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Electron and hole concentrations - Long One.

  1. Apr 1, 2008 #1
    Hey there,
    I was studying Millman and Halkias’ ‘Integrated Electronics’ (Tata McGraw Hill, 1991) when I hit this paragraph and couldn’t proceed any further. I’d really appreciate suggestions of any kind.
    I will first reproduce here the above mentioned paragraph, then write down the assumptions which I believe can be made and then proceed to list my questions.

    Pg 27,
    “2-4 Charge Densities in a Semiconductor
    Equation (2-10) namely np = n(sub i)^2, gives one relationship between the electron n and the hole p concentrations. These densities are further interrelated by the law of electrical neutrality, which we shall now state in algebraic form: Let Nd equal the concentration of donor atoms. Since, as mentioned above, these are practically all ionized (ψ), Nd positive charges per cubic meter are contributed by the donor ions a. Hence the total positive-charge density is Nd + p. Similarly, if Na is the concentration of acceptor ions, these contribute Na negative charges per cubic meter. The total negative-charge density is Na + n (b). Since the semiconductor is electrically neutral, the magnitude of the positive-charge density must equal that of the negative concentration, or

    Nd + p = Na + n (Eq: A)”

    Assumptions:

    1. n refers to the conc of free electrons.
    2. When a donor atom is ionized, it loses one free electron and when an acceptor is ionized it gains one electron.
    3. ALL the donor and acceptor atoms are ionized (from ψ).

    Questions:
    From a: If Nd donor atoms get ionized producing Nd positive ions, wouldn’t that also produce Nd free electrons, thereby increasing the RHS of equation A by Nd?
    Wouldn’t some of these free electrons also fill up some of the p holes, reducing the LHS?

    From b: Since the Na atoms become ionized by absorbing some of the free electrons from n, wouldn’t the total negative charge remain unchanged?

    Thanks
     
  2. jcsd
  3. Apr 23, 2008 #2

    LydiaAC

    User Avatar
    Gold Member

    Hello Thoreau:

    Yes
    No, because "n" stands for all free electrons, Nd electrons are already included in n. You can split n as Nd+(n-Nd), the first term being the electrons from the donor atoms and the second the electrons thermally generated.

    Yes, when an electron fill up a hole, "p" is reduced and therefore the LHS. However, "n" is also reduced and therefore the RHS. You get
    Nd+(p-k)=Na+(n-k)
    which is exactly the same equation.

    Yes. When an acceptor atom ionizes, a free electron is no longer free but bonded to an atom. This reduce the availability of free electrons and therefore, conductivity. However, the charge is only changing places and it is not altered.
     
    Last edited: Apr 23, 2008
  4. Nov 30, 2008 #3
    Thank you LydiaAC. I'm sorry I couldn't thank you sooner than this. I'm just vertical again after the worst motorcycle accident I've ever had so far (hope the record holds!) - they'd have had to scrape my brains from the road if not for my helmet.

    And of course, your explanation (for the question I'd asked) was what I was looking for. At times M&H leaves me a bit boggled.
    Thanks again.
     
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