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How to modify 10K Potentiometer to 1K w/o microcontroller? — Parallax Forums

How to modify 10K Potentiometer to 1K w/o microcontroller?

roboticsrobotics Posts: 90
in General Discussion
Hi,

I have a particular 10K Potentiometer that I would like to use because of it's very small size, but I need it to act as a 1K potentiometer (or preferably a 100 Ohm potentiometer). I would like to accomplish this as simple as possible without the use of a microcontroller. It had been suggested that a simple Op-Amp Voltage Follower circuit would work, but I have not been able to locate a detailed specific op-amp circuit; or, perhaps another method be be suggested. Absolute linearity is not required, but a rough spread of resistance values would be acceptable. In advance, thank you!
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Comments

  • ercoerco Posts: 20,256
    Much depends on the circuit you are connecting it to.
  • roboticsrobotics Posts: 90
    The potentiometer is used as a basic resistive element acting as a volume control for a speaker.
  • Alexander (Sandy) HapgoodAlexander (Sandy) Hapgood Posts: 360
    Have a look at the OPA344 Operational Amplifier by Texas Instruments. It will take the voltage input from your 10k potentiometer and output the same voltage signal but with a little more current behind it.

    They're quite easy to wire up for unity gain but I can't remember off the top of my head how much current they can produce.

    Sandy
  • roboticsrobotics Posts: 90
    Hi Sandy,

    Thank you for your suggestion. As I have never worked with op-amps before, can you suggest a circuit or resource that explains how the op-amp will actually effect the reducing of 1-10K to 1-1K or to 1-100Ohms?

    Thank you
  • tritoniumtritonium Posts: 543
    Hi

    If you post the circuit that you are thinking of using we can make more accurate suggestions.

    Dave
  • jmgjmg Posts: 15,173
    robotics wrote: »

    I have a particular 10K Potentiometer that I would like to use because of it's very small size, but I need it to act as a 1K potentiometer (or preferably a 100 Ohm potentiometer). I would like to accomplish this as simple as possible without the use of a microcontroller.
    The potentiometer is used as a basic resistive element acting as a volume control for a speaker.

    You can certainly avoid a microcontroller, but do you have power available ?
    The conventional design is to position the pot (usually LOG law) as a volume control before the final speaker amplifier.
    In that position a 10K pot is fine.

  • roboticsrobotics Posts: 90
    edited 2015-09-10 00:07
    Hi,

    The figure is attached (to the comment below) showing a 4 ohm voice coil used powered by an Adafruit 2.5W D-amplifier. I tested various pots and found that either a 0-100 ohm (preferred) or a 1K pot produced the optimum sound level spread. A strict 10K was too heavily weighted (resistance) to drown out the sound. As mentioned, there is one particular micro-sized pot that would fit my project the best but it is only available as 10K with a linear taper. I tried placing resisters in parallel to the 10K, but the results were too heavily weighted in the higher omage area so as not to provide too good of a volume spread. That is why a simple circuit to taper down the 10K would be much appreciated.
  • roboticsrobotics Posts: 90
    Here is the figure referenced above: The D-Amp is 2.5W (the image is a bit truncated, sorry)
    potentiometer-schematic_090915_3_upload.jpg
    385 x 164 - 23K
  • ercoerco Posts: 20,256
    edited 2015-09-10 01:14
    A big series resistor on the output line to your speaker is not a proper volume control. Use your 10K pot as a voltage divider to reduce the signal input to the audio amplifier.
  • RDL2004RDL2004 Posts: 2,554
    Yes, what erco said. Connect your input signal to one end of the pot and the other end of the pot to ground. Take the signal from the wiper and connect it to the A+ input on the amplifier module.
  • roboticsrobotics Posts: 90
    As mentioned above, the 10K value damps the volume too much; whereas as determined experimentally, a 1K or 100 ohm value pot is optimum.
  • Peter JakackiPeter Jakacki Posts: 10,193
    edited 2015-09-10 01:45
    robotics wrote: »
    As mentioned above, the 10K value damps the volume too much; whereas as determined experimentally, a 1K or 100 ohm value pot is optimum.

    You are asking the wrong question and ignoring the right answer. DO NOT put the pot in series with the speaker as not only is this the wrong way to do it and besides being downright silly, pots are not designed to handle power, just signals. Use a 10K pot like this:
    SIGNAL------>POT RIGHT ARM
                       POT WIPER------> INPUT OF AMPLIFIER
GROUND----> POT LEFT ARM


  • roboticsrobotics Posts: 90
    Hi, thank you for your reply. I don't understand from your diagram where the speaker is connected. I am also not certain what you are referring to as the 'signal' or 'input of amplifier? Sorry, but I am new at this.
  • Peter JakackiPeter Jakacki Posts: 10,193
    edited 2015-09-10 02:18
    Very new it seems yet you have a amplifier board with a speaker connected and is that it? Of course not, besides power you also have the audio signal being fed into the input of the amplifier.

