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POT range constraint

Discussion in 'General Electronics Discussion' started by HLOW, Apr 5, 2013.

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  1. HLOW

    HLOW

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    Jan 24, 2013
    I have a pot that has a range between zero and open Ohms. I need to constrain it between 5-50K Ohms. I'm looking for a simple solution (diagram.)
    Thanks... J
     
  2. BobK

    BobK

    7,682
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    Jan 5, 2010
    If you are really only worried about the low end, put a 5K resistor in series, the range is now 5 to 55K. If you really want 5 to 50K, put a 390K resistor in parallel with the pot as well. This will change the taper however to slightly logarithmic.

    Bob
     
    Last edited: Apr 5, 2013
  3. HLOW

    HLOW

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    Jan 24, 2013
    Do you mean a 45K in parallel to the POT? 5K + 45K = 50K

    Thanks BobK
     
  4. BobK

    BobK

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    Jan 5, 2010
    No, I mean 390K in parallel with the pot. 50K || 390K = 44.3K . If you put 45K in parallel with 50K the result would be 23.7K.

    The equation for resistors in parallel is:

    1/R = 1/R1 + 1/R2

    Bob
     
  5. HLOW

    HLOW

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    Jan 24, 2013
    Humm... OK. Now Im confused? The POT range is 0 to infinity. See the original post.
     
  6. BobK

    BobK

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    Jan 5, 2010
    Oh, I missed that, I read it as a 50K pot. Are you sure? I have never heard of pot that goes to infinity. What does it measure at half way, infinity / 2?

    Bob
     
  7. CocaCola

    CocaCola

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    Apr 7, 2012
    [​IMG]
     
  8. HLOW

    HLOW

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    Jan 24, 2013
    Well... It's a photoresister. When I measure it in full dark it reads as an open circuit and when I flood it with light it reads 0 Ohms... Thus my need to constrain it.
    Anyway I get that infinity/2 does not make sense. So what would be the proper way to describe it?
    Thanks
     
  9. CocaCola

    CocaCola

    3,635
    5
    Apr 7, 2012
    It's not reading open circuit, you need to adjust your multiplier to a higher range or look at the datasheet to get an 'about' resistance range...

    Most photoresistors are somewhere in the 10K to 10M Ohm range in the dark and swing to about 1 to 150 Ohm respective in the light...
     
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,482
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    Jan 21, 2010
    It is best not to describe it as a pot.

    It will not be open in the dark, but the resistance may be too high for your meter to read.

    Try placing a 5k in series and a 50k in parallel,

    The response won't be linear, but it wasn't anyway, so maybe it won't matter.
     
  11. HLOW

    HLOW

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    Jan 24, 2013
    Thanks for the clarification... As you may have already surmised... I'm just a hacker.
    Thanks again.
     
  12. HLOW

    HLOW

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    Jan 24, 2013
    Thanks... I did that and it works just fine for my application. Thanks again for your insight.
     
  13. BobK

    BobK

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    Jan 5, 2010
    People here are better able to help you if you if you tell us exactly what you circuit is, not something that you think is equivalent.

    Bob
     
  14. HLOW

    HLOW

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    Jan 24, 2013
    Good advice!
    Thanks Bob
     
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