How electron-hole pairs are created in a semiconductor material?

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Q. How electron-hole pairs are created in a semiconductor material?

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Concept understanding — Intrinsic Carrier Concentration

Intrinsic carrier concentration is a foundational idea in semiconductor physics — let’s build it from the ground up, with no prior knowledge of semiconductors needed.

1. Intuition: What does "intrinsic" mean?

Imagine a pure, perfect crystal of silicon — no impurities, no defects. At absolute zero temperature ( $0 \text{ K}$ ), all electrons are tightly bound in the crystal lattice. No current flows.

Now, heat it up. Thermal energy shakes the atoms. Some electrons gain enough energy to break free from their bonds. When an electron leaves, it leaves behind a hole — a missing electron that behaves like a positive charge.

In this pure crystal, every free electron comes from a broken bond, and every broken bond creates one hole. So:

Number of free electrons = Number of holes

This balance is the hallmark of an intrinsic semiconductor.

2. The precise definition

Intrinsic carrier concentration ( $n_i$ ) is the number of free electrons (or holes) per unit volume in a pure, undoped semiconductor at thermal equilibrium.

It is denoted by $n_i$ and has units of $\text{cm}^{-3}$ or $\text{m}^{-3}$ .

Key points:

It depends only on the material and temperature — not on doping.
For silicon at room temperature ( $300 \text{ K}$ ):

n_i \approx 1.5 \times 10^{10} \text{ cm}^{-3}

For germanium: $n_i \approx 2.5 \times 10^{13} \text{ cm}^{-3}$
For gallium arsenide: $n_i \approx 1.8 \times 10^{6} \text{ cm}^{-3}$

3. The formula (for exams)

The precise expression is:

n_i = \sqrt{N_c N_v} \cdot e^{-E_g / (2kT)}

Where:

$N_c$ = effective density of states in the conduction band
$N_v$ = effective density of states in the valence band
$E_g$ = bandgap energy (eV)
$k$ = Boltzmann constant ( $8.617 \times 10^{-5} \text{ eV/K}$ )
$T$ = absolute temperature (K)

Important: The exponential term $e^{-E_g/(2kT)}$ dominates — a small change in $E_g$ or $T$ causes a huge change in $n_i$ .

4. Why does it matter?

It sets the baseline for all semiconductor devices. Doping increases one carrier type, but the product $n \cdot p = n_i^2$ always holds at equilibrium.
Temperature sensitivity: $n_i$ roughly doubles for every $10^\circ\text{C}$ rise in silicon. This is why circuits fail in heat. …

Why this formula?

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:

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

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:

n = N_c \, e^{-(E_c - E_F)/kT}

Similarly, holes in the valence band:

p = N_v \, e^{-(E_F - E_v)/kT}

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:

n = p

Set the two expressions equal:

N_c \, e^{-(E_c - E_F)/kT} = N_v \, e^{-(E_F - E_v)/kT}

Take natural logs and solve for $E_F$ :

-(E_c - E_F) + \ln N_c = -(E_F - E_v) + \ln N_v

E_F = \frac{E_c + E_v}{2} + \frac{kT}{2} \ln\frac{N_v}{N_c}

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 $p$ :

n p = N_c N_v \, e^{-(E_c - E_F)/kT} \, e^{-(E_F - E_v)/kT}

The $E_F$ terms cancel:

n p = N_c N_v \, e^{-(E_c - E_v)/kT} = N_c N_v \, e^{-E_g/kT}

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:

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

Take the square root:

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