Introduction to Infrared with Jeff

Check out our newest addition to the robotics tools tutorial series!

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Over the past couple months we've been building some tutorials on beginner concepts like robotics motors and servo motors with Jeff, one of our educational outreach coordinators. For our most recent addition, Jeff covers the basics of infrared - how to set up different sensors, troubleshooting, and how to hook them up to an Arduino to read the results - to use with robotics projects.

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As always, if you have any questions, comments or ideas for other videos you'd like to see, feel free to leave them below!


Comments 25 comments

  • Member #354383 / about 11 years ago / 2

    Are you guys stalking me? How'd you know I decided to work on my IOIO IR remote replacement project for my RGB LED strips in my room yesterday? I've been researching how to make an app for Android to let me control my LEDs (and potentially my PC if I can figure out how to wake my computer with IR).

  • I'm sorry, I can't help it. This in the background just works for me.

  • Member #115584 / about 11 years ago / 2

    Why does the polarizing filter make the IR light visible to the camera?

    • Kratz / about 11 years ago / 3

      As far as I know, most IR light is always visible to electronic cameras. For instance, point your TV remote at your phone/webcam/etc, and you should see the signal.

      • Member #115584 / about 11 years ago / 2

        In the video, it was stated that the IR was not visible to "Greg's really fancy camera" until a pair of polarizing sunglasses were introduced. It was also stated that the cell phone camera worked because it contains a polarizing filter. I wish this had been explained. I suspect that something other than polarization is going on here.

        • Miskatonic / about 11 years ago / 1

          It may have to do with the response of the CCD in the camera. I suspect that a really good camera has a CCD that is sensitve across a wider spectrum and less likely to pick up infrared. This is a guess.

          • Member #412006 / about 11 years ago / 1

            Sheet polarizers are designed to work for a specific wavelength range - for sunglasses, this is understandably "visible" (400-700nm or thereabouts). In the IR, this polarizer likely has higher transmission, essentially acting like a longpass filter. As an example, check out this commercically available polarizer:http://www.thorlabs.com/NewGroupPage9.cfm?ObjectGroup_ID=4984 . Transmission increases drastically after 800nm from 40% to 90%.

          • Kamiquasi / about 11 years ago / 1

            They're most likely CMOS sensors, but that's not too important for IR (for UV photography, a CCD is preferred). It is likely that the 'really fancy camera' has a better IR cut-off filter (find a SONY with 'nightshot' mode, that flips the IR filter out of the way without the hassle of manually removing one) which would attenuate the IR reaching the sensor better.

            The LEDs used were stated to be 940nm, which is already more efficiently blocked than 850nm (try an 850 on the 'really fancy camera').

            The bit that perplexed me, though (and thus upvotes for those above), is why a polarizing filter made a difference. While the polarizing filter (there's probably not one in the cellphone camera - if there is, you'd readily notice this by pointing it at an LCD screen and rotating the phone, or at 90 degree angles to a clear sky with a low sun) may have affected the view of the LED itself in some circumstances (showing stresses in the plastic, for example), I can't think of any way that the emitted IR, once filtered for polarization (and I'm not sure it's still polarized once it exits the other end of the sunglasses) would cause it to display on the camera where without such polarization filtering it would not. If anything, without that polarization it should be more readily visible. So the question would be whether there's some frequency shift going on (http://www.hindsinstruments.com/wp-content/uploads/Polarized_Light_and_Its_Interaction.pdf page 16 is fascinating, but unless you hid a rotating waveplate somewhere..), or whether the camera's internals throw a fit about polarization (because reasons, that's why), or...?

            If you can replicate this result experimentally, it might be fun to drop it on r/Physics.

            • noworries / about 11 years ago / 1

              The IR cutoff filters on CCD cameras vary in two ways, their cutoff wavelength and how sharp their cutoff is. On expensive cameras, the cutoff wavelength is shorter than on cheap cameras, and the transition band is also likely to be narrower. As to why looking through a polarizing filter helped identify the IR emitter, perhaps it improved the contrast (assuming the IR emitter outputs polarized light).

              Webcams and cheap camera modules are good choices to view IR emitters, since they are forced by economics to use cheap IR filters. Higher priced SLR cameras can afford better filters in their design to improve the color accuracy in daylight photos. (If you have ever looked at a photo taken without an IR filter of a subject in direct sunlight, you know how ghastly it looks).

              For astrophotography use, even the expensive SLR camera's IR blocking filter is a poor performer, since the filters block a significant portion of a the primary hydrogen emission line (Ha). A cottage industry has sprung up for modifying stock SLR cameras by removing the vendor provided IR blocking filter and replacing it with one which has a longer wavelength cutoff and a sharper cutoff slope. The new filter is even more costly, but the results are worthwhile for astrophotography since it allows for much shorter exposure times.

  • The person who posted this has an amazing name

  • Micko / about 11 years ago / 1

    what oscilloscope are you using????

    • saustin / about 11 years ago / 1

      They're using the excellent Parallax oscilloscope. But naturally, they discontinued it. They used to offer it on sale for about $89, and was even at $49 at close-out. It uses your PC for the display. It's not for really high frequency stuff, but for troubleshooting signals between chips (or LEDs, as here), it was fantastic. They're offering a new version, but it's a lot more expensive ($200), and out of stock at present.

      • Member #408458 / about 11 years ago / 1

        Ah, but Parallax is now offering the Propscope, which includes a logic analyzer, a function generator, and a bunch of other neat stuff that I'm still figgering out. It's currently out of stock but expected soon: http://www.parallax.com/StoreSearchResults/tabid/768/txtSearch/propscope/List/0/SortField/4/ProductID/586/Default.aspx

      • Miskatonic / about 11 years ago / 1

        Good eye! The Parallax scope has been my everyday scope for years. It's in my backpack right now, next to the Saleae Logic Analyzer. I've spoken with the good people at Parallax about how well loved this product is, but alas, times change.

        • Micko / about 11 years ago / 1

          Boooo, that its gone!!! but thanks for the heads up

          • saustin / about 11 years ago / 1

            Well, if kind Mr. 408458 (below) is right, there may be hope. That function generator feature is a great addition. And I do know it's faster. Given that the old scope's full retail price was about $160, and went on special at $89, then maybe if Parallax puts this new one on sale occasionally too, I may snap it up. Miskatonic's right, in that it's a good everyday scope.

  • Drun / about 11 years ago / 1

    The IR emitter and receiver pair would be a great addition to the SIK at financially little cost when looking at the increase in learning potential.

    • Miskatonic / about 11 years ago / 1

      D, I think you're right....this will require a bit of wrangling in the Inventory Dept. though.

  • chartle / about 11 years ago / 1

    You should post a link to the IR LED/Detector pair. Like this.

    https://www.sparkfun.com/products/241

    • Jonathan Bruneau / about 11 years ago / 1

      You read my mind. Thanks for posting the link.

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