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I want to drive one or two small (5 or 10 mm) white LEDs in parallel using one or multiple 18650 3.7vV Li-ion cells. The main reason I want to use the 18650's is, I already have them, they are rechargeable and have a pretty good mAh rating.

Is this as simple as putting a resister and a switch and I'm done? What about when the battery is drained and can only output 2.x volts? Will the light simply not light at that point and we charge the batteries?

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  • \$\begingroup\$ The simplest situation is where you use one rechargeable battery and, if paralleling LEDs, use one resistor per LED to provide some modest current limiting. It's just easier that way. The end-point voltage for an 18650 is about \$3\:\text{V}\$. So if your LEDs are red and need only about \$2-2.2\:\text{V}\$, then it may be fine. The more complicated approach is to use a small inductor to boost the voltage. But for this and an 18650, you'd probably want to stack two or three LEDs so you only have to boost the voltage upward. (A buck-boost is harder and to be avoided.) Or, an IC is an option? \$\endgroup\$
    – jonk
    Sep 4 '18 at 22:39
  • \$\begingroup\$ i was going for a white led, which are somewhere in the 3.3v range... \$\endgroup\$
    – Dave
    Sep 4 '18 at 22:50
  • \$\begingroup\$ It would be helpful if you provide a link to datasheet of the LED you have in mind \$\endgroup\$
    – Maple
    Sep 4 '18 at 22:52
  • \$\begingroup\$ Then two of these LEDs in series (with resistor also in series or else an active current limiter with a BJT or two) and an inductor boost circuit might be appropriate and could work with a single battery. Such circuits and their inductor can be easily fabricated at home without an IC. Sometimes just out of spare parts grabbed out of a dead CFL or LED mains lamp, if you have one. How much work are you willing to go to? Would you just prefer to buy an IC? \$\endgroup\$
    – jonk
    Sep 4 '18 at 22:54
  • \$\begingroup\$ Perhaps if your concerned about over discharge, you could replace the one white LED with two RED leds (in series). Then you don't need a resistor. Plus due to red LED forward voltage (1.8V), overdischarge won't be an issue. \$\endgroup\$ Sep 5 '18 at 8:07
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I Want to drive one or two small ( 5mm or 10mm ) LEDs in parallel

is this as simple as putting a resister and a switch and i'm done?

It is even simpler. Put a resistor and LEDs in series and you are done.

What about when the battery is drained and can only output 2.x volts?

If Li-Ion battery is drained to 2V it is as good as dead. You should never let it discharge below 3V, ideally recharging at 3.3-3.4V

Which means it will most likely be higher than double of most LED forward voltage, allowing you to connect diodes in series.

If your LED forward voltage is above 1.7V then you can use them in parallel, but you need individual resistors connected in series with each LED.

So, that takes care of LEDs. Now, if there is no very strong reason to use two batteries in parallel I'd recommend staying with one 18650 cell. There are many potential problems with using Li-Ion cells in parallel. You can search this site and find literally hundreds of questions and answers on this one, if you are curious. Otherwise just stick to single cell.

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