Comunicações Ópticas
RECEPTORES ÓPTICOS
Prof. Dr. Vitaly F. Rodríguez-Esquerre
Diagrama básico de um link óptico
Transmissor  Receptor
Sinal óptico: gerado por um LED ou laser.
Meio de propagação: ar ou guia de ondas (fibra óptica).
Receptor: fotodetector  converte sinal óptico em uma
corrente elétrica.
Signal Path through an Optical Link
Fundamental Receiver Operation
• The first receiver element is a pin or an avalanche photodiode, which
produces an electric current proportional to the received power level.
• Since this electric current typically is very weak, a front-end amplifier
boosts it to a level that can be used by the following electronics.
• After being amplified, the signal passes through a low-pass filter to reduce
the noise that is outside of the signal bandwidth.
• The also filter can reshape (equalize) the pulses that have become distorted
as they traveled through the fiber.
• Together with a clock (timing) recovery circuit, a decision circuit decides
whether a 1 or 0 pulse was received,
Fotodetetor
Dispositivos que convertem a luz incidente em uma
corrente elétrica
Requisitos:
Boa sensibilidade (responsividade) no comprimento de
onda de operação e nula em outros comprimentos de
onda -> Seletivo
Resposta temporal rápida -> largura de banda grande
Dimensões físicas compatíveis com as fibras ópticas
Baixo ruído
Photodiodes
• Due to above requirements, only photodiodes
are used as photo detectors in optical
communication systems
• Positive-Intrinsic-Negative (pin) photodiode
– No internal gain
• Avalanche Photo Diode (APD)
– An internal gain of M due to self multiplication
• Photodiodes are sufficiently reverse biased
during normal operation  no current flow, the
intrinsic region is fully depleted of carriers
Physical Principles of Photodiodes
• As a photon flux Φ penetrates into a semiconductor, it will be
absorbed as it progresses through the material.
• If αs(λ) is the photon absorption coefficient at a wavelength λ,
the power level at a distance x into the material is
Absorbed photons
trigger photocurrent
Ip in the external
circuitry
Photocurrent 
Incident Light Power
Examples of Photon Absorption
Photodetectors
Principle of the p-n junction Photodiode
 Schematic diagram of a reverse
Vr
SiO2
Electrode
Iph
Vout
R
biased p-n junction photodiode
 Photocurrent is depend on number
of EHP and drift velocity.
 The electrode do not inject carriers
but allow excess carriers in the
sample to leave and become
collected by the battery.
 Net space charge across the diode
e–
h+
hv > Eg
n
p+
AR
coating
E
Electrode
r
net
eN d
W
Depletion
region
x
in the depletion region. Nd and Na
are the donor and acceptor
concentrations in the p and n sides.
–eN a
E (x )
 The field in the depletion region.
x
E max
Vr
Principle of pn junction photodiode
(a)
• (a) Reversed biased
junction
photodiode.
• Annular electrode to allow photon
to enter the device.
• Antireflection coating (Si3N4) to
reduce the reflection.
• The p+-side thickness < 1 μm.
• (b) Net space charge distribution,
within SCL.
• (c) The E field across depletion
region.
p+n
SiO 2
Iph
p+
Electrode
h> Eg
Vout
R
h+ e–
n
E
Antireflection
coating
(b)
Electrode
rne t
W
Depletion region
eNd
x
–eNa
E (x)
(c)
x
14
E ma x
Photodetectors
Principle of the p-n junction Photodiode
 Operation of a p-i-n photodiode.
(b) Energy band diagram under reverse bias.
(a) Cross-section view of a p-i-n photodiode.
(c) Carrier absorption characteristics.
Photodetectors
Principle of the p-n junction Photodiode
 A generic photodiode.
Photodetectors
Principle of the p-n junction Photodiode
 Variation of photon flux with distance.
A physical diagram showing the depletion region.
 A plot of the the flux as a function of distance.
 There is a loss due to Fresnel reflection at the surface,
followed by the decaying exponential loss due to absorption.
 The photon penetration depth x0 is defined as the depth at
which the photon flux is reduced to e-1 of its surface value.
Photodetectors
RAMO’s Theorem and External Photocurrent
 An EHP is photogenerated at x = l. The electron and the hole drift in opposite directions with
drift velocities vh and ve.
 The electron arrives at time telectron = (L-l )/ve and the hole arrives at time thole = l/vh.
V
Iphoto(t)
0
Semiconductor
0
l
h+
thole
L-l
l
iphoto(t)
telectron
velectron
vhole
L
x
t
0
evh /L
e–
thole
ielectron(t)
photocurrent
thole
t
eve /L
i (t)
telectron
t
 e vh e ve 



