EE145 Spring 2002
Homework 5 Solution
Prof. Ali Shakouri
Second Edition ( 2001 McGraw-Hill)
Chapter 5
5.1 Bandgap and photodetection
a. Determine the maximum value of the energy gap that a semiconductor, used as a photoconductor,
can have if it is to be sensitive to yellow light (600 nm).
b. A photodetector whose area is 5 × 10-2 cm2 is irradiated with yellow light whose intensity is 2 mW
cm-2. Assuming that each photon generates one electron-hole pair, calculate the number of pairs
generated per second.
c. From the known energy gap of the semiconductor GaAs (Eg = 1.42 eV), calculate the primary
wavelength of photons emitted from this crystal as a result of electron-hole recombination.
d. Is the above wavelength visible?
e. Will a silicon photodetector be sensitive to the radiation from a GaAs laser? Why?
Solution
a
We are given the wavelength λ = 600 nm, therefore we need Eph = hυ = Eg so that,
Eg = hc/λ = (6.626 × 10-34 J s)(3.0 × 108 m s-1) / (600 × 10-9 m)
∴
b
Eg = 3.31 × 10-19 J or 2.07 eV
Area A = 5 × 10-2 cm2 and light intensity Ilight = 2 × 10-3 W/cm2. The received power is:
P = AIlight = (5 × 10-2 cm2)(2 × 10-3 W/cm2) = 1.0 × 10-4 W
Nph = number of photons arriving per second = P/Eph
∴
Nph = (1.0 × 10-4 W) / (3.31 × 10-19 J) = 3.02 × 1014 Photons s-1
Since the each photon contributes one electron-hole pair (EHP), the number of EHPs is then:
NEHP = 3.02 × 1014 EHP s-1
c
For GaAs, Eg = 1.42 eV and the corresponding wavelength is
λ = hc/Eg = (6.626 × 10-34 J s)(3.0 × 108 m s-1) / (1.42 eV × 1.602 × 10-19 J/eV)
∴
λ = 8.74 × 10-7 m or 874 nm
The wavelength of emitted radiation due to electron-hole pair (EHP) recombination is therefore
874 nm.
d
It is not in the visible region (it is in the infrared).
e
is,
From Table 5.1 (in the textbook), for Si, Eg = 1.10 eV and the corresponding cut-off wavelength
λg = hc/Eg = (6.626 × 10-34 J s)(3.0 × 108 m s-1) / (1.1 eV × 1.602 × 10-19 J/eV)
∴
λg = 1.13 × 10-6 m or 1130 nm
5.1
Homework 5 Solution
EE145 Spring 2002
Prof. Ali Shakouri
Since the 874 nm wavelength of the GaAs laser is shorter than the cut-off wavelength of 1130
nm, the Si photodetector can detect the 874 nm radiation (Put differently, the photon energy
corresponding to 874 nm, 1.42 eV, is larger than the Eg, 1.10 eV, of Si which means that the Si
photodetector can indeed detect the 874 nm radiation).
5.2 Minimum conductivity
a. Consider the conductivity of a semiconductor, σ = enµe + epµh. Will doping always increase the
conductivity?
b. Show that the minimum conductivity for Si is obtained when it is p-type doped such that the hole
concentration is
pm = ni
µe
µh
and the corresponding minimum conductivity (maximum resistivity) is
σ min = 2eni µeµ h
c. Calculate pm and σmin for Si and compare with intrinsic values.
Solution
a
Doping does not always increase the conductivity. Suppose that we have an intrinsic sample
with n = p but the hole drift mobility is smaller. If we dope the material very slightly with p-type then p
> n. However, this would decrease the conductivity because it would create more holes with lower
mobility at the expense of electrons with higher mobility. Obviously with further doping p increases
