Problems and solutions

A laser with constant output power is coupled to an external intensity modulator, which in turn is coupled to a single mode fibre with the dispersion parameter β2.. At the output of the

Calculation problems and short questions

from old exams

Solutions and answers

CALCULATION PROBLEMS AND SHORT QUESTIONS 1

2 OPTICAL FIBRES 2

Problems 2

Short questions 5

S OLUTIONS 8

3 OPTICAL SOURCES AND TRANSMITTERS 30

Problems 30

Short questions 33

S OLUTIONS 34

4 OPTICAL DETECTORS AND RECEIVERS 50

Problems 50

Short questions 53

S OLUTIONS 54

5 LIGHTWAVE SYSTEMS 66

Problems 66

Short questions 67

S OLUTIONS 68

6 OPTICAL AMPLIFIERS 73

Problems 73

Short questions 76

S OLUTIONS 77

7 DISPERSION COMPENSATION 93

Problems 93

S OLUTIONS 94

8 MULTICHANNEL SYSTEMS 99

Problems 99

Short questions 99

S OLUTIONS 101

9 SOLITONS 103

Problems 103

Short questions 103

S OLUTIONS 104

10 COHERENT LIGHTWAVE SYSTEMS 106

Problems 106

Short questions 106

S OLUTIONS 108

Almost all of the problems and solutions are translations, from Swedish into English,

of old exams The problems is ordered in chapters, which follow the course book

Calculate numerically also the ratio between the intensity at the centre of the core and the intensity at the surface of the cladding

(Exam 960830)

2.2 A standard single mode fibre is to be characterised The refractive index of the core is 1.44 and the relative refractive index difference is 10-3 The fibre is to be used at 1.7 μm and has the normalised frequency of 2

How large is the diameter of the core? How much of the power is transported in the core? What is the material dispersion parameter? What is the minimal

frequency is ω It can be assumed that the electrical field only has one field component apart from zero and that the field is directed in the x-axis direction Further, the field component is an even function of y and is independent of x Derive the dispersion relation, i.e the relation between the propagation constant

β and the free-space wave number k0 How does the E-field vary in the y-axis and the z-axis directions?

(Exam 010423)

Dispersion in Single-Mode Fibers

2.5 A step-index fibre with a pure silica core has a core refractive index of 1.5000, a cladding refractive index of 1.4967, and is used to transmit light at 1.55 μm The core radius is 5 μm

a) How many modes can propagate in the fibre?

b) Which are the two main causes of dispersion in this fibre?

c) Determine the dispersion due to these two mechanisms

For waveguide dispersion, make the (realistic) assumption that the material

dispersion can be neglected (dn/dω=0)

2.6 A standard single-mode optical fibre has minimum dispersion at λ≈1.3 μm and minimum attenuation at λ≈1.5 μm To be able to make use of the low

attenuation we would like to design a dispersion shifted optical fibre with zero dispersion at 1.5 μm To facilitate coupling of light into the fibre, the core radius should be as large as possible By minimising the core-cladding index

difference, the core radius will be sufficiently large Determine the core radius and the core-cladding index difference The fibre material is SiO2 and the

refractive index is approximately 1.5 Neglect the material dispersion when calculating the contribution of waveguide dispersion

P01   cos  , ε<<1 and Ω/2π is in the MHz-region The signal is transmitted on

a standard fibre, and higher order dispersion can be neglected The power

attenuation coefficient is α How does the detected power vary as a function of the length L of the fibre? The detector is ideal

2.9 A lossless fibre with negligible higher order dispersion than β2 is excited by a Gaussian pulse The power of the pulse has the RMS-width σ0 The spectral width of the optical pulse, i.e the RMS-width which is measured with an optical

envelope or the slowly varying amplitude, has the phase equal to 0 at the

position of the excitation At what distance along the fibre does the complex amplitude show the largest second order derivative of the phase with respect to the time? Calculate this time derivative, and the RMS-width of the power, and the spectral RMS- width at the corresponding position.(Exam 980305)

