The diffusion current density due to diffusion of electrons is given by: Jn=(e)(Dn)(dn/dx) The diffusion current density due to diffusion of holes is given by: Jp=-(e)(Dp)(dp/dx)

I understand that for a semiconductor, Electron diffusion current density and hole diffusion current density are opposite to each other.So they will have the opposite sign. But why is electron diffusion current positive and hole diffusion current negative?(despite the fact that electron charge is -ve and hole charge is +ve)

  • \$\begingroup\$ Err, the charge is negative ? \$\endgroup\$ – Andy aka Nov 15 '16 at 18:20
  • \$\begingroup\$ Because Ben Franklin got it wrong www-spof.gsfc.nasa.gov/Education/woppos.html \$\endgroup\$ – JIm Dearden Nov 15 '16 at 18:27
  • \$\begingroup\$ Electron current versus Conventional Current is confused a lot when we talk about semiconductor and electrons flow. So it is simply a matter of how you interpret the problem. Electron current should be used in this case. \$\endgroup\$ – 12Lappie Nov 15 '16 at 19:07

Diffusion happens because of an unequal concentration of carriers. Carriers diffuse from the higher concentration to the lower concentration. The change of concentration is described by a gradient which points in the direction of the higher (!) concentration. In other words they move in the direction of the negative gradient.

This is where the additional minus sign comes in so that the current becomes positive.

  • \$\begingroup\$ Sir thank you for explaining so well. Sir I would like to ask a basic question. If the concentration of p is given by p(x)=e^(-x), then Jp will be positive, ie hole current density will be in the +ve x direction. On the other hand, if p is given by p(x)=e^(x), the hole current density will be negative implying diffusion of holes in the negative x direction. \$\endgroup\$ – Soumee Nov 16 '16 at 15:37
  • \$\begingroup\$ Sir can you help me with this question electronics.stackexchange.com/q/313330/115973 \$\endgroup\$ – Soumee Jun 28 '17 at 9:57

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.