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Potentiometer transducer and loading error

Short answer

A potentiometer transducer gives output voltage e_o = E_s·x/L proportional to wiper position. With R_p/R_m = 0.2, at 50% travel e_o = 4.76 V (error 0.238 V); at 75% travel e_o = 7.23 V (error 0.271 V). Loading error arises from finite load resistance; reduce by making R_m ≫ R_p or using a buffer amplifier.

(a) Explain the working of a sliding-contact (potentiometer) displacement transducer. State its advantages and limitations.

(b) A 10 kΩ\Omega linear potentiometer with a 10 V supply drives a 50 kΩ\Omega load. Find the output voltage and loading error at 50% and 75% of the travel, and state how the loading error can be reduced.

(a) Potentiometer transducer

A resistance potentiometer (pot) has a resistive element (wire-wound, carbon or conductive plastic) with a sliding contact (wiper) mechanically linked to the moving member. A supply EsE_s is applied across the full length LL, and the output is taken between the wiper and one end. With no load, the output is linear:

eo=EsxLe_o = E_s\frac{x}{L} 
           Es (supply)
   o-------+
           |
          [|]  resistive element, resistance Rp, length L
          [|]
          [|]<--- wiper, moved by displacement x ---o  e_o
          [|]
           |
   o-------+----------------------------------------o  common

Advantages: simple, cheap, high output, no amplifier needed, works on dc, available in linear or rotary form.

Limitations:

  • Wear and friction of the sliding contact; limited life.
  • Finite resolution for wire-wound types (a step of one wire turn).
  • Loading error with a finite load resistance.
  • Limited frequency response because of wiper inertia; noise at the contact.

(b) Loading error

With a load RmR_m across the wiper output, and xx = fractional position:

eoEs=x1+x(1−x)RpRm\frac{e_o}{E_s} = \frac{x}{1 + x(1-x)\dfrac{R_p}{R_m}}

Rp/Rm=10/50=0.2R_p/R_m = 10/50 = 0.2, Es=10E_s = 10 V.

xIdeal (V)Actual eoe_o (V)Error (V)Error % of EsE_s
0.505.00051+0.25(0.2)=4.762\dfrac{5}{1 + 0.25(0.2)} = 4.7620.2382.38
0.757.5007.51+0.1875(0.2)=7.229\dfrac{7.5}{1 + 0.1875(0.2)} = 7.2290.2712.71

Answer: eo=4.76e_o = 4.76 V at 50% (error 0.238 V) and 7.237.23 V at 75% (error 0.271 V).

The loading error is largest at about 2/3 of the travel. To reduce it:

  • Make Rm≫RpR_m \gg R_p (use a high-input-impedance voltmeter or a buffer amplifier).
  • Use a smaller RpR_p (limited by power dissipation).
  • Use a tapped (non-linear) pot to compensate.

✓ Answer checked by Claude Opus 5.5 (high effort)

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