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Digital Signal Transmission
LED/LDTrans.
OpticalAmplifier
PIN PD /APD
Amplifier& Filter
Decision Ckt.& pulse regen.
Signal Proc.equipment
Elec. InputPulses
Optical pulses( att. & dist.)
Ampl. Optical pulses
Current pulseswith noiseVoltage pulses
& ampl. noiseRegenerated o/p pulses
Signal path through an optical data link
Noise Sources
PD Amplifier EqualiserPhoton stream h
•Photon detectionQuantum noise(Poisson func.)
•Bulk dark current•Surface leakage current•Statistical gain fluctuation•Beat noise
•Thermal noise of load resistor•Amplifier noise
vout(t)
vout(t) = A M o P(t) * hB(t)* heq(t)
Error ProbabilityBER = Ne/Nt
vout(t) = A M o P(t) * hB(t)* heq(t)
Error in detection Noises , ISI, Non-zero extinction
1 level
0 level
V p(1/0)
p(0/1)
p1(y)
p0(y)
Threshold
-
p(1/0) = ƒp0(y) dy
V
p(0/1) = ƒp1(y) dy-
V
Pe = p(1) p(0/1) + p(0) p(1/0)
Error probabilityNoise variance mean square fluctuation of
vout(t)
about its mean < vout(t) > GA
p0(y) = (1/2off) exp [ -(vout – boff )2/2off2 ]
p1(y) = (1/2on) exp [ -(bon – vout )2/2on2 ]
BER = Pe = (1/2) [ 1- erf ( Q/2 ) ]
Q = (Vth – boff )/off = (bon – Vth )/on
Pe = 10-9 Q = 6
Pe = ? If boff = 0; bon= V; off = on =
Regenerative repeater
h
Fiber
hLow noise
pre-amplifier& PD bias
control
Amplifier,AGC &
Equalizer
Threshold detection
& regeneration
TimingExtraction
ErrorDetection
Alarm
Drive Circuitfor optical
Source
Fiber
Error in Regeneration Insufficient SNR at decision instant ISI due to dispersion Time and phase jitter EYE PATTERN ANALYSIS
Optical detector
Optical Source
Pre-amplifiers
Preceeds the equalizer
Maximize sensitivity by minimising noise maintaining suitable bandwidth
Possible receiver structuresLow impedance front endHigh impedance front endTransimpedance front end
Low impedance pre-amplifier
PD operates into a low-impedance amplifier ( i.e. 50 ) Bias / load resistance used to match amplifier i/p impedance Bandwidth maximized due to lesser RL
Receiver sensitivity is less due to large thermal noise Suitable only for short-distance applications
h
Rb CaRa
A
RL = Rb || Ra
High impedance pre-amplifier
PD operates into a high-impedance amplifier Bias / load resistance matched to amplifier i/p impedance Receiver sensitivity is increased due to lesser thermal noise Degraded frequency performance Equalization is a must Limited dynamic range
h
Rb CaRa
A
RL = Rb || Ra
Equ.
Trans-impedance pre-amplifier
HOL() = -G Vin / Idet = -G { RL || ( 1/ j CT) } V/A
HCL() Rf / { 1 + (j RfCT / G) } V/A
B G / (2RfCT )
RL = Rb || Ra
h
Rb
Ca
Ra
-G
_
+
Rf
VoutVin
Trans-impedance pre-amplifier
PD operates into a low-noise high-impedance amplifier with (-)ve FB Current mode amplifier, high i/p impedance reduced by NFB Receiver sensitivity is increased due to lesser thermal noise Improved frequency performance Equalization required depending on bit rate Improved dynamic range
RL = Rb || Ra
h
Rb
Ca
Ra
-G
_
+
Rf
VoutVin
Simplified Eye PatternBest sampling time
Distortion atsampling times
Maximum signalVoltage (V2)
Slope gives sensitivity to timing errors
Threshold
Noise margin (V1)
Distortion at zerocrossings (T)
Time interval over whichsignal can be sampled
Eye Pattern Analysis Width of the eye timing interval over which the
signal can be sampled Height of the eye eye closure indicates best
sampling time Noise Margin (%) = (V1/V2 ) x 100 Slope of eye pattern sides timing error sensitivity Timing Jitter (%) ( T/Tb ) x 100 Rise time & fall time measurements Non-linearities in channel characteristics
asymmetric eye pattern
Design of Point-to-Point Links
To determine repeater spacing one should calculate Power budget
Optical power loss due to junctions, connectors and fiber One should be able to estimate required margins with respect of
