I understand that when a pn-junction is forward biased minority carriers accumulate and form an exponential charge concentration in relationship to the distance from the junction, as the figure below shows:

enter image description here

This excess concentration occurs because of the increased diffusion current caused by the lowered potential barrier in the depletion region.

However, in the open-circuit case, that is, without external voltage sources, it is assumed (on the figure taken from Sedra book on Microelectronics) that the concentration in that case ($p_{n0}$ and $n_{p0}$) is uniform with respect to the $x$-axis. Why is that the case if in the open-circuit case we still have a non-zero diffusion current? Why the charge density $p_{n0}$ is not higher near the junction?


1 Answer 1


In the open circuit case, you don't have any current inside the p-n junction since it's in equilibrium...there is an in-built potential barrier in the depletion region, which prevents diffusion of minority carriers across the junction. Since the sample is in equilibrium now, there are no issues as to why the minority carrier concentration is uniform w.r.t. position.


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