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I'm not good at physics, but I'm trying to make a lantern. So I have an LED with this characteristics:

  • N = 3 Watt
  • I = 0.7 Ampere
  • U = 3.4-3.6 Volt

So I connect it to 6V battery, but need ~3.5 V. What type of resistor do I need?

I calculated internal resistance with R = U / I and got 3.6 / 0.7 = 5.14 Ω. And using this formula I = 𝛆 / (R + r) I got total resistance of 6 / 0.7 = 8.57 Ω. Here 8.57 - 5.14 = 3.42 Ω of external resistor. Am I right?

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  • \$\begingroup\$ "U", did you mean forward voltage ? \$\endgroup\$
    – Long Pham
    Commented Aug 19, 2018 at 8:23
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    \$\begingroup\$ @LongPham actually, yes. as I see it's in our country Voltage labeled as U \$\endgroup\$
    – mend4x
    Commented Aug 19, 2018 at 8:25
  • \$\begingroup\$ @LongPham V outside of the US is U, but in this case Vf would minimize the confusion for everyone. \$\endgroup\$
    – winny
    Commented Aug 19, 2018 at 8:39
  • \$\begingroup\$ @winny yeah, my country uses U. \$\endgroup\$
    – Long Pham
    Commented Aug 19, 2018 at 8:40

2 Answers 2

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Powering an LED with a battery and using a current limiting resistor is not recommended. Especially for a lantern where you want a more consistent luminous output.

You could use a buck/boost DC-DC converter to keep the voltage consistent and adjust the output voltage for a high efficiency current liming resistor.

You must first measure the actual Vf then adjust the output voltage

The TI High Efficiency Single Inductor Buck-Boost Converter TPS63030DSKR was made especially for this type of application (i.e. battery powered white LED).

enter image description here


I'm connecting two 3V CR2032 Lithium batteries in series.

A CR2032 has a capacity of only 235 mAh and that capacity is for a much lower load than 700 mA, if you can get 700 mA.
Source: DATASHEET ENERGIZER CR2032

with your proposed design the LED will light up for a few minutes then dim to 20% of the original intensity a few minutes later.

I would recommend using Panasonic or Samsung 18650 Li-ion cell(s).


An LED being driven with 700 mA will need some thermal management. You may do better with a higher efficacy LED driven at a lower current.

Your Vf of 3.5V is too high. I recommend the highest efficacy LEDs available today (August 2018)

  • High power Cree XP-G3 (2.7V-3V) or
  • Mid power Samsung LM301B (2.6V-2.9V)


A constant current regulator made for battery operation is recommended. This does not work with your 6V battery. This circuit was designed for a single or dual cell supply (e.g. AA, NiMH). With this circuit the battery voltage should not exceed the LED's Vf

Microchip MCP1643
Very simple inexpensive ($1 single qty) made to drive a single white LED.
0.5V - 5.0V input, 5.0V, 550 mA output.
Works well with mid and high power LEDs.

MCP1643 is a compact, high-efficiency, fixed frequency, synchronous step-up converter optimized to drive one LED with constant current, that operates from one and two-cell alkaline and NiMH/NiCd batteries. The device can also drive two red/orange/yellow series connection LEDs.

enter image description here
enter image description here

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  • \$\begingroup\$ better late than never) I ended up using Li-Pol battery. and it works fine! thanks \$\endgroup\$
    – mend4x
    Commented Sep 29, 2019 at 17:42
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The calculation is awkward, but correct. A '6V' battery is 6.0V when new, and 4.0V when nearly discharged. So, a resistor can achieve the 0.7A current only while the battery is fresh.

Some non-resistor options exist. AMC7140 is an integrated circuit, one of those will keep the current steady at 0.7A with no resistor required.

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  • \$\begingroup\$ I'm connecting two 3V CR2032 Lithium batteries in series. The light from 3V only is not pretty enough, and two batteries with twice more capacity should be better I guess. \$\endgroup\$
    – mend4x
    Commented Aug 19, 2018 at 8:33
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    \$\begingroup\$ Beware that you are throwing away a lot of power ~40% of your batteries energy is wasted on heating the resistor (Which needs to be ~2 Watt) \$\endgroup\$
    – Oldfart
    Commented Aug 19, 2018 at 8:42
  • \$\begingroup\$ @neomendax 2xCR2032 batteries will not supply 0.7A, 7mA more like. What people usually do with CR2032s and LEDs is not bother with any resistor at all, the internal resistance of the battery limits the current quite effectively. For 0.7A, you need a rechargable battery, perferrably 3x2v lead acid, 5x1.2v NiMH, or lithiums with a constant current driver. AA batteries might manage 0.7A for a short while. \$\endgroup\$
    – Neil_UK
    Commented Aug 19, 2018 at 10:43
  • \$\begingroup\$ @Neil_UK you mean using batteries as in mobile phones etc.? \$\endgroup\$
    – mend4x
    Commented Aug 19, 2018 at 11:49
  • \$\begingroup\$ I like this chip with only a 0.5V drop out. Most CCRs I have seen only work up to 350 mA and have a much higher drop out. Great for high voltage applications. The minimum 5V cuts out a single cell Li-ion. battery. Certainly better than a current limiting resistor for a battery powered LED. \$\endgroup\$
    – Misunderstood
    Commented Aug 19, 2018 at 18:53

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