Download Optical Communications Laboratory

Optical Communications Laboratory
1.
BEHAVIOUR OF ELECTRICAL CIRCUITS AT HIGH SPEED
A major emphasis in practical optical communications (OC) systems is to control fast devices (lasers, LED's, detectors)
with fast electronics. Making things work fast is not easy, for example, consider that the current required to charge a
100 picoFarad (ρF) capacitor with 5V across it in 1 nanosecond (ns) is 0.5A. The voltage induced in a stray inductance
). (Note: a typical wire has an inductance of a few ηH/cm).
of 1 nanoHenry (ηH) is 0.5V! ( V = L dI
dT
Before beginning the experiment, you may want to review the operation of the function (or signal) generator (B&K
4040a) and oscilloscope (Tektronix TDS 3052), their equipment circuits and a brief review of capacitance, inductance
and related concepts in appendix IV.
Use the short (approx. 30cm) wire pair, the 2 BNC to banana plug adapters, and the T-BNC connector to connect the
signal generator FUNCTION OUT to the CHANNEL 1 input of the oscilloscope with the instrument settings as
summarized in fig. 1. While viewing the oscilloscope, adjust the function generator to a 0V to 5V square wave with a
period of 1 kHz. Then set the time base of the oscilloscope to 100 ns. Use the SLOPE and TRIGGER LEVEL features
to look at the square wave rising and falling edges (ask your lab instructor to demonstrate as necessary). Note
important features such as the rise time and fall time of the square wave, the size and periodicity of the "ringing"
following the rising and falling edges and the time decay length of the exponentially decaying ringing signal. Now
place a 50 Ω (ohm) resistor in parallel with the scope input; note any changes. Repeat with the long wire pair (approx.
180 cm). Make any sketches of the waveforms as necessary.
Fig. 1
B&K Precision 4040A
Function Out
Set-up for Observing High Speed Ringing
Tektronix TDS 3052
Ch. 1
50 ohm
Next, replace the twisted wire pair with a 25 metre long coaxial cable, and connect it to the TTL output of the function
generator. Set the pulse delay to the center position, and the pulse width to as low as will still produce a pulse. With a
frequency of 10kHz - 100kHz, sketch and explain the pulses seen with no termination at the oscilloscope, a 50 Ω
termination, and a less than 10 Ω termination. Next, change the function generator pulse width to the center position to
produce a square wave output. Explain the look of the rising and falling edges of the square wave with the three
different terminations.
THEORY
"Ringing" occurs in high speed electronics for two reasons: (1) strong reflections of the signal due to impedance
changes, and (2) transient resonance due to an LRC circuit.
Point 1 can be understood by realizing that a wire pair or coaxial cable has a characteristic impedance caused by the
capacitance and inductance per unit length of the wire. Most high-speed electronics uses 50 Ω transmission lines as a
standard. Just like a wave travelling along a spring connected to a wall will reflect off the hard wall, so will a signal in
a low impedance wire reflect off a high impedance termination (in opposite phase). Also, just like a wave travelling
along a spring connected to a light string will reflect at the interface, so will an electrical signal in wire reflect from a
too low impedance termination (in same phase). A matched termination will, however, dissipate the incoming energy
without reflection. One may get many reflections if the impedance match is very bad.
Point 2 is familiar from work you have done either in electronics or in calculus. The transient solution of a series
LRC circuit (as opposed to a steady state solution) is an exponentially decaying sinusoidal (see Fig. 2).
Fig. 2
Series LRC Circuit and Transient Solution
R
L
V
C
Current
Time
Current Rt
ωd2
t ]
2L
}
e
{C1 sin[
½
4ωo2 – ωd2
Where
ωo
1
.
t ] + C2 cos[
LC
½
4ωo2 –
For short wires with little capacitance or inductance, and a Scope impedance which is much bigger than the Scope
capacitive reactance at the frequencies of interest, the equivalent circuit for this experiment is shown in Fig. 3.
