I want to charge one 3.7 LiIon 18650 battery using a photovoltaic panel.
The panel is rated 6V.
The project has two stages:
a) to charge the battery - no matter how
b) to charge in sunlight and light one white LED during dark time
For the a) I think a 0,5W 4,7V DO35 Zener Diode coupled directly to the panel may suffice.
For the b) stage I need an automatic charger/commuter
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For the b) stage I found only a couple of schematics on internet - 2, attached.
Both are stated to work on 2V panel - which I cant understand, because I could not find a 2V panel, but only 1V, 6V and above). I know I will not get 6V but only 3-4 V under regular daylight and using not-oriented panel, but still a 2V looks excessively pessimistic.
Question is: although these 2 circuits are stated to work on 2V solar panel - is it safe for the 3.7 battery to charge from the 6V panel?
Thank you very much for your kind opinion and for any hint to an alternative simpler schematics!
NiMH is pretty easy to use indeed. Just supply it a voltage with limited current (resistor) and you're done. Can also sustain trickle charging so no need to disconnect etc.
One NiMH is 1.2 V x 2 = 2.4V. Then it can be charged by the circuits I found - is the provided voltage is still 3 V when I connect it to 6 V panel (panel working voltage will be probably around 4V).
If this 2.4 V(in practice I think it will be some 2 V) would b e sufficient to open an white LED, that is to be checked.
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I cant figure out: are those circuits regulated (or limited) to 3 V, or their output will increase with the input voltage? Thank you!
White LEDs tend to have a forward voltage of about 3V. So you'd better take a set of three NiMH batteries, then you have a voltage of about 4.5V (fully charged) dropping to about 3.6V when nearly empty. That should be good enough, and your solar panel can still charge them.
3 x 1.2V = 3.6V (NiMH) - according to the labels of the units I have in hand (perhaps they is another type of 1.5 V, I will have a look in the store).
In any case, I can connect the 3 units directly to the panel - no circuit at all. Even the panel will go as high as 6 V (which I doubt), then the NiMH old sheep will be OK (I hope, communication by smoke is another project :-)).
I measured my 3.7 LiIon units by multimeter and it was 4.2V when fully charged. I did not know it is normal
For this batteries you need a voltage regulator with output 4.2V and current limiting resistor approximately 50 ohm.
4.2V is normal.
For 3 batteries 1,2V you can use also above mentioed voltage regulator
Demo only. So I think time is OK. I guess the real current will be around 200 mA and my unit are 3 x 800 mAh = 2400 mAh. Then charging time (real conditions) might be 12 hrs. Assuming fully discharged of course.
If I use a Zener regulator that will take some current as well (maybe 20 mA? - there is a quite fantastic willy-billy method to label the Zeners so I am reluctant to go for a Data Sheet).
I also assume the variable current will destroy the NiMh within some 200 charging cycles (read 200 days) but this is very good for the kids demo project).
falexandru:
I also assume the variable current will destroy
What you mean.
When batteries are discharged on beginning of charging current will be max 200 mA - that how much can deliver your charger ? - then will drop to 0 mA when batteries voltage reach 5V.
I am talking about voltage depression and smaller "slices" when the panel only partially charges the NiMH battery.
However, at this cost and for demo only, that is not going to be problem. I assume the units will last for at least for few months.
I am almost decided to go for NiMH instead of LiIon.
It may be a nightmare to find a Zener on line because of labeling, so I will go to "mortar-and-brick" store.
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Next step is to see whether my Savonius turbine can go the same way to charge the NiMH. The turbine's output will be perhaps a 2V average, which means that I have to push up the voltage and then to regulate it via the same Zener regulator.
falexandru:
Demo only. So I think time is OK. I guess the real current will be around 200 mA and my unit are 3 x 800 mAh = 2400 mAh. Then charging time (real conditions) might be 12 hrs. Assuming fully discharged of course.
that would be if you place your batteries in parallel, but you'll have them in series so total capacity is still 800 mAh and your charging time at 200 mA about 4 hours.
Right, my mistake. But is very counter-intuitive, I must admit. I have to find a simple way to explain this (first to myself :-)).
Question is whether units for a total of 4.5 V (or a 3 x1.2V = 3.6 V) can be charged by a panel which is rated 6 V, but can easily drop to 4V when in shadow. Maybe the prototype will show how (and whether) it works.
