On Sat, Jun 4, 2011 at 4:34 PM, Oli Glaser wrote: > I would still run the numbers for your estimated current use. > If you use it 5 minutes a day at, say 10mA, then: 0.010*(5/60)*365 =3D > ~300mAh per year > 300mAh/(24*365) =3D ~30uA average constant current. That actually sounds pretty good. I'm estimating 50mA consumption for a maximum of 6 minutes every other day. So let's say 200 days * (6/60) of an hour. So that's 20 hours per year. 50m= A * 20h =3D 1000mAh. There are 8760 hours in a year (Wolfram Alpha), so 1000mAh/8760h =3D 0.114m= A constant current. BUT, realistically, how many lit hours are there in a day? Let's say 6, so that's 1/4 of a day. So that's 2190 lit hours per year. The solar cell migh= t produce 0.1% of rated output (according to Russell), so let's say 20mA*0.00= 1 =3D 0.02mA. 0.02mA * 2190 =3D 43.8mAh. So about 5% of required energy (minimum) to keep the battery charged at 0.1= % solar panel rated output. Will provide 50% of required energy at 1% rated output. So really, all I need is 5% rated output to keep the battery charged, for a few hours a day. That should be more than doable if the unit is placed unde= r a desk lamp, for example, which is perfectly reasonable. Or put on the kitchen table, or near a window where it will get (indirect) daylight. Thanks, Oli!! You put it into perspective. One issue though is that I think I'd need to hook up the panel directly to the battery. Since the panel is rated at 4 volts, and can potentially produce more than 4, I'd probably need to put a 4V or 4.096V zener to prevent more than 4V going into the battery). But then theres the question (sorry I couldn't find a datasheet for that panel) - what will the solar cell voltage be at 1% rated output? Enough to charge the battery? Or would I need to hook it up to the switching regulator? --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .