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pn = ni^2 exp(Va/Vt)

Discussion in 'Electronic Basics' started by PZ, Feb 22, 2006.

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  1. PZ

    PZ Guest

    there is such an equation pn = ni^2 exp(Va/Vt) in junction given in
    textbook. I am wondering how true it is against other equations.
    suppose Va is zero,

    pn = ni^2 exp(Va/Vt) => pn = ni^2

    at same time p is p(0)= Ni^2/Nd, and n(0) = Ni^2/Na,

    so p(0) x n(0) = Ni^4/NaNd, and is supposed to equal ni^2, that would

    Ni^4/NaNd = Ni^2 => NaNd = Ni^2.

    But this Na and Nd, they are the dopant of p and n junction, their
    product doesn't equal to Ni^2.

    Any idea where is wrong?

  2. Guest

    First, let me warn you that I'm not an engineer nor a student -- all
    I've got is this library book called "Microelectronic Devices and
    Circuits" by Clifton G. Fonstad, so I could be very very wrong, but are
    you sure about this line:
    For a p-n junction wouldn't there be two different expressions for n_0
    and p_0, one for each side of the junction? In other words, if you had
    a p-n junction (fixed-width font like Courier required):

    Quasineutral Depletion Quasineutral
    vvvvvvvvvvvv vvvvvvvvv vvvvvvvvvvvv
    | | | | |
    | | |
    A o--| p, N_Ap | | | n, N_Dn |--o B
    | | |
    | | | | |
    -w_p -x_p x_n w_n

    ^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^
    P-Side N-Side

    [Figure 7.3 in the Fonstad book]

    then wouldn't the equations be something like:

    p_0(-x_p) = N_Ap
    p_0(x_n) = ni^2/N_Dn


    n_0(-x_p) = ni^2/N_Ap
    n_0(x_n) = N_Dn

    or something like that?

    Then I think if you multiply them together you get ni^2 in both cases.

    But I dunno... I'm just trying to learn this stuff myself, and a lot
    of it is barely making sense. Maybe somebody who actually knows what
    he's talking about can give you a better (correct) answer.
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