Chapter 14 · Semiconductor Electronics: Materials, Devices and Simple Circuits

14.1 Metals, Semiconductors and Insulators: Energy Bands

Copper conducts about $10^{20}$ times better than glass, and silicon sits in between. The reason is not how many electrons a solid has but how its electron energy levels merge into bands — and how wide the gap is between the full band and the empty one.

The processor in a phone is a sliver of silicon about the size of a fingernail carrying more than ten billion switches. Silicon is not a metal and not an insulator: a pure crystal of it conducts electricity about a hundred billion times worse than copper, yet a trace of the right impurity, or a little light, or a few degrees of warming, changes its conductivity enormously. That controllability is what makes silicon, germanium and gallium arsenide the raw material of every diode, transistor, LED, solar cell and camera sensor — the family of devices this chapter is about.

In Chapter 3 we explained resistivity with free electrons drifting through a lattice of ions ($\rho=m/ne^2\tau$, Section 3.1) and saw that metals and semiconductors respond to heating in opposite ways (Section 3.2). But that picture cannot say why one solid has $10^{28}$ free electrons per cubic metre and another almost none. The answer comes from what happens to the energy levels of atoms when billions of them are packed together in a crystal: the levels spread into energy bands separated by forbidden gaps. In this section we classify solids by their resistivity, build the band picture, and use it to explain conductors, insulators and semiconductors — and why a semiconductor conducts better when it is warmed or illuminated.

Classifying Solids by Resistivity

Measured at room temperature, the resistivities of solids span more than twenty-five powers of ten. They fall roughly into three groups (the boundaries are not sharp):

Group	Resistivity $\rho$ ($\Omega$ m)	Conductivity $\sigma=1/\rho$ (S/m)	Examples
Metals (conductors)	$\sim10^{-8}$ to $10^{-6}$	$\sim10^{6}$ to $10^{8}$	silver, copper, aluminium, nichrome
Semiconductors	$\sim10^{-5}$ to $10^{6}$	$\sim10^{-6}$ to $10^{5}$	germanium, silicon, gallium arsenide
Insulators	$\sim10^{11}$ to $10^{19}$	$\sim10^{-19}$ to $10^{-11}$	glass, quartz, diamond, rubber, PVC

Resistivity alone does not define a semiconductor. What really sets the group apart is how it behaves: its resistivity falls steeply as the temperature rises (a negative temperature coefficient, the opposite of a metal), it drops when light falls on it, and it can be lowered a millionfold by adding a few impurity atoms per million.

📘 Definition 14.1 — Semiconductor

A semiconductor is a solid whose resistivity at room temperature lies between that of metals and insulators (roughly $10^{-5}$ to $10^{6}\ \Omega$ m), whose resistivity decreases as the temperature rises, and whose conductivity can be changed greatly by adding small, controlled amounts of impurity or by illuminating it.

Elemental semiconductors: silicon (Si) and germanium (Ge), both in group 14 with four valence electrons.
Compound semiconductors: inorganic — gallium arsenide (GaAs), indium phosphide (InP), cadmium sulphide (CdS), cadmium selenide (CdSe); organic — anthracene, doped phthalocyanines; and conducting polymers such as polypyrrole and polyaniline.

Almost all integrated circuits are made of silicon, which is abundant (sand is silicon dioxide), forms a superb insulating oxide, and works well up to about $150\,^\circ$C. Compound semiconductors are used where silicon cannot do the job — for example, to emit light.

✏️ Example 1 — Three bars, three kinds of solid

Three bars have the same shape: length $1.0$ cm and square cross-section $1.0\ \text{mm}\times1.0\ \text{mm}$. Their resistivities at room temperature are $\rho_A=1.7\times10^{-8}\ \Omega$ m, $\rho_B=2.3\times10^{3}\ \Omega$ m and $\rho_C=1.0\times10^{14}\ \Omega$ m. (a) Classify each material. (b) Find the resistance of each bar. (c) Find the conductivity of $B$.

💡 Show solution▶

(a) Classify $A$ is in the metal range (it is copper), $B$ is a semiconductor (it is pure silicon), $C$ is an insulator.

