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I'm going crazy trying to figure out if there is an easy way to power a single 3 W LED (3 W - Forward Voltage: DC 3.2-3.4 V - Forward Current: 500-700 mA) with a single 18650 Li-ion 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 this without using a DC/DC buck converter?

I'm trying to change an LED on a torch, so I don't have much space to work in.

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  • \$\begingroup\$ can you make a heatsink? \$\endgroup\$
    – D.A.S.
    Commented Jan 4, 2019 at 18:51
  • \$\begingroup\$ Is it a single led or a string? \$\endgroup\$
    – Electron
    Commented Jan 4, 2019 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
    Commented Jan 4, 2019 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\$
    – Electron
    Commented Jan 4, 2019 at 19:03
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    \$\begingroup\$ If you put 100 Ohms to 5V to LED measure the voltage of LED ~ 50mA and verify it works. Then measure battery. It must be 3.7min. Then report results in question. \$\endgroup\$
    – D.A.S.
    Commented Jan 4, 2019 at 19:41

1 Answer 1

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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.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ a simpler alternative would be to just use a constant current regulator, there are many specificaly designed for this task \$\endgroup\$
    – diegogmx
    Commented Apr 7, 2020 at 21:24
  • \$\begingroup\$ @diegogmx which one do you recommend with say 50~150 mV dropout @ desired current? I know it's just a current sense, Comparator and low RdsOn FET in a 4 pin IC \$\endgroup\$
    – D.A.S.
    Commented Apr 8, 2020 at 0:33
  • \$\begingroup\$ 10,000 Farads? Really? Oh, I think that is part of a model for the LiIon cell. \$\endgroup\$
    – PStechPaul
    Commented Jun 19, 2023 at 19:09

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