---
title: "Si crystal resistance with As then B doping"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:si-arsenic-boron-resistance"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-arsenic-boron-resistance
site: BCT Engineering
---

# Si crystal resistance with As then B doping

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2071 Chaitra, 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-arsenic-boron-resistance
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Find the resistance of 1 cm³ silicon crystal doped with arsenic, the doping density is such that every Arsenic atom sites every 10⁹ silicon atoms. Atomic concentration of silicon is 5×10²² cm⁻³, ni = 1×10¹⁰ cm⁻³, μₑ = 1350 cm²V⁻¹s⁻¹ and μₕ = 450 cm²V⁻¹s⁻¹. Find the resistance if the above silicon sample is further doped with Boron, the doping density is such that every Boron atom sites every 10⁶ silicon atoms.

## Short answer

Arsenic gives Nd=5×10¹³ cm⁻³ → n≈Nd, σ=e n μe≈1.08×10⁻² S cm⁻¹, R=1/σ≈92.6 Ω. Boron gives Na=5×10¹⁶ cm⁻³; compensation yields p≈Na−Nd=4.995×10¹⁶ cm⁻³, σ≈e p μh≈3.6 S cm⁻¹, R≈0.278 Ω.

## Answer

Take the sample as a **1 cm cube** (L = 1 cm, A = 1 cm²), so $R = \rho L/A$ is numerically equal to $\rho$ in Ω cm.

**Given:** Si atomic concentration $5\times10^{22}\ \text{cm}^{-3}$, $n_i = 10^{10}\ \text{cm}^{-3}$, $\mu_e = 1350$, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $e = 1.6\times10^{-19}$ C.

#### Part 1: Arsenic doped (1 As per $10^9$ Si)

$$
N_d = \frac{5\times10^{22}}{10^9} = 5\times10^{13}\ \text{cm}^{-3}
$$

$$
n \approx N_d = 5\times10^{13},\qquad p = \frac{n_i^2}{n} = \frac{10^{20}}{5\times10^{13}} = 2\times10^{6}\ \text{cm}^{-3}
$$

$$
\begin{aligned}
\sigma &= e(n\mu_e + p\mu_h)\\
&= 1.6\times10^{-19}(5\times10^{13}\times1350 + 2\times10^{6}\times450)\\
&= 1.08\times10^{-2}\ \text{S cm}^{-1}\\
R &= \frac{1}{\sigma} = 92.6\ \Omega
\end{aligned}
$$

#### Part 2: Further doped with boron (1 B per $10^6$ Si)

$$
N_a = \frac{5\times10^{22}}{10^6} = 5\times10^{16}\ \text{cm}^{-3}
$$

Since $N_a \gg N_d$, the sample becomes **p-type**:

$$
p = N_a - N_d = 5\times10^{16} - 5\times10^{13} = 4.995\times10^{16}\ \text{cm}^{-3}
$$

$$
n = \frac{n_i^2}{p} = 2.0\times10^{3}\ \text{cm}^{-3}
$$

$$
\begin{aligned}
\sigma &\approx ep\mu_h = 1.6\times10^{-19}\times4.995\times10^{16}\times450\\
&= 3.596\ \text{S cm}^{-1}\\
R &= \frac{1}{3.596} = 0.278\ \Omega
\end{aligned}
$$

| Sample | Majority carrier (cm⁻³) | R (1 cm cube) |
|---|---|---|
| Pure Si (for reference) | $n = p = 10^{10}$ | $3.47\times10^{5}\ \Omega$ |
| As doped | $n = 5\times10^{13}$ | 92.6 Ω |
| As + B doped | $p = 4.995\times10^{16}$ | 0.278 Ω |

**Answer: with arsenic only, $R \approx 92.6\ \Omega$; after boron doping, $R \approx 0.278\ \Omega$** (the sample is now p-type).

## Check yourself

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

1. One As per 10^9 Si atoms with 5 x 10^22 Si atoms/cm^3 gives N_d of?
   A. 5 x 10^13 cm^-3
   B. 5 x 10^31 cm^-3
   C. 5 x 10^16 cm^-3
   D. 10^9 cm^-3
   Answer: A. N_d = 5 x 10^22 / 10^9 = 5 x 10^13. (Key point: Arsenic doping density)

2. What is the resistance of the 1 cm cube with arsenic only?
   A. 0.278 ohm
   B. 92.6 ohm
   C. 3.47 x 10^5 ohm
   D. 2.2 ohm
   Answer: B. sigma = e n mu_e = 1.08 x 10^-2 S/cm, so R = 92.6 ohm. (Key point: Resistance with arsenic)

3. Why does the sample become p-type after adding 5 x 10^16 boron?
   A. Boron is a donor
   B. Arsenic is removed
   C. n_i increases
   D. N_a is much larger than N_d, so p = N_a - N_d
   Answer: D. Acceptors compensate the donors and exceed them, leaving p = 4.995 x 10^16. (Key point: Type after boron)

4. Which equation gives the final conductivity and what is R?
   A. sigma = e p mu_h, R about 0.278 ohm
   B. sigma = e p mu_e, R about 0.09 ohm
   C. sigma = e n mu_e, R = 92.6 ohm
   D. sigma = e p mu_h, R about 3.6 ohm
   Answer: A. sigma = 1.6e-19 x 4.995e16 x 450 = 3.6 S/cm, so R = 0.278 ohm. (Key point: Resistance after boron)

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