# Arduino + Solar Charger Shield V2.2: Battery and Solar panels selection [closed]

first of all I would like to apologize for the, probably very basic and stupid questions for you, but I'm not an electronic expert..

I'm implementing my project using an Arduino Mega 2560 (I need more pins.. this is the most important reason that it drives the Arduino choice) I'll put my project in an external environment without the electricity source. For this reason my idea is to use a battery with a solar panel as power input to charge the battery. I also applied some low-power techniques in order to reduce energy consumption as much as possible. Regarding the solar panel and battery connections I purchased this Arduino Solar Shield.

Solar Charger Shield V2.2 - Seeed Wiki

And here I'm a bit (a lot) confused about the choice of the battery and the solar panel(s) to be purchased. First of all, my idea is to use multiple (2 or 4) solar panels in order to exploit as much as possible the solar light (multiple positions) but I'm not able (I'm not sure) about the solar panel selection to purchase. Considering that the maximum current that the solar shield could be delivered is 700mA can I use 4 solar panels like this https://www.seeedstudio.com/1W-Solar-Panel-80X100.html (applying also 4 applying 4 diodes to avoid reverse current)?

About this the possible number or size of the solar panels Do I need to consider the amount of th Current at Peak Power? For example, using 4 solar panels with 170mA as Current at Peak Power, 4*170 = 680 that it's > of the current of the solar shield. So, is this wrong? Do I reduce the amount of the panels or do I use a different panel?

About the Solar input voltage it's 4.8~6V. These are the specifications about the solar panel:

Dimensions: 100x80x2.5(±0.2) mm Typical voltage: 5.5V Typical current: 170mA Open-circuit voltage: 8.2 V Maximum load voltage: 6.4V Maximum load voltage, Open-circuit voltage are > Solar input voltage. Is it a risk? About the lipo battery, can I use a battery 3.7v and about the mAh value as high as possible (for example this https://www.amazon.it/2100mAh-Lithium-Polymer-Replacement-Bluetooth/dp/B095BTSMYH/?th=1 ?) What about the maximum current value (in this case 5A)? Is it ok? In case, can you suggest me something else?

** *********** EDIT *********** **

As suggested I'm updating my question in order to give additional info about my scenario.

My project will be installed in an external environment (with some atmospheric protections IP67). The location, from solar irradiation point of view, it's a good place: https://globalsolaratlas.info/detail?m=site&c=40.283716,17.586365,9&a=18.056138,39.997805,18.056138,40.420456,18.535172,40.420456,18.535172,39.997805,18.056138,39.997805&s=40.11799,18.382874

• I'll have to take the measurements from more sensors:
• starting from 1 to 10/15 humidity/temperature (DHT22) sensor;
• starting from 1 to 10/15 load cells (4 load cells for each HX711);
• I'll send the data using some transmissions using a LORA Shield.

I don't need to retrieve the measurements continuously or in very short periods.

The idea is to sleep my Arduino (in low power with all power sensors disconnected) and, using a RTC DS3231 module, to wake-up it on regular intervals, read the data, send it via LORA and go to sleep again.

The idea is to wake-up every 15 minutes. The time that I imagine about the measurements and the send will take a few seconds (around 30s in case I can approximate it to 1 minute).

And at the moment I don't have the exact measurements about the used current (because I can connect all here.. but I need to purchase the solar panels and the battery in order to make a real test also)..So, the idea now is to approximate.. In case I can adjust something later.

At the moment, measuring the current I have.

around 0.35mA when the Arduino is sleeping.

I probably can reduce it again because I have to remove 2 power led indicators (1 on my DS3231, one on my LORA shield... and, in case, also the power led on my Arduino).

When the Arduino works (live mode) probably I can consider around 90/100mA (probably this is too much) for 30s).

I'm not expert but, using this values I calculated the average about the used current in 1 hour:

((100mA * 4 minutes) + (0,35mA * 56 minutes)) /60 = 7mA/m 420 mA/h

I'll need 5v about the voltage.

I don't have any limitation about the size or the space.

I think that considering 7days without solar power supply or bad water conditions could be enough (in my opinion it's too much)

This is a question that's better suited for the electrical engineering SO, but I can give you some pointers to reword and re-post your question so it's not as broad.

Right now you have a few unknown variables to juggle, so you should be jotting down your thoughts on all of the following:

• Where will the unit be placed and/or located? How long do you expect it to go without needing to have its battery recharged?
• What is the continuous power consumption? You should know your approximate continuous current consumption and at which voltage you will be running the Arduino and the sensors that you connect to it
• What environmental conditions do you expect the unit to experience? Do you have a limiting factor on size or space?
• How much are you willing to pay?

You can definitely use multiple solar panels, but you'd have to connect them in parallel (so they maintain the same voltage), and even then, the linked charger shield can only go up to 900 mA of charging current. The 600-700 mA is the supported power output of the battery, or the max current that the shield can provide to your system thru the shield. It can be larger if you lower the system's voltage and stay within the maximum limits of the charger shield.

With regards to your second question, yes, current at peak power is a good metric for your approximations. If you go thru my questions, you will know how much your systems consumes in a day and choose a battery that can discharge no more than X amount of percent per day (depending on the type of battery). Your goal is to be able to charge that battery back at least close to its full capacity--your solar panels or the amount of thereof, as well as your charger, will be dictated by that.

For the Voc of the solar panel - ideally, it would be less than the solar input voltage. It's the voltage of the panel without a load connected to it. Of course, you will have a load connected to the solar panel and thus the typical voltage will be around what's stated, but if you don't want to damage your charger, you have to make sure your SP will be properly loaded under most conditions! It will be safer to connect multiple smaller SPs to get the current that will keep the system up and charge your battery.

With regards to the battery current - you have to check which current it is. Is it the charging current? Output current? Be mindful that you are limited by your charger in this case (3W output, 900 mA charging current).

Hope this gives you a start! It's a good question, but you need to do some more reading and you need to have a better idea of the design for your system before we can be more helpful.

Edit:

Per the updated information...

I will use your estimates, but do consider that for sensors that require power, you can add them and make your calculations more accurate. Also, while LoRA is low-power, it will consume some power too, and you can factor that in.

Let's assume your numbers are correct and the average current consumption averages out to ~ 420 mA/h. Then, if you want to keep the system alive for 7 days, you'll need a battery with a capacity of 420 mA/h * 24 hours/day * 7 days -> 70,560 mAh or 70.560 Ah. At 5 volts, your system effectively requires 0.420 A/h * 5 V -> 2.1 Watts/hour (finicky units here, eh?). That would translate to ~353 Watts/week. Also, 5V means you will need a battery that can provide that voltage - a lower voltage battery will not support your system.

The next thing you have to consider then is the battery type - there are a few, but the ones I'd go for are either Li-ion or VRLA (valve-regulated lead acid). There is a tradeoff between cost - Li-ion batteries cost more, but they are typically better in the majority of factors rather than SLA (sealed lead acid) batteries. Doing some googling, you could get a 70-100 Ah SLA/VRLA for ~\$200-300 and a Li-ion for ~\$400-500. Do note, however, that you can discharge Li-ion batteries more deeply than VRLA ones - it's recommended to keep the charge on the VRLA batteries around ~50% and preferably around ~75% to extend its life. You can discharge Li-ion batteries up to 80% or a bit more and that's fine, they don't suffer as extensive reduction in their lifetime like SLAs.

To put that into perspective, if you have an application that requires ~8 Ah, then ideally for a VRLA you'd need to size a battery such that it doesn't go below 50% for your worst-case application scenario. That is, if you think you'll need the device to run for 7 days without any sun, then you'll need a battery of a 16 Ah capacity or more. Even bigger if you want to extend its lifetime. For Li-ion, aiming at a bit higher than 8 Ah should be fine. For your application, that means getting a SLA battery that's > 140 Ah (unless it's a deep discharge SLA battery) or a Li-ion battery that's close to 70 Ah.

Now, it must be noted that most batteries that provide those capacities typically come in 12V, so you would have to add a buck-converter to convert from 12V to the battery voltage your charger needs (3-4.5V). The charger you linked states that it can provide up to 3W out, which is 600 mA max at 5V. That would support your application as your average current draw is less than 600 mA.

For charging, however, you have to keep in mind that your system, on average, draws 420 mA/hour. So the charger's charging capabilities would actually be 900 - 420 -> 480 mA/hour. That means that based on worst-case considered direct solar irradiation hourly data in the link you provided, you'd get ~ 8 hours of direct sunlight (again, worst case!). So 8 hours you can charge -> 480 * 8 = 3840 mAh and you'll lose 6720 (420 * 16) mAh. That is not good, because you're losing the charge on the battery each day.

That means 2 things - you have to get a charger that permits a larger charging current within the maximum charging current of the battery rating AND you have to get enough panels (OR a big enough panel) to be able to provide that current. With the panels that you linked, the max current per panel is ~170 mA for the typical voltage of 5.5V. To provide a charging current of ~ 900 mA, you'd need 5 panels, but as mentioned above, that still wouldn't be enough. So, back to the drawing board...