What determines a battery's charging rate?

I have a power bank which is rated to store 20000mAh of charge and 72Wh of energy. It is recommended by the manufacturer to charge it with a Quick Charge charger which provides 1.5 Amps with 12 Volts. The company claims that using this charger, the battery can be fully charged within 7 hours.

Now, here's what I don't understand. The charger would provide 1.5 x 12 = 18 Watts of continuous power, which means that would provide it with 18Wh of energy after an hour of charging at the same rate. Being charged completely in 7 hours is understandable since the charger would reduce the wattage as the battery fills up. What I don't understand is if the charger is providing the battery with 1.5 Amps, shouldn't it take at least 20/1.5 = 7 hours 20 minutes to provide it with 20Ah of charge, which certainly disagrees with the claim of the company? Where is my reasoning wrong?

• Charging is not a linear process. – MaMba Jun 10 '16 at 9:17
• @MaMba True yet that does not explain what is asked. – Bimpelrekkie Jun 10 '16 at 9:18
• A difference of two-three is usually due to the absorbtion time from 80-100 % charge when the charger is in CV mode, but your factor of 72/18 is even higher. Like FakeMustache said, it could be a power limit in the DC/DC, the manufacturer claimed 1.5 A but it provides less than that or the seven hours is extragged. – winny Jun 10 '16 at 9:23
• When you get to cell sizes in the tens of amps, I think one of the bigger influences will be from the charger's maximum output current. Charging Li-Ions at 1C is not uncommon for fast chargers, but at 20A, that'd be one heck of a charger if you wanted 1C charging so the size of the charger becomes a limiting factor (in addition to all the usual battery characteristics and idiosyncrasies). – Sam Jun 10 '16 at 9:31

You forgot to include the difference between the charger voltage and the battery voltage !

From 72 Wh and 20 Ah I conclude that the battery voltage is 3.6 V, that is the voltage of LiIon battery when its considered empty. When it still has charge it can be up to 4.2 V. I use an average of 4V.

The 12 V, 1.5 A from the charger will be (efficiently) converted by a switching charging circuit to 4 V and 4.5 A.

So although 1.5 A (at 12V) is taken from the power adapter, the battery is charged with 4.5 A (at 4 V) ! So charging 20 Ah at that current will take 5 hours.

Of course some power is lost in the conversion from 12 V to 4 V, some power is also lost in the batteries. So 7 hours is a more realistic value.

• Thank you! I felt that I was certainly missing out on something when thinking about this. Your answer explains it very well. Anyway, could you please shed some light on the switching charger circuit? In a DC circuit, isn't the current constant throughout the circuit, so what I mean is wouldn't the whole circuit have 4.5A as opposed to 1.5A? – Anindya Mahajan Jun 10 '16 at 9:22
• If you're interested in that phenomenon, search for "buck converter" on Google. A buck converter is a voltage regulator which converts power by using an inductor (or coil). The coil stores the energy magnetically and that can be released at a higher current while at a lower voltage. The charger circuit in your power bank will be a buck converter with some extra current sensing and limiting to make it a proper battery charger. There are ICs that do this, an example: pdfserv.maximintegrated.com/en/ds/MAX745.pdf – Bimpelrekkie Jun 10 '16 at 9:29

Your battery is 20A·h and 72W·h, meaning its rated voltage is 3.7V. As such, you cannot charge a battery like this with 12V directly. There is a converter somewhere (either in the charger, or in the battery circuitry), which steps down 12V 1.5A to a lower voltage and higher current.

• Could you please expand upon your answer? As in why would the voltage be needed to be lowered? From what I understand, as long as the external voltage is more than the battery's voltage, the battery should be charged. – Anindya Mahajan Jun 10 '16 at 9:27
• Because the excessive charge voltage would turn into heat inside the battery. I'm not sure your battery pack could manage to dissipate 12 extra Watts. – Dmitry Grigoryev Jun 10 '16 at 9:35