Chapter 1, Problem 1.17P

The hole concentration in silicon is given by
$p\left(x\right)={10}^4+{10}^{15}\exp \left(-x/L_p\right)x\ge 0$
The value of $L_p$ is $10\mu \text{m}$ . The hole diffusion coefficient is $D_p=15{\text{cm}}^{\text{2}}\text{/s}$ . Determine the hole diffusion current density at (a) $x=0$ , (b) $x=10\mu \text{m}$ , and (c) $x=30\mu \text{m}$ .

To determine

The hole diffusion current density at given location.

Answer to Problem 1.17P

The hole diffusion current density at $x=0$ is $2.4\text{A}/{\text{cm}}^2$ .

Explanation of Solution

Given information:

The expression of the hole concentration is $p(x)={10}^4+{10}^{15}\exp \left(\frac{-x}{L_P}\right)\text{where }x\ge \text{0}$

The value of the $L_P=10\text{μm}$ .

The hole diffusion density is $D_p=15{\text{cm}}^2/\text{s}$ .

Location is $x=0$ .

Calculation:

The gradient of the hole concentration is $\frac{dp(x)}{dx}$ .

Differentiating the equation of the hole concentration,

  $\begin{array}{l}\frac{dp(x)}{dx}=\frac{d}{dx}\left({10}^4+{10}^{15}\exp \left(\frac{-x}{L_P}\right)\right)\text{where }x\ge \text{0}\\ \frac{dp(x)}{dx}=\frac{d}{dx}\left(\exp \left(\frac{-x}{L_P}\right)\right)\text{where }x\ge \text{0}\\ \frac{dp(x)}{dx}=\frac{-{10}^{15}}{L_P}{e}^{\frac{-x}{L_P}}\text{where }x\ge \text{0}\end{array}$

The current diffusion density is given by the expression as shown below:

  $J_P=-e{D}_P\frac{dp(x)}{dx}$   (1)

Plugging the given values in the equation (1).

  $\begin{array}{l}{J}_P=-1.6\times {10}^{-19}\times 15\times \frac{-{10}^{15}}{10\times {10}^{-6}\times {10}^2}{e}^{\frac{0}{L_P}}\\ J_P=2.4\times \frac{-{10}^{-3}}{10\times {10}^{-6}\times {10}^2}{e}^{\frac{0}{L_P}}\\ J_P=2.4\text{A}/{\text{cm}}^2\end{array}$

To determine

The hole diffusion current density at given location.

Answer to Problem 1.17P

The hole diffusion current density at $x=10\text{μm}$ is $0.883\text{A}/{\text{cm}}^2$ .

Explanation of Solution

Given information:

The expression of the hole concentration is $p(x)={10}^4+{10}^{15}\exp \left(\frac{-x}{L_P}\right)\text{where }x\ge \text{0}$

The value of the $L_P=10\text{μm}$ .

The hole diffusion density is $D_p=15{\text{cm}}^2/\text{s}$ .

Location is $x=10\text{μm}$ .

Calculation:

The gradient of the hole concentration is $\frac{dp(x)}{dx}$ .

Differentiating the equation of the hole concentration,

  $\begin{array}{l}\frac{dp(x)}{dx}=\frac{d}{dx}\left({10}^4+{10}^{15}\exp \left(\frac{-x}{L_P}\right)\right)\text{where }x\ge \text{0}\\ \frac{dp(x)}{dx}=\frac{d}{dx}\left(\exp \left(\frac{-x}{L_P}\right)\right)\text{where }x\ge \text{0}\\ \frac{dp(x)}{dx}=\frac{-{10}^{15}}{L_P}{e}^{\frac{-x}{L_P}}\text{where }x\ge \text{0}\end{array}$

The current diffusion density is given by the expression as shown below:

  $J_P=-e{D}_P\frac{dp(x)}{dx}\mathrm{..............................}(1)$

Plugging the given values in the equation (1).

  $\begin{array}{l}{J}_P=-1.6\times {10}^{-19}\times 15\times \frac{-{10}^{15}}{10\times {10}^{-6}\times {10}^2}{e}^{\frac{-10\times {10}^{-6}\times {10}^2}{10\times {10}^{-6}\times {10}^2}}\\ J_P=2.4\times \frac{-{10}^{-3}}{10\times {10}^{-6}\times {10}^2}e{-}^{{}^{\frac{-10\times {10}^{-6}\times {10}^2}{10\times {10}^{-6}\times {10}^2}}}\\ J_P=2.4\times \frac{1}{2.7182}\\ J_P=0.883\frac{\text{A}}{{\text{cm}}^{\text{2}}}\end{array}$

To determine

The hole diffusion current density at given location.

Answer to Problem 1.17P

The hole diffusion current density at $x=30\text{μm}$ is $0.1194\text{A}/{\text{cm}}^2$ .

Explanation of Solution

Given:

The expression of the hole concentration is $p(x)={10}^4+{10}^{15}\exp \left(\frac{-x}{L_P}\right)\text{where }x\ge \text{0}$

The value of the $L_P=10\text{μm}$ .

The hole diffusion density is $D_p,D_p=15{\text{cm}}^2/\text{s}$ .

Location is $x=30\text{μm}$ .

Calculation:

The gradient of the hole concentration is $\frac{dp(x)}{dx}$ .

Differentiating the equation of the hole concentration,

  $\begin{array}{l}\frac{dp(x)}{dx}=\frac{d}{dx}\left({10}^4+{10}^{15}\exp \left(\frac{-x}{L_P}\right)\right)\text{where }x\ge \text{0}\\ \frac{dp(x)}{dx}=\frac{d}{dx}\left(\exp \left(\frac{-x}{L_P}\right)\right)\text{where }x\ge \text{0}\\ \frac{dp(x)}{dx}=\frac{-{10}^{15}}{L_P}{e}^{\frac{-x}{L_P}}\text{where }x\ge \text{0}\end{array}$

The current diffusion density is given by the expression as shown below:

  $J_P=-e{D}_P\frac{dp(x)}{dx}\mathrm{..............................}(1)$

Plugging the given values in the equation (1).

  $\begin{array}{l}{J}_P=-1.6\times {10}^{-19}\times 15\times \frac{-{10}^{15}}{10\times {10}^{-6}\times {10}^2}{e}^{\frac{-30\times {10}^{-6}\times {10}^2}{10\times {10}^{-6}\times {10}^2}}\\ J_P=2.4\times \frac{-{10}^{-3}}{10\times {10}^{-6}\times {10}^2}e{-}^{{}^{\frac{-30\times {10}^{-6}\times {10}^2}{10\times {10}^{-6}\times {10}^2}}}\\ J_P=2.4\times 0.0498\\ J_P=0.1194\frac{\text{A}}{{\text{cm}}^{\text{2}}}\end{array}$