How much current can a regular pot take before it's damaged?

[dentaku]

I'm wondering how much DC can safely be sent through a regular pot?
Last night I built a VCO with an LM13700 and need a control voltage source so I just connected a 1M Bourns http://canada.newark.com/bourns/pdb181-e420k-105b/potentiometer-rotary-1mohm-17mm/dp/02J2827 between +9V and -9V then the wiper into the control voltage input of the VCO through a 27K resistor.
It makes a very nice triangle wave by the way.

I'm assuming very little current ever flows through this pot no matter where it's set because the wiper is connected to a 27K resistor before going into the OTA but what's the limit?
How much abuse can a pot like this take before you damage it?

I see the specs say
• Power Rating:    200 mW
• Voltage Rating: 200 V

[c4757p]

Well, 200mW, so 200mW = (I^2)(1M); I = 447uA.

The wiper will have a separate rating, but in general you should avoid drawing current through it. Not to avoid damage, but because its resistance is unspecified.

There's no way 18V/27k = 667uA will hurt it, if you're just concerned with protecting it.

[dentaku]

I get the 18V/27k = 666.667 uA part using Ohm's Law but how does "• Power Rating:200 mW" tell me I = 447uA?

Well, 200mW, so 200mW = (I^2)(1M); I = 447uA.

The wiper will have a separate rating, but in general you should avoid drawing current through it. Not to avoid damage, but because its resistance is unspecified.

There's no way 18V/27k = 667uA will hurt it, if you're just concerned with protecting it.

[robrenz]

All your answers are here in the Bourns potentiometer handbook

[grumpydoc]

I get the 18V/27k = 666.667 uA part using Ohm's Law but how does "• Power Rating:200 mW" tell me I = 447uA?

By combining ohms law and the equation for power

1) V=IR

2) W=IV

1) can also be stated as
1a) I=V/R

Substituting 1) into 2) gives

3) W=I²R

substituting 1a) into 2) gives

4) W=V²/R

Rearranging 3 gives
I=sqrt(W/R)

So 200mW and 1M gives I=sqrt (.2/1000000) or I=0.000447214A or 447.2uA

EDIT: Corrected an error above and made the substitution clearer - if you want to see what a balls up I made of the original it's quoted below.

[electronics man]

its not about the current its about the voltage you put through it its just v = IR and its the power disapated in the pot its like any other resistor.

[woodchips]

A pot is just that, a potential divider. Whilst the actual track can take the power mentioned in the data sheet, you will be very very lucky to find any current rating for the wiper. Unless it is a wire wound pot then don't take any current from the wiper. If you want a variable resistor, where you can take significant current from the wiper, don't use a pot. Pots like conductive plastic have a wiper current rating of less than 1 uA, cermet might be a little higher.

[grumpydoc]

its not about the current its about the voltage you put through it its just v = IR and its the power disapated in the pot its like any other resistor.

Well, for dissipation by the track, it is about the current as that times the voltage gives you power - but you can calculate the power dissipated by a resistor knowing just the voltage and resistor value. Or for that matter just the current and the resistor value.

What c4757p calculated was how much current would cause a 1Mohm resistor to dissipate 200mW which is what dentaku asked. I was just showing how you derive the power given voltage and resistance equation.

Of course getting 447uA to flow in the 1M track would require 447V across the ends.

I think that a 1M pot is too high for this application - the amplifier bias current for the LM13700 is probably useful in the 5-500uA range (the datasheet says 2mA max) so you "load" a divider formed by a 1M pot too much for the resulting control current to be linearly related to the wiper position.

[dentaku]

I e-mailed this to a friend, who unlike me, has done this kind of math in the last 24 years since I was in High-School and he explained the extra steps that lead to the I=sqrt(W/R) equation. Now I see how you got to that point.
It's the I = V/R being expressed as I = (W/I) / R part that I couldn't get to for some reason. After that it all made sense.
Of course, like you say later in this thread "getting 447uA to flow in the 1M track would require 447V across the ends." so this question was more about the math than anything practical

I get the 18V/27k = 666.667 uA part using Ohm's Law but how does "• Power Rating:200 mW" tell me I = 447uA?

By combining ohms law and the equation for power

1) V=I/R

2) W=IV

Rearranging 2 gives V=W/I or I=W/V

Substituting into 1) gives

3) W=I²R
and
4) W=V²/R

Rearranging 3 gives
I=sqrt(W/R)

So 200mW and 1M gives I=sqrt (.2/1000000) or I=0.000447214A or 447.2uA

[robrenz]

Excerpt from the Bourns trimmer primer

[megajocke]

However, you can get the full rated current of the track of 447 uA to flow through part of the track  without having 447 V by running current through the wiper. You need to respect the current rating of the track on both sides of the wiper in addition to the wiper current rating or else the power dissipation density in part of the track will be out of spec.