sufficiently to result in the conductivity increasing with the extent of doping.
b
To find the minimum conductivity, first consider the mass action law:
np = ni2
isolate n:
n = ni2/p
Now substitute for n in the equation for conductivity:
σ = enµe + epµh
∴
eni2 µe
σ=
+ µh ep
p
To find the value of p that gives minimum conductivity (pm), differentiate the above equation
with respect to p and set it equal to zero:
dσ
eni2 µe
=−
+ µh e
2
dp
p
∴
−
eni 2 µ e
+ µh e = 0
pm 2
Isolate pm and simplify,
5.2
Homework 5 Solution
EE145 Spring 2002
pm = ni
Prof. Ali Shakouri
µe
µh
Substituting this expression back into the equation for conductivity will give the minimum
conductivity:
σ min
µe
eni2 µe
eni 2 µe
=
+ µ hep m =
+ µh eni
pm
ni µ e µ h
µh
µh
+ eni µe µ h = eni µe µh + eni µe µh
µe
∴
σ min = eni µe
∴
σ min = 2eni µe µh
c
From Table 5.1, for Si: µe = 1350 cm2 V-1 s-1, µh = 450 cm2 V-1 s-1 and ni = 1.45 × 1010 cm-3.
Substituting into the equations for pm and σmin:
pm = ni
µe
1350 cm 2 V−1 s−1
10
−3
10
= (1.45 × 10 cm )
cm-3
2
−1
−1 = 2.51 × 10
µh
450 cm V s
σ min = 2eni µe µh
∴
σ min = 2(1.602 × 10−19 C)(1.45 × 1010 cm−3 ) (1350 cm2 V−1 s−1 )(450 cm2 V−1 s−1 )
∴
σmin = 3.62 × 10-6 Ω-1 cm-1
The corresponding maximum resistivity is:
ρmax = 1 / σmin = 2.76 × 105 Ω cm
The intrinsic value corresponding to pm is simply ni (= 1.45 × 1010 cm-3). Comparing it to pm:
pm 2.51 × 1010 cm −3
=
= 1.73
ni 1.45 × 1010 cm−3
The intrinsic conductivity is:
σint = eni(µe + µh)
∴
σint = (1.602 × 10-19 C)(1.45 × 1010 cm-3)(1350 cm2 V-1 s-1 + 450 cm2 V-1 s-1)
∴
σint = 4.18 × 10-6 Ω-1 cm-1
Comparing this value to the minimum conductivity:
σ int 3.62 × 10 −6 W −1 cm−1
=
= 0.866
σ min 4.18 × 10 −6 W -1 cm-1
Sufficient p-type doping that increases the hole concentration by 73% decreases the conductivity
by 15% to its minimum value.
5.3 Compensation doping in Si
a. A Si wafer has been doped n-type with 1017 As atoms cm-3.
1. Calculate the conductivity of the sample at 27 °C.
5.3
Homework 5 Solution
EE145 Spring 2002
Prof. Ali Shakouri
2. Where is the Fermi level in this sample at 27 °C with respect to the Fermi level (EFi) in intrinsic
Si?
3. Calculate the conductivity of the sample at 127 °C.
b. The above n-type Si sample is further doped with 9 × 1016 boron atoms (p-type dopant) per
centimeter cubed.
1. Calculate the conductivity of the sample at 27 °C.
2. Where is the Fermi level in this sample with respect to the Fermi level in the sample in (a) at 27
°C? Is this an n-type or p-type Si?
Solution
Given temperature T = 27 °C = 300 K, concentration of donors Nd = 1017 cm-3, and drift mobility
µe ≈ 800 cm2 V-1 s-1 (from Figure 5Q3-1). At room temperature the electron concentration n = Nd >> p
(hole concentration).
a
2000
1000
Holes
Electrons
100
50
1015
1016
1017
1018
Dopant Concentration,
1019
1020
cm-3
Figure 5Q3-1 The variation of the drift mobility with dopant concentration in Si
for electrons and holes at 300 K.
(1)
The conductivity of the sample is:
σ = eNdµe ≈ (1.602 × 10-19 C)(1017 cm-3)(800 cm2 V-1 s-1) = 12.8 Ω-1 cm-1
(2)
In intrinsic Si, EF = EFi,
ni = Ncexp[−(Ec − EFi)/kT]
(1)
In doped Si, n = Nd, EF = EFn,
n = Nd = Ncexp[−(Ec − EFn)/kT]
(2)
Eqn. (2) divided by Eqn. (1) gives,
Nd
 E − EFi 
= exp Fn

