---
title: "Minority carriers suppressed in n-type"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:ntype-minority-suppressed"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/ntype-minority-suppressed
site: BCT Engineering
---

# Minority carriers suppressed in n-type

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2074 Asoj, 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/ntype-minority-suppressed
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Show that in n-type semiconductor minority carrier concentrations are suppressed.

## Short answer

In an n-type semiconductor, electrons are majority carriers (n≈Nd). From np=ni², the hole concentration p=ni²/Nd is much smaller than ni, so minority carriers are suppressed.

## Answer

In an **n-type** semiconductor, donors (P, As, Sb) supply extra electrons, so electrons are the majority carriers. **Minority carrier suppression** means that the hole concentration becomes much **smaller than the intrinsic value** $n_i$ when donors are added.

#### Proof

**1. Mass action law** (valid in thermal equilibrium for a non-degenerate semiconductor):

$$
np = n_i^2
$$

because $n = N_ce^{-(E_c-E_F)/kT}$ and $p = N_ve^{-(E_F-E_v)/kT}$, so $np = N_cN_ve^{-E_g/kT} = n_i^2$, independent of $E_F$ (i.e. of doping).

**2. Charge neutrality:**

$$
n = p + N_d^+ \approx p + N_d
$$

since all donors are ionised at room temperature.

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

$$
n^2 - N_dn - n_i^2 = 0 \;\Rightarrow\; n = \frac{N_d}{2}+\sqrt{\frac{N_d^2}{4}+n_i^2}
$$

For $N_d \gg n_i$:

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

**4. Compare with intrinsic:**

$$
\frac{p}{n_i} = \frac{n_i}{N_d} \ll 1
$$

So $n$ is raised above $n_i$ by the factor $N_d/n_i$, and $p$ is pushed below $n_i$ by the same factor. **The minority holes are suppressed.**

#### Physical reason

The large number of donor electrons greatly increases the chance that a thermally generated hole meets an electron and recombines. The recombination rate is proportional to $np$, while the generation rate depends only on temperature; equilibrium then requires $np$ to stay at $n_i^2$, so a large $n$ forces a small $p$.

#### Numerical example (Si, 300 K)

| Doping $N_d$ (cm⁻³) | n (cm⁻³) | p (cm⁻³) |
|---|---|---|
| 0 (intrinsic) | $1.45\times10^{10}$ | $1.45\times10^{10}$ |
| $10^{14}$ | $10^{14}$ | $2.1\times10^{6}$ |
| $10^{16}$ | $10^{16}$ | $2.1\times10^{4}$ |

Since $p \ll n$, the conductivity is $\sigma \approx eN_d\mu_e$: it is controlled by the donor concentration only.

## Check yourself

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

1. Which law is the key to showing hole suppression?
   A. np = n_i^2, independent of doping
   B. n + p = n_i
   C. n = p always
   D. np = N_d
   Answer: A. The product np stays at n_i^2 in equilibrium, so raising n lowers p. (Key point: Mass action law)

2. For N_d >> n_i, what is the hole concentration?
   A. N_d
   B. n_i
   C. N_d^2 / n_i
   D. n_i^2 / N_d
   Answer: D. With n about N_d, p = n_i^2/N_d. (Key point: Hole concentration)

3. What does p/n_i equal, and what does it show?
   A. n_i/N_d, much less than 1, so holes are suppressed
   B. N_d/n_i, much greater than 1
   C. 1, so nothing changes
   D. N_d n_i
   Answer: A. p/n_i = n_i/N_d << 1, so p is pushed below its intrinsic value. (Key point: Ratio to intrinsic)

4. What is the physical reason holes are reduced when donors are added?
   A. Donors absorb holes permanently
   B. Holes are swept out by the field
   C. Generation rate falls
   D. Many electrons increase recombination of thermally generated holes while generation stays fixed
   Answer: D. Recombination is proportional to np while generation depends only on temperature, so np stays at n_i^2. (Key point: Physical reason)

## Related in Electrical Engineering Material

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- [Compensated semiconductor: electron and hole concentrations](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Acompensated-carrier-concentration) — Electrical Engineering Material, chapter 6, asked 4×
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- [Minority suppression and donor-defined conduction](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aminority-suppression-impurity-donor) — Electrical Engineering Material, chapter 6
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