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I am on a hiking trip an I forgot to pack the recharging device for my penlite batteries. I have enough power with me thanks to my solar panel. I can charge every device that is charging with usb. My garmin etrex vista however uses old fashion penlite (AA) batteries and I forgot to bring my AA charger. In the village where I am, the local shop doesn't sell AA batteries :(. I can use my gps, but only if it is connected to my solar panel, which doesn't make it practical when hiking through the mountains.

Since I do have energy with me in the form of a portable solar panel (10W), I am wondering if there is a Macgyver way to charge AA batteries.

  • We, actually you more than us, need to know the specs of both the solar panel output and the battery. Nevertheless, this can be done, but it's not risk-free. – JoErNanO Nov 4 '14 at 13:18
  • @JoErNano I have added my power source to the OP and I am talking about your average rechargeable AA (1.5V) batteries – user141 Nov 4 '14 at 13:32
  • We actually need the mAh rating of the batteries. i.e. how much current-time the battery provides when fully charged. – JoErNanO Nov 4 '14 at 13:36
  • Ah, that is going to be tricky since I have different AA batteries I am using together at the same time, ranging from 900 mAh to 2700 mAh. I guess based on your comment I should not do that? – user141 Nov 4 '14 at 13:40
  • No no it's just that given a constant setup you will need to modify your charge-time accordingly. – JoErNanO Nov 4 '14 at 13:49
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+100

Safety Disclaimer

There are various way to do this, none of which are risk-free. Any damage to you, your personal property, nature, the Universe is entirely your fault.

You Cannot Fully Charge Your Device

The bottom line however is that you should avoid fully-charging the battery. Instead you should aim for a short charge, giving you enough juice to continue operating your device for a limited time period. In other words the solution outlined below is a temporary fix which is aimed at giving you time to get to the nearest (safe) power supply.

Necessary Equipment

You will need:

  1. A power supply of some sort
  2. Something to use as a resistance to reduce the current flow between your charger and the battery
    • Light bulbs are perfect for this kind of thing
  3. Protective gear of any kind
  4. Loose wiring to connect things together
  5. Basic knowledge of electronics.

Procedure

There is a guide I found here which explains both the underlying method and the safety risks involved. To summarise:

  1. Check your power supply output - both Voltage Vs and Current Is
  2. Check your battery specs - Voltage rating Vb and maximum charge in mAh Ib
  3. Compute the potential difference ΔV = Vs - Vb between your supply and battery
  4. Once you have this data you use Ohm's law V = Ic * R to compute the resistance you need to obtain enough current to charge your battery, but low enough for this procedure to be safe Ic << Ib

    • This computation would work out to something like this: Ic = ΔV / R where R = Sum(r) is the total resistance provided by whatever you are using as a resistor r
    • If you are using bulbs as resistors you would have to measure their resistance, either empirically like the guy in the link I posted did, or mathematically using again Ohm's law (see this as an example calculation)
    • The resistors must be placed in series so that their total resistance is equal to the sum of the individual resistances
  5. At this point you wire the resistances in series

  6. Wear protective clothing: gloves and glasses are a minimum!
  7. You then connect the negative pole of the power supply S to the negative pole of the battery B: -S -------- -B
  8. To charge you now have to connect the positive pole of the power supply S to the bulbs, to the positive pole of the battery B: +S ---- Sum(r) ---- +B

How do you know if this is working? The battery temperature will increase.

Based on your previous computation, since you know the mAh rating of the battery you know how long you have to apply current Ic for it to be fully charged. Also know that the typical charge profile of a battery is not linear, so applying current for half the time required for a full charge will not half-charge your battery. It will most probably charge it more.

3
  1. Check whether your batteries are NiMh (or NiCd) or Lithium. Don't mix them, as they have different charging and discharging requirements. I assume you're not talking alkalines, and that you know you can't charge them :).

  2. Use the 5 V output of the panel, anything higher will risk to blow up your batteries. As it's probably USB, it should be also limited (or it should be possible to limit) to 500 mA, which is actually still pretty high for Lithium batteries, and even more so for Nickel-based batteries.

  3. Put at least 3 batteries in series (connecting the plus of one to the minus of another and so forth...). This will sum up their nominal voltage to about 4.5 V. For your safety, use only similar batteries (in capacity and chemistry) and possibly with the same charge level.

  4. A resistance, as JoErNanO proposes, can be a good idea to limit the current. However, it won't prevent the batteries from overvoltage, and this can be quite dangerous so every once in a while check their charge level and if it's more than - say - 50-60%, stop charging them and use them. A better limiter for the voltage would be a diode, as it drops an approximately constant voltage. Try to also limit the current to less than 0.5 C, possibly less. 1C means that your battery fully charges in an hour (900 mAh -> 900mA).

You can also refer to the Nickel-metal-hydride guide, and Lithium guide at Battery University, they're quite complete and reliable. There is also a guide for charging from a USB port that should serve your purpose.

  • +1 For a good answer! What is C in your last numbered point? – JoErNanO Nov 6 '14 at 10:44
  • @JoErNanO it stands for Coulomb, or Ampère times seconds. But the practical meaning is the proportion of the charge/discharge current to the battery capacity. 1C = 1 hour discharge, 0.5C = 2 hours discharge, and so forth. – clabacchio Nov 6 '14 at 10:47
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    Ok that's what I thought. If that's the case you might want to rephrase limit the current to limit the charge since Coulombs express the battery charge. ;) – JoErNanO Nov 6 '14 at 10:51
  • @JoErNanO well not actually, since it's not charge but charging rate. I know it's confusing (for me too), I'll post a link that explains it better than me. – clabacchio Nov 6 '14 at 10:53
  • @clabacchio Well excuse me, but you can't measure charging rate in C. Charging rate is in A = C s. Current is in A. Your answer looks fine to me, just this needs to be corrected. – yo' Nov 6 '14 at 18:09
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Unfortunately you can't really charge batteries without a blocking diode which acts as a one way valve for the electricity. Judging by the fact that you didn't have access to AA batteries, you probably don't have a spare electronics parts store. Did you try asking for mignon or R6s (regional names for AAs)?

  • @pnuts to regulate the direction of energy. – Stephen P. Nov 1 '14 at 1:32
  • What 'charger'? Isnt the whole point of the question that she doesn't have a charger? And who knows what the voltage produced by the solar cell is, or how variable? It may well be possible for the voltage differential to reverse. – imoatama Nov 2 '14 at 16:58

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