Table 1: Truth Table
| A |
B |
C |
D |
Y |
| 0 |
0 |
0 |
0 |
1 |
| 0 |
0 |
0 |
1 |
0 |
| 0 |
0 |
1 |
0 |
0 |
| 0 |
0 |
1 |
1 |
0 |
| 0 |
1 |
0 |
0 |
1 |
| 0 |
1 |
0 |
1 |
0 |
| 0 |
1 |
1 |
0 |
0 |
| 0 |
1 |
1 |
1 |
0 |
| 1 |
0 |
0 |
0 |
1 |
| 1 |
0 |
0 |
1 |
0 |
| 1 |
0 |
1 |
0 |
0 |
| 1 |
0 |
1 |
1 |
0 |
| 1 |
1 |
0 |
0 |
0 |
| 1 |
1 |
0 |
1 |
0 |
| 1 |
1 |
1 |
0 |
0 |
| 1 |
1 |
1 |
1 |
0 |
Table 2: Karnaugh MapBoolean Equation: Y = B’C’D’ + A’C’D’ = C’D’(A’ + B’)
| Y |
|
A |
A |
|
|
| |
1 |
1 |
0 |
1 |
|
| C |
0 |
0 |
0 |
0 |
|
| C |
0 |
0 |
0 |
0 |
D |
| |
0 |
0 |
0 |
0 |
D |
| |
|
|
B |
B |
|
PMOS transistors pass 1’s when the inputs are 0’s and NMOS transistors pass 0’s when inputs are 1’s. Therefore, when constructing the truth table, we look at parallel and series combinations of the NMOs and PMOS transistors to determine whether a 1 or 0 is being passed. After the truth table, we construct our Karnaugh map and find a minimum Sum of Products equation.