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Potentiometer loading effect and linearity

Short answer

Loading from a meter of resistance Rm makes the output eo/Ei = x/[1 + r·x(1−x)] with r = Rp/Rm, introducing non-linearity. Linearity worsens as Rm decreases; maximum error occurs at x = 2/3, and at midpoint the error is about r/8, e.g. 10% when Rp = Rm.

Explain the loading effect on a potentiometer and explain the effect of load resistance on the linearity of the potentiometer. Also show that the error occurs at the midpoint of the potentiometer.

A resistive potentiometer is a displacement transducer. The wiper position x gives an output voltage proportional to the displacement, provided that no current is drawn from the wiper.

Loading effect

Let the total resistance be Rp, the wiper be at fraction x = Rx/Rp of its travel, the supply be Ei, and the load (meter) resistance be Rm.

 Ei (+) o-----+
              |
             [ ] (Rp - Rx) = (1 - x)Rp
              |
              +<--- wiper ---+------------o  eo
              |              |
             [ ] Rx = xRp   [Rm]  load (meter)
              |              |
 Ei (-) o-----+--------------+------------o

Without load: eo = Ei·x (a straight line).

With load, Rm is in parallel with Rx. The parallel resistance is Rx·Rm/(Rx + Rm), and the total resistance across the supply is:

R = (Rp − Rx) + RxRm/(Rx + Rm)

eo = Ei · [RxRm/(Rx+Rm)] / R

Substituting Rx = xRp and r = Rp/Rm:

eo/Ei = x / [1 + r·x(1 − x)]

So the loaded output is smaller than the ideal value Ei·x. The meter draws current, and the potentiometer output becomes non-linear.

Effect of load resistance on linearity

  • If Rm ≫ Rp (r → 0): eo/Ei → x, and the characteristic is nearly a straight line.
  • As Rm decreases (r becomes large), the curve bends further below the ideal line, with the greatest deviation in the upper-middle region.
  • Error = ideal − actual = Ei·x − eo:

ε/Ei = r·x²(1 − x) / [1 + r·x(1 − x)]

For small r the error ≈ r·x²(1−x). It is zero at x = 0 and x = 1, and the maximum occurs where d/dx[x²(1−x)] = 0, i.e. 2x − 3x² = 0, giving x = 2/3, with error ≈ 0.148 r.

Error at the midpoint

At the mid position x = 0.5:

eo/Ei = 0.5 / (1 + r/4)

ε/Ei = r·(0.25)(0.5)/(1 + r/4) = 0.125 r / (1 + 0.25 r) ≈ r/8

So the midpoint reading is lower than 0.5·Ei by about r/8 of the supply; that is, an error is present at the midpoint. Example: Rp = Rm (r = 1) gives eo = 0.4 Ei instead of 0.5 Ei, an error of 0.1 Ei = 10 % of full scale.

To keep this error low, use Rm ≥ 10 Rp or a high-impedance (buffered) meter.

✓ Answer checked by Claude Opus 5.5 (high effort)

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