Doubt on Depletion Layer Width in Biased pn-Jnction

[Himabindu]

" Depletion Layer is the region in diode which consists of ions( formed due to formation of octet configuration )of p-type and n-type dopants"
What happens to the ions present in the depletion layer when a diode is biased?
why the depletion layer width changes??

 

[wbeaty]

" Depletion Layer is the region in diode which consists of ions( formed due to formation of octet configuration )of p-type and n-type dopants"
What happens to the ions present in the depletion layer when a diode is biased?
why the depletion layer width changes??

Ions are part of the semiconductor crystal lattice. They were placed there during the doping process. They're supposed to not move during biasing. (If the ions move around, it means that your doping profile is diffusing, and the characteristics of your diode/transistor/etc. are being altered.)

Remember that 'insulator' basically means 'a material lacking mobile charge carriers.' The depletion zone is a fair insulator because those ions aren't mobile.

Analogous situation: Metals are composed of a crystalline grid of positive ions immersed in a "sea" of mobile electrons. If it was possible to sweep all the electrons out of a piece of copper, then the copper would become insulating, and we would have all-metal transistors. But now go and estimate the range of potential needed to remove all electrons from even a very thin film of copper. But while copper contains ~1 mobile electron per atom, doped semiconductor contains ~10^-9 mobile charge carrier per atom (or even less for lighter doping.) Thus the required potentials are many, many orders smaller for semiconductor switching, when compared to those required for metals. That's why it's possible to sweep the carriers out of a region of doped silicon, but it's not really feasible to do the same with a metal.