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 :

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. Since statements B and E are purely accurate, the correct grouping option is B, E Only. Remember: LEDs = Forward Bias, Photodiodes = Reverse Bias, Solar Cells = Zero External Bias. Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits

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

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 em

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 :

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. Since statements B and E are purely accurate, the correct grouping option is B, E Only. Remember: LEDs = Forward Bias, Photodiodes = Reverse Bias, Solar Cells = Zero External Bias. Class 12 Physics: Semiconductor Electronics: Materials, Devices and Simple Circuits

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

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 em

Consider the following statements: A. The junction area of solar cell is made very narrow compared to a photo diode. B…

Q48 jee_main_2026_24_january_morning Zener Diode

A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is ____

Ω

.

Numerical Answer. Answer: 125 to 125

Solution

Related Formula

$PZ = VZ IZ$ $P_Z = V_Z I_Z$

Vᵢₙ = IZ RS + VZ

Core Logic

For the Zener diode, the maximum power dissipated is:

PD = 0.4 W

Since $PD = VZ IZ$ $P_D = V_Z I_Z$:

0.4 = 10 \times IZ IZ = 0.04 A

Step 1: Find Protective Series Resistance

The voltage drop across the series protective resistance $R$ $R$ is:

VR = Vᵢₙ - VZ = 15 - 10 = 5 V

Using Ohm's law, $R = (VR)/(IZ)$ $R = \frac{V_R}{I_Z}$:

R = (5)/(0.04) = (500)/(4) = 125 Ω

Pattern Recognition

To secure a Zener against burning out, calculate its max safe current via $Pmax / Vz$ $P_{max} / V_z$. Feed this current into the required voltage drop ( $Vᵢₙ - Vz$ $V_{in} - V_z$) to get the exact protective resistance.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Q39 jee_main_2026_24_january_evening Logic Gates

Identify the correct truth table of the given logical circuit.

Logic Gates diagram for Q39 - JEE Main 2026 Evening — A combination of logic gates resulting in output Y.

A.
A B Y
0 0 0
0 1 1
1 0 1
1 1 0
B.
A B Y
0 0 1
0 1 0
1 0 1
1 1 0
C.
A B Y
0 0 0
0 1 0
1 0 1
1 1 0
D.
A B Y
0 0 1
0 1 0
1 0 1
1 1 0

A	B	Y
0	0	0
0	1	1
1	0	1
1	1	0

A	B	Y
0	0	1
0	1	0
1	0	1
1	1	0

A	B	Y
0	0	0
0	1	0
1	0	1
1	1	0

A	B	Y
0	0	1
0	1	0
1	0	1
1	1	0

Solution

Related Formula

De Morgan's Laws: A \cdot B = A + B

Core Logic

Let's trace the logic line by line. Top branch: A passes through an AND gate with both inputs tied to A, so it remains $A$ $A$. Bottom branch: A and B pass through a NAND gate, yielding $A \cdot B$ $\overline{A \cdot B}$. Then it passes through an AND gate with inputs tied together, so it remains $A \cdot B$ $\overline{A \cdot B}$.

Step 1: Boolean Expression

Finally, the inputs $A$ $A$ and $A \cdot B$ $\overline{A \cdot B}$ are fed into a final AND gate.

Y = A \cdot A \cdot B

Applying De Morgan's Law:

Y = A \cdot ( A + B)

Y = A \cdot A + A \cdot B

Step 2: Simplify

Since $A \cdot A = 0$ $A \cdot \overline{A} = 0$, we have:

Y = 0 + A B = A B

Checking values: If $A=1, B=0$ $A=1, B=0$, then $Y = 1 \cdot 1 = 1$ $Y = 1 \cdot 1 = 1$. For all other combinations, $Y = 0$ $Y = 0$.

