Estimating the actual capacity of power banks

[NikolayTodorov]

Hi guys, new member here. For the past few hours, I've been trying to estimate how many times a power bank can recharge a smartphone, but I'm not confident in my calculations, hence I'm writing this post. By "power bank" I mean those portable battery packs that have a 5V USB port for charging phones/tablets. (Here's one.) I know that a power bank advertised as having a capacity of 10,000mAh has a 3.7V 10,000mAh battery inside it. There's also a voltage booster increasing the 3.7V to 5V, which brings the effective charge capacity of the power bank to 7,400mAh (10Ah x 3.7V = 37Wh and 37Wh divided by 5V = 7.4Ah, if I'm not mistaken). Now comes the part that confuses me. Will these 7,400mAh at 5V completely recharge, for example, a 3,700mAh 3.7V smartphone battery exactly two times (since 7,400 divided by 3,400 = 2), ignoring other factors like the losses caused by the voltage booster and the smartphone's charge regulating circuitry? Or should I do the math in watt-hours? Is there anything else I might be missing? Thanks!

[Fsck]

I assume you mean "(since 7400 divided by 3700 = 2)"
conservation of energy means you don't need to care how you do the math though it's easier just to do watthours
if the world was perfect (batteries can output full power constantly and are temperature invariant, etc.) and lossless, you'd get more than two charges (something like 2.7?).
in the real world, you'd probably still get around 2.

[Phaedrus]

Do the math in watt hours, and assume ~80% efficiency. Keep in mind that the phone will also likely be using power while charging.

[suicidaleggroll]

When mixing battery voltages, all calculations should be done in Wh.  If the battery voltages are all the same, then you can leave voltage alone and just work in Ah.

In your case, both the power bank battery and the phone battery are a nominal 3.7v, so you can just work in Ah or mAh directly, it doesn't matter that the voltage is being stepped up to 5v and back to 3.7v (ignoring the inefficiencies associated with those transitions).

So...
10,000 / 3,700 = 2.7 charges
Or in Wh
(10*3.7) / (3.7*3.7) = 2.7 charges

But as Fsck said, since you probably have an 85-90% efficiency in both voltage changes, your nominal efficiency will be more like 70%, so your 10 Ah will only be able to deliver 7-7.5 Ah to the phone, which would be more like 2 charges.

[NikolayTodorov]

Thank you for the help, guys. Looks like I was wrong to think that the voltage difference did not matter and only the charge capacity in Ah did. Therefore, in a perfect universe, we'd get exactly what suicidaleggroll stated - "(10*3.7) / (3.7*3.7) = 2.7 charges". Have a great weekend!