# Making a solar cell phone charger

Good day everyone,

So I have a spare solar panel that has been laying around and I want to put it to use. Its a small panel, about 8"X12". The specs read 8 Watts and 17 Volts

I was thinking of maybe making a solar charger for phones and other 5VDC USB devices.

I bought a 5VDC regulator kit, soldered it together and got it to work with the panel
This is the kit:
http://www.superdroidrobots.com/shop/item.aspx/5v-regulator-board-kit/508/

Next I want to hook the output up to a set of maybe 2 or 4 USB ports. I can get these off an old computer or other device.

My question is, how do I know how much current this thing will output at 5VDC?
I know my factory cell phone charger outputs 2000ma or 2 amps.
How do I know that this setup will not output more than that and damage the device being charged.
Or is this setup even strong enough to go up to 2 amps?

Is there a way to test for this?

Thank you and sorry for the long post

-Eddie

## Answers and Replies

There are a bunch of designs for this at this site - http://www.instructables.com/howto/solar+panel+charger/

With respect to the power requirements, some of them use the solar panel to charge a battery pack - http://www.instructables.com/id/Solar-USB-Charger-20-21/
I had not thought about charging a battery pack. That might be better, it is not limited to providing power only while under the sun.

I will look at these examples in detail tonight when I get home

Thank you!

Baluncore
The specifications for the PV is; 17V, 8W. The current will be ( 8W / 17V ) = 0.47A.
The regulator is a linear regulator so Iin = Iout.
The maximum current out will therefore be 470 mA.

The voltage drop across the regulator will be 17Vin – 5Vout = 12V
The regulator will therefore dissipate power ( 12V * 0.47A ) = 5.64W
Maybe you should consider a heatsink if the regulator gets hot.

The specifications for the PV is; 17V, 8W. The current will be ( 8W / 17V ) = 0.47A.
The regulator is a linear regulator so Iin = Iout.
The maximum current out will therefore be 470 mA.

The voltage drop across the regulator will be 17Vin – 5Vout = 12V
The regulator will therefore dissipate power ( 12V * 0.47A ) = 5.64W
Maybe you should consider a heatsink if the regulator gets hot.

The regulator chip did did begin to get a little hot so I did attach a small heatsink to it.
I am using it to charge a portable phone charger that itself is charged with 5V

Seems to work just fine now.

Thank you!