db represented a loss, and positive numbers represented a gain. So when we calculate dbm is the reason why we are starting the ratio with 1mW, because that is the amount of power initially transmitted by the cell site? And then we put the amount available at the cell phone in the numerator to create a ratio of power received to power transmitted? So a 0dbm signal would be theoretically perfect?

4. Oct 7, 2012

### Enthalpy

The -80dBm and -120dBm are a power received at the antenna. The 0dBm reference being 1mW, -90dBm means 1pW, yes.

+27dBm and +33dBm is a transmitting power sent to the antenna, corresponding to 0.5W and 2W.

The strength of the received signal is very difficult to predict on Earth. Models exist, they're complicated and unreliable. To a satellite, you can use the free-space formulas, but not for terrestrial links, because the ground attenuates a LOT even if flat and in direct sight. Buildings would make it more complicated.

5. Oct 7, 2012

### vk6kro

Power levels given in dBm are fixed levels, not ratios.

So, relative to 1 milliwatt, a 37 dBm signal would always mean a power of 5 watts. Unlike the dB (decibel) which is always a ratio of two powers.

The origin of dBm was in telephone systems, but it was convenient because a 3 dBm signal could be turned into a 6 dBm signal with a 3 dB amplifier. Easy to calculate. Now, a similar calculation can be used for RF signals.
An antenna with 3dB more gain will produce a 3 dB increase in the dBm level.

1 milliwatt was just an arbitrary figure used as a reference. Originally, it meant 1 milliwatt in 600 ohms, but in RF context it normally means power in 50 ohms. So, in 50 ohms, it is a different voltage compared with what it was in 600 ohms. But the power is the same.

6. Oct 8, 2012

### apmsj

Ok I think I understand now, thank you!

7. Oct 8, 2012

### marcusl

No, I think this is the cell phone's RF transmit power. 1W max is a typical number, but power is always turned down to the minimum required to maintain an acceptable link to the base station. Using minimum power maximizes battery life and also allows a base station to accommodate the largest number of simultaneous mobile users.