---
title: "Doping to raise Fermi level by 0.1 eV"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:raise-fermi-0-1ev"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/raise-fermi-0-1ev
site: BCT Engineering
---

# Doping to raise Fermi level by 0.1 eV

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2072 Chaitra, 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/raise-fermi-0-1ev
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

If it is desired that the Fermi-level is to be raised to 0.1 eV above intrinsic Fermi-level at room temperature, what type of dopant is to be used? Determine its doping level.

## Short answer

A donor (n-type) dopant must be used. Using n=Nd=ni exp(ΔE/kT) with ΔE=0.1 eV, Nd=1.45×10¹⁰ ×exp(0.1/0.02585)≈6.9×10¹¹ cm⁻³.

## Answer

To raise the Fermi level **above** the intrinsic level, the number of electrons must exceed the number of holes, so a **donor (pentavalent, group V) impurity** such as phosphorus, arsenic or antimony must be added, making the material n-type. (The semiconductor is taken as silicon.)

#### Relation used

For a non-degenerate n-type semiconductor with all donors ionised:

$$
n = N_d = n_i\exp\left(\frac{E_{Fn}-E_{Fi}}{kT}\right)
$$

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

#### Calculation

$$
\begin{aligned}
\frac{E_{Fn}-E_{Fi}}{kT} &= \frac{0.1}{0.02585} = 3.868\\
e^{3.868} &= 47.86\\
N_d &= 1.45\times10^{10}\times47.86\\
&= 6.94\times10^{11}\ \text{cm}^{-3}
\end{aligned}
$$

**Answer: a donor (n-type) dopant is needed, with $N_d \approx 6.9\times10^{11}\ \text{cm}^{-3}$** (about 1 donor per $7\times10^{10}$ Si atoms).

#### Check

$N_d$ is about 48 times $n_i$, so $n \approx N_d$ is a fair approximation. (Solving exactly with $n = N_d/2 + \sqrt{N_d^2/4 + n_i^2}$ changes the result by well under 1%.)

```text
 Ec ---------------------------
 EFn .......................... 0.1 eV above EFi
 EFi - - - - - - - - - - - - - -
 Ev ---------------------------
```

The Fermi level rises logarithmically with doping: each tenfold increase in $N_d$ raises $E_F$ by $kT\ln10 \approx 0.06$ eV.

## Check yourself

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

1. Which dopant raises the Fermi level above E_Fi?
   A. Acceptor such as boron
   B. Donor such as phosphorus
   C. Any dopant
   D. Neither, doping cannot move E_F
   Answer: B. Donors add electrons, making the material n-type with E_F above E_Fi. (Key point: Choice of dopant)

2. Which relation links doping to the Fermi shift here?
   A. N_d = n_i exp(-(E_Fn - E_Fi)/kT)
   B. N_d = n_i (E_Fn - E_Fi)/kT
   C. N_d = n_i exp((E_Fn - E_Fi)/kT)
   D. N_d = n_i^2 exp((E_Fn-E_Fi)/kT)
   Answer: C. n = N_d = n_i exp((E_Fn - E_Fi)/kT). (Key point: Relation used)

3. What donor concentration gives 0.1 eV shift at 300 K in Si?
   A. About 6.9 x 10^11 cm^-3
   B. About 6.9 x 10^16 cm^-3
   C. About 1.45 x 10^10 cm^-3
   D. About 4.8 x 10^1 cm^-3
   Answer: A. exp(0.1/0.02585) = 47.9, so N_d = 1.45 x 10^10 x 47.9 = 6.94 x 10^11. (Key point: Doping level)

4. How much does E_F rise for each tenfold increase in N_d?
   A. About 0.6 eV
   B. About 0.0059 eV
   C. About 0.06 eV (kT ln 10)
   D. It does not change
   Answer: C. The shift is logarithmic: kT ln10 is about 0.06 eV. (Key point: Doping versus shift)

## Close variants set in other exams

- [Doping to raise Fermi level by 0.7 eV](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Araise-fermi-0-7ev) — Electrical Engineering Material, chapter 6, asked 2×

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