# How valence electrons cross the field barrier in P-N junction?

The text from book is:

    1. Electron leaves negative terminal of the battery and
enters the right end (N-type material) of the diode.
2. Electron then travels through the N-type material.
3. The electron nears the junction and recombines and
becomes a valence electron.
4. The electron now travels through the P-type material as a valence electron.
5. The electron then leaves the diode and flows back to
the positive terminal of the battery.


The electrons from N-type material combines with the positive ions in junction. And they become Valence electrons. But how do they cross the negative part of junction? Will not the electrostatic repulsion let it not cross it?

One solution I believe if my doubt is correct is that, the barrier electric field produced by depletion layer behaves like a battery with opposite polarity.

And thus the valence electron crosses because of the field and reaches to P-type material.

But this produces a contradiction when forward current keeps flowing and the depletion layer, so the barrier field, reduces after some time. Please correct me.

• Sometimes conduction in the p-type is described as conduction by holes. Holes are created at the connection to the p-type and travel towards the junction. Electrons also travel towards the junction in the n-type. They meet and the electron fills the hole, at the junction. On the other hand putting electrons into the p-type just fills holes, making them not available to conduct, so the resistance is very high. Sep 12 '20 at 10:39
• @Peter This is what leads to formation of depletion layer. The free electrons from valance band of N-type goes to the hole of P-type and becomes negative ions, while the place it left becomes a positive ion. Which produces an electric field. The question is about: What if an another valence electron is sent in this positive ion section, how it will cross the negative ion section? Sep 12 '20 at 11:01
• The short answer is "with great difficulty"! Adding electrons to the p-type merely fills more holes, ie there are fewer places for valence electrons to move into. In the n-type, removing electrons obviously makes it less conducting. The effect is to widen the depletion layer. Of course if the applied potential/field is great enough electrons will be pulled out of the holes, and the diode has broken down (eg Zener diode). Sep 12 '20 at 11:23

"The electrons from N-type material combines with the positive ions in junction. And they become valence electrons. But how do they cross the negative part of junction? Will not the electrostatic repulsion let it not cross it?"

The first sentence above is incorrect. The electron does not recombine with a positive ion. Instead, the electron falls to a missing state in the valence band (it recombines with a hole). What your text calls "becoming a valence electron". This is a culmination of the two processes:

1. holes moving to the junction from the left (and thus a net valence electron moving right from the junction to the left hand contact).
2. electrons moving to the junction from the right.

When the electron and hole recombine in and near the depletion region they have completed the motion of one net electron moving from the right contact to the left contact. These electrons and holes move across the junction by diffusion. The driving force for diffusion, the large gradient in concentration between n and p under forward bias, is much stronger than the opposing electric field of the junction.

I would urge you to discard this text. It's teaching the version of semiconductor physics that says "there are no such things as holes". It also completely ignores diffusion in the explanation. Your question clearly points out how much confusion this approach can have.

We have had an excellent theory of semiconductors since William Shockley's seminal book:

Electrons and holes in semiconductors : with applications to transistor electronics, William Shockley, New York : Van Nostrand, 1950.