# how does a cell phone transmitter work?

I came across a question that is really puzzeling me. I did some basic calculation for a cell phone transmitter. I based on 900Mhz band and assumed a 20Km distance to a cell tower. On top of that I assumed a 99.99% reliability which comes to almost 38 dB for the fade margin. I also assumed a +6 dBi gain for the TX and RX antennas, and reciver power of -100dBm. My calculations showed me a 45 dBm which is almost 23 watts. so I am really confused that how a cell phone with a 5 V battery can operate on these basis and actually run for a long time. am I not having correct assumptions about the hardware design or are they very well designed to perform such a comm link

• Cell phones use adaptive output power, with a maximum of 2W or so. Usually cell towers are much closer than 20km in suburban or urban areas, so average consumption is much less. I'll let you figure out where the other 10dB comes from (error correction?) as I've not looked at this stuff in ages. Apr 13, 2014 at 19:52
• @Caspian How did you actually input the "reliability" into your calculations? What's the coverage at the edge of the cell and the percent of cell's interior in your model (also which model are you using)? Apr 13, 2014 at 20:00
• To answer this we need to know what formula you used for calculating link loss but as @spehro says, 20km is a long distance. Apr 13, 2014 at 20:34
• Also, cell transmission is not necessarily continuous. GSM, which I am most familiar, transmits in bursts of around 600 µs every 4.5 ms. During this burst, the current drain on the battery can exceed 1.5 A. During the rest of the time slot, it may be only a couple hundred mA. Apr 13, 2014 at 21:01
• The GSM standard is documented, in fact I believe there's an open-source base-station project. There's plenty of RF kung-fu going on with it so calculations relating to simple radio links from your average textbook do not really apply. Apr 14, 2014 at 8:27