Q.In an intrinsic semiconductor the energy gap $E_g$ is $1.2\ \text{eV}$. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration $n_i$ is given by

[!FORMULA] $n_i = n_0 \exp\left(-\dfrac{E_g}{2 k_B T}\right)$

where $n_0$ is a constant.

Why Intrinsic Carrier Concentration Has That Formula

The intrinsic carrier concentration $n_i$ is the number of electrons (or holes) per unit volume in a pure, undoped semiconductor at thermal equilibrium. The formula you see in every textbook is:

where $N_c$ and $N_v$ are the effective density of states in the conduction and valence bands, $E_g$ is the bandgap energy, $k$ is Boltzmann's constant, and $T$ is absolute temperature.

This isn't pulled from thin air. It comes from a simple physical balance: in an intrinsic semiconductor, every electron in the conduction band leaves behind a hole in the valence band. So the electron concentration $n$ must equal the hole concentration $p$, and both equal $n_i$.

Step 1: The electron and hole concentrations individually

Electrons in the conduction band follow Fermi-Dirac statistics. For non-degenerate semiconductors (which intrinsic ones are, since the Fermi level lies near midgap), the distribution approximates the Maxwell-Boltzmann tail:

Similarly, holes in the valence band:

Here $E_c$ is the conduction band edge, $E_v$ is the valence band edge, and $E_F$ is the Fermi level. The effective densities $N_c$ and $N_v$ come from integrating the density of states times the Boltzmann factor — they depend on the effective masses of electrons and holes and on temperature.

Step 2: The intrinsic condition

In an intrinsic semiconductor, there are no dopants. Every electron that jumps to the conduction band creates exactly one hole. So:

Set the two expressions equal:

Take natural logs and solve for $E_F$:

The Fermi level in an intrinsic semiconductor sits very close to the middle of the bandgap, shifted slightly by the ratio $N_v/N_c$. For most practical purposes, it's at midgap.

Step 3: Multiply to eliminate $E_F$

Now here's the clever part. Instead of solving for $E_F$ directly, multiply $n$ and $p$:

The $E_F$ terms cancel:

This product $np$ is a constant for a given material at a given temperature — it does not depend on the Fermi level. This is the law of mass action for semiconductors.

Step 4: Apply the intrinsic condition

Since $n = p = n_i$ in an intrinsic semiconductor:

Take the square root:

$n_i = \sqrt{N_c N_v} \, e^{-E_g/2kT}$ …