Help with Solar Battery/USB Charger Circuit

Discussion in 'The Projects Forum' started by Mbrint, May 30, 2012.

  1. Mbrint

    Thread Starter New Member

    May 30, 2012
    4
    0
    I'm trying to create a Solar USB charger.

    Charger Specifications:

    2x 4.5V solar cells (100mAh each) charge an amount of batteries (must be >8V). Here I have 2x AA NiMH, and 1 Sony infoLithium 7.2V battery pack.

    Female USB connector charges any device that requires a USB to charge. Like all USB charged devices, Apple products require 5V (the 7805 voltage reducer is used). The two "digital" ports on the USB are for newer Apple products; they require 2V each, given by the 33kΩ reducers.

    Solar cells have a secondary circuit to directly trickle charge USB.

    Yellow LED = solar only power to USB Red LED = batteries charging Green LED = battery only power to USB

    Voltage must be above 8V for 7805 Voltage Reducer to output the required 5V.

    Does this theoretical circuit work?

    [​IMG]
     
  2. wayneh

    Expert

    Sep 9, 2010
    12,119
    3,043
    You'll be disappointed with the results. The panels are just too small and will have trouble lighting those LEDs, let alone charging the battery or powering a device.

    I assume the 100mAh rating is really 100mA short circuit current? You'll have to put them in serial to barely meet your minimum voltage - in full sun only - so your max current will be less than half of that rating, maybe far less. An iPhone needs 10 times what this will deliver.

    I'd also avoid mixing battery types unless they are similar chemistries and ratings.

    The 7805 is an easy solution to get 5V for USB, but it will waste a lot of your hard-won power.
     
  3. Mbrint

    Thread Starter New Member

    May 30, 2012
    4
    0
    That is disappointing.. to answer your question about the panels, they are each 0.45, and 100mA (not mAh like I said),
    What type of panels would I need in order to power this system?

    I've just started to teach myself how this all works. Am I correct in thinking that if I used 8 NiMH AA (800mAh) batteries, it would take this system:

    8x800mAh=6400mAh, 6400mAh/200mA=3200 hrs to fully charge, and I should look for a way bigger amperage? Or am I off my rocker?
     
  4. wayneh

    Expert

    Sep 9, 2010
    12,119
    3,043
    0.45 ? What units? Oh, you mean 4.5v I guess?

    On the upside, 6400mAh/200mA=32hrs, not 3200. Yay!

    But, 1. the current is not additive when they are in series and 2. you won't get anywhere near 100mA when driving current thru a battery pack. The max panel current rating is usually taken with the panel leads shorted.

    To make this work in a satisfying way, you need to start with the load and work backwards. Maybe a device you want to charge needs 1000mAh, delivered as 500mA for 2 hrs, to fully recharge. That's 5v*0.5A= 2.5 watts, and 5 Whrs. With all the various losses, let's say you need double that amount from your panels. If you want, say, 10 hours of sunlight to provide that charge, you'll need panels that provide 1W, 10Whrs in 10 hours.

    Your current panels might deliver 50mA under charging conditions, maybe 3 volts. 3v*0.05A = 0.15W So you need at least about 7 similar panels to do the job.

    These are just example numbers to give you an idea how to look at the problem.
     
  5. wayneh

    Expert

    Sep 9, 2010
    12,119
    3,043
    Maybe you want one of these.
     
  6. Mbrint

    Thread Starter New Member

    May 30, 2012
    4
    0
    That's pretty slick, but I want to do this as a project.

    From what I understand, my main problem is not getting enough amperage from my panels. Or is it that they will not produce enough volts as well?

    Starting from my load (iPod), I have found that it is a 850mAh battery. Delivered at 100mA (should the amperage value reflect the power source?) for 8.5 hrs, I would be able to fully charge my battery. Mirroring what you've done above, that's 5v*0.1A= 0.5 watts, and 4.2 Whrs. My panels optimally will get 0.45W each, but I should double that to make up for the losses.

    Were you got to "10 hours of sunlight... 1W, 10Whrs in 10hrs" I get confused.
    I understand the next line were 0.15W becomes 7 panels from your 10hrs, but I don't understand were the switch from 2.5W, and 5 hrs came from.

    Doubling your value to compensate for 4 panels, under charging conditions, 50mA and 12V. 12v*.05A = 0.6W

    How does this all sound? And would just increasing the number of panels be sufficient?
     
  7. wayneh

    Expert

    Sep 9, 2010
    12,119
    3,043
    I doubled the 5Wh to 10Wh to account for losses, and then calculated 1W from the fact that I was allowing for 10 hours of charging to provide the 10Wh.
    Yes, the primary solution is more panel surface area. I'm not sure even 4 will be enough but it's probably more than that commercial device. I think you'll want two parallel strings composed of 2 or 3 panels each in series. Most of the solar lights I've looked at have a peak panel voltage about twice the battery voltage. Two of your panels in series peak at 9V, which may not be enough. You may want to experiment with a single string of 3 in series, and only add 3 more if/when you want faster charging.
     
  8. Mbrint

    Thread Starter New Member

    May 30, 2012
    4
    0
    I'm currently waiting for my components to arrive. I'll let you know how it all works out.

    Thank you very much for all your help!
     
  9. EarlAnderson

    Member

    Nov 13, 2011
    166
    4
    You won't get good results unless you use a higher power solar cell. I would use at least 500 mA but I would prefer 1A. As for the USB charger, use two 1K resistors on the first digital pin. I tried putting 2V on both of them, but only got a charging current of 500mA, but when I used two 1K resistors on the first digital pin, I got a charge current of 1A
     
Loading...