Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations KootK on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Positional Accuracy with a Potentiometer

Status
Not open for further replies.

mpgt_aero

Aerospace
Jan 6, 2016
5
Hello EE Brethren,
Can anyone help me understand how to verify the accuracy of a position sensing system using a potentiometer? I'm working with a butterfly valve that has a position indicator switch. As the valve opens/closes, it will trip the switch between the 2° - 7° range (just to signal closed/not closed). I have to test the valve to determine at what angle the switch is actually tripping. I have a test system for the valve which consists of a 1K Ohm pot tied into a test box. The test paperwork says the tolerance of the equipment is +/- 0.5°, but there's no justification given for it, and it was written well before my time. I'm trying to determine if that +/- 0.5° value is legitimate.
Here's what I know: single turn pot. Total Resistance: 1K Ohm +/- 10%; Linearity: +/- 0.5%. Spec Sheet for the pot gives max current for 1K Ohm of 52.4 mA. I don't know the circuitry inside the test box, but the regulated power supply is 3.6 V +/- 0.2V, adjusted so 90° outputs 0.90 V. My circuits education was quite long ago, and I'm drawing a blank on how to verify the stated test tolerance. I would be much obliged for any guidance you could provide.

Thanks!
 
Replies continue below

Recommended for you

Sounds like a weak design from the start.

- System could use the pot as 'guidance', but monitor the drive motor current (the same way power windows in cars often do) to detect actual end of travel.

- System could have a torsional spring in the drive so that it works more on force application than trying to make absolute position work reliably. That'd be my approach, with the pot or similar just used as a monitoring/guidance.

- System could use an optical position sensor and a micro.

In answer to your question, perhaps we can see if the 0.5° makes sense.

I'll have to assume pot has 270° travel (?). Valve travel 90°. So it's likely using the middle section, your 0.9v to perhaps about 2.1v (?). Guesswork, but makes sense so far.

Right off the bat, if the circuit is simple comparator, then your PS varies by 0.2/3.6 or 5.5%. So no way your end point would stay within 0.5°. There are better bridge circuits where the PS variations could be cancelled out. Used here?

Linearity given 0.5% or 'one part in 200'. But the 0.5° spec is basically 'one part in 720' (360° circle), or is it 'one part in 180' (90° travel)? That could be calibrated out, if there's an adjustment.

Seems doubtful that it's designed optimally. Based on your description.


 
Lots of, but still not enough, information.

How is it calibrated?
What's the calibration accuracy?
How is the zero degree position calibrated, and with what accuracy?
How is the angle data measured, presented, and with what resolution?
What voltages are read out at 0º and 180º
Is 90º the wide open position and 0º the closed position?
Why is the calibration point so far away from the actual angle of interest? i.e., why isn't calibrated at 4.5º instead?

Your current makes no sense. 1kohm*52.4mA=52.4V, which is not presented anywhere, and is inconsistent with the 3.6V power supply. Based on your 0.9V at 500ohm, the current is more like 1.8mA

TTFN
I can do absolutely anything. I'm an expert!
homework forum: //faq731-376 forum1529
 
Like stated, not enough info yet.

Volts, ohms, and current info given are irrelevant to your question. Nice to know, but irrelevant.

Most butterfly valve actuators will have span and elevation adjustments to trim the pot for 0 to 90 degree motion; I assume your test box does.

I think the only variable significant to you then is the linearity. Typically the linearity spec is over the full travel, and often precision single turn pots are 360 degrees. So best you can guarantee for is .005*360= +/-1.8 degrees, not +/-0.5 degrees as stated.

In reality the linearity probably is a slowly moving variation, so probably not that bad at only 2-7 degrees off your adjusted 0 point, but you cannot be guaranteed of that without a test sheet showing YOUR pot.



 
So best you can guarantee for is .005*360= +/-1.8

Maybe, but that depends on where the cal points are and the accuracy by which they're calibrated, particularly since the OP's application is quite limited.

TTFN
I can do absolutely anything. I'm an expert!
homework forum: //faq731-376 forum1529
 
Thanks folks.
My calibration tech and electrical engineer have been unavailable, so I came here. I appreciate the feedback. To answer the questions that I can: 0° = closed, 90° = full open. / I don't know what the calibration points are, I just know what was written down on the box, the pot, or the spec sheet. However, the box is a general use lab tool for several valves, so it would not necessarily be calibrated for my specific use. / 52.4 mA was the max current for a 1K OHm resistance value from the spec sheet. Not necessarily the actual current in the system. Sounds like that was irrelevant anyway. / I will speak with the calibration tech on Monday. I appreciate the input.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor