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Solutions Manual
SILICON VLSI TECHNOLOGY 1 © 2000 by Prentice Hall
Fundamentals, Practice and Modeling Upper Saddle River, NJ.
By Plummer, Deal and Griffin
SILICON VLSI TECHNOLOGY
Fundamentals, Practice and Models
Solutions Manual for Instructors
James D. Plummer
Michael D. Deal
Peter B. Griffin
Solutions Manual
SILICON VLSI TECHNOLOGY 2 © 2000 by Prentice Hall
Fundamentals, Practice and Modeling Upper Saddle River, NJ.
By Plummer, Deal and Griffin
Chapter 1 Problems
1.1. Plot the NRTS roadmap data from Table 1.1 (feature size vs. time) on an
expanded scale version of Fig. 1.2. Do all the points lie exactly on a straight
line? If not what reasons can you suggest for any deviations you observe?
Answer:
0
50
100
150
200
250
Year
1997 2007 20122002
Interestingly, the actual data seems to consist of two slopes, with a steeper slope for
the first 2 years of the roadmap. Apparently the writers of the roadmap are more
confident of the industry's ability to make progress in the short term as opposed to
the long term.
1.2. Assuming dopant atoms are uniformly distributed in a silicon crystal, how far
apart are these atoms when the doping concentration is a). 10
15
cm
-3
, b). 10
18
cm
-3
, c). 5x10
20
cm
-3
.
Answer:
The average distance between the dopant atoms would just be one over the cube
root of the dopant concentration:
x = N
A
−
1/3
a)
x = 1x10
15
cm
−3
()
−1
/
3
= 1x10
−5
cm = 0.1μm =100nm
b)
x = 1x10
18
cm
−3
()
−1/3
= 1x10
−6
cm = 0.01μm = 10nm
Solutions Manual
SILICON VLSI TECHNOLOGY 3 © 2000 by Prentice Hall
Fundamentals, Practice and Modeling Upper Saddle River, NJ.
By Plummer, Deal and Griffin
c) x = 5x10
20
cm
−3
()
−1/ 3
=1.3x10
−7
cm = 0.0013μm = 1.3nm
1.3. Consider a piece of pure silicon 100 µm long with a cross-sectional area of 1
µm
2
. How much current would flow through this “resistor” at room
temperature in response to an applied voltage of 1 volt?
Answer:
If the silicon is pure, then the carrier concentration will be simply n
i
. At room
temperature, n
i
≈ 1.45 x 10
10
cm
-3
. Under an applied field, the current will be due to
drift and hence,
I = I
n
+ I
p
= qAn
i
μ
n
+μ
p
(
)
ε
= 1.6x10
−19
coul
()
10
−8
cm
2
()
1.45x10
10
carrierscm
−3
()
2000cm
2
volt
−1
sec
−1
()
1volt
10
−2
cm
⎛
⎝
⎞
⎠
= 4.64x10
−12
amps or 4.64pA
1.4. Estimate the resistivity of pure silicon in
Ω
ohm cm at a) room temperature, b)
77K, and c) 1000 ˚C. You may neglect the temperature dependence of the
carrier mobility in making this estimate.
Answer:
The resistivity of pure silicon is given by Eqn. 1.1 as
ρ=
1
q μ
n
n +μ
p
p
()
=
1
qn
i
μ
n
+μ
p
(
)
Thus the temperature dependence arises because of the change in n
i
with T. Using
Eqn. 1.4 in the text, we can calculate values for n
i
at each of the temperatires of
interest. Thus
n
i
= 3.1x10
16
T
3/ 2
exp −
0.603eV
kT
⎛
⎝
⎞
⎠
which gives values of ≈ 1.45 x 10
10
cm
-3
at room T, 7.34 x 10
-21
cm
-3
at 77K and 5.8
x 10
18
cm
-3
at 1000 ˚C. Taking room temperature values for the mobilities , µ
n
=
1500 cm
2
volt
-1
sec
-1
and , µ
p
= 500 cm
2
volt
-1
sec
-1
, we have,
Solutions Manual
SILICON VLSI TECHNOLOGY 4 © 2000 by Prentice Hall
Fundamentals, Practice and Modeling Upper Saddle River, NJ.
