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Equilibrium concentration of majority and minority carriers

  1. Mar 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Give the equilibrium concentration of majority and minority carriers and resistivity for Silicon which is doped with 3x10[itex]^{15}[/itex] boron atoms/cm[itex]^{3}[/itex] at 27°C.


    2. Relevant equations
    n[itex]_{o}[/itex] = [itex]\frac{N_{d}-N_{a}}{2}[/itex]+[itex]\sqrt{(\frac{N_{d}-N_{a}}{2})^{2}+(n_{i})^{2}}[/itex]
    p[itex]_{o}[/itex] = [itex]\frac{N_{a}-N_{d}}{2}[/itex]+[itex]\sqrt{(\frac{N_{a}-N_{d}}{2})^{2}+(n_{i})^{2}}[/itex]
    n[itex]_{o}[/itex]p[itex]_{o}[/itex] = n[itex]_{i}[/itex][itex]^{2}[/itex]

    3. The attempt at a solution

    DATA
    n[itex]_{o}[/itex] (equilibrium concentration of majority carriers) = ?
    p[itex]_{o}[/itex] (equilibrium concentration of minority carriers) = ?
    [itex]\rho[/itex] (resistivity for Silicon) = ?
    N[itex]_{a}[/itex] = 3x10[itex]^{15}[/itex] atoms/cm[itex]^{3}[/itex]
    T = 27°C+273 = 273K
    n[itex]_{i}[/itex] (for silicon at 300K) = 1.5x10[itex]^{10}[/itex] atoms/cm[itex]^{3}[/itex]

    SOLUTION
    n[itex]_{o}[/itex] = 0 (I calculated this)
    p[itex]_{o}[/itex] = infinity

    I used the above given 1st equation to calculate n[itex]_{o}[/itex]. And used 3rd equation to calculate the p[itex]_{o}[/itex].
    Actually, I am confused whether I extracted right data or not. And I don't know how to calculate resistivity?

    Please tell me where is mistake in the data and Solution.

    Thanks.
     
  2. jcsd
  3. Mar 8, 2013 #2
    Please help me experts.
     
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