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Silicon VLSI Technology:
UV V
Fundamentals, Practice, and
S
Modeling, 1st Edition
IAIA
SOLUTIONS
?_?_
MANUAL
AAP
PPPR
James D. Plummer, Michael D. Deal, Peter B. Griffin
ROOV
Comprehensive Solutions Manual for Instructors and
Students
VE D
||| ||| || ||| || | || ||| | || ||| |
9780130850379
E
© James D. Plummer, Michael D. Deal & Peter B. Griffin. All rights
D? ?
reserved. Reproduction or distribution without permission is
prohibited.
?
© MEDGEEK
, Solutions Manual
ST TU
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
UV V
S
line? If not what reasons can you suggest for any deviations you observe?
Answer:
250
IAIA
200
150
?_?_
100
50
AAP
0
1997 2002 2007 2012
Year
PPPR
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.
ROOV
1.2. Assuming dopant atoms are uniformly distributed in a silicon crystal, how far
apart are these atoms when the doping concentration is a). 1015 cm-3, b). 1018
cm-3, c). 5x1020 cm-3.
Answer:
VE D
The average distance between the dopant atoms would just be one over the cube
root of the dopant concentration:
E
x = N A −
D? ?
( )
−
a) x = 1x1015 cm −3 = 1x10
−5
cm = 0.1μm = 100nm
b) x = (1x10 )
18 −3 − −6
cm = 1x10 cm = 0.01μm = 10nm
?
SILICON VLSI TECHNOLOGY 2 © 2000 by Prentice Hall
Fundamentals, Practice and Modeling Upper Saddle River, NJ.
By Plummer, Deal and Griffin
, Solutions Manual
ST TU
( )
20 −3 − −7
c) x = 5x10 cm = 1.3x10 cm = 0.0013μm = 1.3nm
1.3. Consider a piece of pure silicon 100 µm long with a cross-sectional area of 1
UV V
µm2. How much current would flow through this “resistor” at room
S
temperature in response to an applied voltage of 1 volt?
Answer:
If the silicon is pure, then the carrier concentration will be simply ni. At room
temperature, ni ≈ 1.45 x 1010 cm-3. Under an applied field, the current will be due to
IAIA
drift and hence,
(
I = I n + I p = qAn i μ n + μ p ε )
( )( )(1.45x10 )(2000cm volt )⎛⎝ 101voltcm ⎞⎠
?_?_
− 19 −8 2 10 −3 2 −1 −1
= 1.6x10 coul 10 cm carrierscm sec −2
= 4.64x10 −12 amps or 4.64pA
1.4. Estimate the resistivity of pure silicon in Ω ohm cm at a) room temperature, b)
AAP
77K, and c) 1000 ˚C. You may neglect the temperature dependence of the
carrier mobility in making this estimate.
Answer:
PPPR
The resistivity of pure silicon is given by Eqn. 1.1 as
1 1
ρ= =
(
q μ n n + μp p ) (
qni μ n + μ p )
ROOV
Thus the temperature dependence arises because of the change in ni with T. Using
Eqn. 1.4 in the text, we can calculate values for ni at each of the temperatires of
interest. Thus
⎛ 0.603eV ⎞
n i = 3.1x1016 T 3/ 2 exp⎝ −
VE D
kT ⎠
which gives values of ≈ 1.45 x 1010 cm-3 at room T, 7.34 x 10-21 cm-3 at 77K and 5.8
E
x 1018 cm-3 at 1000 ˚C. Taking room temperature values for the mobilities , µn =
1500 cm2 volt-1 sec-1 and , µp = 500 cm2 volt-1 sec-1, we have,
D? ?
?
SILICON VLSI TECHNOLOGY 3 © 2000 by Prentice Hall
Fundamentals, Practice and Modeling Upper Saddle River, NJ.
By Plummer, Deal and Griffin
, Solutions Manual
ST TU
ρ = 2.15x105 Ωcm at room T
= 4.26x10 35 Ωcm at 77K
= 5.39x10 −4 Ωcm at 1000 ÞC
UV V
S
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 ni value calculated above.
1.5. a). Show that the minimum conductivity of a semiconductor sample occurs
IAIA
μp
when n = ni .
μn
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).
( )
AAP
1
σ= = q μ n n + μp p
ρ
To find the minimum we set the derivative equal to zero.
∂ ⎧ ⎛⎜ n 2i ⎞ ⎫ ⎛ n2 ⎞
∴
∂σ ∂
= {(
q μnn + μpp = )} ⎨q μ n + μ p ⎟ ⎬ = q⎜ μ n + μ p i2 ⎟ = 0
PPPR
∂n ∂n ∂n ⎩ ⎝ n n ⎠⎭ ⎝ n ⎠
μp μp
∴n 2 = n 2i or n = ni
μn μn
ROOV
b). Using the value for n derived above, we have:
⎛ ⎞
⎜ μp n2i ⎟ ⎛ μp μn ⎞
σ min = q ⎜ μn n i + μp = ⎜ μ + μ ⎟ = 2qn i μ nμ p
VE D
⎟ q n n i p n i
μn μp ⎝ μn μp ⎠
⎜ ni ⎟
⎝ μn ⎠
E
D? ?
c). The intrinsic conductivity is given by
(
σ i = qni μ n + μ p )
?
SILICON VLSI TECHNOLOGY 4 © 2000 by Prentice Hall
Fundamentals, Practice and Modeling Upper Saddle River, NJ.
