digital signal transmission led/ld trans. optical amplifier pin pd / apd amplifier & filter...

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

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APD – Multiplication factor

μA235.0105.024.1

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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

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nW 110311106.17.202

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EP

Photodiode – noiseμA 0.6nW 500A/W19.1

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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