Fig. 3
Equivalent LRC Circuit For a Signal Generator with some inductance
driving the capacitance of the Oscilloscope.
50 ohm
Ls
signal generator inductance
20 pF
scope input
capacitance
V
1)
QUESTIONS
What is the equivalent circuit for a lossless cable or wire? Now, assume that a short wire can, in a gross approximation,
be considered as a single inductor and capacitor, and can be added to the diagram of figure 3. The capacitance of the
180 cm wire pair has been measured to be 100 ρF. Using this and the ringing results from the wire pairs, determine the
capacitance and inductance per unit length for a wire pair (in units ρF/m and ηH/m). Finally, determine the signal
generator inductance Le. What does resistance in the wires do to the signal? What causes dispersion of high frequency
signals in cables (i.e., different frequencies propagating at different speeds)? Can you comment on any relationship
between ringing frequency and rise time? Remember that the oscilloscope is limited to a 100Mz bandwidth anyway.
State the advantages and disadvantages of having a low impedance input. From the reflection period in the 25 m cable,
determine the propagation speed of signals in this 50Ω impedance coaxial cable.
2.
LASERS AND LED DRIVERS
Shown in Fig. 4 is a simple circuit to operate a laser diode (LD) or LED (remember that semiconductor lasers and
LED's are just forward-biased PN-junction diodes). Note the large 0.25W or drive power dissipated in the circuit.
Fig. 4
Simple Scheme to operate a Laser Diode or LED
Rs
Take
R
And
R =
V
I
Vf
V = 5V
I = 50mA
V – I R s - Vf
I
Generally, signal sources like the B&K can only supply tens of milliamps with any amount of speed. While this may
marginally be enough for this experiment, we still assume that it is not. Therefore our circuit in Fig. 4 cannot even be
realized. We need a driver!
The simplest scheme for driving an LED or LD requires a transistor as shown in Fig. 5(a). A description of how the
circuit works is given in Figures 5(b) and 5(c).
Fig. 5(a)
A simple LED driver
100uF
10 ohm
IR LED
MF0E100
A
R1
Rb = 680 ohm
Q1
Vin
Q1 is a MPS 3646 npn transistor
Fig. 5(b)
Biasing of LED in the “low” state
+5V
100uF
10 ohm
IR LED
A
Ic = 0
R1
Rb
R2
Vin
(low)
= 0V
Vbe = 0V
transistor is OFF
This is
equivalent to
+5V
100uF
10 ohm
approx. 1.2V
A
I = 5mA
so Rb = 5V - 1.2V - (5mA) (10ohm)
5mA
= 750 ohm
we'll take 680 as the nearest
standard value
Rb
Fig.
NOTE:
5(c)
Biasing of LED in the “high” state
β ≈ 10
+5V
I = 50mA
100uF
10 ohm
1.2V
Vin (high) - Vbe
Ic =
R2
B
= 45mA
A
5mA
R1
Vce
approx. 0.3V
R2
Vin (low)
approx. 5V
so
Vbe
approx. 0.7V
R1
Choose
=
5V – (10Ω)(50mA) – 1.2V- Vce(sat)
45mA
≈ 65Ω
Vin(high) and R2 such that:
Vin(high) – V≈
be 5.5mA
R2
If Vin(high) = 3V , R2 = 418Ω ;
Therefore, take 470Ω as the nearest value.
Discuss the effect of the transistor base charge, and the LD capacitance (data sheets are in appendices). Describe this
only briefly.
