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I am running about 6.5W though a 10W resistor. The ohm rating is 220 ohms, which is correct for the circuit ohms which is calculated to about 225 ohms.

Here is what is running through my 220 ohm 10 watt resistor:

38.4 volts
0.17 amps
225.88 ohms
38.4V * 0.17A = 6.528W

Within a couple minutes it got so hot that it burned me. I'm ok though because I only touched it for a second.

But I was expecting it to stay cool since the resistor is rated at almost double the power that is going through it. The electronics people told me that it shouldn't get hot with double wattage.

Is this normal? Why would the resistor be getting hot? Also, is there a fire risk? p.s. The resistor is resting on brick.

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    \$\begingroup\$ 5W=6.528W?? That's over a 15% difference on a 10W scale. \$\endgroup\$ – Bort Oct 31 '16 at 23:59
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    \$\begingroup\$ @User91232 - No, it's rated at 35% more. Also, if you look at the "Temperature Rise" graph in the datasheet, it appears that at 65% load, there is about a 165C rise in temp. Can someone verify that for me? Also note, that if it were ACTUALLY 50% load, the rise in temp would only be 60C (Big difference eh?). \$\endgroup\$ – Bort Nov 1 '16 at 0:05
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    \$\begingroup\$ 10W is what it can handle before it melts. 6W is still a lot of heat it has to dissipate. \$\endgroup\$ – Majenko Nov 1 '16 at 0:06
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    \$\begingroup\$ Note: your resistor is dissipating around 1/200th of the heat a 1kW space heater dissipates. Compare how large a space heater is vs. the size of your resistor. How do they compare? \$\endgroup\$ – Cort Ammon Nov 1 '16 at 1:39
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    \$\begingroup\$ Regardless of how big, small, great, or terrible the resistor is, you still have 6.5W coming out of it and that heat has to get out the area somehow. \$\endgroup\$ – David Schwartz Nov 1 '16 at 1:43
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First let's do a quick number crunch:

6.528W/10W = 65% (of 10W)

Referring to the datasheet:

enter image description here

There is about a 165C rise in temp. Do not touch!.

As for "Is it a safe temperature for the resistor?", refer to the next figure:

enter image description here

I'll admit that the Derating Curve Graph kinda hurts my head. But, if you follow the 10W curve over to 25C (about room temperautre), the resistor should be able to handle 100% of it's rated power. Note that I'm only assuming the ambient temperature is 25C! If you have it lying on a brick, it should be okay. It appears that the resistor can handle up to about 115C ambient temp @ 65% load. But that would be pushing it to the max.

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  • \$\begingroup\$ THANK YOU! Is it still safe to leave on for several days though? It is resting on brick but in a windowsill. I made sure nothing else is touching it. Would like to put in a tiny insulated metal box if that existed. \$\endgroup\$ – hbsrnbnt Nov 1 '16 at 0:13
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    \$\begingroup\$ I did a little math and research, about 44C is the threshold for pain in touching a hot surface, so around 42C (108F) is safe to touch... according to the chart, the safe-to-touch threshold is at only about 15% load! \$\endgroup\$ – hbsrnbnt Nov 1 '16 at 0:22
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    \$\begingroup\$ @User91232 if you put it in an insulated box it will get even hotter, because the heat still has to go somewhere but it will take a bigger temperature difference to drive that heat flow through the insulation! What you may want to do is to clamp the resistor to a heatsink - you can buy a purpose-made one, but a simple metal plate will work - and then attach that to an insulating base so the heat won't damage the surface you place it on. Remember air has to be able to flow over the surface of the heatsink. Calculating the temperature rise in this situation is a separate question... \$\endgroup\$ – nekomatic Nov 1 '16 at 8:57
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    \$\begingroup\$ @nekomatic's point is a good one. It's perfectly possible for a load resistor to get hot enough to cause problems if it's in an insulated box. This may include melting other components or wire insulation, or even unsoldering its own contacts. These could have significant knock-on effects. \$\endgroup\$ – Chris H Nov 1 '16 at 9:00
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    \$\begingroup\$ You can buy metal-cased resistors with mounting holes for attaching directly to a heatsink. \$\endgroup\$ – nekomatic Nov 1 '16 at 9:02
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This is normal behavior for a power resistor of the size you are using. Just because it is running at 50% of its rating doesn't mean it will run cool. I looked at the data sheet for a similarly sized 10 watt resistor. It had a curve showing temperature rise versus percentage of rated load. For 50% (5 watts), the temperature rise is 125C which is greater than boiling water.

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You are pushing 6W to the resistor. That means 6J of heat per second.

Using calorimetric equation Q=c.m.(T2-T1), where Q is total heat, c is specific heat capacity, m is mass and T are temperatures, one can derive P=dQ/dt=c.dT/dt.

If you use your values, you can see that the resistor's temperature rises according to P/(c.m). Thanks to its small weight, the rise in temperature is really fast.

There is also counter-process: heat dissipation. The higher difference T-Ta, where T is resistor temperature and Ta is ambient temperature, the higher heat dissipation. There are more variables to rule the dissipation and the temperature: Heat capacity of the heatsink (air), mass flow of the air, etc.

Regarding your question:

  • Yes, it is normal and it is caused by power to weight ratio.
  • If insufficiently cooled, it will get really hot.
  • If it will be resting on a brick, there is no fire hazzard. If you put it between two bricks and seal it, it can melt itself and burn. If left with flamable pars, there is fire hazzard.
    Personally, I would add cooling to it, small aluminium radiator shall dissipate 10 W easilly (Foolproof roundup).
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    \$\begingroup\$ heat capacity is irrelevant in this case and is only distracting the reader from the only important factor, which is thermal resistance \$\endgroup\$ – szulat Nov 1 '16 at 11:38
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    \$\begingroup\$ I dont think so. The single resistor weights less than gramm, if you attach same resisor to 1kg lead block and seal both of them, the single resistor will warm up in minutes, the second one will warm up much slowly. The capacity rules the speed of warming. Final temperature is ruled by heat dissipation. Thermal resistance is important, but it is not the only factor definitely. \$\endgroup\$ – Crowley Nov 1 '16 at 12:35
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    \$\begingroup\$ and that's why heat capacity can be completely ignored in this case - because it is so small. and actually in nearly all cases, because we are usually interested in steady state. but your answer lists heat capacity as the main factor. sure, it is all connected, but for people who ask questions like this, your answer is only misleading. \$\endgroup\$ – szulat Nov 1 '16 at 12:54
  • \$\begingroup\$ @szulat I have changed the capacity result as "temperature rise is fast" and yes, the capacity does not affect steady state. Note that small capacity leads to fast process and vice versa. And OP is concerned that it got hot in several minutes. \$\endgroup\$ – Crowley Nov 1 '16 at 13:03
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    \$\begingroup\$ @Crowley Yes, "I am running about 6.5W" sounds like a steady state and in several comments OP expressed concerns about how long it can be left like that which further reinforces the steadiness of the state of OP's resistor. Besides, with the power output changing, one can simply average the power output and work with that value, with heat capacity determining how big the amplitude can be. \$\endgroup\$ – Agent_L Nov 1 '16 at 17:25

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