Introduction to Electronics (Senior 6 Physics): free lessons | Ganzaa

Introduction to Electronics

Lessons

Senior 6 · Physics

Introduction to Electronics

5 lessons

Lesson 1 of 5

Energy Bands and Semiconductors

25 min

One atom in ten million

Every phone, solar panel and computer depends on a strange fact. A block of very pure silicon conducts electricity so poorly that it is closer to an insulator than to a metal. Yet if one silicon atom in every ten million is replaced by a phosphorus atom, a change no chemical test would notice, its resistivity falls by a factor of about four hundred thousand (Worked example 2). The difference between copper, silicon and glass, and the reason a tiny impurity transforms silicon, both come from how electrons are allowed to be arranged in a solid. This unit starts there and builds up, device by device, to the logic gates inside a computer.

Conductors, semiconductors and insulators

Material	Resistivity at 20 °C / Ω m	As temperature rises, resistance…
copper (conductor)	$1.7 \times 10^{-8}$	rises slowly
germanium, pure (semiconductor)	about 0.5	falls steeply
silicon, pure (semiconductor)	about $3 \times 10^{3}$	falls steeply
glass (insulator)	$10^{10}$ to $10^{14}$	stays very high

Semiconductors are not just "medium" conductors. Their resistance falls as they warm up, the opposite of a metal, and it can be changed enormously by adding impurities or by shining light on them. Both facts are explained by energy bands.

From energy levels to energy bands

In a single atom, electrons occupy sharp energy levels, as you saw with X-ray lines in Unit 7. When $N$ atoms are packed into a crystal, their outer orbits overlap, and by the Pauli exclusion principle each level splits into $N$ levels very close together. With about $10^{23}$ atoms in a gram, these levels form practically continuous energy bands, separated by forbidden gaps where no electron states exist.

Figure 1.1: Energy bands in the three kinds of solid (energies not to scale). Dark: filled states; light: empty states.

The valence band holds the electrons that form the bonds between atoms. When it is completely full, its electrons cannot carry a current: there is no empty state for them to move into.
The conduction band is the next band up. Electrons in it can gain small amounts of energy from an electric field and drift, carrying current.
In a metal the bands overlap (or the top band is only partly filled), so there are always free electrons. In an insulator the gap $E_{g}$ is large (5.5 eV for diamond). In a semiconductor it is small: 1.12 eV for silicon and 0.66 eV for germanium.

At room temperature the typical thermal energy is $kT = (8.62 \times 10^{-5}\,\mathrm{eV\,K^{-1}})(300\,\mathrm{K}) = 0.026\,\mathrm{eV}$ , far smaller than the gaps. Only a tiny fraction of electrons, roughly in proportion to $e^{-\frac{E_g}{2kT}}$ , is thrown across. For silicon that fraction is about $4 \times 10^{-10}$ ; for diamond it is about $10^{-46}$ , effectively zero. As the temperature rises, the fraction grows rapidly, so a semiconductor's resistance falls. In a metal, the number of free electrons hardly changes, but the vibrating ions scatter them more, so its resistance rises.

Definition

A semiconductor is a solid whose valence band is full and whose conduction band is empty at absolute zero, separated by a small band gap $E_{g}$ (roughly 0.1 eV to 3 eV), so that it conducts a little at room temperature and its conductivity rises with temperature. Examples: silicon ( $E_g = 1.12\,\mathrm{eV}$ ), germanium (0.66 eV), gallium arsenide (1.42 eV).

Worked example 1: which light can a semiconductor absorb?

A photon can lift an electron from the valence band to the conduction band only if $hf \ge E_g$ . Find the longest wavelength that can do this in (a) silicon and (b) gallium arsenide ( $hc = 1240\,\mathrm{eV\,nm}$ ). (c) Explain why a silicon solar cell responds to all visible light, and why silicon looks opaque but is transparent to infrared of wavelength 1.5 µm.

Solution

(a) $\lambda_{\max} = \frac{hc}{E_g} = \frac{1240}{1.12} = 1.11 \times 10^{3}\,\mathrm{nm} = 1.11\,\mathrm{\mu m}$ .

