SILICON VLSI TECHNOLOGY Fundamentals, Practice and Models Solutions Manual for Instructors

Solutions Manual


SILICON VLSI TECHNOLOGY
Fundamentals, Practice and Models
Solutions Manual for Instructors




James D. Plummer
Michael D. Deal
Peter B. Griffin

, Solutions Manual


Chapter 1 Problems
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:

250

200

150

100

50

0
1997 2002 2007 2012
Year

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.

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:

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  1x1015 cm 3  1x10 5 cm  0.1m  100nm

b) x  1x10 

18
cm 3  1x10 6 cm  0.01m  10nm

, Solutions Manual


 

c) x  5x10 20 cm  3  1.3x10 7 cm  0.0013m  1.3nm

Consider a piece of pure silicon 100 µm long with a cross-sectional area of 1 µm2.
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 ni. At room
temperature, ni ≈ 1.45 x 1010 cm-3. Under an applied field, the current will be due to
drift and hence,


I  I n  I p  qAn i  n   p 
     1volt
 1.6x10 19 coul 10 8 cm 2 1.45x1010 carrierscm 3 2000cm 2 volt  1 sec 1  2 
10 cm 

 4.64x1012 amps or 4.64pA

Estimate the resistivity of pure silicon in  ohm cm at a) room temperature, b)
77K, and c) 1000 ˚C. You may neglect the temperature dependence of
thecarrier mobility in making this estimate.

Answer:

The resistivity of pure silicon is given by Eqn. 1.1 as
1 1
 

q  n n  p p  
qni  n   p 


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
ni  3.1x1016 T3/ 2  0.603eV
exp
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
x 1018 cm-3 at 1000 ˚C. Taking room temperature values for the mobilities , µ n =
1500 cm2 volt-1 sec-1 and , µ p = 500 cm2 volt-1 sec-1, we have,

, Solutions Manual


 2.15x105 cm at room T
 4.26x1035 cm at 77K
 5.39x104 cm at 1000 ÞC

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.

a). Show that the minimum conductivity of a semiconductor sample occurs
p
when n  ni .
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).
1



 q  n n  p p 
To find the minimum we set the derivative equal to zero.
     n2    n2 





q nn  pp  

q  n n  p i 
  q  n  p i   0
n n n  n   n2 

p
n 2  n 2 or n  n p
i
n i n

b). Using the value for n derived above, we have:

 
  
n  2qni
2
min  q   n n i  p  p ni   q nni
p   n
p i  n p
n p   n p 
 ni 
 n 



c). The intrinsic conductivity is given by


 i  qni  n   p 