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Pull up resistor calculation

  1. Mar 22, 2013 #1

    So, if I have a pull up and apply a current source. How would I need to read/calculate resistance?

    I = 5V/8k = .0006A would result in a voltage drop of close to 5V

    Using a current source/Voltage reader I supply .0006A which should equate to ~5V drop

    But when I read my voltage reader I get 9.83V. Is this correct?

    Then my resistance would be 9.83/.0006 ~ 16.4k ?

    These calculations do not seem correct.
  2. jcsd
  3. Mar 22, 2013 #2


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    Staff: Mentor

    Sounds like your current source is not very precise. Keep in mind that real-world current sources need to have some voltage drop across them in order to operate correctly. Typically you need a couple of volts across a current source in order for it to say "in compliance".
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