I am designing a test coil for a reed switch.

I have found a potential solution which is a coil wrapped on a stainless steel tube with PEEK discs at each end.

  • The length of the coil is 18.4mm.
  • 30 AWG enamelled copper wire.
  • 7 layers for a total of 505 windings, coil radial thickness of 1.78mm. Coil assembly diameter of 6.74mm.
  • Coil will be wrapped in Kapton tape and potted into an aluminium tube with a ID of 7.5mm and a wall thickness of 2.25mm.
  • Environment will be around 40C.
  • The current needed to generate the required field strength is 400mA DC.
  • The voltage drop across the coil is 1.15V DC.
  • The power drawn is 0.46W.
  • The coil will only be used for brief periods, around 2 second, very infrequently.

I realise this is a highly complex problem that may require modelling to answer, but I'm hoping there is an experienced electrical engineer that can give an idea if 400mA is feasible in this scenario.

  • \$\begingroup\$ You could estimate the temperature rise: energy dissipated, divided by heat capacity. Since it's "very" infrequently, that means the coil probably has enough time to cool down in between uses. \$\endgroup\$ Commented Jul 12, 2022 at 11:38
  • 1
    \$\begingroup\$ Compare it in size with a 1W resistor. Sounds like it'll be fine. \$\endgroup\$
    – user16324
    Commented Jul 12, 2022 at 11:46

2 Answers 2


Not doing any math, but we're looking at a cylindrical object of 10mm diameter and 20mm length.

Let's pick a 7W resistor, diameter 8mm, length 22mm. The datasheet helpfully provides a graph for the temperature rise.

enter image description here

You can't compare directly with the resistor's maximum dissipation because it is made with high temperature materials (ie, ceramic) and both your potting compound and enamel will certainly not reach these temperatures safely.

However at 5W they give a temperature rise of 225°C (eyeballing the graph) so your cylinder of similar size with a dissipation of 0.5W should have a temperature rise of 22.5°C.

Now I wouldn't wrap the winding in kapton tape because that will create an insulating layer of air and prevent the heat from reaching the aluminium tube which I assume is intended as a heat sink. Your dissipation is low enough that the coil will be absolutely fine without any extra wrapping. If you want it to be more robust, you can impregnate the coil with varnish or epoxy to make it solid. The inner windings will get rid of heat better, and it'll hold together better.

If you're really worried, the resistance of copper has a positive tempco, so by measuring the DC resistance, you'll know the temperature.

Note some stainless steels are ferromagnetic, so it might be a good idea to make sure before using that as the coil former.

  • 1
    \$\begingroup\$ +1. Finding a similar situation for a complex thermal problem. (Seems like all thermal dissipation problems are complex.) \$\endgroup\$
    – Marla
    Commented Jul 12, 2022 at 14:41

Using this online calculator (see below) and plugging in your parameters for a horizontal cylinder of 18.4mm x 6.74mm diameter, I get a temperature rise of 63°C for a cylinder surface temperature of 103°C. I used the complex formulas embedded in this 'calculator' rather than some back-of-envelope calculation because both convection and radiation losses are extremely nonlinear (only Newtonian conduction is linear). Note the contributions of Messrs. Prandlt, Rayleigh and Nusselt from the world of fluid dynamics.

Assuming you don't encapsulate the coil, the inner windings will be a bit hotter than that, but still should be less than 120°C. The rating of ordinary PU insulated wire is typically 130°C so you should be okay for continuous operation with a 40°C maximum ambient. There are higher rated insulations such as polyimide which is good for 200°C.

You may need more voltage than expected since the current will drop as the resistance increases with temperature. If the coil heats to an average of 120°C the resistance will rise about 40% from the resistance at 20°C. If you jack up the voltage proportionally to maintain the same magnetic field, and therefore the same current, that means 40% more power dissipation, of course. Normally relays are designed to pull in with about 70% of nominal voltage to account for this.

enter image description here


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