Diode Lasers and Photonics Integrated Circuits 2nd Ed Coldren, Corzine Mashanovitch Solution Manual

Type:

Solution Manual

Resource:

Diode Lasers and Photonic Integrated Circuits

Edition:

2nd Edition

Author(s):

Larry A. Coldren Scott W. Corzine Milan L. Masanovic

CHAPTER 1

2011 ECE 227A Homework Solution #1

7. In practice, there are two general nonradiative mechanisms for carriers that are important. The first involves nonradiative recombination centers, such as point defects, surfaces, and interfaces, in the active region of the laser. To be effective, these do not require the simultaneous existence of electrons and holes or other particles. Thus, the recombination rate via this path tends to be directly proportional to the carrier density, N. The second mechanism is Auger recombination, in which the electron–hole recombination energy, E21, is given to another electron or hole in the form of kinetic energy.

Thus, again for undoped active regions in which the electron and hole densities are equal, Auger recombination tends to be proportional to N3 because we must simultaneously have the recombining electron–hole pair and the third particle that receives the ionization energy.

9. In direct bandgap double heterostructures, carriers are confined inside the PIN junction, and most of them recombine inside the I region, in transit between P and N. In a regular PN diode, very few carriers recombine in the junction, and they are just swept across the junction by the electric field.

2011 ECE227A Homework #2 Solutions

11. i) bulk material is normally thicker than quantum wells. ii) For quantum wells, carriers are confined in the wells, so the carrier injection efficiency is higher. iii) The density of states are different. Quantum wells have separate energy states, while the energy states of bulk material are quasi-continuous.

12. i) They need to lattice matched to avoid a lot of defects. ii) The cladding material must have a larger bandgap and smaller refraction index at working wavelength, compared to the waveguide material.

13. i) Si has an indirect bandgap, and most of the III-V materials, such as GaN, InP, GaAs, have direct bandgap. For indirect bandgap material, phonons are generally needed to generate photons because of the different crystal momentum between conduction band valance band. So their efficiency to generate light is very low.

14. GaAs/AlGaAs – datacom lasers in the ~700-900nm range, a lot of them vertical cavity.

InGaAsP – in-plane lasers for telecom and sensing, 1.3-1.6um.

15. The main material parameter controlling the current leakage in a double heterostructure or a quantum well is the conduction band offset. The band offset will determine the height of a quantum well in a laser active region. Due to their much lighter effective mass, electrons require much tighter confinement with increasing temperature than holes. Thus, it is important to have high band offset in the conduction band. For GaAs/AlGaAs, 2/3 in the conduction band, for InP/InGaAsP, 40% in the conduction band.

16.

17.

18.

20 wells, because of the energy states splitting. ( Assuming the wave function of the first well can still affect the 10th well.)

19.

Since the electrons in InGaAsP material has quite high mobility and they tend to escape from the quantum at high temperature, decreasing the injection efficiency.

To improve the performance of laser at higher temperature, the electron leakage current can be decreased by increasing the conduction band offset.

20. Blue laser has lower wavelength than the red laser – thus, the minimum spot size that can be resolved on a Blu ray disc is smaller than that of a regular DVD. To get an even higher capacity, we should use an ultra violet laser.

21. Aixtron and Veeco.

23. When interpolating, one must be careful if different bands come into play in the process as discussed in the text. For this case, InAlSb's bandgap changes from direct to indirect for x = 0.535. The plot is shown below.

1.24

1.25

1.26

1.27

1.28

1.29

1.30

1.31

1.32

1.33

1.34

1.35

1.36

CHAPTER 2

2.14 a)

η iα m αi+αm 1  1 α m = ln  L  R ηd =

αi

For La = 500um and Lp=500um: 1  1  αm= ln  = 11.39 / cm 0.1cm  .32  La α ia + L p α ip (500)(15) + (500)(5) = = = 10 / cm La + L p 500 + 500

ηd =

0.7 × 11.39 = 0.3727 11.39 + 10

For La = 500um 1  1  αm= ln  = 22.79 / cm 0.05cm  .32  α i = 15 / cm 0.7 × 22.79 ηd = = 0.4221 22.79 + 15 For La = 250um 1  1  αm= ln  = 45.58 / cm 0.025cm  .32  0.7 × 45.58 ηd = = 0.5266 45.58 + 15 b) Γ g th = α i + α m

