Simple Solar Ni-MH Charger

tjetson

Senior Member
Is this circuit likely to work?


My reasoning for the resistor is to limit the panel's maximum of 400mA to about 100mA to charge the 1000mAH batteries., and the zener is to limit the voltage to around 4.3 volts, because I suspect that anything less than around 4 volts wouldn't charge the battery.

EDIT: Was going to post a different thread about the MAX712, but got my titles confused :(
 

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BeanieBots

Moderator
I have to say, it's painful to see such waste in a system which traditionally is all about conservation.

A solar panel behaves very much like a constant current source, so any series resistance will have little effect except at low light levels which is when you least need it. Same for the zener the way you have it.

If you really do want to use a panel that is too large for the battery, then make the resistor much smaller and use the zener to clamp the panel voltage. This will give you much better efficiency at lower light levels and only "dump" power at higher levels.
 

tjetson

Senior Member
If you really do want to use a panel that is too large for the battery, then make the resistor much smaller and use the zener to clamp the panel voltage. This will give you much better efficiency at lower light levels and only "dump" power at higher levels.
Would it be possible for you to post a circuit? I'm afraid that with my quite limited knowledge of these things (like my inefficient circuit) that I don't quite follow you. Thank you for your advice.
 

BeanieBots

Moderator
Simply remove the zener from your circuit.
Then put a zener directly across the panel to limit its voltage.
 
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tjetson

Senior Member
Simply remove the zener from your circuit.
Then put a zener directly across the panel to limit its voltage.
Is a 4.3 volt zener ok? Also, what value would you suggest for the resistor? I am using a solar cell with 415ma short and the batteries will most likely be 1000mAh.
 

BeanieBots

Moderator
I'd probably go a bit higher but 4v3 will work. (4v7 to 5v6 would be better)
A NiMh will be about 1.4v when full and still being charged at C/10 so your resistor value needs to be quite low. A lot depends on the SPECIFIC zener you have. They are quite sloppy on voltage and have a large tollerance so you would need to experiment. (and take into account resistance of your current meter). The extra ~0.6v lost in the diode takes it very close to the edge if not over.

I ask again, can your zener handle 400mA?
That's ~2W dissipation. Many are only good for a few 10's of mA.

IMHO you would be better off with a low quiescent current LDO 5v regulator feeding a diode + resistor.
 

tjetson

Senior Member
I ask again, can your zener handle 400mA?
That's ~2W dissipation. Many are only good for a few 10's of mA.

IMHO you would be better off with a low quiescent current LDO 5v regulator feeding a diode + resistor.
I don't own any zener diodes. I had merely looked at a few on Jaycar.com and found the 4.3v one. A LDO regulator is an interesting thought. If I use the LDO, what resistor values should I use?

And, would the LM2937 be acceptable?
http://www.futurlec.com/Linear/LM2937.shtml
 
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BeanieBots

Moderator
Not the most efficient but it should be OK.
Use a schottky in series with the battery.
If you have 5v -0.3v going into 3 X 1.4v then you have 0.5v across the resistor.
To limit that to 100mA will require 5 ohms.
I'd go with a 4R7 or 5R1 resistor as a close compromise.

It would be better if you could find a 6v LDO regulator but I don't know the number of a good one off the top of my head.
 

tjetson

Senior Member
Alright, thanks a lot, BeanieBots. Your contributions are most appreciated. Just one problem remains. What would happen if the solar cell didn't put out enough voltage for the LDO, say only 4 volts or so? Would it simply not work, or is it likely to cause damage?
 
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premelec

Senior Member
LDOs tend to get less stable at the dropout point - check your data sheet! However I don't think damage is liikely just some odd behavior...
 

tjetson

Senior Member
I ask because I had the idea of the Picaxe monitoring the solar voltage. When it drops too low, turn off the LDO. Do you think this sort of thing is necessary?
 

BeanieBots

Moderator
If the panel voltage is too low, the battery gets no charge.
That's all there is to it. Job done. (assuming you've fitted a diode in series).
No need to switch it off at low voltage.
 

tjetson

Senior Member
Sorry to bring up an old thread, but is it possible for a mod to rename this thread? I can't seem to. I was thinking for anyone who does a search in the future, the title of this thread doesn't make sense :eek: Also, using all of you guys' advice, would this circuit be correct?

