Diffusion current density in a semiconductor - A/cm^2?

Question

The units for current density is generally $\frac{A}{m^2}$. One particular equation I'm seeing from a textbook (Microelectronics, by Donald A. Neaman) is:

$$J_n=e D_n \frac{dn}{dx}$$

for the diffusion current density due to the diffusion of electrons (for one dimension). In this context, $e \text{ (unit: A)}$ is the electronic charge, $dn/dx$ is the gradient of the electron concentration ($n$ is electron concentration (unit: $cm^-3$), $x$ is distance ($\mu m$)), and $D_n \text{ (unit: cm}^2/\text{s)}$ is the electron diffusion coefficient.

The units are then: $$\text{A}\cdot \frac{\text{cm}^2}{\text{s}}\cdot \frac{1}{\text{cm}^3\cdot \mu m}$$ $$=\text{A}\cdot \frac{\text{cm}^2}{\text{s}}\cdot \frac{1}{\text{cm}^4 \cdot 10^{-4}} $$ $$= \frac{\text{A}}{\text{cm}^2\cdot \text{s}\cdot 10^{-4}}$$

The quantity for $dx$ is converted from $\mu m$ to $cm$ with $10^{-4}$.

What happens to the $1/s$ from $D_n$ from the above equation?

Answer 1

e is the fundamental charge, not the fundamental current, so its units are coulombs, not amps.

And, of course, a coulomb is an ampere-second, so everything works out as expected.

Answer 2

Charge is equal to product of current and time. ... Therefore ( seconds ) time in denominator get cancelled by numerator (seconds) time. Therefore units of diffusion current density =A/cm^2