11.39 + 10 = 534.8 / cm 1 .08 × 2 15 + 22.79 g th (500) = = 472 / cm .08 15 + 45.58 g th (250) = = 757 / cm .08 g th (500 + 500) =

g th

J A wL  J  I th = th = ( Number of Wells ) tr  e g 0 ηi ηi  well 

(3 × 10 cm)(500 × 10 cm) (6 wells)(100 A / cm )e I (500 + 500) = −4

th

−4

534.8 600

= 31.4mA 0.7 472 3 × 10 − 4 cm 500 × 10 − 4 cm I th (500) = (6 wells )(100 A / cm 2 )e 600 = 28.2mA 0.7 757 −4 3 × 10 cm 250 × 10 − 4 cm 2 I th (500) = (6 wells )(100 A / cm )e 600 = 22.7 mA 0.7 2

(

)(

)

(

)(

)

c)  1  Γ vg a fR = η i ( I − I th )   2π  qV 

1/ 2

= 5 × 10 9 Hz @ I − I th = 30mA

( 2π * 5 × 10 Hz )(1.6 × 10 C )(6 * 7 × 10 m * 3 × 10 m * 500 × 10 m) = 1.58 × 10 m a=  3 × 10   (0.7)( 30 × 10 A) (.08)(1 / 2) 4  9

− 19

−9

−6

8



−6

−3



d) ∂ g g0N = = a ∂ N N th J N 2.5 N N g = g 0 J ln = g 0 J ln 2.5 = 2.5 g 0 J ln = g 0 N ln J tr N tr N tr N tr g 0 N = 2.5 g 0 J = 2.5(600 / cm) = 1500 / cm g 1500 N th = 0 N = = 9.49 × 1017 cm − 3 − 15 2 a 1.58 × 10 cm e) N tr = N th e f)

− g th g0 N

= 9.49 × 10 e 17

− 534.8 1500

= 6.64 × 1017 cm − 3

− 19

2

 V a' Γ '  fR '= fR   V ' a Γ  g a' = 0 N N th '

a ' (500) = a ' (250) =

N th ' = N tr e 1500 / cm

1/ 2

g th g0 N

472 − 3 1500

6.64 × 1017 cm e 1500 / cm

757 − 3 1500

= 1.65 × 10 − 15 cm 2 = 1.36 × 10 − 15 cm 2

6.64 × 1017 cm e 1/ 2  1.65 1  f R ' (500) = 5GHz  1 = 7.23 GHz   1.58 0.5   1 1.36 1  f R ' (250) = 5GHz   0.5 1.58 0.5 

1/ 2

= 9.28 GHz

2.15 a)  4 * 90  ln   ln(4h / w) 3  °  ZT = = = 101.6 C / W W π ξL   −4 π . 6  250 × 10 cm  cm − K  b) Ith pulsed = 22.7mA (from 2.14 for 250um Laser) ηd pulsed = 0.527 (from 2.14 for 250um Laser) Pd = Pin = I 2 RS + IV D ; VD = 0.88V Guess Ith = 25 mA CW

(

)

Pd (25mA) = 0.025 2 (50) + (.0258)(.88) = 53.25 mW ∆ T = PD Z T = .05325(101.6) = 5.41° C I th ' = I th e

∆T T0

5.41

= ( 22.7 mA) e 55 = 25mA → close to guess so do not have to reiterate

c) I = 50mA Pin = (.05)2(50) + (.05)(.88) = 0.169 W Pdo = 169mW-Po ≈ 169mW (assuming Po is small) ∆ T = PD Z T = .169(101.6) = 17.2 ° C I th ' = I th e

∆T T0

η d '= η d e PO0 ' =

17.2

= ( 22.7 mA) e 55 = 31.0mA −∆T To

= 0.527e

− 17.2 110

= 0.451

hv η d ' ( I − I th ) = 0.8(0.451)(50 − 31) = 6.85mW q

Now reiterate for more accurate number: ∆ T = PD Z T = (.169 − .00685)(101.6) = 16.5 ° C I th ' = I th e

∆T T0

η d '= η d e PO0 ' =

16.5

= ( 22.7 mA) e 55 = 30.6mA −∆T To

= 0.527e

− 16.5 110

= 0.454

hv η d ' ( I − I th ) = 0.8(0.454)(50 − 30.6) = 7.05mW q

0.035

0.030

0.025

0.020 CW Pulsed

0.015

0.010

0.005

0.000 0.010

0.020

0.030

0.040

0.050

0.060

0.070

0.080

0.090

0.100

0.110

2.16

2.17

CHAPTER 3

3.8

3.9

3.10

3.11

3.12

3.13

3.14

3.15

3.17

3.18

3.19

(a) eta_i = 0.5*3.5/2.5=0.7 or eta_i = 0.6*7/6 =0.7

(b)

