Semiconductor Electronics – JEE Main PhysicsPractice Questions & PYQs

The fundamental principle of Zener diode regulation is that the component maintains a constant voltage across its terminals while operating in the reverse breakdown region. Excess voltage from the input source is dropped across the series resistance to maintain the regulated output. For a 12 V input and an 8 V regulated output, the potential difference across the series resistor is 4 V.

In this circuit, the current flowing through the 400 Ω resistor is determined by the equation $I = \frac{4}{400} = 0.01$ A. Since the two identical Zener diodes are in series and share the total regulated voltage, each diode maintains a voltage drop of 4 V. The power dissipated in each diode is found using the equation $P = 4 \times 0.01 = 0.04$ W, which is equal to 40 mW.

A multimeter measures resistance by passing a small current through the component, and a capacitor naturally draws this current as it charges, causing the meter to register a deflection. Because this charging effect occurs simply by virtue of connecting the component to the multimeter, it is inaccurate to claim that no deflection will be observed. Thus, the statement asserting that a capacitor yields no deflection in either probe orientation is incorrect, as the multimeter will inherently display a response due to the current flow required to charge the device.

Boolean algebra operations allow for the simplification of complex logic expressions by utilizing fundamental laws such as De Morgan's theorem to reduce gate configurations. Applying this principle to the expression $Y = \overline{\overline{AB} \cdot A}$ transforms the operation into the sum of its complemented components, expressed as $\overline{\overline{AB}} + \overline{A}$. By recognizing that the double complement of a term reverts to its original form, the expression simplifies to $AB + \overline{A}$, which, based on the provided derivation, evaluates to $0$.

When a diode is forward biased, it operates at a fixed voltage drop, effectively acting as a consistent load in the circuit. In this series configuration, the total battery EMF must be accounted for by the voltage drops across both the diode and the protective resistor. According to Kirchhoff's voltage law, the sum of these drops must equal the source voltage provided by the battery.

With a battery EMF of 1.5 V and a diode voltage drop of 0.5 V, the potential difference that must exist across the resistor is 1.0 V. To keep the current from exceeding the safe limit of 10 mA, or 0.01 A, we use Ohm's law, expressed as $R = \frac{V}{I}$. Substituting the voltage of 1.0 V and the current of 0.01 A into this equation yields $R = \frac{1.0\,\text{V}}{0.01\,\text{A}}$, which results in a minimum resistance of $R = 100\,\Omega$.

The analysis of a logic gate circuit relies on tracing the inputs $A$ and $B$ through each successive gate by applying the Boolean operation specific to each gate to generate intermediate signals. By substituting these intermediate variables into the final output function and simplifying the expression using fundamental Boolean algebraic identities or De Morgan's laws ($\overline{A \cdot B} = \overline{A} + \overline{B}$), the expression reduces to the standard logic function represented by the signal in the correct option.

The behavior of this network is governed by Ohm's Law, which defines current as the ratio of voltage to total resistance. Since all components are linked in a single series path, the total resistance is the sum of the individual resistors, calculated as $5 + 10 + 5 + 10 = 30$ ohms. When the voltage of 9 V is applied across this equivalent resistance, the current $i$ is determined by the expression $i = \frac{9}{30} = 0.3$. This calculation yields a final current of 0.3 A.

The potential at point A is determined by identifying which diode is forward biased and conducting current, effectively establishing the voltage at the node. A diode in a forward-biased state acts as a conducting element with a characteristic voltage drop equal to its built-in potential, while a reverse-biased diode acts as an open switch. Within this circuit, diode D1 is forward biased, whereas diode D2 is reverse biased. Since the forward-biased D1 has a built-in potential of 0.7 V and is supplied by an input of 12.7 V, the voltage at node A is simply the input potential less this voltage drop. Therefore, calculating $V_A = 12.7 - 0.7 = 12.00$ determines the potential at point A in Volts.

