What's the point of the LM317?

Hi all, first post.

I've recently been researching for my own first build, and I came across the LM317 driver.

My plan to power my laser is to use supercapacitors, purely as an excuse to play with them as well. The thing is, because I'm using a flashlight, I have limited space. For the LM317 driver, I understand I'll need roughly 7.5V for a red diode. This is roughly 3 supercaps in series, which I might not have room for.

My question is this:

If all the LM317 is doing is regulating the current, what's to stop me just using a resistor to drop the remaining volts and limit the current? This would surely eliminate the need for the 3V needed to power the LM317? Which means I could use two supercaps in series, saving space, meaning I could maybe even get in another bank to retain capacitance lost from the caps being in series.

So why can't I just use a resistor? Am I missing something obvious?

Sam

It's just that what you already mentioned. The LM317 "Regulates" current and a resistor "Limits" current.

The LM317 will give a fixed current regardless of input voltage (within limits) while a resistor would let a certain current flow depending on the voltage across it. So once the voltage drops, current will also drop. Which is fine, until you connect a little higher voltage than usual and exceed the diode's rating and blow it. Laser diodes are touchy that they are very sensitive to even minor over currents and voltage spikes.

Like djquan said above, you CAN just use a simple resistor, BUT the current it allows will change with voltage, so it's only recommended if you're going to be underdriving the diode by a fair margin. For example, if you choose a resistor to run the diode at max on fully charged supercapacitors, when you go to charge them, the voltage will be a bit higher, the diode will get a higher current, and you'll pop it. Using a LM317/other driver means you can change the voltage around all you like and the current will always be the same.

I don't know where you got the 7.5V figure from. Vf of a regular red diode is at max about 3V, allowing 1.5V dropout means 4.5V would work fine.

I think the dropout for an LM317 in constant current mode is ~3V so he would need around 6V.

Indeed, forgot about the drop of the resistor which is 1.25V, so it'd be 2.7V. Still, well under 7.5V

Thanks guys, it was extremely late when I posted this and as soon as I woke up I realised what a stupid question it was and worked it out. For some reason last night, I was thinking that the resistor, once I'd worked out the current at a certain voltage, would then maintain that current should the voltage change, just like an lm317.

Just got to look at the dimensions of my host and figure out the best combination of super caps now! Who knows, I just might end up using rechargeable batteries!