IR LED off of 24V rail - Best way to do it?

I have a situation in a piece of equipment where power to a necessary IR LED has malfunctioned. Every other function on this equipment works fine, just for some reason it stopped powering the LED. The only DC power available inside the unit is 24 V straight off the PS.

Using a linear circuit, I have confirmed that powering the LED by bypassing the main board works, unfortunately (obviously) it puts off a lot of heat. I'd like to put a small board inside the unit, with a switching supply to power just the IR LED.

After some poking around, I've determined this circuit will work, and I can assemble for less that $1.50. This is pretty much straight from the LM2576 data sheet, just with my load attached. Data sheet also calls for a much larger inductor, as the regulator is rated at 3A, but I'm only going to be using ~80mA. So I'm planning on using an EC36-101K, which is rated for about 275mA.

Is there a better way to do this, or anything wrong with doing it this way?

 

Thanks for your reply! I did the eBay search, and sure enough, there's a lot of ten of those assembled units for $12. Probably the better way to go.

I had considered a single series resistor, but efficiency is a concern. This will be in a smallish enclosed area that already requires one fan, so I need to keep heat to a minimum.

 

The 5V version of the 2576 was the only version available from where I was getting the parts (also the only buck converter they had), and only $0.58 a piece. Sourcing parts elsewhere increased cost significantly, so glad I was able to get the pre-assembled units for even less than I had counted on making my own!