I can't find it anywhere, what is the thermal time constant or thermal capacitance of a standard discrete 1/2W resistor?
A table with other resistors of different power ratings would be nice, too.
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asked Oct 3, 2014 at 12:14
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\$\begingroup\$ I don't think there is a "standard", you'd have to check manufacturers data sheets. \$\endgroup\$– pjc50Commented Oct 3, 2014 at 12:57
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6\$\begingroup\$ So here's a fun fact. The Heat capacity of everything is about 3 J/(K*cm^3). (+/-50% or so) So measure the volume and you can make a guess. The time constant will depend on the thermal resistance which is material and geometry dependent. Unless you are pulsing the R's for a very short time, you care more about it's thermal conductivity to the outside world, than you do about it's heat capacity. SMD resistors with big copper pours on the ends are good. \$\endgroup\$– George HeroldCommented Oct 3, 2014 at 12:58
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\$\begingroup\$ @GeorgeHerold - "The Heat capacity of everything is about 3 J/(K*cm^3). (+/-50% or so)" That is an amazing rule-of-thumb! Why did my school teacher never tell me that? +1 \$\endgroup\$– gbulmerCommented Oct 3, 2014 at 19:10
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2\$\begingroup\$ @gbulmer, Grin. I've been designing these thermal conductivity, heat capacity stages. (plastics are a bit lower) So I think this is due to the fact that above the debye temperature, every atom in a solid has ~3kT of energy. And then the number density (#atoms/cc) of stuff, is not all that different. (OK I'll post my table as an answer.. and kill it if someone complains.) \$\endgroup\$– George HeroldCommented Oct 3, 2014 at 21:07
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\$\begingroup\$ If you're going for very high peak-to-average power ratios, the internal construction (and its thermal conductivity) may be important. \$\endgroup\$– pericynthionCommented Oct 3, 2014 at 21:25
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1 Answer
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OK a table of volumetric heat capacity for @gbulmer.
Material Heat capacity (J/(K*cm^3)) at 300K. Aluminum 2.42 Copper 3.4 Iron 3.5 Al2O3 3.0 G-10 2.7 Nylon 1.7
Addition: Guesstimate of time constant for through hole resistor.
So let's just assume the whole thing is made from ceramic, say Alumina. Thermal conductivity is k = 30 W/(m* K) = 30mW/(mm* K) (millimeters will be easier for me) And make the diameter 2 mm (area = ~3mm^2) and the length 6mm. Then thermal conductivity (end to end) is
R = 1/k * l/Area = 67 K/W. Then the heat capacity is 3J* volume (cm^3) = 18 mJ/K. Now we need to scale the thermal resistance by some amount. I'll guess 1/2 (say 30K/W) but this is most likely not enough. (I'll over estimate the time constant.) Then the time constant is 30k/W*18mJ/K = 0.54 seconds. So that seems OK, but maybe high.
As an interesting aside I once tried to measure how much of the heat in a through hole resistor comes out through the leads. The answer was basically zero, except with very small heat input... and the numbers were still "in the noise". Through hole resistors are meant to cool by convection around the body.
answered Oct 3, 2014 at 21:14
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\$\begingroup\$ Unfortunately you can't make an HTML table, but there are various ways of getting fixed spacing and preserved spaces. \$\endgroup\$ Commented Oct 3, 2014 at 21:31
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\$\begingroup\$ @ChrisStratton, thanks. The <pre> changes the font and </pre> brings it back? \$\endgroup\$ Commented Oct 4, 2014 at 2:30
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1\$\begingroup\$ Wikipedia has a Table of Specific Heat Capacities, which include even more substances. \$\endgroup\$– gbulmerCommented Oct 4, 2014 at 11:27
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\$\begingroup\$ Actually this is to calculate the time constant so I would need the thermal resistance but since it wasn't actually my question (and that apparently it's geometry dependant, hence has to be specified in datasheets) this one is sorted out by that very useful rule of thumb (+ more precise tables). \$\endgroup\$ Commented Oct 6, 2014 at 14:13
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1\$\begingroup\$ @MisterMystère, What's the resistor. The surface mount ones are on Alumina, mostly and you can get a guesstimate* of the time constant. Heat capacity times thermal resistance. Take the end to end thermal resistance and divide it by 1/2 (maybe) because the heat is generated everywhere. *Hey when did guesstimate become a "real" word? \$\endgroup\$ Commented Oct 6, 2014 at 15:29