LCIE un Cu as a tunction of x which must exist in the semiconductor.s 36 The total current in a semiconductor is constant and equal to J=current is composed of a hole drift current and electron diffusion current. Assume thatthe hole concentration is a constant and equal to 10" cmand assume that the elec-tron concentration is given by n(x) = 2 x 10e cm where L =tron diffusion coefficient is D.= 27 cm/s and the hole mobility is u,Calculate (a) the electron diffusion current density for x>0, (b) the hole drift currentdensity for x> 0, and (c) the required electric field for x>0.10 A/em. The total15 um. The elec-420 cm/V-s.%3D

Given Current density J =-10 A/cm2 Concentration of hole p= 1016 cm-3 Concentration of electron…

Answered: on of x which must exist in the…

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LCIE un Cu as a tunction of x which must exist in the semiconductor.
s 36 The total current in a semiconductor is constant and equal to J=
current is composed of a hole drift current and electron diffusion current. Assume that
the hole concentration is a constant and equal to 10" cmand assume that the elec-
tron concentration is given by n(x) = 2 x 10e cm where L =
tron diffusion coefficient is D.= 27 cm/s and the hole mobility is u,
Calculate (a) the electron diffusion current density for x>0, (b) the hole drift current
density for x> 0, and (c) the required electric field for x>0.
10 A/em. The total
15 um. The elec-
420 cm/V-s.
%3D

Expert Solution

Step 1

Given

Current density J =-10 A/cm²

Concentration of hole p= 10¹⁶cm^-3

Concentration of electron n(x) =2*10¹⁵*e^x/Lcm^-3

Electron diffusion coefficient D_n= 27 cm²/s

Hole mobility $μ_{p}$ = 420 cm²/V-s

Formula for electron diffusion current density

$\begin{array}{rcl} J_{n} & = & e . D_{n} \frac{d n}{d x} \\ J_{n} & = & 1.602 * 10^{- 19} * 27 * \frac{d (2 * 10^{15} * e \frac{x}{15 * 10^{- 6}})}{d x} \\ = & 57.34 * 10^{- 4} \frac{d (e^{\frac{x}{15 * 10^{- 6}}})}{d x} \\ = & 382.32 * e^{\frac{x}{15 * 10^{- 6}}} \end{array}$

$J_{n} = 3.8232 {A/cm}^{2}$

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