2.10 A chirped Gaussian pulse is created in the following way A laser with

constant output power is coupled to an external intensity modulator, which in turn

is coupled to a single mode fibre with the dispersion parameter β2 At the output

of the modulator the light power is Gaussian shaped and the pulse has no chirp (C0=0) At the output of the fibre a Gaussian shaped power with RMS-width σ1, chirp parameter C1 and peak power P1 is desired How long should the fibre be and what are the RMS-width σ0 and the peak power P0 at the modulator output? Assume the fibre attenuation α is zero

(Exam 000901)

2.11 A standard single mode fibre is excited with a pulse with carrier frequency

194 THz The power of the pulse is Gaussian shaped and has an RMS-width of 0.2 ns The optical field is without chirp at the exciting position Where along the fibre has the time derivative of the instantaneous frequency a maximal absolute value and what is the maximal absolute value? Find also the RMS-width of the pulse power for this position

ω0.What is the peak power, the RMS-width of the power, the time delay and the time derivative of the instantaneous frequency of the pulse at the output?

(Exam 010831)

2.13 Estimate the limiting bit rate for a 60 km single-mode fibre link at 1.3 and 1.55 μm wavelength assuming transform-limited 50 ps (FWHM) Gaussian input pulses Assume β2=0 and -20 ps2/km and β3=0.1 ps3/km and 0 at 1.3 and 1.55

μm wavelengths, respectively

2.14 A laser diode is current modulated so that Gaussian shaped pulses with width of T/8 are obtained, where T is the bit slot The laser is assumed to have a Gaussian shaped spectrum when it is unmodulated, the linewidth of the

RMS-spectrum is 5 MHz at the full width half max The linewidth enhancement factor

of the laser is +4 or -4 The pulses are launched into a standard single mode fibre at 1.55 μm How long can the fibre be in the two cases if the dispersion criterion σ/T<1/4 is used? The bit rate is 2.4 Gbit/s

(Exam 950601)

2.15 Consider a single mode fibre system from a dispersion point of view The system works at minimum dispersion parameter D, and the transmitting laser is direct modulated and has a chirp parameter = 6 The unmodulated laser has the linewidth 50 MHz at full width at half height of the spectral density The data rate is 1/T = 10 Gbit/s The launched pulse is Gaussian with the width of the optical intensity 35 ps at full width at half height Determine the maximal length

of the fibre with the condition that the dispersion condition is fulfilled, i.e the received RMS-width of the optical signal <T/4

(Exam 950831)

2.16 The dispersion limit, i.e the relation between the bit rate and the length of the link, for a system with a typical direct modulated light emitting diode and a typical direct modulated single mode laser diode should be compared

numerically In both cases the chirp parameter is assumed to be zero The carrier wavelength is chosen so that the dispersion parameter is zero for the standard single mode waveguide used The transmitted pulse width is assumed to be half

of the permitted width at the receiver Typical values can be found in the

(Exam 020308)

2.18 Consider a single mode fibre system from a dispersion point of view The system works at 1.55 μm and uses a standard fibre The linewidth of the laser is

100 MHz at full width half maximum of the spectrum, which is considered also

to have a Gaussian shape The laser is direct modulated and has a chirp

parameter, C4 The bitrate is 1/T=2.5 Gb/s Assume the transmitted pulse is

a Gaussian pulse with the width equal to 140 ps at full width half maximum of the optical power at the transmitter Calculate the maximum length of the fibre, providing normal dispersion limit criterion is fulfilled

(Exam 020823)

Short questions

2.19 What is waveguide dispersion and what is its cause?

(Exam 970306)

2.20 An advantage with an optical fibre is its immunity against electromagnetic interference even without metallic shield Explain why outer light disturbances, for example lightning, which penetrate into the fibre at midpoint of the fibre link, are harmless