Fig. 6(a)
"just at turn-on" when Vin = Vin(high)
V = Vcc - 1.2
= 5 - 1.2 = 3.8
Ic = 0
Co-b
This is equivalent to
R1
Vb
Co-b
R2
Rb
Vin (high)
R2
Vin (high)
Ci-b
Vb
AND
Vb(0) = 0
Ci-b
0V
The delay time, the time for Vb(α) = VBE(n) (i.e. to turn on the transistor), or
t = -τ ln { 1 - Vbe(m) }
Vin(high)
Now, we must include the charging of minority carriers in the base. From the MPS 3646 data sheets, the charge in the base
"to just turn-on the transistor" is given by Fig. 13. At Ic = 50mA, θA ≈ 90pC. We can define an equivalent base-emitter
junction capacitance as follows:
θA = C = 90pC = 129pF
0.7
Vbe(n)
So τ ≈ ( Cib + Cob + C) R2
≈ (8pF + 5pF + 129pF) 470Ω ≈ 67ns
and t ≈ -67ns ln{ 1 -
0.7
} ≈ 18ns Æ is a good guess at the delay time. The delay time is
3
tabulated for Ic = 300mA, as 10ns
Fig. 6(b): transistor is "on", current begins to rise
Co-b
Vb = 5 - 1.2
= 3.8 V **
dv
dt
Co-b
R1
V (+)
Vin (high)
Rb
R2
( Vin (high) - Vbe )
R2
Vbe (m)
0.7 V
β ≈ 10 in saturation, but we are not saturated yet, so β ≈ 40 – 50
Vin(high) –Vbe(m) + Cob dV β + Cob dV
R2
dt
dt
And therefore:
V(t) = Vb - Vin(high) – Vbe(m)
R2
β · R1 + (1 + β) Cob dV(t) R1
dt
Solving this differential equation with Vb ≈ constant yields:
V(t) = - β { Vin(high) – Vbe(m) }R1 x ( 1 – e(-t/τ)) + Vb
R2
Z = ( 1 + β) Cob · R1
i) we now ignore Cib because it is charged and does not change since Vbe(m) does not change.
**not a bad approximation, since at full current of 50mA, V = 5 -1.2-0.5 = 3.3V
**
Vin(high) – Vbe(m) + Cob dV β
R2
dt
The turn-on time is roughly the time for V(t) ≈ Vce(sat)
(ie. To put the transistor into saturation), or:
t = -ln
The important thing to note
is that the time constant is
governed by (1+β)Cob
– the well-known Miller
Effect
1-
Vb – Vce(sat)
β (Vin – Vbe) (R1/R2)
τ
time β = 50
τ = 51 (5pF)(60Ω) ≈ 15ns
t = -15ns ln 1 –
3.8 – 0.3
.
50 (3 – 0.7) (60/470
= 4ns
How does the LED capacitance come into the calculation for the rise time? The "top" part of the circuit looks like:
100uF
I
R' = 1.2V
+5V
Equivalent cct.
I
10 ohm
100 uF
10 ohms
I1
I2
V
R'
Cd = 100pF
Vb
Note that it is generally difficult to solve for the rise time exactly because R’ varies strongly during the turn-on period.
We can estimate the effect of the LED capacitance by the following argument:
I2 ≈ Cd dV ≈ Cd [ V(5mA) – V(50mA) ]
dt
∆t
∆t ≡ switching time, or for our case, I have measured
V(5mA) – V(50mA) ≈ 0.07V, so I2 = (100pF) (0.07V) = 2.3mA
(3ns)
which is small compared to I, so I1 ≈ I and the current I can rise as fast as the cicuit can allow. Imagine if the LED was
initially off (ie. There was no pre-bias network resistor Rb), so in this case to turn the LED on,
I2 = (100pF) (1.2V) = 40mA !!!!
(3ns)
Thus, the LED capacitance robs the current required to increase the LED current I1. In this case the rise time is limited
by the diode capacitance. This calculation also shows why it is desirable to have a pre-bias network.
Fig. 6(c): "just after the storage time"
Case (a):
Vb = 5 - 1.2
Co-b
dv
= 3.8
dt
Co-b
R2
R1
V
Ic >= 0
Vin (low)
If (dV/dt) is large such that Vbe ≥ 0.7V,
ie. The transistor turns-on and Ic
increases such as to keep Vbe ≈ 0.7V.
