---
title: "Si cube with As and B per million"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:si-cube-boron-million"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-cube-boron-million
site: BCT Engineering
---

# Si cube with As and B per million

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2070 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-cube-boron-million
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Calculate the resistance of pure silicon cubic crystal of 1 cm³ at room temperature. What will be the resistance of the cube when it is doped with 1 arsenic in 10⁹ silicon atoms and 1 boron atom per million silicon atoms? Atomic concentration of silicon is 5×10²² cm⁻³. Use other required data from above given list.

## Short answer

For pure Si, σi=ni e(μe+μh)=4.176×10⁻⁶ S cm⁻¹, R_i=1/σ_i≈2.39×10⁵ Ω. With As (1 per 10⁹) and B (1 per 10⁶), compensation makes p≈Na−Nd=4.995×10¹⁶ cm⁻³, σ≈e p μh=3.596 S cm⁻¹, R≈0.278 Ω.

## Answer

The cube is 1 cm³, so L = 1 cm, A = 1 cm² and $R = \rho L/A$ equals $\rho$ numerically.

**Data used** (the "list" in the paper): $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$, $\mu_e = 1350$, $\mu_h = 450\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}$, $e = 1.6\times10^{-19}$ C, Si atoms $5\times10^{22}\ \text{cm}^{-3}$.

#### Pure silicon

$$
\begin{aligned}
\sigma_i &= n_ie(\mu_e+\mu_h)\\
&= 1.45\times10^{10}\times1.6\times10^{-19}\times(1350+450)\\
&= 4.176\times10^{-6}\ \text{S cm}^{-1}\\
R_i &= \frac{1}{\sigma_i} = 2.39\times10^{5}\ \Omega
\end{aligned}
$$

#### Doped with As and B

$$
N_d = \frac{5\times10^{22}}{10^9} = 5\times10^{13}\ \text{cm}^{-3},\qquad N_a = \frac{5\times10^{22}}{10^6} = 5\times10^{16}\ \text{cm}^{-3}
$$

$N_a \gg N_d$, so the cube is **p-type** (compensated):

$$
p = N_a - N_d = 4.995\times10^{16}\ \text{cm}^{-3},\qquad n = \frac{n_i^2}{p} = 4.2\times10^{3}\ \text{cm}^{-3}
$$

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

| Sample | Carriers (cm⁻³) | R (Ω) |
|---|---|---|
| Pure Si | $n = p = 1.45\times10^{10}$ | $2.39\times10^{5}$ |
| As only (for comparison) | $n = 5\times10^{13}$ | 92.6 |
| As + B | $p = 4.995\times10^{16}$ | 0.278 |

**Answer: pure cube $R \approx 2.39\times10^{5}\ \Omega$; doped cube $R \approx 0.278\ \Omega$** (about $8.6\times10^5$ times smaller). Boron dominates because it is 1000 times more concentrated than arsenic.

## Check yourself

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

1. Which formula gives the conductivity of pure silicon?
   A. n_i e mu_e
   B. e n_i^2 (mu_e + mu_h)
   C. n_i e (mu_e + mu_h)
   D. n_i (mu_e + mu_h)/e
   Answer: C. In intrinsic material n = p = n_i, so sigma = n_i e (mu_e + mu_h) = 4.18 x 10^-6 S/cm. (Key point: Pure Si conductivity)

2. What is R of the pure 1 cm cube?
   A. 2.39 x 10^5 ohm
   B. 0.278 ohm
   C. 92.6 ohm
   D. 4.18 x 10^-6 ohm
   Answer: A. R = 1/sigma_i = 2.39 x 10^5 ohm for L = 1 cm and A = 1 cm^2. (Key point: Resistance of pure cube)

3. With N_d = 5 x 10^13 and N_a = 5 x 10^16, what is the result?
   A. n-type with n = 5 x 10^13
   B. Intrinsic
   C. p-type with p = 4.995 x 10^16
   D. p-type with p = 5.005 x 10^16
   Answer: C. Compensation gives p = N_a - N_d. (Key point: Net carrier type)

4. Why is the doped cube resistance about 0.278 ohm?
   A. Because electrons dominate with mu_e = 1350
   B. Because holes dominate: sigma = e p mu_h = 3.6 S/cm
   C. Because n_i rises
   D. Because mobility rises with doping
   Answer: B. sigma is about 1.6e-19 x 4.995e16 x 450 = 3.596 S/cm, so R = 0.278 ohm, boron dominating as it is 1000 times more concentrated. (Key point: Doped resistance)

## Related in Electrical Engineering Material

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