I received a battery pack that I had ordered today, one of those ones that provide 5 volts using a USB connector, and plug into a USB socket to charge.
I noticed that on the box, it said it was rated for 6000 mAh.
The battery itself said 22.2 Wh, and that the output voltage was 5V DC.
Converting that back to mAh was: (22.2 / 5) x 1000 = 4,400 mAh. About 75% of the 6,000 mAh printed on the box!
Feeling shortchanged, I picked up another (Li-Polimer) battery pack from a different manufacturer I had lying around and saw it's capacity was: 15 Wh, 4,000 mAh. The voltage again being 5 volts.
Converting the Wh reading to mAh was: (15.0 / 5) * 1000 = 3,000 mAh (about 75% of the mAh reading).
My guess is that as the batteries deplete, like the lead-acid batteries of old, the voltage decreases. So if the voltage dropped to 3 volts, it would still deliver the milliamps for the mAh reading, but the Wh for the watt hour reading would be reduced.
I suppose this means that the useful capacity of the battery (the period of time where it actually delivers 5 volts) is even less… perhaps only just half of the mAh rating?
Best Answer
Firstly the battery is not \$5\mathrm{V}\$, it is nominally \$3.7\mathrm{V}\$ as pointed out already.
However the confusion lies in misinterpreting the \$\mathrm{mAh}\$ rating of the battery, and it has nothing to do with converter losses, and everything to do with
$$P=I\times V$$
If the battery is rated at \$6\mathrm{Ah}\$, it means you can draw \$1\mathrm{A}\$ for \$6\mathrm{\space hours}\$. So lets say you are drawing \$1\mathrm{A}\$ from the battery. This means it is delivering:
$$P=I\times V = 1 \times 3.7 = 3.7\mathrm{W}$$
Now lets say you feed this through an ideal boost converter. In this case \$P_{in}=P_{out}\$. So assuming we boost it up to \$5\mathrm{V}\$, this means that the current that must be drawn from the output to discharge the battery at this rate is:
$$I=P/V=3.7/5=0.74\mathrm{A}$$
So what this means is at the higher voltage, you can draw a current \$0.74\mathrm{A}\$ for \$6\mathrm{\space h}\$ for that battery capacity - so the output capacity is \$4.44\mathrm{Ah}\$.
Again, this is not converter losses - in fact with losses the number would be lower. Instead it has to do with the fact that you are sacrificing current capacity for a voltage gain which is how a boost converter works - if this wasn't the case then you would have invented free energy.
Essentially at a higher voltage, the energy delivered by each unit of charge \$\left(\mathrm{V}=\mathrm{J}/\mathrm{C}\right)\$ is higher, hence at a higher voltage but lower current you are still delivering the same energy.