Semiconductor Electronics – JEE Main PhysicsPractice Questions & PYQs

A Zener diode connected in parallel with a load acts to maintain a steady voltage across that branch, meaning the 5.6 V breakdown voltage applies directly to the 800 $\Omega$ resistor. Calculating the current through this load using Ohm's law gives $I_L = \frac{5.6}{800} = 7$ mA.

According to Kirchhoff's current law, the total current entering the node from the series resistor splits between the load and the Zener diode. With the series branch supplying 17 mA to the junction, the current directed through the Zener diode is found by subtracting the load current from the total branch current, resulting in $I_z = 17 - 7 = 10$ mA.

The diode behaves as an open circuit until the voltage across it reaches its breakdown threshold, so calculating the circuit current assuming no breakdown is the first step. The total resistance is the sum of the 5000 and 10000 ohm resistors, leading to a circuit current defined by $I = \frac{120}{15000}$, which is 8 mA. The resulting voltage drop across the 10000 ohm branch is calculated as $V = 8 \times 10000$, equaling 80 V, which exceeds the 50 V breakdown limit and confirms the Zener diode regulates the voltage.

With the Zener diode now maintaining a constant 50 V, the voltage drop across the 5000 ohm series resistor is fixed at 120 V - 50 V = 70 V. The source current flowing through this series resistor is determined by $I_{series} = \frac{70}{5000}$, which is 14 mA, while the 10000 ohm resistor branch draws a current of $I_{parallel} = \frac{50}{10000}$, equal to 5 mA. Applying Kirchhoff's current law, the current passing through the Zener diode is the difference between these two currents, yielding $14 - 5 = 9$ mA.

A Zener diode only transitions into its breakdown region to regulate voltage once the applied potential satisfies $V_{\in} \geq V_{breakdown}$. Because the voltage provided in the circuit is less than the rated 10 V breakdown threshold, the component remains in its cut-off state rather than conducting. In this non-conducting mode, the internal resistance of the Zener diode becomes effectively infinite, causing it to act as an open switch. Consequently, no current flows through the path, leading to a measured value of 0.0 mA.

A transistor enters saturation when the collector-emitter junction acts as a short circuit, causing the entire supply voltage to drop across the collector resistor. Under these conditions, the maximum collector current is defined by $I_C = \frac{V_{CC}}{R_C} = \frac{10\,\text{V}}{1\,\text{k}\Omega} = 10\,\text{mA}$. With the dc current gain fixed at 250, the minimum base current required to sustain this collector current is calculated by dividing the collector current by the gain, resulting \in $I_B = \frac{I_C}{\beta} = \frac{10\,\text{mA}}{250} = 40\,\mu\text{A}$.

Diodes function as unidirectional gates for electrical current; a diode operating in reverse bias acts as an open circuit that inhibits current flow, whereas a forward-biased diode allows current to pass while contributing its own internal resistance to the path. Because diode $D_2$ is positioned in reverse bias within this circuit, it acts as an open switch and effectively removes its branch from the circuit. This leaves current to flow exclusively through the path containing the 100 Ω resistor and the other active components, which result in a total series resistance of 300 Ω. Applying Ohm's Law with the provided battery voltage of 6 V, the current $I$ in the active loop is determined by the equation:

$I = \frac{6}{300} = 0.020 \text{ A}$

Consequently, the current flowing through the 100 Ω resistance is 0.020 A.

Resistivity represents the inherent opposition a material provides to electric current, defined as the mathematical reciprocal of electrical conductivity. In an n-type semiconductor where the electron concentration dominates and the contribution of holes is ignored, conductivity is determined by the product of electron density, the elementary charge of an electron, and electron mobility. This relationship is defined by the formula:
$\sigma = ne\mu$
Since resistivity is the inverse of conductivity, it is expressed by the equation:
$\rho = \frac{1}{ne\mu}$
Substituting the provided values, we use an electron density of 10^19 m^-3, an elementary charge of 1.6 * 10^-19 C, and an electron mobility of 1.6 m^2/(V.s). The denominator simplifies to the product of these values, effectively 1.6 * 1.6, which equals 2.56. Calculating the final value gives:
$\rho = \frac{1}{2.56} \approx 0.4$
The resistivity of the semiconductor is therefore approximately 0.4 Ω m.