And dV = Vbe – Vin(low)
dt
R2 Cob
is the maximum rate of change of V.
Vbe = R2 Cob dV + Vin(low)
dt
Therefore the maximum fall time is of the order of:
dV = Vbe – Vin(low)
dt
R2 Cob
or
dt = dV R2 Cob
Vbe – Vin(low)
dV ≈ Vb - Vce
dt ≈ (3.8 – 0.3)(470Ω)(5x10-12F) ≈ 12ns
0.7 - 0
Case (b): for dV/dt < Vbe – Vin(low) we have something simplar to Fig. 6(a);
R2 Cob
Vb
38V
R1
V
Co-b
With the initial condition that the change on Cob is given by:
q(t = φ) = -(Vbe(m) – Vce(sat))Cob
therefore, V = Vb + R1
[-Vbe(m) + Vce(sat) + Vin(low) ] > -Vb
R1 + R2
exp (-t/τ) , τ = (R1 + R2)Cob
and the time for V = 0.9 Vb is:
t = -τ ln
R2
0.1 Vb
.
(R1/R2) [Vb + Vbe – Vce(sat) – Vin(low) ]
τ ≈ (470 + 60)(5pF) ≈ 2.7ns
t = -2.7ns ln
(0.1)(3.8)
.
60
( /60+470) [ 3.8 + 0.7 - 0.3 – φ ]
≈ 0.6ns , so case (a) will dominate
Fig. 7(a): circuit to measure optical power incident on a reverse-biased detection:
2)
EXPERIMENTAL AND QUESTIONS
(a)
The circuit shown in Fig. 5(a) is built. A visible 680 ηm LD is mounted in the center of the optical laser mount .
The laser must be aligned to the microscope objective. The 10-turn pot that is the variable resistor R1 is preset to the
value for 50 mA "on" current. Check that this is correct by measuring the voltage drop across the 10 Ω resistor using
the oscilloscope. Make sure you are familiar with the parts of the circuit and how it works.
Set the B&K for a 0 to +3V, 100 kHz square wave. Attach to the circuit as Vin and monitor the voltage at point A with a
10x probe (don't forget to calibrate your probe - ask your T.A.) and the signal out of the function generator on the other
scope channel. Always use the probes instead of a direct input to the scope to prevent the reflections you learned about
previously. Also remember that the probe ground will be your earth ground in the circuit. For best results, connect it to
a ground point close to where you are probing your voltage. Check the voltage at point A on the oscilloscope, and
adjust the resistor to the correct "on" current. The peak to peak voltage seen on the oscilloscope should be the
difference in the "on" and "off" currents of the LD times 10 (10 Ω resistor) or about 0.45V. Verify that your circuit
does pulse the LD, by looking at the LD (DO NOT LOOK DIRECTLY AT THE LASER) at a frequency of 2-20 Hz.
(b)
Take a look at the rise and fall times of the B&K signal to the board. Then look at the rise and fall times of the
diode current on the oscilloscope (by measuring point A). Note that the "rising" edge looks like a falling edge on your
scope because when the current increases during an "on" period, it actually drops the voltage at point A.
Do you notice anything unusual about the diode rise times compared to those of the B&K signal? If so, explain this
difference in terms of how a transistor turns on.
(c)
Look at the falling edge on the 50ns time scale with the function generator waveform superimposed, and attach
a 47ρF capacitor across R2. Observe what happens. This is known as a speed-up capacitor, and decreases the storage
time by storing charge to neutralize the base charge "just after turn-off". We can make a rough guess as to how much
charge is required by looking at the value for QT in the transistor data sheets. You can calculate the voltage across the
resistor "just before turn off", and then calculate the minimum capacitance by the Q=CV relationship. Now assume the
speed-up capacitor completely reduces the storage time to zero; what is the experimental storage time of the transistor?
