Recently I got a hi power LED (3.3v, 1500mA) and few tp4056 modules and i would like to build a rechargable flashlight with it.

At first I tried connecting it directly to a 18650 battery and ... well it burned.

I tried to find available 3.3v led drivers for powering it, but all of them either work with 220v ac, or at least 5v dc (buck converters).

Is there a way to do this? For example by simply adding a resistor or any other diy friendly way?


5 Answers 5


You'll need a LED driver, probably a buck (no need for a more complicated buck-boost if you accept some dimming when the battery is close to discharged).

Try googling "flashlight LED driver", there are many such products, which are usually in a round form factor which corresponds to the diameter of a 18650 battery. Here is an example.

enter image description here

There are tons of choices on ebay, aliexpress, etc. If you want reviews and tests, googling "flashlight LED driver reviews" will give you some interesting tests, for example:


You can also look at the forum for these things, go into the DIY section and you'll find plenty of advice.

Also at 5W it will be a little brighter than at 3W, but not that much (our eyes do not perceive light linearly, they work more in a log-scale). The lower run time is not always a good compromise. 1-3W is a nice power, unless you want to go all out of course.

  • \$\begingroup\$ I could not find a 3.3v (min) 1.5A buck driver in your link list. \$\endgroup\$ Commented May 6, 2018 at 22:42

You can use a resistor but it would burn a lot of power through heat. Like you said it's not easy to find a LED driver rated for this input voltage.

I found this which seems to be the perfect fit, but I'm not sure how reliable it is.

Another option would be to find a linear voltage regulator, which should be easy to find for this input voltage, and use it as a current regulator. The efficiency might not be as good as using a led driver. Here's a relevant article on that.

  • \$\begingroup\$ Thanks for that link to the CN5711-based board. Any thoughts where to buy just the IC in North America? (I don't see it on Digikey). Any other, better alternatives come to mind, one year later? I need to independently drive a couple 1A Cree LEDs from a single LiPo cell, with PWM dimming. \$\endgroup\$
    – rkagerer
    Commented Jun 18, 2019 at 7:51


The Cree's XP-G3's Vf will be ≈3.1V, likely less.

I will need to back off on the 0.13Ωn resistor. I now recommend a 0.47Ω

The Cree is spec's at 2.73V @ 350 mA @ 85°C
With a max 2000mA where Vf = 3.06V @ 85°C.

enter image description here

An 18650 at 2000ma will have a max of 4V.
Voltage will drop rather quickly to less than 4V.
Now that I know the Vf (3.06) at max current (2 Amp) of the LED and the max voltage is ≈4V.

97 resistors that will work: Digikey 0.47 3W+ Resistors
I would use the Yageo PSP400JB-0R47

enter image description here

This spot lens is made for a Cree XP LED. LEDil is a very good company.


Here is 108 others: Digikey Spot Lenses

enter image description here

Heat is going to be a big issue here.

I would make a custom PCB for this. These guys do a very good job and small PCBs under 4" x 4" (100mm x 100mm) are very cheap, $5 for 10 PCBS. Shipping to the US is $21 DHL. PCBway

If it is in your budget I would use 2 oz. copper, 2 sided FR4 with the immersion gold plating would be a good option for thermal management. It is much more planar (smoother) than HSAL and will help a lot with thermal conduction between the PCB and heatsink.

Qty 10 with 1 oz copper: $5
Qty 5 with Immersion gold plating over 1 oz copper $39
Qty 5 with HASL over 2 oz copper: $41
Qty 5 with Immersion gold plating over 2 oz copper $66

I would then sandwich the PCB between two plates of copper or aluminum with a hold large enough for the LED and lens. I would flood the PCB top side with copper for the XP-G3's thermal pad. I would flood the bottom side with copper and drill a lot of vias near the LED to connect the top and bottom thermal copper.


An 18650 battery is a very good choice for powering an LED.

An LED with a 3.3 Vf works very well with an 18650 and resistor.

You did not specify the part number of lumens for your LED. My choice would a Cree XP-3G, the most efficient (185 lm/w) high power white LED. Over 600 lumens and 3.1V @ 1500 mA. Cost about $1 in small quantities. At 1500mA, 4.65W, your are driving it hard so the efficacy drops to 130 lm/w.

There are LED drivers made especially for a single Li-ion cell. For example the TI TPS63030DSKR but it has a maximum of 800 mA. Finding a driver with higher amperage will be very difficult.

The current limiting resistor is not as bad as one might think. 3.3v at 1500 mA will run hot. Hot will drop the Vf a little bit. And LED luminous intensity will drop also.

The mid-point in the discharge curve is 3.5v, so that is the value to use when calculating the resistor. That will yield an average efficiency of 94%. At the start of the discharge curve at 4V the efficiency is 83.5%. At 3.4V 97%. You cannot beat 94% with a switcher current source.

enter image description here

enter image description here Source: hobby-hour.com LED Series Resistor Calculator

A good flashlight is very difficult to make. The flashlight I'd like to have is the Coast HP314R, a $500 rechargeable.
The non-rechargeable HP314 is only $330 powered with four D cells.

