---
title: "Electron concentration in an extrinsic semiconductor"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:extrinsic-electron-concentration"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/extrinsic-electron-concentration
site: BCT Engineering
---

# Electron concentration in an extrinsic semiconductor

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2071 Shrawan, 6 marks
- Video: [Charge Carrier Concentration of Doped Semiconductors](https://www.youtube.com/watch?v=icIXyp6GD0w) — Jordan Edmunds Chetty
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/extrinsic-electron-concentration
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Derive the relation for finding the concentration of electron in an extrinsic semi-conductor.

## Short answer

Charge neutrality n+N_a = p+N_d and mass‑action np = n_i² give n² − (N_d−N_a)n − n_i² = 0. The positive root is n = (N_d−N_a)/2 + √[((N_d−N_a)/2)² + n_i²]. For normal n‑type doping (N_d >> n_i), n ≈ N_d.

## Answer

In an extrinsic semiconductor the electron concentration is found from two conditions: **charge neutrality** and the **mass action law**. The result is valid for n-type, p-type and compensated material.

#### Step 1: Carrier concentrations in terms of $E_F$

For a non-degenerate semiconductor,

$$
n = N_c\exp\left[-\frac{E_c-E_F}{kT}\right],\qquad p = N_v\exp\left[-\frac{E_F-E_v}{kT}\right]
$$

#### Step 2: Mass action law

Multiplying,

$$
np = N_cN_v\exp\left(-\frac{E_g}{kT}\right) = n_i^2
$$

This product does not depend on $E_F$, so it holds for doped material as well.

#### Step 3: Charge neutrality

With $N_d$ donors and $N_a$ acceptors, all ionised at room temperature:

$$
n + N_a = p + N_d
$$

#### Step 4: Solve for n

Put $p = n_i^2/n$:

$$
n + N_a = \frac{n_i^2}{n} + N_d \;\Rightarrow\; n^2 - (N_d - N_a)n - n_i^2 = 0
$$

Taking the positive root:

$$
n = \frac{N_d-N_a}{2} + \sqrt{\left(\frac{N_d-N_a}{2}\right)^2 + n_i^2},\qquad p = \frac{n_i^2}{n}
$$

#### Special cases

| Case | Electron concentration |
|---|---|
| n-type, $N_a = 0$, $N_d \gg n_i$ | $n \approx N_d$ |
| Compensated, $N_d - N_a \gg n_i$ | $n \approx N_d - N_a$ |
| p-type, $N_a \gg N_d$ | $n \approx n_i^2/(N_a-N_d)$ |
| $N_d = N_a$ or undoped | $n = n_i$ |

#### Fermi level form

Dividing $n$ by $n_i = N_ce^{-(E_c-E_{Fi})/kT}$:

$$
n = n_i\exp\left(\frac{E_F-E_{Fi}}{kT}\right) \;\Rightarrow\; E_F - E_{Fi} = kT\ln\frac{n}{n_i}
$$

**Example:** Si, $N_d = 10^{16}$, $N_a = 0$: $n = 10^{16}$ cm⁻³, $p = 2.1\times10^{4}$ cm⁻³, $E_F - E_{Fi} = 0.348$ eV.

At low temperature (freeze-out) the donors are not fully ionised and $n \approx \sqrt{N_cN_d/2}\,e^{-\Delta E/2kT}$, where $\Delta E = E_c - E_d$; the formula above applies to the normal extrinsic range.

## Check yourself

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

1. Why does np = ni^2 also hold for doped material?
   A. Doping does not change n
   B. Doping does not change p
   C. Because ni changes with Nd
   D. The product is independent of E_F
   Answer: D. Multiplying the n and p expressions cancels E_F, leaving Nc Nv exp(-Eg/kT). (Key point: Mass action law independence)

2. Which charge-neutrality equation applies with Nd donors and Na acceptors?
   A. n + Na = p + Nd
   B. n + Nd = p + Na
   C. n = p + Nd + Na
   D. n + p = Nd + Na
   Answer: A. Negative charges (electrons and ionised acceptors) equal positive charges (holes and ionised donors). (Key point: Charge neutrality)

3. Which quadratic follows after putting p = ni^2/n?
   A. n^2 + (Nd - Na) n + ni^2 = 0
   B. n^2 - (Nd - Na) n - ni^2 = 0
   C. n^2 - (Nd + Na) n - ni^2 = 0
   D. n^2 - ni^2 = Nd - Na
   Answer: B. n + Na = ni^2/n + Nd rearranges to n^2 - (Nd - Na)n - ni^2 = 0. (Key point: Quadratic result)

4. For a compensated sample with Nd - Na >> ni, n is approximately:
   A. Nd + Na
   B. ni
   C. Nd - Na
   D. ni^2/(Nd - Na)
   Answer: C. The square-root term gives n ~ Nd - Na. (Key point: Special case)

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