Thanks in advance for anyone who can help me with my specific question!

I have many 18v Li-on battery packs (1.5Ah-4Ah) due to the other tools that use this voltage, and I am making a lamp using a specific 12v 36W filament. The spec sheet for the filament states colour temp up to 14.5v input so I know it can take this voltage no problem.

Using P=VA, I know the filament will draw 2.91A and so the resistance of the filament is 4.12 ohms. Hopefully I am correct so far!

I want to reduce the voltage from the battery to 14.5 volts, and have adjustment due to the slight variation of voltage from the battery as it depletes (20v-16v), to enable the colour temp of the filament to stay constant.

What is the best way to do this? Can I use a simple adjustable voltage divider such as a potentiometer, and if so, what rating do I need, 10k, 50k? It's this part that I cannot work out, confused!! Many thanks.

  • 2
    \$\begingroup\$ You would waste a lot of energy, which is going to be dissipated, so you need at least some 10W rated pot. These are rather big and not really cheap, you are better of using something more efficient like a buck converter. \$\endgroup\$
    – PlasmaHH
    Sep 24, 2015 at 10:45
  • \$\begingroup\$ I was just having a look at the costs of 10W+ pots out of interest and a few low resistance ones I found were > $100. You can pickup the buck regulators pre-made on e-bay for $5 odd: ebay.com.au/itm/… \$\endgroup\$
    – PeterJ
    Sep 24, 2015 at 11:13
  • \$\begingroup\$ The power will be more than 36W if you operate it at the higher voltage (but no more than 44W). The color will be more blue, and the lifetime will be drastically reduced (your trade-off to make). Use a buck regulator, not a resistor!! Current will be less then 3A even with 16V in. \$\endgroup\$ Sep 24, 2015 at 14:57
  • \$\begingroup\$ Thanks very much for the info PlasmaHH and Spehro, extremely helpful, I'll use a buck regulator as you stated. And thanks Peter for the link, top man. \$\endgroup\$ Sep 25, 2015 at 12:26

1 Answer 1


Don't use a voltage divider or any other linear method of reducing the voltage. Use a buck regulator. Good ones are usually about 95% efficient at converting from a higher voltage to a lower voltage so with a load of 36 watts, the power dissipation might be ~2 watts.

Here's what I mean (and note that it can be easily made to run from 18V in to produce 12V (or 13 or 14 or 11 or 10 volts out): -

enter image description here

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  • \$\begingroup\$ Thanks very much Andy for the excellent info and circuit diagram, very helpful. I'm off to buy a buck regulator! Have a great day :-) \$\endgroup\$ Sep 25, 2015 at 12:31
  • \$\begingroup\$ @CliveWalpole - if you are happy with the answer (and TBH I would LOL) then please consider marking it as "accepted" - there's a symbol near the upvote/downvote thing that does this. Bear also in mind that this is a surface mount one with a paddle and it may not be that easy to soder down home-brew style. Maybe consider one of the many leaded types? \$\endgroup\$
    – Andy aka
    Sep 25, 2015 at 12:39
  • \$\begingroup\$ I'm very happy thanks, it's exactly what I need, easily available and cheap to buy, spot on! I'll probably secure with epoxy or hot glue, within the body of the hand held De-Walt torch I'm using. Thanks again for your help. \$\endgroup\$ Sep 25, 2015 at 16:18

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