calculation of the limit current through a potentiometer

[[email protected]]

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,
Cédric

 

[John Popelish]

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.

 

[John Larkin]

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,
Cédric

Check the wiper current rating. Often it's so low that overall power
dissipation isn't the limiting factor.

John

 

[Chris]

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,
Cédric

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

 

[Chris]

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

 

[[email protected]]

Thanks to all for your explanations, I'll have to take a 0.5 W pot for
what I planned.
Bye,
Cédric