It just occured to me that using this circuit, wouldn't the load run off the regulator's power supply, not the battery's? Meaning that whenever the LDO shut off, the voltage would drop from 5 volts down to 3.6. This might be a problem. I'd need a way to make sure the load always ran from the battery somehow.
 

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SAborn

Senior Member
Why have the diode after the LDO reg and cause a further drop, i would put it before the reg on the input so the rated output remained the same.
 

BeanieBots

Moderator
The diode is ESSENTIAL where it is, without it, the regulator will conduct in reverse when there is no solar energy. At best, it will drain the battery, at worst, it will kill the regulator.

The load is on the battery, it will run at battery voltage from the battery.
Your load will see a voltage between about 3.3v and 4.2v depending on the charge state.

For no reason other than esthetics and to maintain a continuous common ground line, I'd move the resistor to the top of the diagram to be directly in series with the diode. Makes no difference electrically, just looks/feels nicer.

Worth checking the regulator datasheet to ensure that decoupling caps are not REQUIRED to make the regulator work.
 

boriz

Senior Member
Seems like a crazy waste of power to me. A much cheaper/smaller solar panel will do the job better.
 

tjetson

Senior Member
Thank you, Hippy for the rename and BB for the continued help. While the Picaxe (20x2) probably won't mind that voltage difference 3.3 to 4.2, might other components mind? Here are my other components and thoughts.

http://www.futurlec.com/HP03D.shtml - maximum rating is 4 volts. A Schottky diode in series to drop voltage perhaps?

http://www.futurlec.com/HH10D.shtml - doesn't give a maximum voltage. Schottky also?

DS18B20 - will work at 3 to 5 volts

LEDs - I'll find some that will work

http://jaycar.com.au/products_uploaded/ZW3100(mod).pdf - datasheet says will work at 3 volts, haven't tested yet though

LED and Photodiode from serial mouse to use as anemometer - no idea about voltage :eek:
 

BeanieBots

Moderator
This does seem to be a bit like making a design fit a solution rather than solving a design issue with an appropriate solution.

Many of those devices will WORK at reduced/variable voltages but there consquences. The DS18B20 will function but it's accuracy is compromised.
The HP03D is a 3v device and will be destroyed by anything over 3.6v. Don't know what the effects of running it at lower voltages are but as it's a sensitive device, I'd not take any risks. Diode volt-drop on such a device is not a good idea. It really needs a regulated 3v or 3v3 supply.
Some LEDs such as white and/or blue won't work below around 3.6v, so you will be limited on colour and power.

An option would be to use a LDO 3v3 regulator after the batteries, increase the battery to 4 cells and use a 6v regulator as the charger but it's getting a little silly and as Boriz points out, very inefficient.

How about re-doing the problem statement.
How can I generate a constant 3v3 from a battery backed solar panel?

This can now be approached in a different way rather than solving one issue only to find it leaves you with a different one.

The first thing to seriously consider is a more appropriately sized battery for the chosen panel size. This would then leave charging as being as simple as a series diode. A 6v panel could charge a 5 cell NiMh pack which could then run a 5v regulator. Even a 4 volt pack would give a good 5v for most of it's discharge capacity. The 5v could then drive a 3v3 regulator.
It might (I've not fully checked) be possible to run everything from a 3v3 regulator and a just 3 cells.

Personally, if were doing this, I'd be looking at using a switching regulator.
It would be MUCH more efficient, can be trimmed to give any voltage you want and probably be cheaper and smaller than having to put in all the compromises you will need for a linear design.

Traditionally, switchers have always been scary complex things to build. With the abundance of custom chips now available this is no longer the case and many need little more than an inductor and a few caps. (care is still required on layout).

Have a think.
 

tjetson

Senior Member
A 6v panel could charge a 5 cell NiMh pack which could then run a 5v regulator. Personally, if were doing this, I'd be looking at using a switching regulator.
2 questions here
1. Is "switching" the same thing as "switch-mode" ?
2. You say a 6 volt cell can charge 5 batteries. The voltage of 5 batteries is 1.2*5 = 6. Doesn't the charger need to put more than that into a battery to charge it? Have I missed something here?

An additional query, you wrote
"Even a 4 volt pack would give a good 5v for most of it's discharge capacity. The 5v could then drive a 3v3 regulator."

Did you mean a 4 cell pack, or did you actually mean a 4 volt pack?
If I use 4 NI-MH batteries, wouldn't I still need that resistor in there to limit the current to C/10 to avoid battery overcharge?
 
