Intrinsic Carrier Concentration Calculator

Enter Calculation

Conduction band density of states Nc (cm⁻³)

Valence band density of states Nv (cm⁻³)

Band gap energy Eg (eV)

Temperature T (K)

Results

Intrinsic Carrier Concentration

66.75899E8

cm⁻³

√(Nc·Nv)	1706.458E16 cm⁻³
exp(−Eg / 2kT)	391.2137E-12

What is the Intrinsic Carrier Concentration?

The intrinsic carrier concentration n_i is the number of free electrons (equal to the number of holes) per cubic centimetre in a pure, undoped semiconductor at thermal equilibrium. It is one of the most fundamental quantities in device physics: it sets the leakage current of diodes, the dark current of detectors, and the temperature sensitivity of transistors. Wide-gap materials like silicon carbide have a tiny n_i, while narrow-gap materials like germanium have a much larger one.

Thermal generation of equal numbers of electrons and holes across the band gap in an intrinsic semiconductor.

How to Use the Calculator

Enter the effective density of states in the conduction band (N_c) and valence band (N_v) in cm⁻³, the band gap energy E_g in electron-volts, and the absolute temperature T in kelvin. The calculator returns n_i together with the geometric mean \(\sqrt{\text{N}_c \cdot \text{N}_v}\) and the exponential Boltzmann factor so you can see how each part contributes.

The Formula Explained

The expression is $$n_i = \sqrt{\text{N}_c \cdot \text{N}_v}\;\exp\!\left(-\frac{\text{E}_g}{2\,k\,\text{T}}\right)$$. The prefactor \(\sqrt{\text{N}_c \cdot \text{N}_v}\) reflects the available states near the band edges, while the exponential term — the Boltzmann factor — describes how thermal energy promotes electrons across the gap. Because n_i depends exponentially on \(-\text{E}_g/2k\text{T}\), it rises steeply with temperature and falls sharply for wider gaps. Here \(k = 8.617333262\times10^{-5}\ \text{eV/K}\) so that E_g and kT share the same energy units.

n_i increases exponentially with temperature due to the Boltzmann factor.

Worked Example

For silicon at 300 K with N_c = 2.8×10¹⁹, N_v = 1.04×10¹⁹ cm⁻³ and E_g = 1.12 eV: $$\sqrt{\text{N}_c \cdot \text{N}_v} = \sqrt{2.912\times10^{38}} \approx 1.7065\times10^{19}$$ The exponent is $$-\frac{1.12}{2 \cdot 8.617333262\times10^{-5} \cdot 300} \approx -21.66,$$ giving \(\exp \approx 3.91\times10^{-10}\). Therefore \(n_i \approx 6.68\times10^{9}\ \text{cm}^{-3}\), close to the textbook value of about 10¹⁰ cm⁻³ (small differences come from the chosen effective masses).

FAQ

Why does n_i use 2kT instead of kT? Because the Fermi level sits roughly mid-gap, each carrier is "half" of an electron-hole pair, so the gap energy is shared, giving the factor of 2 in the denominator.

Which units should I use? N_c and N_v in cm⁻³, E_g in eV, and T in kelvin. The result is then in cm⁻³.

Does this work for any semiconductor? Yes — it is a universal physics relation. Just supply the correct N_c, N_v and E_g for your material and temperature.