2.23 Sketch the field distribution in core and cladding for the fundamental mode in

a planar waveguide when the normalised frequency

How do the curves in principle change if the system is upgraded with a dispersion compensating fibre roll incoupled at the receiver Note that the length variable is only referring to the length of the standard fibre

(Exam 010831)

2.27 The dispersion limit in a link with a standard single mode fibre has different slopes in a log(length)-log(bitrate)-diagram for different light sources What is the slope when the light source is a LED and a singlemode laser with small spectral width and without chirp The wavelength is 1.55 μm

(Exam 020408)

2.28 The student with name Osquar makes a fibre optic experiment He increases the transmitter power in steps of 1 dB from an ideal unmodulated laser with narrow spectral width He observes that received power, after transmission through the fibre, also increases in steps of 1 dB initially, but thereafter less What has happened?

(Exam 020408)

2.29 A normal single mode laser is placed in a module and is positioned so that the light is optimally coupled into a single mode fibre After some time the light power is zero in the fibre and it is suspected that the laser became defect

However, an investigation shows that the laser has started to lase in another mode, which has two variations in the lateral direction instead of one, which is normal Is this a possible explanation to the zero power? Motivate you answer (Exam 990309)

Solutions

2.1 The condition for single-mode is the normalised frequency V<2.405

Equation (2.2.35):

4 2 8 0 2

5 1 4 2 8 0 2

The field in the core is given by a Bessel function, equation (2.2.41) The intensity

is the squared absolute value The ratio, Q, is

2 0

0

)(

)0(

2

k n

2 2 2

2 /a p q

2 2 2

2 2

1

2 2 2

0

2

) ( /

V

qa n

n

n k

Answer: a 3 06 μmand Q 3 7

2.2 Equation (2.2.35) gives

μm 16.8 diameter μm

4 8 10 2 44 1 2

7 1 2 2

2 2

2

3 1

1

a w

w / a is achieved from equation (2.2.45)

 /  0 65  1 619  2 3/2 2 879  2 6  1 27

a

w

% 71 61 1

2 exp

0

( E0J0 J0 pa

)0(

)()

)()()

0 0

J a K

pa J K

Use 2 2

0 2 1

2 n k  

2 0 2 2 2

2 2 2

2

/a p q

2 2 2

2 2 1

2 2 2 0 2

) ( /

V

qa n

n

n k

2 

 V b

qa

5292 1 ) 1 (

2.4

According to the assignment E E x e x where E x(x,y,z) can be written as

) exp(

) exp(

k n

q    in accordance with equation (2.2.26) We then achieve

)) 2 / (

where p2 n12k02  2  0 according to equation (2.2.25)

In order to connect the solutions at the boundaries, we use the fact that the tangential components of the E- and H-fields should be continuous For example, Ex should be continuous at y d 2

, , 0 ( 0

E E j

2 cos(

) 2

k n

q   

) exp(

)) 2 ( exp(

) 2 cos(

A

) exp(

)) 2 ( exp(

) 2 cos(

ps 1

2  

2

2 2

dV

Vb Vd c

n W

0

0022 0 1

ps 6 1

W D

ps 20

W D M D tot

19

D

km nm ps/

minimised

dV

Vb d V c

n W

According to Figure 2.9:

 

32 1 for 96 0

max 2

Vb

d

V

3 10 1 6 96

0 5 1

6 10 55 1 8 10 3 6 10 19 96

0 min

10 55 1 32 1 2

2.7 Transmitted power P01 cos t

Transmitted field normalized as fieldstren gth2  power

 

4 1 0

) 0

( cos

1

0

t j e t j e t j e P

t j e t

j e l t

j e l

t j

e

4

) 0 0 (

t j e

l e l t j

e

2 2 1 4

2

2 2 1 4

1 2

) 0 0 (

e

2 2 1

cos 1

2

) 0 0 (

1

i.e the signal power is attenuated as el but has also a dispersive effect

cos  ; 1l corresponds to the group delay

2.8 Start with equation (2.4.16)

) 1 ( exp

) 1 ( )

