CHEM 548: Materials Chemistry

Lecture 20: Semiconductors and the Conductivity of Transition Metal Compounds

Wed 11/11/2026 · Meeting 22

Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 10 §§10.2–10.4

Learning Objectives: Semiconductors

By the end of this lecture, you will be able to:

Explain the origin of low conductivity in intrinsic semiconductors and the role of band gaps in limiting charge carrier concentration
Describe donor and acceptor doping mechanisms and their effects on semiconductor conductivity through band structure modifications
Understand the temperature dependence of semiconductor conductivity including freeze-out, saturation, and intrinsic regimes
Explain P-N junction behavior including rectification, band bending, and the formation of depletion regions
Describe practical applications: photovoltaic cells, light-emitting diodes (LEDs), and metal-oxide-semiconductor field-effect transistors (MOSFETs)

Conductivity of Select Materials

Substance	σ (S/m)	Substance	σ (S/m)
Ag	6.2 × 10⁷	Bi₂Ru₂O₇	2 × 10⁵
Cu	5.9 × 10⁷	LaNiO₃	1 × 10⁵
Al	3.8 × 10⁷	doped polyacetylene	8 × 10⁴
Na	2.1 × 10⁷	Fe₃O₄	2 × 10⁴
ReO₃	1.1 × 10⁷	YBa₂Cu₃O₇	1 × 10²
Ti	2.5 × 10⁶	Ge	2 × 10⁰
La	1.6 × 10⁶	Si	10⁻³
SrMoO₃	1.0 × 10⁶	NiO	10⁻⁸
Bi	7.7 × 10⁵	Al₂O₃	10⁻¹²
Mn	6.2 × 10⁵	S	10⁻¹⁵
NbN	4 × 10⁵	SiO₂ (Quartz)	10⁻¹⁶
TiO	3 × 10⁵	Teflon	10⁻²²

Fermi-Dirac Distribution in a Semiconductor

The Fermi-Dirac distribution plotted across a semiconductor band gap, with holes in the valence band and electrons in the conduction band; a click swaps the bottom formula for the carrier-concentration relation and the silicon one-electron-per-trillion-atoms annotation.

The characteristic feature of a semiconductor is a band gap—an energy gap between filled and empty bands.

Returning to the Fermi-Dirac distribution (shown at the bottom) and examining its effect on occupied crystal orbitals
Although thermal smearing occurs, there are no crystal orbitals near the Fermi level
Only a few missing electrons in the valence band (called holes) and just a few electrons in the conduction band
Semiconductors therefore have far fewer charge carriers than metals
The Fermi-Dirac expression can be simplified — at any given energy, we obtain this relationship

[advance: swap in the carrier-concentration relation]

Electron carrier concentration in an intrinsic semiconductor depends exponentially on the band gap
Silicon, band gap 1.1 eV: approximately 10¹⁶ electrons per cubic meter in the conduction band — against 10²⁸ atoms in that same cubic meter
Just one out of every trillion silicon atoms contributes a conduction electron
Intrinsic conductivity is low because the charge carrier concentration is low

A trillion-to-one carrier ratio explains the low conductivity, and it also tells us the fix: add carriers on purpose.

Donor and Acceptor Doping

To achieve meaningful charge carrier concentrations, we use a process called doping.

This involves replacing host atoms with aliovalent dopants

[advance: zoom in on the donor case]

Phosphorus: one more valence electron than silicon — in the localized bonding picture, this extra electron is not needed to satisfy local bonding requirements and can move through the crystal relatively easily
Band picture: phosphorus fills all valence band states and contributes one additional electron just below the conduction band — this electron can be easily thermally excited into the conduction band, where it moves freely through the crystal

[advance: bring back the acceptor panel]

Acceptor doping: aliovalent substitution with an element having one fewer valence electron than the host — boron is a classic acceptor dopant in silicon
Localized picture: electrons from neighboring atoms can fill the missing electron position (the hole) to complete the local bonding
Band picture: an empty state just above the valence band — an electron moves into it, becoming trapped on the boron atom, and the valence band is no longer completely filled, so the hole can now move through the crystal and carry charge
A hole is a missing electron, so it has a positive charge: under an applied electric field, holes move toward the negative electrode, electrons toward the positive — opposite directions for the two carriers

Before we treat that donor electron as free, we should ask how tightly the phosphorus nucleus holds it.

Hydrogenic type orbital in a donor doped semiconductor

A positively charged dopant in a host lattice with its donated electron orbiting on a large dashed circle.

