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Internel potentiel of PN junction? so complicated

  • #1
internel potentiel of PN junction?? so complicated

homojunction PN (not polarized), externel potentiel difference applied = 0 Volt .

i should demonstrate the equation: e*Vbi + Eg + ( Ec - Ef ) + ( Ef - Ev) = 0
Given: Vbi= internel potential
Eg= energy gap
Ef = Fermi energy level
Ec= Energy of conduction level
Ev= Energy of valence level

then i should deduce, that if the junction are so dopped that : Vbi = Eg/e
____________________________________________________________________________

Ive try a lot to solve the first equation, but i couldnt find the solution. what ive found it that Vbi must equal Zero, and thats impossible cause .
hope someone know how to solve that.
 

Answers and Replies

  • #2
homojunction PN- Need help

Hi all, i need help in my little exercise
a homojunction PN (not polarized), externel potentiel difference applied = 0 Volt .

i should demonstrate the equation: e*Vbi + Eg + ( Ec - Ef ) + ( Ef - Ev) = 0
Given: Vbi= internel potential
Eg= energy gap
Ef = Fermi energy level
Ec= Energy of conduction level
Ev= Energy of valence level

then i should deduce, that if the junction are so dopped that : Vbi = Eg/e
__________________________________________________ __________________________

Ive try a lot to solve the first equation, but i couldnt find the solution. what ive found it that Vbi must equal Zero, and thats impossible cause .
hope someone know how to solve that.
 
  • #3
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
17
Hi all, i need help in my little exercise
a homojunction PN (not polarized), externel potentiel difference applied = 0 Volt .

i should demonstrate the equation: e*Vbi + Eg + ( Ec - Ef ) + ( Ef - Ev) = 0
Given: Vbi= internel potential
Eg= energy gap
Ef = Fermi energy level
Ec= Energy of conduction level
Ev= Energy of valence level
That equation looks wrong!

Simply by inspection (draw the energy digram for the p-n junction), you should be able to see that:

[tex]eV_{b} + (E_c - E_F) + (E_F - E_v) = E_g[/tex]

then i should deduce, that if the junction are so dopped that : Vbi = Eg/e
__________________________________________________
I don't understand what you mean by this - it looks like an incomplete statement.
 
  • #4
its right, the equation is as you've write it:
e*Vbi - Eg + ( Ec - Ef ) + ( Ef - Ev) = 0 there's a minus infront of Eg .
but what i dont understand it, that if i draw the energy level diagram of the PN junction, i found that Eg = Ec - Ev and its the definition of Eg.
if i write the statement like this: e*Vbi - Eg + Ec - Ef + Ef - Ev = 0 , then it will reduce to
e*Vbi - Eg + Ec - Ev = 0, and as known Eg = Ec - Ev, so it should that Vb = 0.
i really dont understand how to see by drawing the energy diagram for the PN junction that the realtion is right.
 
  • #5
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
7,051
17
In the above equation Ec - Ef is measured in the n-type region and Ef - Ev is measured in the p-type region. This should be more explicitly stated in the equation.
 
  • #6
marcusl
Science Advisor
Gold Member
2,720
382
and as known Eg = Ec - Ev
No. I think your statement of the problem is incorrect or missing some information. You probably mean Ec on one side of the junction and Ev on the other side, in which case Eg = Ec - Ev is not correct. Take a look at the energy diagram for the p-n in your text, or here

http://ece-www.colorado.edu/~bart/book/book/chapter4/ch4_2.htm" [Broken]
 
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  • #7
actualy what i dont understand thats the energy gap for a semiconductor is the difference betwen Ec and Ev. but , what will be the energy gap for a homojunction? and inside the barrier and outside. at left and right its the same?
marcusl, ive chequed your link, the dont indicate that.
 
  • #8
10X to all. finally ive understand it. ive just drawn the energy junction diagram, and ive solved that. i think the problem that there's something missing in the equation, Ef-EVp , that's how it should be writen.
 

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