Transistors enter saturation when the collector-emitter junction becomes forward-biased, causing the collector-emitter voltage to effectively reach zero and limiting the maximum current flow through the collector. In this state, the collector current is determined solely by the supply voltage and the load resistance according to the relationship $I_C = \frac{V_{CC}}{R_C}$, which equals $\frac{5\text{ V}}{1\text{ k}\Omega} = 5\text{ mA}$. The minimum base current required to force the transistor into this saturation region depends on the current gain, calculated as $I_B = \frac{I_C}{\beta_{dc}} = \frac{5\text{ mA}}{200} = 25\text{ }\mu\text{A}$. Applying Kirchhoff's voltage law to the base-emitter loop yields the equation $V_{BB} = I_B R_B + V_{BE}$, which evaluates to $V_{BB} = (25\text{ }\mu\text{A} \times 100\text{ k}\Omega) + 1.0\text{ V} = 2.5\text{ V} + 1.0\text{ V} = 3.5\text{ V}$.

A Zener diode connected in parallel with a load maintains a fixed output voltage of 6 V. Because this voltage is constant, the load current remains steady at 1.5 mA, calculated using $I_L = \frac{V_Z}{R_L}$. Any total current supplied through the series resistor that exceeds this load demand must flow through the Zener diode, meaning the Zener current is determined by $I_Z = I - I_L$.

When the source voltage is at its minimum of 8 V, the total current entering the circuit is $I = \frac{V_B - V_Z}{R_i}$, which equals 2 mA. Subtracting the load current from this result gives a minimum Zener current of 0.5 mA. When the input rises to its maximum of 16 V, the total input current becomes 10 mA. Subtracting the load current once again provides a maximum Zener current of 8.5 mA.

To achieve an output of $1$ from a NOR gate, both of its input signals, $P$ and $Q$, must be $0$. Given the logical expressions $P = \bar{x} + y$ and $Q = y + \bar{x}$, this requirement is met only when $y = 0$ and $\bar{x} = 0$, the latter of which indicates that $x = 1$. Therefore, selecting the input values $x = 1$ and $y = 0$ ensures that both intermediate signals are $0$, which forces the final circuit output to $1$.

The power gain in decibels is related to the amplification factor through the equation $60 = 10 \log (\text{Gain})$, which indicates a numerical power gain of $10^6$. Using the amplifier relation $\frac{\beta R_L}{R_{\in}} = 10^6$ and substituting the provided values of $10^4$ ohms for $R_L$ and $100$ ohms for $R_{\in}$, the expression becomes $\frac{\beta \times 10^4}{100} = 10^6$. Simplifying the resistance ratio yields $\beta \times 10^2 = 10^6$, which confirms that the common emitter current gain is $\beta = 10^4$.

The behavior of the logic circuit can be determined by observing that the output $Y$ remains high whenever input $A$ is low, and low whenever input $A$ is high, regardless of the value of $B$. Specifically, the table shows that for input combinations $(0,0)$ and $(0,1)$, the output is $1$, while for $(1,0)$ and $(1,1)$, the output is $0$. This consistent reliance on input $A$ establishes that the gate functions as an inverter, satisfying the relation $Y = \bar{A}$.

In a common emitter transistor amplifier, the input resistance relates the signal voltage applied to the base-emitter junction to the resulting change in base current, governed by the formula $R_i = \frac{\Delta V_{\in}}{\Delta I_B}$. Substituting the provided values of 10 mV and 15 μA yields an input resistance calculation of $R_i = \frac{10 \times 10^{-3}\,\mathrm{V}}{15 \times 10^{-6}\,\mathrm{A}}$, which equals 0.67 kΩ when rounded. To determine the voltage gain, one examines the ratio of the change in output voltage across the load resistance to the input voltage, expressed as $A_v = \frac{\Delta I_C \times R_L}{\Delta V_{\in}}$. Using the specified values of 3 mA for the collector current change, 1 kΩ for the load resistance, and 10 mV for the signal voltage, the gain calculation results in $A_v = \frac{3 \times 10^{-3}\,\mathrm{A} \times 1 \times 10^3\,\Omega}{10 \times 10^{-3}\,\mathrm{V}}$, which yields a voltage gain of 300.

Diodes act as switches that allow current to flow only after the applied voltage exceeds their internal barrier potential, which is 0.3 V for Germanium and 0.7 V for Silicon. In the initial setup, the circuit prioritizes the path with the lower barrier potential, meaning the output voltage is clamped at 0.3 V by the Germanium diode. When the Germanium diode is reversed, it acts as an open circuit, shifting the current path to the Silicon diode which requires 0.7 V to conduct. Consequently, the output voltage increases by an amount calculated as $0.7\text{ V} - 0.3\text{ V} = 0.4\text{ V}$.

