Boolean Algebra — De Morgan's theorem
Semiconductor
11
JEE Qs
8%
Hard
45
min
Master the 'break the bar, change the sign' rule for accurate and efficient application of De Morgan's theorem to simplify complex Boolean expressions.
🧮 Key Formulas
✅ Key Points for JEE
- 1De Morgan's theorems state: The complement of a sum is the product of the complements, and the complement of a product is the sum of the complements.
- 2These theorems are crucial for simplifying complex Boolean expressions and for converting expressions between Sum-of-Products (SOP) and Product-of-Sums (POS) forms.
- 3They are fundamental in demonstrating the universality of NAND and NOR gates, allowing any logic function to be implemented using only one type of gate.
- 4When applying De Morgan's, remember the rule 'break the bar and change the sign' (complement the variables and flip the operator AND <-> OR); this rule generalizes to expressions with more than two variables.
⚠️ Common Mistakes
- ✕Incorrectly distributing the complement without changing the operator, e.g., treating (A+B)' as A' + B' or (A*B)' as A' * B'.
- ✕Forgetting to complement ALL variables and ALL operators under the complement bar when applying the theorem.
- ✕Confusion with basic distributive laws; De Morgan's specifically deals with the complement of an entire expression.
📝 Practice Questions
See allQ27.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
Q43.Which of the following circuits has the same outpur as that of the given circuit? (1) (2) (3) (4)
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
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
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
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)
NCERT Chapters
- Class 12 Physics Ch 14: Semiconductor Electronics: Materials, Devices and Simple Circuits