# 5.1 Bandgap and photodetection Solution ```Tutorial-7
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 &times; 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 &times; 10-34 J s)(3.0 &times; 108 m s-1) / (600 &times; 10-9 m)
∴
Eg = 3.31 &times; 10-19 J or 2.07 eV
b. Area A = 5 &times; 10-2 cm2 and light intensity Ilight = 2 &times; 10-3 W/cm2. The received power is:
P = AIlight = (5 &times; 10-2 cm2)(2 &times; 10-3 W/cm2) = 1.0 &times; 10-4 W
Nph = number of photons arriving per second = P/Eph
∴
Nph = (1.0 &times; 10-4 W) / (3.31 &times; 10-19 J) = 3.02 &times; 1014 Photons s-1
Since each photon contributes one electron-hole pair (EHP), the number of EHPs is then:
NEHP = 3.02 &times; 1014 EHP s-1
c. For GaAs, Eg = 1.42 eV and the corresponding wavelength is
λ = hc/Eg = (6.626 &times; 10-34 J s)(3.0 &times; 108 m s-1) / (1.42 eV &times; 1.602 &times; 10-19
J/eV)
∴
λ = 8.74 &times; 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. From Table 5.1, for Si, Eg = 1.10 eV and the corresponding cut-off wavelength is,
λg = hc/Eg = (6.626 &times; 10-34 J s)(3.0 &times; 108 m s-1) / (1.1 eV &times; 1.602 &times; 10-19
J/eV)
∴
λg = 1.13 &times; 10-6 m or 1130 nm
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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 Intrinsic Ge Using the values of the density of states effective masses me* and
mh* in Table 5.1, calculate the intrinsic concentration in Ge. What is ni if you use Nc and
Nv from Table 5.1? Calculate the intrinsic resistivity of Ge at 300 K.
Solution
From Table 5.1, we get
me∗ = 0.56me
and
mh∗ = 0.4me
Now,
⎛ 2πme∗ kT ⎞
⎟⎟
N c = 2⎜⎜
2
⎠
⎝ h
3/ 2
⎡ 2π (0.56 &times; 9.1&times; 10 −31 kg)(1.38 &times; 10 − 23 J K −1 )(300 K ) ⎤
= 2⎢
⎥
(6.626 &times;10 −34 J s) 2
⎣
⎦
3/ 2
= 1.05&times;1025 m-3 or 1.05 &times;1019 cm-3
⎛ 2πmv∗kT ⎞
⎟⎟
N v = 2⎜⎜
2
h
⎝
⎠
3/ 2
⎡ 2π (0.4 &times; 9.1&times; 10 −31 kg )(1.38 &times;10 − 23 J K −1 )(300 K ) ⎤
= 2⎢
⎥
(6.626 &times; 10 −34 J s) 2
⎣
⎦
3/ 2
= 6.35&times;1024 m-3 or 6.35 &times;1018 cm-3
The intrinsic concentration is
⎛ Eg ⎞
⎟⎟
ni = ( N c N v )1/ 2 exp⎜⎜ −
2
kT
⎠
⎝
∴
[
ni = (1.049 &times; 1019 cm −3 )(6.33 &times; 1018 cm −3 )
]
1/ 2
⎞
⎛
0.66 eV
⎟⎟
exp⎜⎜ −
−5
−1
⎝ 2(8.62 &times;10 eV K )(300 K ) ⎠
= 2.33 &times;1013 cm-3 ≈ 2.3 &times;1013 cm-3
From Table 5.1, we get Nc = 1.04&times;1019 cm-3 and Nv = 6.0&times;1018 cm-3
∴
[
ni = (1.04 &times; 1019 cm −3 )(6.0 &times;1018 cm −3 )
]
1/ 2
⎛
⎞
0.66 eV
⎟⎟
exp⎜⎜ −
−5
−1
⎝ 2(8.62 &times; 10 eV K )(300 K ) ⎠
= 2.26 &times; 1013 cm-3 ≈ 2.3 &times;1013 cm-3
The intrinsic conductivity is σ = en&micro; e + ep&micro; h = eni ( &micro; e + &micro; h )
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∴
cm-1
σ = (1.6 &times; 10 −19 C)(2.34 &times; 1013 cm −3 )(3900 + 1900) cm 2 V −1s −1 = 0.022 Ω-1
And the intrinsic resistivity is ρ = 1/σ = 45.45 Ω cm
5.3 Fermi level in intrinsic semiconductors Using the values of the density of
states effective masses me* and mh* in Table 5.1, find the position of the Fermi energy in
intrinsic Si, Ge and GaAs with respect to the middle of the bandgap (Eg/2).
