Problem Solutions for Chapter 4

2πkBT3/2 3/4  Eg 
4-1. From Eq. (4-1), ni = 2  2  (memh) exp - 2k T
 h   B 

3/ 2
 2π(1.38 × 10−23 J / K) 
=2  T
3 /2
[(.068)(.56)(9.11× 10 −31
kg) ]
2 3/ 4

 (6.63 × 10 J.s) 
−34 2

 (1.55 − 4.3 × 10 −4 T)eV 

× exp −
 2(8.62 × 10 eV / K)T 
−5

3/2  1.55 ×   4.3 10 −4 

= 4.15×1014 T exp −  exp 
 2(8.62 × 10 ) 
−5 −5
 2(8.62 × 10 )T 

3/2  8991
= 5.03×1015 T exp - T 
 

4-2. The electron concentration in a p-type semiconductor is nP = ni = pi

Since both impurity and intrinsic atoms generate conduction holes, the total
conduction-hole concentration pP is
pP = NA + ni = NA + nP
2
From Eq. (4-2) we have that nP = ni /pP . Then
2 2 2
pP = NA + nP = NA + ni /pP or pP - NApP - ni = 0

so that

NA 
2
pP = 2
 4ni
1+ 2 +1

 
 NA 

If ni << NA , which is generally the case, then to a good approximation

2 2
pP ≈ NA and nP = ni /pP ≈ ni /NA

4-3. (a) From Eq. (4-4) we have 1.540 = 1.424 + 1.266x + 0.266x2 or

1
 x2 + 4.759x - 0.436 = 0. Solving this quadratic equation yields (taking the plus

sign only)

1
x = 2 [ - 4.759 + (4.759)2 + 4(.436) ] = 0.090

1.240
The emission wavelength is λ = 1.540 = 805 nm.

(b) Eg = 1.424 + 1.266(0.15) + 0.266(0.15)2 = 1.620 eV, so that

1.240
λ = 1.620 = 766 nm

4-4. (a) The lattice spacings are as follows:

o
a(BC) = a(GaAs) = 5.6536 A
o
a(BD) = a(GaP) = 5.4512 A
o
a(AC) = a(InAs) = 6.0590 A
o
a(AD) = a(InP) = 5.8696 A

a(x,y) = xy 5.6536 + x(1-y) 5.4512 + (1-x)y 6.0590 + (1-x)(1-y)5.8696

= 0.1894y - 0.4184x + 0.0130xy + 5.8696

o
(b) Substituting a(xy) = a(InP) = 5.8696 A into the expression for a(xy) in (a),
we have

0.4184x 0.4184x
y = 0.1894 - 0.0130x ≈ 0.1894 = 2.20x

(c) With x = 0.26 and y = 0.56, we have

Eg = 1.35 + 0.668(.26) - 1.17(.56) + 0.758(.26)2 + 0.18(.56)2

- .069(.26)(.56) - .322(.26)2(.56) + 0.03(.26)(.56)2 = 0.956 eV

4-5. Differentiating the expression for E, we have

2
 hc λ2
∆E = ∆λ or ∆λ = ∆E
λ2 hc

For the same energy difference ∆E, the spectral width ∆λ is proportional to the
wavelength squared. Thus, for example,

∆λ 1550  1550  2
= = 1.40
∆λ 1310  1310 

4-6. (a) From Eq. (4-10), the internal quantum efficiency is

1
ηint = = 0.783, and from Eq. (4-13) the internal power level is
1 + 25/ 90

hc(35 mA )
Pint = (0.783) = 26 mW
q(1310 nm)

(b) From Eq. (4-16),

1
P= 2 26 mW = 0.37 mW
3.5(3.5 + 1)

4-7. Plot of Eq. (4-18). Some representative values of P/P0 are given in the table:

f in MHz P/P0
1 0.999
10 0.954
20 0.847
40 0.623
60 0.469
80 0.370
100 0.303

4-8. The 3-dB optical bandwidth is found from Eq. (4-21). It is the frequency f at
which the expression is equal to -3; that is,

