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Effect of doping level on the width of the depletion layer

  1. Sep 14, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-9-14_22-35-12.png

    2. Relevant equations


    3. The attempt at a solution
    I think before getting the steady state, after doping there are more electrons in n-region and there are more holes in p-region, so the more electrons go towards the p-region and more holes go toward the n - region. Consequently the charges on the two sides of the depletion layer gets increased because of the doping.
    After reaching the steady state, for an electron from n region going towards p-region, the negative charge at the starting of the p-region repels it and the positive charge at the n-region attracts the electron.
    The resultant force on the electron due to these two source of charges in both type of p-n junction( doped and pure) should remain same.
    Since, in case of doped p-n junction, the charges is comparatively more,the distance between the two sources of charges should get increased. Consequently, the width of the depletion layer should get increased.
    The correct option is (iii). Right?
     
  2. jcsd
  3. Sep 14, 2017 #2

    phyzguy

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  4. Sep 14, 2017 #3
    For a p–n junction, [​IMG] and [​IMG] be the concentrations of acceptor and donor atoms respectively, and letting [​IMG] and [​IMG] be the equilibrium concentrations of electrons and holes respectively.
    letting [​IMG] be the total width of the depletion region, we get
    [​IMG] ..........................................................................................................(1)
    [​IMG] can be written as [​IMG], where we have broken up the voltage difference into the equilibrium plus external components.
    q is the magnitude of charge of electron.
    [​IMG] ...............................................................................................................(2)

    In eqn. (2) , it is noted that if we increase doping level, ##\Delta V_0 ## increases logarithmically.
    While in eqn.(1), we see that if we increase doping level ##\frac{C_A +C_D}{C_A C_D} ## decreases faster than increase in ##\Delta V_0 ##. So, as a result, d decreases when the doping level is increased.
    Thus, the correct option is (ii). Right?

    After joining p-type and n-type semiconductors, electrons from the n region near the p–n interface tend to diffuse into the p region leaving behind positively charged ions in the n region and being recombined with holes, forming negatively charged ions in the p region. Likewise, holes from the p-type region near the p–n interface begin to diffuse into the n-type region, leaving behind negatively charged ions in the p region and recombining with electrons, forming positive ions in the n region Template:Explain holes move?. The regions near the p–n interface lose their neutrality and most of their mobile carriers, forming the space charge region or depletion layer (see figure A).

    Charges in the depletion region is due to the fixed ions, not mobile electrons. Right?
     
  5. Sep 14, 2017 #4

    phyzguy

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    Right!

    Correct. There are no mobile carriers in the depletion region. It is "depleted" of mobile carriers, which is why it is called the depletion region. The mobile carriers are swept out of the depletion region by the large electric field.
     
  6. Sep 14, 2017 #5
    Thank you for helping me.
     

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