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Dippy

Moderator
Read up on 'boost' and 'buck-boost' as well as 'buck' for switched mode designs.
Switchers , and 'switching' techniques can be hugely useful - but a tad more expensive.
 

BeanieBots

Moderator
Oops, yes, I meant 4 cell pack, not 4v pack.

In this context, switching = switchmode.
A "6v" solar panel will give about 9v open circuit.

You will still need a resistor (but a different value) to limit the current.
It would make life a lot easier if the capacitiy of your battery was a better match to the panel, then no current limit would be needed.

As suggested by Dippy, have a read about buck/boost (maybe not SEPIC) conversion.

There are also battery management chips available but where would be the fun in getting one of those?
 

tjetson

Senior Member
BeanieBots, you mentioned that it would be better to match a panel to batteries. Upon reflection, I think this is the best option. Which, if any, of these would be most suitable to charge either 1 or 2 (whichever works best) 1000mAH Ni-MH batteries?
http://futurlec.com/Solar_Cell.shtml
http://futurlec.com/Solar_Panel.shtml

To put it a different way, what would be the minimum voltage to charge a 1.2 v battery? I believe that anything above 1.2 will charge it, but how much above is a good value?
 
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BeanieBots

Moderator
Well, it's noramlly done the other way around.
Circuit determines required power.
Size panel at 10X circuit requirements.
Fit battery to suit.

But let's do it the other way for fun;)

Battery = 1000mAh
Maximum continuous charge current = C/10 = 100mA.
(so that's "peak current taken care of).

Fully charged terminal voltage of NiMh still charging at C/10 = 1.4v
So, "peak voltage" must be greater than or equal to 1.4 X number of cells (plus 0.6v for diode drop).

You can use a wildly over voltage panel if you want, the current will simply limit and waste power.

I've used the terms "peak current" and "peak voltage" in the same manner that the datasheets you linked to have. These are not to be confused with maximum, short-circuit or open-circuit values.
 
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lbenson

Senior Member
This does look like fun. Does this mean that three of the Futurlec 1.5V 135mA Solar Cells in series would be suitable (in theory) for charging three 1000mAh Ni-MH AA batteries, or close enough to OK? Would that setup be suitable for directly running a 14M or a 20X2 (with appropriate current needs, of course)?
 

BeanieBots

Moderator
Should be close enough.
The current is a little high but would only be problem if the panel is in full sunlight and giving out the max current AND the battery is full.
Even then, a good quality battery could probably cope. A poor quality one probably couldn't.

The ability to cope depends how much excess material is on the negative plate.
This requires space which ideally should be used for CAPACITY.
Hence, a 1000mAh AAA cell is unlikely to cope, but a 1000mAh AA cell should have sufficient plate for gas recombination.
 
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tjetson

Senior Member
So would two of the 1 volt 175 mA panels in series be acceptable to charge a single battery? Is 175 mA too high? An alternative might be the 2 volt 92mA panel further down the page. Or, if I used a 2000mAh battery, would the 175 do?
http://futurlec.com/Solar_Cell.shtml
 
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lbenson

Senior Member
Hoping this is close enough to the thread subject not to be too much of a hyjacking. Trying to get a grasp on some of the concepts.

Regarding: three of the Futurlec 1.5V 135mA Solar Cells in series for charging three 1000mAh Ni-MH AA batteries

BeanieBots: "The current is a little high but would only be problem if the panel is in full sunlight and giving out the max current AND the battery is full."

Can this problem be solved by using higher capacity NiMHs, for instance, 2000mAh AA eneloops? If the problem would then be undercharging, what issues would arise and over what timeframe? Undercharged batteries should not, per se, be a problem for my circuit.

I would like to have a battery + solar solution for running an 08M (mostly at 32kHz) with two DS18B20s and a 433mHz transmitter through the entire winter in Nova Scotia. Readings would be every 5 minutes or so, and transmissions only with changes of 2 degrees C. Full sun would be spotty and mornings only or afternoon only depending upon placement. The circuit runs for some months on non-rechargeable AAs.

Thanks, tjetson, for bringing BB's focus to bear on this issue, and thanks, BB, for your insights.
 

BeanieBots

Moderator
We are now falling into the murky depths of that "black art" of battery charging:eek:

My first hand knowledge based on many years of experience of battery charging only applies to older technologies. ENELOOP have only been around for a few years (at least in my hands) so my knowledge of them is limited.
So, what I'm about to say is applicable to NiCd and early NiMh technology but might also apply to ENELOOP technology.