Compare with published specifications and explain why there may be a difference. (Hint: re-read about the transistor
as a switch).
The LD capacitance can be a limiting factor in the turn-on time. Just as a rough guide of the switching time to turn on
the LD, assume that this time is just the time to charge the LD capacitance with the collector current and use I = C
dV dT (make sure you know what V to use in this equation!). Therefore to switch faster, generally what must you do?
Thus you trade-off speed for
(fill in the blank).
(d)
The rise, fall, storage and delay times limits the maximum frequency which you can transmit digital
information. Explain this statement. Hint: Use diagrams to illustrate your point by taking Vin as a square wave and
roughly draw the LD optical output (assume for the moment that the optical waveform is identical to the electrical one).
It will become very evident what the problem is if you draw it at 1 GHz.
It is worth noting that even if we had a perfect electrical circuit with 0 rise time, the LD has a finite optical response
time. Why is this? (S.M. Sze, some other text, or a grad student could help).
3.
RECEIVERS - The Transimpedance Amplifier
As you have done in 3E3, you will detect optical signals using a reversed biased diode and a resistor as shown in Fig.
7(a); the voltage across the resistor is just given by the product of the photocurrent and the resistance. The small-signal
equivalent circuit for sinusoidal optical signals incident on the detector, is shown in Fig. 7(b). Note that the voltage at
low frequencies is still given by the V=IR relation, but the diode capacitance rolls off high frequencies with a 3dB point
at f=1/(2_RC) (3dB attenuation means .707 times the voltage, and .5 times the power). So, in optical communications,
you trade off
_
for
_
and vice-versa (fill in the blanks).
Fig. 7(a): circuit to measure optical power incident on a reverse-biased detection:
Vr
I
Zin = V/I = R
I ≡ photocurrent
Zin ≡ impedance looking into the
ground
Z in
V
R
Fig. 7(b): small signal AC equivalent circuit for the circuit in Fig. 7(a):
V
I
Rd
Cd
R
Notes: i) remember, in the small-signal regime, DC voltage sources are shorted to ground
ii) Rd is generally very large,
Rd ≈
Vt
Ileakage
≈
0.025 ≈ 500 kΩ for your detector
50nA
The transimpedance amplifier in this experiment is constructed from a resistor and an op-amp, shown schematically in
Fig. 8(a). It converts current into voltage. An explanation of this is given in Figure 8(b) using the simplified circuit
model for the op amp. An important point is that the input impedance of the transimpedance amplifier is Rf/A, where
A is the frequency dependent open loop gain. This lower effective impedance affects the bandwidth and the dynamic
range of the amplifier.
If we attach the reverse-biased photodiode as shown in Fig. 9(a), the amplifier now gives a voltage V=IR, except that
now the resistance "seen" by the photodiode is Rf/A. Therefore, the op amp gives us A times the bandwidth of Figure
5(a). The op amp has an input capacitance of its own (as well as an impedance and input current), but these are
negligible. The photodiode capacitance is 17pF. The op amp gain A becomes 1 at about 10MHz.
Fig. 8(a): Transimpedance Amplifier:
Rf
V = -IRf
I
V
Fig. 8(b)
Derivation of transimpedance amplifier characteristic
Rf
I
Z in
Va
V
-AVa
I ≡ photocurrent
Zin ≡ Va = IRf
1
I
I+A I
= Rf ≈ Rf
I+A
A
Va = V + IRf
Or
Va = -AVa + IRf
Or (1 + A)Va = IRf
Or -V (1+A) = IRf Æ V = -A IRf
A
1+A
≈ -IRf for A>>1
QUESTION: If you have a circuit with total capacitance of Cd=20fP and feedback resistor of Rf=100 kilo-Ohms. You
have an amplifier where A=1x105 at D.C. and A=1 when f=10 MHz. Is this a good op amp for the maximum possible
frequency response? Also, note that the transimpedance property depends on A; at what value of A does V=IRf deviate
by 10% then if A=_? Does it make sense to define the frequency when A=1 as the bandwidth? Hint: use the exact
expressions in Fig. 8(b).
