Solution Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

Solution Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning




Solution Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

,Solution Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning




SOLUTION SET

to

Exercises in



FUNDAMENTALS OF MODERN VLSI DEVICES, 3rd ed.


published by

Cambridge University Press




Yuan Taur and Tak H. Ning




Solution Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

,Solution Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning




Solutions to Chapter 2 Exercises


2.1. From Eq. (2.4),

f (E − E) =
1 eE / kT
1+ e−E / kT = eE / kT +1 ,
f



and

1
f (Ef + E) = .
1 + eE / kT

Adding the above two equations yields

eE / kT +1
f (E f − E) + f (E f + E) = = 1.
eE / kT +1


2.2. Neglecting the hole (last) term in Eq. (2.19), one obtains

N c e−( Ec − E f )/ kT + 2N c e−( Ec + Ed −2E f )/ kT = N d .

Treating exp(Ef/kT) as an unknown, the above equation is a quadratic equation with the solution

E / kT −1 + 1 + 8(N / N )e( Ec − Ed )/ kT
e f
= d c
.
4e− Ed / kT

Here only the positive root has been kept. For shallow donors with low to moderate
concentration at room temperature, (Nd/Nc)exp(Ec − Ed)/kT  1, and the last equation can be
approximated by

E / kT 4(N / N )e( Ec − Ed )/ kT N E / kT
e f
= d c
− Ed / kT
= de c ,
4e Nc

which is the same as Eq. (2.20). If we compare the above relation with Eq. (2.19), it is clear
that in this case, exp[−(Ed − Ef)/kT] << 1, and Nd+  Nd or complete ionization.

If the condition for low to moderate concentration of shallow donors is not met, then
exp[−(Ed − Ef)/kT] is no longer negligible compared with unity. That means Nd+ < Nd (Eq.
(2.19)) or incomplete ionization (freeze-out). [Note that incomplete ionization never occurs for
shallow impurities: arsenic, boron, phosphorus, and antimony at room temperature, even for
doping concentrations higher than Nc or Nv. This is because in heavily doped silicon, the



Solution Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning

, Solution Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning




impurity level broadens and the ionization energy decreases to zero, as discussed in Subsection
9.1.1.2.]

2.3. (a) Substituting Eqs. (2.5) and (2.3) into the expression for average kinetic energy, one
obtains


=
( E − Ec )3/ 2 e−( E − E f )/ kT dE
K.E. Ec
( E − E )1/ 2 e−( E − E f )/ kT dE .
 Ec c



Applying integration by parts to the numerator yields
( E − E c)1/2 e−( E − E f
 )/ kT
()kT
E
dE
3
K.E. = 
c = kT .

Ec
( E − Ec )1/ 2 e−( E − E f )/ kT dE 2


(b) For a degenerate semiconductor at 0 K, f(E) = 1 if E < Ef and f(E) = 0 if E > Ef. Here
Ef > Ec. Therefore,
Ef

=
( E − Ec )3/2 dE 3
( E − E )1/2 dE = ( Ef − Ec ).
E
K.E. Ec f
5
 Ec c




2.4. With the point charge Q at the center, construct a closed spherical surface S with radius r.
By symmetry, the electric field at every point on S has the same magnitude and points outward
perpendicular to the surface. Therefore,

 S
E  dS = 4r2E ,

where E is the magnitude of the electric field on S. 3-D Gauss’s law then gives

Q
E= ,
4sir 2

which is Coulomb’s law.

Since E =−dV/dr, the electric potential at a point on the sphere is

Q
V= ,
4sir




Solution Manual for Fundamentals of Modern VLSI Devices 3d Edition by Yuan Taur, Tak H. Ning