temperature, aging and stability Rise-time budget
For rise-time budget one should take into account all the rise times in the link (tx, fiber, rx)
If the link does not fit into specifications
more repeaters change components change specifications
Several design iterations possible
Link calculationsSpecifications: transmission distance, data rate (BW), BER
Objectives is then to selectFIBER:
Multimode or Single mode fiber: core size, refractive index profile, bandwidth or dispersion, attenuation, numerical aperture or mode-field diameter
SOURCE:LED or Laser diode optical source: emission wavelength, spectral line width, output power, effective radiating area, emission pattern, number of emitting modes
DETECTOR/RECEIVER:PIN or Avalanche photodiode: responsivity, operating
wavelength,rise time, sensitivity
Bitrate (BW) - Transmission distance product
1-10 m 10-100 m 100-1000 m 1-3 km 3-10 km 10-50 km 50-100 km >100 km<10 Kb/s10-100 Kb/s100-1000 Kb/s1-10 Mb/s10-50 Mb/s50-500 Mb/s500-1000 Mb/s>1 Gb/s
I
II
III IV
V
V
VI
VII
I Region: BL 100 Mb/s SLED with SI MMF
II Region: 100 Mb/s BL 5 Gb/s LED or LD with SI or GI MMF
III Region: BL 100 Mb/s ELE
D or LD with SI MMF
IV Region: 5 Mb/s BL 4 Gb/s ELED or LD with GI MMF
V Region: 10 Mb/s BL 1 Gb/s LD with GI MMF
VI Region: 100 Mb/s BL
100 Gb/s LD with SMF
VII Region: 5 Mb/s BL 100 Mb/s LD with SI or GI MMF
SI: step index, GI: graded index, MMF: multimode fiber, SMF: single mode fiber
Link Power Budget
PS - 2LC - f L - nLs – mLm – A SR /
where:
PS = PowerEmitted from source LC = Coupling loss
L = Distance (total fiber length) f = attn. per unit length
N = no. of splices Ls = splice loss
M = no. of connectors Lm = connector loss
A = System margin = Detector quamtum efficiency
SR = minimum Receiver sensitivity
N
Rise Time Budget tsys = ( ti2 ) ½
i=1 Transmitter rise time ttx
GVD rise time tGVD |D| L
Modal dispersion rise time tmod (ns) 440Lq/Bo (MHz) q mode mixing factor
q = 0.5 steady state modal mixing q = 1 no modal mixing q = 0.7 practical estimate
Bo 3 dB optical bandwidth of 1Km length of fiber Receiver rise time trx (ns) 350 / Brx (MHz) tsys
< 70 % of Tb for NRZ < 35 % of Tb for RZ ( In general 70 % of Ton )
Photodiode - responsivity1. A Si pin photodiode has a quantum
efficiency of 70% at a wavelength of 0.85 m.Calculate its responsivity.
2. Calculate the responsivity of a Germanium diode at 1.6 m where its quantum efficiency is 40%.
3. A particular photodetector has a responsivity of 0.6 A/W for light of wavelength 1.3 m.Calculate its quantum efficiency.
APD – Multiplication factor
A given silicon avalanche photodiode has a quantum efficiency of 65% at a wavelength of 900 nm. Suppose 0.5 W of optical power produces a multiplied photocurrent of 10 A.
What is the multiplication factor?
Photon incident rate A Silicon RAPD,operating at a wavelength
of 0.80 μm, exhibits a quantum efficiency of 90%, a multiplication factor of 800, and a dark current of 2 nA. Calculate the rate at which photons should be incident on the device so that the output current after avalanche gain is greater than the dark current.
Quantum Limit A digital fiber optic link operating at 850
nm requires a maximum BER of 10-9. Determine the quantum limit considering unity quantum efficiency of the detector. Find the energy of the incident photons. Also determine the minimum incident optical power that must fall on the photodetector to achieve a BER of 10-9 at a data rate of 10 Mbps for a simple binary –level signaling scheme.
Quantum limit
Consider a digital fiber optic link operating at a bitrate of 622 Mbit/s at 1550 nm. The InGaAs pin detector has a quantum efficiency of 0.8.