 
1
10 log 1/ 2  = −3
[
 1 + (2πfτ )
2
] 

3
 1
With a 5-ns lifetime, we find f =
2π (5 ns )
(
10 0.6 − 1) = 9.5 MHz

4-9. (a) Using Eq. (4-28) with Γ = 1

1  1 2
gth = ln 0.32 + 10 cm-1 = 55.6 cm-1
0.05 cm  

(b) With R1 = 0.9 and R2 = 0.32,

1  1 
gth = 0.05 cm ln 0.9(0.32) + 10 cm-1 = 34.9 cm-1
 

(c) From Eq. (4-37) ηext = ηi (gth - α )/gth ;

thus for case (a): ηext = 0.65(55.6 - 10)/55.6 = 0.53

For case (b): ηext = 0.65(34.9 - 10)/34.9 = 0.46

4-10. Using Eq. (4-4) to find Eg and Eq. (4-3) to find λ, we have for x = 0.03,

1.24 1.24
λ= E = = 1.462 µm
g 1.424 + 1.266(0.3) + 0.266(0.3)2

From Eq. (4-38)

dP(mW)
ηext = 0.8065 λ(µm) dI(mA)

Taking dI/dP = 0.5 mW/mA, we have ηext = 0.8065 (1.462)(0.5) = 0.590

4-11. (a) From the given values, D = 0.74, so that ΓT = 0.216

Then n 2eff = 10.75 and W = 3.45, yielding ΓL = 0.856

(b) The total confinement factor then is Γ = 0.185

4-12. From Eq. (4-46) the mode spacing is

λ2 (0.80 µm)2
∆λ = 2Ln = = 0.22 nm
2(400 µm)(3.6)

Therefore the number of modes in the range 0.75-to-0.85 µm is

4
 .85 − .75 .1
−3 = .22 ×10 = 455 modes
3
.22 × 10

 (λ - 850 nm)2
4-13. (a) From Eq. (4-44) we have g(λ) = (50 cm-1) exp - 
 2(32 nm)2 

 (λ - 850)2
= (50 cm-1) exp - 2048 
 

(b) On the plot of g(λ) versus λ, drawing a horizontal line at g(λ) = αt

= 32.2 cm-1 shows that lasing occurs in the region 820 nm < λ < 880 nm.

(c) From Eq. (4-47) the mode spacing is

λ2 (850)2
∆λ = 2Ln = = 0.25 nm
2(3.6)(400 µm)

Therefore the number of modes in the range 820-to-880 nm is

880 - 820
N= 0.25 = 240 modes

m
4-14. (a) Let Nm = n/λ = 2L be the wave number (reciprocal wavelength) of mode m.
The difference ∆N between adjacent modes is then

1
∆N = Nm - Nm-1 = 2L (a-1)

We now want to relate ∆N to the change ∆λ in the free-space wavelength. First

differentiate N with respect to λ:

dN d  n 1 dn n 1  dn
=   = - 2 = - 2 n - λ 
dλ dλ λ λ dλ λ λ  dλ

Thus for an incremental change in wavenumber ∆N, we have, in absolute values,

1  dn 
∆N = 2 n - λ  ∆λ (a-2)
λ  dλ

5
 λ2
Equating (a-1) and (a-2) then yields ∆λ =
 dn
2Ln - λ 
 dλ

(.85 µm)2
(b) The mode spacing is ∆λ = = 0.20 nm
2(4.5)(400 µm)

4-15. (a) The reflectivity at the GaAs-air interface is

 n-1 2  3.6-1 2
R1 = R2 = n+1 = 3.6+1 = 0.32
   

1 1 1 
Then Jth = α + ln = 2.65×103 A/cm2
β  2L R1R2 

Therefore

Ith = Jth × l × w = (2.65×103 A/cm2)(250×10-4 cm)(100×10-4 cm) = 663 mA

(b) Ith = (2.65×103 A/cm2)(250×10-4 cm)(10×10-4 cm) = 66.3 mA

4-16. From the given equation

∆E11 = 1.43 eV +
(6.6256 × 10 −34
J ⋅ s) 
2
1
+
1 
8(5 nm )  6.19 × 10 kg 5.10 × 10 kg 
2 −32 −31

= 1.43 eV + 0.25 eV = 1.68 eV

Thus the emission wavelength is λ = hc/E = 1.240/1.68 = 739 nm.

