Instantaneous Light Intensity Function
Name: Emanuel G.
Though I by my own do a bit with some science, I (we) have a problem
that I (we) cannot answer. We are in collaboration with a
photoelectric laboratory in Greece, and there is a problem with
measurement of THE INSTANTANEOUS LIGHT INTENSITY FUNCTION,y(t), of a
fluorescent lamp fed from 50 Hz line via magnetic ballast, the reasons
for which we ALL cannot understand.
When a photoDIODE is used for the electronic measurement of y(t), then
the waveform registered by the oscilloscope is close to
y(t) = A + B|sinwt|
where the time-depending part is the ripple (light flickering), and
when photoSENSOR is used, then the waveform is close to
y(t) = A + Bsin(2wt).
That is, a smoothing of the singular waveform is done in the latter
case, and we do not see why the photoSENSOR (should be also quick
enough) smoothes the wave.
We would be very grateful if this message would come to a physicist
who know such measurements and who would be in contact with us (me)
trying to explain the situation.
Probably the two sensors are sensitive in different wavelength ranges.
Several phosphors are used in fluorescent lights, to make a spectrum
that, overall, looks white; the different phosphors fluoresce at different
wavelengths, and they have different decay times. Generally, the
longer-wavelength phosphors have longer decay times. See
Wilkins, A. J. and Clark, C.; "Modulation of light from fluorescent lamps";
Lighting Research and Technology 22 103-109 (1990).
Yeah! It's out of my area of expertise. If I had to make a guess (I would not
bet a lot of money on it.), I would look into the power supply electronics.
Sounds like the diode circuit that depends upon |sin(wt)| has a signal
inverter that converting the (-) phase into the (+) phase thus doubling the
time dependence. The inverter time constant is probably not the same going
from one polarity to the other, i.e. (-) ----> (+) not equal to (+) ---->
(-). This would double the time constant. In the other case the intensity
does not distinguish whether the electrons are moving (+) --> (-) or (-) -->
(+) thus doubling the frequency of the flicker. I admit this is an ex post
facto "explanation". Sounds like there's a need for an EE major to look at
the circuit diagram.
Emanuel- bad news first: I do not know specifically what kind of device
your "photosensor" is.
In my terminology, "photosensor" is a general term for anything which
converts light levels to a signal in some other medium, such as electric
It includes many things such as photodiodes, photoresistors, photovoltaics
(solar cells, photodiodes used in power-generating mode),
photocathodes in photomultiplier vacuum tubes, phototransistors (merely a
photodiode in parallel with the collector-base junction of a transistor),
and presumably others.
Good news- I recognize "photodiode" as something specific. Photodiodes
tend to be fast and accurate.
They are usually linearly responsive to light intensity over more than 3
orders of magnitude,
and usually have response speeds of < 1msec to < 1nsec, depending size,
load-impedance and bias.
(This supposes that you measure the current through them at zero volts, or
externally apply reverse bias.
If you let them generate their own voltage like a solar cell, and you only
measure that voltage, these advantages do not apply.
I am betting you used it right.)
So I tend to believe your "y(t) = A + B|sinwt|" is the accurate
measurement of light levels in the millisecond time range.
If your fluorescent tubes are new and working well, your sinwt is 50Hz.
The sloppiest, least-often understood kind of photosensor is the
photoresistor. (also called photoconductor)
They are often made of grainy yellow Cadmium Sulfide (CdS) printed on a
between interdigitated many-fingered electrodes of printed silvery metal.
So the CdS looks like a yellow squiggle or zig-zag.
This semiconductor has few free electrons or electron-holes normally, so
there is a high resistance between the two metal electrodes.
When light falling into its crystals creates new electrons and holes,
which live long enough to drift between the metal electrodes
at a speed proportional to the voltage the user has applied.
So it is a conductance proportional to light level, or, a resistance
If you are careful, you will apply a large voltage (say 10v) and read the
current through it by the voltage drop across a smaller resistor (< 1v).
That way the response may be somewhat linear with respect to light level.
However, the proportionality varies with light level. It is more
sensitive at low levels than high.
Guessing- I tend to guess that your "photosensor" is such a photoresistor.
Well, it has lots of differences from silicon photodiodes.
For scientific measurements, it has distinct inferiorities.
Their response speeds are usually around a millisecond, and sometimes at
low light- and carrier-levels over 10 milliseconds.
So it's my guess that this slowness is filling up the deep, sharp valley
between adjacent half-sine waves in the ( Absolute Value of [sinewt] ) wave.
Then it looks like a sine wave of twice the frequency.
(Taking the absolute value and low-pass-filtering is a known way to make a
Another way is to use multipliers or non-linear devices to make the square
of the signal.
Some of that could be happening here, too, but I cannot tell from your
Also, think: the photodiode sees light from blue (0.4um) to near infra-red
The photoresistor only sees light in a narrower band, perhaps 0.45-0.60 um.
Your fluorescent tube emits several different wavelengths in that range,
with different time behaviors.
Colors emitted directly from the electrified mercury gas will track the
absolute value of current fairly accurately.
Colors emitted from phosphors on the wall of the tube are often delayed,
having any exponential decay rate from 1msec to 1 second.
In short, both your photosensor and your fluorescent lamp have complex and
Your photodiode has simple behavior; so it can help you understand the lamp.
But your photosensor can't help until you do much work to understand it in
detail, and perhaps it is not much help then either.
LED's often provide light output proportional to current, and their
response speed is usually faster than 0.01millisec.
Using square-waves applied to LED's of a few colors,
you could check out the response speed of your "photosensor",
the sensitivity change at high and low light levels.
and it sensitivity with respect to wavelength.
Start with a green or yellow LED to match the color range of the photosensor.
Click here to return to the Physics Archives
Update: June 2012