p42_039.pdf

39. (a) The number of electrons in the valence band is

N ev = N v P ( E v )=

N v

e ( E v − E F ) /kT +1 .

Since there are a total of N v states in the valence band, the number of holes in the valence band is

N hv = N v −

N ev = N v 1

−

e ( E v − E F ) /kT +1

N v

e − ( E v − E F ) /kT +1 .

Now, the number of electrons in the conduction band is

N ec = N c P ( E c )=

N c

e ( E c − E F ) /kT +1 ,

Hence, from N ev = N hc ,weget

N v

e − ( E v − E F ) /kT +1 =

N c

e ( E c − E F ) /kT +1 .

(b) In this case, e ( E c − E F ) /kT

1and e − ( E v − E F ) /kT

1. Thus, from the result of part (a),

N c

e ( E c − E F ) /kT ≈

N v

e − ( E v − E F ) /kT

or e ( E v − E c +2 E F ) /kT

≈

N v /N c .Wesolvefor E F :

E F ≈

2 ( E c + E v )+ 1

2 kT ln N v

N c