Understanding the width of the depletion region in a p-n junction

  • Thread starter skweiler
  • Start date
  • #1
7
0
I am studying to become and electrical engineer and am currently taking Electronics I. I am having trouble understanding the change in the depletion region that results from a bias. I understand that at the time of manufacture the free electrons from the n-type semiconductor are attracted to the vacancies in the p-type semiconductor and move until the electric field produced by the newly created ions prohibits further majority carriers from crossing the junction. What I don't understand is exactly which mechanism causes the depletion region to shrink when a diode is forward biased (+ on the p-side, - on the n-side) or vice versa when reversed biased? Is the depletion region defined as the width of the region of ions surrounding the p-n junction or the gap created when a potential "pushes" or "pulls" the majority carriers towards or away from the junction? If it is defined by the ions how does the applied potential fill (or empty) more holes and thus create more (or less) ions? How do the ions and their electrons surrounding the junction react to the applied potential?
 

Answers and Replies

  • #2
309
1
This is how I understand it, with limited knowledge of any quantum physics.

Forward bias: Negative voltage on the N-region, positive on the P-region.

Negative voltage pushes more free electrons into the N-region toward the PN junction, and the positive voltage pulls them out of the P region, in effect pushing holes toward the junction. This makes the depletion region smaller. At about 0.7V, the free electrons and the holes meet in the middle and the diode starts to conduct.

Reverse bias: Negative voltage in the P, positive voltage on the N.

Negative voltage pushed more electrons into the P region, which fill up the holes, restricting how much current can flow through there. Positive voltage pulls the free electrons out of the N region, which reduces its ability to conduct. Increasing the voltages strip out more and more holes and electrons, which widens the depletion region.

With high enough reverse bias though, breakdown occurs and the diode starts to conduct again. That part I have to read about a couple more times to understand.
 
  • #3
309
1
Accidental double post
 
  • #4
7
0
This makes the depletion region smaller.

This is what I don't fully understand. Kindly look at this http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html#c3" in the slide called "Depletion Region." In order for the depletion region to become larger more holes in the P-type need to be filled and more extra electrons in the N-type region need to be gotten rid of. Looking at the diagram the region of ions needs to expand.

Could it be that when the diode is reversed biased the negative potential pushes the minority carriers (electrons) in the P-type towards the junction, to recombine with the holes there in an attempt to enter the valence band and counterbalance, the forces exerted from the negative potential on the one side, and the line of negative ions on the P-side of the P-N junction?

When the diode is forward biased the opposite must occur.
 
Last edited by a moderator:
  • #5
309
1
This is what I don't fully understand. Kindly look at this http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun.html#c3" in the slide called "Depletion Region." In order for the depletion region to become larger more holes in the P-type need to be filled and more extra electrons in the N-type region need to be gotten rid of. Looking at the diagram the region of ions needs to expand.

Could it be that when the diode is reversed biased the negative potential pushes the minority carriers (electrons) in the P-type towards the junction, to recombine with the holes there in an attempt to enter the valence band and counterbalance, the forces exerted from the negative potential on the one side, and the line of negative ions on the P-side of the P-N junction?

When the diode is forward biased the opposite must occur.
That's exactly what I was saying, just in different words. In reverse bias, the majority carriers (holes in P type and free electrons in N type) are all pulled away from the depletion region, making it wider. In forward bias, the majority carriers are pushed toward it. At about 0.7V, the depletion region closes completely.
 
Last edited by a moderator:
  • #6
7
0
Thank you for your help. These different ways of explaining the phenomena are, I believe, practically the same; any majority carriers will react in the opposite manner from the minority carriers because they have opposite charges.
 
  • #7
dlgoff
Science Advisor
Gold Member
3,960
1,936
Welcome to PF skweiler.

It looks like Jiggy-Ninja has answered your question and I see you have found http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html" [Broken] website (I love it).

Now for transistors. Here are a couple of images that I feel are good references.

tran10.gif


tran12.gif


http://hyperphysics.phy-astr.gsu.edu/hbase/solids/trans2.html" [Broken]

Enjoy
 

Attachments

  • tran10.gif
    tran10.gif
    41.1 KB · Views: 610
Last edited by a moderator:

Related Threads on Understanding the width of the depletion region in a p-n junction

Replies
8
Views
12K
A
Replies
3
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
Top