Semiconductor Devices - Theory, Equations, and Examples
Semiconductor Devices

1. Introduction to Semiconductors

Semiconductors are materials with electrical conductivity between conductors (metals) and insulators. Their conductivity can be controlled by impurities, temperature, and electric fields, making them essential for modern electronic devices.

2. Types of Semiconductors

  • Intrinsic Semiconductors: Pure semiconductors like Silicon (Si) and Germanium (Ge) with no impurities.
  • Extrinsic Semiconductors: Doped with impurities to increase conductivity.
    • n-type: Doped with donor impurities (like Phosphorus) which add free electrons.
    • p-type: Doped with acceptor impurities (like Boron) which create holes (positive charge carriers).

3. Energy Band Theory

Semiconductors have two important energy bands:

  • Valence Band: Filled with electrons.
  • Conduction Band: Higher energy band where electrons can move freely, contributing to conduction.

The energy gap \( E_g \) between these bands determines the electrical properties:

\( E_g = E_c - E_v \)

Where \( E_c \) is conduction band energy, \( E_v \) is valence band energy.
For Silicon, \( E_g \approx 1.12 \, \text{eV} \).

4. Charge Carriers

In semiconductors, charge is carried by:

  • Electrons (negative charge) in the conduction band.
  • Holes (positive charge) in the valence band, created when electrons leave vacancies.

5. Semiconductor Devices

Some common semiconductor devices include:

  • Diodes: Allow current flow in one direction.
  • Transistors: Amplify or switch electronic signals.
  • Photodiodes & Solar Cells: Convert light into electrical energy.

6. pn-Junction Diode

A pn-junction diode is formed by joining p-type and n-type semiconductors. At the junction:

  • Electrons and holes recombine creating a depletion region with no free charge carriers.
  • This region acts as a barrier to current flow.
P-Type N-Type Depletion Region Conventional Current

7. Diode Current Equation

The current through a diode is given by the Shockley diode equation:

\( I = I_s \left( e^{\frac{qV}{nkT}} - 1 \right) \)

Where:

  • \( I \) = diode current (A)
  • \( I_s \) = saturation current (A)
  • \( q \) = charge of electron (1.6 × 10-19 C)
  • \( V \) = voltage across diode (V)
  • \( n \) = ideality factor (typically 1 to 2)
  • \( k \) = Boltzmann constant (1.38 × 10-23 J/K)
  • \( T \) = temperature in Kelvin

8. Transistors

Transistors are semiconductor devices used to amplify or switch electronic signals. The two main types are:

  • BJT (Bipolar Junction Transistor): Uses both electrons and holes as charge carriers. Consists of emitter, base, and collector regions.
  • FET (Field Effect Transistor): Uses electric field to control conductivity. Includes MOSFET and JFET types.
Emitter (E) Base (B) Collector (C) Electron Flow Collector Current

9. Example Problems

Example 1: Calculate the diode current when \( V = 0.7\,V \), \( I_s = 10^{-12} A \), \( n=1 \), and \( T=300\,K \).
Solution:
Using Shockley equation:
\[ I = 10^{-12} \times \left( e^{\frac{(1.6 \times 10^{-19})(0.7)}{(1)(1.38 \times 10^{-23})(300)}} - 1 \right) \]
Calculate exponent:
\[ \frac{(1.6 \times 10^{-19})(0.7)}{(1.38 \times 10^{-23})(300)} = \frac{1.12 \times 10^{-19}}{4.14 \times 10^{-21}} \approx 27.05 \]
So,
\[ I \approx 10^{-12} \times (e^{27.05} - 1) \approx 10^{-12} \times (5.32 \times 10^{11}) = 0.532\, A \]
Example 2: Identify the type of semiconductor if it has 5 × 1022 electrons/cm³ and 1 × 1015 holes/cm³.
Solution:
Since electrons >> holes, it is an n-type semiconductor.
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