In the circuit shown here, assuming threshold voltage of diode is negligibly small, then voltage mathrmV_AB$\mathrm{V_{AB}}$ is correctly represented by:

### Core Logic Let's perform Boolean analysis on the configuration steps mapped below: mathrmP = mathrmA cdot barmathrmB$$\mathrm{P} = \mathrm{A} \cdot \bar{\mathrm{B}}$$ mathrmQ = mathrmA cdot mathrmB$$\mathrm{Q} = \mathrm{A} \cdot \mathrm{B}$$ mathrmY = overlinemathrmP + mathrmQ = overlinemathrmA cdot barmathrmB + mathrmA cdot mathrmB$$\mathrm{Y} = \overline{\mathrm{P} + \mathrm{Q}} = \overline{\mathrm{A} \cdot \bar{\mathrm{B}} + \mathrm{A} \cdot \mathrm{B}}$$ Factoring using distributive Boolean rules: mathrmY = overlinemathrmA cdot (mathrmB + barmathrmB) = overlinemathrmA cdot 1$$\mathrm{Y} = \overline{\mathrm{A} \cdot (\mathrm{B} + \bar{\mathrm{B}})} = \overline{\mathrm{A} \cdot 1}$$ mathrmY = barmathrmA$$\mathrm{Y} = \bar{\mathrm{A}}$$ ### 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: (mathrmA text AND NOT mathrmB) text OR (mathrmA text AND mathrmB)$(\mathrm{A} \text{ AND NOT } \mathrm{B}) \text{ OR } (\mathrm{A} \text{ AND } \mathrm{B})$ collapses back into simply input A because operand B covers all possible states. ### Evaluation Rubric / Model Answer null ### 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. Y = (A + B) cdot B = Acdot B + Bcdot B = B(A + 1) = B$$Y = (A + B) \cdot B = A\cdot B + B\cdot B = B(A + 1) = B$$ 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. Y = (A + B) cdot A = Acdot A + Acdot B = A + Acdot B = A(1 + B) = A$$Y = (A + B) \cdot A = A\cdot A + A\cdot B = A + A\cdot B = A(1 + B) = A$$ 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) cdot 0 = 0$Y = (0 + 0) \cdot 0 = 0$. - For A=0, B=1$A=0, B=1$: Y = (0 + 1) cdot 0 = 0$Y = (0 + 1) \cdot 0 = 0$. - For A=1, B=0$A=1, B=0$: Y = (1 + 0) cdot 1 = 1$Y = (1 + 0) \cdot 1 = 1$. - For A=1, B=1$A=1, B=1$: Y = (1 + 1) cdot 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$A(A+B) = A$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits

### Related Formula Mass Action Law: n_e cdot n_h = n_i^2$$n_e \cdot n_h = n_i^2$$ ### 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 n_e n_h = n_i^2$n_e n_h = n_i^2$, 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 (n_e n_h = n_i^2$n_e n_h = n_i^2$) 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics

### Related Formula Given logical expression: Y = AoverlineBC + overlineAoverlineC$$Y = Aoverline{B}C + overline{A}overline{C}$$ By Boolean theorem logic: overlineAoverlineC = overlineA+C quad text(NOR configuration) $$overline{A}overline{C} = overline{A+C} \quad \text{(NOR configuration)} $$ ### Core Logic Let's analyze configuration options A and B: * **Configuration A:** Creates AoverlineBC$Aoverline{B}C$ via one 3-input AND gate and one NOT gate for B$B$. Creates overlineAoverlineC$overline{A}overline{C}$ via one 2-input AND gate and two separate NOT gates for A$A$ and C$C$. Finally combines them using a 2-input OR gate. This perfectly mirrors the terms. (Total gates: one 3-input AND, one 2-input AND, three NOT, one 2-input OR).

### Step 1: Verify Configuration B * **Configuration B:** Creates AoverlineBC$Aoverline{B}C$ via one 3-input AND gate and one NOT gate for B$B$. Re-writes the secondary term overlineAoverlineC$overline{A}overline{C}$ into overlineA+C$overline{A+C}$, which corresponds directly to a single 2-input NOR gate. Merges both sub-tracks using a 2-input OR gate. This is also completely valid.

### Pattern Recognition Always utilize De Morgan's identity triggers to dynamically simplify multiple product complements like overlineAoverlineC$overline{A}overline{C}$ into unified NOR formats. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics

### Core Logic Let us check the gate outputs row-by-row to find the boolean expression or map the truth table values:

beginarrayccc X & Y & textOutput \hline 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \ 1 & 1 & 0 endarray$$\begin{array}{ccc} X & Y & \text{Output} \\hline 0 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 0 \ 1 & 1 & 0 \end{array}$$ The output is low (zero) for configurations (B) X=0, Y=1$X=0, Y=1$, (C) X=1, Y=0$X=1, Y=0$, and (D) X=1, Y=1$X=1, Y=1$. Therefore, options (B), (C) and (D) only are correct. ### Pattern Recognition The truth table profile matches a standard NOR logic configuration where the output is 1 only when all input lines are completely low. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics