Making a Solar Powered USB Charger

[captainam3rica]

So I'm trying to make a solar powered usb charger, below is my circuit diagram. With it set up this way, it charges up fine to about 4.9v but then when I plug my ipod into it, it shows it's charging for a few seconds then the voltage drops to about 4.5v and then drains at a constant rate from there about 0.1v every second or two.

I tried adding in another AA to give me about 6v to try and counter the voltage drop, but when I plug in my ipod nothing happens. Thinking that 6v may be too much for devices that are supposed to use 5v, I tried some different resistors going from the batteries to USB with no luck, the ipod does nothing.

It seems this should be a simple fix but I'm a beginner so I'm stuck. All help is appreciated, thanks.

 

[Resqueline]

Try it without the diodes. Connected as shown they'll prevent both charging the AA's and the Ipod. Besides, not all solar panels needs reverse protection diodes.

 

[captainam3rica]

doh, I drew the diode in the wrong spot again. There is only one diode, coming from the + of the solar panels. (picture fixed above)

 

[Resqueline]

Ok, that looks better. Still, 4.9V is way too low to get a charge on those AA's. 5.9V would have been approximately fully charged. Tell us more about those panels.

 

[captainam3rica]

How is 4.9v too low? 4 AA's at 1.2v each is 4.8v and usb devices charge with about 4.75-5v.

The panels are from little LED garden lights. The two in series give me about 5.88V in the shade on a sunny day. Not sure about the mA, not sure how to measure that with my multimeter.

I tried just charging the batteries up and then hooking the USB port directly to the batteries and it does the same thing, so is there just something not right about my USB layout?

 

[davenn]

How is 4.9v too low? 4 AA's at 1.2v each is 4.8v and usb devices charge with about 4.75-5v.

The panels are from little LED garden lights. The two in series give me about 5.88V in the shade on a sunny day. Not sure about the mA, not sure how to measure that with my multimeter.

I tried just charging the batteries up and then hooking the USB port directly to the batteries and it does the same thing, so is there just something not right about my USB layout?

5.88 V on a sunny day, how did you measure that?
No load/open circuit voltage could well be >5V but once you have a load on there its going to drop significantly. So measure the voltage with your batteries connected only and see what the voltage is. Then disconnect the batteries and connect it to the USB port device and see how much it drops the voltage.
NOTE .... you are also going to loose another 0.7V in the diode. It is possible that the panels are unable to supply the current required for charging and thats why the voltage is also dropping significantly

the charging voltage for a 4.8V set of batteries is going to have to be over 5V. The USB is designed to have 5V on it for whatever is connected to it

cheers
Dave

 

[Resqueline]

1.2V is the average on-load voltage, 1.35V is the rested (& resting) off-load voltage of fully charged cells, and 1.45-1.5V is the end voltage of Ni-Mh cells under charge.

I seem to remember reading that Ipods are somewhat finicky about charging conditions but I haven't checked out if your USB setup actually meets the spec's.

Most solar panels won't produce any current to speak of in the shade, or even in lighly cloudy weather.
Measure both the voltage (without diode & batteries) and the short-circuit current both in direct sunlight and in a light shade.

 

[MagicMatt]

Doesn't the iPod contain a LiPoly battery? It's going to be fussy over voltages, and current, to avoid damage to the cell(s).

It wont charge if the voltage is under 4.2V certainly, as that's the minimum voltage needed to charge the cell. It will need enough current to charge the battery, and power the charging circuitry, so that could well be the full 5V 500mA a USB port would supply (based on a 4 hour charge cycle).

I don't know what panels you have, but in the shade to be giving that output under lead you would have to be looking at at least 2 x 10 watt panels. You need at least 2.5watts to the iPod to charge, and in the shade panels may be only 10% output. You might do better to put a 5V regulator in-line and pushing 7.5V+ into the circuit, if you can generate enough current. Failing that, trickle-charge a sealed lead-acid battery with the panel, and use that to power the USB output, since it will be able to push out 6V at a much higher current, and with a diode you'd drop it to 5.3V, and a simple 1ohm 10watt resistor would drop it to 5V and possibly allow you to fast-charge.