By Plummer, Deal and Griffin
ρ=2.15x10
5
Ωcm at room T
= 4.26x10
35
Ωcm at 77K
=
5.39x10
−4
Ω
c
m
at 1000 Þ
C
Note that the actual resistivity at 77K would be much lower than this value because
trace amounts of donors or acceptors in the silicon would produce carrier
concentrations much higher than the n
i
value calculated above.
1.5. a). Show that the minimum conductivity of a semiconductor sample occurs
when
n = n
i
μ
p
μ
n
.
b). What is the expression for the minimum conductivity?
c). Is this value greatly different than the value calculated in problem 1.2 for the
intrinsic conductivity?
Answer:
a).
σ=
1
ρ
= q μ
n
n +μ
p
p
(
)
To find the minimum we set the derivative equal to zero.
∴
∂σ
∂n
=
∂
∂n
q μ
n
n +μ
p
p
(){}
=
∂
∂n
q μ
n
n +μ
p
n
i
2
n
⎛
⎝
⎜
⎞
⎠
⎟
⎧
⎨
⎩
⎫
⎬
⎭
= q μ
n
+μ
p
n
i
2
n
2
⎛
⎝
⎜
⎞
⎠
⎟
= 0
∴n
2
= n
i
2
μ
p
μ
n
or n = n
i
μ
p
μ
n
b). Using the value for n derived above, we have:
σ
min
= q μ
n
n
i
μ
p
μ
n
+μ
p
n
i
2
n
i
μ
p
μ
n
⎛
⎝
⎜
⎜
⎜
⎞
⎠
⎟
⎟
⎟
= q μ
n
n
i
μ
p
μ
n
+μ
p
n
i
μ
n
μ
p
⎛
⎝
⎜
⎞
⎠
⎟
= 2qn
i
μ
n
μ
p
c). The intrinsic conductivity is given by
σ
i
= qn
i
μ
n
+μ
p
(
)
Solutions Manual
SILICON VLSI TECHNOLOGY 5 © 2000 by Prentice Hall
Fundamentals, Practice and Modeling Upper Saddle River, NJ.
By Plummer, Deal and Griffin
Taking values of n
i
= 1.45 x 10
10
cm
-3
, µ
n
= 1500 cm
2
volt
-1
sec
-1
and , µ
p
= 500
cm
2
volt
-1
sec
-1
, we have:
σ
i
= 4.64x10
−
6
Ωcm and σ
mi
n
= 4.02x10
−
6
Ωcm
Thus there are not large differences between the two.
1.6. When a Au atom sits on a lattice site in a silicon crystal, it can act as either a
donor or an acceptor. E
D
and E
A
levels both exist for the Au and both are close
to the middle of the silicon bandgap. If a small concentration of Au is placed in
an N type silicon crystal, will the Au behave as a donor or an acceptor? Explain.
Answer:
In N type material, the Fermi level will be in the upper half of the bandgap as shown
in the band diagram below. Allowed energy levels below E
F
will in general be
occupied by electrons. Thus the E
D
and E
A
levels will have electrons filling them.
This means the donor level will not have donated its electron whereas the acceptor
level will have accepted an electron. Thus the Au atoms will act as acceptors in N
type material.
•
Free
Electrons
Holes
E
F
E
D
E
A
E
C
E
V
1.7. Show that E
F
is approximately in the middle of the bandgap for intrinsic silicon.
Answer:
Starting with Eqn. 1.9 and 1.10 in the text, we have
n ≅ N
C
exp −
E
C
−
E
F
kT
⎛
⎝
⎞
⎠
and p ≅ N
V
exp −
E
F
−
E
V
kT
⎛
⎝
⎞
⎠
In intrinsic material, n = p = n
I
, so we have
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