How it works
In n-type silicon, phosphorus (or arsenic) atoms from Group V donate one free electron each without adding a hole, so electrons become the majority carriers and holes are minority carriers. Applying the mass action law: if donor concentration ND = 10¹⁶ /cm³, then hole concentration p = ni²/ND ≈ (1.5×10¹⁰)²/10¹⁶ = 2.25×10⁴ /cm³ — tiny compared to electrons. In p-type silicon, boron (Group III) accepts one electron, generating a hole. The Fermi level moves toward the conduction band for n-type and toward the valence band for p-type, and this shift is what creates the built-in potential at a PN junction.
Key points to remember
Donor impurities (phosphorus, arsenic) create n-type material where electrons are majority carriers; acceptor impurities (boron, gallium) create p-type where holes dominate. Typical doping concentrations range from 10¹⁵ to 10¹⁸ /cm³, far exceeding ni = 1.5×10¹⁰ /cm³, so the mass action law minority carrier formula p = ni²/ND is essential for numerical problems. The Fermi level shifts are not just theoretical — they directly produce the 0.7 V built-in barrier in silicon PN junctions. Compensation occurs when both donor and acceptor atoms are present simultaneously, with the net type determined by whichever concentration is higher.
Exam tip
The examiner always asks you to find minority carrier concentration using p = ni²/ND or n = ni²/NA — plug in ND or NA given in the problem, use ni = 1.5×10¹⁰, and square it before dividing.