Digital Signal Transmission LED/LD Trans. Optical Amplifier PIN PD / APD Elec. Input Pulses Ampl. Optical pulses Optical pulses ( att. & dist.) Decision Ckt. & pulse regen. Amplifier & Filter Current pulses with noise Regenerated o/p pulses Voltage pulses & ampl. noise Signal Proc. equipment Signal path through an optical data link

Noise Sources vout(t) vout(t) = A M o P(t) * hB(t)* heq(t) PD Amplifier Equaliser Photon stream h vout(t) Thermal noise of load resistor Amplifier noise Bulk dark current Surface leakage Statistical gain fluctuation Beat noise Photon detection Quantum noise (Poisson func.) vout(t) = A M o P(t) * hB(t)* heq(t)

Error Probability BER = Ne/Nt vout(t) = A M o P(t) * hB(t)* heq(t) Error in detection  Noises , ISI, Non-zero extinction  p(1/0) = ƒp0(y) dy  V p1(y) 1 level p(0/1) = ƒp1(y) dy - V p(1/0) Threshold V p(0/1) Pe = p(1) p(0/1) + p(0) p(1/0) 0 level p0(y) -

Error probability p0(y) = (1/2off) exp [ -(vout – boff )2/2off2 ] Noise variance  mean square fluctuation of vout(t) about its mean < vout(t) >  GA p0(y) = (1/2off) exp [ -(vout – boff )2/2off2 ] p1(y) = (1/2on) exp [ -(bon – vout )2/2on2 ] 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 Low noise pre-amplifier & PD bias control Amplifier, AGC & Equalizer Threshold detection & regeneration Timing Extraction Error Detection Alarm Drive Circuit for optical Source Optical detector Optical Source Error in Regeneration  Insufficient SNR at decision instant ISI due to dispersion Time and phase jitter EYE PATTERN ANALYSIS

Pre-amplifiers Preceeds the equalizer Maximize sensitivity by minimising noise maintaining suitable bandwidth Possible receiver structures Low impedance front end High impedance front end Transimpedance front end

Low impedance pre-amplifier h Rb Ca Ra A RL = Rb || Ra 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

High impedance pre-amplifier Rb Ca Ra A RL = Rb || Ra Equ. 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

Trans-impedance pre-amplifier RL = Rb || Ra h Rb Ca Ra -G _ + Rf Vout Vin HOL() = -G Vin / Idet = -G { RL || ( 1/ j  CT) } V/A HCL()  Rf / { 1 + (j RfCT / G) } V/A B  G / (2RfCT )

Trans-impedance pre-amplifier RL = Rb || Ra h Rb Ca Ra -G _ + Rf Vout Vin 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

Simplified Eye Pattern Best sampling time Distortion at sampling times Maximum signal Voltage (V2) Slope gives sensitivity to timing errors Noise margin (V1) Threshold Distortion at zero crossings (T) Time interval over which signal 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 calculations Specifications: transmission distance, data rate (BW), BER Objectives is then to select FIBER: 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 SI: step index, GI: graded index, MMF: multimode fiber, SMF: single mode fiber

PS - 2LC - f L - nLs – mLm – A  SR /  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 - responsivity A Si pin photodiode has a quantum efficiency of 70% at a wavelength of 0.85 mm. Calculate its responsivity. Calculate the responsivity of a Germanium diode at 1.6 mm where its quantum efficiency is 40%. A particular photodetector has a responsivity of 0.6 A/W for light of wavelength 1.3 mm. 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 mW of optical power produces a multiplied photocurrent of 10 mA. 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, h=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

APD – Multiplication factor

Photon incident rate I = 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 λ B / 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

Photodiode – noise

APD- Optimum gain Max. value of RL = 1/ 2CdB = 635.5  SNR at M=1  Ip2 / {2eBIp + (4KTB/RL) } = 7.91 = 8.98 dB Mop2+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