Semiconductor Electronics – JEE Main PhysicsPractice Questions & PYQs

A single $p$-$n$ junction diode connected to an AC source $V$ functions as a half-wave rectifier because it acts as a low-resistance path when forward-biased and an open circuit when reverse-biased. During one half of the input cycle, the diode allows the flow of current $I$ through the resistor $R$, whereas it effectively blocks any current flow during the remaining half-cycle. This cyclic operation produces a pulsating unidirectional current output across $R$, which corresponds to the wave pattern shown in option B.

An OR gate operates based on the Boolean expression $C = A + B$, which dictates that the output remains high whenever at least one of the inputs is set to 1. The output $C$ in the provided truth table results in 1 for the input pairs $(1, 1)$, $(1, 0)$, and $(0, 1)$, dropping to 0 only when both $A$ and $B$ are 0. This specific pattern of inputs and outputs is the defining characteristic of an OR gate.

Testing a transistor relies on evaluating the conductivity of its internal p-n junctions by measuring the resistance across its terminals. Since current flows only when the common terminal is connected to R, it indicates that R serves as the central base, which functions as the P-type layer in this configuration. This specific junction arrangement confirms that the component is an $npn$ transistor, identifying R as the base while P and Q serve as the collector and emitter.

A p-n junction diode functions as a unidirectional valve that permits current to pass through the load resistor $R_L$ only when the component is forward biased. As the 10 V square input signal alternates, the diode conducts during the positive half-cycle, ensuring that the signal appears at the output. During the negative phase, the diode blocks the current, which causes the output across $R_L$ to drop to zero. This repetitive switching behavior creates the specific rectified output waveform shown in the diagram.

Although carbon, silicon, and germanium all possess four valence electrons, their ability to conduct electricity depends on how tightly those electrons are bound within the atomic structure. Carbon confines its valence electrons to the second orbit, leading to strong atomic binding that renders it an insulator with negligible free electrons at room temperature. In contrast, silicon and germanium function as semiconductors because their valence electrons occupy the third and fourth orbits respectively, which results in weaker binding that allows a significant number of electrons to participate in conduction at room temperature.

A diode enters a reverse-biased state when the electrical potential applied to its p-side is lower than that of its n-side, satisfying the condition $V_p < V_n$. This configuration forces the depletion region to widen by pulling charge carriers away from the junction, effectively preventing current flow. Option D depicts this precise arrangement where the p-side is maintained at a lower potential than the n-side.

The base current amplification factor, denoted as $\beta$, represents the ratio of the collector current to the base current, a relationship derived from the fundamental principle that the total emitter current equals the sum of the collector and base currents. To determine this value, calculate the base current by applying the formula $I_B = I_E - I_C$. Subtracting the collector current of 5.488 mA from the emitter current of 5.60 mA results in a base current of 0.112 mA. The amplification factor is then calculated using the expression $\beta = \frac{I_C}{I_B}$. Substituting the given values into this equation yields $\beta = \frac{5.488}{0.112}$, which simplifies to 49.

Semiconductors, which rely on covalent bonding, exhibit an increase in electrical conductivity as temperature rises because thermal energy liberates additional charge carriers to support current flow. Unlike many ionic crystals that are typically transparent, solids with covalent bonding are often opaque to visible light, making them the distinct choice here. Conversely, metallic bonding results in decreased conductivity as the temperature increases, meaning this mechanism does not align with the behavior observed in the described material.

The atomic spacing within a semiconductor crystal is determined by its lattice constant, which profoundly influences the overlap of atomic orbitals and the resulting energy band structure. When the lattice constant is decreased, the modification in the interatomic potential leads to a shift where both the conduction band edge $E_c$ and the valence band edge $E_v$ move to lower energy levels. This downward displacement of the band edges fundamentally alters the energy separation between them, causing the overall magnitude of the band-gap $E_g$ to increase.

Ideal diodes function as closed switches when forward-biased and open circuits when reverse-biased, meaning the current path is dictated by which diode is conducting. This configuration simplifies the circuit into a single series loop because the reverse-biased diode blocks its respective branch, leaving a total resistance of $R = 4 + 2 = 6$ ohms. Applying Ohm's law to the circuit results in a current of $i = \frac{12}{6} = 2$ amperes.

