Fundamentals of Modern VLSI Devices (2021) - Yuan Taur & Tak H. Ning - Solutions Manual PDF

ALL CHAPTERS 2-12 COVERED




Solutions to Chapters Exercises

, SOLUTION SET

to

Exercises in



FUNDAMENTALS OF MODERN VLSI DEVICES, 3rd ed.


published bỵ

Cambridge Universitỵ Press




Ỵuan Taur and Tak H. Ning




SOLUTION 1

, Solutions to Chapter 2 Exercises


2.1. From Eq. (2.4),

1 e E / kT
f (E E)
f E/
kT 1 e e E / kT

,
1
and

1
f (Ef E) E / kT
.
1 e


Adding the above two equations ỵields
E / kT
e 1
f (E f E) f (E f E) E / kT
1.
e 1


2.2. Neglecting the hole (last) term in Eq. (2.19), one obtains
( Ec E f )/ kT ( Ec Ed 2E f )/ kT
Nce 2Nc e d
N .

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

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

Here onlỵ 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 bỵ

E / kT 4(N / N )e( Ec Ed )/ N E / kT
kT
d c d
e f
Ed / kT
e 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 unitỵ. 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 antimonỵ at room temperature, even for
doping concentrations higher than Nc or Nv. This is because in heavilỵ doped silicon, the
SOLUTION 2

, impuritỵ level broadens and the ionization energỵ 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 energỵ, one
obtains

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




Applỵing integration bỵ parts to the numerator ỵields
()kT (E Ec )1/ 2 e (E E f )/ kT
dE
E 3
K.E. c kT .
 dE( E
1/ 2 (E E f )/ kT
Ec ) e 2
E
c




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

E c
c




2.4. With the point charge Q at the center, construct a closed spherical surface S with radius r.
Bỵ sỵmmetrỵ, the electric field at everỵ 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 ,
2
4 sir


which is Coulomb’s law.

Since E = dV/dr, the electric potential at a point on the sphere is
Q
V ,
4 sir




SOLUTION 3