固体光电子导论课后习题(南邮)

1.2 A 1.6 eV photon is absorbed by a valence band electron in GaAs. If the bandgap of GaAs is 1.41 eV, calculate the energy of the electron and heavy hole produced by the photon absorption. Condition:
***
0001.6 1.410.0670.450.058e h r eV Eg eV m m m m m m ω===== Solution: By
2
222
***
1
1()22e h r
Eg m m m ωω
ω-=
+=
We can get
*22
*
**22
**()0.1642()0.0252e r c e
e h r v h h m E E Eg eV m m m E E Eg eV m m ω
ωωω⎧-==-=⎪⎪⎨⎪-==-=⎪⎩
1.4 The absorption coefficient near the bandedges of GaAs and Si are 4110cm - and 3110cm - respectively. What is the minimum thickness of a sample in each case which can absorb 90% of the incident light? Condition:
31
41()10()10Si cm GaAs cm αα--==
Solution: By
0min 0
()()(190%)x
I x I e
I x I α-⎧=⎪⎨
=-⎪⎩ We can get
min min min 00min ()0.10.1
1
ln 0.1
x x I x I e I e x ααα
--==⇒=⇒=-
So
3
min 4
min () 2.310() 2.310x Si cm
x GaAs cm
--⎧=⨯⎪⎨=⨯⎪⎩
1.8 In a GaAs sample at 300K, equal concentrations of electrons and holes are
injected. If the carrier density is n=p=1017cm -3, Calculate the electron and hole Fermi levels using the Boltzmann and Joyce-Dixon approximations.
Condition:
173
183300,, 4.710710c v When T K for GaAs N cm N cm --==⨯=⨯
Solution:
Boltzmann approximation:
ln 0.0402ln 0.1105n
c c F B p c F v B N E E k T eV n N E E k T eV
p ⎧-==⎪⎪
⎨⎪-==⎪⎩
Joyce-Dixon approximation:
(ln 0.0383(ln 0.1103n
c
F B c p F v B v n E E k T eV N n E E k T eV N ⎧-=-=⎪⎪
⎨
⎪-=-+=⎪⎩
1.12 The radiative lifetime of GaAs sample is 1.0 ns. The sample has a defect at the midgap with a capture cross-section of 10-15cm 2 . At what defect concentration does the non-radiative lifetime become equal to the radiative lifetime at i) 77K and ii) 300K?(分电子、空穴计算) Condition:
**1520
00.0680.47 1.010n p r nr m m m m ns
cm ττσ-=====
Solution:
(1)For electrons
7*277*2730013(77) 2.2710/2213(300) 4.4810/22th n th B K n th B K
th v K cm s m v k T m v k T v K cm s ⎧⎧==⨯⎪=⎪⎪⎪
⇒⎨
⎨⎪⎪===⨯⎪⎪⎩⎩
So
163
16311(77) 4.4110(77)(77)11(300) 2.2310(300)(300)nr t t th nr th nr t
t th nr th N K cm N v K v K N K cm N v K v K τστστστσ--⎧⎧===⨯⎪⎪⎪⎪
⇒⎨
⎨⎪⎪===⨯⎪⎪⎩⎩
(2)For holes
6
*277*2730013(77)8.6310/2213(300) 1.7010/22th p th B K p th B K
th v K cm s m v k T m v k T v K cm s ⎧⎧=⨯⎪=⎪⎪⎪⇒⎨⎨⎪⎪===⨯⎪⎪⎩⎩
So
173
16311(77) 1.1610(77)(77)11(300) 5.8810(300)(300)nr t t th nr th nr t
t th nr th N K cm N v K v K N K cm N v K v K τστστστσ--⎧⎧===⨯⎪⎪⎪⎪
⇒⎨⎨
⎪⎪===⨯⎪⎪⎩⎩
1.13 Electrons are injected into a p-type silicon sample at 300K. The e-h radiative lifetime is 1μs. The sample also has midgap traps with a cross-section of 10-15cm 2 and a density of 1016cm -3. Calculate the diffusion length for the electrons if the diffusion coefficient is 30 cm 2/s. Condition:
1521632*
01101030/ 1.08r t n n s cm N cm D cm s m m τμσ--===== Solution:
By
*27
13 1.1210/22n th B th m v k T v cm s =⇒=⨯
We get
1
8.93nr t th ns N v τσ
=
=
Then
1
1
1
8.85r
nr
ns ττ
ττ=
+
⇒=
So
45.1510n L m -==⨯
2.1 The bandgap of 1x x Hg Cd - alloy is given by the expression
()0.3 1.9()Eg x x eV =-+
Calculate the composition of an alloy which gives a cutoff wavelength of a) 10m μ b) 5.0m μ. Solution: By
1.24
c hc m Eg Eg
λμ=
=
We get
1.24 1.24
10(10)0.220.3 1.91.24 1.24 5.0( 5.0)0.29
0.3 1.9c c c c m x m Eg x
m x m Eg x
λμλμλμλμ=
==⇒==-+===⇒==-+
2.2 Calculate the cutoff wavelength for a GaAs detector. If the cutoff wavelength is to be decreased to 0.7μm, how much AlAs must be added to a GaAs? Assume that the bandgap of 1x x Ga Al As - is given by
