Solids

In a single isolated atom an electron may occupy only certain sharp energy levels. Push 10^{29} atoms together into a crystal and those sharp levels smear out into continuous bands of allowed energy separated by forbidden gaps. One number — the width of the gap above the highest filled band — decides whether the solid is a copper wire, a slab of mica, or the silicon chip inside a mobile phone. This short unit builds that picture and uses it to explain doping.

25.1 Energy bands in solids (qualitative ideas)

An electron in an isolated atom has discrete energies E_1, E_2, E_3, \dots given by quantum theory. The Pauli exclusion principle allows at most two electrons (opposite spin) in any one state.

Now bring N identical atoms close together to build a crystal. When the atoms are far apart, each of the N atoms has its own identical 3s level, so that energy is N-fold repeated. As the separation shrinks, the outer electron clouds overlap; the electron of one atom now feels the nuclei of its neighbours. Pauli's principle forbids N electrons from sharing one identical state, so the single level splits into N very closely spaced levels.

How closely spaced? Take one cubic metre of solid, so N \approx 10^{29}, and let the whole set of split levels spread over about 1\ \text{eV}. The average step from one level to the next is then

No instrument can resolve a step that small, and no experiment can tell it apart from zero. The set of levels therefore behaves as one continuous energy band. Inner-shell electrons are buried deep inside the atom, their orbitals barely overlap, and their levels hardly split at all — only the outermost (valence) levels broaden into wide bands.

Two bands matter:

A current flows only if electrons can gain small amounts of energy from the applied field, i.e. only if there are empty states immediately above occupied states. A completely full band carries no net current however strong the field, because every electron that moves one way is matched by one moving the other. This single sentence explains the whole of the next section.

25.2 Difference between metals, insulators and semi-conductors using band theory

Conductors (metals). The valence band is only partly filled (sodium), or a filled valence band overlaps the empty conduction band (magnesium, zinc). Either way there is no gap: empty states sit immediately above occupied ones, a huge number of free electrons (\sim 10^{28}\ \text{m}^{-3}) is available, and the resistivity is tiny.

Insulators. The valence band is completely full, the conduction band completely empty, and E_g is large — 5.47\ \text{eV} for diamond. At 300\ \text{K} the average thermal energy of a particle is only about kT \approx 0.026\ \text{eV}, so practically no electron can be lifted across.

Semiconductors. The band picture is the same as an insulator's, but E_g is small: 1.1\ \text{eV} for silicon, 0.7\ \text{eV} for germanium. At 0\ \text{K} a semiconductor is a perfect insulator. At room temperature a small fraction of the bonds break, electrons jump the gap into the conduction band and leave holes behind in the valence band, and conduction begins.

Property	Conductor	Semiconductor	Insulator
Forbidden gap E_g	zero (bands overlap)	\leq 3\ \text{eV} (Ge 0.7, Si 1.1)	> 3\ \text{eV} (diamond 5.47)
Resistivity at 300 K	10^{-8}–10^{-6}\ \Omega\,\text{m}	10^{-5}–10^{6}\ \Omega\,\text{m}	10^{8}–10^{16}\ \Omega\,\text{m}
Carrier density	\sim 10^{28}\ \text{m}^{-3}	10^{16}–10^{19}\ \text{m}^{-3}	negligible
Temperature coefficient of resistance	positive	negative	negative
Behaviour at 0\ \text{K}	still conducts	perfect insulator	perfect insulator
Examples	Cu, Ag, Al	Si, Ge, GaAs	diamond, mica, glass

The negative temperature coefficient is the giveaway property of a semiconductor. Heating a metal makes the lattice ions vibrate more, electrons are scattered more often, and resistance rises. Heating a semiconductor does that too, but it also breaks far more bonds. In a pure semiconductor the Fermi level (the energy at which a state is half likely to be occupied) sits in the middle of the gap, so on the Boltzmann scale an electron only has to climb E_g/2 to reach the conduction band. The number of carriers therefore grows like e^{-E_g/2kT}, and this exponential growth easily overwhelms the extra scattering, so the resistivity falls.

25.3 Intrinsic and extrinsic semi-conductors

Intrinsic semiconductors

A pure semiconductor is called intrinsic. Silicon and germanium are tetravalent: each atom shares its four valence electrons in covalent bonds with four neighbours, so at 0\ \text{K} every bond is complete and the crystal does not conduct.

At room temperature thermal energy breaks a few bonds. Each broken bond frees one electron into the conduction band and leaves behind a vacancy in the valence band — a hole. A neighbouring bound electron can hop into the vacancy, moving the vacancy the other way, so the hole behaves like a mobile particle of charge +e.

