• Tags: NAND, NOR, universal, De-Morgan
• Difficulty: Moderate

NAND gate: Y = (A.B)'. NOT of AND. Truth table: (0,0)->1, (0,1)->1, (1,0)->1, (1,1)->0. NOR gate: Y = (A+B)'. NOT of OR. Truth table: (0,0)->1, (0,1)->0, (1,0)->0, (1,1)->0. Both NAND and NOR are called universal gates because any logic function can be built from just one type. NOT from NAND: connect both inputs together: Y = (A.A)' = A'. NOT from NOR: connect both inputs together: Y = (A+A)' = A'. AND from NAND: NAND followed by NOT (another NAND as inverter). OR from NOR: NOR followed by NOT. De Morgan's theorems are essential: (A.B)' = A' + B' (NAND = OR of complements), (A+B)' = A'.B' (NOR = AND of complements). These allow conversion between gate types. JEE asks to identify the output of gate combinations or simplify Boolean expressions.