    My diagram is not showing you the speaker or the power or anything else that is not relevant, just the input, that is all you need to worry about. From the Adafruit page it looks like the one that says A+, does that ring a bell. Maybe the A- is grounded in your instance but you have to supply those details or a photo.
  • ercoerco Posts: 20,256
    Pic attached.
    amp.jpg
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    amp.jpg 25.7K
  • kwinnkwinn Posts: 8,697
    Forget the potentiometer on the output. No way to make that work. Connect the pot on the input as shown.
    amp.png
    396 x 182 - 11K
  • roboticsrobotics Posts: 90
    I want to thank you all, especially Peter, Erco, and Kwinn for clarifying the proper hookup of the pot. I will test this and let you know if any problems. Thanks again!
  • roboticsrobotics Posts: 90
    It worked perfectly! The participants in the Parallax forum are the best!
  • ercoerco Posts: 20,256
    Don't forget RDL2004 at 6:27p!

    Man, how I miss post numbers for easy reference.
  • Peter JakackiPeter Jakacki Posts: 10,193
    erco wrote: »
    Don't forget RDL2004 at 6:27p!

    Man, how I miss post numbers for easy reference.

    Just right click on the "time" just under the tag and copy link like this.

  • ercoerco Posts: 20,256
    Thanks for that Peter. I'd also like to see post numbers (reply numbers?) as the old forum showed (pic).
    oldforum.jpg
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  • roboticsrobotics Posts: 90
    Thank you Erco, of course all assistance let to the solution including those fine pointers by Sandy, Jmg, RDL2004, and Tritonium!
  • kwinnkwinn Posts: 8,697
    robotics wrote: »
    It worked perfectly! The participants in the Parallax forum are the best!

    Good to hear. I appreciate the followup post, and I'm sure the others who replied do so as well.
  • Peter JakackiPeter Jakacki Posts: 10,193
    edited 2015-09-10 13:42
    kwinn wrote: »
    robotics wrote: »
    It worked perfectly! The participants in the Parallax forum are the best!

    Good to hear. I appreciate the followup post, and I'm sure the others who replied do so as well.

    Yes, I appreciate the followup post too, it shows good manners and appreciation.
    @Robotics: My apologies if I sounded "pained" earlier :)

  • roboticsrobotics Posts: 90
    Hi Peter,

    Certainly no apologies necessary, you didn't sound 'pained' to me! You reminded me of one of my early teachers who liked to help you learn how to solve problems by providing solid landmarks. Much appreciated!
  • skylightskylight Posts: 1,915
    I know that the problem has been sorted but it got me thinking about ohms law and parallel resistance and wondering if putting a 100K resistance across the pot winding would have given a 1K equivalent resistance?
    Would the wiper then see 0 to 1K ohms?
    Not having the parts I cannot check but it got me thinking.
  • Heater.Heater. Posts: 21,230
    You don't need any parts

    Series resistance:

    R = r1 + r2 + r3 + ...

    Parallel reistance:

    1/R = 1/r1 + 1/r2 + 1/r3 + ...

    You may need a pencil and paper.
  • skylightskylight Posts: 1,915
    edited 2015-09-18 16:16
    the trouble i can see is that the wiper alters the parallel resistance so for each movement of the wiper the result would change so i dont think thats possible.
    In other words youd have to keep changing the parallel resistance each time.

    I know!!! wire another (1M?) pot across the 10K and adjust both :)
    would it be possible with gearing the two spindles together so they adjust accordingly? (just call me heath robinson or what's his name over there? )

    pencil and paper? i'm lazy i used an app :)
  • Heater.Heater. Posts: 21,230
    It's not clear how you want to wire this but putting a 100K across the 10K of the pot the opening poster describes gives:

    R = 1/r1 + 1/r2 = 1/10,000 + 1/100,000 = 9090.9 ohms.

    A change of about 10 percent. Hardly noticeable.
  • skylightskylight Posts: 1,915
    edited 2015-09-18 16:32
    Thanks Heater, being a friday my brain wasnt functioning properly I believe i should have said 1k in parallel
    but see my post above re the changing resistance, it wouldnt work anyway.

    Edit
    Actually having a rethink using a 1k resistor in parallel with the pots winding or track might work as the wiper would eventually short the resistor giving 0 ohms?
    Is that assumption right?
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