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Considering a PN junction with a difference in fermi level between the P and N type regions given by eV0. Does this V0 correspond to the potential at which the diode "switches on"? My intuition tells me that they probably are as when you apply an external potential of V0 across the junction the fermi levels are the same and at this point the electrons/holes can easily pass across the junction. I have been unable to find sources that either confirm or deny this so links to any papers or textbooks would be greatly appreciated.

(See the first diagram on this page for reference http://www.electronics-tutorials.ws/diode/diode_3.html)

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  • $\begingroup$ What do you mean switch on? Clarify exactly what you are asking here an someone can probably help. $\endgroup$ – boyfarrell Dec 14 '14 at 23:26
  • $\begingroup$ electronics-tutorials.ws/diode/diode_3.html looking at the first diagram on this page, i mean the "knee voltage". Apologies if those terms are not in common usage. I think they are related by the ideality factor of the diode but i am not sure. $\endgroup$ – Chris2807 Dec 14 '14 at 23:30
  • $\begingroup$ I see them now in your title but not the body. Anyway thanks for clarifying. When you apply a forward bias which is equal to the builtin potential you have reduced the builtin field to zero so there is no electrostatic impediment to the current flow. So I think your hunch is correct. Whether or not it's exactly the same I'm not sure but it won't be far off in my opinion. $\endgroup$ – boyfarrell Dec 14 '14 at 23:41

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