ni
kT 
∴
(3)
 N  E − EFi
ln  d  = Fn
kT
 ni 
5.4
Homework 5 Solution
EE145 Spring 2002
∴
Prof. Ali Shakouri
∆EF = EFn − EFi = kT ln(Nd/ni)
(4)
Substituting we find (ni = 1.45 × 1010 cm-3 from Table 5.1 in the textbook),
∆EF = (8.617 × 10-5 eV/K)(300 K)ln[(1017 cm-3)/ (1.45 × 1010 cm-3)]
∴
∆EF = 0.407 eV above Efi
50000
µL∝T–1.5
Ge
Nd =1014
10000
Nd =1016
Nd =1013
Nd =1017
1000
Nd =1018
100
Nd =1019
Si
µΙ ∝T1.5
10
70
100
Temperature (K)
800
Figure 5Q3-2 Log-log plot for drift mobility versus temperature for n-type Ge
and n-type Si samples. Various donor concentrations for Si are shown, Nd are in
cm-3. The upper right insert is the simple theory for lattice limited mobility
whereas the lower left inset is the simple theory for impurity scattering limited
mobility.
(3)
At Ti = 127 °C = 400 K, µe ≈ 450 cm2 V-1 s-1 (from Figure 5Q3-2). The semiconductor is still ntype (check that Nd >> ni at 400 K), then
σ = eNdµe ≈ (1.602 × 10-19 C)(1017 cm-3)(450 cm2 V-1 s-1) = 7.21 Ω-1 cm-1
b
The sample is further doped with Na = 9 × 1016 cm-3 = 0.9 × 1017 cm-3 acceptors. Due to
compensation, the net effect is still an n-type semiconductor but with an electron concentration given
by,
n = Nd − Na = 1017 cm-3 − 0.9 × 1017 cm-3 = 1 × 1016 cm-3 (>> ni)
We note that the electron scattering now occurs from Na + Nd (1.9 × 1017 cm-3) number of
ionized centers so that µe ≈ 700 cm2 V-1 s-1 (Figure 5Q3-1).
(1)
(2)
σ = eNdµe ≈ (1.602 × 10-19 C)(1016 cm-3)(700 cm2 V-1 s-1) = 1.12 Ω-1 cm-1
Using Eqn. (3) with n = Nd − Na we have
Nd − N a
 E ′ − E Fi 
= exp Fn
 kT 
ni
so that
∆E′Φ = EFn′ − EFi = (0.02586 eV)ln[(1016 cm-3) / (1.45 × 1010 cm-3)]
∴
∆E′Φ = 0.348 eV above EFi
The Fermi level from (a) and (b) has shifted “down” by an amount 0.059 eV. Since the
energy is still above the Fermi level, this an n-type Si.
5.5
Homework 5 Solution
EE145 Spring 2002
Prof. Ali Shakouri
5.4 Temperature dependence of conductivity
An n-type Si sample has been doped with 1015 phosphorus atoms cm-3. The donor energy level for P
in Si is 0.045 eV below the conduction band edge energy.
a. Calculate the room temperature conductivity of the sample.
b. Estimate the temperature above which the sample behaves as if intrinsic.
c. Estimate to within 20% the lowest temperature above which all the donors are ionized.
d. Sketch schematically the dependence of the electron concentration in the conduction band on the
temperature as log(n) versus 1/T, and mark the various important regions and critical temperatures.
For each region draw an energy band diagram that clearly shows from where the electrons are
excited into the conduction band.
e. Sketch schematically the dependence of the conductivity on the temperature as log(σ) versus 1/T
and mark the various critical temperatures and other relevant information.
Solution
600°C 400°C 200°C
27°C 0°C
1018
2.4×1013 cm-3
1015
Ge
1012
1.45×1010 cm-3
109
Si
106
2.1×106 cm-3
GaAs
103
1
1.5
2
2.5
3
1000/T (1/K)
3.5
4
Figure 5Q4-1 The temperature dependence of the intrinsic concentration.
a
The conductivity at room temperature T = 300 K is (µe = 1350 × 10-4 m2 V-1 s-1 can be found in
Table 5.1 in the textbook):
σ = eNdµe
∴
σ = (1.602 × 10-19 C)(1 × 1021 m-3)(1350 × 10-4 m2 V-1 s-1) = 21.6 Ω-1 m-1
b
At T = Ti, the intrinsic concentration ni = Nd = 1 × 1015 cm-3. From Figure 5Q4-1, the graph of
ni(T) vs. 1/T, we have:
5.6
Homework 5 Solution
EE145 Spring 2002
Prof. Ali Shakouri
1000 / Ti = 1.9 K-1
∴
Ti = 1000 / (1.9 K-1) = 526 K or 253 °C
c
The ionization region ends at T = Ts when all donors have been ionized, i.e. when n = Nd. From
Example 5.7, at T = Ts:
1
 −∆E 
2
1

