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I'm working on a project where I'm going to charge a battery using a solar cell. The cell is capable of outputting a maximum of 4 A, but only 0.5 volts (2W).

My question is, how can I charge the battery with only one cell? I understand I'll have to step up the voltage in some way, but I'm not sure how. I read that I can use boost converters, but don't these need a steady voltage for them to work properly? The solar cell probably won't be outputting its max voltage output all the time.

Ideally, I can regulate the voltage output of the cell to something like 2.4 volts (a single battery is rated at 1.2 volts).

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  • \$\begingroup\$ What kind of battery are you wanting to charge? When you say "a single battery is rated at 1.2 volts" does that mean that is the voltage that the battery needs to be charging? \$\endgroup\$ – angelatlarge Apr 29 '13 at 7:09
  • \$\begingroup\$ Yes. The battery is a rechargeable AA battery. \$\endgroup\$ – Malfunction Apr 29 '13 at 7:28
  • \$\begingroup\$ Charging from such low voltage sources I believe falls under "energy harvesting". One such IC here: maximintegrated.com/datasheet/index.mvp/id/7183 \$\endgroup\$ – Mark B Apr 29 '13 at 8:02
  • \$\begingroup\$ That sounds like quite a large single cell; is there a reason why you have to use only one cell, rather than a more usual arrangement with a stack of 4 or 5? \$\endgroup\$ – pjc50 Apr 29 '13 at 8:39
  • \$\begingroup\$ Search on "Joule Thief" for some example circuits. They will need some interesting re-engineering to handle 4 amps though! \$\endgroup\$ – Brian Drummond Apr 29 '13 at 9:52
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There are a few boost controllers that run down that low, but in general it's better to have a higher voltage to start with. 0.5V at full tilt is really low, presumably it's designed to be stacked to get a usable voltage.

Here's an example of one of the very low input voltage boost converters. Of course this one doesn't do much current, so it doesn't really fit the bill...

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  • \$\begingroup\$ What exactly do you mean by "doesn't do much current"? I thought a dc-dc converter works by stepping up the voltage at the expense of the current. Wouldn't a high current rating for the cell be better in this case? \$\endgroup\$ – Malfunction Apr 29 '13 at 7:25
  • \$\begingroup\$ That particular device is limited to 400ma. In general, high current = high resistive losses, hence the distribution of power at extremely high voltage. \$\endgroup\$ – pjc50 Apr 29 '13 at 8:38
  • \$\begingroup\$ Oh, that makes sense. \$\endgroup\$ – Malfunction Apr 29 '13 at 13:16

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