A2 · Semiconductor basics

Doping: A Pinch of Impurity Sets the Majority Carrier

Pure silicon barely conducts at all. In this unit you mix in a tiny amount of impurity atoms and choose for yourself whether free electrons or holes become the majority.

The impurity decides the majority carrier

The picture is a silicon lattice in which each atom bonds with its neighbors by sharing electrons in pairs. Swap the center atom for a different element and the carriers it frees drift between the rows of the lattice: the gold dots are free electrons and the blue hollow circles are holes. Switch the center atom and find which choice makes free electrons the majority.

Switch the center atom with the buttons.

Majority-carrier relation

n0 = p0 = ni

Every electron is tied up in a bond. Only a few pairs broken by heat split into an electron and a hole, so there are few of each, in equal numbers.

Held in balance

Pure silicon is nearly an insulator

A silicon atom has four valence electrons and fills all its bonds by pairing with four neighbors two electrons at a time. So at room temperature almost no electrons are free to move. Making a current means breaking bonds, which is the same thing as crossing the band gap you saw in A1.

One pentavalent atom = one free electron

Put in an atom with five valence electrons, such as phosphorus or arsenic, and four go into bonds while one is left over. That leftover electron is so weakly bound to its atom that room-temperature heat alone sets it free. The impurity atom that gave it up becomes a positive ion fixed in place, but together with the freed electron the crystal as a whole stays neutral. This is n-type.

One trivalent atom = one hole, and the product holds

Put in an atom with three valence electrons, such as boron, and one bonding site is left empty. When an electron from a neighboring bond crosses over to fill it, the vacancy moves sideways, so it looks as if a positive charge were moving. This is p-type. Raising one kind of carrier lowers the other by the same factor, so the product of the two carrier concentrations, n p = ni², stays constant at a given temperature regardless of doping.

Observen0 p0 = ni²

Doping does not change the product of the two concentrations.

Choosen0 ≈ ? (n)

In n-type, the donor concentration sets the majority concentration.

Fill inp0 ≈ ni² / ?

The minority concentration is the product divided by the majority.

On your ownp0 ≈ ? (p)

In p-type, the acceptor concentration sets the hole concentration.

Back to the first screen

What made free electrons the majority was the 5-valent donor, that is, n-type. As in a mirror, the 3-valent acceptor makes holes the majority and gives p-type, while pure silicon stays balanced with few electrons and holes in equal numbers. Either way, all that was added was a tiny amount of impurity, yet the majority carrier flipped entirely. Even so, the ions fixed in place exactly balance the charge of the carriers that left, so the crystal stays neutral to the end. The knob that picks the majority carrier is a single number: the impurity's count of valence electrons.

Doping is mixing a tiny amount of an impurity with a different number of valence electrons into a pure semiconductor. A 5-valent donor makes free electrons the majority and gives n-type (n0 ≈ ND); a 3-valent acceptor makes holes the majority and gives p-type (p0 ≈ NA). At a fixed temperature the mass-action law n0 p0 = ni² keeps the product of the two concentrations constant, and the crystal stays electrically neutral.

What comes next

Now you have two kinds of carrier in hand. In the next unit, A3, you see the two ways these carriers move: drift, pushed by an electric field, and diffusion, spreading out from a concentration difference. The place where the two meet at the boundary between n-type and p-type is the depletion layer of the pn junction in A4.