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Understanding Law of Mass Action

  1. Mar 21, 2012 #1
    Hello Friends
    I am having trouble understanding law of mass action for extrinsic semiconductors. How the carrier concentration remain the same after the addition of impurities. I want to know conceptually not mathematically.
  2. jcsd
  3. Mar 21, 2012 #2
    To decrease carrier concentration the dopant impurities need to be deactivated, which probably means being somehow knocked out of the crystal or displaced into non-activation sites (I think the activated sites are usually substitutional and non-activated sites are usually interstitial or random?).

    When you introduce other deep level (energy level far away from conduction or valence band) impurities that don't contribute to doping, do you have reason to believe that they would somehow deactivate the shallow level impurities that contribute to doping that are already in place?
  4. Mar 21, 2012 #3
    Suppose in intrinsic silicon, n0=100 then p0=100 and carrier concentration will be 10,000. Now I add 1000 pentavalent impurity atoms. then intuitively n0=1100 and p0=100 or if some or all of the holes are filled by electrons then n0=1000 and p0=0. This is what i want things to be.... Can't figure out where the trick lies....
  5. Mar 21, 2012 #4
    wait, are you trying to make this in the lab?
    typically when you dope semiconductors with the intent of making a junction or a contact you end up with some gigantic number for n+ or p+. so if you try to dope it n, then you would get something like n=10^18cm^-3, and p=p0=100 (not sure what units you were using)
  6. Mar 21, 2012 #5
    for the sake of understanding and as an example let them be just 'n0 carriers per cubic cm'
  7. Mar 21, 2012 #6
    so typical Silicon intrinsic concentration is 10^10cm^-3. so if you decide to dope it (ion implantation or some other method) n type to say 10^18cm^-3, then you would have n=10^18cm^-3 and p=10^10cm^-3
  8. Mar 21, 2012 #7
    The reason is recombination . The recombination is in such manner that the product of the concentration of holes and electrons remains constant and equal to the sqaure of the intrinsic carrier concentration.
  9. Mar 21, 2012 #8
    If you are doping semiconductor material with a very large donor atoms (pentavalent) the recombination rate increases and thus the value of holes present reduces to such a value which makes the carrier concentration constant.
  10. Mar 21, 2012 #9
    ahh that's right! thanks, I totally forgot about that np=no^2 business :)
  11. Mar 21, 2012 #10
    I guess I was off by 8 orders of magnitude. np=n_i^2

    in the doped case p=(10^10)^2/(10^18)=10^(20-18)=10^2 !
  12. Mar 21, 2012 #11
    welcome :)
  13. Mar 21, 2012 #12
    najeeb, I think I misunderstood your question. Were you asking why pn=ni^2?
    I think that has to do with the fact that in thermodynamic equilibrium there can only be one fermi level. When you dope the semiconductor n type the fermi level moves closer to the conduction band and further away from the conduction band, such that n becomes much larger than p
  14. Mar 21, 2012 #13
    aaahhh! took so much of my time. n=10^10, p=10^10, n_i^2=10^20. dope it 10^18 and doing so will cause recombination so lets approx 10^8 holes recombine with electrons then n=(10^2 +10^18) ≈ 10^18 and p=10^2. then np= (10^18)*(10^2)= n_i^2.

    I hope i am right now?
  15. Mar 21, 2012 #14
    not sure about the recombination part, but your math looks right :)
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