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Transistors and boolean expression

  1. Feb 23, 2013 #1
    Hi.
    If I could get some clarification on the attached circuits that would be great. The question asks to find the boolean expression for C.

    I already know the answers, but I don't quite see how they got it.
    For the first circuit: C=A_bar
    The truth table looks like this:
    A C
    0 1
    1 0

    This is where I'm getting completely lost. Isn't C basically connected to ground?? Wouldn't that mean C is always 0?
     

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  2. jcsd
  3. Feb 23, 2013 #2

    NascentOxygen

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    Staff: Mentor

    In the first circuit, C is solidly connected to ground. C is permanently 0.

    I think your lecturer must have had a slightly different circuit in mind. Or else he has a wicked sense of humour. :wink:

    What is the output of your second circuit?
     
  4. Feb 23, 2013 #3
    Thanks.

    I actually think he was being serious because I remember him justifying it. He said something about the arrow in the transistor represented a diode, so C would be the inverse of A. I guess that makes sense, but C=0 makes sense as well. So I think maybe the circuit was drawn wrong?
     
  5. Feb 23, 2013 #4

    NascentOxygen

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    Staff: Mentor

    I guess he intended that C be at the collector, not the emitter.
     
  6. Feb 23, 2013 #5
    Isn't the diode from the base to the emitter?



    I don't quite understand the second circuit.
    This is what I 'think' happens:
    If A or B is =1, then there would be current flowing through the circuit, thus causing a voltage drop over the top resistor (the one right next to the 5V symbol).
    If both A and B are on, then the above statement holds true as well.
    Otherwise (A and B are both off), then the 2 transistors act as gaps, thus no current would flow in the circuit. Thus, the voltage drop across the top resistor is 0V. Then C=5V.
    This matches the truth table they have:
    A B C
    0 0 1
    0 1 0
    1 0 0
    1 1 0
     
  7. Feb 24, 2013 #6

    NascentOxygen

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    Staff: Mentor

    There is a PN junction there, yes.
    So what Boolean operation does this represent? AND, OR, or what?
     
  8. Feb 24, 2013 #7
    NAND.

    Are my justifications in the previous post correct?
     
  9. Feb 24, 2013 #8

    NascentOxygen

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    Staff: Mentor

    Yes
    They sound right.
     
  10. Feb 24, 2013 #9
    For the first circuit, if we placed the output C right between the transistor and resistor, would that mean C is the inverse of A?
    This is why I think it's that.
    If A is low (0V), then the transistor is OFF, then the circuit becomes an open circuit, so the voltage drop across the resistor would be 0V due to no current flow. Thus C is =5V.
    Else, transistor is ON=>current flows=>V drop across resistor=>C is low.
     
  11. Feb 24, 2013 #10

    NascentOxygen

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    Staff: Mentor

    Correct, with the collector being point C.
     
  12. Feb 24, 2013 #11
    If it was at C wouldn't that mean C is always HIGH(1)=5V?
     
  13. Feb 24, 2013 #12

    NascentOxygen

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    Staff: Mentor

    If the output is taken from the collector, it will be 5V when there is no current through the resistor & transistor.

    When current flows into the base (to the emitter), the voltage between collector and emitter drops to approx. 0V.
     
  14. Feb 24, 2013 #13
    Sorry, I got my notations wrong. I kept thinking that the collector was above the resistor. Normally, I think of the collector as the very top of the circuit in a transistor, but the addition of that resistor changed it.
     
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