Power consumption of mobile phone in watts

probably about half that.

greg

it would also depend on the transmit power which changes depending on terrain.

greg

No, that means the battery capacity is 2.812 Amp-hours of capacity. At 3.7V that's 10.4 Watt-hours of energy.

To find the power consumption you need to know how much current the phone draws in Amps. You then multiply that by the 3.7V and you will know how many Watts it is using. I can tell you this though, it's going to vary over a large range from milliwatts to more than 5 Watts at times.

You could use talk time to work backwards and determine average power usage. If you have a 10 hour talk time, then the phone is dissipating an average of (10.4watt-hours/10hours) = 1.04W. This would equate to an average current of (1.04W/3.7V) = 281mA of current. We can verify this by taking current * voltage * time to come up with the energy requirement. .281A * 3.7V *

10hours = 10.397 Watt-hours the capacity of our example battery.

On a sunny day (8 Mar 2007 06:58:00 -0800) it happened snipped-for-privacy@hotmail.com wrote in :

Depends on how long you can use it without a re-charge. So it is 2.8 W / hour, means it can consume 2.8 W for 1 hour,

1.4 W for 2 hours, etc.

Also the mobile (GSM) phone transmit power depends on the signal strength, distance from the base station.

And it transmit it of course uses a lot more power then in standby.

You phone specs should tell you how long it can standby and how long it can transmit, say if the max transmit (speak time) is 2 hours, then it uses 1.4 W minim during transmit. If you can talk 4 hours it uses .7W to transmit. If it has 100 hours standby time, then it uses 28mW in standby.

Hope you can follow the math.

If it dont workatall battery empty.

I don't have any idea why it says (watts) on the label.

3.7V * 0.76 Ah = 2.8 Watt-hours.

If the battery powered the phone for one hour and then went dead, then the phone used up an average of 2.8Watts for that hour.

More typically that battery will be good for several days of operation. 4 days = 100 hours which would mean average power consumption of 0.028 watts.

Note I said average.

In real-life, the power consumption is very peaky. The phone will not be regularly transmitting unless a call is in progress. Most phones will also be in high power consumption if in idle in a fringe area or if it's constantly searching for service in a non-service area. (Many modern phones have a "power-save" option for the latter case).

It might be drinking a good fraction of a watt during a phone call if you are in a fringe area.

Tim.

Oops, let's try again. The battery capacity is as specified 760mAh. The amount of energy is 2.812Watt-hours.

an average of (2.812watt-hours/10hours)= .281W. This would equate to

an average current of (.281W/3.7V) = 76mA of current (a highly suspicious value). We can verify

requirement .076A * 3.7V * 10hours = 2.812 Watt-hours

Sorry for the stupidity.

It is a certainty that the cell phone industry uses every trick in the book to maximize the battery life of portable cell phones. The various air-interface standards (Analog, GSM, GPRS, Edge, CDMA, etc...) all go to great lengths to minimize current consumption by incorporating various power-saving features into the cell phones and into the air interface signalling being used.

For example, when "ringing" a cell phone, the base station will use a specific timing sequence which will allow the cell phone receiver to go to sleep for the majority of the time it is not actively engaged in communication. (In analog, this was the Overhead Message Paging Channel, or something to that effect.) BTW - This is also related to how the cell site "knows" which particular cell site to use to contact the subscriber, and relates to cell site capacity issues.

Essentially, the base station will send out a synchronization preamble (or its equivalent), and the phones will "wake-up" periodically, sync and listen for their ID. If no match, then back to sleep.

Note: This is a gross over-simplification of a very complex signalling scheme.

Now, for transmit: In all air-interface standards I'm aware of (for PCS and Cellular anyway), the base station controls the RF Output power of the cell phone. If the cell site is having trouble "hearing" the phone (i.e., low received signal at the cell site), it will instruct the cell phone to increase its RF output within certain limitations. If this doesn't solve the problem, the cell site will terminate the call. Similary, if the cell site has too much signal from the mobile phone, it will instruct it to turn down its transmitter power (to save battery life AND more importantly, to avoid overloading the cell site's RF receiver distribution amplifier systems - which are shared by all users using a particular cell site.)

There are many reasons signal can change at either end during a call, which I won't get into now. The point is: the phone's RF Output power may change many times a second during a typical call.

Note: Again, this is an over-simplification of a very complex signalling scheme.

To make matters more interesting, the cell site's output power may also change. This can be caused by active (non-linear) cell site transmitter combining where the more channels (or carriers) are connected (dynamically) to antennas, the fewer watts are available for individual channels or carriers on the air.

Another factor is dynamic frequency re-allocation of the cell provider, which may cause the cell site frequencies and coverage(s) to change dynamically with use. In other words, overloaded Cell Site

-"A" might "borrow" some idle bandwidth from an adjacent (available) Cell Site "B". Because of RF intermodulation and co-channel or adjacent channel interference issues, this can also affect RSSI and cell site performance -- It's a non-optimal solution to congestion, but it's better than dropping the call.(?) I don't know how wide spread the practice of dynamic allocation is, but you can reasonably expect it in conjested areas, or in areas that have infrequent, periodic usage demands - such as sports staduims, etc...

And note: This is all transparent to the average user. Except perhaps the "4-bars" or "5-bars" discussions people sometimes have. This is usually an indication of the strength of the signalling channels, NOT the voice channels. So if the Cell Provider amps up their messaging channel transmitters, this can give a false sense of performance. (I have seen this only one time, and I won't name names.) But this can be one explanation why you sometimes can't make a call even though you have 4-bars of signal. (And more likely, the reason is one of several dozen other things that can go wrong, which is too much to get into here.)

If you want to do some more research on battery-saver features of air interface signalling schemes, I would suggest starting with a much simpler one. Do a google search on "POCSAG", or "GOLAY", which were two widely used signalling schemes used with Pagers (beepers).

These will be "simpler" because those devices operated in a predominately one-way fashion.

Do you see now why you can't go by the battery nameplate?

-mpm

No, it means that the energy consumption is 2.812 watt-hours. How long it will last depends on the actual wattage used, which is the rate of using energy, using up the charge from the batteries.

To find actual power consumption, break the battery circuit and interpose an ammeter (or milliammeter, just be sure it has a high enough range, although it's a sure thing it's less than 2.8A, if the phone lasts an hour), and measure it, in real-time. A cute way to insert an ammeter into a circuit where the batteries are in spring-loaded holders is to take two little brass strips, separate them with a piece of stiff paper, and stick the little sandwich between the battery post and contact. Then put the ammeter (or a switch! :-) ) from one brass strip to the other.

Try going to

and put "power consumption of a typical mobile phone" in the search box, with or without the quotes - have an adventure! :-)

Good Luck! RIch