Past Questions — 2022

Electrochemical breakdown (often associated with chemical degradation or corrosion-induced failure in dielectrics) occurs when the material is exposed to environmental conditions that accelerate chemical reactions. In the context of surrounding air, high humidity provides the necessary moisture (water vapor) to act as an electrolyte or a reactant, while high temperature provides the thermal energy required to increase the rate of these chemical reactions (following the Arrhenius equation). Together, high temperature and high humidity facilitate the migration of ions and the formation of conductive paths (such as dendritic growth or oxidation), leading to the breakdown of the dielectric strength of the material. Low temperature or low humidity would inhibit these chemical processes, making them less likely to cause breakdown.

A Bipolar Junction Transistor (BJT) consists of two PN junctions: the Base-Emitter (BE) junction and the Base-Collector (BC) junction. The operating region of the transistor is determined by the biasing of these two junctions: 1. **Cut-off Region**: Both the BE junction and the BC junction are reverse biased. The transistor acts as an open switch, and virtually no current flows. 2. **Active Region**: The BE junction is forward biased and the BC junction is reverse biased. This is the standard mode for linear amplification. 3. **Saturation Region**: Both the BE junction and the BC junction are forward biased. In this state, the transistor acts as a closed switch, and the collector current reaches its maximum value (saturation), regardless of further increases in base current. 4. **Reverse-Active Region**: The BE junction is reverse biased and the BC junction is forward biased (this is rarely used as it is highly inefficient). Comparing these to the options: - Option (a) describes the Cut-off region. - Option (b) describes the Reverse-Active region. - Option (c) describes the Active region. - Option (d) correctly describes the Saturation region.

The threshold voltage (Vth) of an n-channel MOSFET is the gate-to-source voltage required to create an inversion layer of electrons at the silicon-oxide interface. The formula for Vth is approximately: Vth = VFB + 2φF + (√2εs q N_A (2φF)) / Cox, where N_A is the acceptor dopant concentration of the p-type substrate (channel) and Cox is the oxide capacitance. 1. **Increasing channel dopant concentration (N_A):** As N_A increases, the amount of negative charge (ionized acceptors) in the depletion region increases. To overcome this increased negative charge and achieve inversion, a higher positive gate voltage is required. Therefore, increasing the dopant concentration increases Vth. This makes option (a) correct. 2. **Reducing channel dopant concentration:** This would decrease the charge in the depletion region, making it easier to form the inversion layer, thereby decreasing Vth. Thus, option (b) is incorrect. 3. **Reducing gate-oxide thickness (t_ox):** Reducing the thickness increases the oxide capacitance (Cox = ε_ox / t_ox). Since Cox is in the denominator of the depletion charge term in the Vth equation, increasing Cox reduces the voltage required to deplete the channel, thereby decreasing Vth. Thus, option (c) is incorrect. 4. **Reducing channel length:** While reducing channel length can lead to the 'Short Channel Effect' (which typically causes Vth to decrease, known as Vth roll-off), it is not a primary method used to increase Vth. In standard MOSFET physics, reducing length generally lowers the threshold voltage. Thus, option (d) is incorrect.

The resistance (R) of a wire is given by the formula: R = ρ(L/A), where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area. Given for wire B: Resistance R_B = 1 Ω (Note: The 'W' in the prompt is a typographical error for Ω/Ohms). Let the length of B be L_B and the cross-sectional area be A_B. So, R_B = ρ(L_B / A_B) = 1 Ω. Given for wire A: 1. Length L_A = 20 * L_B 2. Cross-sectional area A_A = (1/2) * A_B Now, calculate the resistance of wire A (R_A): R_A = ρ(L_A / A_A) Substitute the values of L_A and A_A in terms of B: R_A = ρ [ (20 * L_B) / (0.5 * A_B) ] R_A = (20 / 0.5) * ρ(L_B / A_B) R_A = 40 * R_B Since R_B = 1 Ω: R_A = 40 * 1 = 40 Ω. Therefore, the resistance of wire A is 40 Ω. Option (a) is correct.

