(a) Determine the position  of the  Fermi  level  relative to  the  CB bottom  at room temperature in a P-doped Si crystal, nd = 1020 m-3 .

(hint: consider the n-type semiconductor in the saturation regime and a effective density of states NC = 1025 m-3)

(b) A n-type semiconductor is said to be degenerate if the Fermi level is closer than 3kT to EC. What is the threshold dopant concentration for P doping?

(c) Determine the fraction of occupied donor states. Verify that the system is in the

saturation regime at room temperature.

(hint: P donor energy level Ed = 45 meV)

Model answer

(a)

For an n-type semiconductor in the saturation regime, n = nd. The density of dopants is small  enough  that  the  structure  of conduction  band  can  be  thought  of as  being unperturbed.

As a consequence, it is possible to write

n = NCe一(EC 一EF ) / kT .

By equating this to nd, we get an equation for determining the Fermi level:

nd  = NCe一(EC 一EF ) / kT                                or      e(EC 一EF ) / kT  =  ,         which becomes

EC  一 EF  = kTln))|| = 0.0256  eV人 ln ))|| = 0.29  eV.

(b)

The threshold doping concentration above which a semiconductor is considered to be degenerate can be obtained from the formula above by imposing

EC  一 EF  = kTln))|| = 3kT ,

which implies

ln ))|| = 3                or            nd  =  如 5 人1023 m-3 .

(c)

The fraction of occupied donor states is given by the Fermi function

1

f (E) = 1 + e(E 一EF ) / kT     ,

where E = EC – Ed.

As a consequence,