How can the neutral regions in a diode have constant potential?

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SUMMARY

The potential in the neutral regions of a PN junction diode remains constant due to the accumulated effect of charges in the depletion region, not the neutral region itself. The neutral region contains charges that are not mobile, which means they do not contribute to the electric field. The relationship V=Q/4ΠΣr is misapplied when considering the potential in these regions, as the depletion region is modeled as two wide sheets of charge, leading to a uniform potential across the neutral areas. Understanding this concept is essential for accurately analyzing diode behavior.

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  • Understanding of PN junction diodes
  • Familiarity with electric potential and electric fields
  • Knowledge of charge distribution in semiconductor materials
  • Basic principles of electrostatics, particularly regarding sheets of charge
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Physics students, electrical engineers, and anyone studying semiconductor devices will benefit from this discussion, particularly those interested in the behavior of PN junction diodes and electric potential concepts.

Amal Thejus
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Homework Statement


It is said that the potential in the neutral regions of a PN junction diode is CONSTANT.

Homework Equations


V=Q/4ΠΣr

The Attempt at a Solution


It is said in all textbooks to ASSUME that the electric field in the neutral regions as zero.
Two aspects confuse me.
1. What charges cause the CONSTANT potential in the neutral region? Is it the charges in the depletion region or the charges in the neutral region(even though net charge is zero there are still charges in the neutral region,which can a cause a potential,right?)
As, for a point in space to have a potential there must be a charge involved since V=Q/4ΠΣr

2. If the potential is due to the charges in the depletion region, how can it be constant through out the neutral region? Since as we move away from the depletion region charges through the neutral region doesn't 'r' in the equation,V=Q/4ΠΣr vary?

Please don't give an answer like since electric field is zero the potential is constant, since i don't think that is a fundamental way to think about it. Please give an answer in terms of the fundamental DEFINITION of Electric Potential. Thanks in advance.
 
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1. The charges in the doped regeon ensure this. In the neutral regeon, there are charges, but they are not mobile.
2. You are using the relation incorrectly... the doped regeon is treated as being much wider than it is thick. Thus it is modeled as two very wide sheets of charge.The equation you wrote is for only one charge.

Calculate the potential and electric field due to a large sheet of charge...
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html
...

https://en.m.wikipedia.org/wiki/Depletion_region
 
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Simon Bridge said:
1. The charges in the doped regeon ensure this. In the neutral regeon, there are charges, but they are not mobile.
2. You are using the relation incorrectly... the doped regeon is treated as being much wider than it is thick. Thus it is modeled as two very wide sheets of charge.The equation you wrote is for only one charge.

Calculate the potential and electric field due to a large sheet of charge...
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html
...

https://en.m.wikipedia.org/wiki/Depletion_region
So you are saying the CONSTANT potential in the NEUTRAL regions is due to the charges in the NEUTRAL region itself?

If so how does the neutral n region which has more electrons than the neutral p region have a positive potential compared to the p region ?
I have attached a photo showing the variation of ELECTRIC POTENTIAL inside the diode.
 

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[QuoteSo you are saying the CONSTANT potential in the NEUTRAL regions is due to the charges in the NEUTRAL region itself? [/quote]No.
I am saying it is due to the accumulated effect of all the charges in the charged regeon.
To convince yourself of this you really need to compute the potential yourself, from first principles. This is a common exercize for year 2 college physics students.
You could also read the links provided.
 

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