---
title: "Intrinsic Si from effective densities of states"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:intrinsic-si-nc-nv"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/intrinsic-si-nc-nv
site: BCT Engineering
---

# Intrinsic Si from effective densities of states

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2072 Kartik, 8 marks
- Video: [ECE 606 Solid State Devices L12.3: Occupation of States - Intrinsic Carrier Concentration](https://www.youtube.com/watch?v=AgHYxsd0qLE) — nanohubtechtalks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/intrinsic-si-nc-nv
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

The effective density of states at conduction band and valence band are 2.9×10¹⁹ cm⁻³ and 1.1×10¹⁹ cm⁻³ respectively. Calculate the intrinsic concentration and intrinsic resistivity of silicon at 300 K temperature.

## Short answer

n_i = √(NcNv) exp(-E_g/2kT). With E_g=1.1 eV, kT=0.02585 eV, n_i ≈ 1.03×10¹⁰ cm⁻³. Conductivity σ_i = e n_i(μe+μh) = 2.96×10⁻⁶ S/cm, resistivity ρ_i = 3.38×10⁵ Ω·cm.

## Answer

For an intrinsic semiconductor

$$
n_i = \sqrt{N_cN_v}\,\exp\left(-\frac{E_g}{2kT}\right),\qquad \rho_i = \frac{1}{en_i(\mu_e+\mu_h)}
$$

**Given:** $N_c = 2.9\times10^{19}$, $N_v = 1.1\times10^{19}\ \text{cm}^{-3}$, T = 300 K. **Data used (Si):** $E_g = 1.1$ eV, $kT = 0.02585$ eV, $\mu_e = 1350$, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $e = 1.6\times10^{-19}$ C.

#### Intrinsic concentration

$$
\begin{aligned}
\sqrt{N_cN_v} &= \sqrt{2.9\times10^{19}\times1.1\times10^{19}} = 1.786\times10^{19}\ \text{cm}^{-3}\\
\frac{E_g}{2kT} &= \frac{1.1}{2\times0.02585} = 21.28\\
e^{-21.28} &= 5.750\times10^{-10}\\
n_i &= 1.786\times10^{19}\times5.750\times10^{-10}\\
&= 1.03\times10^{10}\ \text{cm}^{-3}
\end{aligned}
$$

#### Intrinsic conductivity and resistivity

$$
\begin{aligned}
\sigma_i &= en_i(\mu_e+\mu_h)\\
&= 1.6\times10^{-19}\times1.027\times10^{10}\times1800\\
&= 2.96\times10^{-6}\ \text{S cm}^{-1}\\
\rho_i &= \frac{1}{\sigma_i} = 3.38\times10^{5}\ \Omega\,\text{cm}
\end{aligned}
$$

**Answer: $n_i \approx 1.03\times10^{10}\ \text{cm}^{-3}$; $\rho_i \approx 3.38\times10^{5}\ \Omega$ cm ($3.38\times10^{3}\ \Omega$ m).**

Note: $n_i$ is very sensitive to $E_g$ and T. Because of the factor $e^{-E_g/2kT}$, using $E_g = 1.12$ eV instead of 1.1 eV would reduce $n_i$ by a factor of about 1.5.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula gives the intrinsic concentration from Nc and Nv?
   A. ni = (Nc + Nv) exp(-Eg/kT)
   B. ni = Nc Nv exp(-Eg/2kT)
   C. ni = sqrt(Nc Nv) exp(-Eg/kT)
   D. ni = sqrt(Nc Nv) exp(-Eg/2kT)
   Answer: D. From ni^2 = Nc Nv exp(-Eg/kT), taking the square root gives exp(-Eg/2kT). (Key point: Formula for ni)

2. What is sqrt(Nc Nv) for Nc = 2.9e19 and Nv = 1.1e19 cm^-3?
   A. About 1.79 x 10^19 cm^-3
   B. About 2.0 x 10^19 cm^-3
   C. About 4.0 x 10^19 cm^-3
   D. About 3.2 x 10^38 cm^-3
   Answer: A. sqrt(3.19e38) is about 1.786e19 cm^-3. (Key point: Geometric mean)

3. With Eg = 1.1 eV and kT = 0.02585 eV, ni is about:
   A. 1.03 x 10^19 cm^-3
   B. 1.03 x 10^10 cm^-3
   C. 5.8 x 10^-10 cm^-3
   D. 2.1 x 10^21 cm^-3
   Answer: B. exp(-21.28) = 5.755e-10; times 1.786e19 gives about 1.03e10 cm^-3. (Key point: Resulting ni)

4. How is the intrinsic resistivity obtained?
   A. rho = e ni (mu_e + mu_h)
   B. rho = 1/[e ni (mu_e - mu_h)]
   C. rho = 1/[e ni (mu_e + mu_h)]
   D. rho = ni/(e mu_e)
   Answer: C. Both carriers contribute, so sigma = e ni (mu_e + mu_h) and rho = 1/sigma, about 3.38 x 10^5 ohm cm. (Key point: Resistivity formula)

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