Equilibrium Carrier Concentration Calculator

Equilibrium Carrier Concentration Calculator

Calculate the equilibrium electron and hole concentrations in a doped semiconductor using the mass‑action law and charge neutrality:
For n‑type (when \(N_D \geq N_A\)): \[ n_0 = \frac{N_D-N_A}{2} + \sqrt{\left(\frac{N_D-N_A}{2}\right)^2+n_i^2} \quad \text{and} \quad p_0 = \frac{n_i^2}{n_0} \]
For p‑type (when \(N_A > N_D\)): \[ p_0 = \frac{N_A-N_D}{2} + \sqrt{\left(\frac{N_A-N_D}{2}\right)^2+n_i^2} \quad \text{and} \quad n_0 = \frac{n_i^2}{p_0} \]

* Enter all values in SI units (m\(^{-3}\)).

Step 1: Enter Parameters

Example: 1×1023 m\(^{-3}\) (set to 0 for p‑type)

Example: 0 m\(^{-3}\) (or a value for p‑type doping)

Example: 1.5×1016 m\(^{-3}\) for silicon at 300 K

Formulas:
For n‑type (when \(N_D \geq N_A\)): \[ n_0 = \frac{N_D-N_A}{2} + \sqrt{\left(\frac{N_D-N_A}{2}\right)^2+n_i^2},\quad p_0 = \frac{n_i^2}{n_0} \]
For p‑type (when \(N_A > N_D\)): \[ p_0 = \frac{N_A-N_D}{2} + \sqrt{\left(\frac{N_A-N_D}{2}\right)^2+n_i^2},\quad n_0 = \frac{n_i^2}{p_0} \]


Practical Example:
For an n‑type silicon sample at 300 K with \(N_D = 1\times10^{23}\) m\(^{-3}\), \(N_A = 0\) m\(^{-3}\), and \(n_i = 1.5\times10^{16}\) m\(^{-3}\),
the equilibrium electron concentration is approximately
\[ n_0 \approx \frac{1\times10^{23}}{2} + \sqrt{\left(\frac{1\times10^{23}}{2}\right)^2 + \left(1.5\times10^{16}\right)^2} \approx 1\times10^{23}\, \text{m}^{-3} \] and the hole concentration is
\[ p_0 = \frac{\left(1.5\times10^{16}\right)^2}{1\times10^{23}} \approx 2.25\times10^{9}\, \text{m}^{-3}. \]