EXPERIMENTAL:
The amplifier circuit shown in Figure 10 is built on the PC board. The circuit consists of the transimpedance amplifier,
a voltage reference, and a reverse biased photodiode as a current source. A current limit is imposed on the photodiode
without changing the RC time constant of the circuit at high frequencies. How is this done? Make sure that you
understand the circuit, and explain it very briefly in your report. It is not necessary for the detector to have a common
ground with the LD driver circuit, though you may have to later due to a limited number of power supplies. Remember
that if you connect both probe grounds to points in your circuits, you are connecting those points together through a
common earth ground.
Fig. 10: Transimpedance amplifier circuit:
2.2pF
Vcc = +15V
R1 = 1 kilo-ohm
2.2pF
RL = 5.1 kilo-ohm
A
C2 =
0.1uF
Rf = 22 kilo-ohms
photodiode
R2 = 1 kilo-ohm
+15 100uF
I
B
2
1
3
4
-15
0.7
10uF
V
100uF
100 kilo-ohm
Rs =
2 kilo-ohm
C1 =
1uf
10uF
0.7
Vee = -15V
Be VERY CAREFUL that the photodiode is connected in the correct polarity in the circuit. The black wire (cathode) is
connected to the positive side in a reverse biased configuration. You can make a preliminary test of your circuit by
setting your LD driver to 100 kHz and placing the detector almost up against the microscope objective. If room lights
significantly affect the output, you can turn them off.
Compare the two waveforms (the detector output at point C of Fig. 10 and LD drive current as before) and compare the
rise and fall times. (Remember that this is an inverting amplifier, so a rising edge will look like a falling edge on the
scope). How do the rise and fall times of the detected pulse compare with the current pulse through the LD and RfCf?
Look at the detected waveform rising and falling edge when one of the 2.2 ρf feedback capacitors is removed. Do you
expect this result? Compare the rise time to the bandwidth of this circuit for both cases of feedback capacitance
(remember that rise time is 2.2RC). Explain all observations and discrepancies.
Increase the frequency up to 6 MHz and examine simultaneously on the scope the electrical pulse of the LD drive
current and compare to the detected pulse as you increase the frequency.
QUESTION:
Examine in detail the frequency "roll-off" of the detected signal by measuring the peak-to-peak amplitude of the
received signal versus frequency. Compare with calculations you have made above on the bandwidth of this
transimpedance amplifier and the rise times you have measured. Also comment on any other important changes you
observe as you increase the frequency up to 6 MHz. Explain why these changes might occur.
As shown in Figure 12, a voltage adjust on the positive op amp input provides an output offset adjustment. Observe
that this adjustment is working. It will be important in the next section.
Fig. 12: Effect of DC offset on the non-inverting terminal of the transimpedance amplifier
Rf
I
Vo
V
it can be shown that Vo = -A (-V + IRf)
1+A
≈ - (-V + IRf) for A >>1
We now make an addition to the receiver, which converts the distorted detected pulses into digital logic signals again.
For this we will be using a LM319N comparator with ground as our decision level. Connect the output of the
transimpedance amplifier to the input of the comparator. Then by using the DC offset of the op amp, position the
amplifier waveform about the decision level (halfway between the high and low voltages) so that the comparator can
make its decision. The comparator circuit is shown in Fig. 13. Connect one scope input to point A and the other to
point B, and both scope inputs should be on DC since we are interested in some absolute DC threshold.
Fig. 13: LM311N comporator with the threshold at ground.
