The Designer's Guide Community Forum
https://designers-guide.org/forum/YaBB.pl Modeling >> Semiconductor Devices >> depletion region ionised charges https://designers-guide.org/forum/YaBB.pl?num=1141652117 Message started by jason_class on Mar 6th, 2006, 5:35am |
Title: depletion region ionised charges Post by jason_class on Mar 6th, 2006, 5:35am Hi Murphy and All It is about depletion region. Always we would say depletion region is location of the ionised dopant say for a pn junction. We account all the ionised acceptor/donor at the junction, right? I do not understand why must all the ionised charge be located in the depletion region and right at the junction. The dopant should be randomly distributed in the material by logical way of thinking. What makes the ionised donor/acceptor only be located at junction? Is it just an assumption or approximation to make calculation easier or there is solid reason? I guess it is because of the mobile carriers moved away from the vicinity of the junction and end up all the mobile charges are way from the junction, leaving all the ionised charge at the contact region. When p and n material are disconnectd, the carriers are distributed randomly again. I am not sure if I am right about this. kindly enlighten I apologise if this question sounds silly Thank you Murphy and all best regards Jason |
Title: Re: depletion region ionised charges Post by Marc Murphy on Mar 6th, 2006, 6:26am The entire semiconducter is doped and has the same concentration throughout. Basically, holes leave the p region and electrons leave the n region in the vicinity of the junction. So, this is where the depletion region is, and it is made up of ionized bound atoms (the region is depleted of carriers). The rest of the semiconductor has dopants, but they aren't ionized. The depletion region is always at the junction, regardless of the doping profile. You'll have to read up on diffusion and drift. Diffusion occurs because of difference in concentration. The p side has lots of holes, so holes go to the n side, and vice versa. This depleted region sets up a field that counteracts the diffusion current. At equilibrium, the diffusion current equals the drift current and the depletion region is held at a constant size. If you disconnected a pn-junction then yes I think the carriers would redistribute equally. However, I'm not certain each piece of material would be neutral. The atoms might still be ionized if there was no way for the electrons to get back to the n-material, for example. Cheers, Marc |
Title: Re: depletion region ionised charges Post by jason_class on Mar 6th, 2006, 3:20pm Hello Murphy That clears my doubt. I will further reread the chapter you mentioned to see if I miss anything out! Thank you Murphy best regards jason |
The Designer's Guide Community Forum » Powered by YaBB 2.2.2! YaBB © 2000-2008. All Rights Reserved. |