December
2004, Issue 173
Light-to-Frequency
Conversion (Part 1)
TSL230R-Based
Pulse Oximeter
AC
VERSUS DC
In
order to be successful, you must be able to measure
the AC component of the sensor’s output. What can you
expect as a signal? Photo 2 shows the TSL230R’s output.
The vertical scope cursors show the minimum and maximum
excursion of the frequency output (with the sensor on
a finger). In /100 mode, a full cycle (DC portion) measures
approximately 3.5 ms, and the AC portion is approximately
215 µs. In /1 mode, a 35-µs cycle has only 2 µs of AC.
|
(Click
here to enlarge)
|
Photo
2—The TSL230R’s frequency output displayed on my
oscilloscope shows a slow frequency jitter marked
by the vertical cursors. The output frequency shifts
with the varying amount of light absorption because
of the blood pulsing within the light’s path. |
There
are two methods for taking samples. The device’s frequency
output is directly related to light intensity falling
on the sensor, so one cycle is sufficient as a sample.
In the first case, you need only measure the period
of one cycle to obtain a sample. Keep in mind that the
output in /1 mode is a period and not a symmetrical
square wave as in /2, /10, and /100 modes. This means
that one must measure a full cycle in /1 mode as opposed
to a minimum half cycle in the other divided modes.
(Or at least be sure you are measuring the right half.)
Although the percentage of AC to DC is the same with
all output modes, a 1-µs clock is useless in this case
with a /1 output (in the aforementioned example). The
most desirable output is one that approaches the sampling
rate but doesn’t create timing interference with other
functions.
Another
way to look at data would be to average the output over
the duration of the sampling period. The TSL230R does
this for you to some extent. For instance, /100 mode
gives an output equal to the sum of 100 cycles. However,
using this method, you want this sum to be for the duration
of the sample period and not a certain number of cycles.
In this case, you want just the opposite of the first
method. You want the fastest output you can count, so
that the DC (and the AC) portion will have the largest
count possible. Because the AC is roughly 6%, you can
expect an AC count of ±3 for every 100 counts of DC.
Using the same numbers as the first method, the 35-µs
output would have approximately 880 counts in one sample
period. Using the /100 output of 3.3 ms would have only
eight counts!
