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6: Link Design
Dr Tahmina [email protected]
CIS049: Microwave and Optical Communication
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Previous Lecture-Photodiodes
Introduction and Basic Requirements
Operation of photodiode Basic structure: a PN junction photodetector, Optical Absorption Process Dark Current, Quantum Efficiency, Responsivity, Absorption Coefficient, Rise Time and
Bandwidth
Types of Photodiodes pn photodiode
p-i-n Photodiode
Avalanche Photodiode
Noise Sources
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In this lecture-Link Design
Power Budget Analysis
Bandwidth or Rise-time Budget Analysis
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Point-to-Point Links
Key system requirements needed to analyze optical fiber links:
1. The desired (or possible) transmission distance
2. The data rate or channel bandwidth
3. The bit-error rate (BER)
(a) Core size(b) Core index profile(c) BW or dispersion(d) Attenuation(e) NA or MFD
MMF or SMFLED or laser pin or APD
(a) Emission wavelength(b) Spectral line width(c) Output power(d) Effective radiating area(e) Emission pattern
(a) Responsivity(b) Operating(c) Speed(d) Sensitivity
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Link Power Budget
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Power Budget Example
Specify a 20-Mb/s data rate and a BER = 109.
With a Si pinphotodiode at 850 nm, the required receiver input signal is42 dBm. Select a GaAlAs LED that couples 50 mW into a 50-m core diameter fiber flylead.
Assume a 1-dB loss occurs at each cable interface and a 6-dB system margin.
The possible transmission distance L = 6 km can be found from
PT= PS PR= 29 dB= 2lc+L +system margin = 2(1 dB) + L + 6 dB
The link power budget can be represented graphically (see the right-hand figure).
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Example1: Spreadsheet Power Budget
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Power Budget
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Cable Length 2.0 2.0
Fiber Type Multimode Singlemode
Wavelength (nm) 850 1300 1310 1550
Fiber Atten. dB/km 3 [3.5] 1 [1.5] 0.4 [0.5] 0.3 [0.5]
Total Fiber Loss 6.0 [7.0] 2.0 [3.0]
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Step 1. Fiber loss at the operating wavelength
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Step 2. Connector Loss
Connector Loss0.3 dB (typical
adhesive/polish conn)0.75 dB (TIA-568 max
acceptable)
Total # of Connectors 5 5
Total Connector Loss 1.5 dB 3.75 dB
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Multimode connectors will have losses of 0.2-0.5 dB typically.
Singlemode connectors, which are factory made and fusion splicedon will have losses of 0.1-0.2 dB. Field terminated singlemodeconnectors may have losses as high as 0.5-1.0 dB. Let's calculateit at both typical and worst case values.Remember that we include all the components in the complete link,including the connectors on each end.
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Step 3. Splice Loss
Typical Splice Loss 0.3 dBTotal # splices 1
Total Splice Loss 0.3 dB
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Multimode splices are usually made with mechanical splices, although somefusion splicing is used. The larger core and multiple layers make fusionsplicing abut the same loss as mechanical splicing, but fusion is more reliablein adverse environments. Figure 0.1-0.5 dB for multimode splices, 0.3 being agood average for an experienced installer. Fusion splicing of singlemode fiber
will typically have less than 0.05 dB (that's right, less than a tenth of a dB!)
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Step 4. Total Passive System Attenuation
Typical TIA 568 Max850 nm 1300 nm 850 nm 1300 nm
Total Fiber Loss (dB) 6.0 2.0 7.0 3.0
Total Connector Loss (dB) 1.5 1.5 3.75 3.75
Total Splice Loss (dB) 0.3 0.3 0.3 0.3
Other (dB) 0 0 0 0
Total Link Loss (dB) 7.8 3.8 11.05 7.05
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Add the fiber loss, connector and splice losses to get the link loss.
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Operating Wavelength (nm) 1300Fiber Type MM
Receiver Sens. (dBm@ required BER) -31
Average Transmitter Output (dBm) -16
Dynamic Range (dB) 15
Recommended Excess Margin (dB) 3
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Equipment Link Loss Budget Calculation: Link lossbudget for network hardware depends on the dynamic
range, the difference between the sensitivity of the receiverand the output of the source into the fiber. You need somemargin for system degradation over time or environment,so subtract that margin (as much as 3dB) to get the lossbudget for the link.