This shows these guys charge $170 for the Li-ion batteries and mini USB charger. Getting 3.25 hours of 1200 lumens out of batteries is not easy. They must be using top of the line Li-ion batteries.

Batteries last 3.25 hours @ 1200 lumens, length=17", weight 1.4 kg,b beam distance 2762 FT / 842 M.
enter image description here

I also like their $150 AR25R and am seriously considering buying one.

enter image description here

You are probably better off buying one than making one. The optics are the key.
LINK: Coast Flashlights
Home Depot has good prices on Coast flashlights.

Be sure to check out Battery University to learn the optimum way to charge a Li-ion battery. For example do not allow the battery to discharge below 2.8V. And for longer cycle life, do not charge over 4V.

  • \$\begingroup\$ This is a great answer. Thanks for being thorough. First few clarifications: 1. Yes, the LED is a cree XP-g3. 2. Using a resistor instead of an LED driver for this spec is easiercand cheaper, but finding a 0.6 ohm resistor??? I am going to power my led by 2 18650 li-lion cells in parallel and a charging protection module and a tp4056 for charging it via micro usb. \$\endgroup\$
    – Euphoria
    Commented May 7, 2018 at 14:16
  • \$\begingroup\$ Also can I ask where did you get the screenshot from? It seems to be a great resource! \$\endgroup\$
    – Euphoria
    Commented May 7, 2018 at 14:27
  • \$\begingroup\$ @Euphoria added source link in answer. \$\endgroup\$ Commented May 8, 2018 at 2:23
  • \$\begingroup\$ Ouch $500 - but it peaked my interest. Apparently BestSun is bigger bang for the buck. A "15000 Lumen Tactical Flashlight Torch with 16x Cree T6 LED" is less than $40 and even "2000 Lumen CREE XM-L T6 LED Flashlight Torch Zoomable" which is more similar is only $10, and they are 'aircraft aluminum'. I imagine some more searching could turn up better for less - $500 is a big budget for a flashlight. \$\endgroup\$
    – Rob
    Commented Feb 3, 2019 at 14:39
  • \$\begingroup\$ @Rob Bestsun appears to be a charlatan vendor. There is no way to get 11,000 lm with 9 Cree XM-L. @2 amp, a XM-L T6 can output 692 lm max, but that would be nearly 60 Watts and the flashlight would get to hot to touch. I would not buy from a non-Prime (no free returns) vendor. Coast flashlights are rated to ANSI/FL1 standards. Beam distance lux is the key, not lumens. The charlatan vendors use high wattage LEDs but do not run them at max intensity. With Coast you are paying $170 for the built-in charger. What I like about Coast is the beam focus and distance. \$\endgroup\$ Commented Feb 4, 2019 at 20:55

A simple solution that also allows for PWM dimming later on, would be the manual solder-friendly low-cost AMC7135 IC, which delivers a constant current (which is what LEDs require) of 350mA and will accept input from a 4.2V (fully charged) Li-Ion or Li-Po rechargeable battery. For 700mA, 1050mA, or 1400mA, one can simply use 2, 3 or 4 of them.


A resistor will burn a little power through heat.

Worst case, 4.2V (battery) - 3.3V (LED) * 1.5A = 1.35W, little more than 1/4 of the LED power, which is probably tolerable for DIY.

R = (4.2-3.3)/1.5 = 0.6 ohm 2W.

As the battery voltage declines, so will the voltage difference, so the brightness will fall off faster than a better solution.

But in view of its simplicity, it's a good starting point, even if you decide you need a better solution long term.

But also you MUST heatsink a 5W LED adequately, following the procedure described in its datasheet, and you say nothing in your question about that.

  • \$\begingroup\$ An 18650 battery is 3.6v nominal. The max charge voltage is 4.2. At 1C discharge it will drop to 4V almost instantly. The better voltage to use for resistor calculation is 3.5v, the mid point of the discharge curve. \$\endgroup\$ Commented May 6, 2018 at 22:41
  • \$\begingroup\$ "better" for what purpose? If 1.5A is the LED's max current you can argue for calculating the resistor at 4V, but if you use 3.5V you will shorten the LED's life. (And that's closer to C/2 not C) \$\endgroup\$
    – user16324
    Commented May 7, 2018 at 0:19
  • \$\begingroup\$ Better for even brightness, throughout the discharge curve. Better efficiency. An 18650 cell can easily run at 2C. 5-10C is not unusual. An LED specified at 3.3V 1.5A will have a higher max current. For example a Luminus 3.25V SST-90 is spec'd at 3A and max is 9A. A Cree XP-3G spec'd at 350mA max 2A. For longer cycle life and more lifespan capacity it is recommend not to charge an Li-ion over 4V. Optimum (for greater cycle life) max charge voltage is 3.92V. The LED will not spend much time at the higher current as the discharge curve is steepest at the beginning. \$\endgroup\$ Commented May 7, 2018 at 0:55

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