(b) Resistance $R=\rho L/A$ with $L/A=\dfrac{1.0\times10^{-2}}{1.0\times10^{-6}}=1.0\times10^{4}\ \text{m}^{-1}$: $R_A=1.7\times10^{-4}\ \Omega$, $R_B=2.3\times10^{7}\ \Omega$, $R_C=1.0\times10^{18}\ \Omega$.

Check $R_B/R_A=\rho_B/\rho_A\approx1.4\times10^{11}$ — the "hundred billion times" of the opening paragraph.

From Energy Levels to Energy Bands

An isolated atom has sharp, discrete energy levels (you will meet those of hydrogen in Chapter 12). Each level can hold only a limited number of electrons (the Pauli exclusion principle, from chemistry), so the electrons fill the levels from the bottom up.

Now bring $N$ identical atoms together to form a crystal. While they are far apart they do not affect one another, and all $N$ atoms have exactly the same levels. As the spacing shrinks to a few tenths of a nanometre, the outer electron clouds overlap and every electron feels the pull of neighbouring nuclei too. The exclusion principle forbids $N$ electrons from sharing one level, so each atomic level splits into $N$ slightly different levels. The inner levels, whose electrons stay close to their own nucleus, hardly split; the outer levels spread out by several electronvolts. Since $N$ is about $10^{22}$ for a cubic centimetre of solid, the $N$ levels are packed so closely that they form, for all practical purposes, a continuous range of allowed energies.

📘 Definition 14.2 — Energy band

An energy band is a range of closely spaced electron energy levels in a crystal, formed when one level of the isolated atom splits into $N$ levels as $N$ atoms are brought together. Ranges of energy that contain no allowed levels lie between the bands.

In silicon each atom has the outer configuration $3s^2\,3p^2$. The $3s$ and $3p$ levels together offer $8$ electron states per atom (2 in $3s$, 6 in $3p$), so $8N$ states in the crystal. At the actual atomic spacing of a silicon crystal these $8N$ states rearrange into two bands of $4N$ states each, with a gap between them. The $4N$ valence electrons exactly fill the lower band; the upper band is empty at absolute zero.

📘 Definition 14.3 — Valence band, conduction band and band gap

The valence band is the band containing the valence electrons; it is the highest band that is filled (or partly filled) at absolute zero. Its top edge is $E_V$.
The conduction band is the band just above the valence band. Electrons in it are free to move through the crystal and carry current. Its bottom edge is $E_C$.
The band gap (forbidden gap, energy gap) is $E_g=E_C-E_V$: no electron in a perfect crystal can have an energy inside it.

The key idea is this: an electron can carry current only if there are empty states just above its energy. An applied field gives every electron a tiny push, i.e. a tiny extra energy; an electron can accept it only if it can move into an unoccupied state of slightly higher energy. In a completely filled band there is no such state, so a full band carries no current — however many electrons it holds. A partly filled band conducts easily.

✏️ Example 2 — How close are the levels in a band?

Silicon has $5.0\times10^{28}$ atoms per cubic metre. Consider a grain of silicon of volume $1.0\ \text{mm}^3$. (a) How many atoms does it contain, and how many states are in its valence band? (b) Taking the valence band to be about $10$ eV wide (an illustrative round figure), estimate the average energy spacing between neighbouring states. (c) Compare with the thermal energy $k_BT\approx0.026$ eV at room temperature.

💡 Show solution▶

(a) Count $N=(5.0\times10^{28})(1.0\times10^{-9})=5.0\times10^{19}$ atoms; the valence band has $4N=2.0\times10^{20}$ states.

(b) Spacing $\Delta E\approx\dfrac{10\ \text{eV}}{2.0\times10^{20}}=5\times10^{-20}$ eV.

(c) Compare $\dfrac{0.026}{5\times10^{-20}}\approx5\times10^{17}$: the spacing is utterly negligible, so a band behaves as a continuous range of energies. A bigger crystal has more levels in the same width — even closer together.

A model crystal of $N$ atoms, each with a lower and an upper outer level. The left graph shows every level of the crystal against the atomic spacing $d$ (in units of the crystal's equilibrium spacing $a_0$): far apart, all $N$ atoms share the same two levels; as they approach, each level fans out into $N$ levels — a band. The ladder on the right shows the levels at the chosen spacing, with the electrons (two per level) filling from the bottom. Press Play to push the atoms together; change $N$ and the type of material. (Illustrative model: the energies are round numbers, not data for a real solid.)

material:atoms $N$:$d/a_0$:1.00

levels per band—

lower band width—

upper band width—

gap $E_g$—

Metals, Insulators and Semiconductors in the Band Picture

Whether a solid conducts now depends on two things: is its highest occupied band full or partly full, and if it is full, how big is the gap to the next band?

