The majority carrier conduction electrons in the n-region of a pn-junction are closer to the the positively charged donor ions than the negatively charged acceptor ions in the p-region. The situation is similar for the majority holes. enter image description here

I don't understand why

A: Majority carrier electrons in the n-region don't flow into the positively charged region when the coulomb force would favor this due to the fact that the positive ions are closer to the n-region than the negative ions.


B: Why the electrons would diffuse across the space charge region when there are empty lower energy conduction band states in the space charge region.

It seems like both diffusion and electric field should favor redistribution into the space charge region, but of course this doesn't happen.

Here is the calculation of electric field using Poisson's equation:

$\nabla ^2 V= \nabla \cdot E = \frac{\rho}{\epsilon} $

In the p-region

$\rho = -eN_a$

and in the n-region

$\rho = eN_d$

Where $N_a$ and $N_d$ are the doping concentrations in the p and n regions, respectively.

both are constant with respect to position, thus

$E_a=\frac{-eN_a}{\epsilon}x + C_a$

$E_d=\frac{eN_d}{\epsilon}x + C_d$

and since we assume the electric field to be zero in the neutral regions, we set $C_d$ and $C_a$ accordingly



Where $-x_p$ and $x_n$ are the borders of the space charge region.

This calculation is fine, but it doesn't answer my question because it just tells you the steady-state solution given assumed boundary conditions and the distribution of charge. I need to know why the charge is distributed this way.

Thanks in advance for any help in clearing this up.

  • $\begingroup$ So, the field-induced drift opposing diffusion is not reasonable to you? $\endgroup$ – Jon Custer Mar 15 at 0:31
  • $\begingroup$ Given the steady-state solution it would be, but I'm not convinced this should be a steady-state solution. It seems like coulomb force and diffusion current should push the charge carriers into the space-charge region. $\endgroup$ – Zulle Mar 16 at 21:52

Your Answer

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

Browse other questions tagged or ask your own question.