How to reduce 20 V to 3.6 V to power flashlights

Question

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.

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

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

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.

Answer 1

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.

Answer 2

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.

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.

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.

Answer 3

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.