📐 Result 14.1 — Classification of solids by energy bands

(i) Metals (conductors). Either the conduction band is partly filled, or the valence and conduction bands overlap (there is no gap). Empty states lie immediately above the highest filled ones, so a very small field sets a huge number of electrons ($\sim10^{28}\ \text{m}^{-3}$) drifting. Resistivity is low.

(ii) Insulators. The valence band is completely full, the conduction band is empty, and the gap is large, $E_g\gt3$ eV (diamond: $E_g\approx5.4$ eV). Thermal energy cannot lift electrons across it, so there are no carriers. Resistivity is enormous.

(iii) Semiconductors. The band structure is like that of an insulator — full valence band, empty conduction band at $0$ K — but the gap is small, $E_g\lt3$ eV (silicon $\approx1.1$ eV, germanium $\approx0.7$ eV, gallium arsenide $\approx1.4$ eV). At $0$ K a semiconductor is an insulator; at room temperature a few electrons gain enough thermal energy to cross the gap. Each leaves behind an empty state in the valence band, so both bands can now carry a small current.

Why heating helps. At temperature $T$ the typical thermal energy of an electron is of order $k_BT$, where $k_B=8.62\times10^{-5}$ eV/K is the Boltzmann constant; at $300$ K, $k_BT\approx0.026$ eV — about $40$ times smaller than the gap of silicon. But thermal energy is shared out randomly, and a tiny fraction of the electrons receive much more than the average. The number that cross the gap rises extremely steeply with temperature and falls extremely steeply with $E_g$. That is why germanium (small gap) has far more free electrons at room temperature than silicon, why diamond has essentially none, and why a semiconductor's resistivity falls as it warms. (A metal is different: its number of free electrons is fixed, and heating only shortens the relaxation time — Section 3.2.)

Why light helps. A photon of energy $h\nu$ (Chapter 11) can lift an electron straight from the valence band into the conduction band, provided $h\nu\ge E_g$. Light of lower frequency passes through without being absorbed this way.

📐 Result 14.2 — Absorption of light across the band gap

A photon can excite an electron from the valence band to the conduction band of a semiconductor or insulator only if $h\nu\ge E_g$. The longest wavelength that can do so is $$\lambda_{\max}=\frac{hc}{E_g},\qquad \lambda_{\max}\ (\text{nm})\approx\frac{1240}{E_g\ (\text{eV})},$$ using $hc=1240$ eV nm. Light with $\lambda\gt\lambda_{\max}$ is not absorbed by this process; a material whose $\lambda_{\max}$ is shorter than $400$ nm is transparent to all visible light.

✏️ Example 3 — Why diamond is clear and silicon is not

Find the longest wavelength that can lift an electron across the gap in (a) silicon, $E_g=1.1$ eV; (b) germanium, $E_g=0.67$ eV; (c) diamond, $E_g=5.4$ eV. Which of them can absorb visible light ($400$–$700$ nm)? Use $hc=1240$ eV nm.

💡 Show solution▶

(a) Si $\lambda_{\max}=\dfrac{1240}{1.1}\approx1130$ nm $\approx1.1\ \mu$m (near infrared).

(b) Ge $\lambda_{\max}=\dfrac{1240}{0.67}\approx1850$ nm $\approx1.9\ \mu$m.

Conclusion Every visible photon has $\lambda\lt1130$ nm, so silicon and germanium absorb all visible light and look dark grey and opaque. Visible photons ($1.8$–$3.1$ eV) cannot cross diamond's $5.4$ eV gap, so diamond is transparent. Check: a silicon camera sensor detects all visible light and some near infrared — it does.

✏️ Example 4 — Thermal energy against the band gap

(a) Find $k_BT$ in eV at $300$ K and at $600$ K, and the ratio $E_g/k_BT$ for silicon ($E_g=1.1$ eV) at each temperature. (b) Pure silicon at $300$ K has about $1.5\times10^{16}$ conduction electrons per cubic metre, and $5.0\times10^{28}$ atoms per cubic metre. How many atoms are there for each conduction electron? Compare with copper, which has about one free electron per atom.

