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Derivation of width of depletion layer in the pn-junction

  1. Feb 13, 2015 #1

    I read a derivation for the width of the depletion region [itex]W[/itex] in "SEMICONDUCTOR DEVICE FUNDAMENTALS" by Robert F. Pierret in which at one point it says:

    Here again for better readability:

    [itex]x_N = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_A}{N_D\cdot(N_A+N_D)}V_{bi}}[/itex]
    [itex]x_P = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_D}{N_A\cdot(N_A+N_D)}V_{bi}}[/itex]
    [itex]W = x_N + x_P = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_D+N_A}{N_A\cdot N_D}V_{bi}}[/itex]

    Which is confusing to me as I would expect the same containing:
    [itex]W = x_N + x_P = \sqrt{\frac{2 K_S \epsilon_0}{q}\frac{N_D^2+N_A^2}{N_A\cdot N_D\cdot(N_A+N_D)}V_{bi}}[/itex]

    This same outcome though is found on various places in the internet.

    Does anyone know what I am missing?

    Kind regards,


    [itex]x_N[/itex] length of n-doped region
    [itex]x_P[/itex] length of p-doped region
    [itex]N_A[/itex]density of acceptors
    [itex]N_D[/itex]density of donors
    [itex]V_{bi}[/itex] builtin potential

    All the others are known constants
    Last edited: Feb 13, 2015
  2. jcsd
  3. Feb 13, 2015 #2


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    Staff: Mentor

    ##\sqrt{x}+\sqrt{y} \neq \sqrt{x+y}##

    Ignoring the common prefactors,
    $$x_N = \sqrt{\frac{N_A^2}{N_D N_A (N_A+N_D)}}$$
    $$x_N = \sqrt{\frac{N_A^2}{N_D N_A (N_A+N_D)}}$$
    $$W = \sqrt{\frac{1}{N_D N_A (N_A+N_D)}} \left(\sqrt{N_A^2} + \sqrt{N_D^2}\right) = \sqrt{\frac{1}{N_D N_A (N_A+N_D)}} \left(N_A + N_D\right) \\= \sqrt{\frac{(N_A+N_D)^2}{N_D N_A (N_A+N_D)}} = \sqrt{\frac{N_A+N_D}{N_D N_A}}$$
  4. Feb 13, 2015 #3
    Shame on me
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