I am planning on using a DeWalt 20 V battery pack to power two Thrunite TH30 flashlights. They need 3.6 volts and 5.18 amps for both of them combined.

The flashlights will be physically connected, but with separate power buttons. I will be using the DeWalt battery pack to power the lights exclusively, they will not be using their own batteries. In theory, I could just solder the wires to the battery contacts in the lights and they would work like that.

Is there a way that I could reduce the voltage while extending the length of runtime?

-The LEDs are Cree XHP70.2 which have a maximum drive current: 4800 mA (6 V), 2400 mA (12 V)
-The batteries are 3100mAh IMR 18650 3.6 V

I am open to any advice or refinements to the idea that you may have.

Here is the list of parts I have.

Buck Converter 3 Amp Fuse 13 kOhms Resister 220 uF Capacitor 10 uF Capacitor 220 uF Capacitor

TL431 2N3906 2N3904 330 ohms 51K Ohms 10K Ohms 22K Ohms

I built the PCB twice, and both times there was a short. The caps are not the problem, and there are no shorts in the actual circuit board. When I solder the buck converter to the rest of the circuit, it then shorts.


Above Here is the circuit diagram. Copper is Orange, The caps and resistor are green. The white dots are the contacts for the converter.

Blank PCB

Above Here is the blank PCB. There are no shorts.

PCB with 10 uF Caps

Above Here are the 10 uF caps on the board, again no shorts. Burnt Converter

Above Here is the converter, It shorted when I added it on above the 10 uF caps and applied power. It melted the black chip on the right.

I think there was a short somewhere in the converter. Maybe I melted something accidently when I was soldering to the leads? I'm assuming that I need to buy another converter. I think the rest of the circuit is sound, so I'll rebuild the circuit when I get the replacement converter.

  • \$\begingroup\$ A DCDC charger converter is need to regulate the output to both loads wired in parallel with suitable current and voltage limits \$\endgroup\$ Feb 15, 2022 at 7:16
  • \$\begingroup\$ Do the flashlights have batteries? Are you going to leave the flashlight batteries in place or get rid of them and use the dewalt battery (with appropriate step-down circuit) as your power source? Tony Stewart is saying you need a charger because he is assuming you are going to keep the batteries in the lights. And I agree with him. UNLESS you are removing the batteries. Then you just need a buck converter to supply around 4 V to the lights. \$\endgroup\$
    – user57037
    Feb 15, 2022 at 7:56
  • \$\begingroup\$ @TonyStewartEE75 The lights have built-in charging ports and Dewalt sells an adapter for their batteries that lets you use two USB cables to charge your phone, etc. But I will not be using the flashlight batteries, so I will need something different. Thanks for your help though, I appreciate it! \$\endgroup\$
    – Mark H.
    Feb 15, 2022 at 19:29
  • \$\begingroup\$ How much current do the flashlights draw from 3.6V? You say the LEDs can take 6V 4.8A which is ridiculously high, this would require a special lithium battery like IMR. But do the flashlight's internal electronics actually deliver all that current? Perhaps it is lower. You must know the current to choose a suitable DC-DC converter. \$\endgroup\$
    – bobflux
    Feb 15, 2022 at 19:41
  • 1
    \$\begingroup\$ Please don't buy the LM2596 modules. They're all counterfeit and they will fail. You can google "counterfeit lm2596" for examples. \$\endgroup\$
    – bobflux
    Feb 16, 2022 at 7:57

3 Answers 3


With such high difference of input and output and the requirement for efficiency (long run times on battery), you should prefer switching regulator rathr than linear.

The type of regulator you need depends on the output current that you need.

If the consumption of your flashlights is less than 3 amps, google "LM2596", you will find some modules (small boards) with this type of regulator. Output voltage can be set using a trimmer (typycaly a blue box with screw-head on top).

For better reliability, the trimmer can be desoldered and replaced by constant resistors (there is usually SMD pad under the trimmer), but that is not necessary.

  • 2
    \$\begingroup\$ By reliability, are you referring to accidentally changing the trimmer? A dab of hot glue on the screw can also take care of that if the soldering is too finicky. \$\endgroup\$ Feb 15, 2022 at 9:53
  • 1
    \$\begingroup\$ I think the trimmer can also be affected by humidity and probably somewhat aging. But the change of output won't be significant for flashlight. \$\endgroup\$ Feb 15, 2022 at 19:22
  • 1
    \$\begingroup\$ Ah yes good point. It may not matter here, but it's always good to remove unneeded complexities that are potential points of failure. \$\endgroup\$ Feb 16, 2022 at 13:04
  • \$\begingroup\$ @Passerby The LM2596 seems like a good idea, except I need around 5.16 Amps for the lights. Thanks for your help though. I forgot to answer you sooner. I'm sorry about that. \$\endgroup\$
    – Mark H.
    Mar 19, 2022 at 0:27

Here's a nice ready-made DC-DC (link) that will suit your needs. It has a maximum current of 9A so you only need one. You can adjust output voltage with a resistor (read datasheet). You can use a 50k potentiometer to adjust, then replace with a fixed resistor of the appropriate value once it is set. This is to avoid the failure mode of the pot wiper disconnecting with age.

The flashlights should be wired in parallel.

Here's a link to the mouser search if you're interested in other models.

Please do not buy the fake LM2596 modules. They're all counterfeit, they will die, and they will destroy the flashlights.

Since the flashlight runs on one lithium battery, you can be sure its maximum input voltage is at least equal to the voltage on a fully charged cell. So I'd suggest setting the DC-DC to 4.2V instead of 3.6V. This will increase efficiency of the flashlight LED driver.

