Hi, I'd like to know if the maximum power dissipation is linear with the resistance when using a potentiometer? For instance, I use a 0.25 W potentiometer of 1k Ohm and I set its resistance to 500 Ohm. Between the two contacts where the resistance is 500 Ohm, will it be able to dissipate 0.25 W or only 0.125 W ( (500/1000)*0.25 ) ? If so, then to each potentiometer would correspond a certain current limit . Have you an idea ? Thanks for your help, C=E9dric
calculation of the limit current through a potentiometer
Dec 10, 2006
5 Replies
The current limit concept is pretty good. When you are using half of the resistance element to provide half the total resistance, all the heat is produced in half the area, compared to when you are using the whole element. But since that half is operating in a cooler environment that when the whole element is producing heat, it can actually handle a little more than half the power. But derating to a constant current (that would produce rated power for the full resistance) is a conservative approach, that will never produce a hot spot that is hotter than operating the full element at full power.
That said, I try to never have any component dissipate more than half its rated power, with a few exceptions.
Check the wiper current rating. Often it's so low that overall power dissipation isn't the limiting factor.
John
Don't calculate -- check the manufacturer's specs to be sure. They'll tell you the maximum current.
I'll go by a rule of thumb that the wiper can handle about 150% the rated current at full wattage. For instance, a 1k 0.25w trimmer is rated for 15.8mA at a rated voltage (leg-to-leg) of 15.8V. I'll feel perfectly safe at a maximum current of 24mA. Of course, you're still limited to 15.8V max. Go whichever is lower. More current than that, I'd look it up. The manufacturer will tell you for sure.
As far as power goes, the maximum current will limit the total power, as well as limiiting the power on the active segment of the resistive element. 1.5X rated current will mean a little more than 2X rated power over the active segment of the element. This potential for a hotspot on the element is the limiting factor (except for large low ohm power rheostats, where wiper temperature becomes a determining factor). But the heat will radiate and spread itself out over the width of the resistive element. Also, for tiny higher ohm tweaker pots, the physical wiper itself will act as a heat sink, making the wiper temperature and the temp of the cermet directly under the wiper actually lower than the adjoining cermet.
I know all of this may seem to be TMI for a newbie, but it's actually a very good question. Many early failures of potentiometers in the field are actually the result of application deficiencies rather than manufacturing defects. And it's an easy mistake.
Good luck Chris
Sorry -- a bit of the above was "inartfully expressed". Need a third cup of coffee this morning.
Under any and all circumstances, for all pot wiper conditions and currents, total power dissipation is limited to 1/4W. Current and voltage come next.
Sensible derating for tweaker pots should be similar to that for resistors, as mentioned by Mr. Popelish. Manufacturers no longer overbuild resistors -- they make them "good enough" because the margins are so small. It might indeed be wise, if you want the pot to last for the life of your product, to cut maximum power dissipation to 1/8W or so. And cut the above voltage and current limits accordingly.
Cheers Chris
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