Question:medium
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
Show Solution
Correct Answer: -35.47
Solution and Explanation
Step 1: Switch to SI units from the start.
Instead of working in cm and converting only at the end, convert every given value to metres up front. Recall $1\ \mu\text{m}=10^{-6}$ m and $1\text{ cm}^{-3}=10^{6}\text{ m}^{-3}$.
Step 2: Convert the concentrations and positions.
\[ n_1=2\times10^{17}\text{ cm}^{-3}=2\times10^{23}\text{ m}^{-3},\quad n_2=1\times10^{16}\text{ cm}^{-3}=1\times10^{22}\text{ m}^{-3} \]
\[ x_1=1\times10^{-6}\text{ m},\quad x_2=4\times10^{-6}\text{ m} \]
Step 3: Compute the gradient in SI units.
\[ \frac{dn}{dx}=\frac{n_2-n_1}{x_2-x_1}=\frac{1\times10^{22}-2\times10^{23}}{4\times10^{-6}-1\times10^{-6}}=\frac{-1.9\times10^{23}}{3\times10^{-6}}=-6.333\times10^{28}\text{ m}^{-4} \]
Step 4: Convert mobility and get Dn in SI units.
Since $1\text{ cm}^2=10^{-4}\text{ m}^2$, the mobility becomes $\mu_n=1400\times10^{-4}=0.14\text{ m}^2/(\text{V}\cdot\text{s})$. Using $D_n=\mu_n V_T$,
\[ D_n=0.14\times0.025=0.0035\text{ m}^2/\text{s} \]
Step 5: Compute the current density in A/m^2.
\[ J_n=qD_n\frac{dn}{dx}=(1.6\times10^{-19})(0.0035)(-6.333\times10^{28}) \]
\[ =(5.6\times10^{-22})(-6.333\times10^{28})=-3.5467\times10^{7}\text{ A/m}^2 \]
Step 6: Convert to A/mm^2.
Since $1\text{ m}^2=10^{6}\text{ mm}^2$, one A/m$^2$ equals $10^{-6}$ A/mm$^2$:
\[ J_n=(-3.5467\times10^{7})\times10^{-6}=-35.467\text{ A/mm}^2 \]
Step 7: Round and conclude.
\[ \boxed{-35.47\ \text{A/mm}^2} \]
This matches the value found by working entirely in cm, a good check that the unit handling was correct either way.
Instead of working in cm and converting only at the end, convert every given value to metres up front. Recall $1\ \mu\text{m}=10^{-6}$ m and $1\text{ cm}^{-3}=10^{6}\text{ m}^{-3}$.
Step 2: Convert the concentrations and positions.
\[ n_1=2\times10^{17}\text{ cm}^{-3}=2\times10^{23}\text{ m}^{-3},\quad n_2=1\times10^{16}\text{ cm}^{-3}=1\times10^{22}\text{ m}^{-3} \]
\[ x_1=1\times10^{-6}\text{ m},\quad x_2=4\times10^{-6}\text{ m} \]
Step 3: Compute the gradient in SI units.
\[ \frac{dn}{dx}=\frac{n_2-n_1}{x_2-x_1}=\frac{1\times10^{22}-2\times10^{23}}{4\times10^{-6}-1\times10^{-6}}=\frac{-1.9\times10^{23}}{3\times10^{-6}}=-6.333\times10^{28}\text{ m}^{-4} \]
Step 4: Convert mobility and get Dn in SI units.
Since $1\text{ cm}^2=10^{-4}\text{ m}^2$, the mobility becomes $\mu_n=1400\times10^{-4}=0.14\text{ m}^2/(\text{V}\cdot\text{s})$. Using $D_n=\mu_n V_T$,
\[ D_n=0.14\times0.025=0.0035\text{ m}^2/\text{s} \]
Step 5: Compute the current density in A/m^2.
\[ J_n=qD_n\frac{dn}{dx}=(1.6\times10^{-19})(0.0035)(-6.333\times10^{28}) \]
\[ =(5.6\times10^{-22})(-6.333\times10^{28})=-3.5467\times10^{7}\text{ A/m}^2 \]
Step 6: Convert to A/mm^2.
Since $1\text{ m}^2=10^{6}\text{ mm}^2$, one A/m$^2$ equals $10^{-6}$ A/mm$^2$:
\[ J_n=(-3.5467\times10^{7})\times10^{-6}=-35.467\text{ A/mm}^2 \]
Step 7: Round and conclude.
\[ \boxed{-35.47\ \text{A/mm}^2} \]
This matches the value found by working entirely in cm, a good check that the unit handling was correct either way.
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