The fundamental expression for electric current in a semiconductor is $I = n e A v$, which relates the current $I$ to the carrier concentration $n$, electronic charge $e$, cross-sectional area $A$, and drift velocity $v$. Because charge and area are common to both electrons and holes, the ratio of the currents is simply the product of the concentration ratio and the drift velocity ratio, expressed as:

$\frac{I_e}{I_h} = \left( \frac{n_e}{n_h} \right) \left( \frac{v_e}{v_h} \right)$

Substituting the provided current ratio of $\frac{7}{4}$ and concentration ratio of $\frac{7}{5}$ into this relationship yields the following equation:

$\frac{7}{4} = \frac{7}{5} \times \frac{v_e}{v_h}$

Solving for the drift velocity ratio $\frac{v_e}{v_h}$ by multiplying both sides by $\frac{5}{7}$ results in a final value of $\frac{5}{4}$.

In a common base amplifier configuration, the input signal is applied to the emitter terminal while the output is taken from the collector terminal, with the base acting as the common junction. Because the transistor action causes the collector current to follow the emitter current without any inherent inversion, the output voltage rises and falls in direct synchronization with the input signal. Consequently, the signals are perfectly in phase, which implies that the phase difference is $\phi = 0$.

The minimum energy required for an electron to transition from the valence band to the conduction band is the band gap energy $E_g$, which relates to the threshold wavelength $\lambda$ of incident radiation by the equation $E_g = \frac{hc}{\lambda}$. Using the constants $h = 6.63 \times 10^{-34} \text{ Js}$, $c = 3 \times 10^8 \text{ m/s}$, and the conversion $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$, the energy in electron volts simplifies to $E_g (\text{eV}) \approx \frac{1240}{\lambda (\text{nm})}$. Substituting the given threshold wavelength of $2480 \text{ nm}$ into the equation yields the band gap value:

$E_g = \frac{1240}{2480} = 0.5 \text{ eV}$

A full-wave rectifier works by converting both the positive and negative half-cycles of the AC input into pulses of a single polarity, which generates two output pulses for every one cycle of the input. Because the pulse count per unit of time is doubled, the fundamental ripple frequency is twice the input frequency, which can be expressed as:
$f_{ripple} = 2f$
Substituting the provided input frequency of $50\ \text{Hz}$ into this formula yields $2 \times 50\ \text{Hz}$, resulting in a final ripple frequency of $100\ \text{Hz}$.

Band structure in solids is a direct consequence of the Pauli exclusion principle, which prohibits any two electrons from occupying the exact same quantum state simultaneously. As numerous atoms interact to form a crystal lattice, their discrete atomic energy levels must split and broaden into dense bands to accommodate the multitude of electrons according to this restriction. This arrangement ensures that electrons distribute themselves across a range of energy states rather than collapsing into a single ground level, ultimately determining the electrical properties of the material.

Copper is a metallic conductor, so cooling it to 77 K diminishes the thermal vibrations of the lattice atoms, which reduces electron collisions and causes the electrical resistance $R$ to drop. Conversely, germanium functions as a semiconductor, where lowering the temperature restricts the thermal energy available to excite charge carriers, significantly reducing their concentration and causing $R$ to rise.

Forward biasing a $p-n$ junction involves connecting the $p$-type material to the positive terminal and the $n$-type material to the negative terminal, which drives the majority charge carriers toward the junction interface. This action pushes the carriers into the depletion zone, causing it to shrink while simultaneously opposing the built-in electric field, which results in a reduction of the barrier height.

In a common emitter amplifier, the current gain is determined by the ratio of the forward current gain to the sum of unity and the product of the output conductance and the load resistance. The relationship is defined by the following expression:
$A_i = \frac{-h_{fe}}{1 + h_{oe} R_L}$
Given that $h_{fe} = 50$, $h_{oe} = 25 \times 10^{-6}$ Siemens, and $R_L = 1000$ ohms, these values are substituted into the equation. The product of the output conductance and the load resistance becomes 0.025, which, when added to 1, creates a denominator of 1.025. Dividing -50 by this value yields a final current gain of -48.78.

In an NPN transistor operating as an amplifier, the forward-biased emitter-base junction acts as a source that injects a large concentration of electrons into the base. Because the base region is designed to be extremely thin and lightly doped, the injected electrons quickly diffuse across it toward the reverse-biased collector junction. Once they reach this region, the strong positive electric field of the collector pulls the electrons from the base into the collector, facilitating the amplification of current through the device.

In a reverse-biased p-n junction, the depletion layer acts as a widened barrier that effectively stops the flow of majority charge carriers across the junction. Because this configuration prevents current from moving through the semiconductor, the electrical state within the junction reaches a condition where no charge transport is occurring. Under these specific constraints, the lack of current flow implies that the electric field is zero at the middle of the depletion layer.