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PIC Micro Output I/O

Discussion in 'General Electronics Discussion' started by Rajinder, Sep 21, 2016.

  1. Rajinder

    Rajinder

    402
    7
    Jan 30, 2016
    Hi,
    I have a question regarding the PIC18F4520 micro. If i connect a 10K pull up to one of the pins and make this as an output. I believe that this will be a logic '1'. If i now make this (still as an output) but output a 0, how can the output be a 0 since it is connected to VCC via a 10K Pull-up.

    Best regards,
     
  2. hevans1944

    hevans1944 Hop - AC8NS

    4,214
    1,994
    Jun 21, 2012
    Go revisit Ohm's Law. What is the resistance to ground at the output when the output is low?
     
    Last edited: Sep 22, 2016
  3. Rajinder

    Rajinder

    402
    7
    Jan 30, 2016
    If we have a push pull I/o line
     
  4. BobK

    BobK

    7,607
    1,648
    Jan 5, 2010
    Connect a 10K resistor from the + to the - terminal of a 1.5V battery. Is the voltage at the - terminal now 1.5V? If not, what is it?

    Bob
     
  5. Colin Mitchell

    Colin Mitchell

    1,419
    314
    Aug 31, 2014
    No.

    The micro makes the output HIGH or LOW. The 10k has absolutely no effect what-so-ever.
     
  6. Minder

    Minder

    2,834
    586
    Apr 24, 2015
    The output either switches to +v or common as per the attached dwg.
    M.
     

    Attached Files:

  7. hevans1944

    hevans1944 Hop - AC8NS

    4,214
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    Jun 21, 2012
    The output is shorted to common by the lower MOSFET when the output is low. The upper MOSFET in the totem-pole (NOT push-pull) output is turned off when the output is 0. The 10 kΩ "pull up" resistor is in parallel with the upper MOSFET (which is NOT conducting) and therefore provides a small current (about 0.5 mA with Vcc = 5 V) through the lower MOSFET (which IS conducting) when the output is 0. This current is NOT sufficient to raise the output to a logic 1 level. In essence, when the output is 0, the 10 kΩ resistor and the conductive resistance of the lower MOSFET form a voltage divider at the output producing an output voltage equal to (Vcc) (Ron/Ron + 10,000). The value of Ron is very low, about a tenth of an ohm or less, so the effect of Vcc acting through the 10 kΩ resistor is negligible.
     
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