Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion

If less amount of p-type material is added then Fermi level will slightly move towards valence band(.3eV to .28eV).
If more amount of p-type material is added then Fermi level will move even closer to valence band (.3eV to .1eV).
If further doping is increased then Fermi level enters valence band (below valence band)

So how can we say that .28eV is correct?

Fermi level of n-type semiconductor is above the valence band if you increase acceptor ions fermi level with is increased that's why Fermil level was close to the valence band.

I am not understanding this. Please explain to me.

@All.

Consider the Fermi Level the roof of the room, and the floor of the room is VB. The more you add acceptor atoms, the closer the roof (FL) will be closer to the floor (VB).

The distance will start decreasing, so first it will be 0.29, then 0.28, then 0.27, then gradually 0.1 after a certain amount of addition. Anyway, the new position of the roof (Fermi Level) is 'likely to be' 0.28. This comes first among the given options. If 0.29 were given, I would go for 0.29.

The shift in the Fermi level is logarithmic, not linear (Ef-Ev=KTln(Nv/Na)).

To move the Fermi Level from 0.3eV to 0.1eV, a massive increase in acceptor atoms is required —degenerate doping.

The more practical approach will only shift it towards 0.28eV.