L 
 L
Area = Charge = e
E
e-
h+
e vh
L
ihole(t)
t

Photodetectors
RAMO’s Theorem and External Photocurrent
 As the electron and hole drift, each generates ielectron(t) and ihole(t).
 The total photocurrent is the sum of hole and electron photocurrents each
lasting a duration th and te respectively.
te t  
L- l
ve
and
t h t  
l
vh
Transit time
Work done  e  E dx  V  ie t  dt
ie  t  
e ve
; t  te
L
ih  t  
E
e vh
; t  th
L
Qcollected   ie t  dt   ih t  dt  e
te
th
0
0
V
L
ve 
dx
dt
Photocurrent
The collected charge is not
2e but just “one electron”.
If a charge q is being drifted with a velocity vd(t) by a field between two biased
electrodes separated by L, the motion of q generates an external current
given by
e v d t 
i (t ) 
; t  t transit Ramo’s Theorem
L
Photodetectors
Absorption Coefficient and Photodiode Materials
 Absorbed Photon create Electron-Hole Pair.
1.24
 g [ m] 
E g [eV ]
Cut-off wavelength
vs. Energy bandgap
 Incident photons become absorbed as they travel in the
semiconductor and light intensity decays exponentially
with distance into the semiconductor.
I ( x )  I 0  e - x
Absorption coefficient

Absorption Coefficient
• Absorption
coefficient α is a
material property.
• Most of the photon
absorption (63%)
occurs over a
distance 1/α (it is
called penetration
depth δ)
21
Photodetectors
Absorption Coefficient and Photodiode Materials
 Absorption
The indirect-gap materials are shown with a broken line.
Photon energy (eV)
54 3
2
1
8
1 10
Ge
7
Absorption Coefficient (m-1)
1 10
1 10
0.8
0.7
In0.7Ga0.3As0.64P0.36
In0.53Ga0.47As
Si
6
0.9
GaAs
InP
5
1 10
a-Si:H
4
1 10
3
1 10
0.2 0.4
0.6 0.8 1.0 1.2 1.4 1.6 1.8
Wavelength (mm)
Absorption Coefficient
• Direct bandgap semiconductors
(GaAs, InAs, InP, GaSb, InGaAs,
GaAsSb), the photon absorption does
not require assistant from lattice
vibrations. The photon is absorbed
and the electron is excited directly
from the VB to CB without a change
in its k-vector (crystal momentum
ħk), since photon momentum is very
small.
kCB - kVB  photon momentum  0

Absorption coefficient α for direct bandgap semiconductors
rise sharply with decreasing wavelength from λg (GaAs and
23
InP).
Absorption Coefficient
• Indirect bandgap
semiconductors (Si and Ge), the
photon absorption requires
assistant from lattice vibrations
(phonon). If K is wave vector
of lattice wave, then ħK
represents the momentum
associated with lattice vibration
 ħK is a phonon momentum.
kCB - kVB  phonon momentum  K

Thus the probability of photon absorption is not as high as in a
direct transition and the λg is not as sharp as for direct bandgap
semiconductors.
24
Photodetectors
Absorption Coefficient and Photodiode Materials
Photon absorption in
a direct bandgap semiconductor.
Photon absorption in
an indirect bandgap semiconductor
E
E
CB
Indirect Bandgap
EC
Direct Bandgap
CB
Eg
Photon
EV
Photon
EV
VB
–k
Eg
EC
VB
k
–k
Phonons
k

