Question

Holes are being steadily injected into a region of n-type silicon (connected to other devices, the details of which are not important for this question). In the steady state, the excess-hole concentration profile shown in fig. P3.10 is established in the n-type silicon region. Here "excess" means over and above the thermal-equilibrium concentration (in the absence of hole injection), denoted pn0. If nd = 1016/cm3, ni =1.5×1010/cm3, dp =12 cm2/s, and w =50 nm, find the density of the current that will flow in the x direction.

Answer 1

Complete Question

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Answer:

The  value  is  $J_n = -0.864 \ A/cm^2$

Explanation:

Generally for an n-type semiconductor the current density is mathematically represented  as

         $J_n = -q * d_p * \frac{d p_{n_o}}{d_w}$

Here  $p_{n_o}$ is mathematically represented as

          $p_{n_o} = \frac{n_i^2}{n_d}$

=>      $p_{n_o} = \frac{(1.5*10^{10})^2}{10^{16}}$

=>      $p_{n_o} = 2.25 *10^{4} \ cm^{-3}$

So

    $d p_{n_o} = P_n0 - p_{n_o}$

From the diagram  $P_n0 = 10^8 * p_{n_o}$

=>  $P_n0 = 10^8 * (2.25 *10^{4} )$

So

   $d p_{n_o} = 10^8 * (2.25 *10^{4} ) - 2.25 *10^{4}$

      $d p_{n_o} = 2.25 *10^{12} cm^{-3}$

So  from  $J_n = -q * d_p * \frac{d p_{n_o}}{d_w}$

substitute

    $1.60 *10^{-19} \ C$ for  q  and  $w_2 =50 nm = 50*10^{-9} m = 5*10^{-6} cm$

and from the diagram  $w_1 =0 \ cm$

So

    $J_n = -1.60 *10^{-19} *12 * \frac{2.25 *10^{12} }{ 5*10^{-6} - 0 }$

    $J_n = -0.864 \ A/cm^2$