,

(

0 2

2 0

2 0 0 0

2 2 0

0 0

iC z i T

t iC iC

z i T

T A t

z

A



Expand the exponent

2 2 2 2 0 2 2 2 2 2 0 2

2 2 2 2 0

2 2

2 0 2 2

2

0

2

2 2

) 1 ( ) 1 ( 2

) 1

(

z zC

T

z zC CT

it t zC zC T

z zC

T

z i zC T

t iC iC

z i

T

t iC

2 2 2 2 2

0

2

T

z zC

T



2 2 0 2 2 2 0 2 2

2 0 2 2 2 0

2 2

z C T

T

Differentiate with respect to z and set to zero

0 2

1 2

! 2 0

2 2 0

2 2 0

2 2 0

2 2

z T

C T

z C dz

2 2

2 0 2

z opt , ok according to equation (2.4.18)

We understand that C2  0 in order to achieve compression

2 2

2 2

2 2

2 0 2 2 2 2

2 0 2 2

2 2

2 0 2 2 2

0

2 2

1

1 1

1

1

C C

C C

T C

C T

T C

C T

 

1 ) 1 ( }

{ 1

~ 2

2 2 0 2

2 2 0 2 2 2 2

2

0

2 2 2 2

z z C z CT z

zC T

z zC CT

opt

i.e at maximal compression (z=zopt) the pulse has no chirp

The spectrum analyser measures A~(z,  )2, i.e the squared absolute value of the Fourier transform of the complex envelope The complex envelope:

i.e the squared absolute value does not change and neither does the RMS-width,

σω At zopt the pulse is transform limited, i.e  (zz opt)  (zz opt) is minimal

2.9 The complex amplitude at the position of excitation is

0 0

2 2

1 exp )

2 2 0 2 2

2 0

0 0 2

2 0 2 2

2 0

0 0

2

exp 2

exp )

,

(

z T

z j T t z

j T

T A z

j T

t z

j T

T A t

2 4

0

2 2

2

)

(

z T

z t

0

2 2

2

z T

z dt

2 2 4 0

2 2 2 2 2 4 0 2 2

z z z

T dt

) sgn(

2

1

2 2 0 2 2

2 0 2 4 0

2

0 2 max

2 0 2

exp

z T

T t



The variance is   2

0 2

0

2 2 4 0

z T

A is the Fourier transform of A(z,  ) While A~(z,)H A~(0,), where H is

the transfer function of the fibre and H 2  1, the spectral width does not change along the fibre That is obvious while the power is preserved and the

“frequencies” are not attenuated differently However, the phase relations changes between them (i.e the “frequencies”) and it is showed by the pulse broadening and chirp

Answer: RMS-width of the intensity: T0  2 0

Optimal position

2

2 0 0

Spectral RMS-width: σω

2.10 Equation (2.4.16) illustrates how the slowly varying amplitude varies in a single mode fibre In (2.4.16) C is the chirp parameter at the input of the fibre and is zero in our case, equation (1) Thus, at the output of the fibre

t z

i T

T A t

z

A

2 2 0

2 2

/ 1 2 2 0

0 0

2 exp )

2 1 1

2

1 exp )

,

(

T

t iC A

2 2 4 0

2 0

2 2

2 / 1 2 2 0

0 0 2

2 4 0

2 4 0 2 2

/ 1 2

2 exp

T

z T

T

z i t

z i T

T A z

T

z i T t

2 2 4 0 2 1

T

z T

z i T

T A A

2 2 0

0 0 1

1 iC

A A

2 1 1

The peak power at the output is A(z, 0 )2 ~P1

And at the input of the fibre 0

2

0 ~ P

A

2 1

0 2

2 2 4

0

2 0 0

1

1 C

P z

T

T P

0 P 1 C

2 1

1 0

2 1 1

2.11 The frequency 194 THZ λ=1.55 μm The dispersion is then dominated by

β2 and β3 can be neglected Typical values of D are in the range 15-18

ps/(km-nm) near 1.55 μm If 16.5 ps/(km-ps/(km-nm) is chosen, then D

21

2  

The power is Gaussian shaped which results in that the slowly varying amplitude

of the field also is Gaussian with the RMS-width T0  2  0 2  109  0 28ns

Equation (2.4.16) with C=0 gives

/ 1 2 2 0

0 0 2

2 0

2 2

/ 1 2 2 0

0 0

2

exp 2

exp )

