Semiconductor Electronics – JEE Main PhysicsPractice Questions & PYQs

In transistor electronics, the small signal current gain describes how effectively changes in the base current control the collector current when the collector-emitter voltage is held constant at 8 V. This amplification factor is determined by the ratio of the change in collector current to the corresponding change in base current, given by the formula $\beta = \frac{\Delta I_C}{\Delta I_B}$.

The collector current changed from 4 mA to 6 mA, resulting in a difference of 2 mA, which equals 0.002 A. Simultaneously, the base current shifted from 20 $\mu$A to 25 $\mu$A, creating a difference of 5 $\mu$A, which is 0.000005 A. Dividing these two values provides the final gain:

$\beta = \frac{0.002}{0.000005} = 400$

An oscillator functions by converting direct current into a periodic alternating signal, a process that requires positive feedback to sustain the oscillation without needing a continuous external input. This mechanism operates by taking a specific fraction of the amplifier's output voltage and routing it back to the input terminal, ensuring that the phase alignment supports the continuous regeneration of the wave. By effectively recycling this power to the input, the circuit maintains its own self-sustaining oscillations, which constitutes the essential function of the feedback loop in oscillator circuitry.

A NAND gate is characterized by an output that remains high for all input combinations except when both inputs are high. Observing the provided table, the output $Y$ is 1 for the pairs 0,0, 0,1, and 1,0, dropping to 0 only when $A=1$ and $B=1$. This inverted relationship follows the boolean expression $Y=(AB)'$, which confirms the component is a NAND gate.

De Morgan's laws allow for the simplification of complex boolean expressions by distributing the complement across individual terms and switching the operator. Applying this to the expression $Y = (A' + B')'$, we invert the literals and convert the OR operation into an AND operation, which results in $Y = (A')' \cdot (B')'$. Since a double complement returns the variable to its original state, the expression simplifies to $Y = A \cdot B$, which is the defining logic function of an AND gate.

Protecting a Zener diode requires limiting the current through it so that its power dissipation remains within the manufacturer's specified maximum. Since electrical power is the product of voltage and current, the highest safe current for the device is determined by dividing the maximum power rating by the Zener breakdown voltage.

The maximum allowable current is $I_{ZM} = \frac{0.4\,\mathrm{W}}{10\,\mathrm{V}} = 0.04\,\mathrm{A}$. To guard the component when the variable power supply is at its peak of $15\,\mathrm{V}$, the series resistor must dissipate the surplus potential, which is the difference between the source voltage and the Zener voltage. This establishes a required voltage drop of $5\,\mathrm{V}$ across the resistor. Dividing this voltage drop by the maximum current yields the necessary resistance:
$R_P = \frac{15\,\mathrm{V} - 10\,\mathrm{V}}{0.04\,\mathrm{A}} = 125\,\Omega$

Zener diodes function as voltage regulators by maintaining a stable breakdown voltage across a variable load, provided the diode operates within its designated current limits. The circuit arrangement divides the input voltage between the series resistance and the parallel Zener-load branch, ensuring the load voltage remains constant at the breakdown level regardless of the load resistance.

To determine the minimum load resistance, we consider the condition where the Zener diode does not conduct any current, as this maximizes the current flowing through the load. With an input voltage of 10 V and a Zener breakdown voltage of 8 V, the voltage drop across the 100 Ω series resistor is 2 V. The total current supplied to the circuit is $I = \frac{2\text{ V}}{100\ \Omega} = 0.02\text{ A}$. Given that all this current must flow through the load, the minimum load resistance is $R_{L,\text{min}} = \frac{8\text{ V}}{0.02\text{ A}} = 400\ \Omega$.

The maximum load resistance occurs when the Zener diode conducts its maximum allowable current of 10 mA. Because the total current supplied by the series resistor remains fixed at 20 mA, the load current becomes the difference between the total current and the Zener current, or $I_L = 20\text{ mA} - 10\text{ mA} = 10\text{ mA}$. Calculating the resistance for this minimal load current yields $R_{L,\text{max}} = \frac{8\text{ V}}{10\text{ mA}} = 800\ \Omega$. Dividing the maximum load resistance by the minimum load resistance results in a ratio of $800\ \Omega / 400\ \Omega = 2$.

