Heads up: I'm not good with electronics and only have a vague idea of it's inner workings.

I have a E3D V6 Extruder rated for 24 V, that i plan to use in my 3D printer. Will there be any problems with it if powered by 12 V? Will it take longer to heat up? Will it be able to heat up enough to melt PLA? Will it work at all for that matter? If there are any other quirks or potential problems that I overlooked, please let me know.

  • $\begingroup$ Thank you for such a detailed answers! I was hoping to get away with mismatched voltage ratings, but oh well. I'll see if i can figure out how to replace the heating element in the extruder; at the worst, ordering a new extruder shouldn't be a problem. $\endgroup$
    – Ivan
    Feb 5, 2019 at 17:37
  • 3
    $\begingroup$ you can use a cheap DC-DC converter to step up the 12v to 24v DC. They even make fixed-voltage-output ones at 24v, since that's a common RV/trucking voltage. $\endgroup$
    – dandavis
    Feb 5, 2019 at 22:12
  • $\begingroup$ @Ivan you could order a 12V heater cartridge from e3D or a reseller, they have those. $\endgroup$
    – Trish
    Feb 7, 2019 at 0:11
  • $\begingroup$ Yep, bought a 12 V heater today, which is way cheaper than an entire extruder. Replacing the old one with it was very simple too; definitely a way to go. $\endgroup$
    – Ivan
    Feb 7, 2019 at 19:03

4 Answers 4


Electrical engineering can be quite complex, but in this case you can save yourself with same simple equations/relations. Using the following formulae:

  • Voltage ($\ U$) equals current ($I$) multiplied by the electrical resistance ($R$)

$$ U=I \times R $$


  • Power ($P$) equals the square of the current multiplied by the electrical resistance

$$ P=I^2 \times R $$ can be rewritten using the first formula to: $$ P= \frac{U^2}{R} $$

Applying these formulae to a 40 Watt, 24 V heater element, the electrical resistance (in $\Omega $) is calculated by: $$ \frac{{(24\ V)}^2}{40\ W}=14.4\ \Omega $$

Running this heater element with 12 V will lead to a power of $$ \frac{{(12\ V)}^2}{14.4\ \Omega}=10\ W $$

The heat produced is proportional to the square of the current multiplied by the electrical resistance, halving the voltage is quartering the heat output. This will heat up very slowly! If it is able to reach the required temperature that is. Calculating the temperature is far more difficult, but if you are interested in doing so, please look into this answer from the Electrical Engineering Stack Exchange.


No, it probably won't work as you want. As explained in another answer, you will only achieve 25% of the expected power. So it will take 4 times as long to heat up, will have a lower 'highest temperature', and most critically will reduce the possible print speed by a factor of around 4 (actually more, since a proportion of the power is lost to the room rather than used to melt filament).

I guess that you could print with this setup as a temporary measure (so long as it's PLA, or some other low-ish temperature filament). It would not be a sensible choice, particularly since the extruder only needs a single component to be swapped out to change between 12 V and 24 V operation (the heater cartridge). All the mechanical parts will be identical between the two versions, and these are the 'expensive' elements in the assembly.


No not by itself.

Also you need to check the wires in the ATX power supply as 16 gauge wire might melt depending on how many amps it needs.

You could on the other hand connect 2 ATX power supplies the plus 12v on power supply 1 to the 12v ground on the second power supply. Then use a volt meter to confirm your getting 24v out. On the 2 leads not connected.

This still could run into problems as you have to be careful with the wire gauge. You need 14g wire for 15 amps, and 12g (thicker) for 20 amps. Finding an ATX power supply with better than 12g wire is highly unlikely.

  • $\begingroup$ The -12V line from an ATX power supply is typically rated for under an amp. Try to power a printer off that, and the best you can hope for is to blow a fuse. $\endgroup$
    – Mark
    May 15, 2019 at 2:23
  • $\begingroup$ @Mark Sorry I meant 12v ground and not negative 12 volts. $\endgroup$
    – cybernard
    May 15, 2019 at 3:01
  • $\begingroup$ Using half the voltage on a resistive heater will quarter the power. $\endgroup$ May 15, 2019 at 9:56
  • $\begingroup$ Sorry but part of this answer is incorrect! Using 24 V on a 14.4 Ohm resistor will give you about 1.7 A, that same resistor on 12 V uses 0.8 A. You cannot choose current, it is a result of applying a voltage to a resistor! It only applies if the heater element is of lower resistance, but that is not what the question asks. $\endgroup$
    – 0scar
    May 18, 2019 at 13:52

It will take longer to heat up. However if you use a boost converter (like I did on my Anet A8 when I upgraded to a Maxiwatt 24 V hot end), then it will work just fine without any further adjustments; to the power supply or the gauges of the wire etc. I set the boost from 12 to 24 volts. Now my A8 heats up in 56 seconds!


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