Amplify Position Sensor Circuit

  • #1
Hi everyone,

I'm working on a college project this year about mountain bike suspension damping. By trade I'm a mechanical engineer however, I'm required to construct a simple sensor circuit to monitor the performance of the damper I am making. Forgive me if my wording or logic is not quite correct on certain aspects.

I picked up a sliding potentiometer to use as a sensor since it was astronomically cheaper than inductive or other resistive options. The system has to be entirely stand-alone since, it will be tested in real world environments. I decided to connect the potentiometer in series with a standard 1.5V AA battery and a 150K resistor since the current rating on the pot is <10uA. As far as observation with a multimeter goes, this is functional. However, the stand-alone voltage data logger I own can only record in increments of 0.01V, this is an issue since the potentiometer voltage ranges between 0 and 0.05V which would only allow me ever to (optimistically) record 5 different values. Not nearly a good enough resolution to demonstrate damping trends.

I believe the solution lies in amplifying the voltage signal from the sensor however, I haven't a clue where to start when it comes to choosing an amplifier and I am even less familiar with how I would go about wiring one to do the required task.

Specifications
Potentiometer - 5K Ohms, <10uA, 200mm travel
Data Logger - 0-32V in increments of 0.01V, USB interface, 400Hz recording speed

Any input would be greatly appreciated!

Cameron.
 

Answers and Replies

  • #2
rbelli1
Gold Member
997
379
Are you sure the current rating is <10uA? That sounds kind of small. What is the part number on that?

BoB
 
  • #3
I'm beginning to doubt the validity of the manufacturer's spec sheet, it was also remarked by someone on another forum that it seemed like an unusually low rating. There is mention of 20mA further down but it's unclear what this represents.
 
  • #4
davenn
Science Advisor
Gold Member
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I'm beginning to doubt the validity of the manufacturer's spec sheet, it was also remarked by someone on another forum that it seemed like an unusually low rating. There is mention of 20mA further down but it's unclear what this represents.

doesn't help when all the spec's are in Chinese ( even after switching to English)
 
  • #5
I suppose that's the price I'm paying for ordering a cheap generic one from China :H

I had to download Google Translate onto my phone and then use the inbuilt OCR software to translate the spec sheet.

Translations of the relevant parts are as follows.

"Current Recommended (uA) - 10"

Then further down (as best the app can manage):

"It can also be converted to a 4 ~ 20mA DC current signal"

Haven't got any data acquisition gismos available so I decided to chance it and hook it up to a 9V battery and 30K resistor, running with a steady voltage across it so far so I think I'll assume that the 10uA is related to something else.
 
  • #6
jim hardy
Science Advisor
Gold Member
Dearly Missed
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Power = Volts2 / Resistance
I'd put 5 volts across it. That's 52/5000 = 1/200th of a watt.
As rbelli hinted, If it's too flimsy to handle that it's useless.

However - because it can take 5 volts end to end does NOT mean it can take 5 volts from its variable terminal to an end.
Identify the terminals using your multimeter.

05149.png


Probably you'll be okay powering it from your logger's 5 volt supply.
 
  • #7
Tom.G
Science Advisor
3,890
2,590
Since the sensor is a film resistance element, that 10uA is almost certainly the wiper current rating.
 

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