Theoretical Solar Question:

Discussion in 'General Electronics Chat' started by ShockBoy, Dec 8, 2009.

  1. ShockBoy

    Thread Starter Active Member

    Oct 27, 2009
    186
    0
    Could this be done: Should this be done:?
    2 solar modules, each consisting of 48 cells, each cell rated at .5V 3.5A.
    One module is completely series to produce 24V and 3.5A and the other module is completely parallel to produce .5V and 171.5A. Total power produced 15.435KW. (3087w * 5hrs.)
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    0.5 Volts x 3.5 Amperes = 1.75 Watts.

    You have 48 solar cells.
    So, you have a total possible power output of 48 x 1.75 Watts = 84 Watts.

    It will take 11.905 hours of bright sunlight for you to reach 1kW.

    If you want to be able to produce 1kW in one hour, you will need 572 of those cells.

    Then you will be able to have ten 100W bulbs burning inside your house, as long as the sun is out.

    If you want to heat up something in the microwave, you will need to turn all the lights off.
     
  3. ShockBoy

    Thread Starter Active Member

    Oct 27, 2009
    186
    0
    Alright, I know you don't like solar..but I'm talking about both panels in parallel, not just one. I understand the one panel is 84 watts. My question is parallel-ing 48 3.5A cells together then parallel-ing both panels i.e. the 24V panel AND the 168A panel/module.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    What do you mean? I like solar energy.
    I'm just trying to give you the facts, as I understand them.

    You're thinking that the 168A is going to be multiplied by the 24v, right?

    Sorry, it does not work that way.

    If you have a total of 96 panels that generate 0.5v, 3.5A each, then you have a potential output of 168 Watts.

    If you tried to connect together two arrays of solar cells, one that was wired in series to generate 24v @ 3.5A, the other wired in parallel to generate 0.5v @ 168A, the series array would "see" practically a dead short. Its' output would be pulled down to 0.5v.

    Then you would have an array of 48 cells that would output about 0.5v at 171.5A, or about 85.75 Watts.
     
  5. ShockBoy

    Thread Starter Active Member

    Oct 27, 2009
    186
    0
    I know I'm new, but I need some clarification. I originally stated connecting the modules in parallel, but after reading your post, it really does not matter. The short that occurs is caused by what. Series and parallel add volts and add amps. This defaults to the lowest voltage/amp feed? I understand that connecting differently rated cells end up the lowest denominator. Why does'nt the highest number win?
     
    Last edited: Dec 8, 2009
  6. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Do you have any of these cells?

    If so, why don't you try your idea with just, say, three connected in parallel, and then three connected in series.

    Then try to connect them together, and see what you wind up with.

    [eta]
    That won't hurt the cells, by the way. It's even OK to operate them when there is a dead short on their outputs.
     
  7. ShockBoy

    Thread Starter Active Member

    Oct 27, 2009
    186
    0
    Yes I do. I will experiment. Thank You! And let you know.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    OK, great!

    Do it scientifically though.

    Measure each cells' output voltage and current before you connect them; record the results.

    Then connect the three in parallel. Measure their combined output voltage and current.

    Then connect another three in series. Measure their combined output voltage and current.

    Then connect the six together. You should read a bit higher voltage than the three in parallel, but nowhere near as high as the three in series.

    When you do the current test, be prepared - remember, when you have three cells in parallel, output current will be 3x3.5a = 10.5A - over the limit for most meters! It is much safer to measure the voltage drop across a 1 Ohm resistor. Make certain the resistor is rated for twice the power that will be dissipated in it.
     
  9. ShockBoy

    Thread Starter Active Member

    Oct 27, 2009
    186
    0
    Thank You. Will do with the 1 Ohm resistor. Let you know tomorrow.
     
  10. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    SgtWookie is correct.

    Another issue with your theory is your power/energy units. Power is measured in W (or kW), while energy is measured in kWhours (or Joules). When you multiply a power in Watts times time in hours, the energy units are kWhours which is a measure of energy. If you multiply power in Watts times time in seconds, you get Joules as energy units.

    It's important to distinguish power and energy. You can collect a low power over a long time and store energy, (e.g. battery charging). Then you can run a high power device for a short time off of the battery.

    The electric company charges you based on your energy usage per kWhour.
     
  11. ShockBoy

    Thread Starter Active Member

    Oct 27, 2009
    186
    0
    No time today, I will post a new thread with the results of the experiment.
     
  12. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    There is no need to start a new thread. This one is yours, and the results of the test are directly related to your initial post.

    If they were to be unrelated, then it would be appropriate to start a new thread.
     
Loading...