Carrier Concentration and Fermi Level

Carrier statistics fix how many electrons/holes a semiconductor has and where the Fermi level sits — the starting point for junction and device analysis. GATE EC tests n, p, mass-action and Fermi-level shifts (1–2 marks).

Key results

• Mass-action law: n·p = n_i². Intrinsic Si n_i ≈ 1.5×10¹⁰ cm⁻³ at 300 K.
• n_i² = N_c N_v e^{−E_g/kT}; n_i rises sharply with temperature.
• n-type: n ≈ N_D, p = n_i²/N_D (N_D >> n_i). p-type: p ≈ N_A, n = n_i²/N_A.
• Fermi level: intrinsic E_F ≈ mid-gap. n-type → E_F toward E_c; p-type → E_F toward E_v.
• E_F − E_i = kT·ln(n/n_i); doping shifts E_F by kT·ln(N/n_i).

Type	Majority	Minority	E_F position
Intrinsic	n = p = n_i	—	mid-gap
n-type	n ≈ N_D	p = n_i²/N_D	near E_c
p-type	p ≈ N_A	n = n_i²/N_A	near E_v

Exam Tricks & Tips

• 🎯 Mass-action n·p = n_i² always holds in equilibrium — find the minority carrier as n_i²/(majority).
• 🎯 In n-type, n ≈ N_D (donor density); minority holes p = n_i²/N_D — often orders of magnitude smaller.
• 🎯 Fermi level moves toward E_c for donors, toward E_v for acceptors; distance = kT·ln(N/n_i).
• 🎯 n_i doubles roughly every ~8–11 °C in Si — temperature dependence dominates via e^{−E_g/2kT}.
• 🎯 A compensated sample: net doping = |N_D − N_A|; the larger determines the type.
• ❌ Common mistake: assuming p = 0 in an n-type sample — minority carriers are small but nonzero (n_i²/N_D).

Expected exam pattern

1–2 marks: compute majority/minority concentrations from doping, apply mass-action, or find the Fermi-level shift / position from n or p.

Quick recap

n·p = n_i² (Si n_i≈1.5×10¹⁰). n-type: n≈N_D, p=n_i²/N_D. E_F toward E_c (n) or E_v (p); shift = kT ln(N/n_i). n_i grows fast with temperature.

Cover the answer, recall, then check. 11 cards on carrier concentration and Fermi level for GATE EC.

Q1. State the mass-action law.
A1. In thermal equilibrium n·p = n_i², independent of doping.

Q2. What is the intrinsic carrier concentration of Si at 300 K?
A2. n_i ≈ 1.5×10¹⁰ cm⁻³.

Q3. For an n-type sample with donor density N_D >> n_i, give n and p.
A3. n ≈ N_D; minority p = n_i²/N_D.

Q4. For a p-type sample with acceptor density N_A, give p and n.
A4. p ≈ N_A; minority n = n_i²/N_A.

Q5. Where is the Fermi level in intrinsic material?
A5. Approximately at mid-gap (E_i).

Q6. Which way does the Fermi level move for n-type and p-type doping?
A6. Toward the conduction band E_c for n-type; toward the valence band E_v for p-type.

Q7. By how much does doping shift the Fermi level from E_i?
A7. E_F − E_i = kT·ln(n/n_i) (positive for n-type, negative for p-type).

Q8. How does n_i depend on temperature and bandgap?
A8. n_i² = N_c N_v e^{−E_g/kT}; n_i rises rapidly with temperature (larger E_g → smaller n_i).

Q9. In a compensated semiconductor, what sets the net type?
A9. The net doping |N_D − N_A|; whichever dopant dominates fixes n- or p-type.

Q10. Are minority carriers ever exactly zero in a doped sample?
A10. No — they equal n_i²/(majority), small but nonzero.

Q11. What happens to a doped semiconductor at very high temperature?
A11. n_i grows until it exceeds the doping, and the material behaves intrinsically.

The Fermi level is the accountant of electrons — its position in the band gap tells you instantly whether a semiconductor is n-type, p-type, or intrinsic, and exactly how many carriers it holds. Doping, temperature, and biasing all move it. GATE loves the mass-action law and the Fermi-level shift formulas because they turn a physics question into a one-line calculation.

Core concept: Carrier concentrations follow from the Fermi level via Boltzmann statistics; the mass-action law n·p = ni² holds at equilibrium, and doping fixes the majority-carrier density.

