Recent content by Amerez

Thank you very much for this. My reply is now late but back at the time when I first read it it clarified things even futher

I'm not studying from Google I'm just using it as a supplement to find answers to holes in formal textbooks (there are astounding holes BTW). I'm formally studying electrical engineering at the university. I very much agree with you that a correct basis in science is only attainable from...

I found the answer to the question in this book: "Muller & Kamins - Device Electronics for Integrated Circuits - 3rd edition" - page 246 The summary of it is that the forward bias current that flows in practice is so small compared to the Drift-Diffusion currents (about 0.1%). So the term...

Added and Image for clarification: https://www.physicsforums.com/data/attachments/56/56624-b0ed937602c8f87174ba11c307ccd2c5.jpg

I'm currently studying the physics of the PN junction. I went though the derivation of the built-in potential in the PN junction under equilibrium: Diffusion current density = Drift current density D_{p}p\frac{dp}{dx} = EU_ppq where D_{p} : Diffusion coefficient for holes p = holes density...

Yes, I've been scanning this site for the past couple of weeks. Most of the sites I've read seem to describe the operation of the PN junction or the diode, but I'm unable to find the answer to this specificity.

lol, it may be the only way. I'm actually reading through Sedra 5th edition on this matter, it states the current is a diffusion current but without giving the reasons. It seems electrical engineering books leave many holes when describing the physics of the devices

I have went through the link you provided, but this doesn't touch on the question. Maybe this thread is more solid-state physics than electrical engineering !?

In a forward-biased PN junction, the potential barrier decreases, allowing more majority carriers from one side to diffuse to the other side where they are minority carriers. After they cross the potential barrier, they form a diffusion current, the drift current of minority carriers is...

Thank you Jim and Baluncore

Right on! This is what I was trying to understand but to no avail after two days of googling! Thanks!

My understanding of the atomic structure is that electrons can gain energy in quantas, so that's why we have orbitals with sharply defined energy around the nucleus. One would logically think that every atom of the same element has the exact same orbitals with the exact same energy of orbital...

In the Image attached, we have two tubes of glass each filled with charged metal and they are in space. As we know, the electric field is only between them and negligible in other places. If we moved an electron worth of charge from - to + we have execute some work on it. In case 1 we have to...

:frown: I know there is no electric field inside a conductor, I thought I could say it "move it against an electric field (of zero when we don't have one)" equally well, sorry for the confusion I still didn't get where the electric field we are working against exists from the explanation...

What I fail to understand is if we move the electron along a path "not" between the plates but "around" the plates. Since there is no field around the plates, where is the force countered responsible for the work done on the electron (or the potential energy increase of the electron)?