[dentaku]

Yup. So you mean when the wiper is set all the way to the "start" and there's very little resistance between that end and the wiper there could possibly be quite allot of current?

Does it matter that the wiper then is connected to a 27K resistor which goes into the LM13700 Input Bias Current pin 16?

However, you can get the full rated current of the track of 447 uA to flow through part of the track  without having 447 V by running current through the wiper. You need to respect the current rating of the track on both sides of the wiper in addition to the wiper current rating or else the power dissipation density in part of the track will be out of spec.

[grumpydoc]

Yup. So you mean when the wiper is set all the way to the "start" and there's very little resistance between that end and the wiper there could possibly be quite allot of current?

It depends on what resistance the wiper is connected to but yes.

The power needs to be dissipated evenly across the track - if the wiper is in the centre each half needs to be limited to 100mW

Suppose you connect pot across a 10V supply the wiper to ground via a 1k resistor and set the wiper so that there is 1k between the +10V rail and the wiper. Now there's 2k across 10V which gives 5mA of current - this is rather more than that bit of track can manage (remember 447uA is the highest current before the track is dissipating too much power) and could conceivably damage it.

As I said above I don't think 1M is a good value for this pot. Generally if you are using a pot as a voltage divider the resistance connected to the wiper needs to be higher than the track resistance so as not to load te divider too much and make the response non-linear.

Looking at various VCO circuits and the datasheet the control current (and remember it is a control current, not voltage) needs to be up to 1mA or so which, as we've seen is technically more than your 1M pot can safely pass.

Also with effectively 27k to ground you'll screw the relationship between wiper position and control voltage.

Consider a 1M pot with one end of the track at 0V and the other at some positive voltage, let's cal lit V_cc and the wiper connected to ground via a 27k resistor.

With the wiper at the V_cc  end the output voltage will be 100% of that. Great.

But move the pot round 5% - now the resistance to V_cc  is 50k and resistance to ground is 950k in parallel with 27K or 26.25k. That means the output voltage will now be 34% of V_cc . You've moved the pot through 5% of its travel but the control voltage has gone through 67% of its range instead of the intended 5%.

I think you need a lower value pot.

[dentaku]

Quite interesting. I never thought a question about not damaging pots would lead to a thread like this.

I just tried it with a 10K then a 50K pot and it all still works. I just happened to have 1M pot on my desk and that's what I used.

The divider with the pot was just something I did so I could test the VCO by sweeping between +9V and -9V.
If I had a MIDI to Volt/octave CV converter I would use it but I don't have anything like that at the moment.

Yup. So you mean when the wiper is set all the way to the "start" and there's very little resistance between that end and the wiper there could possibly be quite allot of current?

It depends on what resistance the wiper is connected to but yes.

The power needs to be dissipated evenly across the track - if the wiper is in the centre each half needs to be limited to 100mW

Suppose you connect pot across a 10V supply the wiper to ground via a 1k resistor and set the wiper so that there is 1k between the +10V rail and the wiper. Now there's 2k across 10V which gives 5mA of current - this is rather more than that bit of track can manage (remember 447uA is the highest current before the track is dissipating too much power) and could conceivably damage it.

As I said above I don't think 1M is a good value for this pot. Generally if you are using a pot as a voltage divider the resistance connected to the wiper needs to be higher than the track resistance so as not to load te divider too much and make the response non-linear.

Looking at various VCO circuits and the datasheet the control current (and remember it is a control current, not voltage) needs to be up to 1mA or so which, as we've seen is technically more than your 1M pot can safely pass.

Also with effectively 27k to ground you'll screw the relationship between wiper position and control voltage.

Consider a 1M pot with one end of the track at 0V and the other at some positive voltage, let's cal lit V_cc and the wiper connected to ground via a 27k resistor.

With the wiper at the V_cc  end the output voltage will be 100% of that. Great.

But move the pot round 5% - now the resistance to V_cc  is 50k and resistance to ground is 950k in parallel with 27K or 26.25k. That means the output voltage will now be 34% of V_cc . You've moved the pot through 5% of its travel but the control voltage has gone through 67% of its range instead of the intended 5%.

I think you need a lower value pot.