1
$\begingroup$

I have seen this and this and a few other. Throughout the web and notes I find the explanation of the workings of a simple diode sort of unconvincing and I am depressed by the fact that I am not able to convince myself. The doubts keep gnawing at me and I'm struggling. I will try to describe what my doubts are in numbered form:

Disclaimer: I have virtually no background in statistical mechanics and quantum physics but textbooks that talk about solid state/semiconductor physics require knowledge about the 2 big topics. Consequently, I cannot relate very well to many concepts and derivations in the books. I'd love to be recommended a one-stop book/resource to learn about the rich history and old experiments and papers of the physics developed in this field, in as much details as necessary.

  1. In reverse bias, why don't electrons just get flushed through the depletion region, filling in all the holes in p-region and jumping their way to n-region and into the wire, even when the applied voltage is not very big? I'd imagine the p-region would then be heavily negatively charged, is that the reason? The conventional explanation that the depletion zone is void of charge carriers thus cannot conduct seems more like a statement than a proper explanation to me. And it also seems to me that there are a bunch of charge carriers in that depletion region which, I am not sure why, are not willing to move. Perhaps I shall ask the same question but in more details in question 2.

  2. A diode in its equilibrium has a thin depletion zone which is basically a charged zone that is positively charged on n-side (due to electrons leaving to p-side) and negatively charged on p-side (due to electrons naturally tending to fill some holes in here). While the electric field produced by them oppose the diffusion, an equilibrium is reached. However, when a reverse bias voltage is applied across the diode, negative charge builds up on p-side and positive on n-side, widening the depletion region. My main concern is what is stopping the electrons from moving on, thus letting the charge build up? I'd imagine that: the electrons on the p-side of pn-junction/depletion region will scoot back to n-side (since n-side is positively charged, that seems natural under the influence of a reverse bias voltage, or does it? unlike capacitor with dielectric in between plates, there's nothing separating n-side and p-side, so aren't the opposite charges on both side likely to 'annihilate'?) neutralizing the pn-junction; at the same time electrons in n-region get attracted to the positive terminal of the voltage source leaving positive charge behind; at the same time electrons from the negative terminal of the voltage source enter the p-region, fill and jump from holes to holes both naturally and also under attraction by positive charge left behind on n-region; the electrons arrive at n-region, get sucked to the positive terminal of voltage source and so on; there's a current !

  3. More question: why is there a (almost) constant voltage drop across a diode during forward bias? Where does it actually come from? Is it because of 1)The charge built up in depletion region making an opposing electric field since equilibrium state. or 2) The negative charge throughout p-region because unlike electron-carrier-rich conductor such as copper wire, p-region is originally neutral but now it is so fed up with electrons using its holes for transportation i.e. there are more mobile electrons than immobile protons in the atoms. or 3) something about band gap energy and fermi level (the origin and significance of which I am currently not very familiar about and comfortable with, but would like to be after I pick up some few prerequisites in thermodynamics, QM and Stat.Mechanics in a while.

Lastly, thanks.

$\endgroup$
  • $\begingroup$ Why will electrons get flushed through the depletion layer when there is an electric field opposing their motion? $\endgroup$ – Jon Custer Mar 13 '18 at 16:18
  • $\begingroup$ Because there is a greater externally applied electric field / voltage across the diode. I believe the electric field that you're referring to is one that is generated by charges built up in the depletion zone during reverse bias. My question is why the charges are allowed to build up in the first place. $\endgroup$ – Y.JQ Mar 13 '18 at 16:38
  • $\begingroup$ Fundamentally, I think the root of your discomfort is not fully understanding (1) electron-hole detailed balance ($np = n_{i}^{2}$, and (2) how doping works. I would suggest finding a professor, TA, or fellow student to review and discuss. $\endgroup$ – Jon Custer Mar 13 '18 at 17:37
1
$\begingroup$

In questions 1 & 2, you seem to assume that the positive terminal connected to the n-type region of the diode in reverse bias is injecting holes into the device, while the negative one connected to the p-type region is injecting electrons it it.

If this were the case, once injected, the minority carriers would start to recombine. If the level of injection were high enough, you are perfectly right that they would start to diffuse towards the depletion region where they would find no obstacle to get across it. In both cases current would flow.

But this doesn't happen because ideal contacts can only inject majority carriers into the device. So the only minority carriers available are the very few electrons already present in the p-type region, and the very few holes already present in the n-type region, which are responsible for the small leakage current.

The reason why the contacts behave in this way is because they are optimized to allow injection of majority carriers when the diode is in direct bias. A good n-contact metal, for example, has its Fermi level aligned as close as possible with the conduction-band edge so that electrons can easily flow. But, as consequence, the Fermi level is then badly aligned with the valence-band edge so that holes can't flow when the bias is reversed. Similarly for the p-contact.

For question 3, you can have a look a this answer: What determines the forward voltage drop for a diode?.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.