Arduino Project powered by Solar or Battery

Disclaimer: I may have made a mistake in my working's in which case I would appreciate if someone could correct me. OP: Take what's below with a pinch of salt.

To be honest, If I try to make a perfect calculation on what you will and won't get, this will just be too complex for me to answer. Instead, I'm going to make very rough approximations.

"The system will be located near Narbonne where the insolation varies from 1.18 (winter) up to 6.44 kWh/m²/day (summer)."

What we need to do first is establish the average amount of energy per day you'll get. After doing a basic mean average of the values in your table, I got:

44.47 / 12 = 3.706 kWh/m²/day

Your solar panel has the following dimensions:

160 mm x 138 mm

We'll ignore the uncertainty and so we have a surface area of(standard form converts to meters):

160*10\$^{-3}\$ m * 138*10\$^{-3}\$ m = 0.02208 m²

Hence we can now see that per day, assuming (I have absolutely no idea of the efficiency of your solar panel) a rather optimistic 20 % efficiency, you'd get:

0.02208 * 3.706 * 0.2 = 0.0163 kWh/day

0.0163 * 1000 * 3600 = 58913 J/day

Continuous Wattage Possible: 58913 / (3600 * 24) = 0.6819 W

Now, at this point we haven't even addressed the fact that in winter you're getting on average 6 times less power from the solar panel than in summer. Potentially this means you have to store a great deal of power from summer and utilise it all throughout winter. Assuming you could run on 0.68W (including the efficiency of the Arduino, etc), The main problem as I see it is that some days you'll have virtually no power whatsoever. Additionally, you may need to step up or step down the power produced from the solar panel, which, in itself will incur efficiency penalties.

I advise that you total up the exact power consumption of your Arduino, then actually conduct real world testing and procure a maximum power consumption under peak load. At the minute, you may need to either add more solar panels(increasing the area) or use solar power in conjunction with grid power.

Edit(Assuming the purported 50 mA current consumption at 7 V):

P = VI

7 * 50x10\$^{-3}\$ = 0.35 W

Now, based on this, you'll be fine if you're perfectly storing every ounce of energy you get from your solar panels and there aren't weeks where there is barely any solar power available. So let's assume that we want the Arduino to function even with the worst winter insolation values.

Taking your minimum of 1.18, redoing the calculations:

1.18 * 0.02208 * 0.2 = 0.00521 kWh / day

0.00521 * 1000 * 3600 = 18759 J / day

Continuous Wattage Possible: 18759 / (3600 * 24) = 0.217 W

So in winter months you'll have around 0.217 W available, but the reality could be worse then that as is the case with all weather based power sources. What does this mean? It means that realistically in order to A. Power the Arduino and B. Have a decent safety margin, you will need (Assuming a safety margin of 2X that required):

0.35 W * 2 = 0.7 W

0.7 / 0.217 = 3.22

Hence you need to increase the surface area of your solar array by 3.22 times. In other word's, you'll want four solar panels, presumably connected in parallel (don't quote me on this :D) in order to power your Arduino through thick or thin.

Final note: Your power booster will incur efficiency penalties as well as the act of energy storage, which is why I took such a large safety margin. Hope this helps.

It's not the battery that's upside down, it's the entire circuit! Typically you want ground potentials on the bottom (only.)

That being said, I think your circuit will mostly work, except you need to move the resistor (R1) to be between the + output of the solar panel and the Zener(D1)/input(U2) pin, and probably also lower its resistance significantly. Dropping from 8V to 5V at 300 mA happens at (8-5)/0.3 == 10 Ohms.

Btw: To learn more about electronics, analyze the circuit until you see why the current position of R1 and D1 makes the U2 always see the full output of SOLAR, and all R1/D1 does in the schematic is wasting current.

Finally, the MCP1700 is just a linear voltage regulator, with a maximum input voltage of 6V. That's a pretty inefficient way of taking advantage of the voltage that comes from your panel. When the weather is overcast, the panel will provide less voltage than you can use, and you may not charge at all. When the sun is bright, the panel will provide a lot more energy than you can use, and you'll burn it off in all of R1, D1, and U1. I would highly recommend using a micropower harvesting circuit, or at least a buck/boost or SEPIC switching controller that can turn a variety of input voltages to useful output voltage (if not current.)

Finally, LiPo batteries do not like being trickle charged while full. Best case, you metalize the Lithium and the battery dies; worst case you overheat it and start a fire. If you're going to "float" the battery, make sure you "float" it below the "nominal recovery" voltage of 4.05V -- somewhere around 3.85-3.90V would probably be safer. Check the data sheet for your particular manufacturer/battery to get a better indication of where to set the limit.

Finally, if there is sufficient back-voltage prevention in the MCP1700, then it should be safe to leave it as indicated even when the panel is not providing (much) energy. However, many linear regulators, especially ULDO ones, do not like back-power, and thus you may need to add a low-drop diode or a P-channel MOSFET or other switch between the regulator and the battery, with the gate controlled by the presence/absence of power into the regulator.