3.20

(b) (a)

kappa_Lg1 = 2.5 from the graph, kappa = 2.5/500um = 50 cm^{-1}

(c)

3.21

4.1

4.2

4.3

4.4

4.5

4.6

4.7

Homework 4 Solutions

4.8 (a) 0.1(1eV )

C(300K) = C0 e− k(300K) = 10−29 cm6 /s 0.1(1eV )

C(400K) = C0 e− k(400K)

(1) (2)

Using (1) to solve for C0 in (2): 0.1(1eV )

0.1(1eV )

C(400K) = (10−29 cm6 /s)e k(300K) e− k(400K) = 2.63x10−29 cm6 /s

(3)

(b) 0.1(2eV )

0.1(2eV )

Using (3), C(400K) = (10−29 cm6 /s)e k(300K) e− k(400K) = 6.92x10−29 cm6 /s

1

4.9

4.10

Homework 8 Solutions

5.10 L1 = 100µm L2 = 200µm κ = 0.0131µm−1 La = 50µm λ0 = 1.55µm ng = 3.8 1 T = 40.0e9 Hz 1 tanh(κL1 ) = 32.988µm Lef f 1 = 2κ 1 Lef f 2 = 2κ tanh(κL2 ) = 37.766µm

Lcav = T2ncg0 = 986.159µm Lactive = Lcav − Lef f 1 − Lef f 2 − La = 865.406µm 5.11 ∆λgain = 3.2nm ∆f = 40GHz λ2 ∆λ = c00 ∆f = 0.32nm Nmodes =

∆λgain = 10 ∆λ

1 ∆τp = Nmodes ∆f = 2.5ps

1

5.12

5.13

5.14

5.15

5.16

Homework 9 Solutions

5.17 R = 0.32 L = 400µum λ0 = 0.86µm Ith = 10mA ηi = 0.78 < αi >= 14cm−1 Va = 5µm3 V = 200µm3 a = 5x10−16 cm2 nsp = 1.3 (a) Ii = 5mA If = 35mA 3 B = 0.8x10−10 cm s For short-wavelength lasers, BN 2 >> AN + CN 3 and: q a τf = ηiqV If B = 605.64ps q q −1 Ii τd = τf [tanh−1 IIth − tanh If ] = 120.38ps f (b) αm = L1 ln R1 = 28.486cm−1 m ηd = ηi αiα+α = 0.523 m 1.23985 hν = = 1.442V q λ0 P0 = ηi ηd hν q (If − Ith ) = 18.849mW (c) ng = 4.2

1

α=3 vg = nc0g From Example 5.6 on page 307: v 2 ηi α2 nsp q

g m ∆ν = (1 + α2 ) 4πη 2 (I −I ) = 0.782M Hz f th d

2

Homework 10 Solutions

5.18 (a) I = 35mA Ith = 10mA ηi = 0.78 Va = 5µm3 V = 200µm3 Vp = V + Va a = 5x10−16 cm2 ng = 4.2 vg = nc0g Above threshold, 2 ωR ≈

ηi vg a(I−Ith ) qVp

fR = ω2πR = 7.326GHz p √ f3dB ≈ fR 1 + 2 = 11.383GHz (b) α=3 p p ∆νmax = fR (1 + α2 ) ln(4) = 27.277GHz

1

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5.20

5.21

5.22

5.23

6.11

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6.11 (b)

6.12 (a)

6.12 (b)

6.13

6.14

7.4

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7.5

7.6

7.7

7.8

7.9

7.10

7.11

7.12

7.13

7.14

7.16

8.5

8.6 (a) (a)

(b)

(c)

8.7 Modulator length: 250 um Waveguide thickness: 350 nm n = 3.45 pad height = 2um er=2.65 50 ohm termination resistor The length will be determined from the required bandwidth of the modulator, that is, from the maximum capacitance of the modulator. The capacitance has two components: the pad capacitance, which is fixed in this case, and the PIN modulator diode capacitance, which is length dependent.

Since the bandwidth is given by 𝑓3𝑑𝐵 =

1 2𝜋𝑅𝐶

the capacitance can be calculated from here, C = 0.127pF. The capacitance of the pad is the same as in Example 8.7, 𝑆

𝐶𝑝𝑎𝑑 = 𝜖0 𝜖𝑤𝑔𝑑 𝑝𝑎𝑑 = 0.025pF. ℎ

Therefore, the PIN capacitance has to be C-Cpad = 0.102 pF. 𝐶𝑝𝑖𝑛 = 𝜖0 𝜖𝑤𝑔𝑑

𝐿𝑤 ⟹ 𝐿 = 113.35𝜇m ℎ

8.8

8.9