---
title: "Si cube with As then 10¹⁴ boron"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:si-cube-arsenic-boron-1e14"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-cube-arsenic-boron-1e14
site: BCT Engineering
---

# Si cube with As then 10¹⁴ boron

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2078 Kartik, 6 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-arsenic-boron-1e14
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Find the resistance of a cubic pure silicon crystal. Find the resistance when the Si-crystal is doped with one Arsenic atom in 10⁹ Silicon atoms. If the sample is further doped with 10¹⁴ Boron atoms what will be the new resistance?

## Short answer

Assuming 1 cm cube, pure Si R ≈ 2.39×10⁵ Ω. Doping with 1 As per 10⁹ Si yields Nd = 5×10¹³ cm⁻³, R ≈ 92.6 Ω. Adding 10¹⁴ B/cm³ gives majority holes, p = Na – Nd = 5×10¹³ cm⁻³, σ ≈ e p μh = 3.6×10⁻³ S/cm, R ≈ 277.8 Ω.

## Answer

The size of the cube is not stated; take the usual **1 cm³ cube** (L = 1 cm, A = 1 cm²), so $R = \rho L/A$ is numerically equal to $\rho$. The boron dose is taken as $10^{14}$ atoms per cm³. Atomic concentration of Si = $5\times10^{22}\ \text{cm}^{-3}$.

**Data used (Si, 300 K):** $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$, $kT = 8.617\times10^{-5}\times300 = 0.02585\ \text{eV}$, $\mu_e = 1350$, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $e = 1.6\times10^{-19}\ \text{C}$.

#### (a) Pure silicon

$$
\begin{aligned}
\sigma_i &= n_ie(\mu_e+\mu_h) = 1.45\times10^{10}\times1.6\times10^{-19}\times1800\\
&= 4.176\times10^{-6}\ \text{S cm}^{-1}\\
R_i &= 1/\sigma_i = 2.39\times10^{5}\ \Omega
\end{aligned}
$$

#### (b) 1 As in $10^9$ Si atoms

$$
N_d = \frac{5\times10^{22}}{10^9} = 5\times10^{13}\ \text{cm}^{-3}, \quad n \approx N_d,\quad p = \frac{n_i^2}{N_d} = 4.2\times10^{6}\ \text{cm}^{-3}
$$

$$
\begin{aligned}
\sigma &\approx eN_d\mu_e = 1.6\times10^{-19}\times5\times10^{13}\times1350\\
&= 1.08\times10^{-2}\ \text{S cm}^{-1}\\
R &= 92.6\ \Omega
\end{aligned}
$$

#### (c) Further doped with $10^{14}$ boron cm⁻³

Boron is an acceptor. Since $N_a > N_d$ the sample becomes **p-type** (compensated):

$$
p = N_a - N_d = 10^{14} - 5\times10^{13} = 5\times10^{13}\ \text{cm}^{-3},\quad n = \frac{n_i^2}{p} = 4.2\times10^{6}\ \text{cm}^{-3}
$$

$$
\begin{aligned}
\sigma &\approx ep\mu_h = 1.6\times10^{-19}\times5\times10^{13}\times450\\
&= 3.6\times10^{-3}\ \text{S cm}^{-1}\\
R &= 277.8\ \Omega
\end{aligned}
$$

| Sample | Carriers (cm⁻³) | R (Ω) |
|---|---|---|
| Pure Si | $n=p=1.45\times10^{10}$ | $2.39\times10^{5}$ |
| As doped | $n = 5\times10^{13}$ | 92.6 |
| As + B | $p = 5\times10^{13}$ | 277.8 |

**Answer: $R_{pure} \approx 2.39\times10^5\ \Omega$, $R_{As} \approx 92.6\ \Omega$, $R_{As+B} \approx 277.8\ \Omega$.** The resistance rises after boron doping because the carrier number is the same but holes have a lower mobility than electrons.

## Check yourself

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

1. What is R of the pure 1 cm cube?
   A. 2.39x10^5 ohm
   B. 92.6 ohm
   C. 277.8 ohm
   D. 4.18x10^-6 ohm
   Answer: A. R = 1/(e ni (mu_e + mu_h)). (Key point: Pure Si)

2. What is the As-doped resistance?
   A. 2.39x10^5 ohm
   B. 92.6 ohm, with Nd = 5x10^13 cm^-3 and sigma = e Nd mu_e
   C. 277.8 ohm
   D. 9.26 ohm
   Answer: B. 1.6e-19 x 5e13 x 1350 = 1.08e-2 S/cm. (Key point: As doped)

3. After adding 10^14 B per cm^3 to the As-doped sample, what are the type and carrier density?
   A. n-type with n = 1.5x10^14
   B. p-type with p = 10^14
   C. p-type with p = Na - Nd = 5x10^13 cm^-3
   D. Intrinsic
   Answer: C. Acceptors exceed donors by 5x10^13. (Key point: Compensation)

4. Why does R rise to 277.8 ohm even though the carrier number is the same?
   A. Holes have higher mobility
   B. Boron reduces the number of carriers to zero
   C. Resistance does not change
   D. Holes have lower mobility (450) than electrons (1350)
   Answer: D. Same carrier number but mu_h is one third of mu_e, so conductivity drops 3 times. (Key point: Resistance rise)

## Related in Electrical Engineering Material

- [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×
- [Si crystal resistance with As then B doping](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-arsenic-boron-resistance) — Electrical Engineering Material, chapter 6
- [Resistance of Si cube doped with Sb](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-antimony) — Electrical Engineering Material, chapter 6
- [Si cube doped 1 As in 10⁶](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-arsenic-1e6) — Electrical Engineering Material, chapter 6, asked 5×
- [Resistance of 8 cm³ Si cube](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-8cm3) — Electrical Engineering Material, chapter 6, asked 5×
- [Si cube with As and B per million](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-boron-million) — Electrical Engineering Material, chapter 6