,

(

z T

z j T t z

j T

T A z

j T

t z

j T

T A t

2 2

2 4 0 2 2

) ( )

(

) (

z t z

T

z t

2

) ( )

(

z T

z t

4 0

 z T

z dz

2 )

(

2 2 2 4 0

2 2 2 2 2 2 4

z z z

2 )

( z T

3800 )

( 21

) ( 08 0

1 2 2 2

2

km ps

ns T

z opt

2 6 2

1 ) sgn(

)

(

4 0 2 4

0 4 0 2

2 0 2 max

s Grad

1 /

The absolute value ≈1 GHz/ns

The RMS-width is derived from the squared absolute value of A(z,t)

2 0

2 0

2 2 4 0

2 2

2 4 0

2 0 2

2 exp 2

) (

1 2

exp )

(

2 2 exp

z

T

z T

t z

T

T t

0  z opt      

Answer: zopt=3800 km, maximum absolute value= 1 GHz/ns, T0=0.28 ns

2.12 The pulse will be attenuated, broadened, chirped and delayed as it propagates through the fibre The propagation can be expressed as

L

2 )

(

exp(   

The power attenuation is given by the factor exp(  L)

The delay is given by the group velocity time τ=L/vg={2.3.1}=Lβ1

It remains to calculate the broadening and the chirping

Use the formulas for chirped Gaussian pulses (without attenuation and delay) Equation (2.4.16) gives

2

exp ) (

)

,

(

2 2 0

2 2

/ 1 2 2 0

0 0

L j T

t L

j T

T A t

0  2 

T , because A0, T0 are field quantities The power P(L,t)  A(L,t)2 and the peak power output

2 2 0 2

0 2

2 4 0

2 0 2 0 1

2 1 ) ( )

T A L

2 2 0 2 2

2

0

2

) ( 2

) (

exp ) (

2

exp

L T

L j T t L

j T

2 2

2 2

0 2 0

2 2 0 2 0

2 2

0 2

1

2

1 2

2 2

 2

2 4

0

2 2

) (

L t

2

) ( 2

1 2

1

L T

L t f

4 0

2 2

2 4 0

2

2 1

1 4

2

1 ) ( 2

T

L f

2 1

2 2 0 2

2 2

0 2 1

4 0 2

2 1

1 4

1 2

where L is the transform-limitation distance,

 is the RMS after propagation and 0 is the RMS before propagation

Case 1

σ

σ0

After fibre propagation

Before fibre

propagation

10 50 2 ln 2

1 1 23

.

21

2 1

1123

21

2 1 2

1 max  

2 0 2 2 2

2 0

2 2

0

2

4 2

1 1

2

1 2

1000 s 10 4 2

From equation (2.3.5):

