Question:
Consider that the concentration of electrons in a semiconductor bar varies linearly from \(2\times10^{17}\text{ cm}^{-3}\) at \(x=1\ \mu\text{m}\) to \(1\times10^{16}\text{ cm}^{-3}\) at \(x=4\ \mu\text{m}\) along the \(x\)-direction. Assume that the concentration of electrons does not vary along the \(y\)- and \(z\)-directions.
Given: the mobility of electron is \(1400\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}\), the thermal voltage is \(25\text{ mV}\) and the electronic charge is \(1.6\times10^{-19}\) Coulomb.
The density of electron diffusion current (in \(\text{A/mm}^2\)) is (rounded off to two decimal places).
Consider that the concentration of electrons in a semiconductor bar varies linearly from \(2\times10^{17}\text{ cm}^{-3}\) at \(x=1\ \mu\text{m}\) to \(1\times10^{16}\text{ cm}^{-3}\) at \(x=4\ \mu\text{m}\) along the \(x\)-direction. Assume that the concentration of electrons does not vary along the \(y\)- and \(z\)-directions.
Given: the mobility of electron is \(1400\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}\), the thermal voltage is \(25\text{ mV}\) and the electronic charge is \(1.6\times10^{-19}\) Coulomb.
The density of electron diffusion current (in \(\text{A/mm}^2\)) is (rounded off to two decimal places).
Given: the mobility of electron is \(1400\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}\), the thermal voltage is \(25\text{ mV}\) and the electronic charge is \(1.6\times10^{-19}\) Coulomb.
The density of electron diffusion current (in \(\text{A/mm}^2\)) is (rounded off to two decimal places).
Show Hint
Get Dn from the Einstein relation Dn = mu_n VT, then use Jn = q Dn (dn/dx) with the gradient in cm and cm^-3 units before converting to A/mm^2.
Updated On: Jul 20, 2026
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Correct Answer: -35.47
Solution and Explanation
Step 1: Write down the diffusion current formula.
When electron concentration \(n\) changes with position, electrons move by diffusion from the region of high concentration to the region of low concentration. The resulting conventional current density carried by this electron diffusion is
\[ J_{n,\text{diff}}=qD_n\frac{dn}{dx} \]
where \(q\) is the electronic charge, \(D_n\) is the electron diffusion coefficient, and \(\dfrac{dn}{dx}\) is the rate of change of electron concentration with position.
Step 2: Find the diffusion coefficient using the Einstein relation.
The Einstein relation connects mobility and the diffusion coefficient through the thermal voltage \(V_T\):
\[ D_n=\mu_nV_T \]
Here \(\mu_n=1400\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}\) and \(V_T=25\text{ mV}=0.025\text{ V}\).
\[ D_n=1400\times0.025=35\ \text{cm}^2/\text{s} \]
Step 3: Compute the concentration gradient.
The concentration changes linearly from \(n_1=2\times10^{17}\text{ cm}^{-3}\) at \(x_1=1\ \mu\text{m}\) to \(n_2=1\times10^{16}\text{ cm}^{-3}\) at \(x_2=4\ \mu\text{m}\). Converting positions to centimeters, \(1\ \mu\text{m}=10^{-4}\text{ cm}\), so \(x_1=1\times10^{-4}\text{ cm}\) and \(x_2=4\times10^{-4}\text{ cm}\).
\[ \frac{dn}{dx}=\frac{n_2-n_1}{x_2-x_1}=\frac{1\times10^{16}-2\times10^{17}}{4\times10^{-4}-1\times10^{-4}} \]
\[ =\frac{-1.9\times10^{17}}{3\times10^{-4}}=-6.333\times10^{20}\ \text{cm}^{-4} \]
Step 4: Substitute into the diffusion current formula.
\[ J_{n,\text{diff}}=qD_n\frac{dn}{dx}=(1.6\times10^{-19})(35)(-6.333\times10^{20}) \]
\[ =(5.6\times10^{-18})(-6.333\times10^{20})=-3546.67\ \text{A/cm}^2 \]
Step 5: Convert to A/mm^2.
Since \(1\text{ cm}=10\text{ mm}\), \(1\text{ cm}^2=100\text{ mm}^2\), so dividing an A/cm^2 value by \(100\) gives A/mm^2:
\[ J_{n,\text{diff}}=\frac{-3546.67}{100}=-35.4667\ \text{A/mm}^2 \]
Step 6: Round the answer.
Rounded to two decimal places, the density of electron diffusion current is
\[ \boxed{-35.47\ \text{A/mm}^2} \]
The negative sign only shows that this diffusion current flows in the negative \(x\)-direction, opposite to the direction along which the concentration is falling.
When electron concentration \(n\) changes with position, electrons move by diffusion from the region of high concentration to the region of low concentration. The resulting conventional current density carried by this electron diffusion is
\[ J_{n,\text{diff}}=qD_n\frac{dn}{dx} \]
where \(q\) is the electronic charge, \(D_n\) is the electron diffusion coefficient, and \(\dfrac{dn}{dx}\) is the rate of change of electron concentration with position.
Step 2: Find the diffusion coefficient using the Einstein relation.
The Einstein relation connects mobility and the diffusion coefficient through the thermal voltage \(V_T\):
\[ D_n=\mu_nV_T \]
Here \(\mu_n=1400\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}\) and \(V_T=25\text{ mV}=0.025\text{ V}\).
\[ D_n=1400\times0.025=35\ \text{cm}^2/\text{s} \]
Step 3: Compute the concentration gradient.
The concentration changes linearly from \(n_1=2\times10^{17}\text{ cm}^{-3}\) at \(x_1=1\ \mu\text{m}\) to \(n_2=1\times10^{16}\text{ cm}^{-3}\) at \(x_2=4\ \mu\text{m}\). Converting positions to centimeters, \(1\ \mu\text{m}=10^{-4}\text{ cm}\), so \(x_1=1\times10^{-4}\text{ cm}\) and \(x_2=4\times10^{-4}\text{ cm}\).
\[ \frac{dn}{dx}=\frac{n_2-n_1}{x_2-x_1}=\frac{1\times10^{16}-2\times10^{17}}{4\times10^{-4}-1\times10^{-4}} \]
\[ =\frac{-1.9\times10^{17}}{3\times10^{-4}}=-6.333\times10^{20}\ \text{cm}^{-4} \]
Step 4: Substitute into the diffusion current formula.
\[ J_{n,\text{diff}}=qD_n\frac{dn}{dx}=(1.6\times10^{-19})(35)(-6.333\times10^{20}) \]
\[ =(5.6\times10^{-18})(-6.333\times10^{20})=-3546.67\ \text{A/cm}^2 \]
Step 5: Convert to A/mm^2.
Since \(1\text{ cm}=10\text{ mm}\), \(1\text{ cm}^2=100\text{ mm}^2\), so dividing an A/cm^2 value by \(100\) gives A/mm^2:
\[ J_{n,\text{diff}}=\frac{-3546.67}{100}=-35.4667\ \text{A/mm}^2 \]
Step 6: Round the answer.
Rounded to two decimal places, the density of electron diffusion current is
\[ \boxed{-35.47\ \text{A/mm}^2} \]
The negative sign only shows that this diffusion current flows in the negative \(x\)-direction, opposite to the direction along which the concentration is falling.
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