1. 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

2. 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)

3. 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) -

4. 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 =  Q = 6
Pe = ? If boff = 0; bon= V; off = on = 

5. Regenerative repeaterh
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

6. 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

7. Low impedance pre-amplifierh 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

8. High impedance pre-amplifierRb 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

9. 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 )

10. 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

11. 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

12. 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

13. 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

14. 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

15. Bitrate (BW) - Transmission distance product
SI: step index, GI: graded index, MMF: multimode fiber, SMF: single mode fiber

16. 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

17. 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 )

18. 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.

19. 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?

20. 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.

21. Quantum Limit
A digital fiber optic link operating at 850 nm requires a maximum BER of 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.

22. 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.

23. 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?

24. 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.

25. 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.

26. Photodiode - responsivity

27. APD – Multiplication factor

28. 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.

29. Quantum Limit __Pr(0) = exp ( -N ’) = 10-9
N ’ = 9 ln 10 = = 21 photons needed per pulse
Energy E = No. of photons x hν = hc = Poτ λ
B / 2 = 1 / τ ; assuming unbiased data
Po = ( x 10 –34 J.s ) ( 3 x 10 8 m/s ) (10 x 10 6 bps )
2 ( 0.85 x 10-6 m )
= dBm

30. Quantum limit

31. Photodiode – noise

32. APD- Optimum gain Max. value of RL = 1/ 2CdB = 635.5 
SNR at M=1  Ip2 / {2eBIp + (4KTB/RL) }
= = 8.98 dB
Mop2+x = {2eIL + (4KT/RL) }/{xe(Ip + ID)}
=  F(M) = Mx ( 0 < x < 1.0)
SNR at M=Mop  1.78 x 103 = 32.5 dB
SNR Improvement  dB

33. 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 = 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 = x 10 –25 A2 / Hz
Noise power in TIA / Noise power in HIA = 10 log = 13 dB