Effect of doping concentration on the depletion width of p-n junction diodes

[Ahsan Khan]

TL;DR

Hi all

I have been reviewing my understanding on the basics of p-n junction diode. Although I have my degree in Mechanical Engineering but I teach introductory Physics covering almost all of pre-university Physics.

My question for now is as follows: Why does a high doping concentration causes a decrease in the thickness of depletion region.

Any help will be highly appreciated!

Thanks a bunch

Regards!

 

[phyzguy]

Have you tried reading this?
https://en.wikipedia.org/wiki/Depletion_region
Note especially the sentence that says:
"When the electric field is sufficiently strong to cease further diffusion of holes and electrons, the depletion region reached the equilibrium." As the doping goes up, how does this affect how far you need to deplete to achieve a certain electric field?

 

[eq1]

It's a somewhat indirect path to see it. Diffusion has to be slowed so that diffusion and drift current can balance and the junction can be in equilibrium. Slowing diffusion requires a given electric field which requires a given amount of bound charge to be revealed. If the doping is high then charge per cm is high which means one need less cm to get the prescribed amount of charge. Hence less thickness (usually people would say length) of the depletion region.

I highly recommend Chenming Hu's, the inventor of the FinFET, text on this subject. His coverage of the background material needed to parse the sentences above (i.e. why is there diffusion in the first place and what is bound charge) at the undergraduate level is second to none, in my opinion. You probably want to look at motion and recombination in Ch2. I think you'll find this stuff a lot more approcable if you read it. Ch2 is only like 25 pages, with illustrations, it amazing how much he covered in that small space. Thanks to Berkeley for making this textbook free!

https://www.chu.berkeley.edu/modern...-integrated-circuits-chenming-calvin-hu-2010/

 

[Ahsan Khan]

Have you tried reading this?
https://en.wikipedia.org/wiki/Depletion_region

Thank you phyzguy, yes I read wikipedia on this.But I am still not satisfied with the answer (not because answer is wrong but because I must be missing something).

As the doping goes up, how does this affect how far you need to deplete to achieve a certain electric field?

I have the following issues with this.
(1) It is said to get the same electric field with the increased doping less space distance in cm is required. My issue is that as E=-dV/dx, for more E, why it dx is necessary to decrease because for more E,dV could increase without a having dx to decrease. I am not denying that a decrease in depletion with would do our purpose but using this logic that for more E, dx must decrease seems incomplete especially when the potential dV is also not going to remain constant with increased doping level.

Regards!

 

[dlgoff]

Here's a reference that may help.

https://www.physics-and-radio-elect...iconductor-diodes/widthofdepletionregion.html

 

[Jody]

The electric field inside of the depletion region is non-zero, but the electric field outside of it is zero . In order for the electric field outside of the depletion to be zero the total charge inside of the depletion region has to be zero (Gauss).

If you increase the charge per volume by doping it and it has the same cross-sectional area, then the only way to counter it or maintain that zero total charge is less length.

 

[Ahsan Khan]

Thank you eq1 for the help and I will surely read the book from the link. But from what you wrote I have some small questions or doubts.

Slowing diffusion requires a given electric field which requires a given amount of bound charge to be revealed.

When here you say "given electric field" does the word given electric filed means the electric filed requires is fixed for a pn junction of given material and is same(given) irrespective of doping level?

Won't a high level of doping results high amont if charge flow from either side therefore should cause stronger electric field with the depletion region.

So when you say a given electric field you mean what ever strong field has developed for the given doping level and is definitely not fixed. Is that what you mean by the word "given electric field "?

If the doping is high then charge per cm is high which means one need less cm to get the prescribed amount of charge. Hence less thickness (usually people would say length) of the depletion region.

Ihighly recommend Chenming Hu's, the inventor of the FinFET, text on this subject. His coverage of the background material needed to parse the sentences above (i.e. why is there diffusion in the first place and what is bound charge) at the undergraduate level is second to none, in my opinion. You probably want to look at motion and recombination in Ch2. I think you'll find this stuff a lot more approcable if you read it. Ch2 is only like 25 pages, with illustrations, it amazing how much he covered in that small space. Thanks to Berkeley for making this textbook free!

https://www.chu.berkeley.edu/modern...-integrated-circuits-chenming-calvin-hu-2010/
[/QUOTE]

 

[eq1]

When here you say "given electric field" does the word given electric filed means the electric filed requires is fixed for a pn junction of given material and is same(given) irrespective of doping level?

I was referring to the built-in potential which is technically the integral of the electric field. Every junction of two dissimilar materials has it. It will be a function of doping levels.

Won't a high level of doping results high amont if charge flow from either side therefore should cause stronger electric field with the depletion region.

Changing the doping will change the built in potential which makes sense. Changing the doping changes the concentration gradient which means the amount of potential needed to counter it will also change. Right now when I read it in words the explanation seems a bit circular but really it's a system of equations, not a single equation, with a couple of unknowns, and that system can be solved. The math is actually pretty straightforward and a better way to describe it. See the Chenming text sections 4.1 and 4.2

 

[eq1]

But from what you wrote I have some small questions or doubts.

Personally I think the key to understanding the PN junction is to be able to answer this question: why does the concentration gradient persist? Via doping we created a concentration gradient, which is pretty obvious, and whenever there is a concentration gradient there is diffusion [1], in any material. Usually in a diffusion process the concentration gradient eventually goes away, the material finds equilibrium, but in a PN junction it does not, the concentration gradient persists. That must mean there is some force, in its equilibrium state, that opposes diffusion. So what is that and where does it come from? When you understand that you'll get it and everything else will just be calculation.

[1] https://en.wikipedia.org/wiki/Molecular_diffusion

 

[Jody]

Was my response in #6 okay or acceptable?

 

[anorlunda]

I highly recommend Chenming Hu's, the inventor of the FinFET, text on this subject.

Wow! Thanks for sharing. I started reading that book thanks to you, and it is excellent.

 

[eq1]

Was my response in #6 okay or acceptable?

I’ve been thinking about this one and my answer is everything you said looks right to me but I don’t see how one goes from doping concentrations to absolute length through Gauss’s law. I think Gauss’s law can give the ratio Xp/Xn as a function of concentrations but not the absolute length. But maybe I’m wrong.

See:
https://ecee.colorado.edu/~bart/book/pnelec.htm#field