Optical Communications
Part 02-Lecture 01
ECE 4 2 3
OPT I CA L COMMU N I CATIONS
DR . S HE R IF HE KA L
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Optical fiber Is a dielectric channel used to guide light by total internal
reflection
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What is optical Fiber?
An optical fiber is a flexible, transparent fiber made by drawing glass
(silica) or plastic to a diameter slightly thicker than that of a human
hair.
Optical fibers are used most often as a means to transmit light
between the two ends of the fiber and find wide usage in fiber-optic
communications
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Optical communication system
Advantages of optical fibers:
Disadvantages of optical fibers:
 Smaller and lighter than other types of cables
 High cost
 Very high tensile strength
 Needs more expensive TX & RX
 Optical fiber has a bandwidth capability of
400MHz/km or greater.
 More difficult and expensive to
splice than wire
 OF
can
provide
data
transmission
performance up to 10Gbps, 40Gbps and even
100Gbps with new hardware that is now
available.
 Lower losses.
 Greater repeater spacing.
 OF cables are immune to electromagnetic
interference.
 Secure Transmissions.
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Optical communication system
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Optical communication system
Input sensor: Converts message into electric signal (baseband signal ), microphone
Modulator: Converts baseband signal into format appropriate for channel transmission ; impresses the signal onto the
optical carrier
Optical carrier: Generates the light-wave onto which the information is carried (laser diode (LD)
or light emitting diode (LED)). The carrier source is intensity modulated which means that as the
input current changes the output optical power changes in the same way
Coupler : Couples light from the source to the fiber channel
Channel: Glass or plastic fibers that are the transmission medium
Receiver coupler : Transfers the optical power from the fiber to the photo-detector
Photo-detector (o/p sensor): Converts optical power to electrical current, ideally the current
should be a replica of the current used to modulate the light source
Message destination : devices such as speakers, video monitors and computers
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Photodetectors
A photoconductor is a semiconductor which exhibits a change in conductance
(resistance) when radiant energy is incident on it.
Consider an n-type S.C. where Po is the incident
Light
radiation power. Then the transmitted power after
penetrating a thickness D is (Po
e-αD).
The absorbed power (“useful”) is
𝑷𝒐 𝟏 − 𝒆−𝜶𝑫
The number of photons absorbed per sec
𝑁𝑝ℎ
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𝑃𝑜 1 − 𝑒 −𝛼𝐷
𝑃𝑜 1 − 𝑒 −𝛼𝐷 λ
=
=
ℎ𝑓
ℎ𝑐
D
W
L
𝐼𝑝ℎ
V
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Photodetectors
Assuming unity internal quantum efficiency, then each of these absorbed photons will generate an
electron-hole pair(EHP) per sec. If Gph is the photogeneration rate of EHP per unit volume then
𝑁𝑝ℎ 𝑃𝑜 1 − 𝑒 −𝛼𝐷 λ
𝐺𝑝ℎ =
=
𝑉𝑜𝑙
ℎ𝑐 × 𝐿𝑊𝐷
In steady state conditions, i.e. constant illumination from the laser, we will reach steady state
when there is no net change in the excess carrier concentration:
𝑑∆𝑝𝑛
∆𝑝𝑛
= 𝐺𝑝ℎ −
= 0 ⇒ ∆𝑝𝑛 = 𝜏𝑟 𝐺𝑝ℎ
𝑑𝑡
𝜏𝑟
Assuming that the injection is high ⇒ 𝜏𝑟 = 1/𝛼𝑟 ∆𝑝𝑛 , thus:
∆𝑝𝑛 = 𝜏𝑟 𝐺𝑝ℎ
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𝐺𝑝ℎ
=
⇒ ∴ ∆𝑝𝑛 =
𝛼𝑟 ∆𝑝𝑛
𝐺𝑝ℎ
=
𝛼𝑟
𝑃𝑜 1 − 𝑒 −𝛼𝐷 λ
𝛼𝑟 × ℎ𝑐 × 𝐿𝑊𝐷
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Photodetectors
The photoconductivity is the increase in the conductivity ∆𝜎 due to the absorbed light:
∆𝜎 = 𝑞𝜇𝑒 ∆𝑛𝑛 + 𝑞𝜇ℎ ∆𝑝𝑛
∆𝜎 = 𝑞∆𝑝𝑛 𝜇𝑒 + 𝜇ℎ = 𝑞 𝜇𝑒 + 𝜇ℎ
𝑃𝑜 1 − 𝑒 −𝛼𝐷 λ
𝛼𝑟 × ℎ𝑐 × 𝐿𝑊𝐷
We can express the increase in conductivity in terms of the light intensity (power density W/m2)
𝐼𝑜 = 𝑃𝑜 /(𝑊𝐿)
∆𝜎 = 𝑞 𝜇𝑒 + 𝜇ℎ
𝐼𝑜 1 − 𝑒 −𝛼𝐷 λ
𝛼𝑟 × ℎ𝑐 × 𝐷
⇒ ∆𝜎 ∝
𝐼𝑜
𝛼 is the absorption coefficient
𝛼𝑟 is the recombination coefficient.
