Why does the electrical resistance of a semiconductor decrease with temperature, unlike metals?

In metals, heating increases lattice thermal vibrations, causing higher electron collision frequency and increasing resistance. In semiconductors, the band gap $E_g$ requires thermal excitation to free bound valence electrons. As temperature $T$ rises, thermal energy creates exponentially more electron-hole pairs ($n_i \propto T^{3/2} e^{-E_g / 2 k_B T}$), dominating over scattering effects and drastically decreasing resistivity.

How can I solve logic gate circuit problems quickly without making errors?

Instead of tracing complex logic levels visually, write the Boolean expression output for every individual gate using canonical forms. Simplify the combined expression using De Morgan's laws ($\overline{A \cdot B} = \bar{A} + \bar{B}$ and $\overline{A + B} = \bar{A} \cdot \bar{B}$). If short on time, evaluate the final output for extreme binary conditions: $(A=0, B=0)$ and $(A=1, B=1)$ to eliminate wrong options instantly.

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Question #1NEET 2023

A Zener diode having breakdown voltage $V_Z = 6\text{ V}$ is used in a voltage regulator circuit. If the unregulated DC input voltage is $V_{in} = 10\text{ V}$ , the series resistance is $R_S = 200\text{ }\Omega$ , and the load resistance is $R_L = 1\text{ k}\Omega$ , calculate the current $I_Z$ flowing through the Zener diode.

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Step-by-Step Solution & Concept

Step 1

Determine the load current $I_L$ passing through load resistor $R_L = 1\text{ k}\Omega = 1000\text{ }\Omega$ clamped at $V_Z = 6\text{ V}$ : $I_L = \frac{V_Z}{R_L} = \frac{6\text{ V}}{1000\text{ }\Omega} = 6 \times 10^{-3}\text{ A} = 6\text{ mA}$ .

Step 2

Calculate the total current $I_S$ supplied from the input source through series resistor $R_S = 200\text{ }\Omega$ : $I_S = \frac{V_{in} - V_Z}{R_S} = \frac{10\text{ V} - 6\text{ V}}{200\text{ }\Omega} = \frac{4}{200}\text{ A} = 20\text{ mA}$ .

Step 3

Apply Kirchhoff's junction law $I_S = I_Z + I_L$ to compute Zener diode current: $I_Z = I_S - I_L = 20\text{ mA} - 6\text{ mA} = 14\text{ mA}$ .

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Question #2JEE Main 2024

In a common-emitter transistor amplifier, the audio signal voltage output across collector resistance $R_C = 2\text{ k}\Omega$ is $2\text{ V}$ . If the base resistance is $R_B = 1\text{ k}\Omega$ and current gain $\beta = 100$ , the input signal voltage $V_{in}$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Express the voltage gain $A_V$ of a common-emitter amplifier in terms of current amplification factor $\beta$ , collector resistance $R_C$ , and base resistance $R_B$ : $A_V = \beta \cdot \frac{R_C}{R_B}$ .

Step 2

Substitute given values $\beta = 100$ , $R_C = 2000\text{ }\Omega$ , and $R_B = 1000\text{ }\Omega$ : $A_V = 100 \times \frac{2000}{1000} = 200$ .

Step 3

Relate voltage gain to output voltage $V_{out} = 2\text{ V}$ and input voltage $V_{in}$ : $A_V = \frac{V_{out}}{V_{in}} \implies 200 = \frac{2\text{ V}}{V_{in}} \implies V_{in} = \frac{2}{200}\text{ V} = 0.01\text{ V} = 10\text{ mV}$ .

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Mastering Semiconductor Electronics: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Mass Action Law

Electrical Conductivity of Semiconductor

Dynamic Resistance of Diode

Transistor Current Relation

Transistor Current Amplification Factor

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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