---
title: "Intrinsic Si concentration (hole mass 0.56me)"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:intrinsic-si-056"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/intrinsic-si-056
site: BCT Engineering
---

# Intrinsic Si concentration (hole mass 0.56me)

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2081 Chaitra (new course), 4 marks; 2073 Chaitra, 8 marks (asked 2 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-056
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

The density of states related effective masses of electrons and holes in silicon are approximately 1.08mₑ and 0.56mₑ respectively. The electron and hole drift mobilities at room temperature are 1350 and 450cm²V⁻¹s⁻¹ respectively. Calculate intrinsic concentration and intrinsic resistivity of silicon. The energy band gap for silicon is 1.1ev.

## Short answer

Calculate effective densities of states: Nc = 2.82×10¹⁹ cm⁻³, Nv = 1.05×10¹⁹ cm⁻³. Then intrinsic concentration ni = sqrt(Nc Nv) exp(–Eg/(2kT)) = 1.0×10¹⁰ cm⁻³. Resistivity ρ = 1/[e ni (μe+μh)] = 3.47×10⁵ Ω cm.

## Answer

For an intrinsic semiconductor,

$$
n_i = \sqrt{N_cN_v}\exp\left(-\frac{E_g}{2kT}\right), \quad N_{c,v} = 2\left(\frac{2\pi m^*_{e,h}kT}{h^2}\right)^{3/2}
$$

**Data:** $m_e^* = 1.08m_e$, $m_h^* = 0.56m_e$, $E_g = 1.1$ eV, $\mu_e = 1350$, $\mu_h = 450$ cm$^2$V$^{-1}$s$^{-1}$. Take $T = 300$ K (room temperature), $kT = 0.02586$ eV, $m_e = 9.109\times10^{-31}$ kg, $h = 6.626\times10^{-34}$ J s, $k = 1.381\times10^{-23}$ J/K.

#### Effective densities of states
$$
\begin{aligned} N_c &= 2\left[\frac{2\pi \times 1.08 \times 9.109\times10^{-31} \times 1.381\times10^{-23} \times 300}{(6.626\times10^{-34})^2}\right]^{3/2} \\ &= 2.82\times10^{25}\ \text{m}^{-3} = 2.82\times10^{19}\ \text{cm}^{-3} \\ N_v &= N_c\left(\frac{0.56}{1.08}\right)^{3/2} = 1.05\times10^{19}\ \text{cm}^{-3} \end{aligned}
$$

#### Intrinsic concentration
$$
\begin{aligned} \sqrt{N_cN_v} &= \sqrt{2.82\times10^{19}\times1.05\times10^{19}} = 1.72\times10^{19}\ \text{cm}^{-3} \\ \frac{E_g}{2kT} &= \frac{1.1}{2\times0.02586} = 21.27, \quad e^{-21.27} = 5.80\times10^{-10} \\ n_i &= 1.72\times10^{19}\times5.80\times10^{-10} = 9.99\times10^{9}\ \text{cm}^{-3} \end{aligned}
$$

#### Intrinsic resistivity
$$
\begin{aligned} \sigma_i &= en_i(\mu_e+\mu_h) = 1.602\times10^{-19}\times9.99\times10^{9}\times1800 \\ &= 2.88\times10^{-6}\ \Omega^{-1}\text{cm}^{-1} \\ \rho_i &= \frac{1}{\sigma_i} = 3.47\times10^{5}\ \Omega\,\text{cm} \end{aligned}
$$

**Answer:** $n_i \approx 1.0\times10^{10}$ cm$^{-3}$; $\rho_i \approx 3.47\times10^{5}$ $\Omega$ cm ($3.47\times10^{3}$ $\Omega$ m).

## Check yourself

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

1. How does the effective density of states depend on effective mass?
   A. N is proportional to m*
   B. N is proportional to (m*)^(1/2)
   C. N is proportional to (m*)^(3/2)
   D. N is independent of m*
   Answer: C. N = 2(2 pi m* kT/h^2)^(3/2), so Nv = Nc (0.56/1.08)^(3/2) = 1.05 x 10^19 cm^-3. (Key point: Density of states formula)

2. What value of ni results for these data?
   A. About 1.0 x 10^10 cm^-3
   B. About 1.0 x 10^19 cm^-3
   C. About 5.8 x 10^-10 cm^-3
   D. About 3.5 x 10^5 cm^-3
   Answer: A. sqrt(Nc Nv) = 1.72e19, times exp(-21.27) = 5.8e-10 gives about 1.0e10 cm^-3. (Key point: Intrinsic concentration)

3. Which expression gives intrinsic conductivity?
   A. sigma = e ni mu_e
   B. sigma = e ni (mu_e + mu_h)
   C. sigma = e ni (mu_e x mu_h)
   D. sigma = ni / e (mu_e + mu_h)
   Answer: B. n = p = ni, so both carriers contribute: sigma = e ni (1350 + 450). (Key point: Conductivity formula)

4. What is the intrinsic resistivity?
   A. About 2.88 x 10^-6 ohm-cm
   B. About 3.47 x 10^5 ohm-cm
   C. About 3.47 x 10^-5 ohm-cm
   D. About 1800 ohm-cm
   Answer: B. rho = 1/sigma = 1/(2.88e-6 ohm^-1 cm^-1) = 3.47e5 ohm-cm. (Key point: Resistivity)

## Close variants set in other exams

- [Intrinsic concentration and resistivity of Si](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aintrinsic-si-effective-masses) — 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 concentration and its temperature dependence](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aintrinsic-concentration-derivation) — Electrical Engineering Material, chapter 6