---
title: "Resistance of Si cube doped with Sb"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:si-cube-antimony"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-cube-antimony
site: BCT Engineering
---

# Resistance of Si cube doped with Sb

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2080 Bhadra, 8 marks
- 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-antimony
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Find the resistance of a cubic pure Si-Crystals at 300K. If this silicon sample is doped with Sb (one Sb in 10⁹ Si-atoms), what will be the new change in its resistance? Take atomic concentration = 5×10²² cm⁻³.

## Short answer

Pure Si: σi = e ni (μe+μh) = 4.18×10⁻⁶ S/cm, R = 1/σi ≈ 2.39×10⁵ Ω. Doping with 1 Sb per 10⁹ Si atoms gives Nd = 5×10¹³ cm⁻³, n ≈ Nd, σ ≈ e Nd μe = 1.08×10⁻² S/cm, R ≈ 92.5 Ω. Resistance drops about 2600 times.

## Answer

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

#### 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 Sb per $10^9$ Si atoms
Sb is a donor:

$$
\begin{aligned} N_d &= \frac{5\times10^{22}}{10^9} = 5\times10^{13}\ \text{cm}^{-3}, \quad n \approx N_d \\ p &= \frac{n_i^2}{N_d} = 4.2\times10^{6}\ \text{cm}^{-3}\ (\text{negligible}) \\ \sigma &= eN_d\mu_e = 1.602\times10^{-19}\times5\times10^{13}\times1350 = 1.081\times10^{-2}\ \Omega^{-1}\text{cm}^{-1} \\ R' &= \frac{1}{\sigma} = 92.5\ \Omega \end{aligned}
$$

#### Change in resistance
$$
\Delta R = 2.39\times10^{5} - 92.5 \approx 2.39\times10^{5}\ \Omega \ (\text{decrease}), \qquad \frac{R}{R'} = 2586
$$

**Answer:** Pure Si: $2.39\times10^{5}$ $\Omega$; doped: 92.5 $\Omega$. Resistance drops by about $2.39\times10^5$ $\Omega$ (about 2600 times), even though only one atom in a billion is replaced.

## Check yourself

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

1. What is the resistance of the 1 cm cube of pure Si and how is it found?
   A. R = 1/(e ni (mu_e + mu_h)) = 2.39x10^5 ohm
   B. R = 1/(e ni mu_e) = 4.6x10^4 ohm
   C. R = e ni (mu_e + mu_h) = 4.18x10^-6 ohm
   D. R = 92.5 ohm
   Answer: A. sigma_i = 4.18x10^-6 S/cm so R = 2.39x10^5 ohm. (Key point: Intrinsic R)

2. For 1 Sb per 10^9 Si atoms, what is Nd?
   A. 5x10^31 cm^-3
   B. 5x10^13 cm^-3
   C. 5x10^22 cm^-3
   D. 2x10^-10 cm^-3
   Answer: B. Nd = 5x10^22 / 10^9 = 5x10^13 cm^-3. (Key point: Donor concentration)

3. What is the resistance after doping and why?
   A. 2.39x10^5 ohm, since 1 in 10^9 changes nothing
   B. 925 ohm, using mu_h
   C. 92.5 ohm, because sigma = e Nd mu_e = 1.08x10^-2 S/cm
   D. 0.0925 ohm
   Answer: C. Electrons from Sb dominate conduction. (Key point: Doped R)

4. By what factor does the resistance fall?
   A. About 2 times
   B. About 10^9 times
   C. It rises about 2600 times
   D. About 2600 times, even with only one atom in a billion replaced
   Answer: D. R/R' = 2.39x10^5 / 92.5 = 2586. (Key point: Size of change)

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