p42 039

background image

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

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

, we g et

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

 1 and 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

+2E

F

)/kT

≈ N

v

/N

c

. We solve for E

F

:

E

F

1

2

(E

c

+ E

v

) +

1

2

kT ln



N

v

N

c



.


Document Outline


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