HW9Sol.pdf

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Chapter 17
17.1 We wish to use a photodiode as a detector for a signal of 9000 Å wavelength.
Which would be the best choice of material for the photodiode, a semiconductor of
bandgap = 0.5 eV, bandgap = 2 eV, or bandgap = 1 eV? Why? (Assume all three
are direct gap and are equivalent in impurity content, etc.)
Solution.
λ1 = 1.24
= 2.48 µm
0.5
λ2 = 1.24
= 6200 Å
2
λ3 = 1.24
= 1.24 µm
1
λ2 cutoff at too short a wavelength
λ1 cutoff at to long a wavelength
(increases bandwidth for noise – reduces S/N).
17.2 To improve the signal-to-noise ratio of the diode in Problem 17.1, we wish to
use a semiconductor low-pass filte which has the following absorption properties at
room temperature:
for 9000 Å radiation α = 0.2cm−1 for 7000 Å radiation α = 103 cm−1 .
How thick must the filte be to attenuate 7000 Å background noise by a factor of
104 ?
By what factor is the signal (at 9000 Å) attenuated by a filte of this thickness?
Neglect reflectio at the surfaces.
Solution.
I
e −αz
I0
1
10 4 = e−103 z
I
I0
z = 9.21 × 10 −3 cm
= e −2×10−1 ×9.21×10−3 = 0.999
17.3 If the minimum useful photocurrent of the diode in Problem 17.1 is 1 µA
(peak pulse value), what is the minimum signal light intensity (peak pulse value)
which must fall on the detector? Assume an internal yield or quantum efficien y
ηq = 0.8 and sensitive area = 10 mm 2 .

Solution.
10−6 coul/sec × 1.25 photons/carrier
1.6 × 10−19 = 7.8 × 10
coul/carrier
12 photons/sec .
7.8 × 1012 photons/sec × 1.38 eV/photon × 1.60 × 10−19 joules/eV
10−1 cm2 = 1.72 × 10 −5 watts/cm 2
17.4 Below is a list of semiconductor materials and their bandgaps.
Eg[eV]
Si 1.1
GaAs 1.4
GaSb 0.81
GaP 2.3
InAs 0.36
a) Based solely on bandgap energy, which of these materials could possibly be
used to make a detector for the light from a GaAs laser?
b) Which of the following ternary compounds could possibly be used to make
a detector for the light from a GaAs laser?
GaAs(1−x)Sbx GaAs(1−x)Px Ga(1−x)InxAs
c) If a reverse biased Si photodiode with a quantum efficien y of η = 80% and
an area of 1 cm 2 is uniformly illuminated with the light from a GaAs laser to an
intensity of 10 mW/cm 2 what is the photocurrent which f ows?
Solution.
a) For eff cient detection you need Eg < Ephoton and for a GaAs laser Ephot � 1.4eV
∴Si, GaSb, InAs will work
GaP will not work
b) GaAs1−xSbx and Ga1−xInxAs would work because GaSb and InAs have bandgaps
less than that of GaAs. GaAs1−xPx would not work because the bandgap of GaP
exceeds that of GaAs.
c) 10 × 10 −3 W/cm 2 × 1cm 2 = 10 −2 W of optical power, each photon has
1.24/0.9 × 1.6 × 10 −19 = 2.20 × 10 −19 joules of optical power ∴ the number of
photons striking the detector is
10−2 2.2 × 1019 = 4.55 × 1016 photons/sec
at 80% quantum efficien y the current is
4.55 × 10 16 × 0.8 × 1.6 × 10 −19 = 5.82 mA
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17.5 a) Determine the maximum value of the energy gap which a semiconductor,
used as a photodetector, can have if it is to be sensitive to light of wavelength
λ0 = 0.600 µm.
b) A photodetector whose area is 5 × 10 −6 m 2 is irradiated with light whose
wavelength is λ0 = 0.600 µm and intensity is 20 W/m 2 . Assuming each photon
generates one electron-hole pair, calculated the number of pairs generated per second.
c) By what factor does the answer to (b) change if the intensity is reduced by
a factor of 1/2?
d) By what factor does the answer to (b) change if the wavelength is reduced by
a factor of 1/2?
Solution.
a) E = hν = hc
λ
= 6.6 × 10−34 × 3 × 108 6000 × 10−10 = 3.3 × 10 −19 joule
= 3.3 × 10−19
= 2.06 eV
1.6 × 10−19 b) 20 Watts
m 2 × 5 × 10−6 m 2 = 10 −4 watts
pairs generated
sec
=
10 −4 joule/sec
3.3 × 10 −19 joules/pair = 3 × 1014 pairs/sec
c) Intensity = 20 Watts/cm2 × 1
2 = 10 Watts/cm2
pairs/sec reduced by factor of 1
2
6000 × 10−10
d) If λ = = 3000 × 10
2
−10
Energy per photon = E = hc
λ = 6.6 × 10−19 joule
10 −4 joules/sec
6.6 × 10 −19 joule
also a factor of 1
2
17.6 We wish to design a waveguide photodiode in GaAs, with the geometry as
shown in Fig. 17.5, for operation at λ0 = 0.900 µm wavelength.
a) If the photodiode has a waveguide thickness and depletion width W = 3 µm
at an applied reverse bias voltage of 40.5 V, what is the magnitude of the change in
effective bandgap due to the electric f eld? (Assume m ∗ = 0.067m0).
b) What length L is required to produce a quantum efficien y of 0.99? (Assume
that scattering loss and free carrier absorption are negligible).

