Semiconductor Electronics – JEE Main PhysicsPractice Questions & PYQs

The law of mass action dictates that for a semiconductor at thermal equilibrium, the product of the electron density and hole density equals the square of the intrinsic carrier density, which is expressed by the formula $n_e n_h = n_i^2$. To determine the electron concentration in the doped semiconductor, we rearrange this expression to $n_e = \frac{n_i^2}{n_h}$. Substituting the known values, the square of the intrinsic carrier density of $1.5 \times 10^{16}$ per cubic meter results in $2.25 \times 10^{32}$ per cubic meter squared. Dividing this numerator by the given hole density of $4.5 \times 10^{22}$ per cubic meter yields the final electron concentration, represented by the equation $n_e = 5 \times 10^9$ per cubic meter.

Logic circuits determine output states by evaluating boolean expressions based on input combinations of $A$ and $B$, which in this configuration simplifies to the XNOR function $Y = AB + \overline{A} \overline{B}$. This function produces a high output of 1 whenever the inputs are identical and a low output of 0 when they differ. Evaluating the states confirms this: inputs $A=0, B=0$ and $A=1, B=1$ both yield $Y=1$, while mixed inputs $A=0, B=1$ and $A=1, B=0$ result in $Y=0$. Consequently, the complete truth table follows the sequential output pattern of 1, 0, 0, 1.

A Zener diode operates as a voltage regulator by conducting excess current to maintain a constant potential difference across its terminals, satisfying Kirchhoff's current law at the junction where the source current branches into the load and the diode. Given the input source of 22 V and a Zener breakdown voltage of 15 V, the voltage drop across the series resistor is determined as $22 - 15 = 7$ V. By applying Ohm's law, the total current flowing from the source is $I = \frac{7}{35} = 0.2$ A, while the current diverted to the load resistor is $I_L = \frac{15}{90} = \frac{1}{6}$ A. The current specifically passing through the Zener diode is the difference between these two values: $I_Z = \frac{1}{5} - \frac{1}{6} = \frac{1}{30}$ A. Finally, the power dissipated is calculated using the formula $P = 15 \times \frac{1}{30} = 0.5$ W, which is equivalent to $5 \times 10^{-1}$ W, identifying the value of $x = 5$.

When a Zener diode operates in the breakdown region, it maintains a steady potential difference across its terminals, which effectively clamps the voltage to its breakdown level. In this circuit, the 2kΩ resistor is positioned in parallel with the Zener diode, meaning the potential difference across it is fixed at 5V.

The current through the resistor is determined by Ohm's Law, using the equation:
$I = \frac{V}{R}$
Substituting the given values, the current calculation becomes $I = \frac{5}{2 \times 10^3}$, which equals 0.0025A. Converting this result into the required form of $x \times 10^{-4}$, we obtain $25 \times 10^{-4}$, which confirms that the value of x is 25.

A NOR gate produces a high output only when both of its inputs are logic low, so the output of the entire circuit depends on how these inversions propagate through the logic chain. Tracing the signals through the first gate creates the shared term $\overline{A+B}$, which is combined with A and B in the second and third gates, respectively. Using boolean simplification, the output expression is $Y = (A + \overline{A+B}) \cdot (B + \overline{A+B})$. Testing this relationship across all four possible input combinations of A and B yields the corresponding output sequence of 1, 0, 0, and 1.

When diodes $D_1$ and $D_2$ operate in a forward-biased configuration, they act as ideal conductors with zero resistance. This condition creates a short circuit that pulls the input potential to 0 V, which consequently reduces the input current to 0 A. Because the circuit path draws no current, the output voltage remains at the potential of the supply, such that $V_0 = 5$ V.

A transistor is a monolithic three-layer semiconductor device, structured as p-n-p or n-p-n to serve as an amplifier or switch, and it cannot be replicated by simply connecting two discrete diodes back-to-back due to fundamental differences in semiconductor fabrication and junction physics. When evaluating the performance of these devices, the amplification factor $\beta$ characterizes the current gain, which is defined by the ratio of the collector current to the base current as $\beta = \frac{I_{C}}{I_{B}}$.

