Calculating the value of resistors to obtain a range of voltages

Discussion in 'Math' started by Ford Prefect, Dec 8, 2018.

  1. Ford Prefect

    Thread Starter Senior Member

    Jun 14, 2010
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    Does anyone know of an easy way of calculating the value of resistors (R1,VR1,R2 in the photo) to obtain a range of voltages.
    In other words... I have a voltage source of 12vDC and want to calculate the resistor values to obtain output voltages at the wiper of a potentiometer of:

    1. voltage output of between 2.0 volts and 5.5 volts

    (2.0 volts when the potentiometer is close to lowest setting / 5.5 volts when the potentiometer is close to highest setting).
    2. voltage output of between 3.5 volts and 4.5 volts
    (3.5 volts when the potentiometer is close to lowest setting / 4.5 volts when the potentiometer is close to highest setting).
    Resistor Divider2.jpg

    I want the potentiometer to be as sensitive as possible to obtain 1/10th volt between the ranges. Eg. as in 1 above..2.0v, 2.1v, 2.2v, 2.3v etc.
     
    Last edited: Dec 8, 2018
  2. ericgibbs

    Moderator

    Jan 29, 2010
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  3. Ford Prefect

    Thread Starter Senior Member

    Jun 14, 2010
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    Hello again Eric..and thanks.
    But do you know of an easy way of calculating the value of the resistors?
    Is there an online calculator or application I can use?
     
  4. ericgibbs

    Moderator

    Jan 29, 2010
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    hi FP,
    Not seen an online calc, will have a look around.
    Remember the pot resistance will have to be a preferred value, eg: 1K 5K 10K etc... this will decide the R1 and R2 values
    E
    This image and LTS shows one way of checking your results.:rolleyes:
    E

    Update:
    Added example for the second range
     
    Last edited: Dec 8, 2018
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  5. Ford Prefect

    Thread Starter Senior Member

    Jun 14, 2010
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    Yes, thanks again Eric, I am aware of the pot resistance will decide the value of R1 and R2 and I have been juggling about with different resistor values trying to find the outputs I want but it's long and tiresome. Maybe I can write and Excel macro to change R1 and R2 by a few ohms each step and fix VR1 to try to get the outputs.
    It is also possible to make VR1 as 1k and with a 1k (or other) fixed resistor in parallel.
     
  6. ericgibbs

    Moderator

    Jan 29, 2010
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    hi,
    You have shown only two voltage range circuits in your opening post are there more.?
    To solve just for two options wouldn't justify writing an Excel routine.
    I have updated post #4 showing the second circuit.
    I chose a 10K pot in order to reduce the current loading on the 12V supply. R1 and R2 can be higher values.
    E
     
  7. ericgibbs

    Moderator

    Jan 29, 2010
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    hi,
    These are the results for a 1K pot, just divide the R1/R2 values by 10.
    E
     
  8. absf

    AAC Fanatic!

    Dec 29, 2010
    1,933
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    This is the way I would do it without much tedious calculation....

    voltage div 2,0 to 5,5.PNG
    Lets use problem 1 as an example.
    Since you need 2.0 to 5.5V and your supply is 12V. So I use 10KΩ/V as the basis of calculation.

    So the circuit would turn up as in the circuit above.
    And you wouldn't be able to get 35K pot of course, so lets use 100K pot just as an example. 100K/35K= 2.857.....
    Use that factor as a constant and multiply R1 and R2 by the same factor and get the nearest value in E24 or 1% range resistors.

    Allen
     
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  9. Ford Prefect

    Thread Starter Senior Member

    Jun 14, 2010
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    Hi Eric,
    I am getting this message when I try to run the simulation in LTSpice...
    PotSub.JPG
     
  10. ericgibbs

    Moderator

    Jan 29, 2010
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    hi.
    A quick fix while I find the pot models.
    Replace the 1k pot with a 1K resistor and measure top and bottom voltages of that resistor.
    I will post the pot models.
    E

    Update:
    Unzip this potzip, place the files in the same folder as the voltage divider you are developing.
    Later on, move the .sub to the LTS folder SUB folder and the .asy to the SYM folder.

    It includes a simple test asc
     
    Last edited: Dec 8, 2018
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  11. Ford Prefect

    Thread Starter Senior Member

    Jun 14, 2010
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    Actually Allen this is good.
    I have used a 25k pot with 68k and 20k for the resistors and getting between 1.8v and 6.4v for the voltage outputs.
    This should be a good basis to start.
     
    Last edited: Dec 8, 2018
  12. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    When you are using a 25K pot instead of a 35K pot, multiply the values by (25/35).
    The 20K will become 14.29K
    the 68K will become 48.57K

    Bertus
     
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  13. Ford Prefect

    Thread Starter Senior Member

    Jun 14, 2010
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    Cool, thanks Bertus :)
     
  14. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    The voltage across POT is:

    Vpot = 5.5V - 2V = 3.5V

    So if your pot is 25kΩ the pot current is Ipot = 3.5V/25kΩ = 140μA.

    By knowing this we can solve for R1 and R2.

    R1 = (12V - 5.5V)/140μA = 46.4kΩ

    R2 = 2V/140μA = 14.29kΩ
     
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