---
title: "Si with 10¹⁶ As: carriers, conductivity, Fermi shift"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:arsenic-1e16-carriers-fermi"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/arsenic-1e16-carriers-fermi
site: BCT Engineering
---

# Si with 10¹⁶ As: carriers, conductivity, Fermi shift

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2073 Shrawan, 6 marks
- Video: [ECE Purdue Semiconductor Fundamentals L1.5: Materials Properties - Free Carriers in Semiconductor](https://www.youtube.com/watch?v=TKCWJLVIwoE) — nanohubtechtalks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/arsenic-1e16-carriers-fermi
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

A silicon ingot is doped with 10¹⁶ arsenic atoms/cm³. Find the carrier concentrations, conductivity of the sample and the shift in Fermi level from its intrinsic Fermi level at 27°C.

## Short answer

n≈Nd=10¹⁶ cm⁻³, p=ni²/n≈2.1×10⁴ cm⁻³. Conductivity σ=e(nμe+pμh)≈2.16 S cm⁻¹. Fermi level shift ΔE=kT ln(n/ni)=0.348 eV above intrinsic.

## Answer

Arsenic is a group V **donor**, so the ingot is n-type.

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

#### Carrier concentrations

All donors are ionised at 27°C and $N_d = 10^{16} \gg n_i$:

$$
\begin{aligned}
n &\approx N_d = 10^{16}\ \text{cm}^{-3}\\
p &= \frac{n_i^2}{n} = \frac{(1.45\times10^{10})^2}{10^{16}} = 2.1\times10^{4}\ \text{cm}^{-3}
\end{aligned}
$$

#### Conductivity

$$
\begin{aligned}
\sigma &= e(n\mu_e + p\mu_h)\\
&= 1.6\times10^{-19}(10^{16}\times1350 + 2.1\times10^{4}\times450)\\
&\approx 1.6\times10^{-19}\times1.35\times10^{19}\\
&= 2.16\ \text{S cm}^{-1}\ (= 216\ \text{S m}^{-1})
\end{aligned}
$$

Resistivity $\rho = 1/\sigma = 0.463\ \Omega$ cm. (If the doped-Si mobility from Kasap's graph, $\mu_e \approx 1200$, is used, $\sigma \approx 1.92$ S cm⁻¹.)

#### Fermi level shift

$$
\begin{aligned}
E_{Fn}-E_{Fi} &= kT\ln\frac{n}{n_i} = 0.02585\ln\frac{10^{16}}{1.45\times10^{10}}\\
&= 0.02585\times13.44 = 0.348\ \text{eV}
\end{aligned}
$$

**Answer: $n = 10^{16}\ \text{cm}^{-3}$, $p = 2.1\times10^{4}\ \text{cm}^{-3}$, $\sigma \approx 2.16$ S cm⁻¹, and the Fermi level lies 0.348 eV above the intrinsic Fermi level.**

```text
 Ec ------------------------
 EFn ....................... (0.348 eV above EFi)
 EFi - - - - - - - - - - - -
 Ev ------------------------
```

## Check yourself

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

1. What type is silicon doped with arsenic and what is n?
   A. p-type, n = 2.1 x 10^4
   B. n-type, n = 10^16 cm^-3
   C. n-type, n = 1.45 x 10^10
   D. p-type, n = 10^16
   Answer: B. Arsenic is a group V donor, fully ionised, so n is about N_d = 10^16. (Key point: Type of semiconductor)

2. How is p found?
   A. p = n n_i
   B. p = N_d - n_i
   C. p = n_i^2 / n = 2.1 x 10^4 cm^-3
   D. p = n_i
   Answer: C. Mass action law np = n_i^2. (Key point: Hole concentration)

3. Which term dominates sigma = e(n mu_e + p mu_h) and what is sigma?
   A. Hole term; 0.1 S/cm
   B. Electron term; about 2.16 S/cm
   C. Electron term; about 216 S/cm
   D. Both equal; 1 S/cm
   Answer: B. n mu_e is about 1.35 x 10^19, giving sigma = 1.6 x 10^-19 x 1.35 x 10^19 = 2.16 S/cm. (Key point: Conductivity)

4. What is E_Fn - E_Fi?
   A. 0.348 eV above E_Fi
   B. 0.348 eV below E_Fi
   C. 0.463 eV above E_Fi
   D. 0.02585 eV above E_Fi
   Answer: A. kT ln(10^16 / 1.45 x 10^10) = 0.02585 x 13.44 = 0.348 eV. (Key point: Fermi level shift)

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