ENE 311
Homework # 3
Due: 23 Jan 2007 (Group 21)
26 Jan 2007 (Group 1)
1. At room temperature the effective density of states in the valence band is 2.66
x 1019 cm-3 for silicon. Find the corresponding effective mass of holes.
Compare this mass with the free-electron mass. Ans: 1.03m0
2. Calculate the location of Ei in silicon at liquid nitrogen temperature at 77K and
at 100ºC. Is it reasonable to assume that Ei is in the center of the forbidden
gap? Given mh = m0 and me = 0.19m0. Ans: 0.583 eV and 0.557 eV
3. A silicon sample at T = 300 K contains an acceptor impurity concentration of
NA = 1016 cm-3. Determine the concentration of donor impurity atoms that
must be added so that the silicon is n-type and the Fermi energy is 0.20 eV
below the conduction band edge. Ans: 2.26 x 1016 cm-3
4. Find the electron and hole concentration and Fermi level in silicon at 300 K
for 1 x 1015 boron atoms/cm3. Ans: 1015 cm-3, 9.3 x 104 cm-3, 0.26 eV
Extra problem: (not required for grading)
5. Calculate the Fermi level of silicon doped with 1015, 1017, and 1019 phosphorus
atoms/cm3 at room temperature, assuming complete ionization. From the
calculated Fermi level, check if the assumption of complete ionization is
justified for each doping. Assume that the ionized donors is given by
n  N D 1  F  ED  
N
1 e
D
( EF  ED ) / kT
Ans: valid for the case of 1015 cm-3
6. Repeat the problem 4 by changing from boron impurities alone to be 3 x 10 16
boron atoms/cm3 and 2.9 x 1016 arsenic atoms/cm3. Ans: 1015 cm-3, 9.3 x 104
cm-3, 0.26 eV
7. For an n-type silicon sample with 1016 cm-3 phosphorus donor impurities and a
donor level at ED = 0.045 eV, find the ratio of the neutral donor density to the
ionized donor density at 77 K where the Fermi level is 0.0459 eV below the
bottom of the conduction band. Ans: 0.876
Solutions
Homework 3
1.
NV = 2(2πmp kT / h2)3/2
The effective mass of holes in Si is
mp = (NV / 2) 2/3 ( h2 / 2πkT )
 2.66  1019  10 6 m 3 

= 
2


2
3
6.625  10 
2 1.38  10 300 
34 2
 23
= 9.4 × 10-31 kg = 1.03 m0.
2.
Calculate the location of Ei in silicon at liquid nitrogen temperature at 77K
and at 100ºC. Is it reasonable to assume that Ei is in the center of the forbidden
gap? Given mh* = m0 and me* = 0.19m0.
Soln
 2 ln N
E i  ( E C  EV ) 2  kT

 EC  EV 
2
V
NC 
 mh* 

ln  * 
4
 me 
3kT
We may find Ei by measuring it from the top of the valence band as Ei – EV OR we
may assume EV = 0 to be our reference. Therefore, we have
Ei  EV 
 EC  EV 
2
 mh* 
3kT  mh* 

ln  *   Eg 
ln  * 
4
4
 me 
 me 
3kT
(1)
Actually, Eg is the function of temperature T. Since the problem did not give Eg(T),
therefore, we may assume Eg(T) is about the same as Eg(300K) = 1.12 eV.
At 77 K
Ei  EV 
1.12 3  1.38  10 23  77  1 
3

ln 
  0.56   8.27  10   0.5683
19
2
4  1.6  10
 0.19 
At 373 K
Ei  EV 
1.12 3  1.38  10 23  373  1 

ln 
  0.56  0.04  0.564
2
4  1.6  10 19
 0.19 
Because the second term on the right-hand side of the Eq.1 is much smaller
compared to the first term, over the above temperature range, it is reasonable to
assume that Ei is in the center of the forbidden gap.
Ec – EF = kT ln [NC / (ND – NA)]
3. From
ND – NA = NC exp [–(EC – EF) / kT ]
which can be rewritten as
ND – NA = 2.86 × 1019 exp(–0.20 / 0.0259) = 1.26 × 1016 cm-3
Then
or
ND= 1.26 × 1016 + NA = 2.26 × 1016 cm-3
A compensated semiconductor can be fabricated to provide a specific Fermi
energy level.
4. The ionization energy for boron in Si is 0.045 eV.
At 300 K, all boron
impurities are ionized. Thus pp = NA = 1015 cm-3
np = ni2 / nA = (9.65 × 109)2 / 1015 = 9.3 × 104 cm-3.
The Fermi level measured from the top of the valence band is given by:
EF – EV = kT ln(NV/ND) = 0.0259 ln (2.66 × 1019 / 1015) = 0.26 eV
5.
n  ni e
 EF  Ei  / kT
n
N 
 kT ln  D 

 ni 
 ni 
EF  Ei  kT ln 
For N D  1015
E F  Ei 
1.38  10
23
1.6  10 19
 1015 
 300 ln 
 0.30 eV
9 
 9.65  10 
For N D  1017
E F  Ei 
1.38  10
1.6  10


  0.42 eV
 9.65  10 
23
19
17
 300 ln 
10

10
9
For N D  1019
E F  Ei 
E
C
1.38  10
1.6  10
23
19

  0.54 eV
 9.65  10 
 300 ln 
19
9
 Ei    EF  Ei    EC  ED   ED  EF
For ND = 1015 cm-3, E  E  
D
F
1.12 
  0.3  0.045  0.215
 2 
For ND = 1017 cm-3, E  E  
1.12 
For ND = 1019 cm-3, E  E  
1.12 
D
D
F
F
  0.42  0.045  0.095
 2 
  0.54  0.045  0.025
 2 
The number of electrons that are ionized is given by
n  ND[1 – F(ED)] = ND / [1 + e  ED  EF  / kT ]
We obtain the number of ionized donors as
n = 1015 cm-3
for ND = 1015 cm-3
17
-3
n = 0.975 × 10 cm
for ND = 1017 cm-3
n = 0.275 × 1019 cm-3
for ND = 1019 cm-3
Therefore, the assumption of complete ionization is valid only for the case of
1015 cm-3.
6. The boron atoms compensate the arsenic atoms; we have
pp = NA – ND = 3 × 1016 – 2.9 × 1016 = 1015 cm-3
Since pp is the same as given in (a), the values for np and EF are the same as
in problem 4. However, the mobilities and resistivities for these two samples
are different.
7. ND+ =
=
1016
1  e ( ED  EF ) / kT
=
1016
1  e 0.135
1016
= 5.33 × 1015 cm-3
1
1
1.145
The neutral donor = 1016 – 5.33 ×1015 cm-3 = 4.67 × 1015 cm-3
N DO
4.76
∴ The ratio of
=
= 0.876 .

5.33
ND