Potentiometer output votlage

Thread Starter

Dritech

Joined Sep 21, 2011
901
Hi all,

I will be using a potentiometer to output voltage as close to 13mV steps. I want the potentiometer to output the same voltage even when the system is reset or when other loads are connected to the same power line.
To help achieving this I was planning to connect one terminal of the pot to a 4.7V zener diode instead of connecting it directly to the 5V supply. This way if a voltage drop occurs (when connecting other loads or for any other reason), the voltage to the pot will remain stable.

Will this work? is it possible to have a repetitive output voltage as close as 13mV?
 

Thread Starter

Dritech

Joined Sep 21, 2011
901
Attached the the schematic. D will be 4.7V and R is for now unknown (Still have to calculate it). The pot will be used for monitoring the motor angle of rotation. I am not going to use an encoder because I am limited with space and budget.

Note: Dividing 4700mV with 360degrees equal 13mV/degree.

Now I have the following questions:

1) Is it possible to get repetitive voltage values for the same angle when using the attached schematic? Are there additional measures that I can take to increase accuracy?

2) The cable length from the wing (signal terminal) to the MCU will be approx. 60cm. Will that have any effect of the signal?

3) Does the pot value make any difference when used for position monitoring?

Thanks in advance.
 

Attachments

bertus

Joined Apr 5, 2008
22,276
Hello,

A standard potmeter will not reach the full 360 degrees.
It will be more like 270 degrees.
Also how often does the angle change?
Wear can be a problem.

Bertus
 

MaxHeadRoom

Joined Jul 18, 2013
28,683
You can get a 360° continuous turn Pot, but you would need to initially 'zero' it.
Pot value should not be that critical, especially if the input impedance is fairly high in comparison to the pot.
Max.
 

MrChips

Joined Oct 2, 2009
30,797
and you can eliminate the resistor R and zener diode D.
If you use a 5V regulated source that should be stable enough for your requirements.
 

Thread Starter

Dritech

Joined Sep 21, 2011
901
and you can eliminate the resistor R and zener diode D.
If you use a 5V regulated source that should be stable enough for your requirements.
Thanks. Will the cable length effect the signal? and does the pot value effect the performance?
 

MaxHeadRoom

Joined Jul 18, 2013
28,683
Personally I would have used a 5k minimum, your resolution will also be a bit finer, but if the input is a micro then the cable and input impedance should not have any effect as far as I see it.
Max.
 

Thread Starter

Dritech

Joined Sep 21, 2011
901

MaxHeadRoom

Joined Jul 18, 2013
28,683
The first, but at 50k you may be tempting a noise issue, the lower impedance would be preferred, I would go 10k max.
I have a new surplus quantity of the second in 91 series 2.5k, you are welcome to one for postage.
Max.
 

AnalogKid

Joined Aug 1, 2013
11,037
Are you doing a ratiometric measurement? That is, are you reading the voltage on the wiper and converting it directly into degrees, or are you looking at the ratio of the wiper voltage and the +5V, and converting that fraction to degrees? If you are going straight from the wiper value to degrees, the absolute value of the +5V is important. As it is, you need it to be regulated to within 0.26% to hold your sampling error to +/- 1 degree, and that is not counting in any errors in the A/D.

ak
 

ramancini8

Joined Jul 18, 2012
473
If I understand the problem an LSB is 13 mV. A 1% change on a 5V supply is 50mV. I don't think a 5V regulated supply or zener diode regulated supply will be accurate enough. This requires a precision reference, and that in turn usually dictates high load impedance, i.e. a high resistance pot.
 

KMoffett

Joined Dec 19, 2007
2,918
for a 360 degree potentometer, why not use a ten turn pot? adjust the body for the desired range of resistance.
The OP never stated the real application for this. If it's 0 to 360 degrees or <0 to >360 degrees, it makes a difference.
Dritech,

It's often easier the help solve your problem than help solve your solution. ;)

Ken
 
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