Pattern Recognition

An AND gate acting on $A$ $A$ and $NAND(A, B)$ $\text{NAND}(A, B)$ inherently acts as a "strictly A and NOT B" checker. The boolean algebra instantly simplifies $A( AB)$ $A(\overline{AB})$ to $A B$ $A\overline{B}$.

Chapter Mix

Class 12 Physics: Semiconductor Electronics

Q35 jee_main_2026_28_january_morning Semiconductor Diode

Assuming in forward bias condition there is a voltage drop of 0.7 V across a silicon diode, the current through diode

D₁

in the circuit is ____ mA. (Assume all diodes in the given circuit are identical)

Circuit with diodes for Q35 — Circuit containing a 12V source, a 0.3kOhm resistor, and three diodes D1, D2, D3 in parallel.

A. $20.15$ $20.15$
B. $11.7$ $11.7$
C. $17.6$ $17.6$
D. $18.8$ $18.8$

Solution

Related Formula

I = (V - Vd)/(R)

Core Logic

Check the polarity of the battery to determine which diodes are forward-biased. Diodes $D₁$ $D_1$ and $D₂$ $D_2$ are forward-biased, while $D₃$ $D_3$ is reverse-biased (acts as an open circuit). Because $D₁$ $D_1$ and $D₂$ $D_2$ are in parallel, the total voltage drop across the parallel combination is just $0.7 ~V$ $0.7 \mathrm{~V}$.

Step 1: Loop Equation

Applying KVL to the main loop containing the forward-biased diodes:

12 - 0.3 \times 10³ Itotal - 0.7 = 0

11.3 = 300 \cdot Itotal

Step 2: Total Current

Itotal = (11.3)/(300) ~A = 37.66 \times 10⁻³ ~A = 37.66 ~mA

Step 3: Current Division

Since $D₁$ $D_1$ and $D₂$ $D_2$ are identical and in parallel, the total current divides equally between them.

ID1 = Itotal2 = (37.66)/(2) ~mA = 18.83 ~mA

Rounding to nearest option gives $18.8 ~mA$ $18.8 \mathrm{~mA}$.

Pattern Recognition

Identical diodes in parallel share the current equally. The voltage drop across the entire parallel diode bank is just the drop of one diode ( $0.7 ~V$ $0.7 \mathrm{~V}$).

Chapter Mix

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

Q30 jee_main_2026_28_january_evening Logic Gates

Two p-n junction diodes

D₁

and

D₂

are connected as shown in figure.

Logic Gates diagram for Q30 - JEE Main 2026 Evening — A logic gate formed using two diodes connected to a 5V supply through a pull-up resistor.

A and B are input signals and C is the output. The given circuit will function as a ____.

A. $OR Gate$ $\text{OR Gate}$
B. $NOR Gate$ $\text{NOR Gate}$
C. $NAND Gate$ $\text{NAND Gate}$
D. $AND Gate$ $\text{AND Gate}$

Solution

Core Logic

The circuit contains two diodes with their n-sides connected to the inputs A and B, and their p-sides tied together and connected to $+5V$ $+5\text{V}$ through a resistor $R$ $R$. The output C is taken from the common p-side junction.

Step 1: Analyzing the Truth Table

If either $A = 0$ $A = 0$ (ground) or $B = 0$ $B = 0$ (ground), the corresponding diode becomes forward biased. Current flows through the resistor $R$ $R$, dropping the voltage at C to near $0V$ $0\text{V}$ (Logic 0). If both $A = 1$ $A = 1$ ( $+5V$ $+5\text{V}$) and $B = 1$ $B = 1$ ( $+5V$ $+5\text{V}$), both diodes are reverse biased. No current flows through $R$ $R$, so the voltage at C remains at $+5V$ $+5\text{V}$ (Logic 1).

Step 2: Conclusion

The output C is 1 ONLY when both inputs A AND B are 1. This corresponds exactly to the truth table of an AND Gate.

