Transcript MS Powerpoint

Introduction to Optical Electronics
Semiconductor Photon Detectors (Ch 18)
Semiconductor Photon Sources (Ch 17)
Lasers (Ch 15)
Laser Amplifiers (Ch 14)
Photons in Semiconductors
(Ch 16)
Photons & Atoms (Ch 13)
Quantum (Photon) Optics (Ch 12)
Resonators (Ch 10)
Electromagnetic Optics (Ch 5)
Wave Optics (Ch 2 & 3)
Ray Optics (Ch 1)
Optics
Physics
Optoelectronics
1
Laser Amplilfiers
Amplifier
Input
light
  d

0
Output
light
z
z  dz
d
2
Gain & Phase Coefficients
Lorentzian Lineshape
 / 2 

 ( )   ( 0 )
2
2
  0     / 2 
2

2
where   0   N  2
 4 t 
sp

E( z )
H ( )
0



E( z ) / z

0
1
 ( )   ( )  j ( )
2
Using the Kramers-Kronig Relationship:
0
  0
Lorentizian:  ( ) 
  

3
Exercise 13.1-1
Attenuation and Gain in a Ruby Laser Amplifier
a)
Consider a ruby crystal with two energy levels separated by an energy
difference corresponding to a free-space wavelength 0 = 694.3 nm, with
a Lorentzian lineshape of width  = 60 GHz. The spontaneous lifetime
is tsp = 3 ms and the refractive index of ruby is n = 1.76. If N1 + N2 = Na =
1022 cm-3, determine the population difference N = N2 – N1 and the
attenuation coefficient at the line center (0) under conditions of thermal
equilibrium (so that the Boltzmann distribution is obeyed) at T = 300 K.
b)
What value should the population difference N assume to achieve a gain
coefficient (0) = 0.5 cm-1 at the central frequency?
c)
How long should the crystal be to provide an overall gain of 4 at the
central frequency when (0) = 0.5 cm-1 ?
4
Rate Equations
Understanding Lifetimes
2
– 1 and 2 are overall lifetimes for atomic
energy levels 1 and 2.
– Lifetime of level 2 has two contributions
(where rates are inversely proportional to
decay times)
and
   
1
2
1
21
2
 21
tsp
 nr
1
1
 20
1
20
 211  tsp1  nr1
 Population densities N1 and N2 will
vanish unless another mechanism is
employed to increase occupation
5
Rate Equations
Absence of Amplifier Radiation
• Pumping Rates – R1 & R2 defined
R2
2
R1
• Rate Equations:
1
d N2
N2
 R2 
dt
2
2
d N1
N N
  R1  1  2
dt
1  21
• Steady-State Conditions
d Ni
0
dt
 21
R2
2
1
R1
 1 
N 0  R2 2 1 
  R1 1
  21 
where N 0  N 2  N1 in steady-state
1
 20
6
Exercise 13.2-1
Optical Pumping
Assume that R1 = 0 and that R2 is realized by exciting atoms from the
ground state E = 0 to level 2 using photons of frequency E2 / h absorbed
with a transition probability W. Assume that 2 ≈ tsp, and 1 << tsp so that
in steady state N1 ≈ 0 and N0 ≈ R2 tsp. If Na is the total population of
levels 0, 1, and 2, show that R2 ≈ (Na – 2N0)W, so that the population
difference is N0 ≈ Na tsp W / (1 + 2 tsp W)
7
Rate Equations
Presence of Amplifier Radiation
•
Pumping Rates
2
d N2
N    I
 R2  2 
 N2  N1 
dt
2
h
R2
d N1
N N    I
  R1  1  2 
 N2  N1 
dt
1  21
h
•
Four Case Studies
Wi
1
 21
2
1
R1
1
 20
(Homogeneous Broadened Transitions)
1. I = 0, R2(t) = R20 u(t), R1(t) = 0
2. 1 = 0, R2(t) = R20 u(t)
3. 1 = 0, R2(t) = R20, I = Pulse
4. Steady State -  / t  0
where Wi      
I
  
h
8
Case 1
I = 0, R1(t) = 0, R2(t) = R20 u(t)
d N2
N2    I
 R20 

 N2  N1 
dt
2
h
d N2
N
 R20  2
dt
2
 N 2  t   R20 2 1  e  t / 2 
d N1
N N    I
  R1  1  2 
 N2  N1 
dt
1  21
h
N1  t  