D c

10 10 10

10 17 10 3 2

10 55

3

2 12 9 3 12 8

2 6 2





From page 50

3 12

6

4 2 8

1000 2

ln 8

10 10 5 2 2

52 20 2 2

4

0 2 3

3 0 2 0

i.e the last term in equation (2.4.23) can be neglected

2 2

2

0

2

2 2

1 2

2 2 0

2 2 2

0

52 2

20 17 52

20 4 1 2

2  L 

L

131 64 13100 4100

  3

0 3 2 2 2 2

3

6 0 2 2 2 2

0

1

2 1

exp

2 0

2 2 0

ps 15 16 37

2 1 15

6 2

6 6 2

3 2 2 2

1 1 4

3 2

4 1

3 2 0 2

4 4

L S L

c S c

8 8

1 2

2

0 2 0

2 2 2 2

6 8

V , and the worst case occurs when 0 is minimal, i.e when B is

maximal We know that the modulation bandwidth of a typical light diode is ≈

100 MHz (see page 91) and B is of the same magnitude, for example 200 Mb/s

  15 10 1 10

3 1

10 3 2 10 200 8

6

3 3 3

2 2 3

3

10 26 10

3 1 10 09 0 128

10 3 2 6 128

) 2 ( 6 128

= 2 6  10 35 km(b/s)3 = 2 6  10 8 km(Gb/s)3

Let us check if Vω2<<1 The linewidth is about 10 MHz, according to page 117,

10 3 10 5





 , i.e when 7

2 0

2 2 2 0 2

0 2 0

2 2

0 2 0

2 0 0

2

4

1 2

2 2

2 2

1 2

2 2

2 2

2 2

2

2

1 2

2 1

2

2 1 2

1

C L

L

C L

L C C

L

L C C

C C

2 2 2

2

2 2

2 2

2

1 1

1 1

2 1

1 2

B

2

2 1 4

1



2.18 Consider the “superformula” (2.4.23) and the dispersion limit rule,

4

T



Further we have to find various values and possibly make some approximations

 The wavelength λ=1.55 μm and standard fibre results in that the β3-term can

be neglected β2=-20 ps2/km, according to page 53

 The linewidth 100MHz

ps 140

1

 indicates that V2can be neglected

Let us calculate σ ω and σ 0

Transmitted power has the formula 2

2 2 0 2 2 2 0 2 2

2 2

2 0 2 2

2

0 2

2

2 2

4 1

1 4

1 1

C C

T C C

C

T

C

C C

2 2 0 2

2

0

1

1 1

4 2

C

C T

1 16 1 10 60 10 5 2 4

1 4

24

2 12

L

59 km

2.19 The waveguide dispersion, DW, is one part of the chromatic dispersion, and is caused by a field redistribution between core and cladding when frequency changes and hence the group velocity changes with frequency

2.20 Light which penetrates into the optical fibre from outside will gradually leak out because the condition for total reflection is not fulfilled

2.21 In a multimode fibre the light propagates in several modes (hundreds) which have different group velocity, and hence a pulse will be smeared out when it propagates trugh the fibre The mode dispersion is much larger than the

intramode dispersion and is dependent on the bandwidth of the pulse The

smallest mode dispersion is achieved with a parabolic attenuation profile

2.22 The two orthonormal modes do not have the same group velocity in a physical single mode fibre Random deformations in the fibre make the modes couple to each other randomly which result in that the dispersion increases as length

2.25 Let the point of intersection be denoted by BxLx If you want to operate at

BxLx, with unchanged power, the eye, i.e the Q-value, is decreased with about 1

dB and the BER is increased Alternatively, the output power can be increased

by about 1 dB and the Q-value and BER are unchanged

2.26 Rayleigh scattering Small “in-frozen” density fluctuations give rise to

variations in index When the light reaches these the light is scattered (similar to light in fog) Attenuation ~1/λ4

2.27 A LED has a large spectral width  

B

L~ 1 the slope is -1 decade/decade

A single mode laser  ~ 12 

B

L -2 decade/decade See equations (2.4.26) and (2.4.30)

2.28 The limit for Brillouin scattering is reached, i.e the light is increasingly

scattered backwards as the power increases

3 Optical sources and transmitters

Problems

3.1 A laser is biased at 40 mA and has the threshold current 20 mA The reflectance

of the mirrors is 0.3 The length of the cavity is 500 μm and the internal loss coefficient is 10 cm-1 The reflectance of the left hand side mirror is increased to 1

by depositing a coating How many percent does the output power increase from the right hand side mirror compared to the first case? The bias current is still 40

mA The spontaneous emission is neglected and the stimulated emission is

proportional to the number of the charge carriers, i.e the transparency value is 0 (Exam 950601)

