I did a charge on this page and could not find the answer I was looking for. Perhaps it was over my head, or perhaps I was using the wrong search terms, so here goes nothing:

I am trying to understand something about the battery pack I've purchased and the number of times it can charge my electronic gadgets. I purchased a RAVPower 15000mAh battery pack, which can be found here:


On the back of the battery pack it says the following:

Capacity: 15000mAh/55.5Wh Input: DC5V/1.5A Output: DC5V/4.5A Total

ON the amazon page it says it can charge an iPhone 6 6.2 times.

My understanding is that an iPhone 6 battery has the following specs:

3.82V 6.91Wh 1809mAh (1000*6.91Wh/3.82V)

When taking the capacity of 15000mAh I figured that I could get to the number of Wh by just multiplying by the 5V, but it appears that the actual battery cells are 3.7V cells, so the 15000*3.7/1000 = 55.5Wh, which makes sense to me, but...

When I take the 55.5Wh capacity and I divide by the capacity of the 6.91Wh iPhone I get to 8 charges, which differs from the company claim of 6.2 charges.

So here is where I am confused.

I can get to the 6.2 charges by taking the 3.82V/5V and multiplying by the battery capacity of 55.5Wh and then dividing the result by the 6.91Wh capacity of the iPhone battery.

It seems as if I've already stepped down the Wh capacity of the battery by multiplying by 3.7V above, yet I'm stepping down the battery capacity again. Or am I really just losing the remaining charges because the output is 5V, but it gets stepped down to 3.82V when it hits the iPhone and the rest is lost due to inefficiency? Or is there some other explanation?

Also, now if I take this battery pack and I want to charge it using the RAVPower 9W solar panel:


This panel tells me that it is a 5V panel capable of producing 1.8A of output. I see that my max input for the battery is 5V at 1.5A, so I believe that I should be able to add 7.5W of battery capacity per hour under full sunlight. My experience has been closer to about 5W of battery capacity per hour, which brings me in the ballpark of my other calculations where I'm actually charging the battery pack at 3.7V*1.5A = ~5.5W.

And what does the 5V on the input and output actually mean? Is the battery voltage stepped up to make sure that I charge any gadget that is less than 5V? Or is there some other reason?

Thank you in advance for your help!


Your calculations look correct at 8 charges, assuming that efficiency is 100%.

Your 15000 mAh pack is at a nominal 3.7 volts, but outputs at 5 volts to charge your phone. It uses a dc-dc boost to get to the 5 volts. Most of these powerbank that I have looked at claim an efficiency of around 85% for the boost. In other words, when your 15000 mAh pack outputs at 5 volts, it loses 15% of its capacity to boosting the voltage to 5 volts.

Then your Iphone needs to get the 5 volts back down to the 3.8 volts or so that it's battery needs. If their buck converter is also 85% efficient, then again you will lose 15% or in other words it will take approximately 15% extra energy to fully charge the iphone.

If you multiply the two efficiencies: 85% (you will have 85% of the original capicity remaining after the boost to 5 volts) x 85% (you will have 85% of the 85% after the drop back down to 3.8 volts to charge to iphone), you get 73% of the original 55 Wh which is essentially the 6.2 charges vs the 8 charges you were expecting.

Why do they use 5 volts? It is not so that the voltage is high enough to charge any battery that they might charge, but because they are using a USB connection. If you happen to connect to a computer's USB which is always 5 volts, everything is compatible (compatible in an electrical sense that the voltage is always the same).

Edit: I see that Russell McMahon answered at the same time, but beat me to posting. He is saying essentially the same thing, but probably explaining it better.

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  • \$\begingroup\$ Thank you for this answer! So I assume this would work the same way with the solar panel? I can only charge the batteries at 3.7V*1.5A minus losses of ~15%? And then if I'm charting a 7.2V battery the same would occur as with the iPhone? I lose 15% going out and then another 15% going in? \$\endgroup\$ – keatz Apr 18 '15 at 14:45
  • \$\begingroup\$ @keatz - yes, same idea for the solar panel, but when the panel is charging the battery bank, it sends the energy in at 5 volts, and the battery bank then converts it down to 3.7 volts, where you will lose the 15%. Yes, the same with charging a 7.2 volt battery, approx 15% loss in and 15% out; however, these 15% numbers are what I have measured for many different powerbanks. It is possible to get better efficiency, but it would cost more. I.E., the 15% loss is not a hard and fast rule, it might be more and might be less depending on the electronics used to do the boost and drop of voltages. \$\endgroup\$ – Filek Apr 19 '15 at 3:09

mAh claims for such packs are often exaggerated - sometimes very much so.
Their very specific and reasonably self consistent claims give some hope that the device does what is claimed.

Battery is 3.7V nominal, 4.2V max, 3V or more minimum. 3.7V is ~~ mean voltage across discharge cycle.

Input is from nominal 5V "USB" source.
Output is nominal 5VDC from a boost converter.

Your 8 x charge calculation is about right at 100% efficiency.
Their 6 x charge claim = 75% of theoretical is encouraging as it sounds like a real world figure reflecting efficiency.

In a well designed supply where $ are not absolutely vital they may use a buck converter to convert 5V in to Vbattery BUT in many units may just use a linear regulator. Either way some input energy will be lost.

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