Transcript ocn 2007 lecture 19

Digital Signal Transmission
Elec.
Input
Pulses
LED/LD
Trans.
Optical
PIN PD /
Amplifier Ampl.
APD
Optical pulses
Optical pulses
( att. & dist.)
Decision Ckt.
Amplifier
& pulse regen.
& Filter
Regenerated
Voltage pulses
o/p pulses
& ampl. noise
Signal Proc.
equipment
Current pulses
with noise
Signal path through an optical data link
Noise Sources
Photon
stream
h
PD
Amplifier
•Bulk dark
current
•Photon
•Surface leakage
detection
Quantum noise current
(Poisson func.) •Statistical gain
fluctuation
•Beat noise
Equaliser
•Thermal noise
of load resistor
•Amplifier noise
vout(t)
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)
V
1 level
p(0/1) = ƒp1(y) dy
p(1/0)
Threshold
-
V
p(0/1)
0 level
p0(y)
-
Pe = p(1) p(0/1)
+ p(0) p(1/0)
Error probability
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
Error
Detection
Optical
detector
h
Fiber
Low noise
pre-amplifier
& PD bias
control
Amplifier,
AGC &
Equalizer
Threshold
detection
&
regeneration
Alarm
Optical
Source
Drive Circuit
for optical
Source
h
Fiber
Timing
Extraction
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
Ra
Ca
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
h
Rb






Ra
Ca
A
Equ.
RL = Rb || Ra
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
Rf
h
_
Rb
Ra
Ca
Vin
-G
+
Vout
RL = Rb || Ra
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
Rf
h
_
Rb






Ra
Ca
Vin
-G
+
Vout
RL = Rb || Ra
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
Maximum signal
Voltage (V2)
Best sampling time
Slope gives sensitivity
to timing errors
Distortion at
sampling times
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



Rise-time 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
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
1-10 m
<10 Kb/s
10-100 Kb/s
100-1000 Kb/s
1-10 Mb/s
10-50 Mb/s
50-500 Mb/s
500-1000 Mb/s
>1 Gb/s
I Region:
10-100 m 100-1000 m 1-3 km
3-10 km
10-50 km 50-100 km >100 km
VII
V
I
V
II
III
VI
IV
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 ELED 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
step index,
graded
index,Gb/s
MMF: multimode LD
fiber, SMF:
fiber
VI Region:SI:100
Mb/sGI:
 BL
 100
with single mode
SMF
VII Region: 5 Mb/s  BL  100 Mb/s
LD with SI or GI MMF
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
Rise Time Budget





tsys
N
= (  ti 2 )
i=1
½
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
1.
2.
3.
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, =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
R
R
 (μm)
1.24
 (μm)
1.24
0.7  0.85

 0.48 A/W
1.24
0.4 1.6

 0.52 A/W
1.24
R 1.24 0.6 1.24


 57%
 (μm)
1.3
APD – Multiplication factor
q
 q
0.65  0.9
6
Ip 
P0 
P0 
0.5 10  0.235 μA
h
hc
1.24
IM
10
M

 43
I p 0.235
Photon incident rate
I
= M IP = M  Pin=
Photon rate= Pin =
I
h c/ λ
Mηe
M η e λ Pin = 2 nA
hc
=1.736 x 107 s-1
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
=
J.s ) ( 3 x 10 8 m/s ) (10 x 10 6 bps )
2 ( 0.85 x 10-6 m )
- 76.2 dBm
–34
Quantum limit
N
e
P ( n)  N n
; P(0)  e  N  109  N  9 ln 10  20.7  21 photons
n!
E  20.7
h

 20.7
hc

 20.7
1.24
 20.7 eV
0.8 1.55
B
P0  E  20.7 1.6 10 19  311 106  1 nW
2
Photodiode – noise
iQ  (2qI p B)1/ 2 ; I p  R  P0 
 
1.24
P0  1.19 A/W  500 nW  0.6 μA
iQ  (2 1.6 1019  0.6 106 150 106 )1/ 2  5.4 nA
iDB  (2qI D B)1/ 2  (2 1.6 1019 1109 150 106 )1/ 2  0.22 nA
iDS  (2qI S B)1/ 2  0
 4kT
iT  
 RL
1/ 2

B 

1/ 2
 4 1.38 10  293
6
 
150 10 
500


 23
 70 nA
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 =

BW of HIA



(4 M x 4 M) / 8 M = 2 M
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