Step 5. Data From Manufacturer's Specification forActive Components (Typical 100 Mb/s link)
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Rise-Time Budget (Bandwidth Budget)
Consider a single pulse:
T
Input:
10%
Output:
t
t
90%
ts
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Non-Return-to-Zero (NRZ) and Return-to Zero (RZ)
t
tp
1 1 1 0 1 0 0 0
T 2T 3T 4T 5T 6T 7T
Power
0
t
tp
Power
1 1 1 0 1 0 0 0
0 T 2T 3T 4T 5T 6T 7T
tp = T/2 pulse width
R = 1/T data rate, b/s
NRZ signal
RZ signal
tp = T pulse width
R = 1/T data rate, b/s
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Spectrum of the (NRZ) Signal:
Frequency
ptTT
11
2
1
PowerSpectralDensity(Watts/Hz)
The required transmission bandwidth is:
22
1 R
TBNRZ
0
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Spectrum of the RZ Signal
Frequency
ptTT
121
PowerSpectralDensity(Watts/Hz)
The required transmission bandwidth is:
RT
BRZ 1
0
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For theNRZ code:t = pulse durationT = Repetition periodR = Bit ratet = T = 1/R
Require the total system rise time to be no more than:
ts = 0.7tThat is, we will allow the total system rise time to be no more than 70% ofthe pulse duration. This is a reasonable restriction.
For theRZ code:t = T/2 = 1/R
Require the total system rise time to be no more than:ts = 0.35t
That is, we will allow the total system rise time to be no more than 35% ofthe pulse duration.
RZ and NRZ rise time constraints
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Rise-Time Budget
A rise-time budget analysisdetermines the dispersion limitation of an optical
fiber link. The total rise time tsysis the root sum square of the rise times from each
contributor tito the pulse rise-time degradation:
The transmitter rise time ttx
The fibre rise time, tf
The group-velocity dispersion (GVD) rise time tGVDof the fibre The modal dispersion rise time tmodof the fibre
The receiver rise time trx
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LightSource
Fiber
Receiver
222
rxftxsys tttt
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Example
Consider the design of an optical link that must transmit 400Mb/s NRZ pulsetrain over a 100-km path with an error rate of 10-9 or better. This must be
accomplished without any repeaters.
For the 400Mb/s NRZ signal, the allowable rise time is
ts = 0.7/4 X 108 = 1.75ns
We will need to choose a suitable fibre, Laser and a Receiver
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Even if the entire system rise time were allotted to the fiber, its pulsespread per unit length would have to be no more than:
D(t/L) = 1.75 ns /100 km = 0.0175 ns/kmD(t/L) = 17.5 ps/km
For multimode SI fibers, and considering only modal distortionD(t/L) ~ 15 ns/km
CIS049: Microwave and OpticalCommunication
tF = Dt
For multimode GRIN fibers, and considering only modal distortion:D(t/L) ~ 1 ns/km
These values are much greater than the 17.5 ps/kmallowed in the rise-time budget.We conclude that these multimode fibers cannot be used.
We will investigate single-mode fibers as a solution.
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For a typical single-mode fiber at 0.8 mm:
D(t/L) ~ 0.5 ns/kmThis limit is mostly caused by the high material dispersion in the firstwindow.We will have to operate either at 1.3 mm or 1.55 mm to achieve smallenough pulse spreading to meet the system specifications.Before continuing with the rise-time (bandwidth) budget, lets look at the
power considerations.
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Compare fiber losses at 1.3 mm and 1.55 mm:At l =1.3 mm, we have loss = 0.5 dB/km
For L = 100 km, the total fiber loss would be:0.5(100) = 50 dB
(We will see later that our power budget is only 37 dB, so the 1.3 mmwavelength will not work.)
At l =1.55 mm, we have loss = 0.25 dB/kmFor L = 100 km, the total fiber loss is:
0.25(100) = 25 dBThis will leave 37 25 = 12 dB for other components.
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Now continue with the rise-time budget calculations. Check the pulsespreading at 1.55 mm.
At 1.55 mm we find:Material Dispersion
M = - 20 ps/(nm x km)Waveguide Dispersion
M9 = 4.5 ps/(nm x km)The total dispersion at 1.55 mm is then:
Mt = M + M9Mt = -20 + 4.5 = -15.5 ps/(nm x km)
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Light SourceTo maintain low dispersion we must choose a light source having a narrow
spectral width. This would have to be a laser diode.Laser diode specifications:
Emission wavelength: l = 1.55 mmMaterial: InGaAsPRise time: tLS = 1 nsSpectral width: Dl = 0.15 nm
(This would be a single longitudinal-mode LD)
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The dispersive pulse spread per unit length is then:IMtI Dl = [-15.5 ps/(nm x km)] x 0.15 nm =2.33 ps/km
And the pulse spread after the signal travels 100 km would be:Dt = L IMtI Dl = 100(2.33) = 233 ps = 0.23 ns
Thus, the fibers rise time is :
tF = Dt = 0.23 nsRecall that the system rise time is limited to:tS = 1.75 ns
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Conclusion:
The single mode fiber and 1.55 mm LD look okay.Could we use an LED in place of the laser diode?Typically for an LED at 1.55 mm the spectral width is on the order of:
Dl = 50 nm
The total pulse spread is then:
Dt = L IMtI Dl = 100(15.5)50Dt = 77,500 ps = 77.5 ns
The fibers rise time is then:tF = Dt = 77.5 ns
We see that:
tF >> tS = 1.75 nsThe fibers rise time exceeds the system limit because of the large
spectral width of the LED.We conclude that LED cannot be used.
