LED light diffraction, scattering

[joeisthedude]

I am constructing a 'white line detector' to use on the underside of a robot. The crucial components are an ultrabright red LED and a light-to-voltage converter. My initial idea is to leave the LED as it is, i.e. unshielded. I have to light-to-voltage converter, which is a phototransistor with a lens (TSL-257), connected nearby and it is shielded using heat-shrink tubing to block out ambient light. it is being used to detect a change in color from black to white.

My question is, do I need to worry about "shielding" the LED to guard against scattering. If I just shield the TSL257, I get a voltage reading of 125mV, and when it crosses over to white, it is about 2V. This is perfectly fine, but the other members of my team want to put a heat-shrink tube over the LED as well because they think it will concentrate the light into more of a "beam". My concern is, the black heat-shrink tube will absorb the LED light and the intensity that is emitted to the floor will be weaker than if it just left alone.

Which is more justified? placing a tube over the LED or just letting it shine?

Thanks!

 

[Valerie Park]

The best approach in this case would be to let the LED shine freely, without any additional shielding. As long as the TSL257 is properly shielded to block out ambient light, the LED should be able to create a bright enough beam of light to detect the color difference between black and white. Heat-shrink tubing over the LED may reduce the intensity of the light emitted and make it more difficult for the light-to-voltage converter to detect the change in color.

 

[astrokreng]

I would say that both options have their justifications and it ultimately depends on the specific needs and goals of your project. Let's break down the potential effects of shielding the LED and leaving it unshielded.

Shielding the LED can potentially concentrate the light into a more focused beam, which may improve the accuracy and sensitivity of your white line detector. However, as you mentioned, the black heat-shrink tube may absorb some of the light, resulting in a weaker intensity being emitted to the floor. This could affect the performance of your detector and may need to be compensated for in your calculations or design.

On the other hand, leaving the LED unshielded may provide a stronger and more uniform light emission, which could also improve the accuracy of your detector. However, it may also lead to more scattering of light, which could potentially affect the readings from your light-to-voltage converter.

In my opinion, the best approach would be to test both options and see which one yields the most accurate and consistent results. You can also try using different types of heat-shrink tubing, such as clear or white, to see if it affects the light intensity and scattering differently. Additionally, you may want to consider using a diffuser to help distribute the light more evenly and reduce scattering.

Ultimately, the decision should be based on the specific requirements and goals of your project, and it may require some trial and error to find the best solution. It's always important to carefully consider all potential effects and make informed decisions based on experimentation and data. Good luck with your white line detector!