1. Find the minimum number of photons in a pulse required for a BER of 10-9.2. Find the corresponding minimum incident power.
Photodiode - noise
An InGaAs pin photodiode has the following parameters at 1550 nm: ID=1.0 nA, =0.95, RL=500 , and the surface leakage current is negligible. The incident optical power is 500 nW (-33 dBm) and the receiver bandwidth is 150 MHz.Compare the noise currents. What BER can be expected with this diode?
APD – Optimum Gain A good silicon APD ( x=0.3) has a
capacitance of 5 pF, negligible dark current and is operating with a post detection bandwidth of 50 MHz. When the photocurrent before gain is 10-7 A and the temperature is 18oC; determine the maximum SNR improvement between M=1 and M=Mop assuming all operating conditions are maintained.
Pre-amplifiers Example A high i/p impedance amplifier which is employed in an
optical fiber receiver has an effective input resistance of 4 M which is matched to a detector bias resistor of the same value. Determine:
The maximum bandwidth that may be obtained without equalization if the total capacitance CT is 6 pF.
The mean square thermal noise current per unit bandwidth generated by this high input impedance amplifier configuration when it is operating at a temperature of 300 K.
Compare the values calculated with those obtained when the high input impedance amplifier is replaced by a transimpedance amplifier with a 100 K feedback resistor and an open loop gain of 400. It may be assumed that Rf << RT, and that the total capacitance remains 6 pF.
Photodiode - responsivity
A/W48.024.1
85.07.0
24.1
)μm(
R
A/W52.024.1
6.14.0
24.1
)μm(
R
%573.1
24.16.0
)μm(
24.1
R
APD – Multiplication factor
μA235.0105.024.1
9.065.0 600p
P
hc
qP
h
qI
43235.0
10
p
M I
IM
Photon incident rateI = M IP = M Pin= M η e λ Pin = 2
nA h c
Photon rate= Pin = I =1.736 x 107 s-1
h c/ λ M η e
The photon rate should be greater than 1.736 x 107 s-1 for the multiplied current to
be greater than the dark current.
Quantum Limit __Pr(0) = exp ( -N ’) = 10-9
N ’ = 9 ln 10 = 20.7 = 21 photons needed per pulse
Energy E = No. of photons x hν = 20.7 hc = Poτ
λB / 2 = 1 / τ ; assuming unbiased data
Po = 20.7 ( 6.626 x 10 –34 J.s ) ( 3 x 10 8 m/s ) (10 x 10 6 bps )
2 ( 0.85 x 10-6 m )= - 76.2 dBm
Quantum limit
photons 217.2010ln910)0(;!
)( 9
NePn
eNnP N
Nn
eV 7.2055.18.0
24.17.207.207.20
hch
E
nW 110311106.17.202
6190 B
EP
Photodiode – noiseμA 0.6nW 500A/W19.1
24.1;)2( 00
2/1
PPRIBqIi ppQ
nA 4.5)10150106.0106.12( 2/16619 Qi
nA 22.0)10150101106.12()2( 2/169192/1 BqIi DDB
0)2( 2/1 BqIi SDS
nA 7010150500
2931038.1442/1
6232/1
BR
kTi
LT
APD- Optimum gain Max. value of RL = 1/ 2CdB = 635.5 SNR at M=1 Ip2 / {2eBIp + (4KTB/RL) }
= 7.91 = 8.98 dB Mop
2+x = {2eIL + (4KT/RL) }/{xe(Ip + ID)}
= 41.54 F(M) = Mx ( 0 < x < 1.0) SNR at M=Mop 1.78 x 103 = 32.5 dB SNR Improvement 23.5 dB
Pre-amplifiers Example RT = (4 M x 4 M) / 8 M = 2 M
BW of HIA B = 1/ ( 2 RT CT ) = 1.33 x 104 Hz = 13 KHz Mean square thermal noise current = 4KT / RT = 8.29 x 10 –27 A2 /
Hz
BW of TIA B = G/ ( 2 Rf CT ) = 1.06 x 108 Hz = 106 MHz Mean square thermal noise current = 4KT / Rf = 1.66 x 10 –25 A2 / Hz
Noise power in TIA / Noise power in HIA = 10 log10 20 = 13 dB