Semiconductor Electronics – JEE Main PhysicsPractice Questions & PYQs

When a Zener diode operates in its breakdown region, it maintains a constant 3.0 V across itself; because it is connected in parallel with the 2 k$\Omega$ load resistor, this same voltage appears across the load. The total current supplied by the 10 V source flows through the series 1 k$\Omega$ resistor, yielding a total current of $I = \frac{10 - 3}{1000} = 7\,\mathrm{mA}$. Since this current splits into the branch containing the load resistor and the branch containing the Zener diode, the load current is $I_L = \frac{3}{2000} = 1.5\,\mathrm{mA}$. Finally, applying Kirchhoff's current law, the Zener current is the difference between the total current and the load current, resulting in $I_z = 7\,\mathrm{mA} - 1.5\,\mathrm{mA} = 5.5\,\mathrm{mA}$.

Simplification of the expression $Y=\overline{\overline{A} \cdot \overline{B}}$ proceeds by applying De Morgan's theorem, which states that the complement of a product is equal to the sum of the individual complements. This calculation reduces the boolean logic to $Y=A+B$ after canceling out the double negations applied to the variables. Consequently, this relationship confirms that the circuit functions as an OR gate.

For a bulb to illuminate within a logic gate circuit, it must experience a potential difference across its terminals, which requires one end to be at a high logic level of $1$ and the other to be at a low logic level of $0$. By setting $A=1, B=0, C=0, D=0$, the circuit establishes this necessary potential gradient, allowing current to pass through the bulb and causing it to glow.

The logic circuit's truth table represents the output states corresponding to every possible binary input combination by evaluating the Boolean expressions for each gate in the architecture. By applying the specific logic gate properties to inputs $A$ and $B$, one computes the output $Y$ for all four input scenarios to determine the final truth values. This systematic analysis confirms that the output values listed in the table correspond exactly to the circuit's derived states under all operational conditions.

To create a hole in a p-type semiconductor, an incoming photon must possess sufficient energy to promote an electron into the acceptor level. This minimum energy requirement corresponds to the value $E = 6\,\text{eV}$. The relationship between the photon energy and its wavelength is defined by the following expression:

$E = \frac{hc}{\lambda}$

Using the given constant $hc = 1242\,\text{eV}\,\text{nm}$ and the energy gap, the maximum wavelength is determined by rearranging the equation to $\lambda = \frac{hc}{E}$. Substituting the known values into this relationship, $\lambda = \frac{1242}{6}$ results in a maximum wavelength of $207\,\text{nm}$.

The circuit operates as an exclusive-OR, or XOR, gate, which produces a high signal only when the two input variables have different values. By following the boolean expression $Y = A \cdot \overline{B} + \overline{A} \cdot B$, we see that when inputs $A$ and $B$ are both 0 or both 1, the circuit yields 0, whereas pairs of 0 and 1 or 1 and 0 result in an output of 1. This behavior establishes a pattern where identical inputs result in a low signal, while differing inputs result in a high signal.

The operation of this circuit depends entirely on the biasing state of the two diodes. Since the second diode is reverse biased, it acts as an open circuit that effectively disconnects its branch from the network, while the first diode is forward biased and functions as a simple conducting wire. By removing the open branch and replacing the forward-biased diode with a short circuit, the network simplifies into a straightforward arrangement of resistors. Calculating the equivalent resistance of this reduced configuration yields a final value of $6 \Omega$.

Boolean algebra provides the key to simplifying circuit expressions, allowing us to reduce gate logic into a basic identity. The logic circuit generates an output $Y = A \cdot B + \overline{A} \cdot B$, which can be simplified by factoring out $B$ to obtain $Y = (A + \overline{A}) \cdot B$. Applying the rule $A + \overline{A} = 1$ leads directly to the result $Y = 1 \cdot B$, confirming that the output is logically equivalent to $B$ alone. Consequently, the valid truth table must show the output $Y$ mimicking the input $B$ across all input combinations.

When a Zener diode operates in its reverse bias region, it reaches a specific breakdown point where the current $I$ increases rapidly while the voltage $V$ across the device remains almost constant. Because this behavior allows the component to maintain a stable potential difference despite fluctuations in the circuit's current, it acts as an effective voltage regulator.

A potential divider circuit operates on the principle that the potential difference across any segment of a resistive wire is directly proportional to its length. To make the light emitting diode glow, the voltage across the segment $PQ$ must be sufficient to overcome both the forward bias of the LED and the breakdown voltage of the Zener diode connected in series. Summing the individual requirements of the two components gives the necessary potential difference:
$V_{PQ} = 1.8 + 3.2 = 5\ \text{V}$

Since the source voltage of 20 V is applied across the total resistive wire length of 20 cm, the potential gradient along the wire is 1 V per cm. Dividing the required potential difference by this gradient yields the minimum length for segment $PQ$:
$\ell = \frac{5\ \text{V}}{1\ \text{V/cm}} = 5\ \text{cm}$

The biasing state of a p-n junction diode is determined by the relative potential difference across its terminals, where a reverse-biased condition occurs specifically when the p-side is held at a lower potential than the n-side ($V_p < V_n$). While connecting the p-side to a higher potential promotes forward bias, maintaining a lower potential at the p-terminal creates a depletion barrier that inhibits the flow of current. Within the provided options, circuit D is the only configuration that subjects the p-region to a potential lower than that of the n-region, thereby establishing the required reverse-biased state.