(b) $\lambda_{\max} = \frac{1240}{1.42} = 873\,\mathrm{nm}$ .

(c) Visible light (400 nm to 700 nm) has photon energies of 1.8 eV to 3.1 eV, all above silicon's 1.12 eV, so every visible photon can create a free electron: this is how a solar cell works. A 1.5 µm photon has only $\frac{1240}{1500} = 0.83\,\mathrm{eV}$ , less than the gap, so it cannot be absorbed this way and passes through.

Intrinsic semiconductors: electrons and holes

Figure 1.2: Flat models of a silicon crystal. Each double line is a covalent bond, a pair of shared electrons.

Silicon has four valence electrons, and in the crystal each atom shares one with each of its four neighbours. When a bond breaks (Figure 1.2(a)), it creates two carriers at once: a free electron in the conduction band, and a gap in the bond, called a hole. A neighbouring bonding electron can jump into the hole, so the hole moves the opposite way; it behaves exactly like a positive charge $+ e$ . A pure semiconductor, in which electrons and holes are made only in such pairs, is an intrinsic semiconductor, and in it the electron and hole concentrations are equal: $n = p = n_{i}$ . For silicon at 300 K, $n_i \approx 1 \times 10^{16}\,\mathrm{m^{-3}}$ , compared with $5.0 \times 10^{28}$ silicon atoms per cubic metre.

Both carriers drift in an electric field, the electrons against the field and the holes along it, so both currents are in the same direction. Each carrier type has a mobility $μ \mu$ , its drift speed per unit field ( $v = \mu E$ ); the conductivity is then

$\sigma = \frac{1}{\rho} = e\,(n\mu_n + p\mu_p)$

For silicon, $\mu_n = 0.14\,\mathrm{m^2\,V^{-1}\,s^{-1}}$ and $\mu_p = 0.045\,\mathrm{m^2\,V^{-1}\,s^{-1}}$ .

Extrinsic semiconductors: doping

Adding a carefully controlled impurity is called doping, and the result is an extrinsic semiconductor.

n-type. A pentavalent atom (group 15: phosphorus, arsenic, antimony) has five valence electrons. Four form bonds; the fifth is held so weakly (about 0.045 eV for phosphorus in silicon) that at room temperature almost every one is freed (Figure 1.2(b)). Such impurities are donors. Electrons are the majority carriers; holes are the minority.
p-type. A trivalent atom (group 13: boron, aluminium, gallium) has only three valence electrons, so one bond is incomplete: a hole (Figure 1.2(c)). A neighbouring electron easily moves in, so the hole wanders through the crystal. Such impurities are acceptors. Holes are the majority carriers.

In band terms, donors add filled levels just below the conduction band and acceptors add empty levels just above the valence band; both gaps are tiny compared with $E_{g}$ . A doped crystal is still electrically neutral: each freed electron leaves behind a fixed positive donor ion, and each hole is balanced by a fixed negative acceptor ion.

Definition

An n-type semiconductor is one doped with donor atoms, so that its majority carriers are electrons; a p-type semiconductor is one doped with acceptor atoms, so that its majority carriers are holes. Example: silicon doped with arsenic is n-type; silicon doped with gallium is p-type.

Worked example 2: the effect of one atom in ten million

Use $n_i = 1.0 \times 10^{16}\,\mathrm{m^{-3}}$ , $\mu_n = 0.14$ and $\mu_p = 0.045\,\mathrm{m^2\,V^{-1}\,s^{-1}}$ , and $5.0 \times 10^{28}$ silicon atoms per m³. (a) Find the resistivity of pure silicon. (b) One atom in $10^{7}$ is replaced by phosphorus, and every donor is ionised. Find the new resistivity, ignoring the holes. (c) By what factor has the resistivity fallen?

Solution

(a) $\sigma = e\,n_i(\mu_n + \mu_p) = (1.60 \times 10^{-19})(1.0 \times 10^{16})(0.185) = 2.96 \times 10^{-4}\,\mathrm{S\,m^{-1}}$ , so $\rho = \frac{1}{\sigma} = 3.4 \times 10^{3}\,\Omega\,\mathrm{m}$ .

(b) $n = \frac{5.0 \times 10^{28}}{10^{7}} = 5.0 \times 10^{21}\,\mathrm{m^{-3}}$ , half a million times $n_{i}$ . $\sigma = (1.60 \times 10^{-19})(5.0 \times 10^{21})(0.14) = 112\,\mathrm{S\,m^{-1}}$ , so $\rho = 8.9 \times 10^{-3}\,\Omega\,\mathrm{m}$ .

Condition: this assumes that the mobilities are unchanged by doping and that all donors are ionised, both good approximations at room temperature for light doping.

Experiment 1.1: a semiconductor and a metal compared

An NTC (negative temperature coefficient) thermistor, made of a semiconducting metal oxide, and a coil of copper wire are placed side by side in a beaker of water with a thermometer. The water is heated slowly and stirred, and the resistance of each is read with a multimeter every 20 °C. Safety: use a hotplate or a water bath, keep the meter leads dry, and do not touch the hot beaker.

θ / °C	20	40	60	80
thermistor R / kΩ	12.5	5.30	2.49	1.27
copper coil R / Ω	10.0	10.8	11.6	12.3
$\frac{1}{T}$ / $10^{-3}\,\mathrm{K^{-1}}$	3.411	3.193	3.002	2.832
ln (R ÷ Ω) for the thermistor	9.44	8.58	7.82	7.15

Illustrative readings. The copper's resistance rises by about 23%; the thermistor's falls almost tenfold. For a semiconductor, $R \propto e^{\frac{E}{2kT}}$ , so a graph of $\ln R$ against $\frac{1}{T}$ (temperature in kelvin) is a straight line of gradient $\frac{E}{2k}$ , from which an energy gap can be estimated.

Worked example 3: an energy from a thermistor

Use the first and last columns of the table to find the gradient of the graph of $\ln R$ against $\frac{1}{T}$ , and hence the energy $E$ ( $k = 8.62 \times 10^{-5}\,\mathrm{eV\,K^{-1}}$ ).

Solution

Gradient $= \frac{9.44 - 7.15}{(3.411 - 2.832) \times 10^{-3}\,\mathrm{K^{-1}}} = \frac{2.29}{0.579 \times 10^{-3}} = 3.96 \times 10^{3}\,\mathrm{K}$ .

$E = 2k \times \text{gradient} = 2 \times 8.62 \times 10^{-5} \times 3955 = 0.68\,\mathrm{eV}$ .

Comment: 0.68 eV is the same order as the band gaps of semiconductors. A metal-oxide thermistor is not a pure, intrinsic crystal, so this is an effective "activation energy" rather than a true band gap, but the steep fall of resistance with temperature has the same cause: more carriers.

Warning

Common mistakes. (1) Saying an n-type semiconductor is negatively charged: it is neutral; "n" names the majority carrier, not the net charge. (2) Saying holes are positrons or protons: a hole is a missing electron in a bond, which moves like a positive charge. (3) Thinking doping adds electrons from outside the material: each donor atom brings its own extra electron. (4) Thinking the band gap is a physical gap in space: it is a range of energies that electrons in the crystal cannot have.

Tip

Phosphorus has five, boron has three: "five gives one away" (n-type, donor), "three needs one more" (p-type, acceptor). And for temperature: metals mind the shaking, semiconductors count the carriers.

Semiconductors 1: intrinsic and extrinsic semiconductors, from Mr Smith's Physics online. Watch for how holes move and how doping with group 13 and group 15 atoms gives p-type and n-type material.

✦

Practice

Using energy bands, explain the difference between a conductor, a semiconductor and an insulator.
Explain why the resistance of a semiconductor falls as its temperature rises, while that of a metal rises.
Germanium has a band gap of 0.66 eV. Find the longest wavelength of light that can create an electron–hole pair in it, and name the region of the spectrum.
State what is meant by a hole, and explain why holes and electrons in a semiconductor both contribute to a current in the same direction.
Silicon is doped with gallium. State whether it becomes n-type or p-type, name the majority carriers, and say whether the crystal is charged.
A silicon sample is doped with $2.0 \times 10^{22}$ arsenic atoms per m³, all ionised. Find its resistivity, ignoring holes ( $\mu_n = 0.14\,\mathrm{m^2\,V^{-1}\,s^{-1}}$ ).
In a doped crystal the product $n p$ equals $n_{i}^{2}$ . For the sample in question 6, find the concentration of holes ( $n_i = 1.0 \times 10^{16}\,\mathrm{m^{-3}}$ ), and comment.
Gatete says: "Diamond and silicon have the same crystal structure, so they should both be semiconductors." Evaluate his statement.

