1.1 Phosphorous donor atoms with a concentration of 1016 cm-3 are added to a pure sample of
silicon. Assume that the phosphorous atoms are distributed homogeneously throughout the
silicon. The atomic weight of phosphorous is 31.
a) What is the sample resistivity at 300 K?
The electron mobility in the silicon equals:
µ − µ min
1414 − 68.5
µ n = µ min + max
= 68.5 +
= 1184 cm2 /V-s
α
16
0. 711
N
10
1+ ( )
1+ (
)
Nr
9.2 × 1016
1
1
1
1
=
= 0.53 Ω cm
ρ= =
≅
−19
σ q( µ n n + µ p p ) q µ n n 1.6 × 10 × 1184 × 1016
b) What proportion by weight does the donor impurity comprise
weight
volume P
Mm A N d
31 × 1.6 ×10 27 × 10 3 ×1016
=
= 2.1 × 10 −7
≅
weight
2.928
density of Si
volume P +Si
c) If 1017 atoms cm-3 of boron are included in addition to phosphorous, and distributed
uniformly, what is the resulting resistivity and conductivity type (i.e., p- or n-type
material)?
The semiconductor is p-type since Na > Nd
The hole density is obtained from:
N a+ − N d−
N + − N d− 2
9 × 1016
9 × 1016 2
+ ( a
) + n i2 =
+ (
) + (1010 ) 2 = 9 x 1016 cm-3
2
2
2
2
and the mobility is calculated from the sum of the donor and acceptor densities
µ − µ min
470.5 − 44.9
µ p = µ min + max
= 44.9 +
= 310.6 cm2 /V-s
α
16
0. 719
N
11 × 10
1+ ( )
1+ (
)
Nr
2.23 × 1017
leading to the conductivity of the material:
1
1
1
1
≅
=
= 0.22 Ω cm
ρ= =
−19
σ q( µ n n + µ p p ) qµ p p 1.6 × 10 × 310.6 × 9 × 1016
p=
d) Sketch the energy-band diagram under the condition of c) and show the position of the
Fermi level.
Nv
1.04 × 1019
=
0
.
0259
ln
=123 meV
E F − E v = kT ln
p
9 × 1016
1.2 Find the equilibrium electron and hole concentrations and the location of the Fermi level for
silicon at 27o C if the silicon contains the following concentrations of shallow dopants.
a) 1 x 1016 cm-3 boron atoms
Na = 1016 cm-3
Since these are shallow acceptors and the material is not compensated, degenerate or
close to intrinsic, the hole density equals the acceptor density
p ≈ 1016 cm-3
Using the mass action law we then find the electron density
n = ni2 /p = 2.1 x 104 cm-3
The Fermi energy is then obtained from:
n
10 4
= 0.0259 ln
= -348 meV
E F − E i = kT ln
ni
1.45 × 1010
b) 3 x 1016 cm-3 arsenic atoms and 2.9 x 1016 cm-3 boron atoms
Na = 2.9 x 1016 cm-3 , Nd = 3 x 1016 cm-3
Since these are shallow acceptors and the material is not degenerate or close to intrinsic,
the electron density equals the difference between the donor and acceptor density
n ≈ Nd – Na = 1015 cm-3
Using the mass action law we then find the hole density
p = ni2 /n = 2.1 x 105 cm-3
The Fermi energy is then obtained from:
n
1015
= 0.0259 ln
= -288 meV
E F − E i = kT ln
ni
1.45 × 1010
1.3 An n-type sample of silicon has a uniform density Nd = 1016 cm-3 atoms cm-3 of arsenic and a
p-type silicon sample has Na = 1015 atoms cm-3 of boron. For each semiconductor material
determine the following:
a) The temperature at which half the impurity atoms are ionized. Assume that all mobile
electrons and holes come from dopant impurities.
N
E − Ec
n = d = N c exp( F
) since half the donors are ionized,
2
kT
N
2π me*kT 3 / 2
E − Ec
or f(Ed ) = ½ so that EF = Ed so that d = 2 [
] exp( d
)
2
2
h
kT
Ec – Ed = 49 meV, Nd = 1016 cm-3
Solution: T = 84.4 K, Nc = 4.19 x 1018 cm-3
Similarly for Ea – Ev = 45 meV, Na = 1015 cm-3
Na
2π m *h kT 3 / 2
E − Ea
= 2[
] exp( v
)
2
2
h
kT
Solution: T = 63.7 K, Nv = 1.79 x 1018 cm-3
b) The temperature at which the intrinsic concentration ni exceeds the impurity density by a
factor of 10. See table 1.4 for ni(T).
− 7014
n i = 10 N d = 1017 = 3.87 ×1016 T 3 / 2 exp(
) , yielding T = 776 K
T
− 7014
n i = 10 N a = 1016 = 3.87 × 1016 T 3 / 2 exp(
) , yielding T = 636 K
T
c) The equilibrium minority concentration at 300 K. Assume full ionization of impurities
n2
(1.45 ×10 10 ) 2
p≅ i =
= 2.1 x 104 cm-3
16
Nd
10
n≅
ni2 (1.45 × 1010 ) 2
=
= 2.1 x 105 cm-3
15
Na
10
d) The Fermi energy referred to the valence-band edge Ev in each material at 300 K. The
Fermi energy if both types of impurities are present in a single sample.
N
1.04 × 1019
= 875 meV
E F − E v = kT ln v = 0.0259 ln
p
2.1× 10 4
N
1.04 × 1019
= 239 meV
E F − E v = kT ln v = 0.0259 ln
p
1015
if Na = 1015 cm-3 and Nd = 1016 cm-3
n i2
(1.45 × 1010 ) 2
p≅
=
= 2.34 x 104 cm-3
15
Nd − Na
9 × 10
N
1.04 × 1019
= 873 meV
E F − E v = kT ln v = 0.0259 ln
p
2.34 × 10 4
1.4 A piece of n-type silicon has a resistivity of 5 Ohm-cm at 27o C. Find the thermal equilibrium
hole concentrations at 27, 100, and 500o C.
µ
− µ min
1
1
and µ = µ min + max
=
α
σ q( µ n n + µ p p )
N
1+( )
Nr
14
-3
ρ = 5 Ohm-cm for Nd = 9.16 x 10 cm and µn = 1365 cm2 /V-s
Nd
Nd 2
ni2
2
no =
+ (
) + ni , p =
2
2
n
ρ=
T (K)
300
373
773
ni (cm-3 )
1.41 x 1010
2.00 x 1012
9.53 x 1016
n (cm-3 )
9.16 x 1014
9.16 x 1014
5.58 x 1016
p (cm-3 )
2.17 x 105
3.94 x 109
9.58 x 1016