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.

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-09-15T05:30:40+02:00

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Answer:

The value is [tex]J_n = -0.864 \ A/cm^2[/tex]

Explanation:

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

[tex]J_n = -q * d_p * \frac{d p_{n_o}}{d_w}[/tex]

Here [tex]p_{n_o}[/tex] is mathematically represented as

[tex]p_{n_o} = \frac{n_i^2}{n_d}[/tex]

=> [tex]p_{n_o} = \frac{(1.5*10^{10})^2}{10^{16}}[/tex]

=> [tex]p_{n_o} = 2.25 *10^{4} \ cm^{-3}[/tex]

So

[tex]d p_{n_o} = P_n0 - p_{n_o}[/tex]

From the diagram [tex]P_n0 = 10^8 * p_{n_o} [/tex]

=> [tex]P_n0 = 10^8 * (2.25 *10^{4} ) [/tex]

So

[tex]d p_{n_o} = 10^8 * (2.25 *10^{4} ) - 2.25 *10^{4} [/tex]

[tex]d p_{n_o} = 2.25 *10^{12} cm^{-3} [/tex]

So from [tex]J_n = -q * d_p * \frac{d p_{n_o}}{d_w}[/tex]

substitute

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

and from the diagram [tex] w_1 =0 \ cm [/tex]

So

[tex]J_n = -1.60 *10^{-19} *12 * \frac{2.25 *10^{12} }{ 5*10^{-6} - 0 }[/tex]

[tex]J_n = -0.864 \ A/cm^2[/tex]