Photodetectors
Quantum Efficiency and Responsivity
 External Quantum Efficiency
Number of EHP geberated and collected I ph e


Number of incidnet photons
P0 h
 Responsivity
I ph
Photocurrent (A)
R

Incident Optical Power (W) P0
R 
e
e

h
hc
Spectral Responsivity
Photodetectors
Responsivity (A/W)
 Responsivity vs. wavelength for a typical Si photodiode.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Ideal Photodiode
QE = 100% (  = 1)
g
Si Photodiode
0
200 400 600 800 1000 1200
Wavelength (nm)
The pin Photodiode SiO
• The pn junction photodiode has two
drawbacks:
2
Electrode
– Depletion layer capacitance is not
(a)
sufficiently small to allow
photodetection at high modulation
frequencies (RC time constant
limitation).
– Narrow SCL (at most a few microns) 
(b)
long wavelengths incident photons are
absorbed outside SCL  low QE
Electrode
p+
i-Si
n+
rnet
eNd
x
• The pin photodiode can significantly -eNa
reduce these problems.
E(x)
 Intrinsic layer has less doping and wider region (5 – 50 μm).
(c)
Eo
28
W
x
Photodetectors
The pin Photodiode
 Reverse-biased p-i-n photodiode
 pin photodiode circuit
 pin energy-band diagram
Photodetectors
The pin Photodiode
 Schematic diagram of pin photodiode
Electrode
In contrast to pn junction
SiO2
E(x) built-in-field is uniform
Electrode
p+
x
i-Si
E0
n+
W
h > Eg
rnet
eNd
E
–
h+ e
x
–eNa
Iph
R Vout
Vr
 Small depletion layer capacitance gives high modulation frequencies.
 High Quantum efficiency.
Photodetectors
The pin Photodiode
 A reverse biased pin photodiode is illuminated with a short wavelength
photon that is absorbed very near the surface.
 The photogenerated electron has to diffuse to the depletion region
where it is swept into the i- layer and drifted across.
p+
h > Eg
Diffusion
e–
h+
l
i-Si
E
Drift
W
Vr
Photodetectors
The pin Photodiode
p-i-n diode
(a) The structure;
(b) equilibrium energy band diagram;
(c) energy band diagram under reverse bias.
Photodetectors
The pin Photodiode
 The responsivity of pin photodiodes
Photodetectors
Photoconductive Detectors and Photoconductive gain
 Quantum efficiency versus wavelength for various photodetectors.

Photodetectors
The pin Photodiode
 Junction capacitance of pin
Cdep 
 0 r A
W
 Electric field of biased pin
Vr Vr
E  E0 

W W
 Small capacitance: High modulation frequency
 RCdep time constant is  50 psec.
 Response time
t drift
W