A photodiode operates optimally under reverse bias because this configuration minimizes the background dark current to an extremely low level, as current conduction is primarily limited to thermally generated minority carriers. When incident light strikes the device, it generates additional electron-hole pairs, which increases the total concentration of these minority charge carriers in the depletion region. Because the original reverse current is already very small, this light-induced increase creates a large fractional change, leading to a significantly higher reverse bias current that allows for the precise detection of optical signals.

Diodes function as switches where a forward-biased diode permits current flow while consuming a fixed potential, whereas a reverse-biased diode behaves as an open switch. In this specific circuit, diode $D_1$ is forward-biased, dropping 0.6 V from the supply, while diode $D_2$ is reverse-biased and disconnected from the current path.

The circuit effectively functions as a simple series loop consisting of the source, the forward-biased diode, and two series resistors. The net potential difference across the resistors is $1 - 0.6 = 0.4$ V, and the total resistance \in the loop is $60 + 40 = 100$ $\Omega$. Applying Ohm's law, the current $i$ flowing through the resistor is determined by $i = \frac{0.4}{100} = 0.004$ A, which equals 4 mA.

The voltage gain of a transistor circuit in a common-emitter configuration quantifies the amplification of the input signal and is calculated using the formula $A_v = \beta \frac{R_C}{R_{\in}}$. In this expression, $\beta$ represents the current gain, which is determined by the ratio of the change in collector current to the change in base current, given by $\beta = \frac{\Delta I_C}{\Delta I_B}$. Based on the transistor characteristics provided, the change in collector current is 5 mA, while the corresponding change in base current is 100 µA, leading to a current gain of $\beta = \frac{5 \times 10^{-3}}{100 \times 10^{-6}} = 50$. Given that the load resistance is 2 kΩ and the input resistance is 0.50 kΩ, we determine the voltage gain by calculating $50 \times \frac{2}{0.50}$. Performing this calculation yields a voltage gain of 200.

The voltage gain of a common-emitter transistor amplifier is defined as the ratio of the output voltage variation to the input voltage variation. Since the output voltage is generated by the collector current flowing through the output resistance and the input voltage is determined by the base current flowing through the input resistance, the gain is expressed by the formula $A_V = \frac{I_C R_o}{I_B R_i}$ Given that the collector current is 10 mA and the base current is 200 \mu A, we determine the gain by applying these values along with the given input resistance of 200 \Omega and output resistance of 60 \Omega . Substituting these into the formula results \in the calculation $\frac{10 \times 10^{-3} \times 60}{200 \times 10^{-6} \times 200} = 15$ confirming that the voltage gain for this experiment is 15.

A NAND gate functions by taking the output of a standard AND operation and immediately inverting it, resulting in a high signal for all input combinations except when both inputs A and B are high. This behavior effectively negates the product of the inputs, meaning that if either input is low, the output remains high, and only when both inputs are simultaneously high does the output drop to low. This specific logic is mathematically defined by the Boolean expression $Y = \overline{A \cdot B}$.

A p-n junction diode functions as a gate that only allows current to pass once the applied voltage reaches its cut-in voltage of 0.6 V. As long as the magnitude of the input voltage $V_{\in}$ remains at or below 0.6 V, the diode stays in an open-circuit state, meaning the output voltage $V_0$ follows the input voltage exactly such that $|V_0| = |V_{\in}|$. When the input signal exceeds this 0.6 V threshold, the diode becomes forward-biased and clamps the output to that fixed voltage, effectively creating a clipped waveform. This behavior results in a signal that mirrors the input within the range of -0.6 V to 0.6 V while flattening the peaks beyond these limits.

A Zener diode connected in parallel with a load resistor maintains a stable breakdown voltage across its terminals, which effectively clamps the voltage across the load. Because the Zener breakdown voltage is 5 V, the voltage across the load resistor $R_L$ is likewise fixed at 5 V. Applying Ohm's Law, defined by the relationship $I_L = \frac{V_L}{R_L}$, allows us to determine the current by dividing the 5 V potential by the 1 kΩ resistance. This calculation confirms that the current flowing through the load is 5 mA.