n = Nd =  Nc Nd  exp 
2
 2 kTs 
∴
∴
Ts =
Ts =
−∆E

2k ln 

1
2
Nd 

N c Nd 
=
−∆E
 2N 
d 
2k ln 
 Nc 
∆E
 N 
k ln c 
 2Nd 
Take Nc = 2.8 × 1019 cm-3 at 300 K from Table 5.1 (in the textbook), and the difference between
the donor energy level and the conduction band energy is ∆E = 0.045 eV. Therefore our first
approximation to Ts is:
Ts =
(0.045 eV)(1.602 × 10 −19 J/eV )
∆E
=
= 54.68 K
19
−3
 N c 

2.8
×
10

cm
(
)

k ln
1.381× 10 −23 J/K)ln 

15
−3
 2Nd  (
 2(10 cm ) 
Find the new Nc at this temperature, Nc′:
3
3
 T 2
 54.68 K  2
= 2.179 × 1018 cm-3
Nc′ = Nc  s  = (2.8 × 1019 cm−3 )

300 K
300
Find a better approximation for Ts by using this new Nc′:
Ts′ =
(0.045 eV)(1.602 × 10 −19 J/eV )
∆E
=
= 74.64 K
18
−3
 N c′ 

2.179
×
10

cm
(
)

k ln
1.381× 10 −23 J/K)ln 

 2Nd  (
2(1015 cm −3 ) 

3
∴
3
 T′ 2
 74.64 K  2
Nc′′ = Nc s  = (2.8 × 1019 cm−3 )
= 3.475 × 1018 cm-3
 300 K 
 300 
A better approximation to Ts is:
Ts′′=
(0.045 eV )(1.602 × 10 −19 J/eV )
∆E
=
= 69.97 K
18
−3
 N c′′ 


3.475
×
10
cm
(
)

k ln

1.381× 10 −23 J/K)ln 
 2Nd  (
2(1015 cm−3 ) 

3
∴
3
 T ′′ 2
 69.97 K  2
Nc′′′= Nc s  = (2.8 × 1019 cm−3 )
= 3.154 × 1018 cm-3
 300 K 
 300 
5.7
Homework 5 Solution
EE145 Spring 2002
∴
Ts′′′=
Prof. Ali Shakouri
(0.045 eV)(1.602 × 10 −19 J/eV )
∆E
=
= 70.89 K
18
−3
 N c′′′

3.154
×
10

cm
(
)

k ln
1.381× 10 −23 J/K)ln 

 2Nd  (
2(1015 cm −3 ) 

