---
title: "Intrinsic concentration and resistivity of Si"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:intrinsic-si-effective-masses"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/intrinsic-si-effective-masses
site: BCT Engineering
---

# Intrinsic concentration and resistivity of Si

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2079 Bhadra, 6 marks; 2078 Bhadra, 8 marks; 2076 Chaitra, 8 marks; 2071 Chaitra, 6 marks (asked 4 times)
- Video: [Intrinsic Concentration in Semiconductor Material | Electronics Devices and Circuits - EDC](https://www.youtube.com/watch?v=SOXrmHw_flM) — Engineering Funda
- 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-effective-masses
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Given that the density of states related effective masses of electrons and holes in Si are approximately 1.08mₑ and 0.60mₑ respectively and the electron and hole drift mobilities at room temperature are 1350 and 450 cm²V⁻¹s⁻¹ respectively. Calculate the intrinsic concentration and intrinsic resistivity of Si. The energy band gap for Si is 1.10eV. (T = 300K)

## Short answer

Calculate Nc and Nv using Nc,v = 2(2πm*kT/h²)^(3/2) with given effective masses; then ni = √(Nc Nv) exp(–Eg/2kT) ≈ 1.05×10¹⁰ cm⁻³. Intrinsic conductivity σ = e ni (μe+μh) yields resistivity ρ = 1/σ ≈ 3.3×10⁵ Ω·cm.

## Answer

In an intrinsic semiconductor, $n = p = n_i$ with

$$
n_i = \sqrt{N_c N_v}\,\exp\left(-\frac{E_g}{2kT}\right), \quad N_c = 2\left(\frac{2\pi m_e^* kT}{h^2}\right)^{3/2}, \quad N_v = 2\left(\frac{2\pi m_h^* kT}{h^2}\right)^{3/2}
$$

**Data:** $m_e^* = 1.08m_e$, $m_h^* = 0.60m_e$, $m_e = 9.109\times10^{-31}$ kg, $k = 1.381\times10^{-23}$ J/K, $h = 6.626\times10^{-34}$ J s, $T = 300$ K, $E_g = 1.10$ eV, $\mu_e = 1350$, $\mu_h = 450$ cm$^2$V$^{-1}$s$^{-1}$. Then $kT = 0.02586$ eV.

#### Effective densities of states
$$
\begin{aligned} N_c &= 2\left[\frac{2\pi (1.08 \times 9.109\times10^{-31})(1.381\times10^{-23})(300)}{(6.626\times10^{-34})^2}\right]^{3/2} \\ &= 2.82\times10^{25}\ \text{m}^{-3} = 2.82\times10^{19}\ \text{cm}^{-3} \\ N_v &= N_c\left(\frac{0.60}{1.08}\right)^{3/2} = 1.17\times10^{19}\ \text{cm}^{-3} \end{aligned}
$$

#### Intrinsic concentration
$$
\begin{aligned} \frac{E_g}{2kT} &= \frac{1.10}{2 \times 0.02586} = 21.27 \\ \sqrt{N_c N_v} &= \sqrt{2.82\times10^{19} \times 1.17\times10^{19}} = 1.813\times10^{19}\ \text{cm}^{-3} \\ n_i &= 1.813\times10^{19} \times e^{-21.27} \\ &= 1.813\times10^{19} \times 5.80\times10^{-10} \\ &= 1.05\times10^{10}\ \text{cm}^{-3} \end{aligned}
$$

#### Intrinsic resistivity
$$
\begin{aligned} \sigma_i &= e n_i (\mu_e + \mu_h) \\ &= 1.602\times10^{-19} \times 1.052\times10^{10} \times (1350 + 450) \\ &= 3.03\times10^{-6}\ \Omega^{-1}\text{cm}^{-1} \\ \rho_i &= \frac{1}{\sigma_i} = 3.30\times10^{5}\ \Omega\,\text{cm} \end{aligned}
$$

**Answer:** $n_i \approx 1.05\times10^{10}$ cm$^{-3}$; $\rho_i \approx 3.3\times10^{5}$ $\Omega$ cm ($3.3\times10^{3}$ $\Omega$ m).

## Check yourself

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

1. Which expression gives the intrinsic carrier concentration used in this answer?
   A. n_i = sqrt(Nc Nv) exp(-Eg/2kT)
   B. n_i = (Nc + Nv) exp(-Eg/kT)
   C. n_i = sqrt(Nc Nv) exp(-Eg/kT)
   D. n_i = Nc Nv exp(-Eg/2kT)
   Answer: A. Electrons and holes are created in pairs, so n_i is the geometric mean of Nc and Nv times exp(-Eg/2kT). (Key point: Intrinsic concentration formula)

2. How is Nv obtained from Nc once Nc is known?
   A. Nv = Nc (0.60/1.08)
   B. Nv = Nc (0.60/1.08)^(3/2)
   C. Nv = Nc (1.08/0.60)^(3/2)
   D. Nv = Nc (0.60/1.08)^2
   Answer: B. N is proportional to (m*)^(3/2), so Nv/Nc = (mh*/me*)^(3/2). (Key point: Nv from Nc)

3. With Eg = 1.10 eV and kT = 0.02586 eV, what is the exponent Eg/2kT?
   A. 42.5
   B. 10.6
   C. 21.27
   D. 1.10
   Answer: C. 1.10/(2 x 0.02586) = 21.27, giving exp(-21.27) = 5.8 x 10^-10. (Key point: Exponent value)

4. Which formula gives the intrinsic conductivity of Si in this answer?
   A. sigma = e n_i (mu_e - mu_h)
   B. sigma = e n_i mu_e only
   C. sigma = e n_i (mu_e + mu_h)
   D. sigma = e n_i^2 (mu_e + mu_h)
   Answer: C. Both electrons and holes (n = p = n_i) carry current and their contributions add; resistivity is then 1/sigma. (Key point: Intrinsic conductivity)

## Close variants set in other exams

- [Intrinsic Si concentration (hole mass 0.56me)](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aintrinsic-si-056) — Electrical Engineering Material, chapter 6, asked 6×

## Related in Electrical Engineering Material

- [Intrinsic Si from effective densities of states](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aintrinsic-si-nc-nv) — Electrical Engineering Material, chapter 6
- [Intrinsic conductivity and resistivity of GaAs](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Agaas-intrinsic-conductivity) — Electrical Engineering Material, chapter 6