This set of input-output combinations describes a NOR gate, where the output $Y$ is $1$ exclusively when both inputs $A$ and $B$ are $0$. By applying De Morgan's Law to the base logic expression $\overline{A+B}$, we arrive at the equivalent form $\overline{A} \cdot \overline{B}$, which confirms that the result aligns with the behavior of an AND operation performed on two inverted input signals.

Zener diodes placed in an anti-series configuration operate as a voltage clipper to limit the signal magnitude. When the input voltage satisfies the condition $|V_{\in}| < 6 \text{ V}$, the diodes remain off, and the output signal directly tracks the input. Once the input magnitude exceeds the breakdown threshold, represented by $|V_{\in}| \geq 6 \text{ V}$, the reverse-biased diode enters the breakdown region to clamp the signal, resulting in a constant output of 6 V or -6 V. This creates a waveform that mimics the input profile during low-amplitude segments but holds flat at the breakdown levels whenever the input exceeds the specified 6 V limit.

Applying Kirchhoff's Current Law at node A involves setting the sum of currents at the junction to zero, represented by the equation $\frac{30-V_A}{10}+\frac{0-V_A}{10}+\frac{0-V_A}{10}=0$. The node connects to a 30 V source through a 10 ohm resistor and to a 0 V ground terminal through two additional 10 ohm resistors. Solving this algebraic expression results in $30 = 3V_A$, meaning the potential at point A is 10 V. Because point B is held at a ground potential of 0 V, the potential difference between points A and B is calculated as 10 V.

An AND gate operates on the principle of logical multiplication, where the output signal only remains high if every input signal is also high. Looking at the truth table, the output is $0$ for all combinations except when both inputs $A$ and $B$ are $1$, resulting in an output of $1$. Since this behavior follows the boolean relationship $C = A \cdot B$, it identifies the circuit as an AND operation.

The small-signal current gain $\beta_{ac}$ of a transistor reflects how sensitively the collector current responds to changes in base current while the collector-emitter voltage remains constant. To determine this value at the operating point defined by $V_{CE} = 10\,\mathrm{V}$ and $I_C = 4.0\,\mathrm{mA}$, we identify the two output characteristic curves that bracket this point, corresponding to base currents of $20\,\mu\mathrm{A}$ and $30\,\mu\mathrm{A}$. At the constant collector-emitter voltage of $10\,\mathrm{V}$, these base currents result in collector currents of $3.0\,\mathrm{mA}$ and $4.5\,\mathrm{mA}$, respectively. The change in collector current is calculated as $\Delta I_{C} = 4.5\,\mathrm{mA} - 3.0\,\mathrm{mA} = 1.5\,\mathrm{mA}$, while the change in base current is $\Delta I_{B} = 30\,\mu\mathrm{A} - 20\,\mu\mathrm{A} = 10\,\mu\mathrm{A}$. Applying the definition of current gain, we find $\beta_{ac} = \frac{\Delta I_{C}}{\Delta I_{B}} = \frac{1.5\,\mathrm{mA}}{10\,\mu\mathrm{A}} = 150$.

When a photodiode operates under reverse bias, the magnitude of the photocurrent $I_{ph}$ depends on the efficiency of charge carrier collection within the depletion region. As the biasing voltage is raised, the electric field across the junction strengthens, which more effectively sweeps out the electron-hole pairs created by incident light, causing the current to rise. Once the voltage is sufficient to collect all available photogenerated carriers before they can recombine, the photocurrent reaches a stable limit, where further increases in voltage result in a saturation effect.

Since neither Zener diode reaches its breakdown point when the input voltage is below 4 V, the output voltage simply tracks the input linearly, following the relation $V_0 = V_i$. Upon reaching 4 V, the first Zener diode enters the breakdown region to clamp the voltage, but because real voltage sources contain internal resistance, the sharp rise in circuit current causes an additional voltage drop across that internal resistance. Consequently, the output voltage deviates from the ideal linear path and levels off as the current increases abruptly; as the input continues to rise toward 6 V, this clamping behavior maintains the output near the threshold of the 4 V Zener.

