Alright, I bought a 3W 1.6v IR LED to make a night vision flashlight (because all digital cameras can see IR to some degree, depending on their filters), thinking I could power it just like I have all my other super bright flash lights, by hooking it up to a battery and a potentiometer in series. And I tried a NiMH AA and an alkaline, and they both supplied only around 30-60mA, but if my basic understanding of electronics is correct, 3W @ 1.6v = 2 amps, not 30-60mA.

So someone helpfully explained to me that this is because high power LEDs actually need driver circuits, to regulate the current when the voltage changes. And apparently the LM317T is one such regulator I have lying around.

Could someone please explain to me how I could use this to power my LED with a suitable high current wall wart (I have a 5v, 2A lying around, I know it won't be able to supply the LED with its max 2A after the voltage buck, and the LM317T's max current is 1.5A without a heatsink, but anything is better than 60ma)? I have found many guides on how to use it to power 2.5-3.5v white LEDs, but not 1.5v IR LEDs. I'm confused as to what vAdj resistor I am supposed to use, and if I need two or one, some schematics I've seen only use 1 while others use 2.

 

You can try using the lm 317 as a current source.

http://users.telenet.be/davshomepage/current-source.htm
http://theparanoidtroll.com/electro...ials/other/constant-current-sourceload-lm317/

Basically with 1 resistor you can adjust the current output . Read about how it works though . You can try adjusting it to give a 1 amp constant output.

 

For a small change in voltage, the current change in the LED can be high and can destroy the LED. That is why the LEDs are being driven by constant current.
It is not necessary and also not advisable to drive the LED with its max. current. Lower current will produce lower intensity. Operation near maximum ratings will reduce the life! If the LED is operated with a current of 1A, the power dissipation on LED would be 1.6W assuming 1.6V drop, and the LED will require a heatsink.

It is not LM317T's max current is 1.5A without a heatsink. It is 1.5A max with or without a heatsink. The power dissipation on the LM317T would be the drop on LM317T X Current, and with 1A and an input of 5V, it would be (5-1.6)X1 = 3.4W. That would require a heatsink too.

A single resistor of 1.2 Ohms 2W, between adj and output is sufficient (Output should be taken on the adj terminal end) will give 1A current. A capacitor of 0.1uf plus some bulk capacitor at the input and ground should be used.

 

For a small change in voltage, the current change in the LED can be high and can destroy the LED. That is why the LEDs are being driven by constant current.
It is not necessary and also not advisable to drive the LED with its max. current. Lower current will produce lower intensity. Operation near maximum ratings will reduce the life! If the LED is operated with a current of 1A, the power dissipation on LED would be 1.6W assuming 1.6V drop, and the LED will require a heatsink.

It is not LM317T's max current is 1.5A without a heatsink. It is 1.5A max with or without a heatsink. The power dissipation on the LM317T would be the drop on LM317T X Current, and with 1A and an input of 5V, it would be (5-1.6)X1 = 3.4W. That would require a heatsink too.

A single resistor of 1.2 Ohms 2W, between adj and output is sufficient (Output should be taken on the adj terminal end) will give 1A current. A capacitor of 0.1uf plus some bulk capacitor at the input and ground should be used.

Okay, I tried a 3 ohm, 3W resistor (cause its what I had lying around) between vOut and vAdj (without the LED connected), but I only got a voltage drop of 1.2v across the resistor. Don't I need 1.6v, because my LED is 1.6v?

And whats the capacitor between vIn and ground for? Is it a power filtering cap to reduce ripple? Why do I care about ripple on an LED?

 

Okay, I tried a 3 ohm, 3W resistor (cause its what I had lying around) between vOut and vAdj (without the LED connected), but I only got a voltage drop of 1.2v across the resistor. Don't I need 1.6v, because my LED is 1.6v?

And whats the capacitor between vIn and ground for? Is it a power filtering cap to reduce ripple? Why do I care about ripple on an LED?

