I have a system that consumes 0.497 Ah, with a voltage of 5V. I did the calculations to feed the system with solar enegeria:

  1. Watt = 5 * 0.497 = 2.485W
  2. Hours = 48 hours of system use
  3. Total = 149.1W consumption

By calculations I would say I need a solar panel that generates 2.07 ah and a battery 12.42 Ah.

Calculate solar panel:

149.1 / 6 (sun hours) / 12 (system voltage) = 2.07 Ah

Calculate battery:

2.07 * 6 (sun hours) = 12.42

That's right ?

  • \$\begingroup\$ I used to base this link:electronics.stackexchange.com/questions/71972/… \$\endgroup\$ Commented Dec 7, 2015 at 13:05
  • 2
    \$\begingroup\$ Your units are all over the place. Current is measured in Ampere, voltage in Volt, power in Watt, energy in Watthours (for practical purposes, otherwise it's Joule), charge is measured in Coulomb or Amperehours (mostly given for batteries). Please learn the difference and use them correctly, it makes things a lot easier. \$\endgroup\$
    – Arsenal
    Commented Dec 7, 2015 at 14:41
  • \$\begingroup\$ 100% efficiency is a bit out of touch with reality. \$\endgroup\$
    – Ecnerwal
    Commented Dec 7, 2015 at 15:05

1 Answer 1


I assume your system "consumes" 0.5 Ampere of current and not 0.5 Amperehours of charge (that would be an unusual measure).

So you have calculated the power of your system to 5 V * 0.5 A = 2.5 W. It will run for 48 hours, so it needs 2.5 W * 48 h = 120 Wh.

So you'd need a battery with 120 Wh, but there is no ideal 5 V battery and no ideal 5 V solar panel.

You chose (I suppose) a 12 V battery and a solar panel with charger for that. And you want to charge the battery in one day so your system can run for two days without recharging.

In an ideal world you need a solar panel of 120 Wh / 6 h = 20 W (that is 12 V with a current rating of 20 W / 12 V = 1.67 A). And a battery with 120 Wh / 12 V = 10 Ah.

But the world is far from ideal:

To get from the 12 V of the battery to 5 V of your system, you'll use a switchmode power supply. Let's say you get a not very well build one (because it's cheap) and it has only an efficiency of 80%. The battery must have more energy to power the losses of the power supply. So with 80% you would need 120 Wh/80% = 150 Wh.

Next thing to consider is that for improved battery life (if that is an issue) you don't want to have it cycle from 0% to 100% often, but more like only from 25% to 90% (or even less). So you only use 65% of the rated capacity. For 150 Wh needed energy you'd use a 230 Wh battery.

The increased need alone now requires the solar panel to deliver at least 25 W. But sizing the solar panel based on peak power and sun hours is asking for trouble (except for some very sunny regions I guess). So to get a better estimate, you look up some statistics of the area where it is supposed to be used. At my place you have a measly 1.5 sun hours in December, and a day is roughly 8.3 hours long. Of course your solar panel will produce some power even when you don't have direct sunlight hitting it, but it's far from the peak power. So maybe it's 100% peak power and 30% power (I made that number up, no idea how much it is) for the rest of the day. So you'd get 25 W * 1.5 h + 30% * 25 W * (8.3-1.5)h = 85 Wh. We need roughly twice that amount. So better go for a 50 W panel.

I haven't even mentioned that you need a maximum power point tracking charger to get that, so the charger will have a certain efficiency which reduces the amount of energy available to charge your battery, think of another 90% efficiency and you need 55 W.

There should be better estimates around on how much solar energy is available. Like this graph:

solar power available

(taken from Wikipedia by SechWatt)

It shows the total energy produces by a 1 kWp (kilowatt peak) solar panel per month somewhere in northern Germany. The average day (from sunrise to sunset) in December is 7.8 hours there, so you have 20 kWh in December, which averages to 20 kWh / 31 days = 645 Wh per day for a 1 kWp panel. With our needed 150 Wh we end up with a (150 Wh)/(645 Wh) * 1kWp = 232 Wp solar panel. So my estimation of 30% was probably way off.

Note: There should be calculators around for this kind of analysis.

If you plan to use the system for several years without replacing the components, you have to factor ageing in as well (battery capacity reduces, solar panel power reduces). So that makes things even larger.


Use a battery rated for 230 Wh (12 V / 19.2 Ah), and a 232 Wp solar panel (12 V / 19.3 A), if you want your system to work in December in northern Germany.

If you plan to use it elsewhere, calculate again.

This should only be considered a rough guideline on what should be considered and I wouldn't consider it a complete analysis.

  • \$\begingroup\$ +1 Exellent explanation! Congratulations for taking your time to answer so fully. \$\endgroup\$ Commented Dec 7, 2015 at 16:03
  • \$\begingroup\$ Thanks for the answer. I tried to do the calculations considering the solar average in my area. 5,00 kWh. But I did not understand the equation of the calculation. I am having very small values. \$\endgroup\$ Commented Dec 7, 2015 at 16:28
  • \$\begingroup\$ @victorhrs I used the value from the chart. It gives the monthly production of a 1 kilowatt peak solar panel. To get to a daily base, I divided by 31 days. I implicitly switched from kWh to Wh because numbers would get small. 1 kWh are 1000 Wh, so your 5 kWh / 31 days = 0.161 kWh or 161 Wh per day. \$\endgroup\$
    – Arsenal
    Commented Dec 7, 2015 at 16:35

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