Solar Panel Battery Charger

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?

Picture of Solar Panel Battery Charger
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.
verence5 years ago
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.