vd
v d  d E
 The speed of pin photodiodes are invariably limited by the transit time of
photogenerated carriers across the i-Si layer.
 For i-Si layer of width 10 m, the drift time is about is about 0.1 nsec.
Photodetectors
The pin Photodiode
Drift velocity (m sec-1)
 Drift velocity vs. electric field for holes and electrons in Silicon.
105
Electron
104
Hole
103
102
104
105
106
Electric field (V m-1)
107
Photodetectors
Absorption Coefficient and Photodiode Materials
 Absorption
The indirect-gap materials are shown with a broken line.
Photon energy (eV)
54 3
2
1
8
1 10
Ge
7
Absorption Coefficient (m-1)
1 10
1 10
0.8
0.7
In0.7Ga0.3As0.64P0.36
In0.53Ga0.47As
Si
6
0.9
GaAs
InP
5
1 10
a-Si:H
4
1 10
3
1 10
0.2 0.4
0.6 0.8 1.0 1.2 1.4 1.6 1.8
Wavelength (mm)
Example
Bandgap and photodetection
(a) Determine the maximum value of the energy gap which a semiconductor, used as a
photoconductor, can have if it is to be sensitive to yellow light (600 nm).
(b) A photodetector whose area is 510-2 cm2 is irradiated with yellow light whose
intensity is 20 mW cm-2. Assuming that each photon generates one electron-hole
pair, calculate the number of pairs generated per second.
Solution
(a)
Given,  = 600 nm, we need Eph = h = Eg so that,
Eg = hc/ = (6.62610-34 J s)(3108 m s-1)/(60010-9 m) = 2.07 eV
(b)
Area = 510-2 cm2 and Ilight = 2010-3 W/cm2.
The received power is
P = Area Ilight = (510-2 cm2)(2010-3 W/cm2) = 10-3 W
Nph = number of photons arriving per second = P/Eph
= (10-3 W)/(2.0591.6021810-19 J/eV)
= 2.97871015 photons s-1 = 2.97871015 EHP s-1.
Example
Bandgap and Photodetection
(c) From the known energy gap of the semiconductor GaAs (Eg = 1.42 eV), calculate the
primary wavelength of photons emitted from this crystal as a result of electron-hole
recombination. Is this wavelength in the visible?
(d) Will a silicon photodetector be sensitive to the radiation from a GaAs laser? Why?
Solution
(c) For GaAs, Eg = 1.42 eV and the corresponding wavelength is
 = hc/ Eg = (6.62610-34 J s)(3108 m s-1)/(1.42 eV1.610-19 J/eV)
= 873 nm (invisible IR)
The wavelength of emitted radiation due to EHP recombination is 873 nm.
(d) For Si, Eg = 1.1 eV and the corresponding cut-off wavelength is,
g = hc/ Eg = (6.62610-34 J s)(3108 m s-1)/(1.1 eV1.610-19 J/eV)
= 1120 nm
Since the 873 nm wavelength is shorter than the cut-off wavelength of 1120 nm, the
Si photodetector can detect the 873 nm radiation (Put differently, the photon energy
corresponding to 873 nm, 1.42 eV, is larger than the Eg, 1.1 eV, of Si which mean
that the Si photodetector can indeed detect the 873 nm radiation)
Example
Absorption coefficient
(a) If d is the thickness of a photodetector material, Io is the intensity of the incoming
radiation, the number of photons absorbed per unit volume of sample is
n ph 
I 0 1 - exp(-  d )
d h
Solution
(a) If I0 is the intensity of incoming radiation (energy flowing per unit area per
second), I0 exp(- d ) is the transmitted intensity through the specimen with
thickness d and thus I0 exp(- d ) is the “absorbed” intensity
Example
(b) What is the thickness of a Ge and In0.53Ga0.47As crystal layer that is needed for
absorbing 90% of the incident radiation at 1.5 m?
For Ge,   5.2  105 m-1 at 1.5 m incident radiation.
For In0.53Ga0.47As,   7.5  105 m-1 at 1.5 m incident radiation.
(b)
For Ge,   5.2  105 m-1 at 1.5 m incident radiation.
 1 - exp(-  d )  0.9
1
 1 
 1 
-6
d  ln 
ln

4
.
428

10
m  4.428m



5
  1 - 0.9  5.2  10  1 - 0.9 
1
For In0.53Ga0.47As,   7.5  105 m-1 at 1.5 m incident radiation.
d
1
 1 
-6
ln

3
.
07

10
m  3.07 m


5
7.5  10  1 - 0.9 
Example
InGaAs pin Photodiodes
Consider a commercial InGaAs pin photodiode whose responsivity is shown in fig.
Its dark current is 5 nA.
(a) What optical power at a wavelength of 1.55 m would give a photocurrent that
is twice the dark current? What is the QE of the photodetector at 1.55 m?
(b) What would be the photocurrent if the incident power in a was at 1.3 m?
What is the QE at 1.3 m operation?
Responsivity (A/W)
1
0.8
0.6
0.4
 The responsivity of an InGaAs
pin photodiode
0.2
0
800 1000 1200 1400 1600 1800
Wavelength (nm)
Solution
(a) At  = 1.55´10-6 m, from the responsivity vs. wavelength curve we
have R  0.87 A/W. From the definition of responsivity,
I ph
Photocurrent ( A)
R

