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LED photo-electric effect

  1. Feb 25, 2005 #1
    Does anybody know why illuminating an LED with a filament lamp would produce a voltage in a simple circuit with just an LED and voltmeter? I'm guessing it's related to the photo-electric effect, but I have no idea what is actually going on.

    If any of you guys could help me out, or point me in the right direction, it would be a great help!

    Thank you in advance!
     
  2. jcsd
  3. Feb 25, 2005 #2

    chroot

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    It would only make sense -- if passing current through an LED produces light, why wouldn't shining light on an LED produce current?

    It is, in fact, the photo-electric effect. An LED is basically just a p-n junction, comprised of two kinds of semiconductors joined together. The region around the junction is normally depleted of charge carriers, because free electrons on the n-side of the junction want to rush over into the electron-starved p-side.

    When you shine light on the depletion region, the photo-electric effect kicks electrons out of atoms. These free electrons then immediately slide down the potential hill toward the p-side of the device, and a current flows.

    - Warren
     
  4. Feb 25, 2005 #3
    Yeh, that seems to make sense. It was the point about the photoelectric effect freeing electrons in the depletion region that I was originally unsure about. Will white light incident on the depletion region definetly free electrons from these otherwise stable atoms?
     
  5. Feb 25, 2005 #4

    chroot

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    The frequency-sensitivity of an LED can be pretty complicated. A good thermal white light source should have enough light of every frequency to stimulate any LED to produce a measurable current.

    - Warren
     
  6. Feb 28, 2005 #5
    Ah ok, so visible or even infra-red light has enough energy to free these electrons. Thank you very much for your help, it really is appreciated!
     
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