basic diode operation doubt

Discussion in 'General Electronics Chat' started by pawankumar, Feb 4, 2010.

  1. pawankumar

    Thread Starter Member

    Oct 28, 2009
    42
    3
    greetings brothers,
    i ve studied in books that when a pn diode is forward biased,the depletion layer is squeezed and thereby,electrons flow from n side to p side what happens after removing the voltage source?how does the layer come back to normal?
    how is a si diode having a barrier potential of 0.7v irrespective of size and current carrying capablity?
    thank you in advance,
    pawan
     
    Last edited: Feb 4, 2010
  2. Shikhar

    New Member

    Dec 21, 2009
    8
    0
    The answer to your question lies in the fact how the depletion region is formed in an unbiased pn junction. In a P-N junction (diode), electrons and holes start moving towards the other end because of the concentration gradient. This continues till a sufficient barrier potential is developed which opposes the flow of majority carriers.
    Now, coming back to ur question, when the forward bias is removed, the concentration gradient is more than should be in an unbiased junction. Thus the majority carrriers continue crossing the junction till the equilibrium state is achieved!!!

    and why is forward bias voltage drop= 0.7V, the answer lies with the nature!!!
     
  3. pawankumar

    Thread Starter Member

    Oct 28, 2009
    42
    3
    thanks...but is there some logic about why it is 0.7v for sillicon and 0.3v for Ge..
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Because band-gap for Si is equal 1.12eV and for Ge 0.67eV.
    The electrons in Si are closer to the centre of a atom. So, its harder to pull out a electron form Si atom then in Ge.
     
  5. pawankumar

    Thread Starter Member

    Oct 28, 2009
    42
    3
    thanks ..............
     
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