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Comments on Understanding of basic semiconductors

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Understanding of basic semiconductors

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Trying to update my understanding of basic semiconductors.

It has been a very long time since I seriously looked at semiconductors. The most basic semiconductor application is a diode.

From Wikipedia, I see a picture: https://en.wikipedia.org/wiki/File:Two_diode_structures.png , and I'm not sure if my understanding of it is correct.

I understand about p and n doping, but I've never seen "p+" nor "n+" nor "n-". From my own thinking, I guess "x+" and "x-" denote over- and under- doped semiconductors, is that correct?

I can guess that the n over-doping is to prevent to accidentally create a Schottky barrier.

If so, the metal to "p+" junction would act as a Schottky barrier, doing the actual work of the diode. Is that true?

I don't know what to think about the "p+" to "n-" junction, and I have absolutely now idea what that "p guard" is good for.

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From my own thinking, I guess "x+" and "x-" denote over- and under- doped semiconductors, is that correct?

"+" stands for heavy doping. "-" stands for light doping. n+ and p+ are heavily doped. n- and p- are lightly doped.

If so, the metal to "p+" junction would act as a Schottky barrier, doing the actual work of the diode. Is that true?

A junction between metal and p- (lightly doped) would be a Schottky junction. A junction between metal and heavily doped semiconductor is effectively ohmic, and conduct the same in both directions.

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Why would a metal to "p+" junction be ohmic? (4 comments)
Why would a metal to "p+" junction be ohmic?
watchmaker‭ wrote 10 months ago

Why would a metal to "p+" junction be ohmic?

With reverse voltage, electrons would be held back on the metallic (super hyper n+++doped) side while any electrons that should accidentally (thermal happening) pop up on the p-side would be instantly re-absorbed by the hungry holes.

Nick Alexeev‭ wrote 10 months ago · edited 10 months ago

The depletion region becomes thin when the p+ side is heavily doped. Holes or electrons (depending on bias) can tunnel through the barrier, which produces an ohmic I-V curve. Here are lecture slides which have step-by-step explanation and illustrations: https://transport.ece.illinois.edu/ECE340F11-Lectures/ECE340Lecture35-MSContacts.pdf [starts from afar, though ]

watchmaker‭ wrote 10 months ago

Nick Alexeev‭ Thank you for the link. On a first, quick scan I spotted one of my mistakes: I was assuming that the abundant electrons in the metal would be leaking into the semiconductor, but the link states it is the other way round.

Trying to understand why, I took a more serious look at the formulas, but got stuck.

On the bottom of page 4 there is the formula $\Phi_S=\chi+(E_C - E_F)_{FB}$ (where, two line above $E_f$ is used and it is not clear if it is a typo).

What is that subscript $(\ldots)_{FB}$?

Nick Alexeev‭ wrote 10 months ago

watchmaker‭ FB stands for flat band condition. It's a simplifying assumption. Explanation with drawings here: https://www.chu.berkeley.edu/wp-content/uploads/2020/01/Chenming-Hu_ch5-1.pdf#page=2 (section 5.1)