Static CMOS switch-level circuit to for a "¬Cout" full-adder circuit

Discussion in 'Homework Help' started by Aromasin, Apr 18, 2018.

  1. Aromasin

    Thread Starter New Member

    Apr 18, 2018
    1
    0
    The full question is as follows: "Design a static CMOS switch-level circuit to implement the complement of the Carry output ‘NCo’ using the fewest NMOS and PMOS transistor switches."

    So far I've derived the Boolean for NCout:

    ¬A.¬B + ¬A.¬C + ¬B.¬C

    That should be correct, and the furthest it can be simplified. From this I derived the following circuit. I'm now at a roadblock. I have no idea how to reduce this system further. I am 100% sure this circuit can be simplified further, seeing as you can make a whole 1-bit Full Adder using only 10 MOSFETs, and I'm currently sat at 22.

    I'm under the impression I need to use De Morgan's to find the PMOS circuit, and the inverse of the above boolean to find the NMOS circuit, but I have almost no experience working with 3 variable Boolean expressions so I'm lost. Would the PMOS be something like ¬(A+B.A+C.B+C)? That just feels incorrect to me; NMOS would be A.B+A.C+B.C I guess?

    I've been wracking my brains over this question for a whole day now and I just want to move on to the next part of the assignment at this point, or I won't be able to complete it before the hand-in. I've never had to turn to a forum before, so I'm definitely at my wits end (hand in a blank piece of paper kind of end). Any and all help is much much appreciated. Explain this like I'm 5 years old if you want. Just anything that will help me understand how to derive CMOS circuits using the fewest transistors, using a Boolean expression
     
Loading...