Discussion
What is the minimum number of two-input NAND gates used to perform the function of two input OR gate ?
Answer: C
Y=A+B. This is the equation of OR gate. We require 3 NAND gates to create OR gate. We can also write
After 1st NAND operation
Y = (A AND B)'
Y= A' + B' (Demorgan's Law)
After 2nd NAND operation
Y= ( A' + B')'
Y= A . B (Demorgan's Law)
After 3rd NAND operation
Y= ( A . B )'
Y= A' + B ' (Demorgan's Law)
So we need 3 NAND gates.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development