---
title: "Acceptor dopants and σ = peμh"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:acceptor-conductivity-pe-mu"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/acceptor-conductivity-pe-mu
site: BCT Engineering
---

# Acceptor dopants and σ = peμh

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2082 Kartik (new course), 4 marks
- 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/acceptor-conductivity-pe-mu
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Explain how acceptor dopants contribute holes in valence band in p-type extrinsic semiconductor. Also prove that σ = peμₕ where symbols have their usual meanings.

## Short answer

Trivalent impurities create acceptor levels just above Ev; room‑temperature electrons from the valence band fill these levels, leaving mobile holes. Under electric field E, hole drift current density J = e p v_d = e p μh E; with negligible electron contribution, conductivity σ = J/E = p e μh.

## Answer

#### Acceptors and holes in the valence band
A trivalent impurity (B, Al, Ga) in Si has only three valence electrons, so one of its four covalent bonds is incomplete. This empty bond creates a localized energy level $E_a$ just above $E_v$ (about 0.045 eV for B in Si). At room temperature, an electron from a neighbouring Si–Si bond (the valence band) easily gains this small energy and moves into the empty bond. The boron becomes a fixed negative ion $B^-$, and the missing electron in the valence band is a free **hole** that can move through the crystal.

```text
 Ec -------------------------
 
 Ea  - - - - (-) (-) (-) - -   ionized acceptors
 Ev -----o-----o-----o-------  holes in VB
```

Each ionized acceptor gives one hole and no free electron, so when $N_a \gg n_i$: $p \approx N_a$ and $n = n_i^2/N_a \ll p$.

#### Proof of $\sigma = pe\mu_h$
Let an electric field $E$ act on a p-type sample of cross-section $A$. Holes drift along $E$ with drift velocity $v_{dh} = \mu_h E$. In time $dt$, the holes within a length $v_{dh}dt$ cross the area, carrying charge $dQ = e\,p\,A\,v_{dh}\,dt$. So the hole current density is
$$
J_h = \frac{1}{A}\frac{dQ}{dt} = e\,p\,v_{dh} = e\,p\,\mu_h E
$$
The few electrons contribute $J_e = e\,n\,\mu_e E$ in the same way, so
$$
\sigma = \frac{J_h + J_e}{E} = e\,p\,\mu_h + e\,n\,\mu_e
$$
Since $n \ll p$ in a p-type semiconductor, the electron term is negligible:
$$
\sigma \approx p e \mu_h \approx N_a e \mu_h
$$

## Check yourself

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

1. How does an acceptor make a hole?
   A. It donates an electron
   B. It captures a conduction electron
   C. It splits a bond
   D. A valence electron moves into the incomplete bond, leaving a hole
   Answer: D. The acceptor becomes a fixed negative ion and a hole is left in the valence band. (Key point: Acceptor action)

2. What is the hole drift velocity?
   A. mu_h E
   B. mu_h/E
   C. E/mu_h
   D. e E
   Answer: A. Mobility is velocity per unit field. (Key point: Drift velocity)

3. How does J = e p v follow?
   A. Charge e A dt crosses
   B. Charge e p A v dt crosses area A in dt
   C. Hole count equals Na
   D. Current equals voltage
   Answer: B. Holes in length v dt carry charge e p A v dt, so J = e p mu E. (Key point: Current density)

4. Why is sigma ~ p e mu_h in p-type?
   A. mu_e = 0
   B. e is small
   C. n << p
   D. Holes are heavier
   Answer: C. The electron term e n mu_e is negligible. (Key point: Neglecting electrons)

## Related in Electrical Engineering Material

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