I have a 24V solar system, and I want to get only 12V from the DC output.

Currently I am taking a wire from only 1 battery, but this causes unbalance in the batteries and I have to check it a lot and switch the wires to balance, also the wire is a bit long and I get a drop and when it is near or below 12V my devices stop functioning.

I thought if I can get the 24V down to 12V at the end of the wire I'd be sure to get constant voltage for the devices, and would not have to worry about the balance of the batteries and it woyld be safe since the charge controller has a deep discharge safety.

The problem is I can't use resistors or transistors (voltage regulator) due to excess heat and a transformer is not an option (nor DC to DC converters - can't get any here,) I am looking for nearly 5A of current.

So I am kind of stuck here.

How can I do this?


Would making a stairs of voltage regulators be ok to use? (24V-19V-15V-12V) or would it be better if I put multiple 12V regulators from 24V?

Efficiency isn't a problem, but I don't have large heat sinks. Or is there a way to drop enough voltage so I can feed it to the 12V voltage regulator? Using a resistor will limit the current and I want to avoid that.

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    \$\begingroup\$ Simple voltage dropping isn't efficient, and efficient voltage dropping isn't simple. \$\endgroup\$ Jun 26, 2015 at 18:08
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    \$\begingroup\$ The proper solution is a 24 to 12 V switching regulator, at the end of the wire, as you have surmised. Probably 5 dollars on eBay, free shipping to anywhere. But if you are on another planet... can you rewire the whole system to work on 12 V? It might be possible to rearrange the panels, batteries are easy, charge controller might be ok. It won't solve the voltage drop problem. \$\endgroup\$
    – tomnexus
    Jun 26, 2015 at 18:18
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    \$\begingroup\$ You could wire up a relay on a timer to swap between the batteries every five minutes or so. This would balance the charge. But as others have said it would probably be cheaper and simpler to buy a DC-DC converter. \$\endgroup\$
    – Jon
    Jun 26, 2015 at 19:57
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    \$\begingroup\$ @tomnexus how can I build a regulator that can withstand that much drop voltage? 5A is not small.. it is 12v device but low battery + long jointed wire = big V-drop, so reaches other end at 10.7v when battery has 11.7v... borders are closed so no shippings get here \$\endgroup\$
    – bakriawad
    Jun 27, 2015 at 0:52
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    \$\begingroup\$ @Umar I have many PC's PSUs and other electronic parts from lights and radios.. I can get some parts as well from a closer technician, not a big varaity but could get around \$\endgroup\$
    – bakriawad
    Jun 27, 2015 at 0:56

2 Answers 2


If you need a high-power load resistor, you can get a long length of wire (enamel/transformer wire is best) and put it in a bucket of water. You now have an incredibly high-power resistor. If you want to drop 12V at 5A (probably less due to regulator overhead) then you want a 2.4 Ohm length of wire. Do you have a multimeter?

If you put this load resistor in series with a 12V regulator, the regulator will keep its output at 12V, but its input voltage will drop as the current increases, so the power dissipated in the regulator will be significantly reduced when at high loads (will be dissipated in the bucket instead).

Best of luck!

  • \$\begingroup\$ Splendid! I can also add salt to the water to control its resistance! But wouldn't this limit the current? It won't be easy getting 5A through.. I don't think i'd be able to pass 2A through that. I am quite new to electronics so I still have done informations missing... \$\endgroup\$
    – bakriawad
    Jun 27, 2015 at 3:17
  • \$\begingroup\$ @bakriawad I think the water is just for cooling. You can use salt water for making a resistor, but it will quickly corrode your electrodes (and produce toxic gases). \$\endgroup\$
    – tomnexus
    Jun 27, 2015 at 8:24
  • \$\begingroup\$ Yes, the water is just for cooling / heat sinking. All of the current should be carried by the wire. \$\endgroup\$
    – Luke Wren
    Jun 27, 2015 at 8:41
  • \$\begingroup\$ Ah I get it.. I thought using water as a resistor, this is still great! I think I can fish a wire with that much resistance. Is aluminum wire good as well? It had more resistance than the others? \$\endgroup\$
    – bakriawad
    Jun 27, 2015 at 17:04
  • \$\begingroup\$ Aluminium has around a 50% higher resistivity than copper, yes. The resistance will depend on the wire's cross-sectional area, as well as its length and resistivity. See here: en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity \$\endgroup\$
    – Luke Wren
    Jun 27, 2015 at 19:00

(Edited) I know this is a two-year old thread, but I still want to answer. Although, you have eliminated DC-DC converters I still feel that's the best option. A buck converter will do the job in the blink of an eye. If you can't get one locally, order from one of the international shopping websites (EBay, AliExpress, Bangood etc.)

I recommend an LTC3780, since you want to handle 5A. It also comes with current limit adjustment and charge level indicator, if you need them in future. I hope this helps.

  • 1
    \$\begingroup\$ @ChrisStratton Fully agreed. Note how the accepted answer is a water-cooled resistor, while the OP also ruled out resistors :P -- BharathRam, your answer looks fine now, have a +1 :) \$\endgroup\$
    – marcelm
    Oct 20, 2017 at 10:43
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    \$\begingroup\$ my first thought was a buck converter, but they were unavailable (still kind of are.. pretty rare), @marcelm linear regulators / resistors tend to withstand low current and a 12V drop on a 5A rules them out.. however the suggestion of a resistant wire inside a bucket of water solves the problem, it might not be an efficient way but it helped me for almost 8 months until i was able to get my hands on a 30A buck converter \$\endgroup\$
    – bakriawad
    Mar 17, 2018 at 15:44

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