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Sizing a Potentiometer and padding resistors

  1. Apr 21, 2004 #1
    Hello I have an Electronic control unit supplying +5volts and 15mA to a Potentiometer. The control unit needs to have padding resistors that need to be sized correctly so the output of the potentiometer is 0.5 volts to 4.5 volts. Can anyone help me figure out how to size the potentiometer and padding resistors? Your help is really appreciated.

  2. jcsd
  3. Apr 21, 2004 #2
    It can be figured simply by ohms law. You will have a resistor on each end of the pot. So from the + terminal of the power supply you will have a resistor, then the pot, then the other resistor connected to the - terminal of the supply. You need to know the ohms value of the pot. From there it is all ohms law. BTW, this only works if the wiper terminal on the pot will not be supplying any current, aka an UNLOADED voltage divider.
  4. Apr 21, 2004 #3
    Thanks for your help. So if I have a 250 ohm pot, how do I figure out the values needed for the resistors?
  5. Apr 21, 2004 #4


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    Is this homework?

    Ohm's law is: Voltage = Current * Resistance

    You could figure the total current with all the resistances added together, then multiply that current by the individual resistance to find the voltage. What do you come up with?

  6. Apr 21, 2004 #5
    Sorry, we would be more specific, but this sounds suspicously like a homework question. What exactly is the application?
  7. Apr 23, 2004 #6
    No it's not a homework question, it's a work question. We are adding an electronic control unit made by Rexroth to one of our machines. I haven't worked with ohms law too much, so I'm getting confused on how to size the padding resistors. If I use a 5k ohm pot, I'm stilled confused on how to find the resistor values. I know the voltage, but not the resistor value or current, so I have two unknowns out of three?? It might be a simply solution, but I'm confused and probably having a brain blockage. The control unit is supplying +5 volts and up to 15mA and the pot. output has to be between 0.5-4.5volts. If you had more specific info in how to size the padding resistors I would greatly appreciate it.

    Thank you,
  8. Apr 23, 2004 #7
    So you will have a total of 4 volts across 5K. That works out to be 8 micro amps. It is a series circuit so each resistor will have the 8 uA flowing through it. Use ohms law to find out how much resistance it takes for 8 uA to burn up .5 volts. Use the formula Cliff gave to come up with 625 ohms for each resistor. Each resistor is going to burn up a half a volt.
    Last edited: Apr 23, 2004
  9. Apr 23, 2004 #8


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    For a 5K pot I come up with .0008A or 800uA. This would be 625 ohms for each resistor.

    1/4 watt resistors would be plenty big for this, a 560 and a 56 or 68 in series would get you each 'resistor' since there is no 625 in regular stock. If you want, a quick check with the multimeter would get you a combination that equals 625 ohms.

    Then check the resistance across the resistors and pot to make sure its around 6.2Kohms no matter the position of the knob BEFORE hooking it up and ask if you need help. In fact, I'd suggest hooking it up to 4 AA batteries or a 9V battery to check to make sure the divider works properly (10% to 90% of the source voltage at the output) before risking it by hooking it up live.

    Good luck.

  10. Apr 27, 2004 #9
    Where I get confused is when you say that with the 5K pot you have 800uAmps, but when you add in the padding resistors, doesn't that change the amperage running thru the resistors and pot? So how can you use the 800uA to find the padding resistor values? Thanks for your help averagesupernova and Cliff.

  11. Apr 27, 2004 #10
    You figure what the current should be with 4 volts across the pot regardless of padding resistors. In the end, you KNOW you will have to have 4 volts across the pot to get the desired voltage swing. You KNOW the value of the pot so you know the current through it. You adjust the padding resistors to match the current. Typical engineering, figure from what end result is required back to what it takes to get it.
  12. Apr 27, 2004 #11


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    E = IR

    E/R = I

    4V/5000 ohms = .0008A

    Now we know the current, now get the right resistor for the drop.

    E/I = R

    .5V/.0008A = 625 ohms


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