A full-wave rectifier generates a pulsating output that requires filtering to produce a stable DC voltage for practical use. Connecting a capacitor in parallel across the load $R_L$ is an effective strategy because the capacitor inhibits the steady DC component from passing through itself, thereby forcing the current through the load to maintain a consistent output. Alternatively, placing an inductor in series with the load $R_L$ works by opposing the time-varying AC ripple current, which effectively suppresses those fluctuations and ensures a steady DC signal reaches the load, confirming that both configuration methods are valid.

Logic gates are identified by their input-output relationships, where the first circuit produces a 5 V output if either or both input voltages $V_A$ and $V_B$ are 5 V and remains at 0 V only when both inputs are 0 V, which is characteristic of an OR gate. The second circuit features a transistor in a common emitter configuration that serves to invert the signal, where a high input voltage produces a low output and a low input voltage produces a high output, confirming the behavior of a NOT gate.

The power gain of a common emitter amplifier is determined by the square of the current gain multiplied by the ratio of the output load resistance to the input circuit resistance. This relationship is expressed by the formula $P = \frac{\beta^2 R}{r}$, where $P$ is the power gain, $\beta$ is the common emitter current gain, $R$ is the output resistance, and $r$ is the input resistance. Substituting the given values of 1,000,000 for the power gain, 10,000 Ohms for the output resistance, and 100 Ohms for the input resistance, the equation becomes $1,000,000 = \frac{\beta^2 \times 10000}{100}$. Simplifying the ratio of resistances on the right side produces $1,000,000 = 100 \beta^2$. Dividing both sides of this equation by 100 results in $\beta^2 = 10000$, which indicates that the common emitter current gain $\beta$ is 100.

Diodes function as polarity-sensitive switches that determine whether current flows through a specific path. When a diode is forward biased, it acts as a resistor with a value of $30\ \Omega$, but when it is reverse biased, it provides infinite resistance, effectively acting as an open circuit that stops all current flow. In this specific circuit, $D_{1}$ and $D_{2}$ are forward biased and conduct current, while $D_{3}$ is reverse biased and completely blocks its branch. By treating the forward-biased diodes as ohmic resistors and applying Kirchhoff's laws to the active loops connected to the $200\ \text{V}$ source, the total current $I_{1}$ is determined by the relationship between the applied voltage and the resulting equivalent resistance. Solving the loop equation for this configuration yields $I_{1} = 2\ \text{A}$.

De Morgan's theorem serves as the fundamental principle here, enabling the conversion of an inverted sum into a product of individual input negations. By applying this rule to the expression $\overline{A+\overline{B}}$, the overall inversion distributes across the terms to transform the OR operation into an AND operation, resulting in $\overline{A} \cdot \overline{\overline{B}}$. Since the double negation of the input $B$ effectively cancels out, the circuit's logic simplifies entirely to $\overline{A} \cdot B$.

The position of the Fermi level in a semiconductor is fundamentally determined by the relative concentration of majority charge carriers. In an n-type semiconductor, the inclusion of donor atoms increases the electron density such that $n > p$, causing the Fermi level to shift upward toward the conduction band to accommodate the increased electron population. Conversely, in a p-type semiconductor, introducing acceptor impurities increases the hole density so that $p > n$, which drives the Fermi level downward toward the valence band to reflect the higher probability of finding holes.

Doping intrinsic silicon with pentavalent impurities, which possess five valence electrons, introduces additional free electrons into the crystal structure, thereby effectively increasing the overall electron density of the material. Despite this surplus of mobile charge carriers, an n-type semiconductor maintains global electrical neutrality because the total positive charge of the ionized donor atoms perfectly balances the total negative charge of the free electrons, rendering the net charge of the entire material zero.

