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Arrangement of solar cells

  1. Jan 7, 2017 #1
    1. The problem statement, all variables and given/known data
    My book states that solar cells are joined in series and in parallel to give a greater power because P=VI is increased by increasing V and I.

    2. Relevant equations


    3. The attempt at a solution
    I assume that the current in the solar cells depend on the intensity of sunlight. The holes and electrons in the p-n junction split apart when there is an incident photon. (I am not sure.)

    -If there are 4 solar cells and they are arranged in series. Assume the voltage created by the depletion region is 1V, the overall voltage will be 4V. If the sunlight produces 1 A current in each cells, the power will be 4V (4I) =16W

    -If there are 4 solar cells and they are arranged in parallel. The overall voltage will be 1V and the overall current will be 4A, the power will be 4W.

    -If the each of the 2 cells are arranged in a series and these two strands of series are arranged parallel, the overall current is 4A and the voltage will be 2V, therefore the power output is 8W.

    Why am I wrong? I can't figure out why a combination of series and parallel arrangement can lead to a higher power output.
     
  2. jcsd
  3. Jan 7, 2017 #2

    cnh1995

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    Correct me if I'm wrong. I have not studied solar cells in detail but does the sunlight create current in the cells? I believe the sunlight only creates voltage and current depends on the load.
    If this is right, you can try a load of 4 ohm in each of the three combinations you mentioned and compute the load power in each case. See which one gives the maximum power.
     
  4. Jan 7, 2017 #3

    cnh1995

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    Maybe I misinterpreted your question.
    Adding the solar cells in series increases the voltage rating and adding them in parallel increases the current rating. That is why, series-parallel combination will increase overall power rating.
     
  5. Jan 7, 2017 #4
    If I add a load in the purely series, the current across the load is 4/4 = 1A. If I add a load in the purely parallel arrangement, the current across the load is 1/4 = 0.25 A, how is the current increases?

    When I put all solar cells in series, although the current is lower, however the voltage is very big, how can I know a series-parallel combination is better than purely series when some cells are removed from the series and being placed parallel which increases the current however decreases the voltage?
     
  6. Jan 7, 2017 #5
    There are some mistakes in your reasoning, the overall maximum power is the same in all three cases and it is equal to 4W.
    In the first case the total current is 1A since the cells are connected in series the currents do not add up, instead it is the same current of 1A that flows each cell.
    In the third case the total current is 2A, each parallel branch has a current of 1A not 2A (again the currents in series DO NOT add up , it is the same current that flows through each cell in series).
    From conservation of energy it follows that no matter how you connect some emf sources, the total maximum power equals the sum of the power of each emf source. What it changes by the various connection schemes is the voltage and/or the current at which they deliver the maximum power.
     
  7. Jan 7, 2017 #6

    sophiecentaur

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    The book is not telling the whole story here. The Power available from a cell depends just on the Power of the incident sunlight. This power can only be used by connecting it to the appropriate Load. So you can connect your cells any way you like (as long as your blocks all have the same number of series or parallel cells).
    The following all assumes that the cells are operated under optimum conditions because the V/I characteristic is not simple. But at the (apparent) level of the question, you have to assume this.
    So, if one cell produces 1V and is rated at 1A, that assumes it's connected across a 1Ω resistor.
    If you want a 12V supply (to fit in with most requirements), you need 12 cells in series and would use a 12Ω load, for optimum output and would get 1A. (=12W)
    If you connect two sets of 6 in series and connect the sets in parallel, you will get 6V and will only get 0.5A into your 12Ω load (=6W)
    With a 6Ω load, you are back up to the full available power (6X2=12W)
    In practice, you use the combination which supplies the Volts you want and the available Power is governed by the actual Area (number of cells) in the array.
    So that statement in the book is actually nonsense because it doesn't include the factor of the load resistance.
    The "rating" only applies with the appropriate load resistance. Connecting the any old way will not affect the maximum power available - you just need to choose the right load value.

    Confusion about this is very similar to the common problem that people have about Transformers and PF gets many questions about this aspect of transformers. The turns ratio governs the voltage out and the load resistance governs the load current. In an ideal transformer, the VIin is the same as the VIout but you cannot 'force' the input current to be a particular value; it's the load resistance that does that.
     
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