I have a complex output function in boolean algebra ( Where '~' means NOT):
F=~( (a c ~d) + (a ~c ~d) + (~a c) + (~ a c d) + (a c ! d b) )
I know this can be simplified down to:
F = (~a ~c) + (a d)
Now the simplified version can be implemented in CMOS using 8 transistors, using custom gates. (i.e. not NAND and NOR exclusively)
If I was to implement the simplified in CMOS using ONLY NAND or NOR gates, how many transistors would there be? Is there an easy way to count by just looking at the function?
I figured that
- 1 AND -> 2 NAND
- 1 NOT -> 1 NAND
- 1 OR -> 3 NAND
- 1 NAND -> 4 transistors
- 1 NOT -> 2 Transistors
Which means the simplified version can be made up of 28 transistors?
So if I use demorgans: F=~( ~(~a ~c) ~(a d) )
- ~(~a ~c) is 8 transistors (2*2 for the inverters + 4 for NAND)
- ~(a d) is 4 transistors ( For a single NAND)
- The above 2 are then combined in 4 transistors ( 1 more NAND)
- Making a total of 8+4+4=16 ?