+5V
+5V
100uF
Connect to output of
transimpedance amplifier
Point C, Fig. 10 Å
Vin
A
5
4
470ohms
11
LN319N
6
12
B
output
Vo
100uF
-5V
Note that we have connected the input Vin to the negative input of the LM311N, because the output from the
transimpedance amplifier is negative (if there is no DC offset):
Vin < 0
Vo > 0
Vin > 0
Vo <0
Set the LD driver frequency to 100 kHz and go ahead and adjust the DC offset. Observe the output waveform out of
the comparator (point B) as you adjust the amplifier offset so that the decision level (0 Volts) goes the full range of the
detected waveform (at point A). Compare the waveforms at points A and B under these conditions. Sweep the
frequency up to 6 MHz, re-adjusting the DC offset as necessary, and again compare waveforms. Explain your
observations - especially why pulses at B can by asymmetric.
4.
RECEIVER PERFORMANCE ANALYSIS: EYE DIAGRAMS, NOISE AND BIT ERROR RATE
Before you proceed further, read Appendix VIII, which introduces the eye-diagram and its use as a quick diagnostic
tool to look at the performance of a receiver. As for bit error rate, we will only be looking at the effects of noise, and
not jitter. Look at the formula for bit error rate due to noise, as you will be using them. Figure 15 basically shows how
an eye diagram is formed, while figure 16 shows the circuit for generating eye diagrams using a 4-bit pseudo-random
number generator. We will, in fact, be using a pre built 16 bit pseudo-random number generator, but the concept is the
same.
Fig 15: Conceptual picture of how an "eye-diagram" is formed
Wavetek 145
(trigger)
sync out
pseudo-random
number generator
function
out
"bits"
in
1
2
3
out
4
LED
1
2
3
4
detector
external trigger
1
2
3
4
(trigger pulses)
scope point
4
3
2
1
Viewed on the screen:
Trigger pulse
Note: this diagram
Is for a return-to-zero
System. Your system
Is different, with an eye
Diagram looking like:
waveform
superposition waveform
1
2
(eye diagram)
3
4
The circuit of figure 16 is built. To use it, set the signal generator for a 100 kHz OV to +5V square
wave (be careful not to exceed 5V - a little less the better). Connect the 5V power supply, and
connect a coax cable from the Sync-Out on the frequency generator to the external trigger on the
oscilloscope for all future triggering. When the diode is driven by the random output, the detected
waveform should be your eye diagram.
QUESTIONS:
(a)
Does the eye diagram look like you expect? Sweep the frequency through to 6 MHz. Does it
seem obvious to you why it is called an eye diagram? Past a frequency point the eye diagram
becomes a mess. Can you explain why? (Hint: try to freeze the pattern at this frequency). Which is
the important "eye" in this mess? Explain why the eye closes. This eye closure may be hastened by
noise on the signal and timing jitter. Briefly explain this last statement.
(b)
The written sheets of Appendix IX give the theoretical formulas for noise in the detector
circuit. Using these, make a theoretical plot of S/N versus light level in dB, and note the receiver
sensitivity for S/N = 16 dB. Sensitivity means the detected power for this required signal level (you
can assume a 1.2 eV Si bandgap, and 30% detection efficiency. This produces 250 A current for
1mW optical power). What bit error rate does this S/N level correspond to (from appendix VIII).
This sensitivity is a common designation in the OC field. Describe the dominant noise sources in
your circuit. At a S/N of 16 dB would you say that your receiver is thermal noise limited?
EXPERIMENTAL:
Ask the demonstrator which fibre you will use to connect the LD output to the detector. Due to the
problem of noise spikes (caused by a capacitive coupling between digital circuitry and our detector),
careful alignment is required to achieve maximum S/N ratio. Borrow the fibre stripper, cleaver and
inspection scope from the Fibre Optics lab (ask the lab technician for this). The T.A. or lab
technician will demonstrate how to use these tools.
