---
title: "Donor dopants and σ = neμe"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:donor-conductivity-ne-mu"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/donor-conductivity-ne-mu
site: BCT Engineering
---

# Donor dopants and σ = neμe

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2081 Chaitra (new course), 4 marks; 2075 Asoj, 8 marks; 2073 Chaitra, 6 marks; 2068 Baisakh, 10 marks (asked 4 times)
- Video: [Semiconductors, Insulators & Conductors, Basic Introduction, N type vs P type Semiconductor](https://www.youtube.com/watch?v=ethnHSgVbHs) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/donor-conductivity-ne-mu
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Explain how donor dopants contribute electrons in conduction band in n-type extrinsic semiconductor. Also prove that σ = neμₑ where symbols have their usual meanings.

## Short answer

Pentavalent donor atoms (e.g., P) replace Si, providing a loosely bound fifth electron that ionizes at room temperature, donating a free electron to the conduction band and leaving a fixed positive ion. Conductivity: drift velocity vd = μe E, current density J = n e vd = n e μe E, so σ = n e μe.

## Answer

#### Donor dopants in n-type silicon
An **n-type** semiconductor is made by adding a small amount of a **pentavalent** (group V) impurity such as P, As or Sb to Si or Ge.

1. The donor atom replaces a Si atom in the lattice. Four of its five valence electrons form covalent bonds with the four neighbouring Si atoms.
2. The fifth electron has no bond to join. It stays only weakly bound to the positive donor ion, like the electron in a hydrogen atom but in a medium of high permittivity ($\varepsilon_r = 11.9$) and with a small effective mass.
3. Its binding (ionisation) energy is therefore tiny, about 0.045 eV for P in Si (hydrogen model: $E_b = 13.6\,\frac{m_e^*/m_e}{\varepsilon_r^2}$ eV).
4. In the band diagram this electron occupies a **donor level** $E_d$ just below $E_c$ (about 0.05 eV).
5. At room temperature $kT = 0.026$ eV is enough to ionise almost every donor, so each donor gives one electron to the conduction band and becomes a fixed positive ion $D^+$. Hence $n \approx N_d$.
6. By the mass action law $np = n_i^2$, the hole concentration falls far below $n_i$, so electrons are the majority carriers.

```text
            Ec ----------------------
                 e-  e-  e-   (free)
            Ed  - + - + - + -  ~0.05 eV
                (ionised donors)

            EF  .......... (near Ec)

            Ev ----------------------
```

```text
     Si -- Si -- Si
     |     |     |
     Si -- P+ -- Si    e- (5th electron,
     |     |     |         loosely bound)
     Si -- Si -- Si
```

#### Proof of $\sigma = ne\mu_e$
Consider a bar of n-type material of cross-section $A$, with electron concentration $n$, in an electric field $E$.

1. The field accelerates electrons, but collisions with the lattice limit them to an average **drift velocity** $v_d$ opposite to $E$. The **drift mobility** is defined as

$$
\mu_e = \frac{v_d}{E} \;\Rightarrow\; v_d = \mu_e E
$$

(From the Drude model, $v_d = \frac{e\tau}{m_e^*}E$, so $\mu_e = \frac{e\tau}{m_e^*}$, where $\tau$ is the mean free time.)

2. In time $\Delta t$, electrons travel $v_d\Delta t$. All electrons in the volume $A v_d \Delta t$ cross a given section. Charge crossing:

$$
\Delta Q = e\,n\,A\,v_d\,\Delta t
$$

3. Current density:

$$
J = \frac{\Delta Q}{A\,\Delta t} = n e v_d = n e \mu_e E
$$

4. Ohm's law in point form is $J = \sigma E$. Comparing:

$$
\sigma = n e \mu_e
$$

If holes are also present, their contribution adds: $\sigma = e(n\mu_e + p\mu_h)$. In n-type material $n \gg p$, so $\sigma \approx ne\mu_e \approx N_d e\mu_e$.

Example: Si with $N_d = 10^{16}$ cm$^{-3}$, $\mu_e = 1350$ cm$^2$/Vs gives $\sigma = 10^{16}\times1.602\times10^{-19}\times1350 = 2.16$ S/cm.

## Check yourself

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

1. Why is the fifth electron of a donor atom so weakly bound (about 0.045 eV)?
   A. It sits in a covalent bond with four Si neighbours
   B. The donor nucleus has a smaller charge than Si
   C. It orbits the donor ion like a hydrogen atom, but in a high-permittivity medium with small effective mass
   D. It is trapped in the valence band
   Answer: C. The hydrogen-like binding energy is scaled down by (m*/m) / er^2, with er = 11.9, giving a tiny ionisation energy. (Key point: Why donor electron is loosely bound)

2. Where is the donor level on the band diagram?
   A. Just below the conduction band edge, about 0.05 eV
   B. Just above the valence band edge
   C. At mid-gap
   D. Inside the conduction band
   Answer: A. Because of the small binding energy, Ed lies just below Ec so thermal energy at room temperature ionises almost every donor. (Key point: Donor level position)

3. How is drift mobility defined?
   A. mu = E / v_d
   B. mu = v_d x E
   C. mu = v_d / E, so v_d = mu E
   D. mu = n e v_d
   Answer: C. Mobility is drift velocity per unit electric field. (Key point: Drift velocity and mobility)

4. How does the proof get sigma = n e mu_e?
   A. J = n e v_d = n e mu_e E, then compare with Ohm's law J = sigma E
   B. Compare V = IR with the diffusion current
   C. Use J = n e mu_e / E and set it equal to sigma
   D. Add the diffusion and drift currents and set them to zero
   Answer: A. Charge crossing per unit area per unit time is n e v_d; substituting v_d = mu_e E and comparing with J = sigma E gives sigma = n e mu_e. (Key point: Proof of sigma = n e mu)

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