2
 /
1
R201 1  1 2 et / 
et / 
 21
1  1 /  2
 1  1 /  2

1
2
9
Solving Differential Equations
y ''(t )  a y '(t )  b y(t )  f (t )
• Obtain Forms
– General Form
– Particular Form
• Homogeneous (Natural) and Particular (Forced) Response
– Particular Solution
• y p (t )  Particular Form
(1)
• y(2)
p (t )  a y p (t )  b y p (t )  f (t )  Coefficients
– Note: initial conditions not set
– Homogeneous
• yh (t )  General Form
Typically, try yh (t )  e t in
y h2 (t )  a yh1 (t )  b yh (t )  0
yielding  2  a   b  0
• Use initial conditions (removes the effect of the particular solution’s i.c.)
–
–
y(0)  yh (0)  y p (0)
y (1) (0)  y h(1) (0)  y (1)
(0)
p
10
Case 1
I = 0, R1(t) = 0, R2(t) = R20 u(t)
N 2  t   R20 2 1  e  t / 2 
N1  t  


2
 /
1
R201 1  1 2 et / 
et / 
 21
1  1 /  2
 1  1 /  2

1
Lifetime Ratio:  2 / 1  2
2
Lifetime Ratio:  2 / 1  0.5
2.5
2.5
R20 2
2
N2
2
N1
1.5
1.5
R2 (t )  R20u t 
1
1
2
R20 1
0.5
 21
0.5
N1
1
0
1
2
time (t /  2 )
3
4
5
N2
1
0
1
2
3
time (t /  2 )
4
115
Case 2
1 = 0, R2(t) = R20 u(t)
d N2
N I
 R20  2    N 2  N1 
dt
 2 h
d N2
N I
 R20  2   N 2
dt
 2 h
t  I 

 1 v  
R20 2
1  e  2  Is  
 N2 t  

1  I / I s 


h
where I s 
 2
d N1
N N    I
  R1  1  2 
 N2  N1 
dt
1  21
h
N1 t   ?
12
Case 2
1 = 0, R2(t) = R20 u(t)
t  I 

 1 v  
R20 2
1  e  2  I s  
N2  t  

1  I / I s 


Is 
h
 2
2.5
2
R20 2
N2  I  0
1.5
R2 (t )  R20u t 
1
0.5
N2  I  1.5I s 
1
0
1
2
time (t /  2 )
3
4
5
13
Case 3
1 = 0, R2(t) = R20, I = Pulse
d N2
N I
 R20  2    N 2  N1 
dt
 2 h
d N2 1    2 
d N 2 1  I 
 1 
I  N 2 
 1   N 2  R20
dt
 2  h 
dt
 2  Is 
t
 '


R20 2
2
 I / I s  e  1
 N2 t  
1  I / I s 



where  2' 
2
1  I0 / I s
and I  t   I 0 u  t   u  t  T  
14
Case 3
1 = 0, R2(t) = R20, I = Pulse
R20 2 
 t /  2'
N2 t  
I / I s  e
 1



1  I / I s 
where  2' 
2
1  I0 / I s
and I  t   I 0 u  t   u  t  T  
2.5
N2  t 
R20 2
2
1.5
2
 2'
1
 I  1.5Is 
0.5
1
0
1
2
time (t /  2 )
3
4
5
15
Approach to Case 4
Steady-State Rate Equations
Rate Equations describe the rates of change of the
population densities N1 and N2 as a result of
pumping, radiative, and nonradiative transitions.
1.
Determine rate equations in the absence of Amplifier
Radiation (i.e., no stimulated emission or absorption)
–
2.
Find steady-state population difference N0 = N2 – N1
Determine rate equations in the presence of Amplifier
Radiation (non-linear interactions)
–
–
Find steady-state population difference N = f(N0)
Determine the saturation time constant s
16
Case 4: Steady State
• Pumping Rates
2
d N2
N
 R2  2  N 2Wi  N1Wi
dt
2
R2
d N1
N N
  R1  1  2  N 2Wi  N1Wi
dt
1  21
  
