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Question

Provide a CMOS logic gate circuit that realizes the function Y=ABC+ABC+ABC.Y=\overline{A} B C+A \overline{B} C+A B \overline{C}. How many transistors are required? Explore the possibility of reducing the number of the transistors required.

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We need to provide a CMOS circuit that will realize the Boolean expression:

Y=ABC+ABC+ABCY=\overline{A}BC+A\overline{B}C+AB\overline{C}

Considering that we have Y (and not Y\overline{Y}), this expression will be realized by the pull-up network\textbf{pull-up network}. The pull-up network will have 9 transistors (for each of the variables). If we group two of the summands we can lower the number of transistors by 1.

Y=ABC+A(BC+BC)Y=\overline{A}BC+A\left(\overline{B}C+B\overline{C}\right)

Now, we will complement the given expression Y:

Y=ABC+A(BC+BC)=(A+B+C)(A+BC+BC)=(A+B+C)(A+BCBC)=(A+B+C)(A+(B+C)(B+C))=(A+B+C)(A+BC+BC)=AA+AB+AC+ABC+BBC+CBC+ABC+BBC+CBC=AB+AC+ABC+BC+BC+ABC=AB+AC+ABC+BC+ABCABC+BC=BCY=AB+AC+BC+ABC=AB+C(A+B)+ABC\begin{align*} \overline{Y}&=\overline{\overline{A}BC+A\left(\overline{B}C+B\overline{C}\right)}\\ &=\left(A+\overline{B}+\overline{C}\right)\cdot \left(\overline{A}+\overline{\overline{B}C+B\overline{C}}\right)\\ &=\left(A+\overline{B}+\overline{C}\right)\cdot \left(\overline{A}+\overline{\overline{B}C}\cdot \overline{B\overline{C}}\right)\\ &=\left(A+\overline{B}+\overline{C}\right)\cdot \left(\overline{A}+\left(B+\overline{C}\right)\cdot \left(\overline{B}+{C}\right)\right)\\ &=\left(A+\overline{B}+\overline{C}\right)\cdot \left(\overline{A}+\overline{B}\cdot \overline{C}+BC\right)\\ &=A\overline{A}+\overline{A}\cdot\overline{B} + \overline{A}\cdot\overline{C}+ A\overline{B}\cdot\overline{C}+ \overline{B}\cdot\overline{B}\cdot\overline{C}+\overline{C}\cdot\overline{B}\cdot\overline{C}+ABC+\overline{B}BC+\overline{C}BC\\ &=\overline{A}\cdot\overline{B} + \overline{A}\cdot\overline{C}+ A\overline{B}\cdot\overline{C}+ \overline{B}\cdot\overline{C}+\overline{B}\cdot\overline{C}+ABC\\ &=\overline{A}\cdot\overline{B} + \overline{A}\cdot\overline{C}+ A\overline{B}\cdot\overline{C}+ \overline{B}\cdot\overline{C}+ABC\\ &A\overline{B}\cdot\overline{C}+ \overline{B}\cdot\overline{C}=\overline{B}\cdot\overline{C}\\ \Rightarrow \overline{Y}&=\overline{A}\cdot\overline{B} + \overline{A}\cdot\overline{C}+\overline{B}\cdot\overline{C}+ABC\\ &=\overline{A}\cdot\overline{B} +\overline{C} \left(\overline{A}+\overline{B}\right)+ABC \end{align*}

This expression will be realized by the pull-down network\textbf{pull-down network}. For that we need 8 transistors. Now we can synthetize the complete CMOS circuit, for which we will need 16 transistors in total.

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