How do I generate less than 0.05 V to heat a NiCr wire, or otherwise prevent the wire from consuming too much power?

I want to heat a length of Nichrome wire in order to cut through very thin pieces of nylon -- basically a "hot knife". I estimate the wire will be about 6 mm long. In order to provide some thermal mass and also prevent the Nichrome from being too weak to cut, I plan on a 16 gauge Nichrome wire, which has a resistance of ~0.26 ohms / foot, which is 0.0051 ohms of resistance for 6 mm. According to this calculator to produce 650 F I need 0.44 W.

0.005 ohms at 12 V is ~28.8 kW watts, which is... more.

How can I power the wire which has so little resistance without generating too much power? From my understanding:

  1. Add a resistor will just end up generating more heat from the resistor than the NiCr wire.
  2. I could regulate the voltage down to a lower voltage, but it seems that voltage regulators only go down to about 1.8 V, which still produces 650 W.
  3. I could run a PWM. If I combine a 1% duty cycle with a 1.8 V regulator then I'm at ~6.5 W, which is still much too high. Also, many PWMs and voltage regulators can't handle the ~10 A of current at that voltage.
  4. I could get smaller NiCr wire (higher gauges) and/or lengthen it to increase the resistance. Even wires as small as 32 AWG still require less than 0.2 V at 6 mm and I'm concerned about the wire not being sturdy enough. I could run the wire back-and-forth a few times (say 10 times for 60 mm in length) but then I have problems keeping the wires from touching when trying to concentrate them on the edge of the "hot knife".
  • 7
    \$\begingroup\$ 16-gauge is really thick for a wire cutter. You're essentially going to melt the full width of the wire. The wider the area, the more thermal energy required and the coarser the melted edge will be (the melted plastic ends up as globs on the cut edges). I would use 22 to 28 gauge. If you are pushing hard enough to break the wire, you're doing it wrong. \$\endgroup\$ Commented Jan 17, 2022 at 8:06
  • 2
    \$\begingroup\$ Make the wire longer and thinner until it is reasonable. Use the calculator. It is OK to make it 25 or 50 mm or even more if you have to. Just keep the nichrome wire in one straight line. No need to weave it back and forth as you are saying. \$\endgroup\$
    – user57037
    Commented Jan 17, 2022 at 8:07
  • 2
    \$\begingroup\$ Dont forget to consider current. You would need around 10 A for 0.5 W on that wire. You better change the wire parameters. \$\endgroup\$
    – Rokta
    Commented Jan 17, 2022 at 8:39
  • 1
    \$\begingroup\$ You'd do better to use Kanthal - it's overall tougher for this application, particularly in tensile strength (since the application will be under tension) and has higher resistivity, which gets you some headroom on the driver. \$\endgroup\$
    – J...
    Commented Jan 17, 2022 at 16:17
  • 2
    \$\begingroup\$ What stops you from using half a foot of wire instead? \$\endgroup\$ Commented Jan 18, 2022 at 7:34

5 Answers 5


This is where the old school solution of a transformer is by far the easiest way to get the low voltage and high current that's needed.

Find a transformer you can put a single secondary turn onto, with as thick copper wire as you can manage, or several pieces in parallel. Use a toroidal transformer with an open hole, or indeed any transformer that's unpotted so you can squeeze some more wires into the winding window. The EI type of transformer offers an advantage here because you can put two 1/2 turns on in parallel (one round each outer leg, paralleled to maintain core balance). Some folks strip the secondary from an old microwave oven transformer, which leaves plenty of room for custom secondary windings.

Your 6 mm is too short. There will be significant heat-sinking to your connections, your power calculator probably doesn't factor that in. There will also not be the space to make clamped connections to the wire. 1.6 mm is also probably too thick, overkill for the stiffness that you think you need.

Here's an example of almost exactly what you want to do

enter image description here

Image from wikipedia

It is an 'instant heat' soldering iron which has a transformer in the body, with a single secondary turn powering the thick wire / thin rod heating portion at the end. Note the very thick connections to carry the high current.

As Phil points out in comments, you could use one of these. They're not expensive.

  • 1
    \$\begingroup\$ Another similar item is the power supply for a resistance soldering station or spot welder. Looks like some people make these out of microwave oven transformers. \$\endgroup\$ Commented Jan 17, 2022 at 13:08
  • 1
    \$\begingroup\$ +1 for several things but especially for the heat losses from the ends. It's so short there's no middle, but that calculator does nothing about end effects. If you drive harder to overcome the losses at the ends and get them up to temperature, the middle will be hotter than you'd like \$\endgroup\$
    – Chris H
    Commented Jan 17, 2022 at 17:27
  • 4
    \$\begingroup\$ +1 Why reinvent the wheel? Such pistol style soldering guns used to come with a variety of tips, one of which was a thin blade that would work for your application. You could also fashion your own tip to wrap around a standard soldering tip. Keeping the tip tinned will extend its life. \$\endgroup\$ Commented Jan 17, 2022 at 19:38
  • 1
    \$\begingroup\$ Use of a transformer will also provide safety isolation -- VERY important if your power source is rectified AC! \$\endgroup\$ Commented Jan 18, 2022 at 4:42

In practice extra length of wire will give you better mounting options, better electrical options, and allow you to fine-tune the voltage (though PWM is also an option there).

I hacked together a rather larger hot wire cutter - for 6"-thick foam that was delivered at the wrong width and I had no time to send it back. My drive was rather easier than yours because of the longer, thinner wire (scrap low-voltage transformer, with an incandescent bulb in series to reduce the primary voltage). It helps to separate the mechanical and electrical connections to the heater wire, and that will make the heater wire longer. Mine was supported on steel bolts (an extra 2 nuts per bolt to clamp the wire) run through insulating material (wood - not the best choice but the shops were shut) with all-metal screw terminals for the connections. There was probably 20mm between the closer mechanical support and the bolt; the supports also allowed me to adjust the tension and straighten the wire. The other support was done differently: one of my screw terminals could be moved along a static length of wire as a power control given a fixed voltage.

Now I don't suggest you copy my design, but there are still some useful lessons from it.

Don't forget that in insulating material the wire can get hotter than in air, depending on how much heat is going into melting (latent heat), but that changes the resistance. So don't assume your numbers are all that accurate


As others have mentioned in the comments, your wire is thick. Making it thinner will increase the resistance, so setting the current will become somewhat easier, as it would decrease the needed current significantly.

An useful (but somewhat dirty) trick my father used to make such things was to run them on a small transformer. This will limit the power just well, in some sense better and simpler than a resistor or some linear regulator would.


Adding to other answers: use of a transformer will also provide safety isolation. Especially important if your power source is rectified AC, as found in your average switch-mode power supply. Assume that you will touch the wire at some point.


Increase your resistance a bit

Half a watt at 5 milliohms is 100 Amps, that's rather a lot, so consider going to a smaller wire size. Let's say you use a fine enough wire that you only need 10 A.

Use a current-limited power source

If you ensure that your power source will not deliver more than about 10 A, the voltage is produced is only important if you care about the energy you waste.

A 12 V battery in series with a (large enough (really, it needs to be large, it's going to disspate 12 W)) 1 ohm resistor will deliver at most 12 A. You can probably get away with one or 2 AA batteries if you have an appropriate resistor.

A benchtop powersupply will typically have a current limiting function. Set that to 10 A and you're good.


Your Answer

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

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