In network theory, the concept of duality allows us to transform one circuit into another by replacing specific elements and parameters while maintaining the same mathematical form of the governing equations. The duality rules are as follows: 1. **Topology Transformation**: A parallel connection becomes a series connection, and vice versa. 2. **Element Transformation**: - Resistance (R) $\leftrightarrow$ Conductance (G). Since both are resistive elements, R remains R (or transforms to G, which is still a resistive component). - Capacitance (C) $\leftrightarrow$ Inductance (L). This is because the voltage-current relationship for a capacitor is $i = C \frac{dv}{dt}$ and for an inductor is $v = L \frac{di}{dt}$. 3. **Parameter Transformation**: Voltage $\leftrightarrow$ Current, and Mesh $\leftrightarrow$ Node. Applying these rules to a 'Parallel R-C circuit': - 'Parallel' becomes 'Series'. - 'R' (Resistor) remains a resistive element 'R'. - 'C' (Capacitor) becomes 'L' (Inductor). Therefore, the dual of a parallel R-C circuit is a series R-L circuit.

A Junction Field Effect Transistor (JFET) is a unipolar device, meaning the current is carried by only one type of charge carrier. The type of carrier depends on the material used for the conducting channel: 1. In an n-channel JFET, the channel is made of n-type semiconductor material, and the majority charge carriers are electrons. 2. In a p-channel JFET, the channel is made of p-type semiconductor material. In p-type materials, the majority charge carriers are holes. Since the question specifically asks about a p-channel JFET, the charge carriers are holes. Analysis of other options: - Option (a) Electrons: These are the carriers for n-channel JFETs. - Option (c) Both electrons and holes: This describes bipolar devices (like BJTs), not unipolar devices like JFETs. - Option (d) Atom: Atoms are neutral particles and do not act as charge carriers for electrical current in semiconductors.

1. **Definition of Convolution**: The general formula for the convolution of two continuous-time signals $x(t)$ and $h(t)$ is given by $y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau$. This is the standard mathematical definition and applies regardless of whether the signals are causal or non-causal. 2. **Definition of Causality**: A signal is said to be causal if it is zero for all $t < 0$. For a system to be causal, its impulse response $h(t)$ must be causal ($h(t) = 0$ for $t < 0$). 3. **Analysis of the Integral Limits**: In practical engineering and competitive exam contexts for 'Signals and Systems', when the convolution integral is presented in its general form, the question often tests the condition under which the integral simplifies or the physical realization of the system. 4. **Evaluating the Options**: - If $x(t)$ and $h(t)$ are both causal, the limits of integration effectively change from $(-\infty, \infty)$ to $(0, t)$. This is because $x(\tau) = 0$ for $\tau < 0$ and $h(t-\tau) = 0$ for $t-\tau < 0$ (which means $\tau > t$). - While the general formula provided in the question is mathematically true for *any* two signals, in the context of standard textbook problems for this specific exam pattern, the expression $y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau$ is the fundamental definition used to derive the output of a Linear Time-Invariant (LTI) system. - For a system to be physically realizable (causal), the impulse response $h(t)$ must be causal. If the input $x(t)$ is also causal (starts at $t=0$), the convolution represents a real-time process. 5. **Conclusion**: Given the options provided, the question asks for the condition associated with this operation in a standard LTI system context. The most appropriate answer in the context of engineering causality is that both the input and the impulse response are causal, allowing the system to operate in real-time.

To determine which option is not associated with a P-N junction, let's analyze each term: 1. **Junction Capacitance (a):** This is a general term referring to the capacitance that occurs at the P-N junction. It is composed of both depletion capacitance and diffusion (charge storage) capacitance. 2. **Charge Storage Capacitance (b):** Also known as Diffusion Capacitance, this occurs when a P-N junction is forward-biased. Minority carriers are injected across the junction and stored in the neutral regions, creating a capacitance effect. 3. **Depletion Capacitance (c):** Also known as Transition Capacitance, this occurs when a P-N junction is reverse-biased. The depletion region acts as a dielectric between the P and N regions (which act as conducting plates), creating a capacitor. 4. **Channel Length Modulation (d):** This is a phenomenon specifically associated with **Field Effect Transistors (FETs)**, such as MOSFETs. It occurs when the drain-source voltage ($V_{DS}$) increases beyond the saturation point, causing the effective length of the conducting channel to decrease, which in turn increases the drain current. It is a property of a three-terminal device with a channel, not a basic two-terminal P-N junction. Therefore, channel length modulation is the only option not associated with a P-N junction.