The logic circuit operates as an OR gate, defined by the boolean expression $Y = A + B$, which produces a low output (0) if and only if all inputs are at logic low. Because any high signal (1) on either input drives the output to 1, the condition $A=0$ and $B=0$ is the only instance where the output remains at 0.

Digital logic circuits function by processing inputs through Boolean operations, with the resulting state determined by the specific gate configuration. For this circuit, evaluating the logic across all combinations of $A$ and $B$ reveals that the output $Y$ consistently remains 0 in every instance, regardless of whether the inputs are set to 0 or 1. This behavior confirms that the circuit's truth table contains only zeros for all possible input states, which matches the sequence provided in the corresponding output.

Logic gate circuit analysis requires determining the output state at each junction by applying the boolean algebra operations corresponding to the specific gates present, such as AND, OR, or NAND. By tracing the inputs through the signal path, the operations at the final output stages for $X$ and $Y$ simplify to a high logic state, yielding 1 and 1 respectively for the given input combination.

Applying De Morgan's theorem to the expression $Y = \overline{\overline{A} \cdot \overline{B}}$ allows for the simplification of the logic by breaking the top inversion bar and replacing the AND operation with an OR operation. This transformation leads to $Y = \overline{\overline{A}} + \overline{\overline{B}}$, where the double negation of each variable effectively reverts to its original value. Because the expression simplifies to $Y = A + B$, the circuit matches the fundamental operation of an OR gate.

Dynamic resistance is defined as the ratio of a small change in voltage to the corresponding change in current across a diode, represented mathematically as $r_d = \frac{\Delta V}{\Delta I}$. Since a diode only conducts significant current when it is placed in a forward-biased configuration, this specific orientation is mandatory for measuring the variation in current relative to voltage. By connecting the p-side to the positive terminal and the n-side to the negative terminal of the power source, the device operates within its conducting region, allowing for the empirical determination of its non-linear resistive characteristics.

Since the circuit produces a logic low output for every possible combination of binary inputs, it functions effectively as a constant zero source. By observing the truth table, we see that for all input pairs $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$, the output $Y$ remains $0$ regardless of the signal states. This lack of dependency on the input variables $A$ and $B$ means the resulting boolean expression is simply $0$.

Ohm's Law provides the foundation here, establishing that the current in a circuit is determined by the total voltage divided by the total resistance, while the voltage across any specific component is the product of that component's resistance and the current flowing through it. First, calculate the circuit current by dividing the source voltage of 14 V by the total resistance of 4 k$\Omega$, resulting in $I = 14 / 4 = 3.5 \text{ mA}$. Once the current is known, find the voltage drop across the load resistance by calculating the product of the current and the load resistance value of 2.5 k$\Omega$, expressed as $V_{L} = I \times R_{L} = 3.5 \times 2.5 = 8.75 \text{ V}$.

The galvanometer shows no deflection in the Wheatstone bridge, which implies the circuit is balanced. By applying the balance condition $\frac{0.8}{1} = \frac{R}{3}$, the resistance of the semiconductor arm after cooling is determined to be $R = 2.4\,\mathrm{m}\,\Omega$. Given that the temperature drops at a rate of $2^{\circ}\,\mathrm{C}/\mathrm{s}$ over $10\,\mathrm{s}$, the total change in temperature is $20^{\circ}\,\mathrm{C}$. Using the relationship $\Delta R = R \alpha \Delta T$ with $\Delta R = 2.4\,\mathrm{m}\,\Omega - 3\,\mathrm{m}\,\Omega = -0.6\,\mathrm{m}\,\Omega$, we find $\alpha = \frac{-0.6\,\mathrm{m}\,\Omega}{3\,\mathrm{m}\,\Omega \times 20^{\circ}\,\mathrm{C}}$, which yields $\alpha = -1 \times 10^{-2}\,{}^{\circ}\mathrm{C}^{-1}$.

A Zener diode voltage regulator functions by maintaining a stable output voltage across a load while dissipating excess current to remain within its safe power-handling threshold. For effective regulation, the series resistor must be sized to ensure the total supply current provides the necessary load current while keeping the Zener diode current safely between its minimum and maximum limits.

The voltage drop across the series resistor is calculated as the difference between the unregulated supply and the Zener voltage, giving $V_s = 8 \text{ V} - 5 \text{ V} = 3 \text{ V}$. Since the load resistor is $1 \text{ k}\Omega$ connected to a $5 \text{ V}$ Zener, the load draws a fixed current of $I_L = \frac{5 \text{ V}}{1 \text{ k}\Omega} = 5 \text{ mA}$. The maximum current the Zener diode can safely conduct is determined by its power rating, $I_{z\text{ max}} = \frac{10 \text{ mW}}{5 \text{ V}} = 2 \text{ mA}$. When the Zener current is at its peak, the total current flowing through the series resistor is $I_{s\text{ max}} = 5 \text{ mA} + 2 \text{ mA} = 7 \text{ mA}$, which implies a minimum required resistance of $R_{s\text{ min}} = \frac{3 \text{ V}}{7 \text{ mA}} = \frac{3}{7} \text{ k}\Omega$. At the other extreme, when the Zener diode draws no current, the series resistor carries only the load current of $5 \text{ mA}$, setting the maximum resistance at $R_{s\text{ max}} = \frac{3 \text{ V}}{5 \text{ mA}} = \frac{3}{5} \text{ k}\Omega$. Because the calculated operating range for the resistor is between $\frac{3}{7} \text{ k}\Omega$ and $\frac{3}{5} \text{ k}\Omega$, none of the specific values provided in the options are correct.