6
\$\begingroup\$

I want to operate a UV LED at continuous 20mA for one week. However, when I measure the LED intensity with a photodiode (see image), I find that the light intensity changes continuously in time. I would like to ask you how I can reduce the light intensity drift that I am observing.

The LED data sheet is: http://static.vcclite.com/pdf/VAOL-5EUV8T4-LED-5mm-UV.pdf. I am operating it in series with a 461.6 ohm resistor and a 12 V AC-DC power supply. The voltage drop around the LED is 3.2 V, the current is 19 mA.

I am measuring the light intensity with a photodiode connected to an Arduino. I attach a picture of the circuit and the graph of light intensity (photodiode voltage output) vs time (hours).

Left: photodiode output as a function of time. Right: circuit to operate LED and measure light intensity

\$\endgroup\$
  • \$\begingroup\$ Have you considered driving it with a constant current driver? \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 3 '16 at 12:40
  • 3
    \$\begingroup\$ I wonder if this has something to do with the Arduino logger, instead (given the time span, ambient temperature and light variations seem unlikely). Have you tried to measure a constant current with the same system? (It would also help to see the measuring circuit's schematics and program). \$\endgroup\$ – Sredni Vashtar Oct 3 '16 at 14:19
10
\$\begingroup\$

If you look at the data sheet for the LED it says that at 20 mA the forward voltage may be between 2.8 V and 3.6 V. Importantly it states that this is at an ambient temperature of 25 degC.

So, how well regulated is the ambient temperature in your experiment? I ask this because you seem to be relying on generating a constant current by using a resistor and a 12 V supply and, if the volt-drop on the UV LED changes (maybe due to temperature or aging) then the current into the device will change and the light output will also change.

Gradual self-heating of the LED cannot be ruled out and neither can drift on the 12 V DC power supply. You should consider a precision constant current supply - there are circuits you can build around simple op-amps that will achieve this.

Now, onto the photodiode - its "gain" will be somewhat sensitive to temperature but less so than for infra red types however, dark current will approximately double for every 10 deg C rise so this might start to affect the accuracy of your experiment if you are using the device in photoconductive mode. If you are using the device in photovoltaic mode there may be temperature effects that make inaccuracies worse compared to photoconductive mode.

Also, what about local ambient lighting conditions changing? In a darkened room I think you should be OK but if there is any "spillage" of the sun through a window, then significant errors could be caused.

\$\endgroup\$
  • 2
    \$\begingroup\$ And who even knows how "constant" the constant voltage supply is in the first place. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 3 '16 at 12:44
  • \$\begingroup\$ @IgnacioVazquez-Abrams haha I added that just now plus self-heating effects. \$\endgroup\$ – Andy aka Oct 3 '16 at 12:46
3
\$\begingroup\$

Thank you all very much. I have provided a constant current supply with LM317 to the LED and a constant voltage of 1.5V to the UV photodiode with another LM317. This voltage is counteracted by the photodiode current and I measure the voltage across the photodiode using Arduino as described here http://provideyourown.com/2012/secret-arduino-voltmeter-measure-battery-voltage/

I have also used two pipette tips to align the LED and the photodiode and maintain them rigidly in place.

The light intensity measured by the photodiode is now constant and the LED is operated at 20mA.

Thank you all very much for your help.

\$\endgroup\$
2
\$\begingroup\$

Your photo and graph are useful. In your graph, at about 7 hours, a large step appears in your photodiode output. What was the cause? It may give you a clue that will help debug your experiment.
Your breadboard setup may look mechanically stable, but very small perturbations in diode or LED position can make a large change in photodiode current.
You can monitor LED current by measuring the voltage drop with a multimeter across that 470 ohm resistor in series with the LED. Ensure that it remains constant during your experiment.
Your photodiode will likely be a silicon photodiode whose wavelength sensitivity is maximum at much longer wavelengths (like 0.9 microns). Your LED on the other hand, puts out most of its light power at a much shorter wavelength (0.38 microns). Your photodiode response at this short wavelength will be far, far lower than at its peak-response. This means that any stray light will have a significant influence on the photodiode. Even a weak incandescent source (whose peak wavelength corresponds to the infrared peak of your photodiode) can contribute current to your photodiode detector.
Your experiment suggests that light output falls with time. Is this experiment repeatable? Does photocurrent continually decrease with time? To do this experiment properly, the photodiode - LED combination should be very firmly fixed into a rigid tube so that they cannot move about. Then the experiment should proceed in a constant-temperature environment, in a light-tight box.

\$\endgroup\$
1
\$\begingroup\$

Since this is a test circuit make a small current supply with lm 317. Use a resistor of 1,25/20 = 625 ohm. You can use 680 ohm for the test. If this turns out to be stable you can change the circuit to a permanent one. Diagram can be found on the internet or in the app sheet

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.