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I'm trying to work out the amount of watts my mobile phone uses and thus how much power it needs to charge back to full power. My handset is the Nokia 820 Lumia and the specs are as follows: Battery 3.7v, 1650mAh.

I have calculated the watts by using the following: (1650mAh x 3.7)/(8 x 1000) = 0.76 with 8 being the talk time in hours when on 3G. However, I have become confused when reading the specs of the handset on CNET where it states the following: Power Consumption Operational: 55.0W Standby/Idle: 3.0W

Can anyone explain what the Power Consumption Operational is? This is what is confusing me. Does this mean that the handset is using 55W when fully operational and if so, how much wattage is needed to charge it (ie if I'm using the phone & it's charging).

Thanks, Ads

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Have you got a link to where it says its 55w? – Dean Jan 9 '13 at 1:41
Apologies, the information I was looking at on CNET was incorrect - it was talking about some form of LCD monitor made by Nokia. Thanks. – Ads Jan 9 '13 at 9:05
The 55w figure is definitely wrong. Go hold onto a 60w light bulb and power it up and see how long you want to hold it - mobiles heat up, but not that much, and probably well less than a watt gets turned into RF. It's possible the screen might emit a few watts as light, but then so does the lightbulb... – Chris Stratton Jan 22 '15 at 4:38
up vote 3 down vote accepted

Your post deals with several different aspects, which I'll clarify:

1) The power necessary to charge the battery can be almost anything. You can charge a battery with 1W or with 5W. The voltage must always be the same. If you use less watts, the resulting lower current means less charge and therefore longer chargning time. There is a maximum current at which a battery can be charged. Ususally this is 1C, where the C is the capacity of the battery which is 1650mAh. So, battery max current is 1650mA. In reality, it can be C/2 or other values depending on the battery chemistry. Chargers must remain below this

2) CNET must be wrong about 55W. At 3.7V you're talking about 14.86A which is very high and a battery of 1650mAh would likely get destroyed. I've worked with Cell Phones and although the peaks are high, they are never this high. Given a 1650mAh battery, you can draw 1650mA in one hour, or double in half an hour. In reality, the rate of discharge ends up changing the nominal capacity, so this isn't linear but an approximation.

As you calculated, the wattage is about 0.76W at 3G.

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Thank you, this has answered my question. Apologies on the 55W, CNET was referring to an LCD monitor made by Nokia (with a very similar name to the Lumia 820). Thanks again. – Ads Jan 9 '13 at 9:07

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