You also need to know if the battery has undervoltage protection. Some power tools brands put that feature in the battery, but some others implement it in the tool. In the later case, your project could over-discharge the battery, which greatly shortens its life. So if the battery does not have undervoltage protection, a TL431 controlling the DC-DC's ENABLE input will do the job.

Note the flashlight's undervoltage protection is not relevant here, because it monitors the output voltage of the DC-DC converter, not the output voltage of your 20V battery.

According to the specs the trim resistor should be 13 kOhms for 4.2V output. So you don't need a pot, just a resistor, unless my calculation is wrong of course.

enter image description here

You should wire the lights in parallel (as drawn). Not in series!

Capacitors circled in green are not optional, and both should have very low ESR. The schematic above gives bare minimum values, but the eval board recommends a lot more, so that's what I'd use...

Considering the high currents, you need proper wiring and grounding. Personally I would either make a PCB, or use a piece of blank copperclad PCB to mount the module on. With a bit of milling with a dremel you can make some big fat copper traces like that, which give low resistance and low inductance, which are both important here.

enter image description here

Copper in blue, capacitors in orange.

Here's the link to the eval board schematics. If you want to do the same, they put 3x10µF X7R 63V ceramics on the input and 5x10µF ceramics on the output, and one polymer cap on both sides.

You can use other caps, no need for 63V since you'll only be using 20V on the input and 4.2V on the input. As usual with ceramics, it's cheaper to get 10 of the same value rather than 3 of one value and 5 of another, so just search for 10µF 35V X7R, pick the cheapest ones in qty 10, buy 10, and just put 5 on the input and 5 on the output.

Suggestions for input polymer cap and output polymer cap.

If you use a 6V3 cap on the output, do not solder it before adjusting the output voltage!

Given the current, you have to use polymer caps, not normal cheap caps. The latter can't handle the ripple current, they'll just overheat and pop like the ones on the fake LM2596 modules. And don't get leaded ceramic caps, they have to be SMD for low inductance.

Here's an undervoltage protection circuit using TL431. When voltage on this chip's REF pin rises above 2.5V, it will draw enough current to turn on both transistors and set ON/OFF pin to 0V, which turns on the switching converter. Voltage can be adjusted by changing the 51k resistor. You can use any cheap small signal transistors, 3904/3906, BC547/557, etc.

enter image description here

  • \$\begingroup\$ Ah, thank you for finding that. I will use it instead of the LM2596's. I believe most DeWalt power tools have Undervoltage protection built in them. My flashlights blink once the battery is (empty), so I think I'm good in that aspect. 4.2 V would work great, as the battery gives that voltage at the start of its output. \$\endgroup\$
    – Mark H.
    Feb 16, 2022 at 8:47
  • \$\begingroup\$ I'm sorry, I am finding it tough to grasp, how do the voltage (and amps) change with the second light being used in parallel? Do I need to double the power, or does twice the amount of power flow through at the regular current? If twice the voltage is going through the wires, there would be too much power if only one light was turned on. \$\endgroup\$
    – Mark H.
    Feb 16, 2022 at 8:55
  • 1
    \$\begingroup\$ Wire the lights in parallel: each light's "+" terminal to the DC-DC's "output +", and each light's "-" terminal to the DC-DC's Ground, they will both receive the same voltage (ie, 4.2V). Each will draw the current it needs. So the total current out of the switching converter is the sum of both flashlights' currents. Do not wire them in series, that won't work. \$\endgroup\$
    – bobflux
    Feb 16, 2022 at 9:59
  • \$\begingroup\$ I edited the answer to add more details. \$\endgroup\$
    – bobflux
    Feb 16, 2022 at 15:04
  • 1
    \$\begingroup\$ Ha, unfortunately my plate is full! \$\endgroup\$
    – bobflux
    Feb 18, 2022 at 7:22

A basic step down switching regulator will work of you don't want to mess with whatever circuit is builtin to the flashlights. At 80 to 95% efficiency it's your best bet.

Not noting the current needs, you probably want to measure the amperage with the existing batteries before you replace them. Then you can find a suitable regulator meeting both the current and voltage requirements.

You will put them both in parallel. You do not want to put them in series.

  • \$\begingroup\$ How would I measure the amperage from the batteries, and is there a way to figure out what amperage is being used by the lights? Thanks. \$\endgroup\$
    – Mark H.
    Feb 15, 2022 at 19:11
  • \$\begingroup\$ @MarkH. Multimeter. \$\endgroup\$
    – winny
    Feb 15, 2022 at 21:33
  • \$\begingroup\$ @mark an ammeter or a multimeter in current mode in-between the battery and the flashlight. You probably want to make sure you are in the 10A range. \$\endgroup\$
    – Passerby
    Feb 16, 2022 at 1:37
  • \$\begingroup\$ @MarkH. Based on your edits nothing really changes. No the switch regulator won't produce any really visible flucations in the lights. If the flashlight has any internal boost led driver then it will smooth it out even more. And either use 2 switching regulators (one each due to current needed) or one high average switching regulator to power both. A resistor is not good for this. 3.6V @ 3 Amps is 9 watts of wasted power so you would need beefy 10 Watt power resistors. And it would mean half the battery capacity/run time. \$\endgroup\$
    – Passerby
    Feb 16, 2022 at 3:19
  • \$\begingroup\$ @Passerby Thanks for the edits. :) Based on the size of the regulators, I can fit two easily in my project. I believe these will work fine. Right? Also, reading a comment about the regulators, if they fail, the flashlights will cook so I plan to add a 4.5 V resistor (approximately) between the converter and the light as a backup in case the converter fails. I will let you guys know how it turns out. Cheers, Mark \$\endgroup\$
    – Mark H.
    Feb 16, 2022 at 4:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.