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The electron concentration in silicon at T = 300 K is given by
n(x) = 10^16 exp (-x/18) cm3
where x is measured in m and is limited to 0< or is equal to x < or is equal to 25 m. The electron diffusion coefficient is Dn = 25 cm2/s and the electron mobility is mun = 960 cm2/V-s. The total electron current density through the semi-conductor is constant and equal to Jn = 40 A/cm2. The electron current has both diffusion and drift current components. Determine the electric field as a function of x-bar which must exist in the semi-conductor.

Sagot :

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Answer:

5.6 × 10⁻²[1 + 4.643 × 10⁻⁷exp (x/18)] V/cm

Explanation:

The total current density, J = drift current density, J' + diffusion current density, J"

J = J' + J"

J' = nμeE where n = electron concentration =  10¹⁶ exp (-x/18) cm³, μ = electron mobility = 960 cm²/V-s, e = electron charge = 1.602 × 10⁻⁹ C and E = electric field

J" = eDdn/dx where e = electron charge = 1.602 × 10⁻¹⁹ C, D = diffusion coefficient = 25 cm²/s and dn/dx = concentration gradient = d10¹⁶ exp (-x/18) cm³/dx = (-10¹⁶/18)exp (-x/18) cm³

So,

J = J' + J"

J = nμeE + eDdn/dx

E = (J - eDdn/dx)/nμe

Since J = 40 A/cm², substituting the values of the variables into the equation, we have

E = (J - eDdn/dx)/nμe

E = [40 A/cm² - 1.602 × 10⁻⁹ C × 960 cm/V-s × (-10¹⁶/18)exp (-x/18) cm²] ÷ (10¹⁶exp (-x/18) cm³ × 960 cm²/V-s × 1.602 × 10⁻⁹ C)

E = [40 A/cm² - 1537.92 × 10⁻⁹ C-cm/V-s × (-10¹⁶/18)exp (-x/18) cm²] ÷ (10¹⁶exp (-x/18) cm² × 1537.92 × 10⁻⁹  C-cm³/V-s)

E = [40 A/cm² + 85.44 × 10⁷ C-cm/V-s exp (-x/18) cm²] ÷ (exp (-x/18) cm² × 1537.92 × 10⁷  C-cm³/V-s)

E = [40 A/cm² ÷ (exp (-x/18) cm² × 1537.92 × 10⁷  C-cm³/V-s)] + [85.44 × 10⁷ C-cm/V-s exp (-x/18) cm² ÷ (exp (-x/18)  × 1537.92 × 10⁷  C-cm²/V-s]

E = [0.026 × 10⁻⁷ A/cm²exp (x/18)    V-s/C-cm³] + 0.0056]

E = [0.026 × 10⁻⁷exp (x/18) V/cm] + 0.0056]

E = 0.056[0.026 × 10⁻⁷exp (x/18) V/cm]/0.056 + 1]

E = 0.056[4.643 × 10⁻⁷exp (x/18) V/cm] + 1]

E = 5.6 × 10⁻²[1 + 4.643 × 10⁻⁷exp (x/18)] V/cm

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