# Depletion region in pn juction

1. Oct 9, 2009

### manofphysics

Regarding the formation of depletion region,it forms because free electrons diffuse into p region from n region. after some recombination of holes, a sufficient potential difference is set up to prevent further diffusion.
Now, consider the region to the left side of juction(whole P region including depletion layer.) only as shown in figure.
It contains layer of negatively charged ions to the right and holes everywhere else.Now, what stops a diffusion current to set up in the P region only, which has the effect of distributing the holes evenly in the P region,(i.e. electrons present in negatively charged ions diffuse further into the P region ) since there is a definite concentration gradient of holes in the P region due to the depletion region.?

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2. Oct 9, 2009

### pseudophonist

Remember that the formation of the depletion region is a dynamic equilibrium. Even though there is a potential in place, this only limits the diffusion current of electrons. There will always be diffusion of electrons into the p-type region, and this is matched by an equal flow of holes flowing into the n-type region.

Diffusion CURRENT (i.e. the overall movement of charge) is proportional to the SLOPE of concentration. Deep in the p-region the concentration is fairly constant and the overall effects of diffusion are zero. Its only near to the junction (i.e. in the depletion region) that the concentration begins to vary and this causes a net diffusion current into the n-region. This current is fed by carriers in the p-region which happen to diffuse into the depletion zone.

Also note that electrons bound to negatively charged ions are essentially fixed in the lattice (unless you massively heat the semiconductor)

3. Oct 9, 2009

### manofphysics

Yeah, but what stops valence electrons in the negatively charged ions ( in p region of the depletion region) from breaking away and recombining with holes in the deeper parts of p region itself?

[If this happens, the behavior of diffusion current of n region changes and the system as a whole will have a constant hole & free electron concentration in a short time.But This does not happen in reality ]

Why? Is there something special about electron bonded to these negative ions?They are the same covalent bonds as in the rest of the p region and I think, they can also gain energy by thermal agitation and so generate holes like the rest of the valence electrons.

Last edited: Oct 9, 2009
4. Oct 9, 2009

### pseudophonist

Consider a boron atom in silicon. This will readily accept an electron from the valence band of silicon and become negatively charged, leaving a hole behind in the silicon. This electron is in the "valence band" of the boron (strictly speaking, a bonding orbital, as opposed to an antibonding orbital). This orbital is lower in energy than the valence band edge of silicon and is degenerate with some states in the band.

There can be recombination of the valence electron to the holes in the p-region, however this leaves behind an empty low energy state which is below the valence band edge. The valence band is so highly populated with electrons that no amount of thermal energy can keep away all the electrons from this unfilled state, and you end up back where you started.

So yes, they are similar to the other covalent bonds, but below the valence band edge in energy, which means even if you can get the electron out it will get replaced almost immediately

EDIT: you may want to check out the first 2-3 pages of "Impurities in type-II staggered InAs/AlSb superlattices" by John D.Dow et al in Superlattices and Microstructures Vol 13 No 4 1993 which gives a pretty detailed explanation of the real meaning of acceptor and donor levels in semiconductors. I found it amazing, but it is quite complicated.

Last edited: Oct 9, 2009
5. Oct 9, 2009

### manofphysics

Thanks, pseudophonist, I'll definitely check out Dow.

6. Oct 29, 2009

### manofphysics

Re: pn juction

I have a related question on pn junctions.
See the attchment.
Now, why is the barrier potential constant on both sides outside the depletion region.
for eg. The negative charges in the depletion region should create a negative potential=f(x), not constant.similarly with positive charges, they should create a positive potential=g(x) to the right of them in the n region, not a constant line.
Can anybody explain this?

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7. Oct 29, 2009

### pseudophonist

a potential gradient only exists when there is an electric field. Away from the depletion region the charge of the electrons is counteracted by the holes and ionised dopants in the n-type region. In the p-type region the charge of the holes is countered by electrons and ionised acceptors. In these regions there is no net electric field and so there is no drop in potential

In the depletion region, the density of carriers is very low, however this does not affect the ionised dopants. This causes a net electric field, which causes a change in potential from one end to the other.

8. Oct 29, 2009

### manofphysics

Since away from the depletion region, the p region and the n region is STILL electrically neutral , How is this electric field due to charges in depletion region in the p and n regions "counteracted" ??

9. Oct 29, 2009

### pseudophonist

10. Oct 30, 2009

### manofphysics

I have studied the mathematical portion of this, the solving of poisson eqn, etc.But still, I cannot grasp the physical interpretation. See, a charge creates a electric field 360 degrees around it.Now there is definitely an electric field between positively-charged ionised donors and negatively-charged ionised acceptors,This is in the depletion region. But these very positively-charged ionised donors and negatively-charged ionised acceptors create a field away from the depletion region ALSO, as charges create electric fields 360 degrees about them. What "counteracts" this field, for the net field to be zero away from the depletion region?

11. Oct 30, 2009

### pseudophonist

If we model the fixed charges of the depletion region as an infinitely tall, infinitely wide, but finitely thick prism with constant charge density then the electric field is not dependant on distance from the charges (just like from an infinite plane of charge).
$$E = \rho\times(thickness)/(2\epsilon)$$
Now, insert another infinitely tall, infinitely wide, finitely thick prism of opposite charge next to it such that the total charge enclosed in each is the same. The electric field generated by this second prism cancels out the first. however when we calculate the field between the prisms, it is non-zero.

Interpretting this physically, the field due to acceptors cancels the field due to donors, outside the depletion region. In a physical device, there will be a small field since the device has no infinite dimensions but for most purposes it is negligable since the two charged regions are very very close together.

Last edited: Oct 30, 2009
12. Oct 30, 2009

### manofphysics

Thanks a LOT, pseudophonist. I think I have understood it now fully.

13. Oct 30, 2009

### pseudophonist

14. Nov 20, 2009

### sudar_dhoni

mr manofphysics
what u said cannot happen
because if the electron from the negative ion wants to move to the hole in the left of the p region it cannot do so
because if it gets ready to move there it cannot as the electric field due to + immobile ions on the right and the negative immobile ions on the left are pushing the electron towards right so it cannot move
check hyperphysics