When an electron travels across a p-n junction, it must overcome the potential barrier, which results in a reduction of its kinetic energy equal to the work done against the potential. Applying the principle of conservation of energy gives the following expression:

$\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 - eV$

Substituting the given values of $m = 9 \times 10^{-31}\ \mathrm{kg}$, $v_i = 6.0 \times 10^5\ \mathrm{m\,s}^{-1}$, $V = 0.4\ \mathrm{V}$, and $e = 1.6 \times 10^{-19}\ \mathrm{C}$, we rearrange the equation for the final velocity as $v_f^2 = v_i^2 - \frac{2eV}{m}$. This results in $v_f^2 = (6.0 \times 10^5)^2 - \frac{2 \times 1.6 \times 10^{-19} \times 0.4}{9 \times 10^{-31}}$. Evaluating the terms yields $v_f^2 = 36 \times 10^{10} - \frac{1.28}{9} \times 10^{12} = (\frac{324 - 128}{9}) \times 10^{10} = \frac{196}{9} \times 10^{10}$. Calculating the square root produces $v_f = \frac{14}{3} \times 10^5\ \mathrm{m\,s}^{-1}$, confirming that $x$ is 14.

Identifying a logic gate requires constructing a truth table by comparing the output state $Y$ to every possible combination of input states $A$ and $B$ observed in the provided waveforms. Upon inspection, the output is high only when both inputs $A$ and $B$ are set to the high state simultaneously, while it remains low for every other possible input configuration. Because this specific pattern of input-output behavior matches the Boolean operation $A \cdot B = Y$, the circuit functions as an AND gate.

Dynamic resistance describes how a diode responds to small voltage fluctuations at a specific operating point and is defined by the ratio of the change in voltage to the change in current, given by the formula:

$R = \frac{\Delta V}{\Delta I}$

At the 2 V operating point, a voltage change of 0.1 V corresponds to a current change of 5 x 10^-3 A, which yields a resistance of 20 ohm. At the 4 V operating point, a voltage change of 0.2 V corresponds to a current change of 50 x 10^-3 A, resulting in a resistance of 4 ohm. Calculating the ratio of the resistance at 2 V to the resistance at 4 V gives the fraction 20/4, which simplifies to 5:1.

The power gain of a transistor in common emitter mode is determined by the square of the current gain multiplied by the ratio of the load resistance to the input resistance. The current gain is calculated by taking the ratio of the change in collector current to the change in base current, yielding $\frac{10 \times 10^{-3}}{100 \times 10^{-6}} = 100$. By substituting this current gain and the given resistance values into the expression $\text{Power Gain} = 100^2 \times \frac{2000}{1000}$, we find the total power gain is 20000, which is equivalent to $2 \times 10^4$, meaning the value of $x$ is 2.

The relationship between these current gain parameters is fundamentally dictated by the conservation of charge within a transistor, which requires the emitter current to equal the sum of the collector and base currents, expressed as $I_E = I_C + I_B$. By substituting this expression for $I_E$ into the definition of $\alpha = \frac{I_C}{I_E}$, we obtain $\alpha = \frac{I_C}{I_C + I_B}$, which simplifies to $\alpha = \frac{1}{1 + \frac{I_B}{I_C}}$ after dividing both terms by $I_C$. Given that $\beta = \frac{I_C}{I_B}$, we can express the ratio as $\alpha = \frac{1}{1 + \frac{1}{\beta}}$, and rearranging this equation algebraically leads to the relation
$\beta = \frac{\alpha}{1 - \alpha}$

Zener breakdown occurs primarily due to the quantum tunneling effect, which is facilitated when a semiconductor junction is heavily doped. This high level of doping drastically reduces the width of the depletion layer, creating an intense electric field across the junction when subjected to reverse bias. This concentrated field exerts sufficient force to pull valence electrons from the $p$-type region directly into the conduction band of the $n$-type region, resulting in a sharp, high-magnitude increase in the Zener current $I_{Z}$ at a specific reverse breakdown voltage $V_{Z}$.

The collector current is determined by applying Ohm's law to the collector resistor, where the voltage drop is 0.6 V and the resistance is 1000 $\Omega$. Dividing these values gives a collector current of $I_C = \frac{0.6}{1000} = 6 \times 10^{-4}$ A. Because the transistor has a current gain $\beta = 24$, we relate the collector current to the base current using the formula $I_B = \frac{I_C}{\beta}$. Applying this relationship results in the following calculation:

$I_B = \frac{6 \times 10^{-4}}{24} = 0.25 \times 10^{-4} \text{ A}$

Converting this result into microamperes by shifting the decimal point six places to the right provides the final base current of 25 microamperes.