---
title: "Fermi level shift with As then B"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:arsenic-1e15-fermi-shift"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/arsenic-1e15-fermi-shift
site: BCT Engineering
---

# Fermi level shift with As then B

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2080 Bhadra, 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/arsenic-1e15-fermi-shift
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

An n-type Si wafer has been doped uniformly with 10¹⁵ Arsenic atoms per cm³. Calculate the position of the Fermi energy level with respect to the intrinsic Fermi energy level EFi at 27 °C. If this sample is further doped with 2×10¹⁶ Boron atoms per cm³, where will the Fermi level be shifted?

## Short answer

EF – EFi = kT ln(n/ni). With 10¹⁵ As/cm³, n = 10¹⁵ cm⁻³, so EF – EFi ≈ 0.288 eV above intrinsic level. After adding 2×10¹⁶ B/cm³, p = Na – Nd = 1.9×10¹⁶ cm⁻³, EFi – EF = kT ln(p/ni) ≈ 0.364 eV, so EF lies 0.364 eV below EFi.

## Answer

For non-degenerate Si,

$$
E_F - E_{Fi} = kT\ln\frac{n}{n_i}, \qquad E_{Fi} - E_F = kT\ln\frac{p}{n_i}
$$

**Assume:** $T = 300$ K ($27^\circ$C), $kT = 0.02586$ eV, $n_i = 1.45\times10^{10}$ cm$^{-3}$, full ionisation.

#### With $10^{15}$ As cm$^{-3}$
$$
\begin{aligned} n &= N_d = 10^{15}\ \text{cm}^{-3} \\ E_F - E_{Fi} &= 0.02586\ln\frac{10^{15}}{1.45\times10^{10}} = 0.02586\times11.14 \\ &= 0.288\ \text{eV} \end{aligned}
$$

$E_F$ is **0.288 eV above** $E_{Fi}$.

#### After adding $2\times10^{16}$ B cm$^{-3}$
$N_a > N_d$, so the sample is now p-type:

$$
\begin{aligned} p &= N_a - N_d = 2\times10^{16} - 10^{15} = 1.9\times10^{16}\ \text{cm}^{-3} \\ E_{Fi} - E_F &= 0.02586\ln\frac{1.9\times10^{16}}{1.45\times10^{10}} = 0.02586\times14.09 \\ &= 0.364\ \text{eV} \end{aligned}
$$

$E_F$ is **0.364 eV below** $E_{Fi}$, so it has shifted down by $0.288 + 0.364 = 0.652$ eV.

```text
 Ec ---------------------------
     EF (As only) .. +0.288 eV
 Ei - - - - - - - - - - - - - -
     EF (As + B)  .. -0.364 eV
 Ev ---------------------------
```

**Answer:** $E_F - E_{Fi} = 0.288$ eV (n-type); after boron, $E_F - E_{Fi} = -0.364$ eV (p-type).

## Check yourself

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

1. Which formula gives the Fermi level of the As-doped sample?
   A. E_F - E_Fi = kT ln(n/ni) with n = Nd
   B. E_F - E_Fi = kT ln(ni/n)
   C. E_F - E_Fi = kT ln(Nd)
   D. E_F - E_Fi = Eg/2
   Answer: A. n = ni exp((E_F - E_Fi)/kT) for non-degenerate material. (Key point: n-type formula)

2. Where is E_F with 10^15 As per cm^3 at 300 K?
   A. 0.288 eV below E_Fi
   B. 0.288 eV above E_Fi
   C. 0.364 eV above E_Fi
   D. At E_Fi
   Answer: B. kT ln(10^15/1.45x10^10) = 0.02586 x 11.14 = 0.288 eV. (Key point: Result 1)

3. After adding 2x10^16 B per cm^3, what is the hole concentration?
   A. p = Na + Nd = 2.1x10^16
   B. p = Nd = 10^15
   C. p = Na - Nd = 1.9x10^16 cm^-3, p-type
   D. n = Na - Nd, n-type
   Answer: C. Acceptors exceed donors, so the net holes equal Na - Nd. (Key point: Compensation)

4. Where does E_F end up?
   A. 0.364 eV above E_Fi
   B. 0.288 eV below E_Fi
   C. 0.076 eV below E_Fi
   D. 0.364 eV below E_Fi, a shift of 0.652 eV downward
   Answer: D. E_Fi - E_F = kT ln(1.9x10^16/1.45x10^10) = 0.364 eV; total shift 0.288 + 0.364. (Key point: Final Fermi level)

## Close variants set in other exams

- [Si with 10¹⁵ As: resistance and Fermi level](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aarsenic-1e15-resistance-fermi) — Electrical Engineering Material, chapter 6, asked 2×

## Related in Electrical Engineering Material

- [Si with 10¹⁵ Sb: carriers, resistance, Fermi level](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aantimony-1e15-resistance-fermi) — Electrical Engineering Material, chapter 6
- [Si with 10¹⁶ As: carriers, conductivity, Fermi shift](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aarsenic-1e16-carriers-fermi) — Electrical Engineering Material, chapter 6
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