I have a few things I wanna do with the resistor wire. They all involve getting approximate temperature using known voltage. I understand that temperature heavily depends on environment, but what I know now is nothing, which is much less that can be known in my opinion.

In the shop they offer wires with these 3 properties:

  • length
  • resistance
  • diameter

I have two general things I want to create

  • low temperature drying device for chemicals
  • high temperature electrical shisha

I wonder how could I calculate approximately which wire do I want. I would then test it to get the exact temperature. I understand this would need some additional electronics - but I'm sure I can first calculate the wire so that it's not burning red instead of slightly warm.


1 Answer 1


This mostly depends on what your powering the resistance wire with. The voltage going in, combined with the resistance of the wire will dictate the current flow and therefore the power dissipated in the resistance wire. All power dissipated in the resistance wire is useful heat, to heat up your materials (chemicals/shisha etc).

So the actual resistance is not relevant if you have absolute control over the voltage being fed in. As a rough indicator, a 10W heater with a 12V voltage source would need a total resistance of 14.4Ohm (Pd = V^2 / R), which with this wire would need to be 3.42m long.

How much thermal power you produce (Pd) will then dictate the temperature, depending on the materials you are heating. Thermal energy will be needed to bring up the materials to the desired temperature, and less energy will be required to keep the materials at the required temperature (due to heat escaping to the environment).

To keep a constant temperature, you need to reach a state of equilibrium where thermal energy added to the system matches thermal energy being lost. Heat loss will increase as the system temperature increases, so for any given heater system it will naturally reach a maximum temperature. However, any change to the system will change the maximum temperature (remove materials, the wire may overheat).

In short, you can make a heater system that does not have a controller, but the required power fed in is highly dependent on the system (what materials you are heating, how much etc). Your best bet is to use a lab power supply, in current control mode, and slower creep up the current until it heats up and stabilizes at the required temperature. You can do calculations to estimate heat loss, but they may not be much more useful than empirical testing.

As a rough guide for somewhere to start, make an educated guess as to the rough order of magnitude of power needed. This device heats about 0.5sq m of soil to around 30C, and is rated at 27W. So we could infer that for temperatures of 30-50C, over an area of about an A4 piece of paper to 0.5sq m, 10-30W would do the trick. This is assuming that the wire is heating something else up first (some medium like sand). This is a Fermi style estimation, to give a rough idea or scales. If you start testing with the ability to generate 10-30W of heat from your wire, you can then ramp up your power supply across the range to see what works.

If you do make a heater without a controller, you still need some kind of safety trip to stop it overheating. If you do some googling, you can find various ways to do this. The easiest is probably a simple thermal fuse in series with the resistance wire, that will disconnect power to the resistance wire if it heats beyond a set temperature.

  • \$\begingroup\$ In the equation, what does V stand for? Volume? \$\endgroup\$ Commented Sep 23, 2014 at 8:28
  • \$\begingroup\$ V stands for voltage, Pd for power dissipation, R for resistance. Ohm's law is a fairly fundamental formula in electronics. \$\endgroup\$
    – Oliver
    Commented Sep 23, 2014 at 8:36
  • \$\begingroup\$ Well, I'll try to figure something out, then I'll buy the wire. Anyway, could you please make one example calculation in your answer so that I would know what am I supposed to calculate? It doesn't matter which values you chose as the given ones. \$\endgroup\$ Commented Sep 23, 2014 at 8:57
  • \$\begingroup\$ I've added a fermi-style estimation for power required, but bear in mind is not a 'answer', more of a guide for the system I roughly described. Use it as a starting point for testing. \$\endgroup\$
    – Oliver
    Commented Sep 23, 2014 at 9:15

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