# [Physics] Diffusion Current Calculation Following Ambipolar Transport Equation Solution

homework-and-exercisessemiconductor-physics

In calculating the diffusion current density in a semiconductor in a case where the minority carrier concentration is found via the ambipolar transport equation, is the minority carrier diffusion coefficient used for finding both the electron diffusion current density and the hole diffusion current density?

That is, say, for a p-type semiconductor where the excess minority carrier concentration $\delta_n$ is the solution to the ambipolar transport equation:

$$D' \frac{d^2(\delta_n)}{dx^2}+ \mu'E \frac{d(\delta_n)}{dx}+g-\frac{\delta_n}{\tau_n}=\frac{d(\delta_n)}{dt}$$

where the ambipolar diffusion coefficient and ambipolar mobility $D'$ and $\mu'$
are equal to the minority carrier diffusion coefficient and mobility, $D_n$ and $\mu_n$, respectively.

If one were then to compute the diffusion current densities of electrons and holes, would they both be found using the ambipolar diffusion coefficient $D'=D_{n}$?

$$J_{n|diff}=eD'\frac{d (\delta n) }{dx}$$
and
$$J_{p|diff}=-eD'\frac{d (\delta p) }{dx}$$

This doesn't make intuitive (or intellectual) sense to me but it corresponds to the answer in the back of the book. (Semiconductor Physics and Devices 4th Edition problem 6.19)

What would make sense to me is the electron and hole diffusion current densities being found with their respective diffusion coefficients:

$$J_{n|diff}=eD_n\frac{d (\delta n) }{dx}$$
and
$$J_{p|diff}=-eD_p\frac{d (\delta p) }{dx}$$

where $\frac{d (\delta n) }{dx}=\frac{d (\delta p) }{dx}$

but maybe there's something going on when excess carriers are involved that I'm not keen on. Is this the case?

Am I right or is the book? If the book is right, why?