Extrinsic Semiconductors and the Fermi Level

The syllabus marks extrinsic semiconductors (qualitative), so this topic explains rather than derives. The one calculation that does appear in exams — minority carrier concentration from the law of mass action — is worked below, because it follows from Topic 1 rather than needing new theory.

The problem doping solves

Topic 1 ended with the striking number: in pure silicon at room temperature, about one atom in ten trillion supplies a conduction electron. That is useless for building anything, and worse, the number depends violently on temperature.

Doping replaces a vanishing, temperature-driven carrier population with a large one you control. Add impurities at around one part per million and the carrier concentration rises by six orders of magnitude — and stays put over the normal operating range.

n-type: donors

Silicon has four valence electrons and forms four covalent bonds. Replace an occasional silicon atom with a pentavalent atom — phosphorus, arsenic, antimony — and four of its five valence electrons complete the bonds. The fifth has no bond to join.

It is still weakly bound to its parent nucleus, but only just: the donor level $E_D$ sits about 0.045 eV below the conduction band in silicon. Compare that with $k_BT = 0.026$ eV at room temperature. The electron is essentially free, and at 300 K virtually every donor is ionised.

The impurity donates an electron without creating a hole, so:

• electrons are the majority carriers, holes the minority carriers;
• $n \approx N_D$, the donor concentration;
• the material is still electrically neutral — the donor atom is fixed in the lattice and left positively charged, balancing the freed electron. n-type material is not negatively charged, a point worth stating because the name invites the opposite conclusion.

p-type: acceptors

Replace a silicon atom with a trivalent atom — boron, aluminium, gallium, indium. Three bonds form; the fourth is incomplete. That vacancy accepts an electron from a neighbouring bond, creating a hole that moves through the lattice.

The acceptor level $E_A$ sits about 0.045 eV above the valence band, so again nearly all acceptors are ionised at room temperature.

• holes are the majority carriers, electrons the minority;
• $p \approx N_A$;
• the acceptor atom becomes a fixed negative ion, and again the material overall is neutral.

Minority carriers and the law of mass action

Doping does not simply add carriers of one kind. It suppresses the other, and by an exactly reciprocal amount, because $np = n_i^2$ holds regardless of doping:

The extra electrons make recombination more likely, so holes are consumed faster and their steady-state population falls. The worked example shows how extreme the resulting asymmetry is.

Where the Fermi level moves

This is the part most exam questions turn on.

The Fermi level is the energy at which the occupancy is one half, so it must sit where the carriers are.

• Intrinsic: near mid-gap.
• n-type: many electrons in the conduction band, so $E_F$ moves up, towards $E_C$. Heavier doping moves it closer.
• p-type: many holes at the top of the valence band, so $E_F$ moves down, towards $E_V$.

Quantitatively, from $n = N_C e^{-(E_C-E_F)/k_BT}$:

Only logarithmic in the doping — so multiplying $N_D$ by ten moves $E_F$ up by just $k_BT\ln 10 = 0.06$ eV. Getting the Fermi level close to the band edge takes very heavy doping indeed.

Two consequences worth carrying forward.

Dope heavily enough and $E_F$ enters the band itself. The material is then degenerate, behaves in some ways like a metal, and the Boltzmann approximation of Topic 1 fails. Both sides of a tunnel diode are degenerate — that is what makes it work.

And the difference in $E_F$ between an n-type and a p-type piece is what drives everything in the next topic. Join them, and the Fermi level must become a single flat line through the whole system. The bands have no choice but to bend to accommodate it, and that bend is the p-n junction.