Solution
⎛ me∗ ⎞
1
3
E Fi = Ev + E g − kT ln⎜⎜ ∗ ⎟⎟
2
4
⎝ mh ⎠
For Si, me∗ = 1.08 me and mh∗ = 0.6 me
∴
E Fi = Ev +
⎛ 1.08 me ⎞
1
3
⎟⎟
E g − (8.62 &times; 10 −5 eVK −1 )(300 K ) ln⎜⎜
2
4
⎝ 0.6 me ⎠
∴
E Fi = Ev +
1
E g − 0.011eV
2
So intrinsic Fermi level of Si is 0.011 eV below the middle of the bandgap (Eg/2).
For Ge, me∗ = 0.56 me and mh∗ = 0.4 me
∴
E Fi = Ev +
⎛ .56 me ⎞
1
3
⎟⎟
E g − (8.62 &times; 10 −5 eVK −1 )(300 K ) ln⎜⎜
2
4
⎝ 0.4 me ⎠
∴
E Fi = Ev +
1
E g − 0.0065 eV
2
So intrinsic Fermi level of Ge is 0.0065 eV below the middle of the bandgap (Eg/2).
For GaAs, me∗ = 0.067 me and mh∗ = 0.5 me
∴
E Fi = Ev +
⎛ 0.067 me ⎞
1
3
⎟⎟
E g − (8.62 &times; 10 −5 eV K −1 )(300 K ) ln⎜⎜
2
4
⎝ 0.5 me ⎠
∴
E Fi = Ev +
1
E g + 0.039 eV
2
So intrinsic Fermi level of GaAs is 0.039 eV above the middle of the bandgap (Eg/2).
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5.4 Extrinsic Si A Si crystal has been doped with P. The donor concentration is 1015
cm-3. Find the conductivity, and resistivity of the crystal.
Solution
Nd = 1015 cm-3
Therefore the conductivity is σ = eN d &micro; e = (1.6 &times;10 −19 C)(1015 cm −3 )(1350 cm 2 V −1s −1 ) =
0.216 Ω-1cm-1
And the resistivity is ρ = 1/σ = 4.63 Ω-1cm-1
5.5 Extrinsic Si Find the concentration of acceptors required for a p-type Si crystal to
have a resistivity of 1 Ω cm.
Solution
The resistivity, ρ =
∴
1
eN a &micro; h
Na =
1
e&micro; h ρ
=
(1.6 &times; 10
−19
1
= 1.38&times;1016 cm-3
2 −1 −1
C)(450 cm V s )(1 Ω cm)
It is assumed that the doping concentration does not affect drifty mobility. At this
concentration, from Figure 5.19, the hole drift mobility is approximately 350 cm2 V-1 s-1
rather than 450 cm2 V-1 s-1. Using 350 cm2 V-1 s-1 we find,
Na =
1
e&micro; h ρ
=
(1.6 &times; 10
−19
1
= 1.78&times;1016 cm-3
C)(350 cm 2 V −1s −1 )(1Ω cm)
Author's Note: Question 5.7 provides a better example for calculating the required dopant
concentration for a given resistivity.
