Electronic – Using a potentiometer to replace an absolute optical encoder

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I am trying to replace a rotary absolute optical encoder (single turn). The resolution of the encoder is 13 bits (8191). The reason for the replacement is to reduce the cost.

My approach is to get a single-turn (electrically) infinite turn (mechanically) precision potentiometer. Use a 16-bit ADC (neglect the last three bits due to noise).

Questions:

  1. Type of sensor element: As I have read, conductive plastic is the best for such an application. Am I right?
  2. Linearity: The least I found was 0.15% for a wirewound potentiometer (which I do not know how much error in degrees it has and how it will affect my result)
  3. Electrical travel: most potentiometer's electrical travel is a bit under 360 degrees (320 to 358). For example: beisensors. So in the last 2 degrees will the system be blind and does not know where it is? Is this common? Does it have a solution?

Please correct my information if I am wrong. Do you think these problems are limiting factors in potentiometers, and they cannot be used to the specific task that I need?

Best Answer

All poteniometers are < 360° because there has to be a gap between the ends of the resistance track. You would need two pots ganged onto the same shaft 180° out of phase.

You got me interested, however, and a search for "360 degree angle sensor potentiometer" threw up some interesting products from Novatechnik. See rotary shaft type and the AW360-ZE where they seem to have this figured out.

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Figure 1. Extract from the AW360-ZE datasheet.

Their "potentiometer" takes a Uref reference signal and gives a DC output proportional to Uref x angle.

They don't go into specifics of how they work but here are my thoughts:

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2. (a) A possible method of fabrication. A circular resistance track with connections at 0° and 180° would give a triangle wave output when rotating continuously in one direction. By adding 4-quadrant opto-detection it would be possible to figure out which quadrant the wiper is in and use that in the Logik circuit. (b) gives what might be a more reliable method. By simultaneously switching SW1 and SW2 they could use one switch position to get an actual voltage reading and the other to determine the quadrant.

  • Using the scheme of Figure 2b with the switches and wiper in the position shown Vout = V+/4. That means is in the bottom half of the pot and it's either 135° from the 12 o'clock position.
  • Throw the switches and Vout is again V+/4. That means it's 135° from the 3 o'clock position.
  • Only one point satisfies both conditions.

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