Hello, I have a battery rated at 3.7v 1000mAh and three different solar panels.
First solar panel is rated at 6v @ 550mAh.
Second solar panel is 10v @ 140mAh.
Third solar panel is 20v @ 70mAh
That's what the specs says and I confirmed it myself at a full sun and no load, just the multimeter.
Which solar panel would be better for this rated of battery? and how did you come to that conclusion?
The solar panel will be placed indoor near a window which receive light, but mostly not direct and not all day. Not always will it receive the sun at it's best.
If I hook up the first solar panel to the 3.7v battery, it would need to be above 4 volts to charge up, which is a lot of sun exposure.
The second solar panel would not need the full sun (I think). Even at 5 volts and around 70mAh which is about half the sun needed, could input voltage to the battery.
Last, the third solar panel, I don't even want to know how long will it take to charge at <70mAh.
What is the best bet to gain the absolute best of the sun?
Additional info, battery if from a cellphone, it ranges from 3.5v to 4.2v. Battery has protection for over-discharge and overcharge. All solar panels have a diode. Battery will power a sleepy Arduino that only wakes up every hour. Sleeping Arduino draws 0.01mAh or 0.10mAh, not certain of which one is the accurate one (multimeter screen broke, Grandmother use it to open a coconut).
3.7v sounds like a LiPo battery. LiPo batteries have a nasty habit of bursting into flames when mistreated. That means you need a proper LiPo charge regulator. You will have to choose the solar panel to match the regulator.
I would be surprised if there is any protection built into the battery. The protection is probably part of the phone. But I could be wrong.
If you can charge the phone from a USB connection then perhaps you could regulate the solar voltage down to 5v and use that to charge the phone.
I too have trouble believing that all the protection circuitry you need is built into the battery. Many of those protection circuits limit the discharge only.
Use the 6V solar panel, with this charger to be safe: USB / DC / Solar Lithium Ion/Polymer charger [Rev C] : ID 390 : $17.50 : Adafruit Industries, Unique & fun DIY electronics and kits. That panel will produce 6V in just about any light, and the most charging current of the selection. The supplied current will be proportional to the light intensity. There is no need to worry about "maximizing exposure" to sunlight, because in full sun the panel theoretically has the capacity to charge a completely dead 3.7 V 1000 mAh battery in 2 hours.
Solar panels are rated by voltage and current (V and mA or A).
Battery capacity is rated by milliampere-hours (mAh).
Robin2:
3.7v sounds like a LiPo battery. LiPo batteries have a nasty habit of bursting into flames when mistreated. That means you need a proper LiPo charge regulator. You will have to choose the solar panel to match the regulator.
Yes, it is a LiPo battery. What do you mean by mistreated? If by it you mean current spikes by the solar panel, would adding a capacitor be of any help?
Robin2:
I would be surprised if there is any protection built into the battery. The protection is probably part of the phone. But I could be wrong.
I have about 4 or 5 identical batteries like this, so I teared down two and tested the protection circuit. I confirmed that it has protection for both things.
jremington:
I too have trouble believing that all the protection circuitry you need is built into the battery. Many of those protection circuits limit the discharge only.
Is there any other cheaper modules that can also do this? Maybe on eBay? I could use this for the final project, but for now it would be great if anyone could point at a cheaper one and or similar. For prototyping purposes of course :).
jremington:
That panel will produce 6V in just about any light, and the most charging current of the selection. The supplied current will be proportional to the light intensity. There is no need to worry about "maximizing exposure" to sunlight, because in full sun the panel theoretically has the capacity to charge a completely dead 3.7 V 1000 mAh battery in 2 hours.
I kind of see where you're getting at, with the module you pointed, but I still don't fully understand.
That's amazing data so far, I haven't read all of it because it's quite a lot to absorb for me right now, but really helpful. I'll take a good look and I'll ask about stuff
Thanks for answering!!!!!!!!!!!!!
I assume the solar cells are all of the same type.
The first panel has the most power output. ~twice as much as the other two.
Q: Did you measure the "open" voltage in the low/average light conditions you are going to use it.
Just a DMM on the panel, nothing else.
From memory, the maximum power point (MPP) is about 80% of that voltage.
If that result is >= than ~4.5volt, you can use that cell.
With a proper charge controller ofcourse.
The second panel might do better in very low light conditions, with a low idle current buck charger.
Leo..
so I teared down two and tested the protection circuit. I confirmed that it has protection for both things.
It would help if you explained exactly how you tested the "protection circuit" and demonstrated that it protects against both overcharge and overdischarge.
jremington:
It would help if you explained exactly how you tested the "protection circuit" and demonstrated that it protects against both overcharge and overdischarge.
I opened the package and took the electronic parts, put the battery aside. I hooked the battery terminals from the battery board to a few capacitors with a similar rating. I applied 3.7v to the input of the battery board. Capacitors got charged @ 3.7v.
