Wish to build a 5w IR led flashlight / illuminator. Noob.

[Monkey brew]

Hi everyone.As I wrote in my introduction, I'm quite new to the world of leds, but have a minor knowledge of electronics. In other words, I can solder stuff and follow directions. I always use calculators, since my brain doesn't always agree with me. Anyway, I'm trying to wrap my head around a project I'm working on at the moment. I'm building a night vision scope and need to have more illumination. I've recently bought a IR led (see specs in pics), and I simply can't figure out how to power that sucker. And not to mention the size of the heatsink. I've look for several days for answers, without much luck. I've tried to figure out what type of driver to use. I've used buck converters for many projects, and someone mentioned that it might be possible to use something in that order. I don't know. I does deliver constant voltage and current, and can be adjusted to any given level of power. The led is to be installed in a relatively small enclosure, but there should be enough room for a 30mm - 50mm heatsink and a 25mm 5v cooling fan. The idea is to power the led with high drain li-ion 18650 batteries. I hope that anyone of You could help to explain the leds power needs, since I've hit a mental wall with explanations of forward current, forward, voltage, power consumption, max current and so forth. I gather that all the information I need, is in the datasheet, but it turns into mush in my brain to apply the numbers to practical use. I've added a link to the led in focus.Thanks in advance. 👍https://www.ebay.com/itm/321106122595

 

[Monkey brew]

Was wondering if a simple led flashlight driver board could solve my problems?

 

[DIWdiver]

Possibly one of the reasons you are confused is that the listing you linked has numbers that just don't add up.

The max voltage is listed at 1.5V, the max current is listed at 1.5A. If you multiply these together, you get power. 1.5V x 1.5A = 2.25W. Yet they claim it is a 5W LED!

Even if you use the max pulse current, 1.5V x 3A = 4.5W. Clearly (as mud) they used this and rounded up.

But not all of this is dissipated as heat. A non-trivial portion of it comes out as light. I wouldn't guess the efficiency of this emitter is impressive, so something in the 15-25% range is probably right.

Anyway, what you want is an LED driver with an output of up to 1.5A. If you want to play fast and loose with the reliability, push to 1.5 or even a bit higher. If you are conservative, back off a bit.

Notice I said nothing about the voltage. LEDs are usually driven with current instead of voltage. Most LED drivers are constant-current regulators, though some are better at it than others. The job of the driver is to take whatever power source you have and deliver a regulated current to the LED.

This is the opposite of how most incandescent lamps are driven, with voltage. 120V for line-driven lighting, 12 or 24 for vehicles and low voltage lighting, 3V for those old-fashioned 2-cell flashlights, etc. When you apply the voltage to a cold bulb, it's resistance is low, and a large inrush current occurs. The filament heats up, the resistance increases, and the current is reduced. So in a sense we can say that the driver (power line, alternator, battery, etc) supplies voltage and the bulb determines the current. But by picking the right bulb, we have a pretty good expectation what that current should be, within a small range.

For reasons I won't explain here, we typically supply LEDs with current, and the LED determines the voltage. If we supply 350 mA (as noted in the link), we expect the bulb will have a voltage around 1.4V. If it's quite cold, the voltage will be a bit higher. As the LED warms up, the voltage will drop a little.

If we supply more current, the Vf will go up; less and it will go down. This listing doesn't tell us how much, but experienced people could make reasonable estimates. Having limited experience here, I would guess 10 mA it might be 1-1.2V, and at the 1.5A limit it would probably be somewhere around 1.8V. Someone with more experience may be able to give you more accurate guess, but the only real way to know is to try it (since you don't have the manufacturer's data).

Now for the driver type. Given the low Vf of the LED compared to the battery, you want one that can reduce the voltage with good efficiency. That counts out linear regulators like the AMC7135. You'd dump more power into the regulator than into the emitter. That's exactly what switching regulators were designed for. Because you need to step the voltage down from the battery to the emitter, you want a buck type. A buck-boost type could work, but is not likely to be as efficient as a pure buck type, given similar expense and build quality.

And yes, there are simple flashlight drivers that fit this bill quite well.

 

[Monkey brew]

Thank you DIWdiver, for that very elaborate answer. I've just written to the seller, requesting more information on the LEDs specs. Finding a solution for the problem as it is, is a bit of a challenge. If the answer is the same as stated in the sellers description, I might reconsider another solution with a 12v array and a 12v battery pack. Somewhat easier.