Sprinkle a few impurity atoms into a pure crystal and its conductivity leaps by orders of magnitude — and you choose which carrier moves the charge. Donors give an n‑type solid a surplus of electrons; acceptors give a p‑type solid a surplus of holes. Through it all the mass‑action law n·p = ni² is untouched: raise one carrier and you crush the other. Rendered, not quoted.
source Shockley, W. — Electrons and Holes in Semiconductors, with Applications to Transistor Electronics, D. Van Nostrand, 1950 · scan: archive.org/details/ElectronsAndHolesInSemiconductors amber · no stable DOI
A semiconductor at thermal equilibrium holds electrons (density n) and holes (p). Generation and recombination balance, pinning the product:
Add donor atoms (density Nd, e.g. P in Si) → each gives up an electron: n rises, so p = ni²/n falls. Add acceptors (Na, e.g. B) → holes rise, electrons fall. Charge neutrality fixes the split exactly:
Solve the pair simultaneously: n = ½(Nd−Na) + √[¼(Nd−Na)² + ni²], then p = ni²/n. Two equations, two unknowns — no free parameters.
A few impurities, a giant change. Doping sets the stage for the rest of the band‑gap neighbourhood:
• the majority/minority split this sphere computes is exactly what the‑pn‑junction joins across a metallurgical boundary — n‑side meets p‑side.
• the‑hall‑effect counts those same carriers and reads their sign, confirming n vs p empirically.
• all of it populates the‑band‑gap: donors sit just below the conduction edge, acceptors just above the valence edge.
A live re‑check of the engine's core invariant. It recomputes n·p for an n‑type sample and confirms it still equals ni². Flip the tamper in window 6 and this badge goes red.
witness …
Intrinsic carrier density (Si @300K, approx): ni = 1.0×10¹⁰ cm⁻³. Choose donor and acceptor concentrations. Mobilities μn=1350, μp=480 cm²/Vs.
The proven result, fixed at boot by the selfcheck:
proving…
"You said a few impurities. One part per million of boron in silicon — that is nothing. It cannot possibly change conductivity a millionfold."
It does, and the machine shows why: 1 ppm of a 5×10²² cm⁻³ lattice is ~5×10¹⁶ carriers — but intrinsic Si has only ~10¹⁰. The dopant outnumbers the native carriers by six orders of magnitude. "A few" is measured against the crystal; against the carriers it is a flood.
Doping adds free charge, so the crystal ends up charged.
→ No. The dopant ion's fixed core charge balances its released carrier — n + Na = p + Nd holds. The solid stays neutral.
More donors just means proportionally more of everything.
→ No. Raising n suppresses p = ni²/n. The product is pinned, not the sum.
The free‑electron / Drude picture explains doping.
→ amber Drude is a classical idealization with a fixed carrier count and no gap. It cannot produce two carrier types, holes, or exp(−Eg/2kT) statistics — band theory is required.
The disclosed planted void. This button swaps the minority‑carrier law from p = ni²/n to the wrong p = ni/n — mass action becomes n·p = ni (not squared). Neutrality and suppression both break. The Witness (7) catches it live.