3.2 Two FP-lasers differs only by having different lengths One of them is 300 μm and the other one is 600 μm The bias current is well above threshold current For

a small change of the bias current of 1 mA, a change of the output power

throughout one of the mirrors is measured of 0.24 mW and 0.19 mW What is the internal efficiency and internal loss coefficient for the optical mode? Assume the reflectivity is 0.32 The wavelength is 1.55 μm

(Exam 970306)

3.3 A Fabry-Perot laser diode shall be used as a pump laser for an optical fibre

amplifier The given bias current I shall optimally be used, so that the optical,

outcoupled power through one of the mirrors is maximum When manufacturing the laser diode it is possible to choose the mirror reflectance within the interval 0

to 1 Both mirrors should be equal What is the maximal outcoupled power to the

fibre amplifier? Use the linear gain model in equation (3.5.3), where G N and N 0

are considered known The charge carrier lifetime is τc Internal losses/(unit

length) is αint Rsp is neglected The group velocity is vg The confinement factor,

Г, and the internal efficiency, η, are 1

(Exam 010309)

3.4 A laser diode (LD), constructed as a Fabry-Perot etalon, is to be used as a pump

laser for an optical fibre amplifier The given bias current, I, is to be converted in

the most beneficial way, with one of the mirrors as the outcoupling mirror, into optical pump power to the fibre amplifier When the laser diode is manufactured the reflectivity of the two different mirrors are chosen in the interval 0 to 1, the two mirrors can be chosen to have different values of the reflectivity What is the maximum outcoupled power to the fibre amplifier? Use a linear laser-gain model where GN and N0 are assumed known The carrier lifetime is c Internal

losses/duration is αint Rsp can be neglected The group velocity is vg Internal efficiency factor, η, is 1

(Exam 020823)

3.5 Calculate the so called turn on time for a semiconductor laser If a laser is biased under threshold and the current is increased by a current step so that the total current exceeds the threshold current, then after certain time, the turn on time, the laser starts to lase This is due to that it takes time to fill up with charge carriers so that the threshold is reached, whereafter the laser can be considered to

immediately start to lase Assume that the spontaneous recombination into the laser mode is negligible Thus, calculate the turn on time for Ibias=5 mA,

Ithreshold=10 mA and Istep=10 mA Further, the life time of the charge carriers is 1 ns and 1 ps for the photons

(Exam 990830)

3.6 Consider an FP-laser with nonlinear gain parameter according to equation

(3.5.15) Assume that the bias current increases very little to a new stationary value This causes a change of the stationary number of photons and charge

carriers Calculate these changes as a function of the current change Assume that the spontaneous emission into the lasing mode is negligible

(Also the frequency is changed because of the linewidth factor This is significant for frequency modulation at low frequencies However this matter is not included

function of the modulation angular frequency

Assume that the nonlinear-gain parameter is zero Further, the laser is biased, with the current Ib, essentially above threshold, It, which results in that the number of charge carriers for transparency and the spontaneous emission into the lasing mode can be neglected

The answer may not contain the variables the number of photons or the number of charge carriers in order to achieve maximal credits for this assignment

(Exam 960309)

3.9 The first part of the assignment is to use small signal analysis to find the transfer function between the number of charge carriers and the current for a

semiconductor laser for low frequencies where resonance phenomena are

negligible The bias values for the number of photons, the number of the charge carriers, current and phase are Pb, Nb, Ib and Φb, respectively The gain is

) (N N0

G

G N  The photon lifetime τp, and the carrier lifetime τc are constant The second part of the assignment is to find (1) The transfer function between the number of photons and the current (2) The transfer function between the

frequency chirp and the current for low modulation frequencies Rsp and εNL are neglected

(Exam 990309)

