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[on:]

hi everyone i made a solar charger using this circuit:http://electroschematics.com/4746/solar-charger-circuit/
all worked out well but the problem is im using two 5v,30ma solar panels which i connected in parallel ,is there any way i can increase the current output to about 300ma,right now its giving 70ma,is there a way to build a circuit that would increase current output?
thanks

[Reply #1 on:]

Answer in a word:  "no"

You can get current flow only within the abilities of your "source".  You can't magically multiply unavailable current.  In this case, your source is limited to being able to produce 30MA... so, yes, if you add two in parallel, you get twice the current capacity.

Your solar cell is sort of just like a battery.  You can get 8000 mAh (max) from a D cell battery, but you can only get 300 mAH max from a AAAA cell.  You are sort of asking... how can I get 2 AAAA batteries to give the the same current as a D cell battery...  which... is not really in the realm of "possible".   

Well, nothing is impossible until you try.  It's nice that you tried and now you know the limits of your design.  You also know how to solve it.  You just need to buy 8 more solar panels.

[Reply #2 on:]

It's not clear, what kind of battery you are trying to charge?
There point is,  If there is a big difference between solar panel output voltage and maximum
voltage (fully charged) of the battery, you  can built more "efficient" charger and get more current.
Circuitry, shown on a  link, is  "linear" voltage regulator, it's mean doesn't matter what surplus voltage on the input, it will be "lost". I'd suggest to build a switching regulator,  DC/DC converter.
( http://www.sparkfun.com/products/9275 or something like that).
For example:
Solar panel output     = 10 V in broad day light;
Fully charged battery = 5 V.
Input current             = 70 mA.
Output current          = Power(out)  / V(bat.) = Power(in) * K(ef.) / V(bat).
                                     where K(ef) is efficiency of switching regulator,  93.6 % (5 V)

I(Output current) = I(in) * V(in) * K / V(bat.) = 70 * 10 * 0.9 / 5.0 = 130 mA.

[Reply #3 on:]

thanks both of u ,very helpful,
@magician so in short ur saying that the excess voltage gets converted into extra current???
is there an easy circuit to do so?



BTW right now i am just charging a small mp3 player,i dont have its battery specs but half battery got charged in 1hr

[Reply #4 on:]

im also a little confused, regarding chargers, for example i checked the current output of a wall adapter which says 12v 800ma,but when i connect it to my multimeter like +to red and -to black node ,it reads 5amps??why is this does the current output depend on the device charging???

pls answer both ques

[Reply #5 on:]

Because you've essentially provided a short circuit to your wall adapter. 800mA is what it's rated for. You pull 5A from it, it won't last very long...

[Reply #6 on:]

oh ok i was just checking it, so the rating means the max it should be used for not the max it can provide?so how can i measure the max output current of my solar charger? by connecting + to red and - to black node??

[Reply #7 on:]

I would think that would be OK, since there's no chance it would try to draw more power from the sun than it was designed for ;-)

Still, I can't speak with any authority on the device, so I'd only connect it that way for a short while. It'd be interesting to measure the output under differing levels of light...

[Reply #8 on:]

yes yes it changes under different light levels,so other question can i convert excess voltage into current
???