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Scaling A Pots Voltage

Discussion in 'Electronic Basics' started by Michael, Oct 13, 2007.

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  1. Michael

    Michael Guest


    I have a motor controller board that uses a 5K pot for speed

    The problem I have is that my joystick only moves the pot 20% either side of
    center so I only have 40% of the total range.

    Is there anyway to scale this voltage up to the full range? I was thinking
    about using an amplifier but don't know of one that will increase a voltage
    above a certain point and decrease it below another.

    Can anyone suggest something suitable?

    Thanks in advance,

  2. Ninja

    Ninja Guest

    If the amplifier approach appeals to you, it's simply gain and offset that
    you need to do the job. See
    for an application note on how to design the amplifier.
  3. An op-amp circuit would be easier to adjust, but there is an old trick
    used by early R/C model builders back when they had to build their own
    joysticks: get a wirewound pot and bypass part of the element with silver
    paint. The paint was a General Cement product intended for repairing
    printed circuit boards. I suspect that getting it right would take as
    long as building an op-amp scaler with trimpots for gain and offset.
  4. John Fields

    John Fields Guest

    The brute-force way would be to use two supplies: (View in Courier)

    .. E1
    .. |
    .. [R1]
    .. |
    .. E2--+
    .. |
    .. [R2]<--- EOUT
    .. |
    .. E3--+
    .. |
    .. [R3]
    .. |
    .. E4

    E1 and E4 are the supply voltages, EOUT is the wiper voltage (which
    can vary from E2 to E3), R1 and R3 are the inaccessible regions of
    the pot, and R2 is the 40% of the total resistance of the pot you
    can access.

    Since you have a 5k pot and you can only use 40% of it, that
    resistance is:

    R2 = 0.4 Rt = 0.4 * 5000R = 2000 ohms.

    That makes:

    Rt - 2000R 3000R
    R1 = R3 = ------------ = ------- = 1500 ohms
    2 2

    Let's say that your motor control voltage needs to be 0 to 10V.

    Then that means your circuit now looks like this:

    .. E1
    .. |
    .. [1500R] R1
    .. |
    .. E2--+---10V
    .. |
    .. R2 [2000R]<--- EOUT
    .. |
    .. E3--+---0V
    .. |
    .. [1500R] R3]
    .. |
    .. E4

    Since you want 10V across 5k, that means (with no load on the wiper)
    that the current through the pot will be:

    E2 - E3 10V
    I = --------- = ------- = 0.005A = 5 milliamperes.
    R2 2000R

    Now, since the current in a series circuit is everywhere the same,
    R1 and R3 will also have 5mA through them and the voltages they will
    each drop will be:

    E = IR = 0.005A * 1500R = 7.5V

    So, E1 needs to be 7.5V more positive than E2 and E4 needs to be
    7.5V more negative than E3, reducing the circuit to:

    .. 17.5V
    .. |
    .. [R1]<--- EOUT
    .. |
    .. -7.5V

    Finally, since there's 25V across the pot and 5mA through it, it'll
    be dissipating:

    P = IE 0.005A * 25V = 0.125 watt, which is very little.
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