L
[email protected]
- Jan 1, 1970
- 0
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
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