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Photodetectors
The change in conductance ∆𝐺can be given by
𝐴∆𝜎 𝑊𝐷∆𝜎
∆𝐺 =
=
𝐿
𝐿
∴ 𝑃ℎ𝑜𝑡𝑜𝑐𝑢𝑟𝑟𝑒𝑛𝑡 = 𝐼𝑝ℎ𝑜𝑡𝑜 = ∆𝐼 = 𝑉∆𝐺
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Photodetectors
Example: a photoconductive film of n-type GaAs doped with 1013 cm-3 donors, L = 2 mm,
W = 1 mm, D = 5 μm. The sample has electrodes attached to its ends which are
connected to 1 V supply. The GaAs photoconductor is uniformly illuminated over the
surface area (2 mm x 1 mm) with a 1 mW laser radiation of wavelength λ = 800 nm. 𝛼𝑟
for GaAs is 7.21x10-16 m-3s-1. 𝛼 is 5x103 cm-1. 𝜇𝑒 = 8000 𝑐𝑚2 /Vs and 𝜇ℎ = 400 𝑐𝑚2 /Vs.
the energy gap is Eg = 1.43eV
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Photodetectors
Solution: substituting in the above formulas, we get:
𝑁𝑝ℎ = 3.69 × 1015 photons per second
𝐺𝑝ℎ = 3.69 × 1026 𝑚−3 𝑠 −1
∆𝑝𝑛 = 7.15 × 1020 𝑚−3
∆𝜎 = 96 Ω/𝑚
∆𝐺 = 2.48 × 10−4 Ω−1
𝐼𝑝ℎ𝑜𝑡𝑜 = 0.24 𝑚𝐴
𝑃𝑒𝑙𝑒𝑐𝑡𝑟𝑐𝑖𝑎𝑙 = 0.24 𝑚𝑊
𝑃𝑡ℎ𝑒𝑟𝑚𝑎𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 = 𝑁𝑝ℎ 𝐸𝑝ℎ − 𝐸𝑔 = 0.075 𝑚𝑊
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pin Photodetector
w
The high electric field present in the depletion region causes photo-generated carriers to
Separate and be collected across the reverse –biased junction. This give rise to a current
Flow in an external circuit, known as photocurrent.
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Energy-Band diagram for a pin photodiode
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Photocurrent
Optical power absorbed,
power,P(x):
P0 in the depletion region can be written in terms of incident optical
P( x)  P0 (1  e  s (  ) x )
Absorption coefficient  s ( ) strongly depends on wavelength. The upper wavelength cutoff for
any semiconductor can be determined by its energy gap as follows:
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c ( m) 
E g (eV)
Taking entrance face reflectivity into consideration, the absorbed power in the width of depletion
region, w, becomes:
(1  R f ) P(w)  P0 (1  e  s ( ) w )(1  R f )
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Responsivity
The primary photocurrent resulting from absorption is:
q
Ip 
P0 (1  e  s (  ) w )(1  R f )
h
Quantum Efficiency:
Responsivity:
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# of electron - hole photogenerated pairs

# of incident photons
I /q
 P
P0 / h
IP
q


P0
h
[A/W]
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Construction of optical fiber cable
An optical fiber (American spelling) or fibre (British spelling): is cylindrical dielectric waveguide that
transmits light along its axis, by the process of total internal reflection
 The fiber consists of a denser core surrounded by a
cladding layer
 For total internal reflection to confine the optical
signal in the core, the refractive index of the core
must be greater than that of the cladding
 The boundary between the core and cladding may
either be abrupt, in step-index fiber, or gradual, in
graded index fiber
 The fiber is encased in an insulating jacket which
protects it from moisture and provide some
mechanical strength
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What is optical Fiber?