I applied few LED's from the input charging port and let the caps discharge. This particular battery board cut the current @ 2.7v, and there was no more output until higher voltage was applied. I checked the caps voltage and they were 2.7v. I charge them again with 5v input and the board stop the intake voltage to the caps @ 4.2v. Checked the caps and they were 4.2v.
I did the same test with the original battery and no alterations. Took a lot longer but, it was the same results.
I've tested another protection boards from different battery brands and some didn't had overcharge protection.
jremington:
What were the cutoff voltages?
2.7v and 4.2v with the board I'm using. It varies depending on the brand.
jremington:
Try your best to convince us!
Why do I have to convince you? xD The info is out there now for people to use and conduct their own test.
Wawa:
Q: Did you measure the "open" voltage in the low/average light conditions you are going to use it.
Yes. I measure the open voltage and the average volts in an average light conditions with option 1 were 3.5v. to 4.5v.
With the other options I got more voltage, but, I'm concerned about the solar actually putting juice to the battery due the low mAh.
Wawa:
Just a DMM on the panel, nothing else.
What's that?
Wawa:
With a proper charge controller ofcourse.
Any recommended controller? a Cheap one if possible. I've already been recommended to use the one a Ada website
Wawa:
The second panel might do better in very low light conditions, with a low idle current buck charger.
Leo..
Solar panels and buck chargers? I've heard that those are not recommended due low efficiency.
Why do I have to convince you? xD The info is out there now for people to use and conduct their own test.
You haven't told us exactly which battery, or where you bought it, so the experiment cannot be repeated by someone else. The "information" you have posted so far has no value at all.
DMM stands for digital multimeter.
You measure the voltage with just the DMM. You get the panel's "open" voltage.
75-80% of that voltage is were the solar panel will have it's maximum power point (energy transfer).
If you turn that around. A 13.8volt (charging voltage) lead/acid battery idealy will need a 17.25-18.4volt panel.
If your panel produces 4volt (open) in the light condition you have.
And you use a schottky diode between panel and battery to stop the battery from draining into the panel when it's dark.
Then you only have 3.8volt to work with. That will just dribble a bit into a flat (3.7v) battery.
Small (500mA) buck chargers are very efficent ~90%, and can have very low idle current ~200uA. Google "Pololu".
But you ideally need a charge controler. A chip that looks at the battery voltage as well as the panel's MPP voltage.
Google is your friend.
Leo..
jremington:
You haven't told us exactly which battery, or where you bought it, so the experiment cannot be repeated by someone else. The "information" you have posted so far has no value at all.
I'm sorry to hear that. Why you have to be so mean? Were you dropped as a child? Just kidding, I love you.
Regarding to my original question, is there a cheaper module like the one at adaFruit?
Looks promising but expensive for me right now. Maybe an eBay module?
I'm not trying to be mean. I'm just pointing out that we have no idea what battery you have, so the results of your "tests" are useless to anyone who reads this forum post. Here is what you stated:
Hello, I have a battery rated at 3.7v 1000mAh and three different solar panels.
Wawa:
DMM stands for digital multimeter.
You measure the voltage with just the DMM. You get the panel's "open" voltage.
75-80% of that voltage is were the solar panel will have it's maximum power point (energy transfer).
If you turn that around. A 13.8volt (charging voltage) lead/acid battery idealy will need a 17.25-18.4volt panel.
If your panel produces 4volt (open) in the light condition you have.
And you use a schottky diode between panel and battery to stop the battery from draining into the panel when it's dark.
Then you only have 3.8volt to work with. That will just dribble a bit into a flat (3.7v) battery.
Small (500mA) buck chargers are very efficent ~90%, and can have very low idle current ~200uA. Google "Pololu".
But you ideally need a charge controler. A chip that looks at the battery voltage as well as the panel's MPP voltage.
Google is your friend.
Leo..
Correct me if I'm wrong.
Establishing some parameters:
1x Lipo Battery 3.7v @ 1000mAh.
1x Solar panel (DMM = 6v @ 580mAh).
1x Diode
Lets start and say the battery is at 3.0v of charge.
The solar panel is not on direct sunlight and it is producing 4.5v @ ~430mAh-ish, "open" voltage.
80% Percent of that is were the solar panel has it's MPP = 3.6v @ ~344mAh-ish.
3.6v minus a diode's draw lets say 3.4v, that would definitively push some juice into the battery, right?
Also this is connecting the solar panel and the battery directly, no charger controller in between.
A typical silicon diode has voltage drop of 0.6 to 0.7 V.
mAh = milliampere hours, a measure of battery capacity.
mA = current. The current supplied by a solar panel is proportional to the llght intensity, which if not in direct sunlight, won't be much.
I assume you use schottky diodes for low voltage solar panels.
4.5volt open solar voltage, -0.2v diode = 4.3volt on the battery.
That will even overcharge a full LiPo, if the panel would produce current at that voltage.
At the MPP of 3.4volt, only a flat battery would receive all the power from the solar.
But this is all theory.
I assume you have a DMM with voltage AND current.
Leo..
Were you dropped as a child? Just kidding, I love you.