High power IR LED control w/arduino and IR camera

[Sameer]

Hello everyone. I'm very new to EE and circuits so please excuse the stupid questions.

I'm trying to build a high-intensity IR source and test the ability of an IR camera to track that IR source in different environments. Perhaps I'll attach a gimbal that will move the camera around to track the IR LED source. I am thinking I will use these LEDs I found at Adafruit:

https://www.adafruit.com/products/388

and I've been googling around to figure out how to control and power the LEDs and I'm somewhat lost. It appears that some recommend using an LM317 constant-current source so that I can feed the LED with 1A. (I plan on using the Arduino to cycle the LED on/off so I can get maximum brightness at its peak 1A rather than limiting the current to the constant 100 mA that it is rated for.. I'd cycle the LEDs on/off at a rate of 33ms on/33ms off so with a 30fps camera I'd see one frame on/one frame off. My theory is also that cycling it in this way will help me distinguish my signal from ambient/background IR.. particularly in daylight.)

Adafruit's page however suggests the MOSFET:

http://www.adafruit.com/products/355

It appears that the constant-current source would be relatively easy where I would use the arduino to control a transistor which switches my power connected to the INPUT of the LM317 and a resistor across OUTPUT and ADJUST to fix the output current to 1A. However I don't know how I would use the MOSFET to power the LED.

Also, I am wondering about where one can find an IR camera that I can connect to my PC so I can process the IR feed with OpenCV? Ideally I could get a setup which has a filter on it so that it could be specifically tuned to look for the frequency of the IR LED. (940nm)

Any ideas?

Thanks for your help.

[Dan Lavin]

To power your LED with the MOSFET --

The constant current current circuit you described goes from +power to
the positive lead of the LED.
Connect the drain of the MOSFET to the negative lead of the LED and the
source to ground. Connect the gate to your a digital output of your
arduino via a 1k resistor. (The resistor value is not critical, except
if you are very concerned about turn on and turn off time, in which case
read the gate capacitance spec.) A high digital output will turn the
LED on and a low value with turn it off. If you let the gate float --
not connected to an active high or low-- you may get random turn on. If
this is a problem, connect a resistor in the 47k to 220k range (value
not critical, but should be at least 10x the other resistor) from the
gate to ground.

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[Hugh Pyle]

Careful with the peak current. The spec sheet
(https://www.adafruit.com/datasheets/IR333_A_datasheet.pdf) says for
1A peak current: "pulse width <= 100uS and duty <= 1%". That's not
a good de-rating -- you may actually get more infrared light output by
sticking to a 100mA max, and continuous drive or something close. If
you're using LM317 that translates to a resistor around the 12 ohm
range.

How many LEDs can you wire in series? The more the better -- it'll
increase your efficiency (as "proportionally" less heat goes out your
resistor). You'll need the supply-voltage to be at least 1.4v for
each LED.

-Hugh

[Sameer]

Thanks everyone for the help. I ended up getting these LEDs from superbrightleds.com 

http://www.superbrightleds.com/moreinfo/component-leds/315mw-850nm-infrared-led-with-star-heatsink-125-degree-viewing-angle/653/

which support 700 mA sustained. I plugged in two 1ohm 10W resistors so I think I'm getting 600 mA through the LED now... because I couldn't find a 1.8 Ohm 1W resistor at Radio Shack as recommended by the LM317 calculator

and I now have a bright LED that I can control with my arduino-- i took a quick pic with my phone and it is quite bright even though you with the naked eye it looks like a dull red glow. If I put a visible light filter on my phone then maybe it will be more clear that we have a nice bright IR source.

It works with the LEDs in series nicely too off my 11.1 LiPo battery. I can only fit 7 in series but if I put a bunch of these circuits in parallel that will be quite nice, I can put out quite a bit of IR.

[Sameer]

Hi everyone-- question about the LM317 --

I've been using the LM317 to power the LEDs just fine, when last night at Resistor it was suggested that I actually measure how much current I was putting through the LEDs. I did that and it turns out that when I have all 7 LEDs in series I'm only running about 180 mA through each LED. Now the consensus was that this is probably fine because with all the LEDs running at full current it could actually cause vision damage (and because it's IR you won't have a reflex to close your eyes). 

However, I am curious why this is, does anyone have ideas?

When I run the circuit with no load, I get 704 mA. With just one LED I get just about 698 mA. I don't think it is because the battery has any internal resistance/current limiting. I'm using a 2200 mAh 30C LiPo, which means it can spit out 66A right? So that should be more than enough I would think.

Thanks,

-s

[Nick Vermeer]

Hi Sameer,

I worked with you a bit on this yesterday.  Can you send a schematic of how you have the LM317 hooked up?

Raph thought you had the LM317 hooked up as a constant current source, but I want to verify that circuit is correct.  If it is constant voltage, then each LED is adding it's voltage drop to the chain, which will cause a drop in current when the power supply is constant voltage.

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[Ross Williamson]

Hi all

(sorry if this was in an earlier message)

You have enough voltage across the regulator for 7 LEDs with a voltage
drop of 1.5ish V for each don't you? (about 12-13V)

R

On Fri, Aug 24, 2012 at 9:46 AM, Sameer <par...@gmail.com> wrote:

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[Sameer]

Ross-- yes, I'm pretty sure I have enough voltage. I noted earlier that when I string 8 LEDs in series they don't light at all. So I figured that my battery supported the drop for 7 LEDs but not 8.

Nick, thanks, I am wiring them according to this web page:

http://www.reuk.co.uk/LM317-Current-Calculator.htm

I used the calculator to come up with a resistor value of 1.8 ohms to get me 700 mA. I have the + of the LED wired to the output pin and the resistor between output and adjust. (And power from the battery is wired to +) Then the - of the resistor is wired to a MOSFET (IRLB8721PbF) source pin. The mosfet drain goes to ground and the gate is hard-wired to power/11V. I have the mosfet in there so that I can add a control signal, but for now I want to just leave it on all the time, so this seemed like the best way to do that.

Is that clear? I can upload a drawing if that's not clear.

Thanks

-s

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[Sameer]

On the MOSFET I actually have source connected to ground and drain connected to the - of the LED chain.

-s

[Nick Vermeer]

Ok.. You should check the voltage drop across one of the IR LEDs.  If the voltage drop multiplied by the number of LEDs is close to your battery voltage then the LM317 could be hitting it's dropout.  The maximum voltage the 317 will be able to supply is around 2-3v less than the input voltage. If this is the problem, you can use a low dropout regulator instead.
 

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[Guan Yang]

Or you could use a current mode switching regulator, or a simple bipolar transistor as a current regulator, which would have a dropout around 0.7V.

I have good experiences with this circuit:

http://www.pcbheaven.com/userpages/LED_driving_and_controlling_methods/?topic=worklog&p=2

But I usually use a forward biased red LED instead of the reverse biased Zener, to reduce emitter voltage. Yes, if you don't need a lot of precision, LEDs are great voltage references :-)

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