Write the truth table for the circuits given in Fig. 14.48 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.
The circuit (a) is as follows:
The circuit gives the output (A + A)’ = A’.
So, the output is Y = A’ and the circuit acts as a NOT gate.
The truth table for the circuit is as follows:
Input (A) | Output(Y) |
0 | 1 |
1 | 0 |
The circuit (b) is as follows:
Similar to the circuit (a), the first two gates give inverted output as A’ and B’. The third NOR gate gives the output (A’ + B’)’.
Using De Morgan’s law,
(A’ + B’)’ = (A’)’∙(B’)’ = A∙B (Since, (A’)’ = A)
Hence, the output is Y = A∙B and the circuit acts as an AND gate.
The truth table is as follows:
Input (A) | Input (B) | Output (Y) |
0 | 0 | 0 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
NOTE: NOR gate is a universal gate because it can be used to implement any Boolean function without using any other gates.