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Current through p-type silicon

  1. Apr 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A 1-cm cube of p-type silicon (ρ = 0.1Ω-cm) acquires a linear electron distribution in the x-direction, such that n = 1014/cm3 at one side and n = 105/cm3 at the opposite side.

    Wires are attached to the sides of the cube via ohmic contacts, and a 0.1mV voltage source is applied. Find the values of the electron, hole, and total currents that flow in the external circuit.

    2. Relevant equations

    ρ = R[itex]\frac{A}{L}[/itex] = [itex]\frac{E}{J}[/itex]
    E = [itex]\frac{V}{L}[/itex]
    Je = qnμeE + qDe[itex]\frac{dn}{dx}[/itex]

    Jh = qpμhE + qDh[itex]\frac{dp}{dx}[/itex]
    i = [itex]\frac{V}{R}[/itex]
    n0p0 = n2i

    3. The attempt at a solution
    I started by finding R:

    R = ρ[itex]\frac{L}{A}[/itex] = (0.1Ω-cm)[itex]\frac{(1cm)}{(1cm^{2})}[/itex] = 0.1Ω

    Then, the total current is:

    I = [itex]\frac{V}{R}[/itex] = [itex]\frac{(0.1mV)}{(0.1Ω)}[/itex] = 1mA

    p0 = [itex]\frac{n^{2}_{i}}{n_{0}}[/itex] = [itex]\frac{(1.5x10^{10})^{2}}{10^{14}}[/itex] = 225x104/cm3

    p1 = [itex]\frac{n^{2}_{i}}{n_{1}}[/itex] = [itex]\frac{(1.5x10^{10})^{2}}{10^{5}}[/itex] = 225x1013/cm3

    Now, [itex]\frac{dp}{dx}[/itex] = 225x1013/cm4 and [itex]\frac{dn}{dx}[/itex] = -1x1014/cm4, since dx = 1cm.

    Which values would I use for n and p in the following equations, assuming the above steps are correct?

    Je = qnμeE + qDe[itex]\frac{dn}{dx}[/itex]

    Jh = qpμhE + qDh[itex]\frac{dp}{dx}[/itex]
     
    Last edited: Apr 12, 2013
  2. jcsd
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