I have a device that operates on average 0.6 watts. I want to design a battery that will leave this device on for 17 hours straight. I know that charge(Q) = current * time. My device has an operating voltage of 5 volts. So my calculations are:

(0.6 watts / 5 volts) * 17 hrs = 2000 mAh

I also want to charge the battery using a solar panel for 7 hrs of sunlight. So i wanted to compute the required wattage from the solar panel to charge this battery to max capacity in 7 hrs:

(2000 mah / 7 h) * 3.7 v = 1.05 watts

(3.7 volts battery)

Can someone please verify that my math is correct?

  • \$\begingroup\$ I assume you'll use a boost regulator to get that 3.7V up to 5V to run the device. Might be a good idea to include the losses from the regulator into your estimated power usage if it isn't already (Not sure where you're pulling the 0.6W value from). \$\endgroup\$
    – I. Wolfe
    Apr 30, 2015 at 18:30

2 Answers 2


Your question makes some wrong assumptions but also it's much easier to just think in terms of power (W) and energy (Whr) in your question.

If your device uses 0.6 watts for 17 hours, then its total energy usage is 10.2 Whr. Given a small (20%) buffer, you'll probably want something that provides 12.24 Whr of energy in a real-life project.

Now if your battery is Li-Ion based (which is where I assume you get the 3.7V from), you'll want (12.24 Whr / 3.7V) ~3300 mAH of battery capacity.

The question about charging capacity is a little bit more complicated. Li-Ion batteries do not charge linearly. You can charge them very fast from empty but you MUST slow down as you approach full or risk overheating and damaging the battery.

Given that, you may choose to use an overrated battery (so that you can charge faster because battery isn't full) or you can oversize your solar array (so that you can charge faster when the battery is empty)

  • \$\begingroup\$ What effects (losses, inefficiencies, etc.) in particular led you to provide a 20% buffer on the total energy usage? \$\endgroup\$
    – photon
    May 1, 2015 at 2:39
  • \$\begingroup\$ Battery lifetime is a big one. The battery capacity will degrade with charge cycles, so getting a bigger battery in the first place well mean that you will still get the required run time after a number of cycles. \$\endgroup\$ May 1, 2015 at 8:36

To the question of inefficiencies, it sounds like you are using some sort of USB powerbank or USB powered or charged device. Since the battery runs at a nominal 3.7 volts, you need to boost from 3.7 to 5 volts to get the energy out of the device and the device will drop the 5 volts down to 3.7 volts when charging it. I have measured the losses at these two points in phone charging and powerbank charging/discharging and have found a typical loss of 15% on average each time the voltage has to be boosted or dropped.

Of course it will be a different efficiency for every device, but 15-20% is likely to be accurate unless you are using a rather expensive device that might be more efficient.

Also, to give you an idea of how long it will take to charge your lithium ion battery from empty to full: take the capacity and divide by the charging rate and multiply by 1.4.

I.E. A 2000 mAh battery charged at 1000 mA will take 2 x 1.4 = 2.8 hours approx to charge.


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