Once fully charged, any energy which is subsequently put into a battery does nothing except break down the electrolyte into gas. If left unchecked, this would eventually build up enough pressure to pop the release valve (or explode the battery) and discharge electrolyte into the atmosphere thus reducing the capacity of the battery until it eventually dries out and becomes useless. To prevent this, the manufacturers make the negative electrode bigger than it needs to be and a few extra chamicals that recombine the gas back into the electrolyte. However, this process has a limit and is usually only capable of working at currents under C/10.

The larger the battery capacity for the same physical size means less free space for excess negative electrode. So, larger capacity batteries can only handle SMALLER charge and over charge currents. (the opposite of what many people assume).

Most manufacturers state that an indefinite charge current of C/10 will cause no damage but this is worth checking for the specific battery you have.

In applications such memory or clock battery backup where the battery is nearly always on charge, in the past I have recommended much lower charge currents. However, with recent batteries where more and more has been crammed in to obtain higher capacities, it has been done at the expense of insulation layer between the electrodes. When charged very slowly, the crystals grow larger than at higher currents. These (known as dendrites) can grow large enough to puncture the insulation and cause internal shorts at low charge currents.

So, to put it simply, there is no simple answer. That's why there are a whole host of very clever battery charger chips out there and many different oppinions on how it should be done. Bottom line, it depends on your specific battery.

As you have already looked at ENELOOP cells, I'm guessing that you are also aware of the (relatively) fast self discharge rate of NiMh. Standard NiMh will self discharge at around C/200 to C/300. ENELOOP are several orders of magintude better. So, there are now two factors which determine MAXIMUM battery size and one which determines MINIMUM battery size for a simple constant current(ish) setup.

I've never tested ENELOOPs at long term constant current.
However, I'd bet that they are safe anywhere between C/10 and C/20.

Go on, make an intelligent charger. You know you want to!
 

lbenson

Senior Member
>Go on, make an intelligent charger. You know you want to!

Actually, I want to make a charger which is as unintelligent as possible for being able to do the job. I only mentioned the eneloop because the headline (certainly not a datasheet) on Amazon called them NiMH.

Thank you for your explanation. I see that more research is called for.
 

BeanieBots

Moderator
I'm quite sure a simple panel -> diode -> battery arrangement will give many years of faultless service as long as the current is >C/20 and <C/10.

Eneloop IS the latest generation of NiMh, and from my observations so far, very reliable and robust too. My only word of caution is be very wary of AA cells claiming capacities > 2Ah. They are VERY fragile.
 

lbenson

Senior Member
>I'm quite sure a simple panel -> diode -> battery arrangement will give many years of faultless service as long as the current is >C/20 and <C/10.

OK, I'm going in.
 

BeanieBots

Moderator
Please report back on how it goes.
I'd be paricularly intrested in how the Eneloops fair after a few months on trickle charge over a long period. Once they have cycled about 5C of energy, any "oddities" should become apparent.
 

lbenson

Senior Member
Let me formulate what I think I have learned, to see if I am in the ball park.

NiMH batteries charged at the appropriate voltage from a solar panel (approximately the voltage of the batteries plus a diode drop) will charge if the current is at least C/20, and will not overcharge if the current is no more than C/10. No more is needed to do this than a Schottky diode between plus of the panel and plus of the batteries, with the 0V connected. The load is taken off the batteries as in a non-charging circuit.

So for 3 eneloop AA batteries with a capacity of 2000mAh and a total nominal voltage of 3.6 volts, the output from 3 1.5V, 135mA solar cells fed through a BAT85 diode (30V, 200mA) should charge safely at the highest output (C/20=100 < 135mAh < C/10=200).

Is the issue of charging the batteries in series (instead of individually, as in a charger), not critical?

The load would typically be in the microamp range, rising to ten or tens of milliamps for several seconds every five minutes, and again perhaps every 30 minutes for a 433mHz transmission. This load is sufficient to draw down standard AAs in about 4-5 months or less.
 
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BeanieBots

Moderator
That's about it.
The panel voltage needs a little care.
Make sure it's the voltage specified at peak power that is >1.5v per cell.
If it is open circuit voltage, then the current at 1.5v would be almost zero.
Too much voltage from the panel is not a problem. A solar panel 'looks like' a constant current source, so excess voltage is not important but there must be enough to do the charging.

Charging in series is fine, potentially better than individual.

Sounds like you're set to go:)
 
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