The theory

Beginner — intrinsic and the mass-action law

In intrinsic material n = p = ni (Si ni ≈ 1.5×10¹⁰ cm⁻³ at 300 K), and the Fermi level EF sits near mid-gap (EFi). At equilibrium, for any doping:
n·p = ni² (mass-action law).
ni² = Nc·Nv·e^(−Eg/kT), so ni rises sharply with temperature and falls with wider Eg.

Intermediate — doping and majority carriers

• n-type (donor density ND): n ≈ ND, and p = ni²/ND (minority holes).
• p-type (acceptor density NA): p ≈ NA, and n = ni²/NA.
  This assumes full ionisation and ND, NA ≫ ni (typical at room temperature). The Fermi level moves toward the conduction band for n-type, toward the valence band for p-type.

Advanced — Fermi-level position and statistics

The exact occupancy uses Fermi-Dirac statistics f(E) = 1/(1 + e^((E−EF)/kT)); for E − EF ≫ kT this reduces to the Boltzmann approximation. Carrier concentrations:

• n = ni·e^((EF − EFi)/kT)
• p = ni·e^((EFi − EF)/kT).
  So the Fermi-level shift from mid-gap directly gives the carrier ratio: EF − EFi = kT·ln(n/ni) = kT·ln(ND/ni) for n-type. At higher temperature EF drifts back toward mid-gap (the material becomes more intrinsic); heavy doping can push EF into the band (degenerate semiconductor), where Boltzmann fails.

Worked example

An n-type Si sample is doped ND = 10¹⁷ cm⁻³ (ni = 1.5×10¹⁰). Find minority hole concentration and the Fermi-level shift above mid-gap at 300 K.
Majority electrons n ≈ ND = 10¹⁷ cm⁻³.
Minority holes p = ni²/ND = (1.5×10¹⁰)²/10¹⁷ = 2.25×10²⁰/10¹⁷ = 2250 cm⁻³.
EF − EFi = kT·ln(ND/ni) = 0.02585·ln(10¹⁷/1.5×10¹⁰) = 0.02585·ln(6.67×10⁶) = 0.02585·15.71 ≈ 0.406 eV above mid-gap.

GATE relevance

1–2 mark NAT/MCQ: majority/minority carrier concentration via mass-action law, Fermi-level position from doping, temperature dependence of ni, and n = ni·e^((EF−EFi)/kT). Directly sets up junction built-in potential calculations.

Exam tricks and shortcuts

• n·p = ni² always at equilibrium — get one carrier, the other follows.
• Majority ≈ doping; minority = ni²/doping (tiny).
• Fermi shift: EF − EFi = kT·ln(majority/ni); positive for n-type, negative for p-type.
• Mnemonic: "n·p is a constant" (mass action) — doping trades one carrier for the other.

Key formulas

• Mass action law: n·p = nᵢ².
• Intrinsic: nᵢ = √(N_c N_v)·e^(−E_g/2kT).
• n-type: n ≈ N_D, p = nᵢ²/N_D; p-type: p ≈ N_A, n = nᵢ²/N_A.
• Fermi level: E_F − E_i = kT ln(n/nᵢ) (n-type, above E_i); E_i − E_F = kT ln(p/nᵢ) (p-type).
• Intrinsic Fermi E_i ≈ midgap.

Doping shifts the Fermi level toward the majority band; the np product stays fixed at nᵢ².

Worked Example

Problem: Silicon at T = 300 K has intrinsic concentration n_i = 1.5 × 10¹⁰ cm⁻³ and thermal voltage kT = 0.026 eV. The sample is doped p-type with N_a = 10¹⁷ cm⁻³. Find the equilibrium hole and electron concentrations and the position of the Fermi level relative to the intrinsic level E_i.

Solution:
For a p-type sample with N_a ≫ n_i and complete ionization, the majority (hole) concentration is:
p ≈ N_a = 10¹⁷ cm⁻³.

Use the mass-action law n·p = n_i² to get the minority (electron) concentration:
n = n_i²/p = (1.5 × 10¹⁰)² / 10¹⁷ = (2.25 × 10²⁰)/10¹⁷ = 2.25 × 10³ = 2250 cm⁻³.

Fermi level position — for p-type material E_F lies below E_i by:
E_i − E_F = kT·ln(p/n_i)
= 0.026 × ln(10¹⁷ / 1.5 × 10¹⁰)
= 0.026 × ln(6.67 × 10⁶)
= 0.026 × 15.71 ≈ 0.408 eV.

Answer: p ≈ 10¹⁷ cm⁻³, n ≈ 2250 cm⁻³, and the Fermi level lies about 0.41 eV below the intrinsic level E_i.