5.6 Minimum conductivity
a. Consider the conductivity of a semiconductor, σ = en&micro;e + ep&micro;h. 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
&micro;e
&micro;h
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and the corresponding minimum conductivity (maximum resistivity) is
σ min = 2eni &micro; e &micro; 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 &gt; 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 = en&micro;e + ep&micro;h
en &micro;
σ = i e + &micro; h ep
p
2
∴
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:
en &micro;
dσ
= − i 2 e + &micro;he
dp
p
2
en &micro;
− i 2 e + &micro;he = 0
pm
2
∴
Isolate pm and simplify,
pm = ni
&micro;e
&micro;h
Substituting this expression back into the equation for conductivity will give the
minimum conductivity:
σ min =
∴
eni &micro; e
&micro;e
eni &micro; e
+ &micro; h epm =
+ &micro; h eni
pm
&micro;h
ni &micro; e &micro; h
2
σ min = eni &micro;e
2
&micro;h
+ eni &micro; e &micro; h = eni &micro; e &micro; h + eni &micro; e &micro; h
&micro;e
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∴
σ min = 2eni &micro; e &micro; h
c. From Table 5.1, for Si: &micro;e = 1350 cm2 V-1 s-1, &micro;h = 450 cm2 V-1 s-1 and ni = 1.0 &times; 1010
cm-3. Substituting into the equations for pm and σmin:
pm = ni
&micro;e
1350 cm 2 V −1 s −1
= (1.0 &times; 1010 cm −3 )
= 1.73 &times; 1010 cm-3
2
−1 −1
&micro;h
450 cm V s
σ min = 2eni &micro; e &micro; h
∴
σ min = 2(1.602 &times;10 −19 C )(1.0 &times; 1010 cm −3 ) (1350 cm 2 V −1 s −1 )(450 cm 2 V −1 s −1 )
∴
σmin = 2.5 &times; 10-6 Ω-1 cm-1
The corresponding maximum resistivity is:
ρmax = 1 / σmin = 4 &times; 105 Ω cm
The intrinsic value corresponding to pm is simply ni (= 1.0 &times; 1010 cm-3). Comparing it to
pm:
pm 1.73 &times;1010 cm −3
=
= 1.73
ni
1.0 &times; 1010 cm −3
The intrinsic conductivity is:
σint = eni(&micro;e + &micro;h)
∴
∴
σint = (1.602 &times; 10-19 C)(1.0 &times; 1010 cm-3)(1350 cm2 V-1 s-1 + 450 cm2 V-1 s-1)
σint = 2.88 &times; 10-6 Ω-1 cm-1
Comparing this value to the minimum conductivity:
σ min 2.5 &times;10 −6 Ω −1 cm −1
=
= 0.87
σ int 2.88 &times;10 −6 Ω -1 cm -1
Sufficient p-type doping that increases the hole concentration by 73% decreases the
conductivity by 15% to its minimum value.
5.9 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 &deg;C.
2. Where is the Fermi level in this sample at 27 &deg;C with respect to the Fermi level
(EFi) in intrinsic Si?
3. Calculate the conductivity of the sample at 127 &deg;C.
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b. The above n-type Si sample is further doped with 9 &times; 1016 boron atoms (p-type
dopant) per centimeter cubed.
1. Calculate the conductivity of the sample at 27 &deg;C.
2. Where is the Fermi level in this sample with respect to the Fermi level in the
sample in (a) at 27 &deg;C? Is this an n-type or p-type Si?
Solution
a. Given temperature T = 27 &deg;C = 300 K, concentration of donors Nd = 1017 cm-3, and
drift mobility &micro;e ≈ 800 cm2 V-1 s-1 (from Figure 5.19). At room temperature the electron
concentration n = Nd &gt;&gt; p (hole concentration).
Figure 5.19: 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 = eNd&micro;e ≈ (1.602 &times; 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 − E Fi ⎞
= exp⎜ Fn
⎟
ni
⎝ kT
⎠
(3)
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∴
⎛N
ln⎜⎜ d
⎝ ni
∴
∆EF = EFn − EFi = kT ln(Nd/ni)
⎞ E Fn − E Fi
⎟⎟ =
kT
⎠
Substituting we find (ni = 1.0 &times; 1010 cm-3 from Table 5.1),
∆EF = (8.617 &times; 10-5 eV/K)(300 K)ln[(1017 cm-3)/ (1.0 &times; 1010 cm-3)]
∴
∆EF = 0.416 eV above EFi
Figure 5.18: 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 &deg;C = 400 K, &micro;e ≈ 450 cm2 V-1 s-1 (from Figure 5.18). The semiconductor
is still n-type (check that Nd &gt;&gt; ni at 400 K), then
σ = eNd&micro;e ≈ (1.602 &times; 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 &times; 1016 cm-3 = 0.9 &times; 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 &times; 1017 cm-3 = 1 &times; 1016 cm-3 (&gt;&gt; ni)
We note that the electron scattering now occurs from Na + Nd (1.9 &times; 1017 cm-3) number of
ionized centers so that &micro;e ≈ 700 cm2 V-1 s-1 (Figure 5.18).
(1)
σ = eNd&micro;e ≈ (1.602 &times; 10-19 C)(1016 cm-3)(700 cm2 V-1 s-1) = 1.12 Ω-1 cm-1
(4)
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(2) Using Eqn. (3) with n = Nd − Na we have,
n Nd − Na
⎛ E ′ − EFi ⎞
=
= exp⎜ Fn
⎟
ni
ni
⎝ kT
⎠
so that
∆E′F = E′Fn− EFi = (0.02586 eV)ln[(1016 cm-3) / (1.0 &times; 1010 cm-3)]
∴
∆E′F = 0.357 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 is an n-type Si.
