How pin diode works?

Question

I dont't understand how pin diode reach the equilibrium state... I can consider the pin diode as a duble junction p-i and i-n (with ideal intrinsic layer without any dopant). First i consider only the p- i junction...when the two parts are in contact, holes move for diffusion from p-semiconductor to i-semiconductor, but ionized dopants make drift curruent...so the equilibrium is reached when the two components are equal. The equilibrium condition make the equation NaXp=NiXi, but Ni=0 (ideal condition) so Xi tends to infinite. What means? holes are placed all along the i-semiconductor? If consider i-n junction the problem is the same....If consider all pin diode at equilibrium i-region remains intrinsic:why? Because holes and electrons recombine in i-region or because they reach opposite iones and recombine with them? I'm looking for something about,but nothing i've found.. Thants

Phil Hobbs · Accepted Answer

Think of a PIN diode as an ordinary PN junction with a really really thick depletion zone. You store a lot of charge there, so it takes quite awhile for it to be swept out when you apply reverse bias. The net effect is that the amount of bias current required for a given linearity level is much reduced.

Cheers,

Phil Hobbs

How pin diode works?

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