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Solar Panel Battery Charger Answered

I'm doing a solar panel cell phone battery charger project for school using a 9V, 3.3W solar panel that I bought from here: http://www.ebay.com/itm/3-3-WATT-9-VOLT-SOLAR-PANEL-3-3W-9V-Monocrystalline-Solar-Panel-/130612072684?pt=LH_DefaultDomain_0&hash=item1e691610ec
(I can get around 8V under a light bulb and 10.5V under sunlight.)

The circuit is shown in the picture with four, 1/4W resistors because the output is going to a USB cable.

The regulator I tried at first was a 7805 from radioshack I noticed I was getting exactly 5V on the output, but only getting 1.5mA of current on the output of the regulator. I've tried the MURATA 78SR, Fairchild LM317T, N.S. LM1804IS, and a few others specifying 5V and 1-1.5A of current but I still only get 1.5mA of current on the output (I need at least 500mA). The phones I'm testing will charge, including a ipod classic, but I'm afraid that the 1.5mA of current is far too small and will damage the battery.

I have yet to try using 3W or 5W resistors but I still don't think it would help, I'm also considering using a boost converter to get the current up (also the voltage) and then feed that output into the 7805.

Am I doing something wrong? Can I get around 1A of output current from the 7805 by doing something else?

2 Replies

mpilchfamily (author)2012-02-15

If the phones will charge then you have no issues. Charging at a lower current isn't going to hurt the batteries. It will only make charge times longer.

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verence (author)2012-02-15

Your solar cell will deliver maximun 3.3W at 9V, so it will give you 3.3W/9V = 367mA of current. That's all. Even with an ideal linear voltage regulator (Iout=Iin), you will never get more than that 367mA. The 78xx, LM317 etc are linear regulators, they keep the current flowing, but 'destroy' excess voltage by converting the power in (Uin-Uout)*I to heat.

You could try to use a switching regulator (buck type). They (in theory) transform the voltage down while transforming the current up, so Uin*Iin = Uout*iOut - so, if Uout < Uin, Iout can be > Iin. In reality, only about 80
% (depending on the design) of the input power will be available at the output. So, you might get 0.8*3.3W/5V = 528mA. With an ideal switching regulator (100% efficiency) 1*3.3W/5V = 660mA are possible. That's the limit. But a good switching regulator design is tricky.

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