💡 Show solution▶

(a) 300 K $k_BT=(8.62\times10^{-5})(300)\approx0.0259$ eV; $\dfrac{E_g}{k_BT}=\dfrac{1.1}{0.0259}\approx42$.

(a) 600 K $k_BT\approx0.0517$ eV; $\dfrac{E_g}{k_BT}\approx21$. The ratio halves — and because the number of excited electrons depends on this ratio exponentially, it increases by a huge factor (Interactive 2 gives about $10^{5}$).

(b) Atoms per electron $\dfrac{5.0\times10^{28}}{1.5\times10^{16}}\approx3.3\times10^{12}$: only one atom in about three million million contributes a conduction electron. In copper every atom does — so copper has about $10^{12}$ times as many carriers.

✏️ Example 5 — Choosing a material from its band data

Three solids have these band structures at room temperature: $X$ — the valence and conduction bands overlap; $Y$ — a full valence band and an empty conduction band separated by $2.3$ eV; $Z$ — the same, but with a gap of $6.0$ eV. (a) Classify each. (b) Can $Y$ be used in a sensor that responds to red light of wavelength $650$ nm? (c) What is the longest wavelength $Y$ can detect?

💡 Show solution▶

(a) Classify $X$: metal (no gap). $Y$: semiconductor ($E_g\lt3$ eV). $Z$: insulator ($E_g\gt3$ eV).

(b) Red light $E=\dfrac{1240}{650}\approx1.91$ eV $\lt2.3$ eV: a red photon cannot cross the gap, so $Y$ does not respond to it.

Left: the band diagram of the chosen solid, drawn to scale in eV. Filled dots in the conduction band are free electrons; open circles in the valence band are the empty states they leave. The number of dots is symbolic (it follows the logarithm of the carrier density); the readouts give the actual numbers. Right: the density of conduction electrons against temperature on a logarithmic scale. Press Play to sweep the temperature, or drag the slider. (Silicon, germanium and copper use measured room-temperature values; the temperature dependence and the GaAs and diamond values come from the simple model $n_i\propto T^{3/2}e^{-E_g/2k_BT}$ — illustrative estimates.)

solid:$T$ (K):300

$k_BT$—

$E_g/k_BT$—

free electrons $n$ ($\text{m}^{-3}$)—

atoms per free electron—

⚠️ Common mistakes with energy bands

1. "More electrons means a better conductor." Diamond has more electrons per cubic metre than copper. What matters is whether there are empty states just above the occupied ones: a full band conducts nothing.

2. Putting electrons in the gap. In a perfect crystal no electron can have an energy inside the band gap. (Impurities can add a few levels there — Section 14.2.)

3. Comparing $E_g$ with the average thermal energy only. $k_BT\approx0.026$ eV is far smaller than $1.1$ eV, yet silicon still has $10^{16}$ free electrons per cubic metre: a tiny fraction of a huge number of electrons gets far more than the average.

4. Using $3$ eV as a sharp law. It is a convenient dividing line between semiconductors and insulators, not a law of nature.

5. Wrong units in $\lambda=hc/E_g$. With $hc=1240$ eV nm, put $E_g$ in eV to get $\lambda$ in nm; with $E_g$ in joules use $hc=1.99\times10^{-25}$ J m.

🚀 Beyond the board syllabus (JEE Main · NEET) — band gaps at work: LEDs, photodiodes and solar cells

Syllabus note: these devices are no longer in the NCERT book or the CBSE syllabus, and they are not in JEE Advanced, but they are listed in the NEET and JEE Main syllabi. Each is a p–n junction (Section 14.3), and each works because of the band gap.

Light-emitting diode (LED). In a forward-biased junction, conduction electrons fall back into empty valence-band states and give out the energy as a photon, $h\nu\approx E_g$. So the colour is set by the material: GaAs ($1.42$ eV) emits infrared near $870$ nm, gallium arsenide phosphide with $E_g\approx1.9$ eV emits red light near $650$ nm, and blue LEDs use gallium nitride-based alloys with $E_g\approx2.7$ eV. Silicon and germanium do not emit light efficiently, so LEDs need compound semiconductors with $E_g$ between about $1.8$ and $3$ eV for visible light. LEDs operate at low voltage, switch fast, last long and waste little power as heat.

