# Light therapy device

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

1. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
It's not a physical reason, and there's no magic in them having 4 pins.

2. ### mattfara

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Mar 16, 2017
So I found this LED at Mouser. Here are some specs:

I gather that the input power = 1.3V * 0.1A = 0.13W
Here is the table which I assume the output power comes:

So the radiant power would be 0.013, thus the efficiency would be 10%. Sounds reasonable.

Were I shooting for a power density of about 25 mW/cm^2, and I knew I'd be making a 20cm^2 board filled with LEDs, then I would need to buy 40 LEDs:

# LEDs = 25 mW/cm^2 * 20cm^2/13mW.

Does that sound right?

This would be the schematic:

3. ### mattfara

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Mar 16, 2017
And I guess these would be good resistors for the job. Anything special I'd need to know about putting this board together? Any good guides you can recommend for making something like this?

4. ### Audioguru

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Sep 24, 2016
Philips produces products that are bought by everybody like lighting products. They also make extremely expensive medical equipment. But everybody does not buy IR LED products and the medical equipment uses x-rays or ordinary heating, not IR energy.

5. ### Audioguru

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Sep 24, 2016
Your LED Wizard does not know that the forward voltage is a range of voltages then its calculated resistor value is only for "typical" LEDs that you cannot buy unless you buy thousands of them, test them all and hope you get 40 typical ones. Maybe you should buy 40 LEDs, test them all then calculate a suitable resistor value yourself for each string to avoid having some very bright strings that might burn out soon and some strings that are dim.

The Philips Luxeon LEDs in 4-pins packages are made for very bright lighting therefore they have a high maximum continuous current rating (cooled by the 4-pins) which makes high brightness and they have a very wide viewing angle. The Vishay IR LEDs in an ordinary 5mm package have a high current allowed only in short duration pulses like in a remote control.

6. ### mattfara

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Mar 16, 2017
Wow that's a lot of good info Thank you. For this application, I'd need to be running the LEDs for at least 10 minutes at a time, so I'll start looking into 4-pin IR LEDs. I just hope that they wouldn't be so powerful as to exceed the 100 mW/cm^2 limit the literature pushes.

7. ### Audioguru

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Sep 24, 2016
The IR power produced by the LEDs is simply determined by their actual forward voltage (that your wizard wrongly assumed is the typical voltage), the supply voltage, the number of LEDs in series and the resistor value.

4-pins LEDs are made for visible lighting and probably are not made to produce IR.

8. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Whilst this is true to some extent (as it is true of resistors, capacitors, and transistors, and any other component), the problem with the array as shown is that the resistor drops only a very small fraction of the total voltage (0.27V).

Assuming a range of forward voltages which are randomly but normally distributed, one would expect the string would have a distribution of Vf with a smaller standard deviation from the mean.

However there main issue is the small voltage drop across the resistor. Variations of Vf with temperature could easily lower the Vf of the string of diodes by a quarter of a volt which could mean the difference between almost zero current (if the array is very cold) to a doubling of current (if it gets very warm). The later is more of an issue because the LEDs will get warm during operation and the feedback causes them to get hotter (I.E. the feedback is positive).

You are far better off using a constant current source for each string. An alternative is to put fewer LEDs in each string (the feedback will still be positive, but the feedback will be less). And yes, there are always other alternatives

If you're aiming at 20mW/cm^2, with a maximum safe exposure of 100mW/cm^2, I think you would be pretty safe in that these LEDs would likely have self destructed prior to achieving a sustained output near that maximum.

However, as an aside, I have some bags of cheap LEDs I bought for a project that didn't go ahead. These are cheap LEDs from a supplier that I would not consider trustworthy. I think I have time today to check the Vf of a large number of these LEDs at some reasonable, and constant current in order to see what variation I see.

9. ### Audioguru

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Sep 24, 2016
If a distributor runs out of LEDs and orders more, the new ones might come from the same recent batch and will not have random forward voltage but all might be low or all might be high. Your current limiting must allow for this.

10. ### mattfara

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Mar 16, 2017
Great. Thx for the food for thought. Would I be able to use a bunch of current regulating diodes leading to each string, all of which are drawing from the same battery?

11. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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My current limiting will not know what batch they're from. It will, however, know what a constant current is.

12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Yeah, that's possible. They have a fairly wide tolerance, and require a minimum voltage to achieve their stated current, but the variation in current between devices would not be hugely significant. The question is whether you can get these in the current that you require, and their price. These will fail short-circuit, so you also need to ensure that you have them adequately rated and probably on a small heatsink.

13. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Using a constant 17.3mA current source (which varied from 17.29mA to 17.34mA during the test) I tested 56 LEDs (A small sample, but enough to give an indicative estimate.

These were cheap while LEDs, no guarantee that they are part of the same batch.

The graph of voltage vs count is shown here (note that the voltage was recorded to 2 decimal paces and the results binned to give the following histogram).

The bins are 0.02V wide and are named according to the upper limit of each bin.

It is possible that this set of LEDs has most of its members in two (perhaps three) batches.

14. ### mattfara

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Mar 16, 2017
OK. It sounds like using CRDs would be hard, then. Is there something else you can recommend?

15. ### mattfara

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Mar 16, 2017
Hmmm, so combining what you and audioguru have said, I might test all my LEDs and try to group them into strings by Vf?

16. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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If you use a constant current device you don't need to.

If you use a resistor and the resistor drops a significant proportion of the voltage you don't need to.

These are operating a relatively high currents, so a constant current driver is the better option (well, it always is, but the extra complexity is warranted in this case)

17. ### mattfara

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Mar 16, 2017
I see. I've never used a constant current device before. Can you point me to a good guide for beginners on how to work with them? Also, do you understand how energy density (J/cm2) is derived from power density (mW/cm2)? These parameters come up often in the light therapy literature, and I understand power density, but energy density isn't explained.

18. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Look in the resources section for the resource on driving LEDs. It has a section on constant current drivers. You also mentioned constant current "diodes" which are an option not discussed there.

Watts are simply Joules per second, so if you want 0.15 Joules and the power of your array is 35mW, then you will get this energy after 0.15/0.035 seconds (about 4.3 seconds). Note that the "/cm^2" will cancel out of both values are "per square cm".

19. ### mattfara

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Mar 16, 2017
I'm in the process of doing my homework. It seems like all the research in light therapy uses mW to describe the output power of the LEDs, whereas many if not most of the datasheets I read on LEDs use mW/sr. Is there a way to convert?

EDIT: I came across a post asking a similar quesitons here. I think my device would be at about 1 cm from the skin, maybe less. So I can just take the value given on a datasheet in mW/sr as mW/cm^2?

Last edited: Apr 4, 2017
20. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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mW/sr is yet another measure of intensity. Whilst it will be higher for a more powerful device, it will also be higher for a device with a more focused beam.