Because carriers are created only in pairs,

where n_i is the intrinsic carrier concentration: 1.5\times10^{16}\ \text{m}^{-3} for Si and 2.4\times10^{19}\ \text{m}^{-3} for Ge at 300\ \text{K}. Compare this with roughly 5\times10^{28} silicon atoms per cubic metre: only about one atom in 10^{12} contributes a carrier.

Here \mu_e and \mu_h are the electron and hole mobilities (drift velocity per unit field, in m² V⁻¹ s⁻¹). Electrons move through the lattice more easily than holes, so \mu_e > \mu_h always.

Extrinsic semiconductors

Intrinsic material is far too resistive to be useful. Adding a controlled trace of impurity — typically one atom in 10^{6} to 10^{8} — is called doping, and the result is an extrinsic semiconductor.

n-type. Dope silicon with a pentavalent atom (P, As, Sb, Bi). The impurity takes the place of a silicon atom in the lattice. Four of its five valence electrons pair up with electrons of the four silicon neighbours and make four ordinary covalent bonds. The fifth electron has no partner to bond with. It is held only weakly by its parent atom, and about 0.045\ \text{eV} in silicon (about 0.01\ \text{eV} in germanium) is enough to set it free — far less than the 1.1\ \text{eV} needed to break a bond. Since kT \approx 0.026\ \text{eV} at 300\ \text{K}, practically every impurity atom has already given up its fifth electron at room temperature. The impurity donates an electron, so it is called a donor, and electrons become the majority carriers.

p-type. Dope silicon with a trivalent atom (B, Al, Ga, In). This atom brings only three valence electrons, so it can complete only three of the four bonds with its silicon neighbours. The fourth bond is left with a single electron instead of a pair. That empty half-bond is a hole. A bound electron from a neighbouring bond needs only about 0.05\ \text{eV} to jump across and fill it, which moves the hole to where that electron came from. Because the impurity accepts an electron it is called an acceptor, and holes become the majority carriers.

In band language the two dopants create new allowed levels inside the forbidden gap. A donor level sits only 0.045\ \text{eV} below the bottom of the conduction band, so its electron needs almost no help to get in. An acceptor level sits only about 0.05\ \text{eV} above the top of the valence band, so a valence electron can hop up into it and leave a hole behind.

So doping to raise one carrier density automatically suppresses the other: push n_e up by a factor of a million and n_h falls by the same factor.

Feature	n-type	p-type
Dopant valency	pentavalent (P, As, Sb)	trivalent (B, Al, Ga, In)
Bonds formed by dopant	4 complete bonds + 1 spare electron	3 complete bonds + 1 hole
Impurity level created	donor, just below CB	acceptor, just above VB
Majority carriers	electrons	holes
Minority carriers	holes	electrons
Carrier relation	n_e \gg n_h, n_e \approx N_d	n_h \gg n_e, n_h \approx N_a
Net charge on crystal	zero	zero

Chapter summary

• Bringing N atoms together splits each sharp atomic level into N levels so close (\sim 10^{-29}\ \text{eV} apart) that they form a continuous energy band; between bands lie forbidden gaps.
• The valence band is the highest band filled at 0\ \text{K}; the conduction band is the empty band above it; their separation is E_g.
• A completely filled band carries no current. Conduction needs empty states just above occupied ones.
• Conductor: bands overlap, E_g = 0. Semiconductor: E_g \leq 3\ \text{eV} (Ge 0.7 eV, Si 1.1 eV). Insulator: E_g > 3\ \text{eV} (diamond 5.47 eV).
• Semiconductors have a negative temperature coefficient of resistance because carrier number grows like e^{-E_g/2kT}.
• Intrinsic: n_e = n_h = n_i, and \sigma = e(n_e\mu_e + n_h\mu_h) reduces to \sigma_i = e\,n_i(\mu_e+\mu_h), with \rho = 1/\sigma.
• Extrinsic: a pentavalent dopant makes four bonds and leaves one spare electron (n-type, donor level below CB, electrons majority); a trivalent dopant makes three bonds and leaves one hole (p-type, acceptor level above VB, holes majority). Both crystals remain electrically neutral.
• Law of mass action: n_e n_h = n_i^{2} at a given temperature.

Practice questions

Group A — Multiple choice (1 mark each)

1. In an intrinsic semiconductor at room temperature [1]an_e > n_hbn_e < n_hcn_e = n_hdn_h = 0
2. The forbidden energy gap of germanium is about [1]a0.7\ \text{eV}b1.1\ \text{eV}c3.0\ \text{eV}d5.5\ \text{eV}
3. On heating a semiconductor its resistance [1]aincreasesbdecreasescstays constantdfirst rises then falls
4. Silicon doped with indium becomes [1]an-typebp-typecintrinsicdan insulator
5. A pure semiconductor at 0\ \text{K} behaves as [1]aa conductorba superconductorca perfect insulatordan n-type semiconductor
6. In n-type silicon the phosphorus atom forms [1]a5 covalent bondsb4 covalent bonds and has 1 spare electronc3 covalent bonds and 1 holedno bonds

Answers to Group A

1. (c) — carriers are produced only in electron–hole pairs.