Black dot: positively charged dopant (e.g. P_Si•). Open circles: host lattice atom (e.g. Si). Dashed circle: the donated e⁻.

r_d = ε_r (m_e/m*) 0.053 nm

Although the electron Although the electron from a donor dopant like phosphorus might seem free to move anywhere, remember that the phosphorus nucleus has one more positive charge than a silicon nucleus.

The positively charged dopant attracts its donated electron — a bound state analogous to a hydrogen atom
Hydrogen-like treatment with two key modifications
First: bound state radius enlarged by the dielectric constant of the semiconductor, which screens the dopant's positive charge
Second: the electron's effective mass in the crystal
Result: bound states with radii up to ~10 nm — quite large compared to atomic scales
Acceptor dopants behave analogously, with the hole attracted to the negatively charged acceptor

That bound state has an energy we can place right on the band diagram.

Energy Levels in a Doped Semiconductor

Band diagram showing donor and acceptor levels inside the gap with their ionization energies, and a table of donor and acceptor ionization energies for P, As, Sb, B, Al, and Ga in silicon and germanium.

Carrier Concentration vs. Temperature

Carrier concentration over donor concentration plotted against temperature, showing the freeze-out regime, the flat saturation regime where n approximately equals N_D, and the steep intrinsic regime.

This gives doped semiconductors an unusual temperature dependence.

Dashed line: conductivity increases exponentially at high temperature
Unlike metals, whose conductivity decreases with temperature — here the carrier concentration grows exponentially, outweighing the scattering from lattice vibrations that dominates in metals
Very low temperatures: not enough thermal energy to free electrons from donors — the freeze-out regime, low conductivity
Modest temperatures (often below room temperature): all dopant electrons thermally activated — the saturation regime
There: nearly all dopant carriers active, valence-to-conduction excitation minimal — little temperature dependence, advantageous for many devices
Still higher: carriers excited directly across the band gap begin to dominate — the intrinsic regime, conductivity again increases rapidly

With the carriers accounted for, we can write the conductivity itself.

Mobility and Conductivity

	E_g (eV)	μ_e (m²V⁻¹s⁻¹)	μ_h (m²V⁻¹s⁻¹)
Elements
Si	1.11 (indirect)	0.19	0.050
Ge	0.67 (indirect)	0.38	0.182
III–V
GaAs	1.43 (direct)	0.90	0.050
InAs	0.36 (direct)	3.30	0.046
InSb	0.18 (direct)	8.00	0.075
II–VI
ZnS	3.6 (direct)	0.012	0.0005
ZnSe	2.58 (direct)	0.053	0.0016
CdTe	1.50 (direct)	0.030	0.0065

σ = σ_n + σ_p = nμ_ee + pμ_he

To describe the conductivity of a semiconductor, we use the expressionσ = n q μₙ + p q μₚwhere conductivity is the product of carrier concentration, charge, and mobility.

Both electrons and holes contribute, so their conductivities add together
Most semiconductors are intentionally doped n-type or p-type — in practice, one carrier dominates
Representative mobilities: III–V semiconductors (like GaAs) much higher than elemental (like Si, Ge) or II–VI
Higher because their crystal structures produce lower effective masses for charge carriers and weaker scattering from lattice vibrations and impurities, allowing electrons to move more freely through the material
General rule: electrons have higher mobility than holes

Let's pause and check your understanding with the first Gradescope quiz.

See Gradescope Quiz 1

1. What is the primary reason for the low conductivity of an intrinsic (pure) semiconductor like silicon?

2. When an acceptor dopant like boron (which has one fewer valence electron than silicon) is added, what type of charge carrier is created?

3. How does the conductivity of a semiconductor generally change as temperature increases (in the intrinsic regime)?

4.In a doped semiconductor, what is the "saturation regime"?

The p-n Junction

Separate p-type and n-type semiconductors before equilibrium, with their Fermi levels and mobile holes and electrons; a click joins them, aligning the Fermi levels, bending the bands, and opening the depletion region.

Now let’s talk about a few simple but important semiconductor architectures.

The p–n junction — arguably one of the most important devices in all of science
One side doped p-type (holes in the valence band), the other n-type (electrons in the conduction band); brought together, they form the p–n junction

[advance: join the two sides and come to equilibrium]

Mobile electrons from the n-side find the mobile holes on the p-side and recombine — the electron fills the hole, both disappear as mobile carriers
The ionized dopant atoms left behind are fixed in the lattice — a depletion region with very few mobile carriers, similar to an intrinsic semiconductor; the remaining ionized donors and acceptors create an electric field across the junction
Fermi level: near the conduction band in n-type, near the valence band in p-type; joined, it must align across both sides — band bending across the depletion region
Imagine being an electron near the p-side conduction band: you would feel the electric field pulling you toward the n-side
In these band diagrams: electrons move downhill, holes move uphill

Now watch what this junction does when we apply a voltage.

p-n Junction Under Bias

Band diagrams of a p-n junction under reverse and forward bias with the corresponding circuits, and the rectifying current-voltage curve.