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Construction of optical fiber cable
Jacket:
◦ Determines the mechanical robustness of the cable.
◦ Usually plastic, PVC etc.
Buffer (or coating):
◦ Protects the fiber optic from physical damage. Fiber identification etc.
Cladding:
◦ Dielectric material, with index of refraction less than the core material. Made of glass or
plastic.
Core:
◦ Light propagation medium. Dielectric material (usually glass). Conducts no electricity.
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Total internal reflection (TIR)
The light which is transmitted usually changes direction
when it enters the second material.
This bending of light is called refraction and it depends
upon the fact that light travels at one speed in one
material and at a different speed in a different material
𝑛=
𝑐
1/ 𝜇𝑜 𝜀𝑜
=
𝑣 1/ 𝜇𝑜 𝜇𝑟 𝜀𝑜 𝜀𝑟
𝑛 = 𝜇𝑟 𝜀𝑟
where
n is the refractive index
c is the speed of light in a vacuum
v is the speed of light in the material
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Total internal reflection (TIR)
Critical angle
Form Snell’s law n1sinθ1 = n2sinθ2
Since n1>n2 then sinθ2 =(n1/n2)sinθ1
θ2 > θ1
when θ1 = θc ; the critical angle at which TIR occurs
then: n1sinθc = n2sin90=n2
hence sin θc = n2/n1
if θ1 > θc then the ray is completely reflected in medium 1.
Hence conditions for light ray propagating through
an optical fiber core are :
1. the core index is dense and greater than the cladding index
2. incident angle of ray with normal to the core- cladding
interface θ1 > θc
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Total internal reflection (TIR)
Acceptance angle is the maximum angle θimax which gives just total internal
reflection at the core gladding interface
When θi = θimax  θ = θc
When θi < θimax  θ > θc  The ray propagate through TIR
When θi > θimax  θ < θc  The ray penetrates the cladding and lost
𝑛𝑎 = 1
θimax =
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Total internal reflection (TIR)
𝑛𝑎 sin 𝜃𝑖 = 𝑛1 sin 𝜃𝑡
sin 𝜃𝑖 = 𝑛1 sin 𝜃𝑡
The minimum value of 𝜃 gives reflection is
𝜃𝑐 which gives maximum value of 𝜃𝑖
sin 𝜃𝑖𝑚𝑎𝑥 = 𝑛1 cos 𝜃𝑐 = 𝑛1 1 − sin2 𝜃𝑐
2
= 𝑛1
1 − 𝑛2 /𝑛1
= 𝑛1
(𝑛1 −𝑛2 )(𝑛1 +𝑛2 )/ 𝑛1
≈ 𝑛1
(𝑛1 −𝑛2 )(2𝑛1 )/ 𝑛1
≈ 𝑛1
2(𝑛1 − 𝑛2 )/𝑛1 = 𝑛1 2∆
∆= (𝑛1 − 𝑛2 )/𝑛1
2
2
the refractive index fractional difference
Or
The relative refractive index difference
Numerical aperture: measure the light collecting ability of the fiber
NA = 𝑛𝑎 sin 𝜃𝑖𝑚𝑎𝑥
NA = 𝑛1 2∆=
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(𝑛12 − 𝑛22 )
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Total internal reflection (TIR)
Ex1: Derive an expression for the maximum value of n2 as a function of n1 that permits all light
incident on the end face of a fiber to be propagated. Assume na = 1 and that the end face is
perpendicular to the fiber axis. Calculate this limiting value of n2 when n1 = 1.46
Solution
NA = 𝑛𝑎 sin 𝜃𝑖𝑚𝑎𝑥 =
(𝑛12 − 𝑛22 )
Under this condition 𝜃𝑖𝑚𝑎𝑥 = 90  sin 𝜃𝑖𝑚𝑎𝑥 = 1  (𝑛12 − 𝑛22 ) = 1
𝑛2 =
(𝑛12 − 1)
If n1 = 1.46 then n2 = 1.0638
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