Photodiode. A junction operated in reverse bias. Photons with $h\nu\gt E_g$ create electron–hole pairs near the junction; the field there sweeps them apart and the small reverse current rises in proportion to the light intensity — a light meter or optical-fibre receiver.

Solar cell. A large-area junction with no applied bias. Light creates electron–hole pairs, the junction field separates them, and the cell becomes a source of emf (about $0.5$–$0.6$ V for silicon) that drives current through a load — its I–V curve lies in the fourth quadrant. Materials with $E_g$ near $1.0$–$1.5$ eV (Si, GaAs, CdTe) match the solar spectrum best.

Exercises

Classifying solids by resistivity.

1 Pure silicon at $300$ K has a resistivity of $2.3\times10^{3}\ \Omega$ m. Find its conductivity.
2 A bar of pure germanium ($\rho=0.46\ \Omega$ m) is $2.0$ cm long with a square cross-section of side $1.0$ mm. Find its resistance.
3 Classify materials with room-temperature resistivities (a) $1.0\times10^{-7}\ \Omega$ m, (b) $0.50\ \Omega$ m, (c) $1.0\times10^{15}\ \Omega$ m.
4 Name two elemental and two inorganic compound semiconductors, and one organic polymer semiconductor.
5 Taking $\rho=1.7\times10^{-8}\ \Omega$ m for copper and $1.0\times10^{12}\ \Omega$ m for a glass, find the ratio of their resistivities.
6 State two properties, other than the value of its resistivity, that distinguish a semiconductor from a metal.
7 A copper wire $1.0$ km long has a cross-section of $1.0\ \text{mm}^2$. What length of pure silicon ($\rho=2.3\times10^{3}\ \Omega$ m) of the same cross-section has the same resistance?

💡 Show solutions▶

1 $\sigma=1/\rho=1/(2.3\times10^{3})\approx4.3\times10^{-4}$ S/m.

2 $R=\dfrac{\rho L}{A}=\dfrac{(0.46)(0.020)}{1.0\times10^{-6}}$ $\approx9.2\times10^{3}\ \Omega=9.2$ k$\Omega$.

3 (a) metal (conductor); (b) semiconductor; (c) insulator.

4 Elemental: Si, Ge. Inorganic compounds: e.g. GaAs, CdS (also InP, CdSe). Polymer: e.g. polypyrrole or polyaniline.

5 $\dfrac{1.0\times10^{12}}{1.7\times10^{-8}}\approx5.9\times10^{19}$.

6 Its resistivity falls as the temperature rises (a metal's rises), and its conductivity can be increased enormously by tiny amounts of impurity or by light.

7 Equal $R$ and equal $A$ need $\rho_1L_1=\rho_2L_2$: $L=\dfrac{(1.7\times10^{-8})(1000)}{2.3\times10^{3}}\approx7.4\times10^{-9}$ m — about $7$ nm, a few dozen atoms thick.

Energy bands. Silicon has $5.0\times10^{28}$ atoms per cubic metre.

1 What happens to one energy level of an isolated atom when $N$ identical atoms are brought close together to form a crystal?
2 A silicon crystal has a volume of $1.0\ \text{cm}^3$. How many states are in its valence band, and how many valence electrons fill them?
3 Define the band gap. Why can no electron in a perfect crystal have an energy inside it?
4 A sodium atom has one $3s$ electron, but the $3s$ level can hold two. Explain with bands why sodium is a metal.
5 A band $5.0$ eV wide in a crystal contains $1.0\times10^{22}$ levels. Estimate the spacing between neighbouring levels.
6 Magnesium atoms have two $3s$ electrons, which would fill the $3s$ band completely. Yet magnesium is a metal. Suggest why.
7 Sketch and label the band diagram of a pure semiconductor at $0$ K and at $300$ K.

💡 Show solutions▶

1 It splits into $N$ closely spaced levels — an energy band. The outer levels split most, because their electron clouds overlap most.

