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    #1

    Help from pot tweakers needed!

    Hi all,

    I'm working on a small electronic project, where I want to deliver 1-~90 mA using a 9 V battery. To do this, I have chosen a linear 10k pot. My problem is, since this pot is linear, that the first 90% of turning the knob, hardly does anything, and the last 10% the current jumps up to 87 mA or so.

    Since finding a reverse logarithmic pot (which I believe is what I actually would need), for all in this range of specs, is extremely difficult, I've been searching the internet for possible ways to tweak my pot to be a little less linear, and found a couple of pages where this is explained. I've tried to connect a resistor parallel to the pot, achieving some small changes, but not quite the one I want... Now I would like to hear if anyone here has experience/knowledge to help me out...

    Problem: I have 9V, 90 mA regulated, and want to be able to control this current steadily rising from ca 1 - 90 mA.

    Solution: ?
    Tags: None
    #2
    Re: Help from pot tweakers needed!

    use a 1k pot, linear.

    it's pretty easy to understand that your minimum desired current determines the maximum resistance.
    since U=RxI, 1000 ohms give you 9/1000 amps or 0.9mA. (that's why nothing happens in the first 90% of your 10k pot until you hit 1k).
    add a series resistor for the maximum current limit. the same law gives you a value of 100 ohms for 90mA.
    "Every normal man must be tempted at times to spit on his hands, hoist the black flag, and begin slitting throats." - H.L. Mencken

    Comment

      #3
      Re: Help from pot tweakers needed!

      Originally posted by kikkoman
      use a 1k pot, linear.

      it's pretty easy to understand that your minimum desired current determines the maximum resistance.
      since U=RxI, 1000 ohms give you 9/1000 amps or 0.9mA. (that's why nothing happens in the first 90% of your 10k pot until you hit 1k).
      add a series resistor for the maximum current limit. the same law gives you a value of 100 ohms for 90mA.
      Better check your calculator batteries. 1K and a 9 volt battery will give 9 ma, not .9 ma. Actually, since you are suggesting a total resistance of 1.1K, the current will be 8.2 ma

      PlainBill
      For a number of reasons, both health and personal, I will no longer be active on this board. Any PMs asking for assistance will be ignored.

      Never be afraid to try something new. Remember, amateurs built the ark. Professionals built the Titanic.

      Comment

        #4
        Re: Help from pot tweakers needed!

        9v @ 0.001a => 9 kohms [9000 ohms]
        9v @ 0.090a => 100 ohms

        A 10k linear pot is about perfect but the behavior sounds like your pot isn't linear.

        Put a 100 kohm resistor in parallel with the pot.
        That will make it's max 9090 ohms and flatten the response curve over the pots whole range.
        .
        Mann-Made Global Warming.
        - We should be more concerned about the Intellectual Climate.
        -
        Be who you are and say what you feel, because those who mind don't matter and those who matter don't mind.
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        -
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        Comment

          #5
          Re: Help from pot tweakers needed!

          Nuther thought.
          You need a range of at least 100 ohm to 9 kohm to get what you want.
          The low end on your pot may not be zero.
          It's range may be something like 5k to 10k in which case you need a different pot or to do magic tricks with resistors.
          .
          Mann-Made Global Warming.
          - We should be more concerned about the Intellectual Climate.
          -
          Be who you are and say what you feel, because those who mind don't matter and those who matter don't mind.
          - Dr Seuss
          -
          You can teach a man to fish and feed him for life, but if he can't handle sushi you must also teach him to cook.
          -

          Comment

            #6
            Re: Help from pot tweakers needed!

            Or you could try this circuit. You might even be able to get a free sample.

            PlainBill
            For a number of reasons, both health and personal, I will no longer be active on this board. Any PMs asking for assistance will be ignored.

            Never be afraid to try something new. Remember, amateurs built the ark. Professionals built the Titanic.

            Comment

              #7
              Re: Help from pot tweakers needed!

              Originally posted by PCBONEZ
              A 10k linear pot is about perfect but the behavior sounds like your pot isn't linear.
              Well that's what I also thought at first, but when you think it over, a linear pot means you have 5k in the middle, resulting in 9v/5000Ohm = 0,0018 = 1,8 mA. At 10k I would have 9/10000 = 0,9 mA. This is very logical once you get your mind clear, which took me quite some time. What I believe I would need is the behaviour of a reversed logarithmic pot.
              Originally posted by PCBONEZ
              Put a 100 kohm resistor in parallel with the pot.
              That will make it's max 9090 ohms and flatten the response curve over the pots whole range.
              .
              Will it really? Hmm... Tried, after reading some guides on how to tweak pots with external resistors, to put 300 Ohm - 2.2 k in parallel with the pot. This changed the behaviour a little bit, but mostly resulted in raising the current at its lowest point to ~10 mA. This would not be a problem if I would have gotten the wanted slope between 10-90 mA, but I didn't.