Solution.
a) The fiel in the depletion layer is
E = 40.5
3 × 10−4 V/cm = 1.35 × 105 V/cm or 1.35 × 10 7 From (9.19)
V/m
∆E = 3
2 (m∗ ) −1/3 (qEh) 2/3
= 3
2 (9.1 × 10−31 × 0.067 kg) −1/3 [(1.6 × 10 −19 × 1.35 × 10 7 V/m)
× (1.05 × 10 −34 Js)] 2/3
= 3
2 · (3.72 × 10−31 )m∗ kg4/3 /s2 (3.94 × 10−11 ) kg1/3 = 1.42 × 10 −20 J
= 1.42 × 10−20
eV
1.6 × 10−19 = 0.089 eV
b) From Fig.9.2, α = 10 4 cm −1
ηq = 1 − e −αL
0.99 = 1 − e −104 ·L
− 10 4 · L = ln 0.01 L = 4.6. × 10 −4 cm
L = 4.6. µm
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80
Chapter 18
18.1 Explain the difference between a multiple quantum well structure and a superlattice.
Solution. See text Sect. 16.1.
18.2 For a deep quantum well of GaAs with a thickness Lz = 70 Å, L x = L y =
100 µm, calculate the magnitude of the steps in the conduction band cumulative
density of states function and the energies at which they occur relative to the bottom
of the well for the n = 1, 2 and 3 levels. (Assume that the well has infinitel high
sidewalls.)
Solution.
∆ϱ(E) me 4meπ
=
πh2 h2 = 4(0.08)(9.1 × 10−31 )π
(6.625 × 10−34 ) 2
= 2.08 × 10 36
K y = Kx = nπ
10−4 n = 1, 2, 3
En = h2
� �2 nπ
+
2me Lz
h2
(K
2me
2 x + K 2 y ) n = 1, 2, 3
En = (1.054 × 10−34 ) 2
2(0.08)9.1 × 10−31 �
nπ
7 × 10−9 �2 + (1.504 × 10−34 ) 2
2(0.08)9.1 × 10−31 �� nπ
10−4 �2 �
nπ
+
10−4 � �
2
En = 7.36 × 10 −38
�
nπ
7 × 10−9 �2 + 7.63 × 10 −38 �
nπ
(2)
10−4 �2 for n = 1
E1 = 7.63 × 10 −38 (2.01 × 10 17 ) + 7.63 × 10 −38 (2)(9.87 × 10 8 )
= 1.53 × 10 −20 + 1.51 × 10 −28 = 1.53 × 10 −20 Joule = 0.0956 eV
E2 = 1.53 × 10 −20 (4) + 1.51 × 10 −28 (9) = 6.12 × 10 −20 =0.383 eV
E3 = 1.53 × 10 −20 (9) + 1.51 × 10 −28 (9) = 1.38 × 10 −19 =0.860 eV
18.3 If the quantum well of Problem 18.2 were incorporated into a diode laser,
what would be the emission wavelength for electron transitions between the n = 1
level in the conduction band and the n = 1 level in the valence band? (Assume 300 K
operation.)
Solution. From problem 16.2 the energy of a conduction band electron in n = 1
level
E1 = 0.0956 eV

For a hole in the valence band n = 1 level, the equation is the same except that
me = 0.08 m0 is replaced by mh = 0.5 m0, thus for the hole
0.0956 eV
E1 = 0.0956 0.08
0.5
0.0153 eV
The photon energy is
= 0.0153 eV
Eg = 1.38 eV
Eph = 0.0956 + 1.38 + 0.0152 = 1.49 eV
The emission wavelength is
λ = 1.24
Eph
= 1.24
= 0.832 µm
1.49
18.4 a) Explain how a single p-i-n MQW diode can function as a self-electro-opticeffect
device (SEED).
b) Why are SEEDs envisioned as being possible digital logic elements in
Solution. See text Sect. 16.5.
18.5 A single-quantum well of GaAs with a thickness of 60 Å is incorporated into
a GaAlAs heterojunction laser. The laser has a stripe geometry, being 200 µm long
and 10 µm wide.
a.) What is the emission wavelength if the energy bandgap is Eg = 1.52 eV in
the emitting region and electron transitions occur between the n = 1 level in the
conduction band and the n = 1 level in the valence band?
b). If the gain coefficien of the laser in part (a) is g = 80 cm−1 , what would be
the gain coefficien of a multiple quantum well laser with 20 identical wells?
Solution.
a.) En = �2
� �2 nπ
+
2me Lz
�2
(k
2me
2 x + k2y )
where kx = nπ/L x and ky = nπ/L y
taking n = 1 and realizing that k2 x and k2y are negligible since L x, L y ≫ Lz
E1 = �2
� �2 π
(1.055 × 10
=
2me Lz
−34 ) 2π 2
2 × 0.08 × 9.1 × 10−31 (6 × 10−9 = 2.123 × 10−2
) 2
2.123 × 10−20
E1 = = 0.13 eV
1.60 × 10−19 E1
Ec
Ev
E1
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for holes m ∗ changes from 0.08 me to 0.5 me then
E1 = 0.13 × 0.08
0.5
The photon energy is
= 0.02 eV
Ephot = �w = Eg + E1C + E1V = 1.52 + 0.13 + 0.02 = 1.67 eV
The emission wavelength is λ0 = 1.24/1.67 = 0.743 µm
b). 20 × 80 = 1600 cm −1