Incident Optical Power (W ) P0
2 I dark 2  5  10-9 ( A)


 11.5 nW
we have P0 
R
R
0.87 A / W )
From the definitions
of quantum efficiency  and responsivity,

I ph
R 
e
e

h
hc
hcR (6.62  10-34 J  sec)(3  108 m / s )(0.87 A / W )


 0.70 (70 %)
-19
-6
e
(1.6  10 coul )(1.55  10 m )
Note the following dimensional identities: A = C s-1 and W = J s-1 so that A W-1 = C J-1.
Thus, responsivity in terms of photocurrent per unit incident optical power is also charge
collected per unit incident energy.
Solution
(b) At  = 1.310-6 m, from the responsivity vs. wavelength curve, R = 0.82 A/W.
Since Po is the same and 11.5 nW as in (a),
I ph  R  P0  (0.82 A / W )(1.15 nW )  9.43 nA
The QE at  = 1.3 m is
hcR (6.62  10-34 J  sec)(3  108 m / s )(0.82 A / W )


 0.78 (78 %)
-19
-6
e
(1.6  10 coul )(1.3  10 m )
Comparisons of pin Photodiodes
NOTE: The values were derived from various vendor data
sheets and from performance numbers reported in the
literature. They are guidelines for comparison purposes.
Detailed values on specific devices for particular applications
can be obtained from photodetector and receiver module
suppliers.
Avalanche Photodiode (APD)
• APD has an internal gain obtained by having a
high electric field that energizes photo-generated
electrons and holes
• These electrons and holes ionize bound electrons
in the valence band upon colliding with them
• This mechanism is known as impact ionization
• The newly generated electrons and holes are also
accelerated by the high electric field and they gain
enough energy to cause further impact ionization
• This phenomena is called the avalanche effect
Photodetectors
Avalanche Photodiode (APD)
Electrode
Iphoto
SiO2
h > Eg
n+ p
e – h+

 Impact ionization processes
resulting avalanche multiplication
R
E
p+
h+
e–
E
rnet
Electrode
n+

p
Avalanche region
e-
x
Ec
Ev
E(x)
h+
x
Absorption
region
Avalanche region
 Impact of an energetic electron's kinetic
energy excites VB electron to the CV.
Photodetectors
Avalanche Photodiode (APD)
 Schematic diagram of typical Si APD.
Electrode
SiO2
Antireflection coating
n+
p
Guard ring

Avalanche breakdown
n
n+
p

p+
p+
Substrate
Substrate
Electrode
Electrode
Si APD structure without a
guard ring
n
More practical Si APD
 Breakdown voltage around periphery is higher and avalanche is
confined more to illuminated region (n+p junction).
Comparisons of APDs
NOTE: The values were derived from various vendor data sheets
and from performance numbers reported in the literature. They
are guidelines for comparison purposes. Detailed values on
specific devices for particular applications can be obtained from
photodetector and receiver module suppliers.
Photodetectors
Heterojunction Photodiode
Separate Absorption and Multiplication (SAM) APD
InGaAs-InP heterostructure Separate Absorption and Multiplication APD
Ip h
Electrode
InP
InP
E
h
Vr
R
InGaAs
e–
h+
E
P+
P and N refer to p- and
n-type wider-bandgap
semiconductor.
E (x)
Avalanche
region
N
n
n+
Absorption
region
x
Vout
Photodetectors
Heterojunction Photodiode
Separate Absorption and Multiplication (SAM) APD
E
Ec
InP
Ev
e–
Ec
DEv
(a) Energy band diagram for a SAM
heterojunction APD where there is
a valence band step DEv from
InGaAs to InP that slows hole
entry into the InP layer.
InGaAs
h+
Ev
InP
Ev
InGaAsP grading layer
InGaAs
h+
Ev
(b) An interposing grading layer
(InGaAsP) with an intermediate
bandgap breaks DEv and makes it
easier for the hole to pass to the InP
layer.
Photogenerated electron concentration
exp( -  x) at time t = 0
v
de
x
A
h > E
B
W
E
g
h+
e–
i
V
r
ph
R