Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Unusual battery charging system

  1. Jul 18, 2004 #1
    I'm going to be going into rural Nepal for 8 months starting this winter. I would like to be able to take pictures for this time. However, there will be no electricity. I've got a nikon coolpix 4500, which uses a Li-ion rechargeable battery (#EN-EL1) which is 7.4V / 680mA. The proprietary charger (#MH-53) takes AC at 100-240V and outputs DC 8.4V / 600mA. The MH-53 supplies only a slightly greater power than the EN-EL1 (5.04W vs. 5.032W)

    Anyway, I'd like to design a design a charger that acts as a substitute for the EN-EL1. I plan on doing this with a solar panel, which will probably end up being 7-10W, outputting, ideally at the .6A that the charger would. However, I suspect that the panel will end up being quite a variable source, and, as I don't want to damage the battery, I'll need to stabilize it to outputting at 8.4V/.6A.

    Now, I've never done this sort of thing before, but I've taken an E&M class recently and would be confidant putting anything that's relatively simple together. Having done some research already I think that if the panel put off a constant flow at .6A, then I could use something like the LM338 in conjunction with a few resistors (and perhaps a potentiometer, in case I'd like to re-use the system for something else) to regulate the voltage to the 8.4V. However, this will probably not be the case. http://www.national.com/ds/LM/LM138.pdf details a 'adjustable current regulator' (on page 12) that looks to be something, but I don't know. (help?)

    Ultimately, is there a self-contained object that I can build that, having a variable source put into it, outputs a pre-set current at a pre-set voltage? Or is there no way to get around not having a constant current to start with?

    Thanks for your help.
     
  2. jcsd
  3. Jul 18, 2004 #2

    Cliff_J

    User Avatar
    Science Advisor

    My suggestion would be to model the battery as a voltage source and not a current source. If the camera is in a low power standby mode that doesn't draw much current, a constant current supply will ramp up the voltage.....could be diasterous. A battery will only offer so much voltage potential and will drop in voltage as you increase the current draw. A battery will offer as much or little current as is allowed to flow in a circuit. A voltage source will also only offer as much current as needed except it will likely be more stable than a battery.

    Something like this would be more suitable:
    http://cache.national.com/ds/LM/LM117.pdf

    And then to offer protection to the camera, you could put a 800mA inline fuse on the output of the regulator and either a set of 12 1N4001 diodes in series or even better a 5W 8.2V zener diode acorss the output to clamp any voltage that might stray too high and protect your camera. Its likely ok to 8.4V but pushing it anymore....

    Your better option would be to build a solar battery re-charger. If something were to go wrong, you're only out a battery.

    Some solar cells are 40V or so in bright sunlight. A linear regulator will get hot dropping that much voltage with your current demands, proper heatsinking would be needed and the life of the regulator will be shortened. A solar cell that generates a voltage closer to your needs (like 12V-16V in bright sunlight) would work better but have a narrow saftey margin for times with a lack of direct sunlight making it usable for a much shorter time period in the day.

    Sizing will be likely you're biggest compromise, and using the solar cell as a battery charger is much better because you will be able to use the best part of the day to charge and use that energy as you wish at any other time. Biggest trick here would be to match the factory battery charger's performance over a long enough time during a typical day to get the battery(s) charged. Also,once the current flowing into the battery reaches a preset limit to disconnect and stop charging the battery to avoid damage from overcharging.

    On overcast days or at any time too far from noon, if the power output is too low from the solar cell the voltage AND current could be too low to be usable. Again, sizing and testing will offer answers (take into account the sun's angle at the time of year you'll be there) because 1V and 1mA is still only .001W and not going to be of much use. If this is really important, an oversized cell would give you more flexibility and would simply charge the battery faster. More to transport, but better than not being able to charge the battery during a week of cloudy weather.

    Lastly, I hope your voltage and current measurements were taken under load. Otherwise, they are only approximations that don't mean much and you'll want to measure again with all the needed parts in action.

    Cliff
     
  4. Jul 18, 2004 #3

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your charger is not going to be able to maintain 0.6A throughout the charge by the very nature of how batteries work. That's most likely its peak current capability.

