1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    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]

    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.

  2. jcsd
  3. Mar 8, 2013 #2
    Please help me experts.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted