Dublin Institute of Technology
School of Electronic and
Communications Engineering
Optical Communications Systems
Optical Receivers
Dr. Yuliya Semenova
Unauthorised usage or reproduction strictly prohibited
Copyright 2003, Dr. Yuliya Semenova, Dublin Institute of Technology
Optical Receiver Block Diagram
The purpose of
the receiver
is:
1) To convert
optical signal
into electrical;
2) Recover the
data.
+
–
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Photodetectors
The most critical component is the photodetector.
photodetector It should
have:
• high sensitivity
• fast response time
• low noise
• size compatible with fibres
• high reliability
This means that semiconductor materials are exclusively
used in lightwave systems. In these, photons are absorbed to
generate electron-hole (e-h) pairs producing a photo-current.
A basic requirement is that the detector material bandgap
energy (Eg) must be smaller than the photon energy (hf).
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Photodetectors (Types)
Most communication systems use reverse-biased p-n junctions
photodiodes) of two main types:
p-i-n photodiodes
No internal gain
Low bias voltage [10-50 V @λ
λ = 850 nm, 5-15 V @ λ = 1300 –1550 nm]
Highly linear
Low dark current
Most widely used
Avalanche photodiodes (APD)
Internal gain (increased sensitivity)
Best for high speed and highly sensitive receivers
Strong temperature dependence
High bias voltage[250 V @ λ = 850 nm, 20-30 V @ λ = 1300 –1550 nm]
Costly
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Optical System – Noise Sources
Source Noise
Modal noise – Due to interaction of (constructive &
destructive) multiple coherent modes,
resulting in intensity modulation.
Photodetector Noise √
Preamplifier (receiver) Noise √
Distortion due to Non-linearity
Crosstalk and Reflection in the Couplers
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Receiver Noise
Noise is a term generally used to refer to any spurious or
undesired disturbances that mask the received signal in a
communication system.
In optical communication system we are generally concerned with
noise due to spontaneous fluctuations rather than erratic
disturbances which may be a feature of copper-based systems.
There are three main types of noise due to spontaneous
fluctuations in optical communication systems:
thermal noise;
dark current noise;
quantum noise.
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Thermal Noise
This is spontaneous fluctuation due to thermal interaction
between, say, the free electrons and the vibrating ions in a
conducting medium, and it is especially prevalent in resistors at
room temperature.
This thermal noise current it in a resistor R:
it2 =
4 KTBe
R
(1)
where K is Boltzman’s constant, T is the absolute temperature and
Be is the electrical bandwidth of the system (assuming the resistor
is in the optical receiver).
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Dark Current Noise
When there is no optical power incident on the photodetector a
small reverse leakage current still flows from the device terminals.
This dark current contributes to the total system noise and gives
random fluctuations about the average particle flow of the
photocurrent. It manifests itself as shot a noise on the
photocurrent.
The dark current noise id2 is given by:
id2 = 2eBe I d
(2)
where e is the charge on an electron and Id is the dark current.
It may be reduced by careful design and fabrication of the
detector.
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Quantum Noise
The detection of light by a photodiode is a discrete process, and
the signal emerging from the detector is dictated by the statistics
of photon arrivals.
The average number of electron-hole pairs per bit:
P0ητ
η τ
N=
P(t )dt =
∫
hf 0
hf
(3)
It is found that the probability P(n) of detecting n photons in time
period τ when it is expected on average to detect N photons obeys
the Poisson distribution:
n
N exp(− N )
(4)
P ( n) =
n!
The electron rate re generated by incident photons is:
re = ηP0 / hf
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Quantum Noise
The number of electrons
generated in time τ is equal to
the
average
number
of
photons detected over this
time period. Therefore:
ηP0τ
N=
hf
P (n)
Standard deviation:
σ= N
(5)
N
n
Incoherent light is emitted by independent atoms and therefore
there is no phase relationship between the emitted photons.
This property dictates exponential intensity distribution for
incoherent light:
n
N
(6)
P (n) =
n +1
(1 + N )
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Probability Distributions
P(n)
Poisson distribution
(coherent light)
N=1000
10-1
n
5x10-2
P ( n) =
0
0
500
1000
1500
N exp(− N )
n!
n
P(n)
Statistical fluctuations
(incoherent light)
N=1000
10-1
n
5x10-2
0
500
1000
1500
N
P ( n) =
(1 + N ) n +1
n
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Digital Signalling Quantum Noise
1.5
1
1
0
1
0
0
1
1
0
1
1
Decision times
1
Threshold
level
0.5
0
-0.5
0
1
2
3
4
5
6
7
8
9
10
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Digital Signalling Quantum Noise
An ideal receiver has a sufficiently low amplifier noise to detect
the displacement current of a single electron-hole pair generated
within the detector (i.e. an individual photon may be detected).
Thus in the absence of light, and neglecting dark current, no
current will flow. Therefore the only way an error can occur is if a
light pulse is present and no electron-hole pairs are generated.
The probability of no pairs being generated when a light pulse is
present (the system error probability) is given by (7).
An absolute receiver sensitivity allows the
determination of a fundamental limit in
digital optical communications.
P (0 / 1) =
= P (e) = exp( − N )
(7)
This is the minimum pulse energy required
to maintain a given bit error rate (BER)
which any practical receiver must satisfy
and is known as the quantum limit.
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Example
A digital optical fibre communication system operating at a wavelength of 1 µm requires a
maximum it error rate of 10-9. Determine:
a)
The theoretical quantum limit at the receiver in terms of the quantum efficiency of the
detector and the energy of an incident photon;
b)
The minimum incident optical power required at the detector in order to achieve the
above bit error rate when the system is employing ideal binary signalling at 10 Mbits/sec,
and assuming the detector is ideal.
SOLUTION: a)
From (7) the probability of error
P(e) = exp(− N ) = 10 −9
and thus N=20.7.
N corresponds to an average number of photons detected in a time period τ for a BER of 10-9.
From (5):
ηP0τ
N=
= 20.7
hf
Hence the minimum pulse energy of quantum limit
E min = P0τ =
20.7hf
η
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Example
b) From part (a) the minimum pulse energy:
P0τ =
20.7 hf
η
Therefore the average received optical power required the minimum pulse energy is:
P0 =
20.7hf
τη
However, for ideal binary signalling there are an equal number of ones and zeros. Thus the
averege received optical power may be considered to arrive over two bits periods, and
P0 ( binary ) =
20.7hf 20.7 hfBT
=
2τη
2η
where BT is a bit rate. At a wavelength of 1 µm, f = 2.998x1014 and assuming an ideal detector, η
= 1. Hence
P0(binary )
20.7 × 6.626 ×10 −34 × 2.998 ×1014 ×107
=
= 20.6 pW
2
P0 = 10 log 10 2.06 × 10 −8 = 3.14 − 80 = −79.9dBm
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A Note on dB and dBm
dB
• optical signals:
P
10 log 1 
 P2 
• electrical signals:
 V1 
 I1 
 V1 I1 

20 log  = 20 log  = 10 log
 V2 
 I2 
 V2 I 2 
•
Popt ∝ I el ∝ Pel
→ electrical dB = 2 x optical dB
dBm
• absolute power value (with 1 mW as reference)
• power level in dBm: 10 log
P 

 1mW 
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Analog Transmission Quantum
Noise
In analog optical fibre systems quantum noise manifests itself as a shot
noise which also has Poisson statistics. The shot noise current is:
2
i s = 2eBe I p
Neglecting other sources of noise the SNR at the receiver:
I p2
S
=
N
i s2
2
Substituting for i s
from Eq.(9) gives:
The photoreceiver current is:
Ip
S
=
N 2eBe
ηP0 e
I p = re e =
hf
S
ηP0e
ηP0
=
=
N hf 2eBe 2hfBe
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(8)
(9)
(10)
(11)
(12)
17
Example
An analog optical fibre system operating at a wavelength of 1 µm has a post detection
bandwidth of 5 MHz. Assuming an ideal detector and considering only quantum noise on
the signal, calculate the incident optical power necessary to achieve an SNR of 50 dB at
the receiver.
SOLUTION:
 S  2hfBe
P0 =  
N η
For S/N = 50 dB, when considering signal and noise powers:
10 log 10
and therefore S/N = 105.
S
= 50
N
At 1 µm, f = 2.998x1014 Hz. For an ideal detector η = 1 and, thus the incident optical power:
In dBm
10 5 × 2 × 6.626 × 10 −34 × 5 × 10 6
P0 =
= 198.6nW
1
P0 = 10 log 10 198.6 × 10 −6 = −40 + 2.98 = −37.0dBm
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Receiver Front-End Noise
(Photodetector + Preamplifier)
Amplifier
Photodetector
Optical
signal
Photodetection
Avalanche
gain
Detector
load bias
Electronic
gain
Electrical
signal
Noise:
•Quantum
•Dark current
Noise:
•Excess due to
random gain
mechanism
•Background
•Beat (from
incoherent carrier)
Noise:
•thermal
Noise:
•Thermal (input
resistance)
•Device (active
elements)
•Surface leakage
currents
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p-n and p-i-n Photodiode Receiver
hf
Cd
RL
Detector
Ra
Ca
Amp
Amplifier
Noise contributions:
Dark current, quantum noise and background
induced photocurrent:
iTS2 = 2eBe ( I p + I d + I b )
it2 =
Thermal noise from the detector load resistor:
4 KTB
R
B
Total noise associated with the amplifier:
S
=
N
i
= ∫ (ia2 + va2 Y )df
2
0
I p2
2 eB e ( I p + I d ) +
2
amp
4 KTB e
2
+ i amp
RL
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Receiver Capacitance and
Bandwidth
Considering the equivalent circuit, the total capacitance for the
front end of an optical receiver CT is given by:
(13)
C =C +C
T
d
a
where Cd is the detector capacitance and Ca is the amplifier input
capacitance.
It is important that this total capacitance is minimised not only to
reduce the noise but also from the bandwidth penalty which is
incurred due to the time constant of CT and the load resistance.
The post detection bandwidth:
1
≥B
2πR L CT
(14)
To increase B is necessary to reduce it, but this introduces a
thermal noise penalty.
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Avalanche Photodiode Receiver
The internal gain mechanism in an APD increases the signal current into the
amplifier and so improves the SNR.
However, the random gain mechanism introduces excess noise into the receiver in
terms of increased shot noise above the total level that would result from
amplifying only the primary shot noise.
Then if the photocurrent is increased by a factor M (mean avalanche multiplication
factor), then the shot noise is also increased by an excess noise factor M:
2
i SA
= 2eB ( I p + I d ) M 2+ x
where x is between 0.3 and 0.5 for silicon APD and between 0.7 and 1.0 for
germanium.
The SNR for the avalanche photodiode:
S
=
N
M 2 I p2
 4 KTBe
2 

2eBe ( I p + I d ) M + 
+ iamp  M − 2

 RL
x
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Optimum Value of M
S
=
N
M 2 I p2
 4 KTBe
2 

+ iamp  M − 2
2eBe ( I p + I d ) M + 
 RL

x
d
( nominator ) = 0
dM
optimal if
2eB( I p + I d ) M opx
 4 KTBe 2  − 2

+ iamp  M
R


L
M op2+ x
=
2
x
 4 KTBe 2 

+ iamp 
RL


=
xeRL ( I p + I d )
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Receiver Structures
idet
it
Cd
RL
Equalizer
iamp
iTS
Ra
Amp
Ca
Vout
Detector and bias
Amplifier
Digital optical receiver equivalent circuit
Three basic amplifier configurations arefrequently used in optical receivers:
•
Low impedance front end
•
High impedance (integrating) front end
•
Transimpedance front end
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Low Impedance Front End
hf
Rb
Ra
Detector and bias
Voltage amplifiers
The modified total resistance:
RTL =
Rb Ra
Rb + Ra
To achieve an optimum bandwidth both Rb and Ra must be minimized.
This leads to a low impedance front end design for the receiver amplifier.
However, this design allows thermal noise to dominate, which may severely limit
its sensitivity.
Therefore the structure demands a trade-off between bandwidth and sensitivity.
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High Impedance Front End
Equalizer
hf
Rb
Detector and bias
Ra
High input impedance
voltage amplifier
This structure tends to give a degraded frequency response as the bandwidth
relationship is not maintained for wideband operation.
The detector output is effectively integrated over a large time constant and must
be restored by differentiation.
This may be performed by the correct equalization at a later stage. Therefore the
high impedance (integrating) front end structure gives a significant
improvement in sensitivity over the low impedance front end design, but it
creates a heavy demand in equalization and has problems of limited dynamic
range.
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Transimpedance Front End
It 2
Equivalent amplifier
voltage noise
Feedback resistor
noise
Rf
Equivalent amplifier
input resistance
V = R f MI
vna2
MI
MI
RTL
CT
-G
Ia2
Vin
Signal
current
Total capacitance
Equivalent amplifier
current noise
Vout
I is the photodiode current. In the
absence of APD M is set to 1.
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Transfer Function
Assuming an amplifier with a gain of –A.
Determine the transfer function by equating
the currents at the amplifier input.
Vin = −
MI +
Thus
V
A
V − Vin
1

= Vin  + j 2πfC 
Rf
R

MI = −

V  1
A
1
+
+
+ j 2πfC 

A  R R f R f

 1
j 2πfC 
1
1

MI = −V
+
+
+
 AR R
AR f
A 
f

V=
− MI
 1
1
1
j 2πfC 

+
+
+
 AR R
AR f
A 
f

V=
− R f MI
 Rf
1 j 2πfGR f

+1+ +
A
A
 AR
Now if A >> 1 and R >> Rf then
V =
− R f MI
j 2πfCR f

1 +
A

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





28
Analysis of Transfer Function
3 dB frequency
Transfer function below
the 3 dB frequency
A
f =
2πCR f
V ≈ − R f MI
As long as the frequency is below the 3 dB frequency, the transfer
function is flat and Rf appears to be in parallel with the photodiode
The intrinsic advantage of the transimpedance front end is that it
is possible to increase Rf, reducing the noise, while keeping the
bandwidth constant by increasing the gain, to offset the effect of
Rf on the bandwidth
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Integrated Front End Example:
Maxim MAX3664
• Low-power transimpedance
preamplifier for 622Mbps DH/SONET
applications
• Nominal 6 kΩ transimpedance
• Differential 100 Ω output
• Paraphrase amplifier converts
single-sided signal to differential
output
• Incorporates noise filter for
photodiode supply
• DC cancellation amplifier removes
DC component, improving dynamic
range
• Sensitivity is at least –28 dBm
(average optical power of 1.6 µW)
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Maxim MAX3664 Performance
622 Mbits/s, approx 2.5 µW average
622 Mbits/s, approx 75 µW average
optical power
optical power
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Maxim MAX3664 Application circuit
Application shows complete SDH 622 Mbits/s receiver
Uses Maxim MAX3664 limiting amplifier and clock recovery IC
Note the use of differential drive and filtering on the photodiode supply
to reduce unwanted PSU noise on input signal
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