4-17. Plots of the external quantum efficiency and power output of a MQW laser.

4-18. From Eq. (4-48a) the effective refractive index is

mλB 2(1570 nm)

ne = = 2(460 nm) = 3.4
2Λ

Then, from Eq. (4-48b), for m = 0

6
 2
λB 1 (1.57 µm)(1570 nm)
λ = λB ± 2n L 2 = 1570 nm ± = 1570 nm ± 1.20 nm
e   4(3.4)(300 µm)

Therefore for m = 1, λ = λB ± 3(1.20 nm) = 1570 nm ± 3.60 nm

For m = 2, λ = λB ± 5(1.20 nm) = 1570 nm ± 6.0 nm

4-19. (a) Integrate the carrier-pair-density versus time equation from time 0 to td (time
for onset of stimulated emission). In this time the injected carrier pair density
changes from 0 to nth.

n= n th
td n th
1  J n  J 
t d = ∫ dt = ∫ dn = −τ  − = τ ln  
J n  qd τ  J − J th 
0 0 − n= 0
qd τ

 Ip 
where J = Ip/A and Jth = Ith/A. Therefore td = τ ln I - I 
 p th

(b) At time t = 0 we have n = nB, and at t = td we have n = nth. Therefore,

 J − nB 
td n th
1  qd τ
td = ⌠
⌡ dt = ∫ dn = τ ln J n 
J n 
0 nB − − th 
qd τ  qd τ 

In the steady state before a pulse is applied, nB = JBτ/qd. When a pulse is applied,
the current density becomes I/A = J = JB + Jp = (IB + Ip)/A

 I - IB   Ip 
Therefore, td = τ ln I - I  = τ ln I I - I 
 th  p + B th

4-20. A common-emitter transistor configuration:

4-21. Laser transmitter design.

7
4-22. Since the dc component of x(t) is 0.2, its range is -2.36 < x(t) < 2.76. The power
has the form P(t) = P0[1 + mx(t)] where we need to find m and P0. The average
value is
< P(t)> = P0[1 + 0.2m] = 1 mW

The minimum value is

1
P(t) = P0[1 - 2.36m] ≥ 0 which implies m ≤ 2.36 = 0.42

Therefore for the average value we have < P(t)> = P0[1 + 0.2(0.42)] ≤ 1 mW,
which implies

1
P0 = 1.084 = 0.92 mW so thatP(t) = 0.92[1 + 0.42x(t)] mW and

i(t) = 10 P(t) = 9.2[1 + 0.42x(t)] mA

4-23. Substitute x(t) into y(t):

y(t) = a1b1 cos ω1t + a1b2 cos ω2t

+ a2(b12 cos2 ω1t + 2b1b2 cos ω1t cos ω2t + b 22 cos2 ω2t)

+ a3(b13 cos3 ω1t + 3b12 b2 cos2 ω1t cos ω2t + 3b1 b 22 cos ω1t cos2 ω2t+ b 32 cos3 ω2t)

+a4(b14 cos4 ω1t + 4b13 b2 cos3 ω1t cos ω2t + 6b12 b 22 cos2 ω1t cos2 ω2t
+ 4b1 b 32 cos ω1t cos3 ω2t + b 42 cos4 ω2t)

Use the following trigonometric relationships:

1
i) cos2 x = 2 (1 + cos 2x)
1
ii) cos3 x = 4 (cos 3x + 3cos x)
1
iii) cos4 x = 8 (cos 4x + 4cos 2x + 3)

iv) 2cos x cos y = cos (x+y) + cos (x-y)

1
v) cos2 x cos y = 4 [cos (2x+y) + 2cos y + cos (2x-y)]

