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1 August, 12:27

What is the non-linearity error, as a percentage of full range, produced when a 1K Ω potentiometer has a load of 10KΩ & is at one-third of its maximum displacement?

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  1. 1 August, 12:40
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    2.28%

    Explanation:

    Being at one third of its maximum range a potentiometer should output V0/3.

    However if this 1kΩ potentiometer has a 10kΩ load:

    (1) I1 = I2 + I3

    (2) V0 = I1 * 2/3*Rp + I3 * 1/*Rp

    (3) Vp = I2 * Rl

    (4) Vp = I3 * 1/3 * Rp

    Where

    I1: current entering the potentiometer

    I2: current going to the load

    I3: current going to the other leg of the potentiometer

    V0: supply voltage

    Vp: output voltage of the potentiometer

    Rp: total resistance of the potentiometer

    Rl: load resistance

    First we determine the intensity of I3 in function of supply power

    I3 = 3 * Vp / Rp = 3 * Vp / 1000 = 0.003*Vp

    Then the load current

    I2 = Vp / Rl = Vp / 10000 = 0.0001*Vp

    With these we determine I1

    I1 = 0.003*Vp + 0.0001*Vp = 0.0031*Vp

    Then

    V0 = 0.0031 * Vp * 2/3*Rp + 0.003*Vp * 1/3*Rp

    V0 = 0.00207 * Vp * Rp + 0.001 * Vp * Rp

    V0 = 0.00307 * Vp * 1000

    V0 = 3.07 * Vp

    Vp = V0 / 3.07

    Vp = 0.3257 * V0

    Now the percentage error is:

    (100 * (0.3333 - 0.3257)) / 0.3333 = 2.28 %
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