    Note that your camera's Li-Ion system already contains some quite sophisticated power-management circuitry. Li-Ion batteries are in fact quite hazardous and require rather complex control circuitry. Your camera already has all the difficult parts done. All you need to do is supply something around 8.4V, with a fuse to limit its current to ~0.6A. Cliff has given you many good suggestions -- a 9V zener is pretty much all you need.

    I would suggest bringing spares of all the power-handling parts of the circuit, like the zener and regulator.

    BTW, I think you're going to have a bigger problem in storing all those digital pics -- flash memory is expensive, and hard drives require lots of power. It would probably make more sense to buy a fully-manual SLR camera and bring along a truckload of film with you.

    - Warren
     
  5. Jul 18, 2004 #4
    Thanks for your ideas. Your second one (the solar battery re-charger) is the one that I'm actually trying to accomplish. I guess I was a little vague. The problem that I see is that I will need to regulate the voltage and current coming from the solar panel so that it matches that which the household charger would put out (DC .6A at 8.4V). I know that I should be able to regulate the voltage (the LM117 looks good). I don't know, however, if I can regulate the current and voltage at the same time to get them down (or up) to what the charger would give out. I think the batteries the camera uses are fairly finicky, and I want to mimic the ideal conditions as closely as possible.

    Oh, and I did not actually take the measurements for the voltage and current--I checked the spec in the manual. Now that you mention it though, I will measure them tomorrow under load and see if the numbers compare.
     
  6. Jul 18, 2004 #5

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You cannot regulate both voltage AND current. You can only regulate one at a time; the other will be as required by Ohm's law.

    If you can produce a supply that will provide 8-9V with a maximum of 0.6A, you won't have any problems at all.

    - Warren
     
  7. Jul 18, 2004 #6
    Yes, I figured one could not regulate both directly, but I thought it might be possible with an intermediate storage step. However, if charging with any current <.6A will work to recharge the battery (though I assume that, as the change in V from the charger to the battery lowers, then the current should as well), then it's not required.

    So the conclusion is that the best thing to do is try to find a panel that will produce >10V at a low current and then build a device with a fuse (and voltage regulator to prevent going over .6A) which will then feed to the battery? I'll probably need to find some sort of plans for the circuit, but it sounds feasible.

    Oh, and about the storage, the camera accepts microdrives which go up to 2 (or 4) gigs. The only problem with them is that they are quite slow. I think I'll go with 2-3 256meg flash (which I already have) and 1 or two of the microdrives. It all will be pretty pricey, but with the cost of developing the equivalent number of pictures on film, I don't know. Also, I could probably re-sell the microdrives if I don't need them anymore, assuming they still work by the end.
     
  8. Jul 18, 2004 #7

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Any current at all will charge the battery. The larger the current, the faster will be the charge. You don't want to exceed 0.6A for safety reasons. As I've already said, your camera already contains sophisticated charging circuitry.
    A voltage regulator limits voltage, not current. It does not explicitly limit current. The circuitry is really dirt simple. Look at the application notes in the back of any voltage regulator datasheet for sample circuits.

    - Warren
     
  9. Jul 19, 2004 #8

    GENIERE

    User Avatar
    Science Advisor

    The best bet is not to charge the small capacity batteries directly from the solar panel. I suggest you purchase a 12-volt, lead-acid gel cell and a solar panel as in THIS link.. A large capacity lead-acid cell will easily handle the solar panel’s output even at high noon. In the evening use the gel cell and appropriate current regulator(s) to charge the small capacity cells.

    If you also buy THIS, you can use your existing charger and also power other AC devices. Judge the capacity of the gel cell and solar panel by what you might wish to power.

    I’d package it in a suitably sized rigid suitcase, flip the lid to expose the solar panel.
     
    Last edited: Jul 19, 2004
  10. Jul 19, 2004 #9

    Cliff_J

    User Avatar
    Science Advisor

    While on this, your charging current will likely be less than 600mA by a long shot. That's a bunch of current for small battery, it could get warm or even HOT! I've read that 1/6th the current in the Ah rating is a good fast charging rate, but I know little about it so this is to be taken as such.

    But if you're output is only 8.4V the battery (if it behaves like a lead acid) will only accept so much current depending on how depleted it is. Hence why I mentioned measuring. Here's a thought on doing it.