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Now consider the rise time of the photodetector:
PD
2 2 2S LS F
t t t t
PD2 2
t 1.75 - 1 - 0.23
tPD = 1.4 ns
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The photodetectors rise time has two components:tTR = transit-time-limited rise time
tRC = circuit-limited rise timeThe circuit-limited rise time is given by:
tRC = 2.19 RLCdThen we can write:
PD
2 2RC TR
t t t
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Assume we know from supplied data that:
tTR = 0.5 nsPhotodiode capacitance: Cd = 1 pF
Then
2
2
L d1.4 2.19 R C 0.5
Solving for the load resistance yields:RL = 594 ohms
The load resistance can be no larger than this (unless a highimpedance receiver is used).
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Assume we know the following:LD power P = 3.2 mWP = 5 dBmSource-fiber coupling loss = 3 dB
Connector loss2 connectors at 1 dB each = 2 dB50 splices at 0.1 dB/splice = 5 dB
Fiber loss = 100 (0.25) = 25 dBTotal system loss = 35 dB
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Power Budget
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The power arriving at the receiver is then:P = 5 - 35 = -30 dBm
For R = 400 Mb/s and Pe = 10-9 we can find the following receiversensitivities:(1) APD: - 40 dBmThis receiver will operate with a 10 dB margin.(2) High-Impedance PINFET: - 32 dBmThis receiver will operate with a 2 dB margin.
(3) Thermal-noise limited PIN: - 28.4 dBmThis receiver will not work.
The design is now complete. We have satisfied both the bandwidthbudget and power budget with reasonable solutions.
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Example 2
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Short-Wavelength Band
Attenuation and dispersion limits on the transmission distance vs. data rate for a770910-nm LED/pincombination.
The BER was 109 ; the fiber-coupled power was13 dBm up to 200 Mb/s.
The attenuation limit curve was derived by using a fiber loss of 3.5 dB/km
The receiver sensitivities shown in the left figure
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Attenuation-Limited Distances for Two Single-Mode Links
1. A DFB laser that has a fiber-coupled output of 0 dBm at 1550 nm.
2. At 1550 nm the single-mode fiber has a 0.20-dB/km attenuation.
3. The receiver has a load resistor RL = 200 and T = 300K.4. At a 1012 BER a value of Q = 7 is needed.
5. The InGaAs pin and APD photodiodes have a responsivity of 0.95 A/W.
6. The gain of the APD is M = 10 and the noise figure F(M ) = 5 dB.
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Problems
1. Consider a plastic fibre having a loss of 50dB/km. Power budget for the system allows a
maximum 24dB of fibre loss. Compute the maximum length of fibre permitted.2. A 100km optical link is designed with a data rate of 200Mb/s and a bit error rate of 10-9. A
GaAlAs LED that can couple 50W optical power in a fibre and a silicon PIN photodiode with a
sensitivity of -32 dBm is used. Calculate the maximum allowable fibre attenuation in dB/km if apower margin of 7 dB is needed. Assume connector loss of 1dB at each end.
3. For the problem 2, the LED and receiver rise times are given as 20ns and 14ns respectively.Can a multimode fibre with modal dispersion of 2ns/km be used in the system? Suggestchanges in the system to make it operable.
4. A fibre system is to be designed at 1510nm. Fibre cables for the link have losses of 0.2dB/kmand are available in 4km lengths. The laser diode emits 5mW and the receiver has asensitivity of -40dBm. A 5dB power margin is specified to account for system degradations.Assume coupling and splicing losses equal to 2dB each. Calculate the maximum allowablelength of link and plot the link budget.
5. A 1550-nm single-mode digital fibre optic link needs to operate at 622Mb/s over 80 kmwithout amplifiers. A single-mode InGaAsP laser launches an average power of 13dBm intothe fibre. The fibre has a loss of 0.35 dB/km, and there is a splice with a loss of 0.1dB everykilometre. The coupling loss at the receiver is 0.5dB, and the receiver uses an InGaAs APDwith a sensitivity of -39dBm. Set-up an optical power budget for this link and find the systemmargin.
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