# Calculate output voltage and current of a boost converter

If I have two AA rechargeable battery (1.2 V, 2000 mAh) connected in series, and I then connect it to a boost converter (DC to DC converter) so that the output voltage is 5V, what is the maximum current I can get? Can you please show how to calculate this?

Just to give you a bit more info, I have one of these so called "mobile boosters" which can charge your phone from 2 AA batteries. At the back, there is a small label saying that the maximum output is 5 V @ 500 mA . However when I plug it to my Blackberry, I turn on the engineering screen, I can see that the charging current is 1250 mA which is what I normally get when I use the wall charger. When I plug it to a computer, the charging current become 500 mA which makes sense as USB2 port can supply up to 500 mA. What I don't get is why does it say 1250 mA when the source can only supply 500 mA. So is the Blackberry gives out false information or the mobile booster actually deliver more than 500 mA?

Please try to keep the answer simple, don't get too technical, I'm not that great with physics :D

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## migrated from physics.stackexchange.comAug 23 '12 at 16:53

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There's the Law. The Laws of Thermodynamics, to be precise. One of the things following from them is that you can't create energy out of nothing. So you'll have to do it with the energy stored in the batteries. Energy is power x time, and for the batteries that will be

$U = 2 \times 1.2 V \times 2000 mAh = 4800 mWh$

So if your switcher has a 100 % efficiency you should be able to get

$\dfrac{4800 mWh}{5 V} = 960 mAh$

from it. Forget 100 % efficiency, 85 % will be nice. So that's 15 % less, or 816 mAh. It's up to you how to use that. If your device uses 10 mA you can power it for 81 hours, that's more than 3 days continuously. If it needs 100 mA the batteries will be gone in 8 hours.

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I would say your "maximum mA", i.e., maximum current, depends on the resistance of your load (although we can assume it is zero for maximum current), internal resistance of the batteries, and some characteristics of the converter (efficiency, internal resistance, maybe something else). So it looks like there is not enough data to answer your question.

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Assuming your DC-DC convertor is 100% efficient, and assuming the current you're drawing is small, the current on the 5V side will simply be 2.4/5 times the current on the battery side.

If the current from the battery is $I_{bat}$ then the power being generated by the battery is just voltage times current or $2.4I_{bat}$. Similarly if the current from the DC convertor $I_{conv}$ the power is $5I_{conv}$. If the convertor is 100% efficient no power is lost so the two powers must be equal:

$$5I_{conv} = 2.4I_{bat}$$

so

$$I_{conv} = \frac{2.4}{5}I_{bat}$$

This argument assumes the currents are low, because as you raise the current you get a voltage drop within the battery due to the battery's internal resistance. Similarly your DC-DC convertor will also have an internal resistance and this will lower the voltage you get on the convertor side.

However, when a manufacturer specifies a maximum current this is normally a safe maximum current rather than the current you'd get if you just shorted the battery. The safe maximum current is typically low enough that you can ignore internal resistance, and you can simply use the equation above.

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