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Help with picking resistors in proportional circuit

Discussion in 'General Electronics Discussion' started by swooshcmk, Apr 5, 2012.

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  1. swooshcmk

    swooshcmk

    2
    0
    Apr 5, 2012
    Hello all,
    I just need some help to get this project kicked off. It's a tad more difficult than a standard series/parallel so I'm not quite sure how to handle it. I'm adding the circuit description and the basic schematic for you guys to follow it. Most industrial joystick controllers use 10k potentiometers so I'm assuming this is similar use of a 10k Potentiometer which has the wiper centered at the 5k/side location. The description states the voltages and currents seen across the 60ohm load coil, so given that information I need to know the what size of resistor I should be using to tie each side of the potentiometer (after the diodes) to ground to accomplish what is needed.

    Thanks for any and all help!

    Chris

    [​IMG]

    [​IMG]
     
  2. timothy48342

    timothy48342

    218
    0
    Nov 28, 2011
    Just looking at the paths current takes from the pot to ground, I redrew the schematic and untangled it a bit.
    [​IMG]
    Now to simplify:
    1- If the power supply is large enough, you can ignore the small voltage drop of the diodes, so lets short those in the drawing.
    2- Since we know the voltage across the coil when the pot is turned all the way in one direction, lets draw the schematic, with a 10k ohm resister and the source placed at one side. (Imagine the slider moving all the way to the left.)
    3- Since 10k ohm is much large than 60 ohms, you can assume that any current flow in the 10k won't affect the current flow through the 60 much, so let's just cut that path out.
    4- Since one of the mystery R's is just a path to ground, it won't affect current through the 60 ohm as long as it doesn't load the power supply much.
    so let's ingore that current path, but remember later to check that the supply isn't loaded much.
    Redawing with all that gives me:
    [​IMG]
    So, now it's pretty simple. We just need to know the supply voltage.
    If V+ is 12V and the 60 ohms drops 3.8V, then the mystery resister drops the other 8.2V
    3.8V / 60 ohm is about 63mA and 8.2V / 63mA is about 130 ohm.
    So that would be 130 ohm for each of the mystery R's.
    But still consider the draw on the source.
    One path through 130 ohm draws about 90mA
    And one path throgh 210 ohm (60+130) 60mA
    (the third path through the 10K can still be ignored)
    So the source needs to handle about 150mA. Can it?

    That's just an example with a 12V supply. Plug in a different supply voltage and you get different values for the R's and also a different total current draw.
    If the supply can't easily hadle the current then some of my earlier assumptions are no longer valid.
    Also if the supply is much lower, (like arround 6V) then my assumption that the diodes can be ignored is invalid also.

    --tim
     
  3. swooshcmk

    swooshcmk

    2
    0
    Apr 5, 2012
    THANK YOU! Your walk through is very much appreciated. I guess I was over complicating it in how I was looking at the circuit flow. Works like a charm!
     
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