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I'm prototyping an autonomous remote system which is composed of the following components:

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

Questions:

  • How do I make sure the system will be able to operate continuously?
  • Which formula(s) can I use to guesstimate?
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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.

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    \$\begingroup\$ I think that storing the energy in the summer for the winter is nearly impossible: I'd rather design the system to work with the average power of the winter, so it's guaranteed to work in summer too. \$\endgroup\$
    – clabacchio
    Aug 20 '12 at 22:58
  • \$\begingroup\$ Well have you actually calculated how much power your Arduino will consume? Until we know that, this is all a bit arbitrary and pointless. In any case, imo it doesn't look good that including summer months you average at 0.68W. You're going to have to use the Arduino in a seriously low power manner(I don't know the consumption of Arduino's FYI) \$\endgroup\$ Aug 20 '12 at 23:20
  • \$\begingroup\$ What if I told you the Arduino consumes about 50mAh ? \$\endgroup\$
    – ZogStriP
    Aug 20 '12 at 23:24
  • \$\begingroup\$ Well mAh is usually used for power sources. So 50mAh would mean that the power source could provide X voltage at 50milliAmps for an hour. So I don't think you can really say that it 'consumes' 50mAh as mAh, Ah etc are units of charge really. \$\endgroup\$ Aug 20 '12 at 23:31
  • \$\begingroup\$ Sorry that was 50mA. Source: gammon.com.au/forum/?id=11497 \$\endgroup\$
    – ZogStriP
    Aug 20 '12 at 23:49
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The following isn't complete, but this is what I just remember

The typical voltage/current [5.5V, 540mA or ~3Wp] is for standard conditions, which are:

  • Light intensity: 1000 W/m² (!)
  • Temperature: 25°C
  • Angle of light beam: 90°
  • Light-Spectrum: AM 1.5

You can look these up here: http://en.wikipedia.org/wiki/Solar_panel (section 'Module Performance and Lifetime')

Things not to forget about:

  • Assume worst condition for season (=1.18 kWh/m²/day, winter)
  • You worst case (winter) is 1.18 kWh/m²/day with ~10 hours sun
  • Average Energy: 0.118 kWh/m² = 118 Wh/m²; o: ~11.8% of 'typical' energy
  • 3 W * 11.8% = 0.354 W (avg); or 0.354 Wh each hour during daylight in winter (1day: 3.54 Wh)
  • Know that panels are aging: At ~80% of original efficiency it should still work.
  • At ~80% efficiency your panel will give you ~2.832 Wh per winter day
  • Note: at night it's dark, your battery should be able to hold enough energy that period
  • Also: These are averages. There may be some longer period of rainy or cloudy days..
  • You will not be able to store all energy into your battery, even with enough capacity.
  • Battery-Efficiency (charge/discharge): about 80-90% for LiIon
  • Storing 2.832 Wh into the battery gives you only 2.27 Wh back discharging
  • This still does not take the efficiency of the charging circuit into account!
  • The up-converter circuit that brings 3.7 V from battery up to 5V: ~90% efficiency.
  • 2.043 Wh per day left...
  • We have still not taken care of suboptimal angle of the light...

Ok, let’s assume we have 2 Wh per day (which is higher than to be expected). This is how you could check if this is enough:

  • Check your circuit’s energy consumption. Measure it..
  • Or: Take a fully charged battery of known capacity, check how long it lasts (e.g. 1 Wh capacity for 20 hours)
  • The test battery should have a lower or equal capacity
  • Now you can estimate how long your circuit will survive with 2 Wh
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  • \$\begingroup\$ PS: You should, if possible, put your arduino into sleep mode, if there is nothing to be done. This is very effincient. arduino.cc/playground/Learning/ArduinoSleepCode Also: donalmorrissey.blogspot.de/2010/04/… \$\endgroup\$
    – SDwarfs
    Aug 21 '12 at 11:44
  • \$\begingroup\$ PPS: You can also reduce your ardiuno to a single chip in some cases. You can get a ATTiny16PU for <3 EUR for example. Works also with lower voltages... but TTL level will be also lower then. See: blog.makezine.com/2011/10/10/… \$\endgroup\$
    – SDwarfs
    Aug 21 '12 at 11:47
  • \$\begingroup\$ PS: Is your device always in the sun? Know: There be shadows of trees, houses during the day. If it's a robot, it may move into shadow... \$\endgroup\$
    – SDwarfs
    Aug 21 '12 at 11:49
  • \$\begingroup\$ The station will always be in the sun (won't be moving) \$\endgroup\$
    – ZogStriP
    Aug 21 '12 at 14:46
  • \$\begingroup\$ Whats the kind of application? Something like a parking ticket vending machine? Does it have to be work continuously or is it sufficient to do something quickly every second (e.g. needs only 10ms) and can be put to sleep mode afterwards (for 990ms each second)? \$\endgroup\$
    – SDwarfs
    Aug 21 '12 at 14:54
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You can calculate the Power needed \$P=U*I \$ (voltage times current for DC). Do this for every power-consuming device (e.g. the arduino, the peripherals etc.) and you know how much power you need by adding them up. Then calculate:

  • How much power you can get out of the solar charger
  • How much power you can store in the battery pack

BTW: You battery pack only provides 3.7V, but 7V are recommended arduino. So you need a boost converter, which in turn "consumes power" due to lossy conversion.

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  • \$\begingroup\$ @susha Isn't the Solar Charger Shield supposed to handle the shift up to 5V? From what I understand, the shield is using either the battery or the solar panel to power the arduino. \$\endgroup\$
    – ZogStriP
    Aug 20 '12 at 22:05

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