Carefully inspect the cleaved end with the scope to ensure a good cleave. Place the fibre in the
brass fibre chuck and place the fibre chuck into the fibre positioner. Slide the fibre chuck so the end
of the fibre is about 2 mm from the objective lens. Using the X and Y positioner thumbscrews, adjust
the fibre position to maximize the light output.
Place the output end of the fibre on the fibre holder and place it near the large area detector. Precise
alignment is not too important because of the large surface area. The arrangement is shown
conceptionally in Fig. 11.
As with the last section where you looked at the LD pulse directly, look at the pulses coming
through the fibre. Adjust the detected signal to about 40 mV peak to peak by adjusting the position
of the fibre. By comparing with the maximum signal seen without the fibre, determine the coupling
efficiency (in percent).
We will next measure the noise generated by our amplifier system for zero light level and compare
to the theoretical value. To do this, take the output of the amplifier (point C in Fig. 10) and connect
it to an 11 times voltage gain block. This circuit is shown in Fig. 17, and is probably familiar to
most students. Check if this amplifier is reducing your frequency response. If it is, note the new
3dB point. You should consider this output as your new detector output when monitoring the
detector signal. At a frequency of 1 kHz examine the output of this amplifier with the oscilloscope.
if necessary, set the scope input to AC coupling and increase the voltage sensitivity until the "noise"
is seen. If you decrease the time base to the ms scale, you can see a very noisy harmonic signal; this
is just the 60 Hz pickup from the room light. Shut off the lights and look at the noise again. The
RMS noise is given roughly by 1/6 to 1/7 the peak-to-peak noise voltage since this is assumed to be
a Gaussian (random) noise source. Compare to your theoretical predictions of the total noise
voltage. It may not be surprising if your results are in disagreement because of a large resistor noise
from the poor quality carbon film feedback resistor. Looking back at previous calculations for noise
sources, is this a possible explanation?
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Fig. 17: NON-Inverting Amplifier 11x GAIN
+15V
signal from
transimpedance
amplifier
7
3
100uF
6
output
4
2
100uF
-15V
R1 22 K
R2
2.2 K
Gain = R1 + R2
R2
MORE EXPERIMENTAL:
Now we will look qualitatively at the dependence of BER (bit error rate) on S/N. Connect the
circuit as shown in Fig. 18. The random number generator will be running the LD driver. A brief
explanation as well as a logic schematic is shown. Use the TTL output, set to a 200 ns pulse, at a
frequency equal to that of the signal generator as a decision gate. The lowest voltage within this
gate will determine whether the detected pulse is a 1 (>0V) or a 0 (<0V). You will want to set this
gate position to the center of the detected pulses, where there is the least chances of an error due to
the closing of the "eye" at high frequencies. Again, we have the pseudo-random number generator
driving the LD.
Set the frequency generator to 100 kHz. Adjust the amplifier offset so that the 0V decision level is
in the center of the detected waveform. Adjust the decision gate. The error detection circuit will
pulse ones during errors. Use the frequency counter to determine the error frequency. Quote this as
a BER by changing this to an error fraction. For offset values all the way from the high to a low
level, graph the BER. Knowing your noise level, does this correspond to theory?
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REPORT:
Especially with a lab like this with many questions asked, you must be extra careful with
organization. I suggest that you try to understand all questions and sections before you write the lab.
Be concise in wording. If you say nothing, the marker will assume you know nothing, but if you
copy pages of long and drawn out theory from a text, he/she will likewise assume that you know
nothing. Understanding is the key to a concise, correct and complete report. Remember, you can
assume that you are writing for someone with similar schooling to yourself (though not necessarily
identical). Therefore, the reader will understand basic undergraduate physics and electronics. You
may refer to the lab manual for some things, but anything which is in the manual, yet is vital to the
understanding of your report must be included in your report. Don't go overboard, again, be short
and concise. Make sure your report is in proper scientific report style. Make sure you know how to
write a good abstract!
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