Which of the following circuits has the same output as that of the given circuit?

Q: Which of the following circuits has the same output as that of the given circuit?

### 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) collapses back into simply input A because operand B covers all possible states. ### Evaluation Rubric / Model Answer null ### Chapt

Q16 jee_main_2025_04_april_evening Extrinsic Semiconductors

Consider a n-type semiconductor in which

nₑ

and

nh

are number of electrons and holes, respectively. (A) Holes are minority carriers (B) The dopant is a pentavalent atom (C)

nₑnh\neq nᵢ²

(where

nᵢ

is number of electrons or holes in semiconductor when it is in intrinsic form) (D)

nₑnh\geq nᵢ²

(E) The holes are not generated due to the donors Choose the correct answer from the options given below:

A. (A), (C), (D) only
B. (A), (C), (E) only
C. (A), (B), (E) only
D. (A), (B), (C) only

Solution

Related Formula

Mass Action Law:

nₑ \cdot nh = nᵢ²

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

Q jee_main_2025_04_april_morning Logic Gates

The Boolean expression

Y=AoverlineBC+overlineAoverlineC

can be realised with which of the following gate configurations. A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate B. One 3-input AND gate, 1 NOT gate, One 2-input NOR gate and one 2-input OR gate C. 3-input OR gate, 3 NOT gates and one 2-input AND gate Choose the correct answer from the options given below

A. B, C Only
B. A, B Only
C. A, B, C Only
D. A, C Only

Solution

Related Formula

Given logical expression:

Y = A BC + A C

By De Morgan's laws:

A\cdot C = A+C (NOR configuration)

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).

Logic circuit layout A for Q15 - JEE Main 2025 Morning

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 ( $A\cdot B = A+B$ ) 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

Q15 jee_main_2025_24_jan_evening Logic Gates

The output of the circuit is low (zero) for :

Digital logic gate circuit diagram with inputs X and Y Q15 — The figure contains a digital combination logic gate configuration with input variables X and Y.

(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:

A. (A), (C) and (D) only
B. (A), (B) and (C) only
C. (B), (C) and (D) only
D. (A), (B) and (D) only

Solution

Core Logic

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

Truth table matrix visualization for Q15 — The figure contains a digital combination logic gate configuration with input variables X and Y.

arrayccc X & Y & Output 0 & 0 & 1 0 & 1 & 0 1 & 0 & 0 1 & 1 & 0 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.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Q6 jee_main_2025_24_jan_morning Optoelectronic Junction Devices

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 :

A. B, D, E Only
B. A, C Only
C. A, C, E Only
D. B, E Only

Solution

Core Logic

Let's analyze each statement conceptually: Statement A: Solar cells require a wide surface layer area to intercept maximum sunlight illumination, so junction area is large.

Statement B: True. Solar cells operate spontaneously to provide power to loads without requiring external bias voltage.
Statement C: False. LEDs are made of heavily doped junctions to maximize recombination probability.
Statement D: False. Beyond a critical limit, high currents cause heating that drops efficiency, so emission intensity does not increase infinitely.
Statement E: True. Forward biasing allows minority injection leading to radiative recombination.

Step 1: Selecting Option

Since statements B and E are purely accurate, the correct grouping option is B, E Only.

Pattern Recognition

Remember: LEDs = Forward Bias, Photodiodes = Reverse Bias, Solar Cells = Zero External Bias.

Chapter Mix

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

Q jee_main_2025_28_jan_evening Diode Rectifiers

In the circuit shown here, assuming threshold voltage of diode is negligibly small, then voltage

VAB

is correctly represented by:

Diode Rectifiers diagram for Q15 - JEE Main 2025 Evening — An AC source connected across an orientation circuit involving an ideal junction diode.

A. $VAB would be zero at all times$
B.
C.
D.

Solution

Core Logic

Analyze the cycle profile behavior of the input voltage waveform $V = V₀ ω t$ $V = V_0 \sin \omega t$:

Positive Half Cycle: Node A achieves a positive potential relative to node B. Under this configuration, the diode enters a Reverse Biased (R.B.) state, acting as an open switch circuit block. Since no current conducts across the resistive path, the potential difference tracked directly mirrors the input wave voltage.
Negative Half Cycle: Node A goes negative relative to node B. This transitions the diode into a Forward Biased (F.B.) condition, acting as a closed short-circuit bypass path. Consequently, the potential settles down immediately to zero.

This behavior is visualized through the input/output tracking waveforms below:

Diode Rectifiers solution step diagram for Q15 — An AC source connected across an orientation circuit involving an ideal junction diode.

Step 1: Selection

Matching this half-wave rectified configuration precisely selects option (4).

Pattern Recognition

When solving diode waveform problems, replace the diode mentally with an open circuit during reverse bias and a short circuit during forward bias to quickly observe the resulting output profile.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Which of the following circuits has the same output as that of the given circuit? A combination gate circuit configuration evaluated for total Boolean output expressions.

Solution & Explanation

Core Logic

Step 1: Final Reduction

Pattern Recognition

Chapter Mix

More Semiconductor Electronics: Materials, Devices and Simple Circuits Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Assemble Correct Set

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Verify Configuration B

Step 2: Verify Configuration C

Pattern Recognition

Evaluation Rubric / Model Answer

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Solution

Core Logic

Pattern Recognition

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Solution

Core Logic

Step 1: Selecting Option

Pattern Recognition

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Solution

Core Logic

Step 1: Selection

Pattern Recognition

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Which of the following circuits has the same output as that of the given circuit?
A combination gate circuit configuration evaluated for total Boolean output expressions.