LED driving mad

Hi, newbie here please help me understand, i am trying to drive 2 LEDs with 1 or 2 Li-ion 18650.
LEDs have Vf=3.4-3.8V If=700mA (that's max i assume).

Driving them in series, i need 6.8-7.6V supply and about 650mA current.

So i fed 8.6V to an LM317 (in current control configuration with 2 Ohm resistance) which worked for 1 LED but not for 2.

To my surprise, the LM317 seems to be presenting a resistance of about 10.5 Ohm, making the 8.6V insufficient for driving both LEDs.

it also would consume close to 1W alone and at 650mA it heets up a lot, making the use of an additional heat sink necessary.

I then fed from the same source to the LEDs with just a 1.43 Ohm resistor in between (plus about 1.45 Ohm which seems to be my cables/connections resistance) and worked just fine, with about 1.07W consumption (apart from the LEDs) and with 7x10R resistors in parallel everything was pretty cool to the touch.

So no heatsink needed, no higher voltage and the power consumption the same.

What am i not getting here ? What's the use of the current controller then ?

Is this a "proper" way to power the LED's ? What would be ?

This is supposed to be a handheld flashlight, so size of components matters.

I apologize for the post being so long

The LM317 has a dropout voltage of 2V or so like most standard linear regulators. You simply have
no voltage overhead for such a regulator in that setup.

You could try searching for an ultra-low dropout voltage regulator, or use a constant current
LED boost-converter from a single cell, or a buck regulator down to about 4V and a small
series resistor to limit current.

Hi MarkT,

"dropout voltage of 2V" that's what i didn't know indeed.

So, leaving it without regulation, with just a resistor in series is OK?

Otherwise i'll go for a single cell with boost converter.

Thanks :slight_smile:

In this case (particularly if you're just using a single cell) the easiest approach would be to put the LEDs in parallel and use a current limiting resistor.