2 $N=(5.0\times10^{28})(1.0\times10^{-6})=5.0\times10^{22}$ atoms; the valence band has $4N=2.0\times10^{23}$ states, exactly filled by the $4N=2.0\times10^{23}$ valence electrons.

3 $E_g=E_C-E_V$, the energy range between the top of the valence band and the bottom of the conduction band. It contains no allowed energy levels of the crystal, so there is no state an electron could occupy there.

4 The $3s$ band has $2N$ states but only $N$ electrons: it is half full. Empty states lie just above the occupied ones, so electrons can gain energy from a field and conduct — the hallmark of a metal.

5 $\dfrac{5.0}{1.0\times10^{22}}=5.0\times10^{-22}$ eV — effectively continuous.

6 The $3s$ band overlaps the (empty) $3p$ band, so there is no gap: electrons at the top of the filled states have empty $3p$ states right next to them in energy.

7 At $0$ K: full valence band (top $E_V$), empty conduction band (bottom $E_C$), gap $E_g$ between. At $300$ K: the same, plus a few electrons (dots) at the bottom of the conduction band and an equal number of vacancies (open circles) at the top of the valence band.

III

Thermal and optical excitation. Use $hc=1240$ eV nm and $k_B=8.62\times10^{-5}$ eV/K.

1 Diamond has $E_g=5.4$ eV. What is the longest wavelength that can excite an electron across its gap?
2 Find the energy of a $600$ nm photon. Can it excite an electron across the gap of (a) silicon ($1.1$ eV), (b) gallium nitride ($3.4$ eV)?
3 Find $k_BT$ at $77$ K (liquid nitrogen) and the ratio $E_g/k_BT$ for germanium ($E_g=0.67$ eV).
4 At what temperature is $k_BT$ equal to one-fortieth of the band gap of silicon ($1.1$ eV)?
5 A semiconductor is transparent to light of wavelength longer than $870$ nm but absorbs shorter wavelengths. Find its band gap.
6 Explain why pure silicon is an insulator at $0$ K.
7 Classify solids whose band gaps are (a) $0.17$ eV, (b) $4.0$ eV, and (c) a solid whose valence and conduction bands overlap.
8 At room temperature pure germanium has about a thousand times as many free electrons per cubic metre as pure silicon. Explain why using band gaps.

💡 Show solutions▶

1 $\lambda_{\max}=1240/5.4\approx230$ nm (ultraviolet).

2 $E=1240/600\approx2.07$ eV. (a) Yes, $2.07\gt1.1$. (b) No, $2.07\lt3.4$.

3 $k_BT=(8.62\times10^{-5})(77)\approx6.6\times10^{-3}$ eV; $E_g/k_BT=0.67/0.00664\approx100$.

4 $k_BT=1.1/40=0.0275$ eV, so $T=\dfrac{0.0275}{8.62\times10^{-5}}\approx319$ K $\approx320$ K.

5 $E_g=\dfrac{1240}{870}\approx1.43$ eV (this is gallium arsenide).

6 At $0$ K no electron has any thermal energy to spare: the valence band is completely full and the conduction band completely empty. A full band carries no current and the empty one has no carriers.

7 (a) semiconductor (it is indium antimonide); (b) insulator ($\gt3$ eV); (c) metal.

8 Germanium's gap ($0.67$ eV) is smaller than silicon's ($1.1$ eV). The number of electrons thermally excited across the gap falls extremely steeply as $E_g$ increases, so the smaller gap gives far more carriers at the same temperature.

Applications and longer problems. $hc=1240$ eV nm, $e=1.6\times10^{-19}$ C.

1 A TV remote control sends infrared light of wavelength $940$ nm. Can a silicon camera sensor ($E_g=1.1$ eV) detect it? (Try it: point a remote at a phone camera.)
2 Optical-fibre links use light of wavelength $1550$ nm. What is the largest band gap a detector material may have? Would silicon or germanium ($0.67$ eV) work?
3 Window glass is transparent to all visible light ($400$–$700$ nm). What can you say about its band gap?
4 A $3.0$ eV photon is absorbed by silicon and lifts one electron across the gap. How much of its energy is in excess of $E_g$, and what fraction of the photon's energy is that? (The excess ends up heating the crystal.)
5 A thermistor made of a semiconducting oxide has a resistance that falls when it is warmed. Explain this with the band picture, and say why a metal wire behaves oppositely.
6 Modelling. A silicon light sensor absorbs $1.0$ mW of light of wavelength $800$ nm. Assuming every photon creates one conduction electron that is collected, find the number of photons absorbed per second and the resulting current.

