Question
Download Solution PDFThe minimum number of 2-input NAND gate required to implement Boolean function F(A, B, C) = AB’+ BC+ AC is (assuming only normal inputs are available) :
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The given Boolean function is
We are to implement this function using only 2-input NAND gates and assuming only normal (i.e., uncomplemented) inputs are available.
Step-by-step NAND Implementation:
1. Generate B’:
Using NAND gate:
2. Generate AB’:
Use NAND to AND:
3. Generate BC:
4. Generate AC:
5. ORing all three terms:
To implement
Three terms OR can be done in 3 NAND gates optimally.
Total NAND Gates Required:
- 1 (B’)
- 2 (AB’)
- 2 (BC)
- 2 (AC)
- 3 (Final OR)
Total = 1 + 2 + 2 + 2 + 3 = 10 gates
Optimization:
With gate sharing and smart logic restructuring, it is possible to reduce the count. The minimum number of 2-input NAND gates required after such optimization is 4.
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