Extrinsic Semiconductors N-Type

Every silicon transistor you have ever used runs on doped silicon, not pure silicon. N-type doping is what turns an insulator into a conductor that obeys your circuit.

Core Concept

Pure silicon has four valence electrons. When you add a Group V element like phosphorus, each phosphorus atom brings five valence electrons into the lattice. Four of those form covalent bonds with neighbouring silicon atoms. The fifth electron is loosely bound and breaks free at room temperature with very little thermal energy.

This process is called donor doping, because the impurity atom donates a free electron to the crystal. The phosphorus atom becomes a fixed positive ion inside the lattice. It cannot move. Only the freed electron moves, so charge neutrality of the crystal is maintained overall.

The result is a semiconductor where electrons are the majority carriers and holes are the minority carriers. A typical N-type sample doped with ND = 10^16 cm^-3 of phosphorus has an electron concentration of roughly 10^16 cm^-3, while the hole concentration drops to around 10^4 cm^-3 due to the law of mass action. This principle is what makes an NPN transistor like the BC547 work.

Key Equations

Electron concentration in N-type (complete ionisation assumed):

n ≈ ND where ND is the donor concentration in cm^-3

Law of mass action (always holds in thermal equilibrium):

n * p = ni^2 where ni = 1.5 × 10^10 cm^-3 for silicon at 300K

Minority hole concentration in N-type:

p = ni^2 / ND

Fermi level position (N-type shifts Fermi level toward conduction band):

EF - Ei = kT * ln(ND / ni) where kT = 26 meV at 300K and Ei is the intrinsic Fermi level

Given:
  Donor concentration ND = 10^16 cm^-3
  Intrinsic carrier concentration ni = 1.5 × 10^10 cm^-3
  Temperature T = 300K, so kT = 26 meV = 0.026 eV

Why this formula:
  In N-type with complete ionisation, nearly all donor atoms ionise,
  so electron concentration equals donor concentration.

Formula:
  n = ND
  p = ni^2 / ND
  EF - Ei = kT * ln(ND / ni)

Substitution:
  n = 10^16 cm^-3
  p = (1.5 × 10^10)^2 / 10^16
  p = 2.25 × 10^20 / 10^16
  EF - Ei = 0.026 * ln(10^16 / 1.5 × 10^10)

Calculation:
  p = 2.25 × 10^4 cm^-3
  ND / ni = 10^16 / 1.5 × 10^10 = 6.67 × 10^5
  ln(6.67 × 10^5) = 13.41
  EF - Ei = 0.026 × 13.41 = 0.3487 eV

Final Answer:
  Electron concentration n = 10^16 cm^-3
  Hole concentration p = 2.25 × 10^4 cm^-3
  Fermi level is 0.349 eV above the intrinsic level (toward conduction band)

Exam Tip: GATE frequently asks you to find minority carrier concentration in a doped semiconductor. Students forget that n * p = ni^2 always holds at thermal equilibrium regardless of doping. If ND = 10^16 cm^-3, the hole concentration is NOT zero. It is ni^2 / ND = 2.25 × 10^4 cm^-3. Also remember: the donor atom itself becomes a positive ion and is immobile. It does not contribute to conduction. Only the released electron moves.

Key Properties

• Majority carriers in N-type are electrons. Their concentration is approximately equal to ND when ND >> ni.
• Common donor impurities for silicon are Phosphorus (P) and Arsenic (As), both from Group V of the periodic table.
• The Fermi level in N-type silicon lies above the intrinsic Fermi level Ei and moves closer to the conduction band as doping increases.
• At room temperature (300K), donor atoms in silicon are almost completely ionised because their ionisation energy is only about 0.045 eV.
• Resistivity of N-type silicon drops sharply with doping. At ND = 10^16 cm^-3, resistivity is roughly 0.5 Ω·cm, compared to 2300 Ω·cm for intrinsic silicon.
• Minority carrier holes still exist in N-type, generated thermally. Their concentration is ni^2 / ND.
• N-type doping is used for the emitter and collector regions of NPN BJTs like the BC547 and 2N3904.

Quick Revision

• N-type: doped with Group V elements (P, As). Majority carriers are electrons.
• Donor atoms are ionised at room temperature and become fixed positive ions.
• n ≈ ND (for complete ionisation). p = ni^2 / ND.
• ni for silicon = 1.5 × 10^10 cm^-3 at 300K.
• Fermi level shifts toward conduction band as ND increases.
• Crystal remains electrically neutral overall despite free electrons.
• Resistivity decreases as doping concentration increases.
• Exam trap: Students set hole concentration to zero in N-type. It is never zero. Use p = ni^2 / ND to find the minority carrier concentration.