---
title: "p-type minority carrier suppression and conductivity"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:ptype-minority-suppression"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/ptype-minority-suppression
site: BCT Engineering
---

# p-type minority carrier suppression and conductivity

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2076 Asoj, 8 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/ptype-minority-suppression
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Define p-type semiconductor. Derive an expression for minority carrier suppression and hence prove that the conductivity in p-type semiconductor is mainly due to the hole.

## Short answer

p-type: doped with acceptors, majority holes. From np = n_i² and p ≈ N_a, minority electron concentration n = n_i²/N_a ≪ n_i (suppression). Conductivity σ ≈ e N_a μ_h, because hole term dominates; electron contribution is negligible.

## Answer

A **p-type semiconductor** is an intrinsic semiconductor (Si, Ge) doped with a small amount of a **trivalent (group III) impurity** such as boron, aluminium, gallium or indium. Each impurity atom forms only three covalent bonds, so one bond is incomplete; it easily accepts an electron from the valence band, creating a **hole**. Such impurities are called **acceptors** ($N_a$). Holes are the majority carriers and electrons the minority carriers.

```text
 Ec ------------------------
 EFi - - - - - - - - - - - -
 EFp .......................
 Ea  - o - o - o - (acceptors, ~0.05 eV)
 Ev ------------------------
```

#### Derivation of minority carrier suppression

**1. Mass action law.** In equilibrium, for any non-degenerate semiconductor,

$$
np = n_i^2
$$

**2. Charge neutrality.** Positive charges = negative charges:

$$
p = n + N_a^-
$$

At room temperature all acceptors are ionised, $N_a^- = N_a$.

**3. Solve.** Put $n = n_i^2/p$:

$$
p^2 - N_ap - n_i^2 = 0 \;\Rightarrow\; p = \frac{N_a}{2} + \sqrt{\frac{N_a^2}{4}+n_i^2}
$$

For normal doping $N_a \gg n_i$, so

$$
p \approx N_a,\qquad n = \frac{n_i^2}{N_a}
$$

**4. Suppression.** Compare with the intrinsic value:

$$
\frac{n}{n_i} = \frac{n_i}{N_a} \ll 1
$$

Adding acceptors raises $p$ above $n_i$ by the factor $N_a/n_i$ and lowers $n$ below $n_i$ by the same factor. The extra holes recombine with the thermally generated electrons. This reduction of the minority electron concentration is called **minority carrier suppression**.

Example: Si with $N_a = 10^{16}$: $p = 10^{16}$, $n = (1.45\times10^{10})^2/10^{16} = 2.1\times10^{4}\ \text{cm}^{-3}$.

#### Conductivity is due to holes

$$
\sigma = e(n\mu_e + p\mu_h) = e\left(\frac{n_i^2}{N_a}\mu_e + N_a\mu_h\right)
$$

Ratio of the two terms:

$$
\frac{n\mu_e}{p\mu_h} = \frac{n_i^2}{N_a^2}\cdot\frac{\mu_e}{\mu_h}
$$

For the example: $(1.45\times10^{10}/10^{16})^2\times3 \approx 6\times10^{-12}$. Even though $\mu_e \approx 3\mu_h$, the electron term is negligible, so

$$
\sigma \approx eN_a\mu_h
$$

Hence the conductivity of a p-type semiconductor is **almost entirely due to holes** and is fixed by the acceptor concentration.

## Check yourself

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

1. What makes a semiconductor p-type?
   A. A trivalent impurity (B, Al, Ga, In) that accepts an electron, creating holes
   B. A pentavalent donor
   C. Removing silicon atoms
   D. Heating only
   Answer: A. Acceptors leave an incomplete bond, giving a hole. (Key point: p-type doping)

2. Which two equations give the carrier densities?
   A. np = Na and p = n
   B. np = ni^2 and charge neutrality p = n + Na-
   C. n + p = ni
   D. n = p + Na
   Answer: B. Solving gives p about Na and n = ni^2/Na. (Key point: Mass action)

3. What is minority carrier suppression?
   A. n rises above ni
   B. n equals Na
   C. n falls below ni by the factor Na/ni because extra holes recombine with thermal electrons
   D. Holes disappear
   Answer: C. n/ni = ni/Na << 1. (Key point: Suppression)

4. Why is p-type conductivity mostly due to holes?
   A. Because mu_h > mu_e
   B. Because n = Na
   C. Because electrons have zero mobility
   D. The electron term is smaller by (ni/Na)^2 x mu_e/mu_h, about 6x10^-12, so sigma = e Na mu_h
   Answer: D. n mu_e / p mu_h is negligible even though mu_e is about 3 mu_h. (Key point: Conductivity)

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