A question on p-n junction depletion layer

[cjs94]

Hi,

I'm studying the p-n junction and I'm a little confused about the depletion layer that forms. I understand how electrons and holes diffuse across the junction, recombine and leave charged ions but what I don't understand is why the ions remain fixed around the junction. Surely the positive ions in the n-type region will attract the free electrons remaining in that region further away from the junction, effectively causing the ions to diffuse evenly throughout the n-type region?

Chris

 

[nsaspook]

Hi,

I'm studying the p-n junction and I'm a little confused about the depletion layer that forms. I understand how electrons and holes diffuse across the junction, recombine and leave charged ions but what I don't understand is why the ions remain fixed around the junction. Surely the positive ions in the n-type region will attract the free electrons remaining in that region further away from the junction, effectively causing the ions to diffuse evenly throughout the n-type region?

Chris

The junction is at thermal equilibrium after the movement of charges when the junction is created. An electric field is created from the charge separation that stops further recombination but the device as a whole is neutral.
http://www.optique-ingenieur.org/en/courses/OPI_ang_M05_C02/co/Contenu_05.html

 

[marcusl]

Surely the positive ions in the n-type region will attract the free electrons remaining in that region further away from the junction, effectively causing the ions to diffuse evenly throughout the n-type region?

Think in terms of two potentials, one is the chemical potential the other electric. The chemical potential drives free electrons from the region of high concentration (n side) to low concentration (p side). This action increases until the electric potential created by the exposed, fixed positive ions on the n side is strong enough to balance the diffusion arising from the chemical potential gradient. At this point, the depletion zone is stable (in equilibrium).