() 1.43 1.25()Eg x x eV =+ Solution: By
1.24
c hc m Eg Eg
λμ=
=
We get
1.24 1.24
0.7(0.7)0.271.43 1.25c c m x m Eg x
λμλμ=
==⇒==+
2.4 An optical power density of 1W/cm 2 is incident on a GaAs Sample. The photon energy is 2.0eV and there is no reflection from the surface. Calculate the excess electron-hole carrier densities at the surface and 0.5μm from the surface. The e-h recombination time is 10-8s. Condition:
241
8(0)1/() 2.1110 2.00
10op P W cm GaAs cm w eV R s ατ--==⨯===
Solution: (1)At the surface
2231()(0)
(0) 6.5910op L GaAs P G cm s w
α--=
=⨯
So
143(0)(0)(0) 6.5910L n x p x G cm δδτ-=====⨯
(2)At 0.5m μ from the surface
()2(0.5)(0)0.35/GaAs x
op op P x m P e
W cm αμ-=== 2231()(0.5)
(0.5) 2.3110op L GaAs P x m G x m cm s w
αμμ--===
=⨯
So
143(0.5)(0.5)(0.5) 2.3110L n x m p x m G x m cm δμδμμτ-======⨯
2.5 Assume that all the photons in an optical beam produce an electron-hole pair in a Ge detector. If all the carriers are collected calculate the responsivity for photon energies of a)0.7eV b)1.0eV c)2.0eV. Condition:
1Q η= Solution: By
Q ph
hv
R e
η=
We get
111
(0.7) 1.43( 1.0)1( 2.0)0.5Q ph Q ph Q ph e R hv eV V hv e R hv eV V hv e R hv eV V hv
ηηη---========
=
2.6 Consider a long Si p-n junction with a reverse bias of 1V at 300K. The diode has the following parameters:
2
173173
227
7
3
1
2
13101012/8/10101010/ 1.7a d n p n p op A cm N cm N cm D cm s
D cm s
s
s
cm
P W cm
w eV
ττα-----==⨯========
Calculate the photocurrent in the diode. Solution: By
22313.6810op
L P G cm s w
α--=
=⨯
10.95n L m μ==
8.94p L m μ==
2ln()0.846a d B bi i
N N k T
V V e n =
= 1/202{
()()}0.18r a d bi a d
N N W V V m e N N εεμ+=+= So
()11.8L L n p I eAG W L L A =++=
2.7 Consider a long Si p-n junction solar cell with an area of 4cm 2 at 300K. The solar cell has the following parameters:
173183
227
7
3101015/7.5/1010 1.0 1.25
a d n p n p L N cm N cm D cm s D cm s
s
s I A
m ττ----=⨯=======
Calculate the open circuit voltage of the diode. If the fill factor is 0.75,calculate the maximum power output. Solution:
By
1202
012.258.667.1410n p n p p n p n i n p n a p d L m L m
I A eD n eD p D D I A Aen L L L N L N μμ-⎧⎪
==⎪⎪
=⇒=⨯⎨⎪
⎛⎫⎛⎫⎪=+=+ ⎪ ⎪⎪ ⎪ ⎪⎝⎭⎝⎭⎩ We get
ln(1)0.83L oc I mkT
V V e I =
+= And
0.6225
m f oc L P F V I W ==
2.8 Consider the solar cell of problem 2.7.A solar system is to be developed from such cells to deliver a power of 15W at a voltage level of 5V. Calculate the total number of solar cells needed. Condition:
10.830.75sc oc f I A V V F ===
Solution:
For each solar cell:
0.870.72m sc m oc I A V V
⎧==⎪
⎨
==⎪⎩ We can get
5
()[
]1715
()[]14
5m
m
N series V N para I =+==+=
So the total number of cells
()()28Tot N N series N para =⋅=
2.10 Consider a silicon photoconductor at 300Kwith the following parameters:
1532267
4
2
101200/400/1051010100d n p n p N cm cm V s
cm V s s
s
A cm
L m
μμττμ----==⋅=⋅==⨯==
A bias of 5V is applied to the detector. Calculate the dark current. If light falls on the detector to produce a generation rate of 1021cm -3s -1,calculate the excess concentration, the photoconductivity, and device gain. Solution:
(1) The dark current
00000153
02530()()109.6/ 2.2510B d
n p n p d d i d V I FA e n p AF e n p A L
n N cm
I mA p n N cm σμμμμ--⎧
==+=+⎪⎪⎪==⇒=⎨⎪==⨯⎪⎪⎩
(2) The excess carrier denisty
143510L p p n G cm δδτ-===⨯