Pattern Recognition

Diodes pointing away from the inputs with a pull-up resistor (connected to $+Vcc$ $+V_{cc}$) form an AND gate. If diodes point towards the inputs with a pull-down resistor to ground, it's an OR gate.

Chapter Mix

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

Q13 jee_main_2025_02_april_evening Logic Gates

In the digital circuit shown in the figure, for the given inputs the P and Q values are :

Digital logic gate circuit diagram with inputs 1 and 1 — The circuit has two inputs equal to 1, passing through multiple gates to produce outputs P and Q.

A. $P = 1, Q = 1$ $\mathrm{P} = 1, \mathrm{Q} = 1$
B. $P = 0, Q = 0$ $\mathrm{P} = 0, \mathrm{Q} = 0$
C. $P = 0, Q = 1$ $\mathrm{P} = 0, \mathrm{Q} = 1$
D. $P = 1, Q = 0$ $\mathrm{P} = 1, \mathrm{Q} = 0$

Solution

Related Formula

Truth relations of basic logic operations:

NAND operation: $Y = A \cdot B$ $Y = \overline{A \cdot B}$
NOR operation: $Y = A + B$ $Y = \overline{A + B}$
NOT operation: $Y = A$ $Y = \overline{A}$
OR operation: $Y = A + B$ $Y = A + B$

Core Logic

The inputs are:

Top input = $1$ $1$
Bottom input = $1$ $1$

Let's analyze step-by-step from left to right:

First Gate (NAND gate at the top-left):
Inputs are $1$ $1$ and $1$ $1$.
Output = $1 \cdot 1 = 0$ $\overline{1 \cdot 1} = 0$.
Bottom-left path with NOT gates:
Top input ( $1$ $1$) goes to a NOT gate, producing $0$ $0$.
Bottom input ( $1$ $1$) goes to a NOT gate, producing $0$ $0$.
These two $0$ $0$ values feed into the OR gate:
Output = $0 + 0 = 0$ $0 + 0 = 0$.

Step 1: Calculate output P

Now trace the path to $P$ $P$:

The inputs to the top-right AND gate are:
Output of the top-left NAND gate = $0$ $0$
Output of the bottom-left OR gate = $0$ $0$
Therefore, output $P$ $P$ is:

P = 0 \cdot 0 = 0

Step 2: Calculate output Q

Now trace the path to $Q$ $Q$:

The gate at the bottom-right is a NOR gate with two inputs:
Input 1: Output of the top-left NAND gate ( $0$ $0$) inverted by a NOT gate = $0 = 1$ $\overline{0} = 1$.
Input 2: Output of the bottom-left OR gate ( $0$ $0$).
Passing these inputs ( $1$ $1$ and $0$ $0$) through the final NOR gate:

Q = 1 + 0 = 1 = 0

Thus, both $P = 0$ $P = 0$ and $Q = 0$ $Q = 0$.

Pattern Recognition

Sees: Combinational trace with inverted nodes. Trap: Missing bubbles (NOT gates) representing inversion on internal circuit paths. Shortcut: The first NAND gate output is $0$ $0$ (since both inputs are 1). This $0$ $0$ directly goes to the upper AND gate, immediately guaranteeing output $P = 0$ $P = 0$ (eliminates options 1 and 4). Now, you only need to evaluate $Q$ $Q$ to choose between options 2 and 3.

Chapter Mix

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

Solution & Explanation

Core Logic

Step 1: Selecting Option

Pattern Recognition

Chapter Mix

Reference Study Guides

More Semiconductor Electronics: Materials, Devices and Simple Circuits Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Find Protective Series Resistance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Boolean Expression

Step 2: Simplify

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Loop Equation

Step 2: Total Current

Step 3: Current Division

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Analyzing the Truth Table

Step 2: Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate output P

Step 2: Calculate output Q

Pattern Recognition

Chapter Mix

More Semiconductor Electronics: Materials, Devices and Simple Circuits Questions — jee_main_2025_24_jan_morning

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