Understand PN Junction Diode: Diffusion & Electrons

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Discussion Overview

The discussion centers on the behavior of a PN junction diode, specifically focusing on the diffusion of charge carriers (holes and electrons) when the diode is forward biased. Participants explore concepts related to doping levels, recombination of charge carriers, and the implications for the flow of current in the diode.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions whether electrons will reach the positive terminal of the battery if all charge carriers have recombined and become neutral.
  • Another participant notes that the amount of doping in the P and N junctions influences the extent of recombination and the width of the depletion region under forward bias.
  • A different viewpoint emphasizes that doping concentrations are small relative to the total number of atoms, suggesting that there will always be free electrons available.
  • One participant corrects a previous statement, clarifying that atoms become ions rather than neutralizing, and explains the role of ions in creating electric fields and voltage differences.
  • Another participant reiterates that ions remain ions and that only holes and electrons recombine, not the ions themselves.

Areas of Agreement / Disagreement

Participants express differing views on the nature of recombination and the status of ions in the diode, indicating that multiple competing perspectives remain without a clear consensus.

Contextual Notes

There are unresolved assumptions regarding the definitions of terms like "neutral" and "ion," as well as the implications of doping levels on the behavior of the diode.

M.Kalai vanan
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If p-type semiconductor and n-type semiconductor of a diode are equally doped, and if the diode is forward biased, then holes will move toward the n-type semiconductor and electrons will move toward the p-type semiconductor and they will diffuse with each other and the ions become neutral atoms since the hole and electron have disappeared. Then will there be any electron that will go to the positive terminal of the battery if all of them have diffused with each other? I can't understand, please help me!
 
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The amount of doping in the P and N junctions determines whether ALL or SOME of them recombine and become neutral. Also, the width of the depletion region gets thinner as forward bias voltage is applied across the diode, and this process too determines whether all or some of the dominant charge carriers in the semiconductor junction actually recombine to neutrality.
 
Doping concentrations are tiny compared to the total number of atoms - there are always electrons left.

and the ions become neutral atoms since the hole and electron have disappeared.
It is the other way round, the atoms become ions. Holes are not ions - they are spots where an additional electron can be bound. And free electrons come from atoms where one electron is not bound, so it will easily form an ion.
Those ions are then the source for the electric fields and therefore the voltage difference, by the way.
 
then it becomes neutral
 

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