where  s   2   1 1  2 
  21 
  
and N 0  R2 2 1  1   R1 1
 2 
• Steady-state Population Differences
• N = N2-N1
• N0 = N2-N1 w/o amp. rad.
• s – Saturation Time Constant
2
1
 20
1
R1
• Steady State
d Ni
N0
0 N 
dt
1   sWi
 21
Wi -1
N
N0
N0
2
0.1
s
1
s
Wi
10
s
17
Rate Equations in the Absence of Amplifier Radiation
• Four-Level Pumping Schemes
3
Rapid decay
Short-lived level
 32
Long-lived level
2
R
Wi
Laser
Pump
2
-1
 21
Short-lived level
1
Rapid decay
0
 20
1
Ground state
• Three-Level Pumping Schemes
3
Rapid decay
R
Short-lived level
 32
Long-lived level
2
Pump
Laser
1
Wi
2
-1
 21
Ground state
18
Rate Equations in the Absence of Amplifier Radiation
• Four-Level Pumping Schemes
3
Rapid decay
N0
N
1   sWi
N0 
s 
tsp N aW
1  tspW
tsp
1  tspW
Short-lived level
 32
Long-lived level
2
R
Wi
Laser
Pump
2
-1
 21
Short-lived level
1
Rapid decay
0
 20
1
Ground state
• Three-Level Pumping Schemes
3
N0 
s 
N a  tspW  1
R
1  tspW
Pump
2tsp
1  tspW
Rapid decay
Short-lived level
 32
Long-lived level
2
Laser
1
Wi
2
-1
 21
Ground state
19
Population Inversion
N  N2  N1
Population
Difference
Four-Level Laser
N0
N
1   sWi
Three-Level Laser
N0
N
1   sWi
Steady-State
Difference
N0 
N0 
tsp NaW
1  tspW
N a  tspW  1
1  tspW
Saturation Time
Constant*
s 
s 
tsp
1  tspW
2tsp
1  tspW
*What is the small-signal approximation?
20
Exercise 13.2-3
Pumping Powers in Three- and Four-Level Systems
a)
Determine the pumping transition probability W required to achieve a
zero population difference in a three- and a four-level laser amplifier
b)
If the pumping transition probability W = 2 / tsp in the three-level
system and W = 1 / 2 tsp in the four-level system, show that N0 = Na / 3.
Compare the pumping powers required to achieve this population
difference.
21
Amplifier Nonlinearity
Gain Coefficient
 ( ) 
 0 ( )
1   / s ( )
2
where  0 ( )  N 0 ( )  N 0
g ( )
8 tsp
Note: 0() is called the smallsignal gain coefficient. Why?
0
1
0.5
0.01
0.1
1
10
100
s
22
Amplifier Nonlinearity
s
   z    z     (0)  (0) 


ln
  ln
  0z
s
s   s
s 

10
d
Gain 12
8
Y  X e 0d
 0d
 ln Y  Y   ln X  X    0 d
Output Y
6
4
2
  0
 d 
where X 
and Y 
s
s
YX
0
1
2
3
Input X
4
0
5
e 0 d
6
d
5
4
Gain
Gain
0
7
 (d ) Y

 (0) X
6
s
3
2
0.001
0.01
Input
0.1
0
1
s
23
10
Saturable Absorbers
Output Y  (d )
Transmittance =
 
Input
X  (0)
where  ( )  0 (i.e., attenuation)
0
0.8
d
0.6
Transmittance Y X
0.7
0.5
0.4
0.3
e 0 d
0.2
0.001
0.01
0.1
Input X
1
0
s
10
24
Gain Coefficient
Gain Coefficient
Inhomogeneously Broadened Medium
  
 s

0

0
1
25