How to connect this potentiometer?

Discussion in 'General Electronics Chat' started by deki, Oct 10, 2011.

  1. deki

    Thread Starter New Member

    Sep 6, 2011
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    I'm not sure how I'd connect this potentiometer to the two diodes? http://www.555-timer-circuits.com/motor-pwm.html
    Would the outputs of the two diodes just go to the 'wiper' part of the potentiometer? Or do they go to the input terminal of the potentiometer?
     
  2. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Most single-turn pots are made like this.
    One of the diodes is backward in the schematic you posted. I posted a correction below. If the schematic is tiny when you open it, click on it to enlarge it.
    Connect the anode of one diode to A and the cathode of the other one to B. The wiper (w) connects to pins 2 and 6 of the 555.
    If you still don't know how to connect your pot, post a picture of it.
    [​IMG]
     
    Last edited: Oct 10, 2011
  3. deki

    Thread Starter New Member

    Sep 6, 2011
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    I knew something was up with those diodes! Thanks heaps for that :)
     
  4. deki

    Thread Starter New Member

    Sep 6, 2011
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    Actually just another question: How would I go about to measure the resistance of the potentiometer?

    Would I just place the negative probe on either A or B, and the positive one on the wiper? I tried that, and I get either 0k or 10k, depending on whether the negative probe is on A or B.
     
  5. BMorse

    Senior Member

    Sep 26, 2009
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    it really does not matter which side the red or black probe is on, just connect one probe to one of the terminals, and connect the other to the wiper, now slowly turn the pot one way or the other and you should see the value change, it will get lower when wiper is turned towards the terminal with the probe, and higher as you move it closer to the other terminal which should have nothing connected to it. and also, you must do this while pot is out of the circuit and disconnected.
     
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  6. deki

    Thread Starter New Member

    Sep 6, 2011
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    ^^Actually thing is, when I put the red probe on the wiper, and the black on say 'A', I get ~9k, yet when the black is on 'B' I get ~1k? (with the dial turned as in the image, more to the left).

    Also when I just want to use it as a simple variable resistor in a circuit: do the A and B connections go to Vs and Ground? And the wiper just goes into wherever I'd like the resistance to be?
     
    Last edited: Oct 12, 2011
  7. BMorse

    Senior Member

    Sep 26, 2009
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    I take it that you are not moving the wiper while probes are in place, and since you took measurements from both sides and they equal out to 10K, I would suspect that the pot is a 10K pot....
     
  8. deki

    Thread Starter New Member

    Sep 6, 2011
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    Yes it is a 10k pot just that I don't understand why I get the two different readings depending on where I set the ground probe.
     
  9. Adjuster

    Well-Known Member

    Dec 26, 2010
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    That is simple enough to explain. If we ignore any contact resistance, which is normally pretty negligible, a potentiometer behaves like two resistors in series joined at the wiper, where the total is the full pot resistance.

    If the whole resistance of the pot end-to-end is R, and we define the wiper position as "A", a percentage of its maximum travel*, then one resistance is given by R times A/100, the other is R times (100-A)/100.

    Thus if you have a 10kΩ pot, and the wiper is at 10% of its travel, you get one resistance of 10kΩx10/100 = 1kΩ, and the other is 10kΩx(100-10)/100 = 9kΩ.

    *Electrical travel along the track resistance that is. The mechanical position may be different, allowing for various errors, and the fact that some potentiometers are made with a deliberately non-linear characteristic, such as "logarithmic " or audio tapers.
     
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  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I am a graphically-oriented person. Maybe you are too.
     
  11. deki

    Thread Starter New Member

    Sep 6, 2011
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    I get it now, thanks :)
     
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