Q13Engineering Physics

Question

For intrinsic semiconductor with a band gap $E_{g} = 0.7 eV$ , calculate the density of electrons and holes at $300 K$ .

Answer 1

In a pure, perfect (intrinsic) semiconductor crystal, at absolute zero ( $0 K$ ), the valence band is completely full and the conduction band is entirely empty; the material acts as a perfect insulator. However, as the temperature rises above $0 K$ , thermal energy causes some electrons to break their covalent bonds and jump across the forbidden energy gap ( $E_{g}$ ) into the conduction band. Every electron that makes this jump leaves behind a positively charged vacancy in the valence band, known as a 'hole'. Because they are created in pairs, the concentration of electrons in the conduction band ( $n$ ) is always exactly equal to the concentration of holes in the valence band ( $p$ ). This fundamental concentration is called the intrinsic carrier density, $n_{i}$ , where $n = p = n_{i}$ .

The Mathematical Model

By applying Fermi-Dirac statistics to the density of states in the conduction and valence bands, solid-state physics provides a rigorous formula for $n_{i}$ as a function of temperature ( $T$ ) and bandgap energy ( $E_{g}$ ):

n_{i} = A \cdot T^{3/2} \cdot exp (- \frac{E _{g}}{2 k T})

Where: - $A$ is a material-dependent constant derived from the effective masses of electrons and holes (for Germanium, $A \approx 1.76 \times 1 0^{22} m^{- 3} K^{- 3/2}$ ). - $T$ is the absolute temperature ( $300 K$ ). - $k$ is the Boltzmann constant ( $1.38 \times 1 0^{- 23} J/K$ or $8.617 \times 1 0^{- 5} eV/K$ ). - $E_{g}$ is the bandgap energy ( $0.7 eV$ ).

Step-by-Step Calculation

First, we evaluate the thermal energy parameter $k T$ at room temperature ( $300 K$ ):

k T = (8.617 \times 1 0^{- 5} eV/K) \times (300 K) \approx 0.02585 eV

Next, we calculate the exponent factor. The exponent determines the probability of an electron acquiring enough thermal energy to cross the gap:

\frac{E _{g}}{2 k T} = \frac{0.7}{2 \times 0.02585} = \frac{0.7}{0.0517} \approx 13.54

Now, we evaluate the exponential term:

exp (- 13.54) \approx 1.31 \times 1 0^{- 6}

Finally, we calculate the pre-exponential factor and multiply them together. Given the known constant $A$ for Germanium, we get:

n_{i} = (1.76 \times 1 0^{22}) \times (300)^{1.5} \times (1.31 \times 1 0^{- 6})

n_{i} \approx (1.76 \times 1 0^{22}) \times (5196) \times (1.31 \times 1 0^{- 6}) \approx 2.5 \times 1 0^{19} electrons/m^{3}

This result highlights a critical engineering reality: even at room temperature, thermal generation produces a massive $2.5 \times 1 0^{19}$ charge carriers per cubic meter in Germanium. This is why Silicon ( $E_{g} = 1.1 eV$ , yielding a much lower $n_{i}$ ) is vastly preferred over Germanium for modern integrated circuits, as it suffers from significantly lower leakage currents at high operating temperatures.

Q13Engineering Physics

Question

Answer

The Mathematical Model

Step-by-Step Calculation