    Get two DMMs. One in series for current, one in parallel for voltage. Use the camera to drain the battery about as far as you would expect. Monitor the current/voltage as the charger charges the battery like every minute so you have an idea what the manf worked out as the best charging curve. Likely the voltage is steady and current is varied (automatically) by the internal resistance of the battery, but a more sophisticated circuit would not be surprising to me.

    Oh, and micro-drives in the past sometimes had issues with altitude - I'd read of a couple mountain climbers who found they didn't work at the peak but fine again near ground level. No idea of the validity again, take it as such.

    Also the battery consumption is something to consider, on some cameras the microdrives can severely limit the life of a battery charge.

    Cliff
     
  11. Jul 19, 2004 #10

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    For those who are interested, here is some information on how Li-Ion batteries must be charged:

    http://www.powerstream.com/li.htm

    Here is a Li-Ion charge controller (National Semi LM3420):

    http://cache.national.com/ds/LM/LM3420.pdf

    You read the applications section in the back for a lot of good info on how Li-Ion chargers are constructed. Your camera probably has a circuit that's very similar to one of the sample circuits in this datasheet.

    None of this really matters for your purposes, though Bucephalus -- if you create a regulated solar panel that mimics the output of your camera's power brick, you'll do fine.

    - Warren
     
  12. Jul 19, 2004 #11

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    BTW Cliff_J,

    Li-Ion batteries can be charged as quickly as 0.7 times their capacity. In other words, a 1.2 A-hr battery could be charged at 0.6 A. Bucephalus didn't tell us the battery's capacity, but 0.6 A peak charging current is not unreasonable.

    - Warren
     
  13. Jul 19, 2004 #12

    Cliff_J

    User Avatar
    Science Advisor

    Learn at least one new thing each day....

    Thanks for the info about LI charging.
    Cliff
     
  14. Jul 21, 2004 #13

    megashawn

    User Avatar
    Science Advisor

  15. Jul 21, 2004 #14
    Build a shaker. Have a tube with a coil in it. Put a magnet in the center. On the end of the tube, put magents with their poles the same as the magnet in the middle. North------south South------North North------South
    FIXED FREE FIXED
    COIL COIL COIL
    So the free coil, when shaked will ossilate through the coil, and then be repelled by the fixed magnets. Have this charge a capacitor. Then discharge the capacitor through the camera, when you want to take a picture.
     
  16. Sep 4, 2004 #15
    http://www.discovercircuits.com/P/pwr-dctodc.htm

    Bunch of dc-to-dc circuits here.

    From the PV panel run a 1:1 transformer chopper inverter. A quick and dirty version would be nothing but a doorbell buzzer in series with a 1:1 transformer primary winding.
    Take back emf off the secondary winding using a single diode.

    BACK EMF ONLY from the secondary winding is used to charge
    a large capacitance through a single diode. The more capacitance
    the better.

    If you just let it run, the back emf would take the capacitor voltage
    up way too high for your purposes. So, you need a simple, voltage controlled
    dump load to keep the capacitor from overcharging. No complex regulation is needed.

    Get handfull of 9v zener diodes and connect them all in parallel. Then connect a bunch of 6v flashlight bulbs in parallel. Now wire the zener cluster in series with the bulb cluster across the capacitor terminals.

    Whenever the capacitor voltage starts to go above 9v, the excess voltage breaks over the zener diodes and is dissipated through the bulbs (load resistance).
    The more of these zener/bulb sets you add in parallel, the more
    stable the capacitor voltage will be at 9v and the cooler all the diodes
    will remain (long life). You could just use resistors instead of bulbs. But
    flashlight bulbs can be found in almost any store, world-wide.
    Use what's convenient and you have built-in survivability.

    You can now take a pretty clean and steady 9v directly from the capacitor
    to replace the regular adaptor output.

    Just make sure you have enough zeners and bulbs in parallel to handle the
    peak output of the PV panel when no other loading (battery charging) is present.
    When you have enough in parallel, none of them will light up at peak PV output.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Unusual battery charging system
  1. Battery Charging (Replies: 3)

  2. Charging Batteries (Replies: 3)

  3. Charging a battery? (Replies: 4)

  4. Charge battery (Replies: 10)

Loading...