Q.Doping a semiconductor results in

Why N-Type Semiconductor Doping Works: The Physics Behind the Formula

When you dope a pure (intrinsic) semiconductor like silicon with a pentavalent impurity — an element from Group V of the periodic table, such as phosphorus, arsenic, or antimony — you create an n-type semiconductor. The "n" stands for negative, because the majority charge carriers are negatively charged electrons.

The key formula that governs n-type doping is:

where $n$ is the concentration of free electrons in the conduction band, and $N_D$ is the concentration of donor atoms introduced.

Let's understand why this simple relation holds, step by step.

Step 1: What happens at the atomic level?

Silicon has four valence electrons. It forms four covalent bonds with neighbouring silicon atoms, achieving a stable octet configuration. Now, introduce a phosphorus atom — it has five valence electrons.

Four of phosphorus's electrons form normal covalent bonds with adjacent silicon atoms. The fifth electron has no place in the bonding structure. It is only very weakly bound to the phosphorus nucleus — the binding energy is tiny, about 0.045 eV for phosphorus in silicon (compared to the 1.1 eV band gap of silicon).

Each phosphorus atom that loses its extra electron becomes a positively charged ion (fixed in the crystal lattice), but the freed electron is mobile and contributes to electrical conduction.

Step 2: Why $n \approx N_D$ and not exactly $N_D$?

The reasoning is straightforward:

• Every donor atom contributes one free electron when ionised.
• At room temperature, the ionisation is nearly complete — the donor energy level lies just below the conduction band edge (about 0.045 eV), so thermal energy easily kicks the electron into the conduction band.
• Therefore, the number of free electrons $n$ is approximately equal to the number of donor atoms $N_D$.

But why "approximately" and not exactly? Two reasons:

1. Intrinsic carriers still exist: Even in doped silicon, a small number of electron-hole pairs are thermally generated. The intrinsic carrier concentration $n_i$ (about $1.5 \times 10^{10} \text{ cm}^{-3}$ for silicon at 300 K) adds to the electron count. However, for typical doping levels ($N_D \approx 10^{15}$ to $10^{18} \text{ cm}^{-3}$), $n_i$ is negligible — so $n \approx N_D$ is an excellent approximation.

2. Incomplete ionisation at very low temperatures: At extremely low temperatures (near 0 K), some donor atoms may not ionise. But for all practical operating temperatures of electronic devices, ionisation is essentially 100%.

Step 3: What about the hole concentration?

In an n-type semiconductor, electrons are the majority carriers, and holes are the minority carriers. The product of electron and hole concentrations is always constant for a given semiconductor at a fixed temperature — this is the law of mass action:

Since $n \approx N_D$, we get:

This tells you that as you increase doping ($N_D$), the hole concentration $p$ decreases — because more electrons mean more recombination, reducing the number of holes.

Step 4: Where does the Fermi level go?

The position of the Fermi level $E_F$ shifts upward (toward the conduction band) in n-type material. The formula is:

where $N_C$ is the effective density of states in the conduction band. The derivation comes from the fact that:

Setting $n = N_D$ and solving for $E_F$ gives the expression above. The Fermi level moves closer to the conduction band as doping increases — exactly what you'd expect when electrons become abundant.

The Big Picture: Why This Matters

The formula $n \approx N_D$ is not just a number — it's a statement that doping gives you direct control over carrier concentration. By choosing how many donor atoms to add, you set the electron concentration, and therefore the conductivity:

where $\mu_n$ is the electron mobility. This is why n-type silicon is the foundation of MOSFETs, bipolar transistors, and virtually all modern electronics — you can engineer the conductivity precisely by controlling the doping level.