The p-n Rectifying Junction

Photovoltaic Cell

Another important application of a p–n junction is in photovoltaic cells and light-emitting diodes (LEDs).

Photovoltaic cell: incoming photons excite electrons from the valence band to the conduction band
If within the depletion region, the built-in electric field separates the electron and hole before they can recombine — electron to the n-type side, hole to the p-type side
Design trade-off: a larger band gap gives higher voltage — the maximum potential drop cannot exceed the band gap energy
But a larger gap also means fewer photons have enough energy to be absorbed
Balance: small enough to absorb sunlight efficiently, large enough for useful voltage — optimal is typically around 1.5 electron volts

And if light in gives current out, we can run the whole thing in reverse.

Light-Emitting Diode

A forward-biased p-n junction driving electrons and holes into the depletion region where they recombine and emit light across the band gap.

LED Colors and the Band Gap

Band diagrams for InGaN and AlGaAs LEDs emitting blue, green, and red, with emission wavelengths for GaN and GaAs alloy compositions listed below.

Metal oxide field effect transistor (MOSFET)

Cross-section of a MOSFET with source, drain, gate electrode, gate dielectric, and channel labeled.

Cross-section labels: source electrode, gate electrode, gate dielectric of thickness d, drain electrode, source n-Si, channel, drain n-Si, p-Si substrate, channel length L.

2006 MOSFET	2010 MOSFET
L = 65 nm	L = 32 nm
d = 1.2 nm	d = ~7 nm
Gate dielectric: SiO₂ (ε_r = 3.9)	Gate dielectric: HfO₂ (ε_r = 30)
Gate electrode: n-doped Si	Gate electrode: TiN

Switching a MOSFET

A MOSFET on a p-type substrate with n-doped source and drain and no applied gate voltage; a click adds the gate-voltage-on panel where an n-type channel forms under the gate dielectric.

MOSFET Scaling and High-k Dielectrics

Metal oxide field effect transistor (MOSFET)

The same MOSFET cross-section, now the anchor for the story of shrinking channels and high-k gate dielectrics.

Cross-section labels: source electrode, gate electrode, gate dielectric of thickness d, drain electrode, source n-Si, channel, drain n-Si, p-Si substrate, channel length L.

2006 MOSFET	2010 MOSFET
L = 65 nm	L = 32 nm
d = 1.2 nm	d = ~7 nm
Gate dielectric: SiO₂ (ε_r = 3.9)	Gate dielectric: HfO₂ (ε_r = 30)
Gate electrode: n-doped Si	Gate electrode: TiN

About 10 or 15 years ago, there was a major change in transistor architecture and materials.

For decades the strategy was simple: shrink the transistor
Smaller lateral dimension L → smaller gate-capacitor area; capacitance decreases with area, so engineers compensated by making the gate dielectric thinner
By around 2006, at about 65 nanometers, the dielectric was down to 12 angstroms (1.2 nanometers) — quantum tunneling across it became a serious problem, making further thinning impossible
For many decades the dielectric was SiO₂: easy to form by oxidizing silicon, but a relatively low dielectric constant (~4)
Solution: amorphous hafnium oxide (HfO₂), dielectric constant about 30 — a physically thicker layer at the same capacitance

Before we close this half, look at our conductivity table one more time.

Conductivity of Select Materials: NiO and TiO

The conductivity table again, now with red boxes around NiO at ten to the minus eight and TiO at three times ten to the five siemens per meter.

See Gradescope Quiz 2

1. In a p-n junction at equilibrium (no external voltage), what is the fundamental origin of the built-in electric field within the depletion region?

2. When a p-n junction is formed, the conduction and valence bands "bend" across the depletion region. What is the primary physical reason for this band bending?

3. Applying a reverse bias to a p-n junction (e.g., positive voltage to the n-type side) stops significant current flow. How does it accomplish this?

4. In an n-channel MOSFET built on a p-type substrate, how does applying a strong positive voltage to the gate electrode "turn on" the transistor?

5. Why was it necessary for the semiconductor industry to switch from silicon dioxide (SiO₂) to a high-k dielectric like hafnium oxide (HfO₂) for the gate?

See Gradescope Quiz 2 In a p-n junction at equilibrium (no external voltage), what is the fundamental origin of the built-in electric field within the depletion region?