This circuit functions as an AND gate, which produces a high output of 1 only when both inputs A and B are high simultaneously. For any other combination, the output Z defaults to 0. When inputs A and B are 1 and 0, the AND operation $Z = A \cdot B$ yields 0, and similarly, when inputs are 0 and 0, the output remains 0. When both inputs are 1 and 1, the output rises to 1, while the final set of inputs 0 and 1 results in an output of 0. These individual computations generate the sequence 0, 0, 1, 0.

A logical operation is classified as reversible if the input states can be uniquely recovered from the output states, meaning the transformation preserves all information without overlap. In the configuration for option B, the mapping between the input variables and the output is strictly one-to-one, ensuring that every specific output value corresponds to a unique and identifiable input state. By contrast, the scenarios represented by options A, C, and D fail this requirement because they produce ambiguous mappings where multiple input combinations yield the same output, making it impossible to reconstruct the original state from the result. This strictly unique mapping confirms the second operation as the only structure capable of maintaining full reversibility.

When a charged capacitor is left in isolation with no connected discharge path, the charge on its plates cannot redistribute or escape, meaning the initial stored charge remains constant indefinitely. Because no current can flow in the first circuit arrangement, the charge on capacitor A stays at its initial value of $Q_A = CV$.

In contrast, the second circuit forms an RC discharging loop where the charge decays over time based on the relation $q(t) = CV e^{-\frac{t}{RC}}$. By substituting the specific time interval $t = CR$ into this decay formula, the exponent becomes $-1$, resulting in a final charge of $Q_B = CV e^{-1}$, which is mathematically equivalent to $\frac{VC}{e}$.

In a Zener diode regulator circuit, the Zener diode clamps the load voltage to its rated value, resulting in a constant load current of $I_L = \frac{6\,\mathrm{V}}{4\,\mathrm{k}\Omega} = 1.5\,\mathrm{mA}$ regardless of input fluctuations. When the unregulated input voltage rises to its peak of $16\,\mathrm{V}$, the total current supplied through the $2\,\mathrm{k}\Omega$ series resistor reaches its maximum value of $I_{\text{total}} = \frac{16\,\mathrm{V} - 6\,\mathrm{V}}{2\,\mathrm{k}\Omega} = 5\,\mathrm{mA}$. Since the Zener diode must conduct the remaining excess current that is not consumed by the load, the maximum Zener current is calculated as $I_Z = 5\,\mathrm{mA} - 1.5\,\mathrm{mA} = 3.5\,\mathrm{mA}$.

De Morgan's theorem establishes that an OR operation is logically equivalent to a NAND operation performed on inverted inputs. By passing the signals $A$ and $B$ through NOT gates to obtain $\overline{A}$ and $\overline{B}$, and then processing these through a NAND function, the resulting expression is $\overline{\overline{A} \cdot \overline{B}}$. Simplifying this expression mathematically yields $A + B$, which confirms that the entire configuration functions as an OR gate.

Transistor amplification relies on the current gain, $\beta$, which is the ratio of the output collector current to the input base current, effectively modulated by the relationship between the output and input resistances. First, determine the current gain using the formula $\beta = \frac{I_C}{I_B}$, yielding 50. Next, establish the resistance ratio, which is the quotient of the output resistance of 100 kΩ and the input resistance of 100 Ω, resulting in a factor of 1000.

The voltage gain is determined by the formula $V_g = \beta \times \frac{R_o}{R_i}$. Applying the calculated values gives $50 \times 1000 = 5 \times 10^4$. Because power gain is the product of the square of the current gain and the resistance ratio, represented by $P_g = \beta^2 \times \frac{R_o}{R_i}$, the computation follows as $50^2 \times 1000 = 2.5 \times 10^6$.