I am building a circuit for some induction coils on a coilgun, and it will draw about 1.75A from a 12V power source (I know it probably won't be very powerful, but it is difficult to find resistors that will allow a higher current). To keep the current at 1.75A, I am planning on using a PWR221T resistor, which is rated for 50W at 25 degrees C. The details sheet also says that, without a heatsink, the resistor is only rated for 2.25W. I am going to be building my circuit on a breadboard, and I'm worried that I will run out of space if I add a heatsink (It also has two Darlington transistors that will probably require heatsinks and a few other circuits). However, each inductor coil is going to be switched on for only a split second, so will I the resistor (and the Darlington transistors) actually heat up enough to necesitate a heatsink? My original plan was to have 1 resistor for both coils, but I could also use two so that they're used separately.

  • \$\begingroup\$ No thermal mass = very quickly. \$\endgroup\$
    – DKNguyen
    Commented Apr 7, 2020 at 2:39
  • \$\begingroup\$ Even though the power as you say is drawn for a short time, that impulse of power is enough to blow a poorly spec'd resistor. The constraint to prioritize should not be "breadboard space" for a project like this. At your bare minimum, the resistor should be able to handle the load you are providing which is given by P=IV =1.75*12 = 21W \$\endgroup\$ Commented Apr 7, 2020 at 2:43
  • \$\begingroup\$ I took some time to do the math for this particular resistor, and taking into account worst-case power dissipation (about 24W max, since I'm relying on a converter), it seems I would need a heatsink with a really low thermal resistance (about 1.3C/W), which, regardless of space, is really expensive. What would happen if I used, say, a 3 C/W aluminum sink, a fan, and had the circuit on for a few seconds? \$\endgroup\$ Commented Apr 14, 2020 at 20:10

2 Answers 2


Far easier: use an automotive 12 volt, 2 ampere incandescent lamp ("bulb") as a series resistor. This has a number of advantages:

  1. For short duty cycles, the cold resistance is less than a tenth of the hot resistance of the lamp, 6 ohms. This provides more current to the circuit.
  2. If the circuit draws too much current, resistance rises, limiting the total to less than 2 amps.
  3. No heat sink is needed, if the lamp is in open air.
  4. You have a visual indication of current drawn.
  5. These lamps are cheap and readily available.

If 2 amps is a bit too much, add a small series resistor, which does not have to handle high power, since most is being dissipated by the lamp.

Gearloose & Assistant

You'd need to know the mass of the resistor, the specific heat capacity of its materials (e.g., ceramic) and the duty-cycle of the current (percentage of the time it's on). That said, the resistor would work for some seconds or minutes, and then act as a "slow-blow" fuse.

  • \$\begingroup\$ If you are going to copy cartoons into your answer, please provide a link or citation for the original creator of the graphic. \$\endgroup\$ Commented Apr 7, 2020 at 13:02

The specific heat of silicon is 1.6 picoJoules per degree Centigrade per cubic micron.

If your resistor body is 1cm cubed, that is 10,000 * 10,000 * 10,000 microns, or 1 Trillion cubic microns. In such a case, the resistor body will heat up by 1 degree C per second if one watt is dissipated.

  • \$\begingroup\$ You think the resistor is made of silicon? \$\endgroup\$ Commented Apr 7, 2020 at 13:02
  • \$\begingroup\$ If it's Sir Amec, ~SiO2, specific heat ~700 J/kg.K. Then there's nichrome, an alloy of Ni, Cr, etc. It seems we can't be so specific about heat... \$\endgroup\$ Commented Apr 7, 2020 at 18:01

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