Charge lower voltage, higher mAh with higher voltage, lower mAh battery

So let's hypothetically say we have a 12V 500mAh battery and we want to charge a 4V 2000mAh battery with it, using a (hypothetical) 100% efficiency converter.

One could say that, as the charge is constant in a circuit, one can best case scenario, charge the 4V battery to 25% of its capacity.

However, if one looks at it in energy terms, the first battery could provide, let's say 500mA for one hour @ 12V, this would be converted (with 100% efficiency) to 1500mA @ 4V, so it would (provided the 4V battery could handle the current) charge the 4V battery to 75% of its capacity.

At least one of these two statements is clearly false. I would like to know which one and why. Or if both then what is the real answer.

• Apart from the fact you can't take 100% of the energy from a battery or get 100% efficiency in conversion,or get 100% of the energy you supply to charging a battery into the battery, one principle rules - conservation of energy. I would say both your statements are incorrect (the first one more incorrect than the last) – JIm Dearden Sep 3 '15 at 8:29

Ignoring charging losses (and I'll get to that), the first statement is wrong and the second statement is correct.

Converter efficiency is measured in terms of power, not current. So, for instance, a 12-volt input at 1 amp would give a 4-volt output at 3 amps, and that's why the 75% figure is correct - it is 3 times greater than the first.

With that said, even if you have a perfect converter, batteries do not charge with perfect efficiency. At the very least, they start getting hot towards the end of the charge cycle. Even when charged, a battery's capacity has to be measured by a discharge cycle, so charging efficiency is actually efficiency through a charge/discharge cycle. Wikipedia suggests a charge efficiency for li-ion batteries of 80-90%. A standard 16-hour @ 0.1C charge cycle for NiCads produced a maximum efficiency of 62%.

• Good answer. Based on his commentd, I think the OP understands the concept that 100% efficiency won't happen in real life, but to make life easier in trying to understand Amp hours versus Watt-hours he assumes 100% efficiency. So yes, when comparing batteries, it is the Watt-hours that tells you the comparable amount of energy. Wnen the voltages different comparing amp-hours doesn't tell you a lot – Filek Sep 4 '15 at 4:25

Assuming everything is magically efficient, both statements are wrong.

Your problem is that you will lose energy as the battery is charged. This is usually converted to heat, so your second statement is wrong because even if the voltage conversions are perfectly efficient, the battery will burn energy in the charging process.

Worse, some battery chemistry require voltages higher nominal output to charge - NiMH for example, output ~1.2V, but charge at 1.4-1.6V.

Statement 1 is wrong because of your perfectly efficient clause. In a very bad circuit, you could charge a 4V battery with a 12V battery like that. You wouldn't, you'd be wasting massive amounts of energy and your 4V battery might explode. (4V sounds like lithium to me. Do you really want to know what happens if you charge a lithium cell at 12V? No, I don't want to either...)

So the true answer is "Somewhere in the middle."

• Thanks for your input, i may have explained it incorrectly. I didnt propose connecting them directly. Of course there has to be a buck converter in the middle. The efficiency of these converter can be in a real scenario around 90% so assuming 100% is not a crazy idea and i just did it for ease of calculations. Disregarding losses my underlying question is: is energy conserved as in my second statement or is charge as in my first. Or did i make a mistake (apart from efficiency) – psg Sep 3 '15 at 13:17