2. (a) — Ge 0.7 eV, Si 1.1 eV.

3. (b) — the exponential rise in carrier number beats the extra scattering.

4. (b) — indium is trivalent, so it supplies acceptor levels.

5. (c) — every covalent bond is intact, so both bands are full or empty.

6. (b) — silicon offers only four neighbours, so the fifth electron of P has no bond to join.

Group B — Short answer (5 marks each)

1. Explain qualitatively how the discrete energy levels of isolated atoms become energy bands in a solid. [5]
2. Using band diagrams, distinguish between a conductor, a semiconductor and an insulator. [5]
3. Why does the resistance of a metal increase but that of a semiconductor decrease with rise in temperature? [5]
4. The energy gap of a certain semiconductor is 0.72\ \text{eV}. Calculate the maximum wavelength of radiation it can absorb. (h = 6.63\times10^{-34}\ \text{J s}) [5]
5. Intrinsic silicon has n_i = 1.5\times10^{16}\ \text{m}^{-3}. After doping, the hole concentration becomes 4.5\times10^{21}\ \text{m}^{-3}. Find the electron concentration and name the type of the semiconductor. [5]
6. Draw the covalent-bond diagram of silicon doped with a pentavalent atom and use it to explain why the material is called n-type although the crystal as a whole is neutral. [5]

Answers to Group B

1. Outline: sharp levels in an isolated atom → Pauli exclusion forbids N identical states → overlapping of outer orbitals on bringing N atoms together splits each level into N sub-levels → with N \sim 10^{29}\ \text{m}^{-3} the spacing is \sim 10^{-29}\ \text{eV}, so the set is effectively a continuous band, separated from the next by a forbidden gap.

3. Outline: in a metal the carrier number is fixed, so more violent lattice vibrations scatter electrons more and \rho rises. In a semiconductor the carrier number grows like e^{-E_g/2kT}; this exponential increase dominates the extra scattering, so \rho falls (negative temperature coefficient).

4. First convert the gap to joules:

E_g = 0.72\times1.6\times10^{-19} = 1.152\times10^{-19}\ \text{J}

Then \lambda_{\max} = hc/E_g with hc = 1.989\times10^{-25}\ \text{J m}:

\lambda_{\max} = \frac{1.989\times10^{-25}}{1.152\times10^{-19}} = 1.73\times10^{-6}\ \text{m} = 1730\ \text{nm}

5. By the law of mass action, n_e = n_i^2 / n_h. Square n_i first:

n_i^{2} = (1.5\times10^{16})^{2} = 2.25\times10^{32}\ \text{m}^{-6}

Then divide:

n_e = \frac{2.25\times10^{32}}{4.5\times10^{21}} = 5\times10^{10}\ \text{m}^{-3}

Holes greatly outnumber electrons, so the material is p-type.

6. Outline: draw the lattice of the n-type figure — P on a silicon site, four double-line covalent bonds to four Si neighbours, one extra electron drawn outside any bond. That spare electron is easily freed (0.045 eV), so electrons are the majority carriers and the material is called n-type. But the phosphorus atom that lost the electron is left behind as a fixed positive ion inside the lattice, so the positive and negative charges still balance exactly and the block is neutral.

Group C — Long answer (8 marks each)

1. (a) Define valence band, conduction band and forbidden energy gap. [3](b) Draw band diagrams for a conductor, a semiconductor and an insulator and use them to explain the enormous difference in their resistivities. [5]
2. (a) Distinguish between intrinsic and extrinsic semiconductors. [2](b) Draw the covalent-bond lattice of n-type and p-type silicon, showing the free electron and the hole, and name the majority carriers in each. [4](c) A germanium sample has n_i = 2.4\times10^{19}\ \text{m}^{-3}, \mu_e = 0.39 and \mu_h = 0.19\ \text{m}^{2}\text{V}^{-1}\text{s}^{-1}. Find the resistivity of the pure sample. [2]

Answer to Group C question 2(c)

Add the mobilities: \mu_e + \mu_h = 0.39+0.19 = 0.58. Then

\sigma = e\,n_i(\mu_e+\mu_h) = 1.6\times10^{-19}\times2.4\times10^{19}\times0.58

\sigma = 3.84\times0.58 = 2.23\ \Omega^{-1}\text{m}^{-1}

\rho = \frac{1}{\sigma} = \frac{1}{2.23} = 0.45\ \Omega\,\text{m}