Semiconductor Electronics – JEE Main PhysicsPractice Questions & PYQs

The circuit arrangement shown represents an exclusive-OR (XOR) logic gate, which operates on the principle that the output is high only when the input values are different. By applying this logic to all possible combinations, the output $Y$ yields $0$ for matching inputs of $(0, 0)$ or $(1, 1)$, and $1$ for mismatched inputs of $(0, 1)$ or $(1, 0)$, which corresponds directly to the truth table provided in the final choice.

A Zener diode acts as a voltage regulator by maintaining a constant output voltage equivalent to its breakdown rating, but it requires a series resistor to limit the current and keep power dissipation within safe operating limits. To determine this resistance, first calculate the maximum allowable current the diode can handle by dividing its power rating by the breakdown voltage, resulting in $I = \frac{1.6}{8} = 0.2\,\mathrm{A}$. To ensure safety during the highest voltage fluctuation of $10\,\mathrm{V}$, the series resistor must drop the excess voltage while limiting the current to this maximum value, which leads to the calculation:
$R = \frac{V_{max} - V_z}{I} = \frac{10 - 8}{0.2} = 10\,\mathrm{\Omega}$

The current gain of an n-p-n transistor in a common emitter configuration is defined as the ratio of the change in collector current to the change in base current, expressed as $\beta = \Delta I_C / \Delta I_B$. Evaluating the changes in these parameters reveals that the collector current increases from 5 mA to 16 mA, resulting in $\Delta I_C = 11 mA$, while the base current increases from 100 µA to 200 µA, resulting in $\Delta I_B = 100 \mu A$.

Calculating the gain involves substituting these changes into the formula using consistent units of Amperes:

$\beta = \frac{11 \times 10^{-3}}{100 \times 10^{-6}} = \frac{11000}{100} = 110$

This quotient represents the dimensionless current gain of the transistor.

The power gain of a transistor in a Common Emitter configuration is determined by multiplying the square of the current gain by the ratio of the output resistance to the input resistance. We first calculate the current gain $\beta$ by taking the ratio of the change in collector current of 10 mA to the change in base current of 100 $\mu$A, resulting in $\beta = 100$. Using the formula $\text{Power Gain} = \beta^2 \times \frac{R_C}{R_B}$, we substitute the resistance values of 1 k$\Omega$ and 10 k$\Omega$ to solve for the gain as $100^2 \times \frac{1}{10} = 1000$. Given that the power gain is represented as $10^x$, we set the equation $10^x = 1000$, confirming that $x = 3$.

Logic gate circuits operate by processing input signals $A$ and $B$ through sequential gates to produce an output $X$, where the final state is determined by Boolean algebraic reduction or step-by-step signal propagation. By examining the truth table in option B, we observe that the output $X$ is $1$ when $B=0$ (regardless of $A$) and $X$ is $0$ when $B=1$ (regardless of $A$), which mathematically corresponds to the Boolean function $X = \bar{B}$. This specific mapping occurs because the circuit configuration effectively ignores the signal from input $A$ and performs an inversion operation solely on the signal from input $B$, validating that option B correctly represents the output logic for all possible input combinations.

Photodiodes are designed to operate specifically in a reverse-bias configuration to maximize sensitivity to incident light, making the claim that they function in forward bias factually incorrect. In contrast, a p-n junction diode consistently demonstrates significantly higher current flow in forward bias once the potential difference exceeds the threshold voltage $V_0$, relative to the minimal leakage current observed in reverse bias before reaching the breakdown voltage $V_z$. Therefore, the premise regarding the device's operational mode is false, while the statement detailing the fundamental rectification characteristics of the diode remains physically accurate.

A NAND gate operates by performing the Boolean AND operation followed by a logical inversion, meaning the output is low only when both inputs are at a high state. The relationship is defined by the Boolean expression $Y = \overline{A \cdot B}$, which dictates that the result remains high for all input combinations except when both A and B are simultaneously at logic level 1. By applying this rule to the provided input table, you can identify the correct output for each row by confirming whether both inputs are high, which produces a logic 0, while all other combinations produce a logic 1.

A transistor functions as a binary switch by transitioning between two extreme operating conditions to represent "on" and "off" states. When forced into the saturation region, the device conducts current fully like a closed switch, whereas it acts as an open circuit blocking all current when placed in the cut-off region. Consequently, the combination of these two specific states allows the transistor to serve its purpose as an electronic switch.

In a common emitter NPN transistor amplifier, the output voltage is determined by the circuit's voltage gain, which is defined as the product of the current gain $\beta$ and the ratio of the output resistance to the input resistance. With a given current gain $\beta = 100$, an output resistance of 10 kΩ, and an input resistance of 1 kΩ, the voltage gain is calculated by multiplying these components. Given an input voltage of 1 mV, which is equivalent to 0.001 V, the output voltage can be determined using the following relationship:

$V_{\text{out}} = \beta \times \frac{R_{\text{out}}}{R_{\text{\in}}} \times V_{\text{\in}} = 100 \times \frac{10 \times 10^3}{10^3} \times 10^{-3}$

This expression simplifies to 100 multiplied by 10 multiplied by 0.001, which results in a final output voltage of 1.0 V.

