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Using a potentiometer to constrain voltage between -9 and 9V

  1. May 21, 2013 #1
    I am stumped on a problem I have been working on. I am told to build a circuit that will have an output voltage between -9 and 9V.

    We are given two 12 V batteries as power sources, and a choice between 3 potentiometers with values of 10k ohm, 20k ohm, and 50k ohm. We are allowed to use any resistor within the range of 10 ohms to 100Mohms.

    How the heck to I use the potentiometer to get a negative voltage? I used voltage division to get a resistor value for a positive 9 volts. I'm stumped
  2. jcsd
  3. May 21, 2013 #2


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    hi there
    welcome to PF

    if you connect the 2 12V batteries in series, then you can produce a -12V, 0V and +12V rails


    now figure out how you will drop it to + and - 9V :)


    Attached Files:

  4. May 21, 2013 #3
    so I need to add my potentiometer and resistor. If I put them in series and leave R as a variable setting it equal to 9V, I can solve for the resistance? I chose the 10k pot first, and I calculated using voltage division, the resistance to be 3333.33 ohms. Does this sound right?
  5. May 21, 2013 #4


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    No, you would need two resistors with a pot between them. You are almost right though.

    This is to get the positive and negative 9 volts relative to the centre point of the two batteries.
  6. May 21, 2013 #5
    okay, so I think I see what your saying. I would use the same valued resistor on the opposite side (relative to the center pt of 0V) which would be my voltage divider for the -12V. I am a little confused on how that battery is producing a negative voltage though. Should I switch the polarity of the bottom one? I guess from KVL that would give me a negative voltage.
  7. May 21, 2013 #6


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    If you measure relative to the junction of the two batteries, the other terminal of the lower battery is negative 12 volts.

    That is all they mean.
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