You will actually get about 1.25 volts across the resistor. That's the normal voltage across that resistor when the LM317 is operating correctly.

So, 1.25 volts/3 ohms = 0.4167 amps.
Measure the voltage across the LED. What do you get ??

 

You will actually get about 1.25 volts across the resistor. That's the normal voltage across that resistor when the LM317 is operating correctly.

So, 1.25 volts/3 ohms = 0.4167 amps.
Measure the voltage across the LED. What do you get ??

Okay, using my high wattage 3 ohm resistor (which is apparently only 2.7 ohms, according to my DMM), I get 1.2v over the resistor, 1.4v over the LED. (had a hard time measuring the LED, cause I was using alligator clips, and I didn't realise the metal backing of the LED was shorted to ground, but once I realised that, it measured 1.4v).

I also measure 0.50 amps coming out of the negative terminal of the LED. Qualitatively, the LED still isn't very bright. It's brighter than it was before, when I tried hooking it up directly to an AA battery, but not by much, and still nowhere near bright enough to be a nightvision flash light. I'm not sure if it is because the voltage is too low (1.4v vs a recommended 1.5-1.7v), or the current is too low (but 0.5A seems like it should be more than enough for decent brightness). Do I just need a smaller resistor? Unfortunately I don't have any other low-resistance, high wattage resistors. All I have is some 50 and 100ohm, 5w resistors, but only two of each, not enough to make a 2 ohm resistor. I'd appreciate any help!

 

Yep, you will need a smaller resistor. The LED voltage should increase as the current is increased.

 

I don't suppose there's some kind of high-wattage, 0.25ohm-3ohm potentiometer I could use to adjust the LED brightness on the fly? I would imagine not, otherwise these schematics would list a pot, and not say "try the following range of resistors"

 

Okay, using two 0.75ohm, 5w resistors in series to make 1.5ohms (its all the store had in high-wattage, low ohm), I get 1.72v over the LED, and 1.3A current draw. I'm now at the max of the LM317, and I'll probably have to switch to the 3 amp LM350. Everything is getting very very hot. The LM317 has a 2" square piece of 1mm copper as a heatsink with thermal compound, and it gets hot. The LED came mounted on a common aluminum "star", and it gets "ow that hurts" hot. Guess the aluminum star isn't really a heatsink.

Anywho, even though I am now running 2.2W through this 3W LED, it still isn't as bright as the comments on the product page (lights up entire room, can read text in pitch black from 12 feet away), so I think the fault is with my digital cameras. I've tried two so far, and the best I get is "looks really bright when pointed straight at camera, but needs to be within 1 foot to use it to read text in pitch black". Looks like I'm going to have to find one that doesn't have an IR filter built in, or try and remove the filter.

Also, this is really weird, apparently the LED is also a solar cell? It generates between 50mv and a whopping 1 volt all on its own, with absolutely nothing but the DMM connected to the LED's + and - terminals. And it is definitely based on light, it jumps from 50mv to 1v the moment I point a flashlight at it. WTF. I had no idea IR LEDs were solar panels.

 

Alright, I bought a 3W 1.6v IR LED to make a night vision flashlight ...
...
... how I could use this (LM317) to power my LED with a suitable high current wall wart (I have a 5v, 2A lying around,
...

As an alternative, if your 5v 2A wall wart is voltage regulated (ie the 5v remains very close to 5.0v at all times) all you need is a resistor and the wall wart.

The resistor will drop 3.4v (5v - 1.6v) and have 1.87A at the LED max spec. So the resistor value is R=E/I = 3.4v/1.87 = 1.82 ohms. It will need to be a large 10W power resistor as the power dissipated by the resistor is P=E*I = 3.4v*1.87A = 6.4W.

If you use a LM317 it will give a tiny improvement in current regulation over just a resistor, maybe 5% better current regulation, which is meaningless for an IR illuminator. Please be aware the LM317 would also dissipate (waste) 6.4W as heat, the same as the resistor solution.