RankLab
Back to Concepts
PhysicsEasyClass 12

Logic Gates + Boolean Algebra

Semiconductor

64

JEE Qs

0%

Hard

75

min

Prioritize understanding truth tables and mastering Boolean algebra simplification techniques, especially De Morgan's theorems, to efficiently solve problems involving logic gates.

🧮 Key Formulas

AND: Y = A.B
OR: Y = A+B
NOT: Y = A'
NAND: Y = (A.B)'
NOR: Y = (A+B)'
XOR: Y = A(+)B = A.B' + A'.B
XNOR: Y = A(XNOR)B = A.B + A'.B'
De Morgan's First Theorem: (A.B)' = A' + B'
De Morgan's Second Theorem: (A+B)' = A' . B'
Idempotence: A+A=A, A.A=A
Commutative: A+B=B+A, A.B=B.A
Associative: A+(B+C)=(A+B)+C, A.(B.C)=(A.B).C
Distributive: A.(B+C)=A.B+A.C, A+(B.C)=(A+B).(A+C)
Identity: A+0=A, A+1=1, A.0=0, A.1=A
Complement: A+A'=1, A.A'=0
Absorption: A+A.B=A, A.(A+B)=A
Double Negation: (A')'=A
Consensus Theorem: AB + A'C + BC = AB + A'C

✅ Key Points for JEE

  • 1Mastering the truth tables for all basic (AND, OR, NOT) and derived (NAND, NOR, XOR, XNOR) logic gates is fundamental for problem-solving.
  • 2De Morgan's theorems are crucial for simplifying Boolean expressions, converting between gate types, and analyzing complex circuits.
  • 3NAND and NOR gates are 'universal gates' — any logic function can be realized using only NAND gates or only NOR gates.
  • 4Focus on simplifying Boolean expressions using identities and De Morgan's theorems to find the minimal equivalent logic circuit.
  • 5Be proficient in converting between different representations: truth table, Boolean expression, and logic circuit diagram.

⚠️ Common Mistakes

  • ✕Confusing the truth tables or symbols for XOR and XNOR gates, leading to incorrect output calculations.
  • ✕Errors in applying De Morgan's theorems, such as neglecting double negation or incorrectly swapping AND/OR operations.
  • ✕Mistakes in simplifying complex Boolean expressions, often due to overlooking basic identities or incomplete application of theorems.
  • ✕Failing to recognize the universal gate property of NAND/NOR when asked to convert circuits using only one type of gate.

📝 Practice Questions

See all

Q27.Consider the following statements: A. The junction area of solar cell is made very narrow compared to a photo diode. B. Solar cells are not connected with any external bias. C. LED is made of lightly doped p-n junction. D. Increase of forward current results in continuous increase of LED light intensity. E. LEDs have to be connected in forward bias for emission of light. Choose the correct answer from the options given below: (1) B, E Only (2) B, D, E Only (3) A, C Only (4) A, C, E Only

2025·MCQMedium

Q43.Which of the following circuits has the same outpur as that of the given circuit? (1) (2) (3) (4)

2025·MCQMedium

Q33.Which of the following circuits represents a forward biased diode? Choose the correct answer from the options given below : (1) (A) and (D) only (2) (B), (D) and (E) only (3) (C) and (E) only (4) (B), (C) and (E) only

2025·ConceptualEasy

Q26. A B Y 0 0 1 0 1 1 1 0 0 1 1 1 To obtain the given truth table, following logic gate should be placed at G: 2025 (22 Jan Shift 2) JEE Main Previous Year Paper (1) OR Gate (2) AND Gate (3) NOR Gate (4) NAND Gate

2025·MCQEasy

Q28.The output of the circuit is low (zero) for : (A) X = 0, Y = 0 (B) X = 0, Y = 1 (C) X = 1, Y = 0 (D) X = 1, Y = 1 Choose the correct answer from the options given below : (1) (B), (C) and (D) only (2) (A), (B) and (C) only (3) (A), (C) and (D) only (4) (A), (B) and (D) only

2025·MCQEasy

Q45. In the circuit shown here, assuming threshold voltage of diode is negligibly small, then voltage VAB is correctly represented by : (1) VAB would be zero at all times (2) (3) (4)

2025·MCQMedium

NCERT Chapters

  • Class 12 Physics Ch 14: Semiconductor Electronics: Materials, Devices and Simple Circuits