Chapter 19
19.1 If a silicon rod of 1 mm diameter is exposed to a tensional force by supporting
a 3 g weight,
a. What axial stress does the rod experience?
b. What is the axial strain?
c. What is the longitudinal strain?
Solution.
a. G = F/π(θ/2) 2
(1lb/in 2 = 703.1kg/m 2 )
= 3 × 10−3 kg
π(10−3 /2) m2 = 3819 kg/m2 = 5.43 lb/in 2
b. G = εE
εE = G
E =
=−1.97 × 10 −7
↑ tension
c. ν =−εl/εa
5.43 lb/in 2
190 GPa × 1.45 × 10 5� lb
in 2 /GPa �
εl =−ν × εa = (−0.28 ×−1.97 × 10 −7 )
= 5.5 × 10 −8
19.2 A square membrane of silicon, 7 mm on a side and 150 µm thick, is exposed
uniformly over its surface to a pressure of 1 × 10 4 kg/m 2 .
a. What is the deflectio at the center point of the membrane?
b. What is the maximum longitudinal stress?
c. What is the maximum transverse stress?
d. What is the resonant frequency of the membrane?
Solution.
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84
a. D = EA 3 /[12(1 − ν 2 )]
= 190 GPa × (150 × 10−6 ) 3
[12(1 − 0.282 )]
= 6.96 × 10 −10 GPa m 2
wmax = 0.001265 Pa 4 /D
=
= 6.41 × 10−10
(1 − 0.28 2 )
0.001265 kg/m 2 × (7 × 10 −3 ) 4 m 4
6.96 × 10 −10 GPa m 2 × 1.02 × 10 8 (kg/m 2 )/GPa
= 6.41 × 10−10
.9216
= 4.28 × 10 −11 P meters = 4.28 × 10 −5 P µm with P in kg/m 2
take P = 10 4 kg/m 2
wmax = 4.28 × 10 −1 µm = 0.428 µm
b. Gl = 0.3081 P(a/t) 2
= 0.3081 × 10 4 (7/0.15) 2 = 6.71 × 10 6 kg/m 2
c. Gt = νGl = 0.28 × 6.71 × 10 6 = 1.88 × 10 6 kg/m 2
d. F0 = (1.65 t/a 2 )[E/e(1 − ν 2 )] 1/2
× 1.654 × 150 × 10−6
(7 × 10 −3 ) 2
� 190 × 1.02 × 10 8
2.3 × 10 3 (1 − 0.28)
= 5.063 × 9.14 × 10 6 = 4.63 × 10 7 = 46.3MHz
19.3 A silicon cantilever beam, 1.5 mm long, 500 µm wide and 250 µm thick, is
loaded with a uniformly distributed force per unit width of 100 kg/m 2 .
a. What is the deflectio at the point 0.5 mm from the free end?
b. What is the maximum stress?
c. What is the frequency of the fundamental vibrational mode?
Solution.
�1/2

a. W(P, x) = (Px 2 /24 EI)(6L 2 − 4Lx + x 2 )
I = at 3 /12 = 500 × 10−6 (250 × 10−6 ) 3
= 6.51 × 10
12
−16
W(P, 1 × 10 −3 P × 1 × 10
) =
−6
24 × 190 × 1.02 × 10 +8 × 6.51 × 10−16 × � 6(1.5 × 10 −3 ) 2 − 42x − x 2 )
= P × 3.30 × 10 −3 (1.35 × 10 −5 − 4 × 1.5 × 10 −3
× 1 × 10 −3 × 1 × 10 −6 )
= P × 3.30 × 10 −3 (6.5 × 10 −6 )
= P × 2.1 × 10 −8
let P = 100 kg/m
w = 2.1 × 10 −6 m = 2.1 µm
b. Gmax = PL2t 4I = 100 × (1.5 × 10−3 ) 2 (250 × 10−6 )
4 × 6.51 × 10−16 = 2.16 × 10 7 kg/m 2
c. F0 = 0.161(t/L 2 )(E/C) 1/2
� −6 250 × 10
= 0.161
(1.5 × 10−3 ) 2
�� 8
190 × 1.02 × 10
2.3 × 103 = 3.23 × 10 5 = 323 kHz
�1/2
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