The output of the circuit defined by $\overline{(\overline{A}+\overline{B})\overline{A}}$ is determined by applying algebraic identities such as Boolean distribution and De Morgan's laws. The expression simplifies step-by-step through the sequence $\overline{\overline{A} + \overline{A}B}$ and $\overline{A(\overline{A}B)}$, which expands to $A(A+\overline{B})$ and results in $A + \overline{AB}$. This specific progression of logic ultimately resolves to $\overline{AB}$.

The voltage gain of a common emitter amplifier is defined as the ratio of the change in output voltage to the change in input voltage, given by $A_v = \frac{\Delta I_C R_L}{\Delta I_B R_{BE}} = \frac{\beta R_L}{R_{BE}}$. The current gain, representing the ratio of change in collector current to the change in base current, is expressed as $\beta = \frac{\Delta I_C}{\Delta I_B}$. Finally, the power gain is calculated as the product of the current gain and the voltage gain, resulting in $A_p = \beta \cdot \left( \frac{\beta R_L}{R_{BE}} \right) = \frac{\beta^2 R_L}{R_{BE}}$.

A forward-biased silicon diode maintains a fixed potential drop of $0.7V$ across its terminals, effectively subtracting this amount from the total voltage supplied to the circuit. This principle allows us to determine the remaining potential available to drive current through any series-connected resistive components.

Given a supply of $3V$, subtracting the $0.7V$ barrier potential leaves an effective voltage of $2.3V$ across the $200\Omega$ resistor. Applying Ohm's law, $I = \frac{V}{R}$, the current flowing through this circuit is calculated as $I = \frac{2.3V}{200\Omega} = 0.0115 A$. Converting this value into milliamperes results in a final ammeter reading of $11.5 mA$.

A digital circuit's output is determined by the Boolean function implemented by its gate architecture, which must be evaluated for all $2^3 = 8$ possible input combinations of $x$, $y$, and $z$. Systematically calculating the output node value for each input state from $000$ to $111$ yields the unique truth table sequence that characterizes the circuit's logical behavior. This sequence aligns with the data in option A, as it accurately maps the circuit's output for every input permutation based on the logical evaluation of the internal gates.

The current through a Zener diode in a regulation circuit is determined by applying Kirchhoff's current law at the node between the series resistance and the Zener diode, where the total current from the source divides into the Zener current and the load current, expressed as $I_Z = I_S - I_L$. The current supplied through the series resistor $R_S$ is calculated using the potential drop across it, $I_S = \frac{V_S - V_Z}{R_S}$, where $V_S$ is the source voltage and $V_Z$ is the breakdown voltage of the Zener diode. The current flowing through the load resistance $R_L$ is determined by the constant Zener voltage maintained across it, $I_L = \frac{V_Z}{R_L}$. Combining these relationships, the Zener current is derived as $I_Z = \frac{V_S - V_Z}{R_S} - \frac{V_Z}{R_L}$, which yields $3.3 \text{ mA}$ for the given circuit parameters.

The relationship between emitter current $I_e$, collector current $I_c$, and current gain $\beta$ in a common emitter configuration is $I_e = I_c + I_b$, where the base current $I_b$ is defined as $\frac{I_c}{\beta}$. Substituting the expression for $I_b$ into the emitter current equation gives $I_e = I_c(1 + \frac{1}{\beta}) = I_c(\frac{\beta+1}{\beta})$, which isolates the collector current as $I_c = I_e(\frac{\beta}{\beta+1})$. Applying the given values $I_e = 7.0$ mA and $\beta = 69$ leads to $I_c = 7.0(\frac{69}{70}) = 6.9$ mA.

Resistance is defined as the ratio of voltage to current, $R = \frac{V}{I}$, which quantifies a component's opposition to electric flow in different operating modes. For the forward bias region of the diode, applying a voltage of 0.1 V with a corresponding current of $10 \times 10^{-3}$ A yields a forward resistance of $R_f = \frac{0.1}{10 \times 10^{-3}} = 10$ ohms. In the reverse bias region, the diode behaves differently, where a voltage of 10 V and a current of $10^{-6}$ A result in a reverse resistance of $R_r = \frac{10}{10^{-6}} = 10^7$ ohms. Calculating the ratio of forward bias resistance to reverse bias resistance gives $\frac{R_f}{R_r} = \frac{10}{10^7} = 10^{-6}$.