Re: calculation of the limit current through a potentiometer
- From: John Popelish <jpopelish@xxxxxxxx>
- Date: Sun, 10 Dec 2006 11:17:34 -0500
cedric_libre@xxxxxxxx wrote:
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 .
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.
.
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