5.10 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 percent 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
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Figure 5.16: The temperature dependence of the intrinsic concentration.
a. The conductivity at room temperature T = 300 K is (&micro;e = 1350 &times; 10-4 m2 V-1 s-1 can be
found in Table 5.1):
σ = eNd&micro;e
∴
σ = (1.602 &times; 10-19 C)(1 &times; 1021 m-3)(1350 &times; 10-4 m2 V-1 s-1) = 21.6 Ω-1 m-1
b. At T = Ti, the intrinsic concentration ni = Nd = 1 &times; 1015 cm-3. From Figure 5.16, the
graph of ni(T) versus 1/T, we have:
1000 / Ti = 1.85 K-1
∴
Ti = 1000 / (1.85 K-1) = 540 K or 267 &deg;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 ⎞
⎛1
⎞2
⎟⎟
n = N d = ⎜ N c N d ⎟ exp⎜⎜
⎝2
⎠
⎝ 2kTs ⎠
∴
Ts =
− ∆E
⎛
2k ln⎜
⎜
⎝
Nd
1
2 Nc Nd
⎞
⎟
⎟
⎠
=
− ∆E
⎛ 2Nd
2k ln⎜⎜
⎝ Nc
⎞
⎟
⎟
⎠
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∴
Ts =
∆E
⎛ N
k ln⎜⎜ c
⎝ 2Nd
⎞
⎟⎟
⎠
Take Nc = 2.8 &times; 1019 cm-3 at 300 K from Table 5.1, 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 =
∆E
⎛ N
k ln⎜⎜ c
⎝ 2Nd
⎞
⎟⎟
⎠
=
(0.045 eV )(1.602 &times;10−19 J/eV )
(1.381&times;10
− 23
) ( (
⎛ 2.8 &times; 1019 cm −3
J/K ln⎜⎜
−3
15
⎝ 2 10 cm
)
) ⎞⎟
= 54.68 K
⎟
⎠
Find the new Nc at this temperature, Nc′:
3
(
)
3
⎛ T ⎞2
⎛ 54.68 K ⎞ 2
18
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N c′ = N c ⎜ s ⎟ = 2.8 &times; 1019 cm −3 ⎜
⎟ = 2.179 &times; 10 cm
⎝ 300 ⎠
⎝ 300 K ⎠
Find a better approximation for Ts by using this new Nc′:
Ts′ =
∆E
⎛ N′
k ln⎜⎜ c
⎝ 2Nd
⎞
⎟⎟
⎠
=
3
∴
(0.045 eV )(1.602 &times;10−19 J/eV )
(1.381&times;10
− 23
⎛ (2.179 &times; 1018 cm −3 ) ⎞
⎟⎟
J/K )ln⎜⎜
15
−3
(
)
2
10
cm
⎝
⎠
= 74.64 K
3
⎛ T′ ⎞2
⎛ 74.64 K ⎞ 2
18
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N c′′ = N c ⎜ s ⎟ = 2.8 &times; 1019 cm −3 ⎜
⎟ = 3.475 &times; 10 cm
300
300
K
⎝
⎠
⎝
⎠
(
A better approximation to Ts is:
Ts′′ =
∆E
⎛ N ′′
k ln⎜⎜ c
⎝ 2Nd
⎞
⎟⎟
⎠
=
3
)
(0.045 eV )(1.602 &times;10−19 J/eV )
(1.381&times;10
− 23
⎛ (3.475 &times;1018 cm −3 ) ⎞
⎟⎟
J/K )ln⎜⎜
15
−3
(
)
2
10
cm
⎝
⎠
= 69.97 K
3
∴
⎛ T ′′ ⎞ 2
⎛ 69.97 K ⎞ 2
18
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N c′′′ = N c ⎜ s ⎟ = 2.8 &times; 1019 cm −3 ⎜
⎟ = 3.154 &times; 10 cm
⎝ 300 ⎠
⎝ 300 K ⎠
∴
Ts′′′=
∆E
⎛ N ′′′
k ln⎜⎜ c
⎝ 2Nd
⎞
⎟⎟
⎠
(
=
)
(0.045 eV )(1.602 &times;10−19 J/eV )
(1.381&times;10
− 23
⎛ (3.154 &times; 1018 cm −3 ) ⎞
⎟⎟
J/K )ln⎜⎜
15
−3
(
)
2
10
cm
⎝
⎠
= 70.89 K
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 &deg;C.
d. and e.
See Figures 5.15 and 5.20
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Figure 5.15: The temperature dependence of the electron concentration in an n-type
semiconductor.
Figure 5.20: Schematic illustration of the temperature dependence of electrical
conductivity for a doped (n-type) semiconductor.
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