By Plummer, Deal and Griffin
ST TU
Silicon VLSI Technology:
UV V
Fundamentals, Practice, and
S
Modeling, 1st Edition
IAIA
SOLUTIONS
?_?_
MANUAL
AAP
PPPR
James D. Plummer, Michael D. Deal, Peter B. Griffin
ROOV
Comprehensive Solutions Manual for Instructors and
Students
VE D
||| ||| || ||| || | || ||| | || ||| |
9780130850379
E
© James D. Plummer, Michael D. Deal & Peter B. Griffin. All rights
D? ?
reserved. Reproduction or distribution without permission is
prohibited.
?
© MEDGEEK
, Solutions Manual
ST TU
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
UV V
S
line? If not what reasons can you suggest for any deviations you observe?
Answer:
250
IAIA
200
150
?_?_
100
50
AAP
0
1997 2002 2007 2012
Year
PPPR
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.
ROOV
1.2. Assuming dopant atoms are uniformly distributed in a silicon crystal, how far
apart are these atoms when the doping concentration is a). 1015 cm-3, b). 1018
cm-3, c). 5x1020 cm-3.
Answer:
VE D
The average distance between the dopant atoms would just be one over the cube
root of the dopant concentration:
E
x = N A −
D? ?
( )
−
a) x = 1x1015 cm −3 = 1x10
−5
cm = 0.1μm = 100nm
b) x = (1x10 )
18 −3 − −6
cm = 1x10 cm = 0.01μm = 10nm
?
SILICON VLSI TECHNOLOGY 2 © 2000 by Prentice Hall
Fundamentals, Practice and Modeling Upper Saddle River, NJ.
By Plummer, Deal and Griffin
, Solutions Manual
ST TU
( )
20 −3 − −7
c) x = 5x10 cm = 1.3x10 cm = 0.0013μm = 1.3nm
1.3. Consider a piece of pure silicon 100 µm long with a cross-sectional area of 1
UV V
µm2. How much current would flow through this “resistor” at room
S
temperature in response to an applied voltage of 1 volt?
Answer:
If the silicon is pure, then the carrier concentration will be simply ni. At room
temperature, ni ≈ 1.45 x 1010 cm-3. Under an applied field, the current will be due to
IAIA
drift and hence,
(
I = I n + I p = qAn i μ n + μ p ε )
( )( )(1.45x10 )(2000cm volt )⎛⎝ 101voltcm ⎞⎠
?_?_
− 19 −8 2 10 −3 2 −1 −1
= 1.6x10 coul 10 cm carrierscm sec −2
= 4.64x10 −12 amps or 4.64pA
1.4. Estimate the resistivity of pure silicon in Ω ohm cm at a) room temperature, b)
AAP
77K, and c) 1000 ˚C. You may neglect the temperature dependence of the
carrier mobility in making this estimate.
Answer:
PPPR
The resistivity of pure silicon is given by Eqn. 1.1 as
1 1
ρ= =
(
q μ n n + μp p ) (
qni μ n + μ p )
ROOV
Thus the temperature dependence arises because of the change in ni with T. Using
Eqn. 1.4 in the text, we can calculate values for ni at each of the temperatires of
interest. Thus
⎛ 0.603eV ⎞
n i = 3.1x1016 T 3/ 2 exp⎝ −
VE D
kT ⎠
which gives values of ≈ 1.45 x 1010 cm-3 at room T, 7.34 x 10-21 cm-3 at 77K and 5.8
E
x 1018 cm-3 at 1000 ˚C. Taking room temperature values for the mobilities , µn =
1500 cm2 volt-1 sec-1 and , µp = 500 cm2 volt-1 sec-1, we have,
D? ?
?
SILICON VLSI TECHNOLOGY 3 © 2000 by Prentice Hall
Fundamentals, Practice and Modeling Upper Saddle River, NJ.
By Plummer, Deal and Griffin
, Solutions Manual
ST TU
ρ = 2.15x105 Ωcm at room T
= 4.26x10 35 Ωcm at 77K
= 5.39x10 −4 Ωcm at 1000 ÞC
UV V
S
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 ni value calculated above.
1.5. a). Show that the minimum conductivity of a semiconductor sample occurs
IAIA
μp
when n = ni .
μn
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).
( )
AAP
1
σ= = q μ n n + μp p
ρ
To find the minimum we set the derivative equal to zero.
∂ ⎧ ⎛⎜ n 2i ⎞ ⎫ ⎛ n2 ⎞
∴
∂σ ∂
= {(
q μnn + μpp = )} ⎨q μ n + μ p ⎟ ⎬ = q⎜ μ n + μ p i2 ⎟ = 0
PPPR
∂n ∂n ∂n ⎩ ⎝ n n ⎠⎭ ⎝ n ⎠
μp μp
∴n 2 = n 2i or n = ni
μn μn
ROOV
b). Using the value for n derived above, we have:
⎛ ⎞
⎜ μp n2i ⎟ ⎛ μp μn ⎞
σ min = q ⎜ μn n i + μp = ⎜ μ + μ ⎟ = 2qn i μ nμ p
VE D
⎟ q n n i p n i
μn μp ⎝ μn μp ⎠
⎜ ni ⎟
⎝ μn ⎠
E
D? ?
c). The intrinsic conductivity is given by
(
σ i = qni μ n + μ p )
?
SILICON VLSI TECHNOLOGY 4 © 2000 by Prentice Hall
Fundamentals, Practice and Modeling Upper Saddle River, NJ.
By Plummer, Deal and Griffin
