A2 · Semiconductor basics

Doping: A Pinch of Impurity Sets the Majority Carrier

Pure silicon barely conducts. By mixing in just a pinch of impurity atoms, choose for yourself whether free electrons or holes become the majority.

Pick the impurity and set the majority carrier

Below is a silicon lattice; each atom pairs two electrons with every neighbor in a bond. Swap one central atom for a different one and see which free carrier is released. Which choice makes free electrons the majority?

Tap to change the central atom.

Majority-carrier relation

n0 = p0 = ni

Every electron is locked in a bond. Only a few pairs broken by heat remain, so electrons and holes are equally few.

Held in balance

Pure silicon is nearly an insulator

A silicon atom has four valence electrons and fills its bonds by pairing two with each of four neighbors. So at room temperature almost no electron is free. Making current means breaking a bond, which is the same act as crossing the band gap of A1.

One pentavalent atom = one free electron

Add an atom with five valence electrons, like phosphorus or arsenic: four go into bonds and one is left over. This leftover is barely bound and frees up with the slightest warmth. The impurity atom that gave it up becomes a fixed positive ion, yet the crystal stays neutral. This is n-type.

One trivalent atom = one hole, and the product holds

Add an atom with three valence electrons, like boron, and one bond seat is empty. An electron from a neighboring bond fills it and the vacancy hops sideways, looking just like a moving positive charge. This is p-type. Raise one carrier and the other drops, so the product of the two concentrations, n p = ni², stays fixed at a given temperature regardless of doping.

Observen0 p0 = ni²

Doping leaves the product of the two carrier densities unchanged.

Choosen0 ≈ ? (n)

In n-type, the majority is set by the donor density.

Fill inp0 ≈ ni² / ?

The minority is the product divided by the majority.

On your ownp0 ≈ ? (p)

In p-type, the acceptor density sets the hole majority.

Back to the first screen

What made free electrons the majority was the pentavalent donor, that is, n-type. As a mirror image, the trivalent acceptor makes holes the majority and gives p-type, while pure silicon holds both equally few in balance. Either way you added only a pinch of impurity, yet the whole majority carrier flips. And throughout, the fixed ions exactly repay the charge that left, so the crystal stays neutral end to end. The handle that sets the majority is a single number: the valence count of the impurity.

Doping mixes a pinch of impurity with a different valence count into a pure semiconductor. A pentavalent donor makes free electrons the majority, giving n-type (n0 ≈ ND); a trivalent acceptor makes holes the majority, giving p-type (p0 ≈ NA). At a fixed temperature the mass-action law n0 p0 = ni² keeps the product of the two densities constant. The crystal stays electrically neutral.

What comes next

Now you hold two kinds of carrier. The next unit, A3, shows the two ways they move: drift, pushed by an electric field, and diffusion, spread by a difference in concentration. Where those two meet at the border between n-type and p-type is exactly the depletion layer of the pn junction in A4.