---
title: "Electron concentration with equal P and B doping"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:equal-doping-electron-concentration"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/equal-doping-electron-concentration
site: BCT Engineering
---

# Electron concentration with equal P and B doping

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2070 Chaitra, 4 marks
- Video: [Comprehensive Guide To Calculating Minority And Majority Carriers | GATE Electronic Devices Part 1](https://www.youtube.com/watch?v=7iW2oo5pypc) — 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/equal-doping-electron-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 10¹⁶ cm⁻³ boron atoms. Calculate the electron concentration in the semiconductor.

## Short answer

Electron concentration equals intrinsic concentration n_i ≈ 1.45×10^10 cm^−3 because equal donor and acceptor doping fully compensates, leaving n = p = n_i.

## Answer

Phosphorus is a donor and boron an acceptor. When both are present, the electrons given by the donors fill the acceptor states — this is **compensation**.

**Given:** $N_d = 10^{16}\ \text{cm}^{-3}$, $N_a = 10^{16}\ \text{cm}^{-3}$, silicon at 300 K, $n_i = 1.45\times10^{10}\ \text{cm}^{-3}$.

#### Charge neutrality and mass action

$$
n + N_a = p + N_d,\qquad np = n_i^2
$$

Since $N_a = N_d$, the neutrality equation gives $n = p$. Then

$$
n^2 = n_i^2 \;\Rightarrow\; n = n_i
$$

$$
n = p = 1.45\times10^{10}\ \text{cm}^{-3}
$$

**Answer: electron concentration $n = n_i \approx 1.45\times10^{10}\ \text{cm}^{-3}$** (equal to the hole concentration).

#### Remarks

- The sample is **fully compensated**: it behaves like intrinsic material in carrier numbers, and its Fermi level returns to $E_{Fi}$.
- It is not identical to pure Si: it contains $2\times10^{16}$ ionised impurities per cm³, which scatter carriers, so the mobilities (and hence conductivity) are lower than in truly intrinsic silicon.

## Check yourself

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

1. What happens when equal donor and acceptor densities are present?
   A. Electrons from donors fill acceptor states, compensating each other
   B. Both double the carriers
   C. The material becomes metallic
   D. Holes are removed only
   Answer: A. This is compensation: donor electrons recombine with acceptor states. (Key point: Compensation)

2. With N_a = N_d, what does n + N_a = p + N_d give?
   A. n = N_d
   B. n = p
   C. p = 0
   D. n = 2N_d
   Answer: B. N_a cancels N_d, so n = p. (Key point: Neutrality equation)

3. What is n?
   A. 10^16 cm^-3
   B. 2 x 10^16 cm^-3
   C. n_i = 1.45 x 10^10 cm^-3
   D. 0
   Answer: C. n = p and np = n_i^2 give n = n_i. (Key point: Electron concentration)

4. How does this sample differ from truly intrinsic Si?
   A. It has more carriers
   B. It has lower mobility because of 2 x 10^16 ionised impurities scattering carriers
   C. Its Fermi level is different
   D. It is p-type
   Answer: B. Carrier numbers equal intrinsic, but ionised impurity scattering reduces mobility and conductivity. (Key point: Difference from pure Si)

## Close variants set in other exams

- [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×

## 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
- [Electron concentration in an extrinsic semiconductor](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aextrinsic-electron-concentration) — Electrical Engineering Material, chapter 6
- [Si crystal resistance with As then B doping](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-arsenic-boron-resistance) — Electrical Engineering Material, chapter 6
- [Si cube doped 1 As in 10⁶](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-arsenic-1e6) — Electrical Engineering Material, chapter 6, asked 5×
- [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
- [Resistance of Si cube doped with As and B](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-cube-arsenic-boron-billion) — Electrical Engineering Material, chapter 6, asked 6×