---
title: "Compensated semiconductor: electron and hole concentrations"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:compensated-carrier-concentration"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/compensated-carrier-concentration
site: BCT Engineering
---

# Compensated semiconductor: electron and hole concentrations

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2081 Chaitra (new course), 3 marks; 2076 Asoj, 4 marks; 2070 Asar, 4 marks (asked 3 times)
- Video: [Understanding Compensated Semiconductors: GATE Electronic Devices Numerical 1](https://www.youtube.com/watch?v=gkF8_ruL_Ng) — Ekeeda GATE & ESE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/compensated-carrier-concentration
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

An n-type semiconductor doped with 10¹⁶ cm⁻³ phosphorus atoms has been doped with 10¹⁷ cm⁻³ boron atoms. Calculate the electron and hole concentrations in the semiconductor.

## Short answer

Since Na > Nd, the sample becomes p-type with p = Na - Nd = 9×10^16 cm^-3. Electron concentration n = ni^2/p ≈ 2.3×10^3 cm^-3.

## Answer

When both donors and acceptors are present, they compensate each other and the larger one decides the type.

**Given:** $N_d = 10^{16}$ cm$^{-3}$ (P), $N_a = 10^{17}$ cm$^{-3}$ (B). Assume Si at 300 K, $n_i = 1.45\times10^{10}$ cm$^{-3}$, full ionisation.

Since $N_a > N_d$, the sample becomes **p-type**. The donors' electrons fill $10^{16}$ acceptors; the rest of the acceptors create holes.

$$
\begin{aligned} p &= N_a - N_d = 10^{17} - 10^{16} = 9\times10^{16}\ \text{cm}^{-3} \\ n &= \frac{n_i^2}{p} = \frac{(1.45\times10^{10})^2}{9\times10^{16}} = 2.34\times10^{3}\ \text{cm}^{-3} \end{aligned}
$$

**Answer:** $p = 9\times10^{16}$ cm$^{-3}$, $n \approx 2.3\times10^{3}$ cm$^{-3}$ (the sample is now p-type).

## Check yourself

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

1. 10^16 cm^-3 P donors and 10^17 cm^-3 B acceptors are present. What type is the sample?
   A. n-type, because phosphorus was added first
   B. Intrinsic
   C. n-type with n = 10^17
   D. p-type, because Na > Nd
   Answer: D. The larger doping decides the type; Na exceeds Nd, so the sample is p-type. (Key point: Type after compensation)

2. What is the hole concentration?
   A. 10^17 cm^-3
   B. 9 x 10^16 cm^-3
   C. 1.1 x 10^17 cm^-3
   D. 10^16 cm^-3
   Answer: B. p = Na - Nd = 10^17 - 10^16 = 9 x 10^16 cm^-3. (Key point: Majority concentration)

3. How is the electron concentration found?
   A. n = Nd = 10^16 cm^-3
   B. n = p - ni
   C. n = ni^2 / p, about 2.3 x 10^3 cm^-3
   D. n = ni x p
   Answer: C. By np = ni^2, n = (1.45e10)^2 / 9e16 = 2.3e3 cm^-3. (Key point: Minority from mass action)

## Close variants set in other exams

- [Electron concentration with equal P and B doping](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aequal-doping-electron-concentration) — Electrical Engineering Material, chapter 6, asked 4×

## Related in Electrical Engineering Material

- [Compensated semiconductor concentrations and conductivity](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Acompensated-conductivity) — Electrical Engineering Material, chapter 6
- [Si cube with As then 10¹⁴ boron](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-arsenic-boron-1e14) — Electrical Engineering Material, chapter 6