A NAND gate performs the logical AND operation followed by a logical NOT operation. Its output is the inverse of an AND gate's output.
The logical expression for a 2-input NAND gate is $Y = \overline{A \cdot B}$. For multiple inputs ($A_1, A_2, \ldots, A_n$), it is $Y = \overline{A_1 \cdot A_2 \cdot \ldots \cdot A_n}$.
The output ($Y$) of a NAND gate is LOW (0) only in one specific scenario: when all of its inputs are HIGH (1).
Let's look at the truth table for a standard 2-input NAND gate:
| Input A | Input B | Output Y = NOT (A AND B) |
|---|---|---|
| 0 (LOW) | 0 (LOW) | 1 (HIGH) |
| 0 (LOW) | 1 (HIGH) | 1 (HIGH) |
| 1 (HIGH) | 0 (LOW) | 1 (HIGH) |
| 1 (HIGH) | 1 (HIGH) | 0 (LOW) |
From the truth table and the definition, we can see that the output ($Y$) is LOW (0) only when Input A is HIGH (1) AND Input B is HIGH (1).
This principle extends to NAND gates with more inputs. The output remains HIGH if any input is LOW, and only drops to LOW when every single input is HIGH.
Therefore, the condition for the NAND gate output to be LOW is when all of the inputs are HIGH.
A two-input logic gate is giving high output only when both the inputs are high. For all other input conditions, the output is low. Select the correct logic gate.
The output is high only if one of the input is high. The above statement represents _____
The number of gate inputs, required to realize expression ABC + AB̅CD + EF̅ + AD is
Which of the following is logically equivalent?
A. ¬p → (q → r) and q → (p ∨ r)
B. (p → q) → r and p → (q → r)
C. (p → q) → (r → s) and (p → r) → (q → s)
Choose the correct answer from the options given below :
Consider the expression Y = P ⨁ Q ⨁ R where P, Q, R are the input variables and Y is the output variable. Y will be logic 0 if
(A) an odd number of input variables are 1
(B) an even number of input variables are 1
(C) an odd number of inputs variables are 0
(D) an even number of input variable are 0
(E) an odd number of input variable between 0 and 1
Choose the correct answer from the options given below: