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Resistance of CMOS

  1. Apr 16, 2012 #1
    1. The problem statement, all variables and given/known data

    I drew a NAND in the picture.

    302tlld.jpg



    3. The attempt at a solution

    I know when A and B are both high ( value 1), resistance will be Rn + Rn because those 2 NMOS will be turned on and resistance will added up since they are in series.

    Also when A and B are both low ( value 0), both PMOS will be turned on. My first question is, 1.) is the resistance become Rp*Rp/(Rp + Rp) because they are in parallel?

    My second question is, if A is high (value 1) and B is low ( value 0) , output should be VDD ( value 1). But what will be resistance be? I think they are neither in series nor parallel. So will the resistance only become Rp ?

    Hope somebody can share their ideas. Thank you
     
  2. jcsd
  3. Apr 16, 2012 #2

    NascentOxygen

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    Staff: Mentor

    Hi nobrainer612! Those would be my answers, too.

    Can you simplify this expression: Rp*Rp/(Rp + Rp) :wink:
     
  4. Apr 16, 2012 #3
    sorry but I don't get what you mean.

    Can you tell me if those I assumed are correct? because I am interested what the resistance is .

    simplify this expression: Rp*Rp/(Rp + Rp) : isn't that equal (Rp^2) / 2*Rp = Rp/2?


    So what I assumed:

    if A is high (value 1) and B is low ( value 0) -----> resistance = Rp only? :bugeye:
    Also when A and B are both low ( value 0) -----> resistance = Rp/2 ? :shy:
     
    Last edited: Apr 16, 2012
  5. Apr 16, 2012 #4

    NascentOxygen

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    Staff: Mentor

    :smile:
     
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