In the given circuit, the breakdown voltage of the Zener diode is 3.0 ~V. What is the value of Iz? MAINIMG

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

Q jee_main_2025_29_jan_morning Logic Gates

Logic Gates diagram for Q15 - JEE Main 2025 Morning — The image features a digital circuit blueprint constructed from basic logic gates with inputs A and B mapped to an output Y.

For the circuit shown above, equivalent GATE is :

A. OR gate
B. NOT gate
C. AND gate
D. NAND gate

Solution

Core Logic

Evaluating the given logic gate diagram combination step-by-step for all input permutations yields the following truth table :

Input A	Input B	Output Y
0	0	0
0	1	1
1	0	1
1	1	1

This behavior matches an OR Gate configuration perfectly.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Q jee_main_2024_01_february_morning Zener Diode

In the given circuit if the power rating of Zener diode is

10~mW

, the value of series resistance

Rₛ

to regulate the input unregulated supply is:

Zener Diode voltage regulation circuit for Q33 - JEE Main 2024 Morning — The diagram illustrates a Zener diode stabilizer network with an unregulated input supply of 8V, a Zener voltage of 5V, a series resistor Rs, and a load resistor RL of 1 kOhm.

A. $5~kΩ$ $5\mathrm{~k}\Omega$
B. $10~Ω$ $10\mathrm{~\Omega}$
C. $1~kΩ$ $1\mathrm{~k}\Omega$
D. $10~kΩ$ $10\mathrm{~k}\Omega$

Solution

Related Formula

Voltage drop across series resistor:

Vₛ = Vᵢₙ - VZ

Load current:

IL = (VZ)/(RL)

Maximum Zener current:

IZmax = (PZ)/(VZ)

Core Logic

Given values: $Vᵢₙ = 8~V$ $V_{\text{in}} = 8\mathrm{~V}$, $VZ = 5~V$ $V_Z = 5\mathrm{~V}$, $RL = 1~kΩ$ $R_L = 1\mathrm{~k}\Omega$, $PZ = 10~mW$ $P_Z = 10\mathrm{~mW}$.

Voltage across $Rₛ$ $R_s$:

VRₛ = 8 - 5 = 3~V

Current through the load resistor:

IL = (5)/(1 \times 10³) = 5~mA

Maximum current allowed through the Zener diode:

IZmax = 10 \times 10⁻³5 = 2~mA

Step 1: Determine the Range of Resistance

Total current through the series loop: $Iₛ = IZ + IL$ $I_s = I_Z + I_L$

For maximum safety configuration (Zener operating at peak current):

Ismax = IZmax + IL = 2~mA + 5~mA = 7~mA

Rsmin = VRₛIsmax = 3~V7~mA = (3)/(7)~kΩ \approx 428.6~Ω

For minimum Zener current requirement ( $IZ \to 0$ $I_Z \to 0$):

Ismin = 0 + 5~mA = 5~mA

Rsmax = VRₛIsmin = 3~V5~mA = (3)/(5)~kΩ = 600~Ω

Therefore, the required window for regulation is:

(3)/(7)~kΩ < Rₛ < (3)/(5)~kΩ

Step 2: Note on Official Key

None of the given multiple-choice options fall strictly within the stable bounds $[428.6~Ω, 600~Ω]$ . Officially, the answer key evaluates option (3) as correct, though the problem functions fundamentally as a bonus candidate under rigorous design tolerances.

Pattern Recognition

Always solve the current constraints at both boundaries ( $IZ = 0$ $I_{Z} = 0$ and $IZ = Imax$ $I_{Z} = I_{\text{max}}$) to bracket the allowable series resistor zone.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Total Current

Step 2: Calculate Load Current

Step 3: Calculate Zener Current

Pattern Recognition

Chapter Mix

More Semiconductor Electronics: Materials, Devices and Simple Circuits Previous-Year Questions — Page 5

Solution

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Selecting Option

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Selection

Pattern Recognition

Chapter Mix

Solution

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine the Range of Resistance

Step 2: Note on Official Key

Pattern Recognition

Chapter Mix

More Semiconductor Electronics: Materials, Devices and Simple Circuits Questions — jee_main_2024_29_jan_morning

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