The Feynman Lectures on Physics vol III, section 14-4 shows the following for a pn junction (a), with (b) representing potential and (c) representing carrier density:

                                    Potential in a semiconductor

Why does the carrier density vary as shown in (c)? Images of the depletion region show a negative charge associated with the p-type side of the region, and a positive charge associated with the n-type side, so surely there should be a greater electron density in the p-type side of the region and a greater hole density in the n-type side?

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    $\begingroup$ The actual shape depends on the doping profiles in the $n$ and $p$ regions. What you have shown is a simple didactic representation - don't read too much into it. $\endgroup$
    – Jon Custer
    Oct 22, 2019 at 17:22

1 Answer 1


The carrier densities far away from the depletion region will be the equilibrium densities of n and p type carriers respectively. However in the depletion region because the p-type and n-type materials are in contact the majority carriers (let's take the case of p type) will be attracted and hence diffuse to the n type material. This process will stop when the positive charge builds up in the n type material (as the electrons would be diffusing (lost) to the p type) and vice versa for the other side.
The possibility of electron density on the p-type cannot extend beyond a certain limit and certainly not be more than the majority carriers under transient or equilibrium conditions. The majority carriers will still be donor sites i.e,holes.


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