# Convert 24V battery output to 12V (step down)

I want to convert the output of my stronger Li battery (24 V) to some devices that require 12 V

24 V Battery Specs:

• Output V: 21-29,4 V
• Capacity: 15 Ah
• Max Current peak: 36 A
• Max Current: 18 A

12 V Devices:

• Jigsaw
• Driller

The target device is designed to work with batteries of 12 V 2,4 A (2 Ah) or 12 V 2,4 A (5 Ah) In the specs of batteries or devices there is no technical info about the maximum consumption.

A step down circuit can do the task but how can I ensure that it doesn't get burned?

Installing a 2,5 A fuse to save the stepdown circuit will work but I will probably need to replace the fuse too frequently. What alternatives are available?

• A step-down converter that can deliver sufficient current should work fine. If you have no spec about current consumption, simply measure it with a multimeter before you do anything else? Commented Mar 15, 2022 at 9:00
• Does the drill have a variable speed control or just an on/off switch? Is it universal motor or something fancy like brushless? Commented Mar 15, 2022 at 9:03
• The tricky bit will be finding a stepdown converter that handles the surge current of a drill or jigsaw starting. I've just fitted LFP batteries to a 15 year old cordless drill (original 12V batteries died) ... the 30A BMS trips on start! But if I squeeze gently, i.e. use the speed control as a soft start, it does what I want, so I'm happy.
– user16324
Commented Mar 15, 2022 at 9:45
• Measuring the peak current is easy. Just measure motor resistance at steady condition. Turn the rotor slowly, it's position may affect measurement value. Take the smallest resistance. 12V divide by measured value. So the DC-DC converter should handle that current for short time. Find that converter is tricky. Commented Mar 15, 2022 at 10:19

It is feasible, but one should be careful with picking a part.

As a course of action, I'd recommend the following:

Filter converters by output voltage. Fixed output voltage will make the circuit simpler, and 12V is a standard voltage, you should have plenty of choice with fixed parts.

Filter by maximum output current (sustained), you want something with overhead, no need to push things to the limit. Over 3A output should work, but no need to shy away from 5A converters. Make sure the converter will have no problem supplying the necessary voltage when the device is idle - low output currents can sometimes be a problem for converters. You should measure minimum current consumption ideally as well and find that current on the efficiency graph of the converter IC.

Check the capacitance of the battery line. That has to be within the spec of the output capacitance of the voltage converter to make sure things are stable. You don't want any oscillating surprises. And you shouldn't add any capacitance there, because the following circuit expects certain capacitance. You can look up if the next components in the circuit can take different input capacitance, but it seems to me like an unnecessary hastle. Just stay within existing specs (if possible).

Check the efficiency of the voltage converter at the expected currents to make sure heat is not going to be a problem. It shouldn't be, but it's not a reason to ignore that. The chip must have thermal resistance rating (C/W degrees C per watt) and efficiency percentage. From that you can calculate power losses in the IC and how hot it can get. +20 (to ambient) degrees should be acceptable, just remember, the chip is probably going to be in enclosed space and with other warm components, so its ambient temperature is not your room temperature. Temperature affects efficiency as well. So don't skip this step. It's not something you can precisely measure (because ambient temperature is hard to estimate), but you should do some napkin math and it should look safe by a good margin.

Of course, you need to make sure the converter can handle peak currents (which are unknown, I take it? Nothing follows from 12V battery spec), you don't want to trip the ICs overcurrent protection. You can always go for higher output current part - again, as low as low currents won't be a problem.

• As the device is a drill, or a jigsaw, either one, I think your suggested currents are probably an order of magnitude off.
– user16324
Commented Mar 15, 2022 at 9:47
• @user_1818839 all we have is a 12V battery peak current spec. You could be right about peak current tho. It's a big fat and very important question mark here.
– Ilya
Commented Mar 15, 2022 at 9:48
• As i have the jigsaw and a 2Ah battery i can try to measure current, but i assume that using a basic multimeter I will not be able to read quick peaks Commented Mar 15, 2022 at 11:12
• @DanielPerez, yep, you need some shunt resistor and a scope. But peak current for a few millisec can easily be 5 times sustained max current. Even if the battery max sustained current has some headroom and it's not maxed out - which is a reasonable assumption - peak current could exceed 10A for a ms-scale moment. And that could trip overcurrent protection on the IC (which you absolutely must have)
– Ilya
Commented Mar 15, 2022 at 11:16
• besides installing a fuse, Is there any way to limit the max current supplied independently of how much the device demands? Commented Aug 24, 2023 at 17:34