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Light therapy device

Discussion in 'LEDs and Optoelectronics' started by mattfara, Mar 30, 2017.

  1. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Oh, and these LEDs would require a heatsink to draw the heat away from them. I would probably go for an aluminum cored PCB mounted on a heatsink.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    What you should be looking for is a high power IR led with a relatively wide beam width.

    A 10W IR LED will probably produce 1W of IR. Placed in a hexagonal pattern about 3.5cm between centres, and used at a distance such that the beam covers at least a 4cm diameter circle (2cotθ where θ is the half angle) should yield about 100mW/cm^2

    Each of these LEDs will be dissipating 9W of heat. You will require either a huge passive heatsink, forced air cooling, or possibly liquid cooling depending be size of your array.

    The smallest battery you could likely use would be a car battery and I would strongly recommend switch mode constant current drivers.

    Oh, and wear glasses that don't pass IR if you value your eyesight.

    (Run them at half power to get 50mW/cm^2)
     
    Last edited: Apr 8, 2017
  3. mattfara

    mattfara

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    Mar 16, 2017
    Wow. This is a very thorough analysis. I appreciate it very much. However, I can't stop the feeling that I'm doing something wrong. The devices I read about in the literature don't use car batteries! I can't seem to figure the right starting point. Is it that it is hard to find IREDs which can get this job done, delivering something in the range of 50-100mW/cm2 over a 20cm2 or so area of tissue? Some of the literature mentions lasers, but LEDs are reportedly used more frequently these days.

    So I guess at this point, I'd ask "If you were me, bent on making this thing, with the specs I need, where would you start?" This is just too hard for me to get off the ground.
     
  4. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Well, 50mW/cm2 over 20cm2 is 1W. Given the approx radiant efficiency of IR LEDs appears to be in the order of 10%, you need 10W of LEDs.

    You can do that with less than a car battery :) (my earlier estimate was for about 200cm2)

    I would suggest that people may be designing things like this using the only power rather than the radiant power.
     
  5. mattfara

    mattfara

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    Mar 16, 2017
    I guess the question is, trying to simplify it for myself, what LED should be used, given that I need a power density at the treatment of between 50 and 100 mW/cm2, the treatment area on the skin should be about 20cm2, and preferably a peak wavelength of 810nm. I keep getting stuck at this point - I either choose an LED that is too weak or too strong, making the power source prohibitive. Can you suggest a good starting point?
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Well, you could try a TSHG5510-ND from Digikey. It has a wide angle, emits 55mW at 100mA at a Vf between 1.5V and 1.7V. the peak emission is at 830nm.

    These are about $1.50 each and would be good for about 1cm^2. You would require 20 of them with a total input power of around 3.2W. Because of the wide beam width, you could place them as close as 1cm from your skin. The wide beam width would also make them far more eye-safe.

    The LEDs are far more efficient than others you've found.

    Because you're running them very near their maximum power, you would want to use a constant current driver (or 2).

    Two strings of ten LEDs would require a little more than 20V at 200mA. 10 AA cells in series would be a reasonable power source. A switch mode (as opposed to linear) constant current source would probably extend your battery life by around 25%

    Looking up the Energizer E91 datasheet, you would expect a run time of about 3 hours with a linear regulator, and around 4.5 hours with a switch mode regulator. The difference is more than I estimated above because the power you can get out of the batteries increases significantly as the current you draw is lowered.
     
  7. mattfara

    mattfara

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    Mar 16, 2017
    This is great info. I really appreciate all the consideration you've given my quest here.

    I'm wondering whether an AC driver might be good, considering the relatively short battery life. I would be using this device quite frequently, after all. I think audioguru mentioned that AC drivers are less consistent - is that a big concern in the setup you are suggesting here?
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Just correcting an error. With batteries, you'd probably want 15 (not 10) AA cells in series.

    If you run it from a plugpack, you could probably get away with something rated for 24V at 500mA or more (go for 1A). A simple linear current regulator (one for each string) would probably be fine because you don't have to worry so much about efficiency.
     
  9. mattfara

    mattfara

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    Mar 16, 2017
    Let me see if I understand the basics of why you are recommending this. The max Vf of one LED is 1.7V. In the ten-LED series, the Vtotal = 10*1.7=17 V. The parallel string should be roughly the same, so the total voltage required for the two would be about 17V. You suggest getting a plugpack that handles 24V, which should more than cover the dropout voltage of the two drivers, which is usually 2V each.

    Each string of 10 LEDs will need 100mA current if the max current on one LED is 100mA, and the two in parallel will need 200mA total. You recommend the plugpack handle 1000mA. I'm not sure why you suggest that the plugpack cover 5x the required current of the LED array. Does it have something to do with the drivers?

    I remember coming across something like a current mirror to account for the range of voltage readings for the the LEDs. Do you think that is called for in this case?

    Lastly, I'm not sure how to choose the two linear current regulators. Do I need to consider anything special about them, considering that they would seem to be in parallel? Would I just choose a 100mA, 17V regulator for each string?
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Sorry, I multiplied the 100mA by 2 twice. 500mA should be fine. Don't go lower than 250mA.

    Look in the LED resource. The is a constant current driver in there that will do the trick. It consists of 2 transistors and 2 resistors. One of the transistors will need to have a small heatsink.

    If you tell me where you purchase components I may be able to point out some appropriate transistors.
     
  11. mattfara

    mattfara

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    Mar 16, 2017
    In the case of the LEDs, digikey is considerably cheaper than mouser.

    Just out of curiosity, how do the infrared security cameras stack up in terms of radiant power, etc.? I was wondering if they could be re-purposed for this situation.
     
  12. mattfara

    mattfara

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    Mar 16, 2017
    THIS for example. Very cheap. Just need to buy a wall plug for it for 12V DC. I don't see much info about irradiance, etc, though the wavelength looks suitable.
     
  13. Audioguru

    Audioguru

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    The very cheap IR light with hundreds of LEDs has no detailed spec's. Some might produce a much higher output power than others or they all might produce only a tiny amount of output power. But it is Cheeeep!
     
  14. mattfara

    mattfara

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    Mar 16, 2017
    Yeah thought I might be able to contact the manufacturer to get some data, but it's a Chinese company. Can't find any emails....

    I didn't see anything in the LED resource about current drivers. Sorry, could you tell me the page?
     
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Section 3. The table of contents terms you that constant current drivers are there
     
  16. mattfara

    mattfara

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    Mar 16, 2017
    Sorry, Steve. Hate to sound this dense, but could you paste a link to the resource you're referring to? I can't seem to find it. I looked at DigiKey's page and the datasheet they link.
     
  17. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I linked to it in the message you quoted.

    See the word "section".

    It is a resource on this site.
     
  18. mattfara

    mattfara

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    Mar 16, 2017
    k gotcha. You mean the schematic shown in figure 3.1? Where would the heat sink be placed? You mentioned that certain transistors might be in order. I'll be putting together a final materials list for the device soon. I'm not sure how strong the resistors in the two linear current regulators should be, which transistors to use, and what heat sink to use.

    Glad this is coming together now. I'm looking forward to the build!
     
  19. mattfara

    mattfara

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    Mar 16, 2017
    You mean that I'll need two heat sinks total, one each for one of the transistors in the two CCDs? I'm not sure which one will heat up most and require the sink.

    Also, what do you think about THESE? Looks like it would be useful in other projects.
     
  20. Audioguru

    Audioguru

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    Sep 24, 2016
    A power supply with more than one voltage will almost always be accidently set to the wrong voltage which will blow up what it is powering.
     
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