(3) The photoconductivity
()10.128n p e n p s cm σμδμδ-∆=+=⋅
(4) The device gain
8(1)401.6710p
p L ph Lp
tr n ph tr n I G I t G L t s F τμμμ-⎧==+⎪⎪⇒=⎨
⎪==⨯⎪⎩
2.13 Consider a GaAs p-i-n detector with an intrinsic layer width of 1.0 μm. Optical power density (photon energy 1.6eV) of 0.1W/cm 2 impinges upon the detector. The absorption coefficient for the active region is 104cm -1. Calculate the prompt photocurrent of device. The device area is 10-4cm 2. Condition:
241421.0 1.60.1
/1010op W m w eV
P W cm cm A cm μα---=====
Solution:
17216(0) 3.9110 3.9510(0)[1exp()]op ph L L ph P J cm s I A I eAJ W ωα---⎧
=
=⨯⎪⇒=⨯⎨
⎪=--⎩
2.14 Consider a silicon p-i-n detector in which the i layer is 10 μm thick. Calculate the maximum quantum efficiency of this detector if only light absorbed in the undoped region contributes to the photocurrent. The absorption coefficient for the active region is 103cm -1. Also calculate the minimum thickness of the i-region needed to ensure a quantum efficiency of 0.8. There are no reflection losses. Condition:
31
10100.8
0Q W m
cm R μαη-====
Solution: (1) By
()2max 1exp 10.86412
[,]
W e W ηαηαα=--⎧⎪
⇒=-=⎨∈⎪⎩
(2) By
()min min 31
1exp 0.8
16.110W W m cm
ηαμα-⎧=--=⎪⇒=⎨=⎪⎩
2.16 An avalanche photodetector has an avalanche region of 0. 5μm and the electric field is such that αimp =βimp =104cm -1. Calculate the multiplication factor of the device. Solution:
121e imp M W
α=
=-
3.8 Consider a GaAs p-n + junction LED with the following parameters at 300K:
22173
163
25/12/510101010n p d a n p D cm s
D cm s
N cm N cm ns ns
ττ--===⨯===
Calculate the injection efficiency of the LED assuming no trap related recombination. Solution: By
(1)(1)
B B eV
n p k T
n n eV
p n k T p p eD n J e L eD p J e L ⎧=-⎪⎪
⎨⎪
=-⎪⎩
And
2
432
63
41053.46810i p n a i p n d n n cm L m
N n L m p cm N μμ----⎧==⨯⎪⎧==⎪⎪⎨⎨
==⎪⎪⎩
==⨯⎪⎩
We can get
0.986n p
n n
inj n p p n
n p n p
eD n J L eD n eD p J J L L γ=
==++
3.9 The diode in Problem 3.8 is to be used to generate an optical power of 1mW. The diode area is 1 mm 2 and the radiative efficiency is 20%. Calculate the forward bias voltage required. Condition:
2
1120%out Qr P mW
A mm η===
Solution:
By
53.54410out ph ph n n
Qr
ph Qr
P I w
I e I A I I e ηη-=⋅⎧⋅⎪
⇒==⨯⎨=⋅⎪⎩
And
exp 1n p n n n B AeD n eV
I J A L k T
⎡⎤
⎛⎫=⋅=
-⎢⎥ ⎪⎝⎭⎣⎦
So
ln 1 1.02n p B n n AeD n k T V V e I L ⎡⎤=+≈⎢⎥⎣⎦
3.12 The light from a GaAs LED is coupled into an optical fiber which has refractive indices of 1.51 and 1.47 for the core and the cladding layers. Calculate the maximum angle of acceptance for the fiber. The LED has a Lambertian(cosine) output. Calculate the coupling efficiency for the diode. Condition:
121.51 1.47r r n n == Solution:
1221/2122
sin ()20.2sin 0.12
A r r fiber A n n θηθ-=-==
=
3.15 Consider an AlGaAs/GaAs heterojunction LED. The injection densities for electrons and holes are equal and are both 1017 cm -3 In the active GaAs region. Calculate the shift in the peak position of the emission peak energy if E g (GaAs) is 1.43eV. Calculate the shift in the peak position if the injection density is increased to 1018cm -3 . The temperature is 300K. Condition:
173
10 1.43n p cm Eg eV -===
Solution:
(1) For injection density of 17310n p cm -==
0.026E kT eV ∆==
11
1.4432
p g E E E eV ≈+∆=
(2) For injection density of 18310n p cm -==
0.055c
n
E kT eV N ∆=
= 21
1.4582
p g E E E eV ≈+∆=