3.10 The modulation bandwidth for direct modulation of the intensity in a

semiconductor laser is approximately the same as the resonance frequency The resonance frequency is here defined as the frequency when the absolute value of a transfer function is maximal Derive an expression for the resonance frequency with given parameters beginning with the rate equations Assume the spontaneous emission in the lasing mode is zero Similarly the so-called “nonlinear-gain

parameter” equal to zero Determine a numerical value of the resonance frequency when the lifetime of the charge carriers is 1 ns, the photon lifetime is 1 ps and the laser is biased at the double threshold current GG N(NN0) where G N=6*103 s-1and N0=108

(Exam 000901)

3.11 Small signal analysis for frequencies essentially lower than the resonant

frequency gives a good insight of how, for example, the clamping of the charge carriers works The first part of the assignment is to use small signal analysis to find the resonance frequency of the transfer function between the number of photons and the current for a semiconductor laser expressed in bias current Ib, photon lifetime τp, carrier lifetime τc and differential gain GN The second part of the assignment is to find, for low frequencies where the resonance phenomena are negligible, the following: (1) The transfer function between the number of photons and the current (2) The transfer function between the number of charge carriers and the current GG N(NN0), where N0for simplification is set to 0 Further, τp

and τc are constants and τp << τc Ib is twice the threshold current Rsp can be neglected

(Exam 020308)

3.12 For a laser the curves below have been measured

Use these in order to determine the life time of the charge carriers τc and of the photons τp Use a simplified gain model, GG N N, neglect Rsp and εNL

at 50 mA bias current

m g

p

R R L

2

1with

R L m

m m

R L I

ln2

1and

1ln2

1

int 1

int 2 2

m g out th

p m g

q P

I I q

int 2

int 1 1 2 1

2

2

th m

m th

m m

m m

out

out

I I

I I

cm243.0

1ln05.0

13

.0

1ln

2

2 1

2

1024

1012

int 1

65.0204065.0

12

1 ln

d

/ 1 ln

1 k

1 m where L k m /

1 ln

1 1

1

int int

int int

1 8 0 2

1 2

hc

d 



μm 600 for 10 2 19 0

1 8 0 2

1 2

1 μm

300

67 1

3.3 Equation (3.5.9) gives

) (

th

G N q I





1 0

Equation (3.5.4):  1  (   int)

mir g

c mir

mir e

G

q v qN I q

h

From equation (1) we see that when αmir = 0 (R=1) minimal Ith is achieved, but also minimal differential efficiency If αmir is very large (R 0) the differential efficiency is maximised, but then Ith is large Thus, an optimal αmir exists

g 



mir mir

int

ac ab

d

dP

mir mir

int int

int

2 int int

int int

c

b ac c

b c c

b c

c b c

b

a

P e

2 int int

int 2

int int

c b c

1 2

c c

e

G

q v qN I

qN I q

h P

q v qN I







int 0

G qv

qN I

i.e I must be at least as large so that the gain compensates for αint Then αmir=0 and Ith is minimal

3.4 Consider equation (3.5.9), which can be modified to be valid for different mirrors

2

1

2 1 2

1

R R L mir     

th

G N

G

q qN

I q

I q

h

P

Let a

q

h 

, I qN b c





int int 1

int 1

c

b c

b

2 int int

int

int int

c

b c b

c b c

c c

e

G

q qN

I

qN I q

Before the current step ,    

c

bias bias N q

I dt

N bias biasc





If there was no threshold, N  N bias N step, at t 

But N clamps at the threshold

q

I N

1

step bias

step c

c step

step bias

th

I I I

I T

T N

N N

N

ln )

10 ln

b NL b

N

P P P P P P N

N N

 P PP P

N N N G N N q

I

I

b b

NL b

N c

b NL b

N

P P P N

G P P P

N N

P N

NL b

b NL b

NL b

N

b b NL N

N N

P N

P N N G

P P G

1 )

b NL p

P q

P

N N q