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Also, why does width of depletion layer decreases and electric field increases if doping increases

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closed as too broad by Voltage Spike, winny, clabacchio Jul 18 '17 at 7:15

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ One question at a time, and provide a background on the question please. A well thought out question that is specific and includes a few paragraphs garners respect and good answers. A short question will proabably be closed, please edit your question. A bonus would be to provide the electric field equation for diodes and and explanation of how you think it behaves. \$\endgroup\$ – Voltage Spike Jul 9 '17 at 5:26
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    \$\begingroup\$ @laptop2d In general yes, but that isn't really two separate questions. If you know the answer to one you have the answer to the other. \$\endgroup\$ – Matt Jul 9 '17 at 13:33
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When you form a PN junction you get diffusion of charge carriers from one side into the other. This causes depletion of the carriers from both semiconductors. The semiconductor outside the depletion region is charge neutral, and the semiconductor inside the depletion region is assumed to have a constant charge density equal to the dopant concentration. This is due to the ionization of the dopant atoms, each contributing one electrons or hole worth of charge. Both the p and the n sides of the junction must have equal total depletion charge in the depletion region.

If you increase the dopant concentration of one side of the junction, two things happen. First, you now must deplete more charge to account for the increase in potential between the sides of the junction. Second, the charge density in the higher doped semiconductor has increased. This leads to an overall reduction in the thickness of the depletion region on the higher doped side since you need to deplete less of the semiconductor to get the required depletion charge.

Electric field is the integral of charge density. This reduction in depletion width thickness causes an increase in electric field because the charge density has increased and the depletion width has decreased.

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  • \$\begingroup\$ I don't know if it's the difficulty of the content or your answer, but I had to read link below and then your answer 4 times to refresh my understanding of what was going on. en.wikipedia.org/wiki/Depletion_region \$\endgroup\$ – horta Jul 9 '17 at 14:12
  • \$\begingroup\$ If you increase the doping on both sides of the junction, does that cause the depletion layer to remain constant or do both sides shrink? I'm trying to figure out if this depletion region shrinking is due to a constant contact potential or because of the relative doping of the other side. \$\endgroup\$ – horta Jul 9 '17 at 14:27
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    \$\begingroup\$ @horta Both sides shrink. I didn't want to get too in depth with my answer so I left that out. But the increase in charge density more than offsets the increase in V_bi of the junction. \$\endgroup\$ – Matt Jul 9 '17 at 14:29

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