# How fast will a resistor heat up under high power?

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.

• No thermal mass = very quickly. Commented Apr 7, 2020 at 2:39
• 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 Commented Apr 7, 2020 at 2:43
• 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? Commented Apr 14, 2020 at 20:10

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.

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.

• If you are going to copy cartoons into your answer, please provide a link or citation for the original creator of the graphic. 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.

• You think the resistor is made of silicon? Commented Apr 7, 2020 at 13:02
• 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... Commented Apr 7, 2020 at 18:01