Extrinsic Semiconductor - p-type Semiconductor

A p-type semiconductor is a semiconductor obtained by doping pure silicon or germanium with a trivalent impurity so that holes become the majority charge carriers. 

If hole concentration is high, then:

This is especially important for numerical problems in board examinations and entrance tests. 

Pure silicon and germanium are tetravalent elements, which means each atom forms four covalent bonds. When a trivalent impurity atom such as boron is added, it can form only three covalent bonds, leaving one bond incomplete. This incomplete bond appears as a hole. 

Dopants Used

• Boron (B) 
• Aluminium (Al) 
• Indium (In) 

During doping:

1. A trivalent impurity replaces a silicon or germanium atom in the crystal lattice. 
2. Only three covalent bonds are completed. 
3. One bond remains deficient by one electron. 
4. This deficiency behaves like a hole. 
5. As a result, the semiconductor becomes p-type. 

The trivalent impurity has only three valence electrons, but the lattice requires four electrons for bonding. Therefore, a vacancy is created in one covalent bond. This vacancy can accept an electron from a neighbouring bond and is therefore associated with an acceptor impurity.

• Trivalent impurities in p-type semiconductors are called acceptor impurities because they accept an electron.
• The missing electron position is called a hole.

In a p-type semiconductor, holes are present in very large numbers because of doping, while a small number of electrons still exist due to thermal generation. Therefore, holes are the majority carriers and electrons are the minority carriers. 

Very Important Relation

For a p-type semiconductor:

• n_h ≫ n_e
• n_en_h = \[n_i^2\]

Where:

• n_h​ = hole concentration
• n_e​ = electron concentration
• n_i​ = intrinsic carrier concentration

The energy band description of a p-type semiconductor shows that the trivalent impurity introduces an acceptor energy level slightly above the valence band. Because this level is close to the valence band, electrons can easily move from the valence band to the acceptor level, leaving behind holes in the valence band.

Band Interpretation

• The valence band contains a large number of holes after doping.
• The acceptor level lies just above the valence band.
• Electrons require only a small amount of energy to shift to the acceptor level.
• Hole conduction becomes dominant. 

Although holes appear to move, the actual particles that move are electrons. When one electron fills a hole, it leaves another hole behind. In this way, the hole appears to move through the crystal.

Stepwise explanation

1. A neighbouring electron jumps into a hole.
2. A new hole is created where that electron came from.
3. Repetition of this process makes the hole appear to move.
4. Thus, the current in a p-type semiconductor is mainly due to hole motion.

Analogy

Think of a row of seats with one empty seat. When one person shifts into the empty seat, a new empty seat appears at the old position. The empty seat seems to move, although only people actually move. In the same way, holes appear to move, though electrons are the real moving particles.

• Formed by doping pure silicon or germanium with a trivalent impurity.
• Holes are the majority carriers.
• Electrons are the minority carriers.
• The crystal remains electrically neutral as a whole.
• The acceptor level lies close to the valence band.
• Conductivity increases relative to that of the intrinsic semiconductor.

Given:

• n_i = 1.5 × 10¹⁶ m⁻³
• n_h = 4.5 × 10²² m⁻³

Solution:

1. Use the mass action law: n_en_h = \[n_i^2\]
2. Rearrange to find electron concentration: n_e = \[\frac {n_i^2}{n_h}\]
3. Substitute the given values: n_e = \[\frac{(1.5\times10^{16})^2}{4.5\times10^{22}}\]
4. Calculate: n_e = 5 × 10⁹ m⁻³

Answer: n_e = 5 × 10⁹ m⁻³

Given:

• Number of Si atoms  = 5 × 10²⁸ m⁻³
• Doping concentration = 1 ppm
• n_i = 1.5 × 10¹⁶ m⁻³

Solution:

1. Calculate donor concentration:ND = 5 × 10²⁸ × 10⁻⁶ = 5 × 10²² m⁻³
2. Since the semiconductor is heavily doped, n_e ≈ N_D = 5 × 10²² m⁻³
3. Use the mass action law: n_en_h = \[n_i^2}\]
4. Calculate hole concentration:
   n_h = \[\frac{(1.5\times10^{16})^2}{5\times10^{22}}=4.5\times10^9\mathrm{~m}^{-3}\]

Answer: n_e = 5 × 10²² m⁻³, n_h = 4.5 × 10⁹ m⁻³