This circuit functions as an AND gate, where a high output of 1 is produced only when both inputs are supplied with a potential of 5 V, representing the state where $A=1$ and $B=1$. In any other scenario where either or both inputs are at 0 V, the output remains 0. This behavior corresponds to the truth table listed in the first option, which correctly represents the boolean operation $Y = A \cdot B$.

The current-carrying capacity of a transistor is fundamentally determined by the mobility of its majority charge carriers, as mobility directly dictates how efficiently these carriers can drift through the semiconductor lattice. In an n-p-n transistor, the majority carriers are electrons, while in a p-n-p transistor, the majority carriers are holes. Because electrons exhibit significantly higher mobility than holes, they can traverse the device more rapidly under the influence of an electric field, leading to a higher overall current flow in n-p-n devices. This inherent difference in carrier mobility is the direct physical cause that explains why an n-p-n transistor is capable of permitting more current than a p-n-p transistor.

A solar cell acts as an active power source, which requires its current-voltage ($I-V$) characteristics to be represented in the fourth quadrant of the Cartesian plane. The curve begins at the short-circuit current ($I_{SC}$) on the vertical axis, where voltage is zero, and slopes downward until it reaches the open-circuit voltage ($V_{OC}$) on the horizontal axis, where current drops to zero. This distinctive downward-sloping shape highlights the device's operational efficiency, with the maximum power point located at the "knee" of the curve where the product of voltage and current is greatest.

Diodes operate fundamentally as one-way valves for electrical current, a behavior determined by the polarity applied across their two terminals. When testing with a multimeter, the component allows current to pass through when forward-biased, but blocks the flow entirely when reverse-biased. This distinctive switching characteristic confirms that a diode is a two-terminal device that permits current flow in only one direction.

An LED functions by emitting light when electrons transition across the forbidden energy gap of a semiconductor, releasing photons with energy directly equal to that band gap. The relationship between the energy of a photon and its wavelength is $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.

To calculate the energy for a wavelength of 4000 Å, substitute the constant values and wavelength into the expression $E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{4000 \times 10^{-10} \times 1.6 \times 10^{-19}}$. Performing this calculation yields $E = 3.1$ eV. Rounding this value to the nearest integer provides an energy band gap of 3 eV.

The electric field intensity within the depletion region of a p-n junction is defined by the ratio of the potential barrier to the width of the depletion layer, which is represented by the formula $E = \frac{V}{d}$. Substituting the potential barrier of 0.6 V and the depletion width of 6 x 10^-6 m into this expression results in the following calculation:

$E = \frac{0.6}{6 \times 10^{-6}}$

Solving this division yields an electric field intensity of 1 x 10^5 N/C.

Ideal diodes act as perfect conductors when forward-biased, essentially behaving like short circuits that permit current to flow without offering any resistance. In this circuit, current passes through two 20 $\Omega$ resistors positioned in parallel, which yields an effective resistance of $\frac{20 \times 20}{40} = 10\ \Omega$. Since this parallel section is connected in series with a 15 $\Omega$ resistor, determining the total equivalent resistance simply requires adding these components to obtain 25 $\Omega$.

The total current drawn from the battery is governed by Ohm's Law, which states that $I = \frac{V}{R_{eq}}$. First, calculate the equivalent resistance of the two 16 ohm resistors connected in parallel, which results in 8 ohms. By adding this equivalent resistance in series with the 2 ohm resistor, the total circuit resistance becomes $R_{eq} = 8 + 2 = 10$. With the 10 V source applied across this total resistance, the battery current is $I = \frac{10}{10} = 1$ ampere.

The function of a Zener diode voltage regulator is to maintain a fixed breakdown voltage across the load, with the series resistor serving to limit the total circuit current. To ensure the diode operates safely under all conditions, the series resistance must be set at its minimum value such that the current does not exceed the diode's maximum capacity of $25 \text{ mA}$ when the load is disconnected. The voltage drop across the series resistor is determined by the difference between the supply voltage and the Zener breakdown voltage, which calculates to $20 \text{ V} - 8 \text{ V} = 12 \text{ V}$. Applying Ohm's law, the minimum resistance required is:

$R = \frac{12 \text{ V}}{25 \times 10^{-3} \text{ A}} = 480 \Omega$

The Zener diode current is governed by Kirchhoff's Current Law, which states that the source current entering the circuit splits into the current flowing through the load and the current passing through the Zener diode. Begin by calculating the total source current by dividing the source voltage by the series resistance, expressed as $i_s = \frac{60}{4 \times 10^3} = 15 \mathrm{mA}$. Next, find the current flowing through the load resistor using the relationship $i_L = \frac{60}{10 \times 10^3} = 6 \mathrm{mA}$. Since the source current is the sum of the load current and the Zener current, subtracting the load current from the total source current reveals that the Zener diode current is 9 mA.

The voltage gain of a transistor in common-emitter mode is defined by the formula $A_v = \beta \times \frac{R_L}{r_i}$, where $\beta$ represents the current gain, $R_L$ is the load resistance, and $r_i$ is the input resistance. The input resistance $r_i$ is calculated by dividing the base-emitter voltage change by the base current change, which is 10 mV / 10 μA, resulting in 1000 Ω. The current gain $\beta$ is determined by the ratio of the collector current change to the base current change, which is 1.5 mA / 10 μA, giving a value of 150. By substituting the current gain and the ratio of the load resistance 5000 Ω to the input resistance into the gain equation $A_v = 150 \times \frac{5000}{1000}$, the final voltage gain is 750.