Ohm's law dictates that current equals voltage divided by resistance, which serves as the basis for evaluating individual circuit paths. Dividing 50 by 1000 yields $I = \frac{50}{1000} = 50$ mA, while dividing 50 by 2000 results in $I = \frac{50}{2000} = 25$ mA. Determining $I_z$ involves finding the difference between these two currents, so subtracting 25 mA from 50 mA provides a final value of 25 mA.

Diodes behave as polarity-sensitive components that allow current to pass when forward biased but function as an open circuit when reverse biased. In this configuration, $D_1$ is forward biased and contributes its internal forward resistance of 50 ohms, while $D_2$ is reverse biased and effectively acts as an open switch that blocks current in its branch. The active path comprises the 6 V battery, the forward resistance of $D_1$, the 130 ohm resistor, and the 120 ohm resistor connected in series. The total resistance of this series loop is $R = 50 + 130 + 120 = 300$ ohms. Applying Ohm's law, the current flowing through this circuit is calculated as $I = \frac{6}{300} = 0.02$ A. Converting this value to milliamperes, the current passing through the 120 ohm resistor is 20 mA.

A Zener diode connected in parallel with a load acts as a voltage regulator by maintaining a constant voltage across the load. In this setup, the voltage across the 5 kΩ load is 10 V, which leads to a load current defined by $I_L = \frac{10}{5000} = 2$ mA. The voltage drop across the 1 kΩ series resistor is $V' = 24 - 10 = 14$ V. The total current flowing from the supply through this resistor is $I = \frac{14}{1000} = 14$ mA. Since the total current is the sum of the currents through the load and the Zener diode, the diode current is $I_Z = 14 - 2 = 12$ mA. Finally, the power dissipated by the diode is $P = 10 \times 12 = 120$ mW.

The common emitter current gain, $\beta$, depends on the common base current gain, $\alpha$, which is the ratio of the change in collector current to the change in emitter current. Dividing 3.5 mA by 4 mA yields $\alpha = \frac{3.5}{4} = 0.875$. The relationship between these two parameters is defined by the following expression:

$\beta = \frac{\alpha}{1 - \alpha}$

Substituting 0.875 into this equation results in $\beta = \frac{0.875}{1 - 0.875} = \frac{0.875}{0.125}$, which simplifies to 7.

Combining the logic gates shown in the circuit leads to the Boolean expression $Y = \overline{\overline{A} + B}$. By applying De Morgan's theorem, we simplify this expression to $Y = A \cdot \overline{B}$, which demonstrates that the output is high only when signal $A$ is high and signal $B$ is low.

We can determine the signal pattern by evaluating these inputs across each time interval. From 0 to 1 seconds, $A = 1$ and $B = 0$, so $\overline{B} = 1$, which results in $Y = 1 \cdot 1 = 1$. During the 1 to 2 seconds interval, $A = 1$ and $B = 1$, so $\overline{B} = 0$, yielding $Y = 1 \cdot 0 = 0$. In the 2 to 3 seconds window, $A = 0$ and $B = 0$, leading to $Y = 0 \cdot 1 = 0$. When we reach the 3 to 4 seconds interval, $A = 1$ and $B = 1$, which results in $Y = 1 \cdot 0 = 0$. Finally, from 4 to 5 seconds, $A = 1$ and $B = 0$, meaning $\overline{B} = 1$ and $Y = 1 \cdot 1 = 1$.

A Zener diode is designed to maintain a specific voltage across its terminals, but it can only handle a finite amount of power. Since power is the product of voltage and current, the maximum current the Zener diode can safely sustain is found using $I = P/V$, resulting in $I = 0.5/8 = 1/16$ A. The circuit utilizes a source of 20 V, where the voltage drop across the protective resistance must be the difference between the source voltage and the Zener voltage, which equals $20 - 8 = 12$ V. Applying Ohm's law, $R_p = V_R / I$, the protective resistance is calculated as $R_p = 12 / (1/16) = 192$ $\Omega$.