Solar power, battery charging, and inverters

[Stavale8099]

Hello,
I'm trying to teach myself ...still learning...but I'm stumbling.

I have a 15W solar panel whose output is 5V and 2.1A. I wish to charge a 12V battery (98Wh/8800mAh), but am unsure how to do this calculation as approximate time it would take (Assume sunny days - I know it's an inefficient system).
Second, I have a 110v inverter that plugs into this battery. I believe it's rated at 0.9A, 100W max. I wish to power (or at least figure out limitations) a 165W self-ballasted mercury vapor bulb. Is this possible? If so, how fast would I drain the battery? The bulb is rated at 1.63Arms.
Thank you so much
Eric

 

[anorlunda]

The calculations are easier than making it work.

5V x 2.1A = 10.5 watts
10.5 watts times X hours = 98 Wh ; solve for X=98/10.5 = 9.33 hours.

But that assumes 100% efficiency. 50% might be more realistic so double that time to about 18 hours. I usually figure that my solar panels produce for 5 hours per day, so 3.5 sunny days. If it's cloudy, double that again to 7 days.

The same kind of calculations apply to the bulb. 165watts is 165Wh per hour. So 98/165=0.6 hours to drain the battery. However, it is probably not practical to drain the battery more than 25%, so 0.6/4 = 015 hours or about 9 minutes. If the inverter is not 100% efficient, maybe 7 or 8 minutes.

But before that calculation, you must boost the voltage to something more than 12V. Current will not flow from 5V to 12V. You can buy DC to DC converter chips. Do you think you could use a circuit board or a breadboard to wire up an integrated circuit chip?

If you need a light that you can use every night and recharge during the day using solar power, you must use an LED light that uses very much less than 165 watts. I think you would have to start by telling us how many lumens of light do you need, for how many hours per day.

 

[Stavale8099]

Thanks for the reply...
So, the current won't flow from 5V to 12V meaning that unless I have a converter chip then I really won't be able to charge this battery. There is a solar panel (more expensive of course - was looking to get away with using the smaller one) that was recommended for this battery, and its output is 14-22V, up to 1.3A (20W max). Now I have some more clarity as to how this works...I guess that would be, at 50% efficiency, ~10h of charge time in direct sunlight for this battery.

Regarding the light - the 165W mercury vapor bulb was a shot in the dark; I'll mainly be using it to recharge my GPS, phone, and perhaps my computer. I'm doing a biodiversity study in Panama, so was looking for the best way to have a very bright, wide color range bulb to attract insects and the like (merc. vap a great but inefficient way to go). I can get away with using a black light DC 15W bulb (I guess this wouldbe 1.5h of run time) or some UV LED bulbs (~40mW per bulb) for much better efficiency, if I'm reading your calculations right...I guess I could daisy-chain the batteries but now I'm getting ridiculous with the weights and cost.

Thank you very much!

 

[CWatters]

There are modules like this that can be adjusted to convert 5V to the 13.8V approx needed to charge the battery.

http://www.banggood.com/DC-DC-Adjustable-Step-Up-Power-Module-p-971005.html

If your panel is only 15w and you want to power 165w load then the charging time will be at least 165/15 times as long as the discharge time. Eg if you want to run the bulb for 10 mins you will have to charge the battery for at least 10*165/15 mins at the very least and probably 25% longer to allow for losses in the battery.

 

[Stavale8099]

There are modules like this that can be adjusted to convert 5V to the 13.8V approx needed to charge the battery.

http://www.banggood.com/DC-DC-Adjustable-Step-Up-Power-Module-p-971005.html

If your panel is only 15w and you want to power 165w load then the charging time will be at least 165/15 times as long as the discharge time. Eg if you want to run the bulb for 10 mins you will have to charge the battery for at least 10*165/15 mins at the very least and probably 25% longer to allow for losses in the battery.

Great - thank you very much for this.