I'm getting crazy trying to figure out if there is any easy way to power up a SINGLE 3w led
3w - Forward Voltage: DC 3.2-3.4V - Forward Current: 500-700mA

with a single 18650 li-on battery

Minimum Capacity: 2250mAh (0.54A discharge at 20°C)
Typical Capacity: 2100mAh (0.54A discharge at 20°C)
Nominal Voltage: 3.6V
Discharge End Voltage: 2.5V
Standard Charging Current: 1.1A
Charging Voltage: 4.20±0.03V
Standard Charging Time: 4.0hours
Max. Continuous Discharging Current: 2.2A
Internal Resistance: less than 35mΩ
Weight: less than 47.0g

Is there anyway to do it without using a dc-dc buck converter? I'm trying to change a led on a torch, so I have not so much space to work in.


  • \$\begingroup\$ can you make a heatsink? \$\endgroup\$ – Sunnyskyguy EE75 Jan 4 at 18:51
  • \$\begingroup\$ Is it a single led or a string? \$\endgroup\$ – Satish Singupuram Jan 4 at 18:54
  • \$\begingroup\$ It's a single 3w led, and I have already a small heatsink on the torch, and one pcb heatsink on the led \$\endgroup\$ – mosfettis Jan 4 at 19:00
  • \$\begingroup\$ You can achieve that by using a simple current limit resistor of require wattage between battery and led, but the brightness of the led will decay with the voltage of battery and you cannot use the battery to its maximum capacity... is it ok? \$\endgroup\$ – Satish Singupuram Jan 4 at 19:03
  • \$\begingroup\$ I already tried with 1ohm, 1,2 ohm, but the led is not powering up. Am I doing wrong calculations? \$\endgroup\$ – mosfettis Jan 4 at 19:06

A buck-boost regulator will give you constant brightness and constant current but involves more design.

A Thermal sensor that rises in resistance with temperature will work well with the right part.

I would choose a 350mA to 500mA PTC and attach it to the LED heatsink. They are rated by Holding current and trip current is 2x but if also heated by LED, it will trip earlier depending on thermal design.

Given the wide variation in Ri (internal resistance) of the 3W LED from 3.2-3.4V @ some rated current like 500mA this is equivalent to Vf=2.8V + If * R1 for Ri = 0.8 to 1.2Ω adding 0.5 Ω parts in series/parallel as battery drops in voltage then bypass. Normally a 3W LED is Ri= 1/3 Ω.

Your specs may vary but it appears to be a 1W LED with 3W pulse capability. 2.6V is dim 2.8V is 10% max current.


simulate this circuit – Schematic created using CircuitLab


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