0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Can you elaborate on what you mean by "ambipolar transport equation"? $\endgroup$ – freecharly Sep 21 '16 at 21:01
  • $\begingroup$ @freecharly, in this context I believe he just means diffusion with electrons and holes (two different charge carriers). $\endgroup$ – leastaction Sep 21 '16 at 21:18
  • $\begingroup$ yes, that's what I mean. Sorry if there is a more general case - its my first class on the topic. $\endgroup$ – Malcolm Regan Sep 21 '16 at 21:19
  • $\begingroup$ Thanks for including the mentioned equation. In the meantime I looked into the book by Neaman, where this term is used for the ambipolar diffusion equation. $\endgroup$ – freecharly Sep 21 '16 at 22:15
  • $\begingroup$ you mean the equation is called something else in that book or are you saying that the diffusion current densities are indeed calculated with the ambipolar diffusion current? $\endgroup$ – Malcolm Regan Sep 21 '16 at 22:25
0
$\begingroup$

The answer to this paradox is, as far as I can see, the following. The so called ambipolar transport equation is derived in Neaman's book by assuming (as an approximation) strict charge neutrality so that the local excess carrier densities for electrons and holes become exactly equal. Let us consider the relevant case of a p-type semiconductor with low electron injection (minority carriers). If you have a local excess electron density the assumption means that you have exactly the same excess hole density in order to keep neutrality. As the ambipolar diffusion is dominated by the excess electron density giving the diffusion coefficient for electrons in this approximation you get exactly the same spatial and temporal behavior of the excess hole concentration. You cannot use the diffusion current density equation for holes anymore because it contradicts the strict neutrality assumption. In reality you have only quasi-neutrality, the spatial shape of excess carrier densities is not exactly the same so that a space charge is created leading to internal fields so that the hole (and electron) current has also a drift component in addition to the diffusion component.

$\endgroup$
  • $\begingroup$ gah forgot all about the low injection condition and quasi neutrality and stuff. I understand it much better now! Thanks a lot. I tried to upvote you but I haven't the reputation. I owe you my A. $\endgroup$ – Malcolm Regan Sep 21 '16 at 23:00
  • $\begingroup$ I wish you good luck! $\endgroup$ – freecharly Sep 22 '16 at 0:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.