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Optical absorption in Semiconductors

  1. Feb 19, 2008 #1

    Defennder

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    [SOLVED] Optical absorption in Semiconductors

    1. The problem statement, all variables and given/known data
    I'm doing a lab report on the optical absorption of semiconductors. More specifically it was to determine the bandgap values of the semiconductor samples. This is the experimental setup.

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    One of the questions which was asked was why a chopped light source was used instead of a constant incident light source. I can't quite figure out why.



    3. The attempt at a solution

    I did some searching on the internet and found that chopping the light source allows one to convert a dc light signal into an ac (pulsed) signal, and the relevant ac signal can be extracted by using a locked-in amplifier. But I still don't understand why a dc light source can't be used instead.
     
  2. jcsd
  3. Feb 19, 2008 #2

    mgb_phys

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    You are trying to measure small changes in a light level due to absorption in the sample.
    But you are also measuring changes in the brightness of the source and the response of the detector, a chopper lets you take these out.

    By using a chopper you are taking the ratio of the signal with the light present and the light blocked a very short time before, any variations in the system slower than the chopper frequency are automatically removed.
    You normally pick a chopper frequency at least 10-20x faster than any inteference you suspect - such as 60Hz line noise.
     
  4. Feb 19, 2008 #3

    Defennder

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    Is the change in brightness of the light source caused by optical absorption of the semiconductor when the frequency of the source approaches the bandgap value of the semiconductor?
     
  5. Feb 20, 2008 #4

    mgb_phys

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    the change in brightness of the detected light is caused by interactions between the light and the semiconductor.
    Changes in brightness of the source you try and avoid - there will be some just because lamps aren't perfect.
     
  6. Feb 24, 2008 #5

    Defennder

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    I see, thanks a lot for your help.
     
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