So
120.015p p E E E eV ∆≈-=
3.18 A heterojunction LED based on GaAs is biased at 100Acm 2 current density at 300K. The active layer of LED is 0.5μm. Calculate the cutoff frequency of the diode. Condition:
2
1000.5J A cm d m μ=⋅=
Solution: By
r
d
J nev ne τ==
25311.2510r
n
J
cm s ed
τ--⇒
=
=⨯⋅ From the Fig 1639910810r
n cm s τ--⨯⨯
So
119.92c r
f MHz πτ=
≈
3.21 A GaAs LED is to be designed with an output power of 5.0W. The maximum device area that can be allowed 100 μm 2.Estimate the thickness of the active region needed. The efficiency of the Device is 20%. The maximum injection density is 1018cm -3. Condition:
2
1835.0
10020%
10out Tot P W A m n cm μη-====
Solution:
ph r
I ndA
P I e ωωη
ηωτ===
2.2r
d m nA
μηω=
=
3.22 Consider a GaAs p-n + junction LED with the following parameters at 300K:
110.2 3.0 2.0inj opt r nr P mW
ns ns γηττ=====
Calculate:
a) the total quantum efficiency of LED ηtotal ;
b) the number of photon generated per second I ph ; c) the injected current I inj . Solution:
(1) 1
0.40.081
1
r
Qr Tot inj Qr opt r
nr
τηηηηηττ=
=⇒==+
(2) 1611610.43710 2.18510opt ph ph opt P P I s I s w w ηη--=
=⨯⇒==⨯⋅ (3) 8.74ph
inj Tot
eI I mA η=
=
4.1 Consider a GaAs laser with a cavity length of 75μm. Calculate the number of
allowed longitudinal modes in an energy width 1
2
g B E k T ± where Eg=4.43eV and
T=300K. Solution:
115.5102r c
Hz n L
ν∆=
=⨯ 112E N h ν∆⎡⎤=+=⎢⎥∆⎣⎦
4.3 Consider a Fabry-Perot cavity of length 100μm. The mirror reflectivity is 0.33 and the absorption loss in the cavity is 120cm -.Calculate the photon lifetime ph τ.The refractive index in the cavity is 3.6. Condition:
1
1000.33
20 3.6loss r L m
R cm n μα-====
Solution:
-1ln 130.866cm Tot loss R L
αα=-=
139.1710r
ph Tot n s c
τα-=
=⨯
4.8 Two GaAs/AlGaAs double heterostructure laser are fabricated with active region thickness of 2.0μm and 0.5μm. The optical confinement factors are 1.0 and 0.8, respectively. The carrier injection density needed to cause lasing is 1831.010cm -⨯ in the first laser and 1831.110cm -⨯ in the second laser.The radiative recombination times are 1.5ns.Calculate the threshold current densities for the two lasers. Solution:
42
1132
22 2.1310/5.8710/las
th th r
las
th th r
d J n e
A cm d J n e
A cm ττ==⨯==⨯
5.1 A GaAs/AlGaAs laser has a threshold current density of 200A/cm -2. In a large signal switching, the laser is switched from zero current to 4 times the threshold current . Calculate the time delay before photon emission if the carrier lifetime is 2ns. Solution:
ln(
)0.575d th
J
t ns J J τ==-
5.2 A 1.55μm quantum well laser has a threshold carrier density of 2∙1012 cm -2.The radiative lifetime is 2.5ns and Auger coefficient is 10-28cm 6s -1.Calculate the time delay for a large signal switching in which the current density isswitched from 0.5J th to 3.0J th . Assume that the Auger rate can be evaluated using the carrier density n th . The active layer thickness is 10nm. Condition:
122
2861
2210 2.51010D r las n cm ns
F cm s d nm τ---=⨯===
Solution:
1832210D
th las
n n cm d -=
=⨯ 21
() 2.5nr th Fn ns τ-⇒==
111
() 1.25r nr ns τττ---⇒=+=
So
()
ln(
)0.28d th
J J i t ns J J τ-==- 6.1 Condition:
GaAs/AlGaAs TW amplifier
158221
0 1.43510410 1.0200 1.0m in dg
eV
A cm cm ns
dn
g cm P mW
ωτ---==⨯=⨯===
Calculation: saturation power P s , the gain Solution:
2.86m
s A P mW dg dn
ωτ==
10
148.21s
g g cm P P -=
=+。