Extrinsic Semiconductors P-Type

The base region of every PNP transistor and the anode of every 1N4007 diode is made of P-type silicon. Without P-type doping, the p-n junction does not exist and rectification is impossible.

Core Concept

Boron has only three valence electrons. When it sits inside a silicon lattice, it can only form three of the four covalent bonds that silicon needs. The fourth bond has a missing electron. This vacancy is called a hole, and it behaves like a positive charge carrier.

The boron atom is called an acceptor impurity because it accepts a nearby electron to complete its bond. When it does, it becomes a fixed negative ion. The electron it accepted came from a neighbouring bond, which now has a hole. The hole appears to move in the opposite direction to electron motion.

In P-type silicon, holes are the majority carriers and electrons are minority carriers. This is the material used for the collector and base of PNP transistors and the anode side of rectifier diodes like the 1N4007. A typical P-type sample with NA = 10^15 cm^-3 of boron has a hole concentration of about 10^15 cm^-3 and an electron concentration of only about 2.25 × 10^5 cm^-3.

Key Equations

Hole concentration in P-type (complete ionisation):

p ≈ NA where NA is the acceptor concentration in cm^-3

Law of mass action (holds at thermal equilibrium):

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

Minority electron concentration:

n = ni^2 / NA

Fermi level position (P-type shifts Fermi level toward valence band):

Ei - EF = kT * ln(NA / ni) where kT = 26 meV at 300K

Given:
  Acceptor concentration NA = 5 × 10^15 cm^-3
  Intrinsic carrier concentration ni = 1.5 × 10^10 cm^-3
  Temperature T = 300K, kT = 0.026 eV

Why this formula:
  Acceptors are fully ionised at 300K, so hole concentration
  equals the acceptor doping level.

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

Substitution:
  p = 5 × 10^15 cm^-3
  n = (1.5 × 10^10)^2 / (5 × 10^15)
  Ei - EF = 0.026 * ln(5 × 10^15 / 1.5 × 10^10)

Calculation:
  n = 2.25 × 10^20 / 5 × 10^15 = 4.5 × 10^4 cm^-3
  NA / ni = 5 × 10^15 / 1.5 × 10^10 = 3.33 × 10^5
  ln(3.33 × 10^5) = 12.72
  Ei - EF = 0.026 × 12.72 = 0.3307 eV

Final Answer:
  Hole concentration p = 5 × 10^15 cm^-3
  Electron concentration n = 4.5 × 10^4 cm^-3
  Fermi level is 0.331 eV below intrinsic level (toward valence band)

Exam Tip: GATE problems on P-type semiconductors often ask for Fermi level position. The formula is Ei - EF = kT * ln(NA / ni), NOT EF - Ei. The sign matters. In P-type, EF is below Ei, so Ei - EF is positive. Also remember: holes are majority carriers but they are NOT the ions. The boron ions are fixed. Only the holes move. Students confuse the fixed acceptor ion with the mobile hole in MCQ options.

Key Properties

• Majority carriers in P-type are holes, with concentration p ≈ NA at room temperature.
• Common acceptor dopants for silicon are Boron (B) and Gallium (Ga), both Group III elements.
• The Fermi level in P-type silicon lies below the intrinsic Fermi level Ei, closer to the valence band.
• Acceptor ionisation energy in silicon is about 0.045 eV for boron, so complete ionisation occurs at 300K.
• Minority electron concentration is ni^2 / NA. For NA = 10^15 cm^-3, this is only 2.25 × 10^5 cm^-3.
• P-type silicon forms the anode of p-n junction diodes and the base and collector of PNP transistors like the 2N3906.
• Hole mobility in silicon (μp ≈ 450 cm^2/V·s) is lower than electron mobility (μn ≈ 1350 cm^2/V·s), so P-type is slightly more resistive at the same doping.

Quick Revision

• P-type: doped with Group III elements (B, Ga). Majority carriers are holes.
• Acceptor atoms become fixed negative ions after accepting an electron.
• p ≈ NA, n = ni^2 / NA at thermal equilibrium.
• Fermi level is below Ei. Ei - EF = kT * ln(NA / ni).
• Hole mobility is 450 cm^2/V·s vs 1350 cm^2/V·s for electrons in silicon.
• Crystal remains charge-neutral: mobile holes are balanced by fixed negative acceptor ions.
• ni = 1.5 × 10^10 cm^-3 for Si at 300K.
• Exam trap: Students write EF - Ei as positive for P-type. It is negative. EF is below Ei in P-type, so Ei - EF is the positive quantity to use.