MSE 630
Homework #1
Solutions
Chapter 1
Problem #
2. 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).
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:

b) x  1x10
c) x  5x10
a) x  1x1015 cm 3
18
3.
cm 3
20



cm3
x  N A 1 / 3
1 / 3
 1x10 5 cm  0.1m 100nm
1 / 3
 1x10 6 cm  0.01m  10nm
1 / 3
1.3x10 7 cm  0.0013m  1.3nm
1.3. 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 


 1.6x10 19 coul 10 8 cm 2 1.45x1010 carrierscm 3 2000cm 2 volt  1 sec 1
101voltcm 
2
 4.64x10 12 amps or 4.64pA
5.
1.5. 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.



 
2 

2 
 n q n n  p nni  q n   p nn i2   0
 

q nn  pp
n n

p
n 2  n 2i
n

p
or n  ni
n
b). Using the value for n derived above, we have:


p
n2i
 min  q n n i
 p
p
n

ni
n

c).

 
p
n 

q


n


n
 2qn i  n p
n
i
p
i

n
 p 



The intrinsic conductivity is given by

 i  qni  n   p

Taking values of ni = 1.45 x 1010 cm-3, µn = 1500 cm2 volt-1 sec-1 and , µp = 500
cm2 volt-1 sec-1, we have:
 i  4.64x10 6 cm and  min  4.02x10 6 cm
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. ED and EA 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 EF will in general be
occupied by electrons. Thus the ED and EA 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.
1.8. Construct a diagram similar to Figure 1-27b for P-type material. Explain
physically, using this diagram, why the capture of the minority carrier electrons
in P-type material is the rate limiting step in recombination.
Answer:
In P type material (on the right), there are many more holes than electrons. Thus the
probability of capturing a hole by the trap is high compared to electrons. So the
normal state of occupancy of the trap level will be with a hole present. The
recombination rate will then be limited by the capture of the "scarce" species which is
the minority carrier or electron in this case.
Stated another way, since EF is below the trap level in P type material, ET will
normally not be occupied by an electron. Thus the limiting process is the electron
capture.
1.9. A silicon diode has doping concentrations on the N and P sides of N D = 1 x 1019
cm-3 and of NA = 1 x 1015 cm-3. Calculate the process temperature at which the
two sides of the diode become intrinsic. (Intrinsic is defined as ni = ND or NA.)
Answer:
Each side of the diode will become intrinsic at the temperature at which n I = ND or NA.
We can estimate these temperatures by looking at the graph in Fig. 1-16. From the
graph, the N side of the diode will become intrinsic around 1400K or 1100˚C. The P
side will become intrinsic at a lower temperature since NA is smaller than ND. From
the figure, the P side becomes intrinsic at about 500K or 200˚C.
More accurately, we solve Eqn. 1.4 to find the exact temperature. Thus
 0.603eV 
n i  3.1x1016 T3 / 2 exp 
kT 
Setting nI = 1019 cm-3 and 1015 cm-3 respectively, we and solving by iteration, we find
that the N side becomes intrinsic at about 1380K or about 1110˚C and the P side
becomes intrinsic at about 545K or about 270˚C.