---
title: "Si cube doped 1 As in 10⁶"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:si-cube-arsenic-1e6"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-cube-arsenic-1e6
site: BCT Engineering
---

# Si cube doped 1 As in 10⁶

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2081 Bhadra, 6 marks
- Video: [Tutorial: Doping](https://www.youtube.com/watch?v=k12GMjtN8aA) — MIT OpenCourseWare
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-cube-arsenic-1e6
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Calculate the resistance of pure silicon cubic crystal of 1cm³ at room temperature (27°C). What will be its new resistance if it is doped with 1 arsenic in 10⁶ silicon atoms. The atomic concentration of silicon is 5×10²² cm⁻³.

## Short answer

Pure Si cube: σ = e ni (μe+μh) = 4.18×10⁻⁶ Ω⁻¹cm⁻¹, so R = 2.39×10⁵ Ω. With 1 As per 10⁶ Si atoms, Nd = 5×10¹⁶ cm⁻³, σ = e Nd μe = 10.81 Ω⁻¹cm⁻¹, giving R = 0.0925 Ω.

## Answer

**Assume:** Si at 300 K, $n_i = 1.45\times10^{10}$ cm$^{-3}$, $\mu_e = 1350$ and $\mu_h = 450$ cm$^2$V$^{-1}$s$^{-1}$. A 1 cm$^3$ cube has $L = 1$ cm and $A = 1$ cm$^2$, so $R$ (in $\Omega$) $= \rho$ (in $\Omega$ cm).

#### Pure silicon
$$
\begin{aligned} \sigma_i &= en_i(\mu_e+\mu_h) = 1.602\times10^{-19}\times1.45\times10^{10}\times(1350+450) \\ &= 4.18\times10^{-6}\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{\rho L}{A} = \frac{1}{\sigma_i}\times\frac{1\ \text{cm}}{1\ \text{cm}^2} = 2.39\times10^{5}\ \Omega \end{aligned}
$$

#### Doped with 1 As in $10^6$ Si atoms
Arsenic is a donor:

$$
\begin{aligned} N_d &= \frac{5\times10^{22}}{10^6} = 5\times10^{16}\ \text{cm}^{-3}, \quad n \approx N_d \\ p &= \frac{n_i^2}{N_d} = \frac{(1.45\times10^{10})^2}{5\times10^{16}} = 4.2\times10^{3}\ \text{cm}^{-3}\ (\text{negligible}) \\ \sigma &= eN_d\mu_e = 1.602\times10^{-19}\times5\times10^{16}\times1350 = 10.81\ \Omega^{-1}\text{cm}^{-1} \\ R &= \frac{1}{10.81} = 0.0925\ \Omega \end{aligned}
$$

The resistance falls by a factor of about $2.6\times10^{6}$. (At this doping, impurity scattering lowers $\mu_e$ to roughly 1000 cm$^2$/Vs in practice, giving about 0.12 $\Omega$; the textbook answer uses the pure-Si mobility.)

**Answer:** Pure Si: $R = 2.39\times10^{5}$ $\Omega$. Doped: $R \approx 0.0925$ $\Omega$.

## Check yourself

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

1. For a 1 cm^3 cube, how is R related to rho?
   A. R in ohm equals rho in ohm cm, since L = 1 cm and A = 1 cm^2
   B. R = rho x 1 cm^3
   C. R = rho / (1 cm^2) only
   D. R is 6 times rho
   Answer: A. R = rho L / A = rho x 1/1. (Key point: Resistance of cube)

2. Which formula gives the conductivity of pure Si?
   A. sigma_i = e ni (mu_e - mu_h)
   B. sigma_i = e ni (mu_e + mu_h)
   C. sigma_i = e Nd mu_e
   D. sigma_i = e ni^2 (mu_e + mu_h)
   Answer: B. In pure Si n = p = ni, so both electrons and holes contribute. (Key point: Intrinsic conductivity)

3. How is Nd found for 1 As in 10^6 Si atoms with 5x10^22 Si per cm^3?
   A. Nd = 5x10^22 x 10^6
   B. Nd = 10^6 / 5x10^22
   C. Nd = 5x10^22 / 10^6 = 5x10^16 cm^-3
   D. Nd = ni^2 / 5x10^22
   Answer: C. One donor per 10^6 atoms means divide the atomic concentration by 10^6. (Key point: Donor concentration)

4. Why can the hole term be ignored after As doping, and what is R about?
   A. p equals Nd, so both terms matter and R is about 2.4x10^5 ohm
   B. Holes have zero mobility so R is 0
   C. p is larger than n, giving R about 10 ohm
   D. p = ni^2/Nd is only about 4x10^3 cm^-3, so sigma = e Nd mu_e and R is about 0.0925 ohm
   Answer: D. Mass action makes minority holes negligible; sigma = 10.8 S/cm gives R about 0.0925 ohm. (Key point: Doped resistance)

## Close variants set in other exams

- [Resistance of Si cube doped with As and B](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-arsenic-boron-billion) — Electrical Engineering Material, chapter 6, asked 6×

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