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प्रश्न
Give the truth table and circuit symbol for NAND gate
What is a NAND gate? Write its truth table.
उत्तर १
The output is the inverted version of the output of AND gate. The symbol for NAND gate is given below
Truth table
| I | O | O/P |
| A | B | Y |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
उत्तर २
NAND gate is the combination of AND gate and NOT gate i.e. NAND gate is nothing but it is AND gate with an inverted output.
The Logical statement of NAND gate is if input A is true and B is true then the output if false otherwise
true.
The Boolean equation of NAND gate is `Y = bar(A.B)`
| Input | Output | |
| A | B | `Y = bar(A.B)` |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
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संबंधित प्रश्न
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(a)
(b)
Draw the schematic symbols for AND, OR, NOT and NAND gate
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(i) Write the truth tables of the logic gates marked P and Q in the given circuit.
(ii) Write the truth table for the circuit.
Draw a diagram to show how NAND gates can be combined to obtain an OR gate. (Truth table is not, required)
Useful Constants and Relations:
| 1. | Charge of a proton | (e) | =1.6 × 10-19C |
| 2. | Planck's constant | (h) | = 6·6 × 10-34 Js |
| 3. | Mass of an electron | (m) | = 9·1× 10-31 kg |
| 4. | Permittivity of vacuum | (∈0) | =8 · 85 × 10-12 Fm-1 |
| 5. | `(1/(4pi∈_0))` | =9 ×109 mF-1 | |
| 6. | Permeability of vacuum | (μ0) | = 4π × 10-7 Hm-1 |
| 7. | `((mu_0)/(4pi))` | =1 × 10-7 Hm-1 | |
| 8. | Speed of light in vacuum | (c) | = 3× 108 ms-1 |
| 9. | Unified atomic mass unit | (u) |
= 931 MeV |
| 10. | Electron volt | (leV) | = 1.6 × 10-19 J |
With the help of a diagram, show how you can use several NAND gates to obtain an OR gate.
Useful Constants and Relations :
| 1. Speed of Light in Vacuum | (c) = 3.00 x 108 m/s |
| 2. Charge of a proton | (e) = 1.60 x 10-19C |
| 3. Planck's Constant | (h) = 6.6 x 10-34 Js |
| 4. Permeability of vacuum | (μ0) = 4π x 10-7 Hm-1 |
| 5. Electron Volt | (1eV ) = 1.6 x 10 |
| 6. Unified Atomic Mass Unit | (1u) = 931 MeV |
| (π) = 3.14 | |
| ( ln 2 ) = 0.693 |
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| A | B | Y |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
It is represented by following logic relation `Y = barA.B + A.barB`. Build this gate using AND, OR and NOT gates.
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