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Open curcuit voltage of photovoltaic devices

  1. Feb 16, 2009 #1
    I have some questions about the open circuit voltage of the photodiodes. Physically, what is the origin of the open circuit voltage?

    In the book "Physics of Solar cells": the open circuit voltage Voc defines the difference between the Fermi energies at which the total recombination rate in the cell is equal to the total generation rate given by the absorbed photocurrent.

    However, what always confusing me is why there is no open circuit voltage of a photodiode when no light illuminate on the device, but actually there is a p-n junction or Schottky barrier (built in potential) exists. But when light is illuminates on the diode, we can measure the open circuit voltage. So how to determine the Voc in a band diagram and what is the theoretically limit of the Voc?

    I appreciate any advice on it.
     
  2. jcsd
  3. Feb 16, 2009 #2

    dlgoff

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    Welcome to PF chenhon5. I think you may be wondering about http://en.wikipedia.org/wiki/Dark_current_(physics)" [Broken].
     
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  4. Feb 16, 2009 #3
    If there are no electron holes being created by the presence of photons, then how will there be a voltage potential? If there was, then this would certainly violate conservation of energy.

    I think where you are confused is with the dark current that dlgoff mentioned. Dark current only exists when the photodiode is in a reverse bias mode. That is when an external voltage potential is being applied to it. In a reverse bias mode, when there are no photons present (dark), there will be a small amount of current going through the photodiode which is why its called "dark current". If there is no voltage being applied across the diode then there will be no dark current.
     
  5. Feb 16, 2009 #4

    Redbelly98

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    I think chenhon's question is for an unbiased photodiode. What's Voc across a photodiode with nothing else connected to it.

    With no light on the photodiode, it behaves just as a normal diode would. So VOC=0.
     
  6. Feb 16, 2009 #5

    Thank you for you two's reply.

    It seems I did misunderstand something. Say we did not apply any bias and light on a Shottky barrier based photodiode, the work function of the metal and Fermi level of the semiconductor will align. So there is not any voltage potential (Voc) created. (I mix it up with the built-in potential.)

    I try to understand it from the energy band diagram view:

    But when the light illuminates, we got a open circuit voltage Voc, is that mean the Fermi level of the semiconductor shift in this condition?(difference between work function and Fermi level will be the Voc). To my understanding, the Fermi level may shift with illumination, since the light will generate excessive carriers, then the equlibrium condition will be broken and shift the Fermi level. Therefore the maximum Voc for a p-n photodiode is the bandgap Eg as said in some books.Is that right?

    However I did not know what is relationship between the Voc and the dark current as you two mentioned. I did know the Voc=kT/q*log(1+Isc/Idark) Could you give more details. Thanks a lot.
     
  7. Feb 16, 2009 #6

    dlgoff

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    I found these notes/paper on silicon photodiodes very well done. You will need Acrobat Reader to open the file.
    http://www.optics.arizona.edu/Palmer/OPTI400/SuppDocs/UDTapp_notes_02.pdf" [Broken]
     
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  8. Feb 16, 2009 #7

    Redbelly98

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    Thanks Don. Just wondering, what are people's opinions of the graph on p. 9, right-hand column?

    My understanding is that the different curves should be displaced vertically from one another, rather than displaced diagonally (downward & rightward) as shown. How difficult can it be to enter the equation, shown above the graph, into some plotting software and show a more representative graph?

    Regards,

    Mark
     
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  9. Feb 16, 2009 #8
    Yes, I agree. The curves 9 show the IV curves with photoresponse with different light power density. As we know, the Isc is proportional to P (power density), however, the Voc is proportiaon to log(P), therefore....
     
  10. Feb 16, 2009 #9
    Thanks a lot, Don. So do you agree that the Fermi level of semiconductor will be shifted with light illumination in order to generate the Voc?
     
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  11. Feb 17, 2009 #10

    dlgoff

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    I'm not sure I would agree with this. From page 5 of the paper, they say that the built-in voltage is equal to the difference of the Fermi levels in the N and P-type regons. Wouldn't that mean the Fermi levels are constants? Not sure.
     
  12. Feb 17, 2009 #11

    dlgoff

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    It I-V curve should look more like this one:
    http://www.rp-photonics.com/img/photodiode.png" [Broken]
    that I found in this thread:
    https://www.physicsforums.com/showthread.php?t=233453"
     
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  13. Feb 17, 2009 #12

    Redbelly98

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    Last edited by a moderator: May 4, 2017
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