8
 1 1 1
vi) cos2 x cos2 y = 4 [1 + cos 2x+ cos 2y + 2 cos(2x+2y) + 2 cos(2x-2y)]
1
vii) cos3 x cos y = 8 [cos (3x+y) + cos (3x-y) + 3cos (x+y) + 3cos (x-y)]

then
1 2 2 3 4 2 2 3 4 constant
y(t) = 2 a2b1 + a2b2 + 4 a4b1 + 3a4b1b2 + 4 a4b2
  terms

fundamental
+ 4a3b1 + 2a3b1b2 cos ω1t + 4 a3b2 + 2b1b2 cos ω2t
3 3 2 3 3 2
terms

2 2
b1 b2
+ 2 a2 + a4b1 + 3a4b2 cos 2ω1t + 2 a2 + a4b2 + 3a4b1 cos 2ω2t
2 2 2 2

2nd-order harmonic terms

1 3 1 3
+ 4 a3b1 cos 3ω1t + 4 a3b2 cos 3ω2t 3rd-order harmonic terms

1 4 1 4
+ 8 a4b1 cos 4ω1t + 8 a4b2 cos 4ω2t 4th-order harmonic terms

+ a2b1b2 + 2 a4b1b2 + 2 a4b1b2[ cos (ω1+ω2)t + cos (ω1-ω2)t]

3 3 3 3

2nd-order intermodulation terms

3 2 3 2
+ 4 a3b1 b2 [ cos(2ω1+ω2)t + cos(2ω1-ω2)t] + 4 a3b1b2 [ cos(2ω2+ω1)t + cos(2ω2-ω1)t]
3rd-order intermodulation terms

1 3
+ 2 a4b1 b2 [ cos(3ω1+ω2)t + cos(3ω1-ω2)t]
3 2 2
+ 4 a4b1 b2 [ cos(2ω1+2ω2)t + cos(2ω1-2ω2)t]
1 3
+ 2 a4b1b2 [ cos(3ω2+ω1)t + cos(3ω2-ω1)t] 4th-order intermodulation
terms

This output is of the form

y(t) = A0 + A1(ω1) cos ω1t + A2(ω1) cos 2ω1t + A3(ω1) cos 3ω1t

+ A4(ω1) cos 4ω1t + A1(ω2) cos ω2t + A2(ω2) cos 2ω2t

9
 + A3(ω2) cos 3ω2t + A4(ω2) cos 4ω2t + ∑ ∑ Bmn cos(mω1+nω2)t
m n
where An(ωj) is the coefficient for the cos(nωj)t term.

-t/τ
4-24. From Eq. (4-58) P = P0 e m where P0 = 1 mW and τm = 2(5×104 hrs) = 105
hrs.

(a) 1 month = 720 hours. Therefore:

P(1 month) = (1 mW) exp(-720/105) = 0.99 mW

(b) 1 year = 8760 hours. Therefore

P(1 year) = (1 mW) exp(-8760/105) = 0.92 mW

(c) 5 years = 5×8760 hours = 43,800 hours. Therefore

P(5 years) = (1 mW) exp(-43800/105) = 0.65 mW

E /k T
4-25. From Eq. (4-60) τs = K e A B or ln τs = ln K + EA/kBT

where kB = 1.38×10-23 J/°K = 8.625×10-5 eV/°K

At T = 60°C = 333°K, we have

ln 4×104 = ln K + EA/[(8.625×10-5 eV)(333)]

or 10.60 = ln K + 34.82 EA (1)

At T = 90°C = 363°K, we have

ln 6500 = ln K + EA/[(8.625×10-5 eV)(363)]

or 8.78 = ln K + 31.94 EA (2)

Solving (1) and (2) for EA and K yields

EA = 0.63 eV and k = 1.11×10-5 hrs

Thus at T = 20°C = 293°K

10
τ = 1.11×10-5 exp{0.63/[(8.625×10-5)(293)]} = 7.45×105 hrs
s

11