💡 Show solutions▶

1 $E=1240/940\approx1.32$ eV $\gt1.1$ eV, so yes — the remote's flash shows up on the phone screen (eyes cannot see it).

2 $E_g\le\dfrac{1240}{1550}=0.80$ eV. Silicon ($1.1$ eV) cannot detect it; germanium ($0.67$ eV) can.

3 Even violet ($400$ nm, $3.1$ eV) is not absorbed, so $E_g\gt3.1$ eV — glass is an insulator.

4 Excess $=3.0-1.1=1.9$ eV, a fraction $1.9/3.0\approx0.63$ (63%) of the photon's energy.

5 Warming excites more electrons across the gap, so the number of carriers $n$ rises steeply and $\rho=m/ne^2\tau$ falls. In a metal $n$ is fixed; heating only increases collisions (smaller $\tau$), so $\rho$ rises.

6 Photon energy $=1240/800=1.55$ eV $=2.48\times10^{-19}$ J. Photons per second $=\dfrac{1.0\times10^{-3}}{2.48\times10^{-19}}\approx4.0\times10^{15}\ \text{s}^{-1}$. Current $=(4.03\times10^{15})(1.6\times10^{-19})\approx6.5\times10^{-4}$ A $=0.65$ mA.

🎯 Exam corner — energy bands BoardJEE MainNEET

E1. Board · 3 marks Using energy-band diagrams, distinguish between a metal, an insulator and a semiconductor. Why does the conductivity of a semiconductor increase with temperature?

💡 Show solution▶

Diagrams Metal: conduction band partly filled, or valence and conduction bands overlapping — no gap. Insulator: full valence band, empty conduction band, wide gap $E_g\gt3$ eV. Semiconductor: same as an insulator but with a narrow gap $E_g\lt3$ eV, and a few electrons in the conduction band at room temperature.

Temperature Raising $T$ gives more valence electrons enough energy to cross the small gap, so the number of carriers (conduction electrons and the vacancies they leave) increases rapidly, and the conductivity rises.

Exam tip One mark is for the three labelled diagrams ($E_C$, $E_V$, $E_g$ marked). Draw them side by side with the same energy axis.

E2. Board · 1 mark Assertion (A): Pure silicon behaves as an insulator at absolute zero. Reason (R): At absolute zero the valence band of silicon is completely filled and the conduction band is completely empty. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

💡 Show solution▶

Concept A full band cannot carry current and an empty band has no carriers, so at $0$ K there is no conduction at all.

Answer (A).

E3. JEE Main A pure semiconductor absorbs light only of wavelengths shorter than $1240$ nm. Its band gap is (A) $0.5$ eV (B) $1.0$ eV (C) $1.24$ eV (D) $2.0$ eV.

💡 Show solution▶

Concept $E_g=\dfrac{hc}{\lambda_{\max}}=\dfrac{1240\ \text{eV nm}}{1240\ \text{nm}}=1.0$ eV.

Answer (B). Exam tip: memorise $hc\approx1240$ eV nm; it turns every photon question into one division.

E4. NEET Carbon (as diamond), silicon and germanium all have four valence electrons per atom. Their band gaps satisfy (A) $E_{g,\text{Ge}}\gt E_{g,\text{Si}}\gt E_{g,\text{C}}$ (B) $E_{g,\text{C}}\gt E_{g,\text{Si}}\gt E_{g,\text{Ge}}$ (C) $E_{g,\text{Si}}\gt E_{g,\text{C}}\gt E_{g,\text{Ge}}$ (D) all three are equal.

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Values Diamond $\approx5.4$ eV (insulator), Si $\approx1.1$ eV, Ge $\approx0.7$ eV. Going down group 14 the atoms get larger, the outer electrons are less tightly bound and the gap shrinks.

Answer (B).

E5. NEETJEE Main (Beyond NCERT: LEDs.) A red LED emits light of wavelength $620$ nm. The band gap of its semiconductor is about (A) $1.2$ eV (B) $1.6$ eV (C) $2.0$ eV (D) $2.5$ eV.

