Let’s Get Technical
fiber equals 0.67 times the speed of
light ( 3 x 108 meters/sec) in a vacuum.
Now, if we make the length of the
fiber significantly longer than the wire
used in the three-inverter ring oscillator,
the travel time through the fiber will
determine the period of the oscillator,
not the gate delay of the transceiver
inverter. For a 100 meter length of
fiber, the travel time through the fiber
is calculated as follows:
Hopefully, you agree that 498
nanoseconds is significantly larger
than the two to four nanosecond
delay specified in the transceiver’s
data sheet. So, since the fiber has to
fill up with light, then empty itself out
for each cycle of the output signal,
the period of the waveform will
be twice the Tfiber time or 996
nanoseconds. This translates to a
frequency of 1,004,016 Hz.
To appreciate what this means,
consider this: the 100 meter fiber
loop stretches all the way around the
atrium of my campus technology
building. It takes 45 seconds for me
to walk a group of students around
the atrium once. The light in the fiber
zips around the atrium over
1,000,000 times each second. For a
nine meter loop of fiber, the frequency
increases to over 11 MHz.
The Fiber Optic Ring Oscillator
was designed to allow students to
accurately calculate the speed of
light. They do this by measuring the
frequency of oscillation and working
backwards (knowing the VOP of the
fiber) to calculate the speed of light.
The schematic for the oscillator is
shown in Figure 3. It is very important
to provide filter power supply voltages
to the transmitter and receiver sides
of the HFBR-5103 fiber transceiver.
The transmitter will pull around 150
mA when it turns on, which could
affect the operation of the receiver if
the filtering is not adequate. In
fact, when I first breadboarded the
oscillator, I ignored the filter caps and
inductors, driving the HFBR-5103
directly from the + 5 V supply.
That was a mistake that I spent
MAY 2004
Figure 3. This is the input/output signal biasing and power supply filtering circuitry
required to operate the fiber transceiver module that forms the inverter of the
Fiber Optic Ring Oscillator.
several days learning my lesson from
— racking my brain, trying to
understand why the circuit was
oscillating faster than expected. My
friend and colleague — Michael
Coppola — consulted with me and
explained how important it is to
properly filter the power supplies.
Michael has a great deal of
experience working with fiber and
had already “been there, done that”
with power supply filters in high
speed fiber circuitry. Michael also
provided a critical review of the printed circuit board designed by Don
McCarty — our Electrical Engineering
Technology Department Technician
— who has worked with me on many
electronic projects. Michael and Don
discussed how thick the traces should
be (signals thin, power thick) and
even the placement of components
on the board to minimize noise and
coupling issues. Don built two
oscillator boards and both worked the
first time.
Figure 4 shows the circuit used to
shape the oscillator output signal and
divide it by 8,192 down to a frequency
we can hear (123 Hz for a 100 meter
fiber and 1,363 Hz for a nine meter
fiber). The frequency of oscillation
(Fosc) signal from the TD output of
the Ring Oscillator (pin eight of the
HFBR-5103) is conditioned by the
100 Ω resistor and 0.01 µF capacitor
to remove the DC offset of the oscillator
signal and provide a path to ground
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