              Can you explain why the 100k resistor would change the behaviour in a better way? (would love to just put one there and try it, but I don't have that many different resistors in stock, will have to go buy)

              Thanx for your time and efforts.

              Comment

                #8
                Re: Help from pot tweakers needed!

                Originally posted by PlainBill
                Or you could try this circuit. You might even be able to get a free sample.

                PlainBill
                Hm, well, thanx for the idea, but I believe this solution is a little more than I would like to invest into this project as far as parts and time goes. My solution also doesn't have to be that accurate, so if there is a easier solution, I will go for that.

                Comment

                  #9
                  Re: Help from pot tweakers needed!

                  >> a linear pot means you have 5k in the middle <<
                  No, not all linear pots are created equal and I dunno what you have.

                  At first you said you wanted 1 mA at the low end and now you say 10 mA ???

                  My calculator has the night off. - Play with this:
                  http://www.1728.com/resistrs.htm

                  Check:
                  pot[highest value] + added resistor.
                  pot[lowest value] + added resistor.
                  Then calculate both currents.
                  .
                  Mann-Made Global Warming.
                  - We should be more concerned about the Intellectual Climate.
                  -
                  Be who you are and say what you feel, because those who mind don't matter and those who matter don't mind.
                  - Dr Seuss
                  -
                  You can teach a man to fish and feed him for life, but if he can't handle sushi you must also teach him to cook.
                  -

                  Comment

                    #10
                    Re: Help from pot tweakers needed!

                    Originally posted by PCBONEZ
                    >> a linear pot means you have 5k in the middle <<
                    No, not all linear pots are created equal and I dunno what you have.
                    Hmm, really? I thought that was the definition of linear...?

                    Originally posted by PCBONEZ
                    At first you said you wanted 1 mA at the low end and now you say 10 mA ???
                    No no no, I would like to have 1 mA, but 10 mA would also do, if that would mean I could get a linear slope between the lowest and highest current.

                    Originally posted by PCBONEZ
                    My calculator has the night off. - Play with this:
                    http://www.1728.com/resistrs.htm

                    Check:
                    pot[highest value] + added resistor.
                    pot[lowest value] + added resistor.
                    Then calculate both currents.
                    .
                    Very nice tool, will save that page. Don't know exactly how to count, have to think it over.

                    Comment

                      #11
                      Re: Help from pot tweakers needed!

                      Linear just means the amount of adjustment [movement] in the dial or slide paddle is 1-to-1 to the change in resistance for the whole scale.
                      They can built them for whatever they want for min-max values.
                      If non-linear they can put the 'fast' end at the high or low end.

                      Lets say it's a 100k pot with a range of 0-100k and a dial with 100 'clicks' rotation.

                      Linear.
                      10 Clicks from 1-10 = 10k change
                      10 Clicks from 90-100 = 10k change
                      ... 1-1 response over the whole range

                      Non-linear
                      10 Clicks from 1-10 = 1k change
                      10 Clicks from 90-100 = 30k change
                      ... response curve is similar to a logarithmic curve.

                      .
                      Mann-Made Global Warming.
                      - We should be more concerned about the Intellectual Climate.
                      -
                      Be who you are and say what you feel, because those who mind don't matter and those who matter don't mind.
                      - Dr Seuss
                      -
                      You can teach a man to fish and feed him for life, but if he can't handle sushi you must also teach him to cook.
                      -

                      Comment

                        #12
                        Re: Help from pot tweakers needed!

                        Originally posted by PCBONEZ
                        Linear just means the amount of adjustment [movement] in the dial or slide paddle is 1-to-1 to the change in resistance for the whole scale.
                        They can built them for whatever they want for min-max values.
                        If non-linear they can put the 'fast' end at the high or low end.

                        Lets say it's a 100k pot with a range of 0-100k and a dial with 100 'clicks' rotation.

                        Linear.
                        10 Clicks from 1-10 = 10k change
                        10 Clicks from 90-100 = 10k change
                        ... 1-1 response over the whole range

                        Non-linear
                        10 Clicks from 1-10 = 1k change
                        10 Clicks from 90-100 = 30k change
                        ... response curve is similar to a logarithmic curve.

                        .
                        This is what I meant, a linear pot with 0 - 100k will have 50k in the middle since it is...linear.

                        Comment

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