Solution Manual for Modern Semiconductor devices for Integrated Circuit. By Chenming Calvin Hu. With Verified Solutions.

Solution Manual
Modern Semiconductor devices for Integrated Circuit.

By Chenming Calvin Hu




Chapter 1


Visualization of the Silicon Crystal
1.1 (a) Please refer to Figure 1-2. The 8 corner atoms are shared by 8 unit cells and
therefore contribute 1 atom. Similarly, the 6 face atoms are each shared by 2 unit
cells and contribute 3 atoms. And, 4 atoms are located inside the unit cell.
Hence, there are total 8 silicon atoms in each unit cell.

(b) The volume of the unit cell is

Vunit cell  5.43 A3  5.43 108 cm  1.60 1022 cm3 ,
3




and one unit cell contains 8 silicon atoms. The atomic density of silicon is

8 silicon atoms
NSi   5.00 1022 (silicon atoms) cm3 .
Vunit cell

Hence, there are 5.001022 silicon atoms in one cubic centimeter.

(c) In order to find the density of silicon, we need to calculate how heavy an
individual silicon atom is

28.1 g/mole
Mass1 Si atom   4.67 1023g/atom.
6.02 10 23 atoms/mole 

Therefore, the density of silicon (Si) in g/cm3 is

ρ Si  N Si  Mass 1 Si atom  2.33 g / cm 3 .

,Fermi Function
1.2 (a) Assume E = Ef in Equation (1.7.1), f(E) becomes ½. Hence, the probability is ½.

(b) Set E = Ec + kT and Ef = Ec in Equation (1.7.1):
1
f(E)   1     0.27 .
Ec kT Ec /kT
1 e 1 e 1


The probability of finding electrons in states at Ec + kT is 0.27.

, * For Problem 1.2 Part (b), we cannot use approximations such as Equations (1.7.2)
or (1.7.3) since E-Ef is neither much larger than kT nor much smaller than -kT.

(c) f(E) is the probability of a state being filled at E, and 1-f(E) is the probability of a
state being empty at E. Using Equation (1.7.1), we can rewrite the problem as

f(E  Ec  kT)  1  f(E  Ec  3kT)
1 1
  1  
1  e Ec  kT  E f 1  e Ec 3kT  E f
 




/kT /kT



where

1 1  eEc 3kT  E f /kT  1 eEc 3kT  E f /kT
1  
1  eEc 3kT  E f /kT 1  eEc13kT  E f /kT 1  eEc 3kT  E f /kT
   .
1  e Ec 3kT  E f /kT



Now, the equation becomes
1 1
     .
Ec  kT  E f /kT  Ec 3kT  E f /kT
1 e 1 e

This is true if and only if

Ec  kT  Ef  Ec  3kT  E f .

Solving the equation above, we find

E f  Ec  2kT .

1.3 (a) Assume E = Ef and T > 0K in Equation (1.7.1). f(E) becomes ½. Hence, the
probability is ½.

(b) f(E) is the probability of a state being filled at E, and 1-f(E) is the probability of a
state being empty at E. Using Equation (1.7.1), we can rewrite the problem as

f(E  Ec )  1  f(E  Ev )
1 1
Ec  E f /kT  1  Ev  E f /kT
1 e 1 e

where

, 1
1 1  eEv  E f /kT  1 eEv  E f /kT 
1
.
 
1  eEc  E f /kT 1 e Ev  E f /kT
1 e Ev  E f /kT 1  e Ev  E f /kT

Now, the equation becomes

1 1
Ec  E f /kT  Ev  E f /kT .
1 e 1 e

This is true if and only if

Ec  E f  Ev  E f .

Solving the equation above, we find

Ec  Ev
Ef  .
2

(c) The plot of the Fermi-Dirac distribution and the Maxwell-Boltzmann distribution
is shown below.
Probability 1

0.9

0.8 Fermi-Dirac
0.7
Distribution
0.6

0.5 Maxwell-Boltzmann
0.4
Distribution
0.3

0.2

0.1

0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
(E-Ef)/kT

The Boltzmann distribution considerably overestimates the Fermi distribution for
small (E-Ef)/kT. If we set (E-Ef)/kT = A in Equations (1.7.1) and (1.7.2), we
have
 1 
e A  1.10 .
1  eA 

Solving for A, we find

1  eA
e   e A 10.11  A  ln10.11 2.31.
A
1.10

Therefore, the Boltzmann approximation is accurate to within 10% for (E-Ef)/kT
 2.31.