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 :

Core Logic

Let's perform Boolean analysis on the configuration steps mapped below:

Factoring using distributive Boolean rules:

Step 1: Final Reduction

The expression reduces to a simple inverter (NOT gate) processing input A. This aligns with Circuit (1), matching option (1).

Pattern Recognition

Identify standard combinations: (A AND NOT B) OR (A AND B)$(\mathrm{A} \text{ AND NOT } \mathrm{B}) \text{ OR } (\mathrm{A} \text{ AND } \mathrm{B})$ collapses back into simply input A because operand B covers all possible states.

Chapter Mix

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

Related Formula

Let us inspect the Boolean expression for the output Y$Y$ from the truth table. From the table:

• If A=0$A=0$, Y=0$Y=0$ regardless of B$B$.
• If A=1$A=1$, Y=1$Y=1$ regardless of B$B$.

Thus, the truth table is represented by the simple direct logical equation: Y = A$Y = A$

Core Logic

Let's check the Boolean output of the options shown in the question paper:

• Circuit (1): Inputs A$A$ and B$B$ go into an OR gate, outputting (A + B)$(A + B)$. This output and B$B$ then go to an AND gate.

This gives Y = B$Y = B$ (Not matching table).

• Circuit (2): Inputs A$A$ and B$B$ go into an OR gate, outputting (A + B)$(A + B)$. This and A$A$ then go into an AND gate.

This gives Y = A$Y = A$ (Perfect match to the truth table where Y$Y$ exactly copies A$A$).

Step 1: Verification of Circuit (2)

Let's double-check the truth table values for Circuit (2):

• For A=0, B=0$A=0, B=0$: Y = (0 + 0) · 0 = 0$Y = (0 + 0) \cdot 0 = 0$.
• For A=0, B=1$A=0, B=1$: Y = (0 + 1) · 0 = 0$Y = (0 + 1) \cdot 0 = 0$.
• For A=1, B=0$A=1, B=0$: Y = (1 + 0) · 1 = 1$Y = (1 + 0) \cdot 1 = 1$.
• For A=1, B=1$A=1, B=1$: Y = (1 + 1) · 1 = 1$Y = (1 + 1) \cdot 1 = 1$.

This perfectly matches the given truth table. Therefore, Circuit (2) is correct.

Pattern Recognition

Identify the logic expression directly from the truth table first! Notice that Y$Y$ is completely independent of B$B$ and strictly equals A$A$. This immediately points to any Boolean simplification that collapses to A$A$ (such as absorption law: A(A+B) = A$A(A+B) = A$).

Chapter Mix

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

Related Formula

Mass Action Law:

Core Logic

Let's analyze each statement for an n-type semiconductor:

• (A) Holes are minority carriers: True, electrons are the majority carriers.
• (B) The dopant is a pentavalent atom: True (like Phosphorus, Arsenic) which provides extra free electrons.
• (C) and (D) contradict the fundamental mass action law nₑ nh = nᵢ²$n_e n_h = n_i^2$, so they are False.
• (E) Holes are generated purely due to thermal excitation, not due to donor atoms: True.

Step 1: Assemble Correct Set

Statements (A), (B), and (E) are explicitly correct.

Pattern Recognition

Mass action law (nₑ nh = nᵢ²$n_e n_h = n_i^2$) holds uniformly for both doped types at thermal equilibrium. In n-type systems, donors directly inject electrons only; holes emerge solely from thermal breakages of lattice bonds.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Related Formula

Given logical expression:

By De Morgan's laws:

Core Logic

Let's analyze configurations A and B:

• Configuration A: Generates A BC$A\overline{B}C$ using one 3-input AND gate and one NOT gate for input B$B$. Generates A C$\overline{A}\overline{C}$ using one 2-input AND gate and two separate NOT gates for inputs A$A$ and C$C$. Combines both terms using a 2-input OR gate.

(Total: one 3-input AND, one 2-input AND, three NOT gates, one 2-input OR gate).

Step 1: Verify Configuration B

• Configuration B: Generates A BC$A\overline{B}C$ using one 3-input AND gate and one NOT gate for input B$B$. Simplifies the second term A C$\overline{A}\overline{C}$ into A+C$\overline{A+C}$, realized directly with a single 2-input NOR gate. Combines both sub-circuits using a 2-input OR gate.

(Total: one 3-input AND, one 2-input NOR, one NOT gate, one 2-input OR gate).

Step 2: Verify Configuration C

• Configuration C: Specifies a 3-input OR gate and a 2-input AND gate at the output, which implements a product-of-sums form rather than the required sum-of-products expression. Hence, Configuration C is invalid.

Both configurations A and B correctly realize the logic function.

Pattern Recognition

Apply De Morgan's theorem (A· B = A+B$\overline{A}\cdot\overline{B} = \overline{A+B}$) to convert negated AND products into standard NOR gate structures, reducing the total gate count.

Evaluation Rubric / Model Answer

Option B: A, B Only

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Core Logic

Let the intermediate outputs of the first layers be Y₁$Y_1$ and Y₂$Y_2$ [cite: 747]:

• Top gate is an AND gate with inputs A$A$ and B$B$, so Y₁ = A · B$Y_1 = A \cdot B$ [cite: 747].
• Bottom gate is an OR gate where one input is B$B$ and the other is A$\bar{A}$ via a NOT gate, so Y₂ = A + B$Y_2 = \bar{A} + B$ [cite: 747].
• The final layer is a NAND gate with inputs Y₁$Y_1$ and Y₂$Y_2$, so Y = Y₁ · Y₂$Y = \overline{Y_1 \cdot Y_2}$[cite: 748].

Step 1: Constructing the Truth Table

Let's compute the output Y$Y$ for all binary input pairs (A, B)$(A, B)$ [cite: 757]:

• For A=0, B=0 Y₁ = 0, Y₂ = 1 Y = 0 · 1 = 1$A=0, B=0 \implies Y_1 = 0, Y_2 = 1 \implies Y = \overline{0 \cdot 1} = 1$ [cite: 757].
• For A=1, B=0 Y₁ = 0, Y₂ = 0 Y = 0 · 0 = 1$A=1, B=0 \implies Y_1 = 0, Y_2 = 0 \implies Y = \overline{0 \cdot 0} = 1$ [cite: 757].
• For A=0, B=1 Y₁ = 0, Y₂ = 1 Y = 0 · 1 = 1$A=0, B=1 \implies Y_1 = 0, Y_2 = 1 \implies Y = \overline{0 \cdot 1} = 1$ [cite: 757].
• For A=1, B=1 Y₁ = 1, Y₂ = 1 Y = 1 · 1 = 0$A=1, B=1 \implies Y_1 = 1, Y_2 = 1 \implies Y = \overline{1 \cdot 1} = 0$ [cite: 757].

Thus, Y=0$Y=0$ uniquely when A=1$A=1$ and B=1$B=1$[cite: 89, 90, 757].

Pattern Recognition

A NAND gate produces an output of 0$0$ if and only if all its inputs are 1$1$. Working backward, this instantly sets Y₁=1$Y_1=1$ and Y₂=1$Y_2=1$. For Y₁ = A · B = 1$Y_1 = A \cdot B = 1$, we must have A=1$A=1$ and B=1$B=1$ simultaneously.

Chapter Mix

Class 12 Physics: Semiconductors