We can see that the change in Ts is very small, and for all practical purposes we can consider the
calculation as converged. Therefore Ts = 70.9 K = −202.1 °C.
d and e See Figures 5Q4-2 and 5Q4-3.
ln(n)
Intrinsic
slope = –E g /2k
Extrinsic
ln(N d )
Ts
Ionization
slope = –∆E/2k
Ti
ni(T)
1/T
Figure 5Q4-2 The temperature dependence of the electron concentration in an n-type semiconductor.
log(n )
log(σ )
Semiconductor
INTRINSIC
Metal
T
EXTRINSIC
Lattice
scattering
log(µ )
IONIZATION
µ∝T
–3/2
µ∝T 3/2
Impurity
scattering
High Temperature
1/T
Low Temperature
Figure 5Q4-3 Schematic illustration of the temperature dependence of electrical
conductivity for a doped (n-type) semiconductor.
5.5 GaAs
Ga has a valency of III and As has V. When Ga and As atoms are brought together to form the GaAs
crystal, as depicted in Figure 5Q5-1, the 3 valence electrons in each Ga and the 5 valence electrons in
each As are all shared to form four covalent bonds per atom. In the GaAs crystal with some 1023 or so
equal numbers of Ga and As atoms, we have an average of four valence electrons per atom, whether
Ga or As, so we would expect the bonding to be similar to that in the Si crystal: four bonds per atom.
The crystal structure, however, is not that of diamond but rather that of zinc blende (Chapter 1 of the
textbook).
5.8
Homework 5 Solution
EE145 Spring 2002
Prof. Ali Shakouri
a. What is the average number of valence electrons per atom for a pair of Ga and As atoms and in the
GaAs crystal?
b. What will happen if Se or Te, from Group VI, are substituted for an As atom in the GaAs crystal?
c. What will happen if Zn or Cd, from Group II, are substituted for a Ga atom in the GaAs crystal?
d. What will happen if Si, from Group IV, is substituted for an As atom in the GaAs crystal?
e. What will happen if Si, from Group IV, is substituted for a Ga atom in the GaAs crystal? What do
you think amphoteric dopant means?
f. Based on the above discussion ,what do you think the crystal structures of the III-V compound
semiconductors AlAs, GaP, InAs, InP, and InSb will be?
Ga
As
Ga atom (Valency III)
As atom (Valency V)
Ga
As
Ga
As
As
Ga
As
Ga
Ga
As
Ga
As
As
Ga
As
Ga
Figure 5Q5-1 The GaAs crystal structure in two dimensions. Average number
of valence electrons per atom is four. Each Ga atom covalently bonds with
four neighboring As atoms and vice versa.
Solution
As
As atom (Valency V)
ψhyb orbitals
Valence
electron
As ion core (+5e)
Ga
Ga atom (Valency III)
ψhyb orbitals
Valence
electron
Ga ion core (+3e)
Ga
As
Ga
As
As
Ga
As
Ga
Ga
As
Ga
As
As
Ga
As
Ga
Explanation of bonding in GaAs: The one s and three p orbitals hybridize to form 4 ψhyb
orbitals. In As there are 5 valence electrons. One ψhyb has two paired electrons and 3 ψhyb have 1
electron each as shown. In Ga there are 3 electrons so one ψhyb is empty. This empty ψhyb of Ga can
overlap the full ψhyb of As. The overlapped orbital, the bonding orbital, then has two paired electrons.
This is a bond between Ga and As even though the electrons come from As (this type of bonding is
called dative bonding). It is a bond because the electrons in the overlapped orbital are shared by both
As and Ga. The other 3 ψhyb of As can overlap 3 ψhyb of neighboring Ga to form "normal bonds".
5.9
EE145 Spring 2002
Homework 5 Solution
Prof. Ali Shakouri
Repeating this in three dimensions generates the GaAs crystal where each atom bonds to four
neighboring atoms as shown. Because all the bonding orbitals are full, the valence band formed from
these orbitals is also full. The crystal structure is reminiscent of that of Si. GaAs is a semiconductor.
a
The average number of valence electrons is 4 electrons per atom.
b
Se or Te replacing As will have one additional electron that cannot be involved in any of the
four bonds. Hence Se and Te will act as a donor.
c
Zn or Cd replacing Ga will have one less electron than the substituted Ga atom. This creates a
hole in a bond. Zn and Cd will act as acceptors.
d
The Si atom has 1 less electron than the As atom and when it substitutes for an As atom in GaAs
there is a "hole" in one of the four bonds. This creates a hole, or the Si atom acts as an acceptor.
e
The Si atom has 1 more electron than the Ga atom and when it substitutes for a Ga atom in GaAs
there is an additional electron that cannot enter any of the four bonds and is therefore donated into the
CB (given sufficiently large temperature). Si substituting for Ga therefore acts as a donor.
f
All these compounds (AlAs, GaP, InAs, InP, InSb) are compounds of III elements and V
elements so they will follow the example of GaAs.
5.10