The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at VZ = 5V$V_{Z} = 5V$ and the desired current in load is 5 mA. The unregulated voltage source can supply up to 25V. Considering the Zener diode can withstand four times of the load current, the value of resistor RS$R_{S}$ (shown in circuit) should be ____ Ω$\Omega$.

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