---
title: "Fermi level with 10¹⁶ Sb then 2×10¹⁷ B"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:antimony-1e16-boron-2e17"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/antimony-1e16-boron-2e17
site: BCT Engineering
---

# Fermi level with 10¹⁶ Sb then 2×10¹⁷ B

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2069 Asar, 6 marks; 2068 Shrawan, 8 marks (asked 2 times)
- Video: [Fermi Energy Level in Intrinsic and Extrinsic Semiconductors](https://www.youtube.com/watch?v=SxLX65vVMtg) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/antimony-1e16-boron-2e17
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

An n type silicon wafer is uniformly doped with 10¹⁶ antimony atoms per cm³. Where will be the Fermi level compared to its intrinsic Fermi level? Where will Fermi level be shifted if the sample is further doped with 2×10¹⁷ boron atoms per cm³?

## Short answer

Initially EF – EFi = 0.02586 ln(10¹⁶/1.45×10¹⁰) = 0.348 eV above. After additional 2×10¹⁷ cm⁻³ boron, p = 1.9×10¹⁷ cm⁻³, EFi – EF = 0.02586 ln(p/ni) = 0.424 eV below.

## Answer

For a non-degenerate semiconductor, the Fermi level shift from the intrinsic level is

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

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

#### (a) Doped with $10^{16}$ Sb cm$^{-3}$
Sb is a donor, so $n = N_d = 10^{16}$ cm$^{-3}$.

$$
\begin{aligned} E_F - E_{Fi} &= 0.02586\ln\frac{10^{16}}{1.45\times10^{10}} \\ &= 0.02586\times13.44 = 0.348\ \text{eV} \end{aligned}
$$

$E_F$ is **0.348 eV above** $E_{Fi}$.

#### (b) Further doped with $2\times10^{17}$ B cm$^{-3}$
B is an acceptor. Since $N_a > N_d$, the sample becomes **p-type**:

$$
\begin{aligned} p &= N_a - N_d = 2\times10^{17} - 10^{16} = 1.9\times10^{17}\ \text{cm}^{-3} \\ E_{Fi} - E_F &= 0.02586\ln\frac{1.9\times10^{17}}{1.45\times10^{10}} \\ &= 0.02586\times16.39 = 0.424\ \text{eV} \end{aligned}
$$

$E_F$ is now **0.424 eV below** $E_{Fi}$; it has moved down by $0.348 + 0.424 = 0.772$ eV.

```text
 Ec ---------------------------
     EF (a) ....  0.348 eV above
 Ei - - - - - - - - - - - - - -
     EF (b) ....  0.424 eV below
 Ev ---------------------------
```

**Answer:** (a) $E_F - E_{Fi} = +0.348$ eV; (b) $E_F - E_{Fi} = -0.424$ eV (p-type).

## Check yourself

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

1. How is the Fermi level shift in n-type material found?
   A. EF - EFi = kT ln(n/ni)
   B. EF - EFi = kT ln(ni/n)
   C. EF - EFi = kT (n/ni)
   D. EF - EFi = ln(n) / kT
   Answer: A. From n = ni exp[(EF - EFi)/kT]. (Key point: Fermi shift formula)

2. With 10^16 cm^-3 Sb, where is EF relative to EFi?
   A. 0.348 eV below
   B. 0.348 eV above
   C. At the same level
   D. 0.424 eV above
   Answer: B. Sb is a donor, n = 10^16, and 0.02586 ln(10^16/1.45e10) = 0.348 eV above EFi. (Key point: Antimony doped value)

3. After adding 2 x 10^17 cm^-3 B, what is the hole concentration?
   A. 2 x 10^17 cm^-3
   B. 2.1 x 10^17 cm^-3
   C. 10^16 cm^-3
   D. 1.9 x 10^17 cm^-3
   Answer: D. Na > Nd, so p = Na - Nd = 2e17 - 1e16 = 1.9e17 cm^-3 and the sample is p-type. (Key point: Net doping after boron)

4. Where is EF now, and how far has it moved?
   A. 0.424 eV below EFi; it moved down by 0.772 eV
   B. 0.424 eV above EFi; it moved down by 0.348 eV
   C. 0.348 eV below EFi; it moved down by 0.348 eV
   D. At EFi; moved down 0.348 eV
   Answer: A. EFi - EF = 0.02586 ln(1.9e17/1.45e10) = 0.424 eV, so total shift is 0.348 + 0.424 = 0.772 eV. (Key point: Fermi level after compensation)

## Close variants set in other exams

- [Fermi level with 10¹⁰ Sb then 2×10¹⁵ B](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aantimony-1e10-boron-2e15) — Electrical Engineering Material, chapter 6, asked 6×
- [Fermi level with 10¹⁷ Sb then further doping](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aantimony-1e17-fermi-shift) — Electrical Engineering Material, chapter 6, asked 6×
- [Fermi level with 10¹⁶ B then 10¹⁷ Sb](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aboron-1e16-antimony-1e17) — Electrical Engineering Material, chapter 6, asked 5×
- [Fermi level of Si with 10¹⁶ Sb](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aantimony-1e16-fermi) — Electrical Engineering Material, chapter 6, asked 5×

## 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
- [Fermi level shift with As then B](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aarsenic-1e15-fermi-shift) — Electrical Engineering Material, chapter 6, asked 2×
- [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×
- [Si cube with As then 10¹⁴ boron](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-arsenic-boron-1e14) — 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