You are given the two circuits as shown in Fig. 14.44. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.
The circuit (a) is as follows:
STEPS:
1. The output of the NOR gate is the complement of A + B i.e., (A + B)’.
2. The NOT gate gives the complement of this output.
3. So, the output Y = A + B.
The truth table for this circuit is as follows:
Input (A) | Input (B) | (A + B)’ | Output(Y) |
0 | 0 | 1 | 0 |
0 | 1 | 0 | 1 |
1 | 0 | 0 | 1 |
1 | 1 | 0 | 1 |
The circuit (b) is as follows:
STEPS:
1. The NOT gates give the complements of the inputs A and B i.e., A’ and B’.
2. The NOR gate gives the output as (A’ + B’)’.
3. Using De Morgan’s law, (A’ + B’)’ = ((A∙B)’)’ = A∙B
4. So, the output Y = A∙B.
Input (A) | Input (B) | A’ | B’ | Output (Y) |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
1 | 1 | 0 | 0 | 1 |
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