Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pull up resistor calculation

  1. Mar 22, 2013 #1
    http://i.imgur.com/TbYIHnb.png?1

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

    V=IR
    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

    berkeman

    User Avatar

    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".
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Pull up resistor calculation
  1. Pull up resistors. (Replies: 3)

  2. Pull up resistors (Replies: 1)

Loading...