Show answersHide answers

Conductor: valence and conduction bands overlap (or a band is partly filled), so many free electrons. Semiconductor: full valence band, empty conduction band and a small gap (about 1 eV), so a few electrons are excited across at room temperature. Insulator: large gap (over about 5 eV), so practically none.
In a semiconductor, heating excites many more electrons across the gap (the number rises roughly as $e^{-\frac{E_g}{2kT}}$ ), creating electron–hole pairs; more carriers outweigh the extra scattering. In a metal, the number of free electrons is fixed, and stronger lattice vibrations scatter them more.
$\lambda = \frac{1240}{0.66} = 1.9 \times 10^{3}\,\mathrm{nm} = 1.9\,\mathrm{\mu m}$ , in the infrared.
A hole is a vacancy left in a covalent bond when an electron leaves it; it behaves as a positive charge $+ e$ . In a field, electrons drift against the field and holes along it; negative charge moving one way and positive charge moving the other both give conventional current in the direction of the field.
Gallium is in group 13 (three valence electrons): p-type, majority carriers holes. The crystal is neutral: each hole is balanced by a fixed negative gallium ion.
$\sigma = (1.60 \times 10^{-19})(2.0 \times 10^{22})(0.14) = 448\,\mathrm{S\,m^{-1}}$ ; $\rho = \frac{1}{448} = 2.2 \times 10^{-3}\,\Omega\,\mathrm{m}$ .
$p = \frac{n_i^2}{n} = \frac{(1.0 \times 10^{16})^2}{2.0 \times 10^{22}} = 5.0 \times 10^{9}\,\mathrm{m^{-3}}$ : far fewer holes than in pure silicon, because many of the extra electrons recombine with holes. This is why holes are called minority carriers and can be ignored for the conductivity.
Wrong. Structure alone does not decide it; the band gap does. Diamond's gap (5.5 eV) is so large that almost no electrons cross it at room temperature (fraction about $10^{-46}$ ), so it is an insulator; silicon's 1.12 eV gap makes it a semiconductor.

Key points

In a solid, atomic levels spread into bands; conduction needs electrons in a partly filled band. Conductor: bands overlap; semiconductor: small gap (Si 1.12 eV); insulator: large gap.
Intrinsic semiconductor: electron–hole pairs, $n = p = n_{i}$ , resistance falls as temperature rises; $\sigma = e(n\mu_n + p\mu_p)$ .
n-type: group 15 donors, electrons are majority carriers. p-type: group 13 acceptors, holes are majority carriers. Both are electrically neutral.
Photons with $hf \ge E_g$ create electron–hole pairs: $\lambda_{\max} = \frac{hc}{E_g}$ .

Sources and credits

9.6: Band Theory of Solids and 9.7: Semiconductors and Doping, from University Physics Volume 3 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Monocrystalline silicon ingot by ArticCynda, Wikimedia Commons, CC0.
Photo: NTC Thermistor by Soumyapatra13, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Semiconductors 1: intrinsic & extrinsic semiconductors (Higher Physics), Mr Smith's Physics online, embedded from YouTube.

The two diagrams, the experiment readings (illustrative), the worked examples and the practice exercises are written by Ganzaa. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1One atom in ten million

2Conductors, semiconductors and insulators

3From energy levels to energy bands

4Intrinsic semiconductors: electrons and holes

5Extrinsic semiconductors: doping

6Experiment 1.1: a semiconductor and a metal compared

One atom in ten million

Conductors, semiconductors and insulators

From energy levels to energy bands

Intrinsic semiconductors: electrons and holes

Extrinsic semiconductors: doping

Experiment 1.1: a semiconductor and a metal compared