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Is this the right truth table for a negative-logic OR? confused!

  1. Oct 16, 2005 #1
    Hello everyone, this is the first time programming a digital board without hardwiring. I'm confused on what the truth table of a "negative-logic OR" would look like:

    The diagram looks like
    Two inputs and 1 output, its an OR gate but both the inputs are sent through an inverter and the output is also inverted so i got the following boolean expression:
    _____
    'A + 'B = AB
    so using Demorgans theroem i get AB,
    so if this is the case, the truth table would be
    A B Output
    0 0 0
    0 1 0
    1 0 0
    1 1 1

    is that right?
     
  2. jcsd
  3. Oct 16, 2005 #2
    Code (Text):

    A  B  Inv  A'  B'  Or  O  Inv  O'
    0  0       1   1       1       0
    0  1       1   0       1       0
    1  0       0   1       1       0
    1  1       0   0       0       1
     
    Assuming that I'm following the process correctly (Invert the inputs -> OR -> Invert the output) I agree with you.
     
  4. Oct 16, 2005 #3
    Grogs, thanks for the responce, but what i don't get is, the book says...Since the output of the OR gate is driving the decimal LED signal on the left digit of the two seven-segment display should be off. Which makes sense, since 0 0 is output 0. But then it says Sinc ethe buttons are active low, and the BNOR2 gate also has active low inputs and output, hitting either button should turn on the LED. This makes no sense becuase 1 0 should be 0, 0 1 should be 0, meaning the light shouldn't turn on if just 1 Pushbutton is pressed.
     
  5. Oct 16, 2005 #4

    Claude Bile

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    Science Advisor

    An OR gate with active low inputs still functions as an OR gate (i.e. it has the same truth table).

    Claude.
     
  6. Oct 17, 2005 #5
    Thank you!:biggrin:
     
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