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Concept LEDs are beyond the CBSE and JEE Advanced syllabi but in NEET and JEE Main. The emitted photon has $h\nu\approx E_g$: $E_g\approx\dfrac{1240}{620}=2.0$ eV.

Answer (C).

E6. Board · Case study Phone cameras use silicon sensors ($E_g=1.1$ eV). Each pixel absorbs photons that lift electrons across the band gap, and the charge collected measures the brightness. (i) (2 marks) Find the longest wavelength a silicon pixel can detect, and decide whether it can detect $940$ nm infrared from a TV remote. (ii) (1 mark) In dim light, images from a hot phone show more random "noise". Suggest why. (iii) (1 mark) Which of germanium ($0.67$ eV) and gallium nitride ($3.4$ eV) could be used for a sensor at $1550$ nm?

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(i) $\lambda_{\max}=1240/1.1\approx1130$ nm. Since $940\lt1130$ nm (photon energy $1.32$ eV $\gt1.1$ eV), yes, it is detected.

(ii) At higher temperature more electrons are excited across the gap thermally, without any light; they are counted as if they were signal.

(iii) A $1550$ nm photon has $0.80$ eV: only germanium ($0.67\lt0.80$) can absorb it. Exam tip: in case studies, write the inequality ($h\nu\ge E_g$) explicitly — it carries the mark.

🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Which semiconductor should an engineer choose to detect light in a pulse oximeter, an optical-fibre receiver and an ultraviolet (UV) index meter — and why can no single material do all three well?

Why it matters. Pulse oximeters, clipped to a fingertip in hospitals worldwide, compare how blood absorbs red ($660$ nm) and infrared ($940$ nm) light to measure oxygen levels. The internet's undersea fibre cables carry light at $1310$ and $1550$ nm. UV index meters for public-health warnings in Australia, India and Europe must respond to UV ($280$–$400$ nm) but not to the far brighter visible sunlight. A detector absorbs a photon only if $h\nu\ge E_g$ — so the band gap decides everything.

What to do.

For silicon ($1.12$ eV), germanium ($0.66$ eV), indium gallium arsenide ($0.75$ eV), gallium arsenide ($1.42$ eV), silicon carbide ($3.26$ eV) and gallium nitride ($3.4$ eV), compute $\lambda_{\max}=hc/E_g$ with $hc=1240$ eV nm. Tabulate.
For each application, list the materials that can absorb the wavelengths needed.
For the UV meter, you also need a material that does not absorb visible light. Which materials qualify?
A material with a small gap absorbs more wavelengths but also has more thermally excited electrons ("dark current"). Use the band picture to explain why fibre-optic detectors are sometimes cooled.

Tools: Calculator or spreadsheet; manufacturers' data sheets for photodetectors (optional).

Hand in: The table, a recommended material for each application with a one-sentence justification, and your answer to step 4.

Going further. Indium gallium arsenide is an alloy: changing the proportion of indium changes $E_g$. Find out how engineers "tune" band gaps by alloying, and why that matters for LEDs as well as detectors.

Instructor note: $\lambda_{\max}\approx1107$ nm (Si), $1879$ nm (Ge), $1653$ nm (InGaAs), $873$ nm (GaAs), $380$ nm (SiC), $365$ nm (GaN). Oximeter: Si works for both $660$ and $940$ nm (GaAs only for $660$ nm; $940\gt873$). Fibre: Ge or InGaAs (Si and GaAs fail). UV meter: SiC or GaN — they absorb UV but are blind to visible light; Si absorbs UV too but would be swamped by visible light. Small gap → more thermally generated carriers (dark current/noise); cooling reduces $k_BT$ and hence the noise.

🧪 Self-check quiz

0/5 correct

Q1. When $N$ identical atoms come together to form a crystal, one energy level of the isolated atom

Q2. In an insulator at room temperature

Q3. The longest wavelength that can excite an electron across a band gap of $2.0$ eV is about

Q4. At absolute zero, pure germanium is

Q5. When a pure semiconductor is heated, its resistivity

← Chapter 14 — Semiconductor Electronics: Materials, Devices and Simple Circuits 14.2 Intrinsic and Extrinsic Semiconductors →