2. When a p-n junction is formed, the conduction and valence bands "bend" across the depletion region. What is the primary physical reason for this band bending?
3. Applying a reverse bias to a p-n junction (e.g., positive voltage to the n-type side) stops significant current flow. How does it accomplish this?
4. In an n-channel MOSFET built on a p-type substrate, how does applying a strong positive voltage to the gate electrode "turn on" the transistor?
5. Why was it necessary for the semiconductor industry to switch from silicon dioxide (SiO2) to a high-k dielectric like hafnium oxide (HfO2) for the gate?

After the quiz, here is your homework for this half.

Learning Objectives: Conductivity of Transition Metal Compounds

After this lecture, you will be able to:

Explain how electron-electron repulsions determine conductivity in transition metal compounds
Apply the Hubbard model to predict metallic vs. insulating behavior by comparing bandwidth (W) and on-site repulsion (U)
Identify Mott-Hubbard insulators and explain their origin
Predict conductivity trends across rock salt structure oxides based on d-orbital contraction and metal-metal distance
Analyze conductivity patterns in perovskite structures and explain differences between 3+ and 4+ oxidation states

Conductivity of Select Materials: NiO and TiO

Substance	σ (S/m)	Substance	σ (S/m)
Ag	6.2 × 10⁷	Bi₂Ru₂O₇	2 × 10⁵
Cu	5.9 × 10⁷	LaNiO₃	1 × 10⁵
Al	3.8 × 10⁷	doped polyacetylene	8 × 10⁴
Na	2.1 × 10⁷	Fe₃O₄	2 × 10⁴
ReO₃	1.1 × 10⁷	YBa₂Cu₃O₇	1 × 10²
Ti	2.5 × 10⁶	Ge	2 × 10⁰
La	1.6 × 10⁶	Si	10⁻³
SrMoO₃	1.0 × 10⁶	NiO	10⁻⁸
Bi	7.7 × 10⁵	Al₂O₃	10⁻¹²
Mn	6.2 × 10⁵	S	10⁻¹⁵
NbN	4 × 10⁵	SiO₂ (Quartz)	10⁻¹⁶
TiO	3 × 10⁵	Teflon	10⁻²²

Conduction on a Ti³⁺ (d¹) Lattice

A square lattice of Ti3+ ions each carrying one d electron; moving an electron between sites creates a Ti2+ and Ti4+ pair, shown before and after the hop.

Hubbard U

The Hubbard model considers interactions (repulsions) between electrons on the same site, while neglecting longer range electron-electron interactions. The strength of the onsite electron-electron interactions is given by the Hubbard U, which we can estimate for gas phase ions.

Ti³⁺(g) → Ti⁴⁺(g) + e⁻	IE of Ti³⁺ = 4th IE of Ti = 43.3 eV
Ti³⁺(g) + e⁻ → Ti²⁺(g)	H_eg of Ti³⁺ = −(3rd IE of Ti) = −29.3 eV
2 Ti³⁺(g) → Ti⁴⁺(g) + Ti²⁺(g)	U = IE + H_eg = 14.0 eV

In a solid U is smaller (2-5 eV for 3d TM ions) because:

the surrounding lattice is polarized by the charge of the site
the electron wavefunctions are more spread out than they are on a gas phase ion

H_eg (electron gain enthalpy) is equivalent to electron affinity (with the opposite sign)

How can we model these electron-electron repulsions?

The Hubbard model: only repulsions between electrons on the same site, neglecting longer-range electron-electron interactions
Gas phase atoms: we can determine the on-site repulsion from ionization energies
Ti³⁺ → Ti⁴⁺ + e⁻ requires the ionization energy of Ti³⁺: 43.3 eV; at the receiving site, Ti³⁺ → Ti²⁺ is the electron affinity (or electron gain enthalpy), the negative of the third ionization energy: -29.3 eV
Sum: 2Ti³⁺ → Ti⁴⁺ + Ti²⁺ costs 14 eV — substantial; compare silicon's 1 eV band gap
In a solid like titanium oxide, U is much smaller — typically 2-5 eV — for two reasons
First: the surrounding lattice can relax or polarize — oxygen ions can shift closer to Ti⁴⁺ and away from Ti²⁺, reducing the penalty
Second: electron wavefunctions are more spread out in extended solids than in isolated gas phase atoms — together these lower U from 14 eV to just a few eV

U is only half the comparison; the other half is the band width.

The $d_{xy}$ Band in a Square TiN Plane

The d_xy band of a square TiN plane dispersing from bonding at Gamma to antibonding at M, with the crystal-orbital pictures at both points.

Mott-Hubbard Insulator

Two limits of the Hubbard model: weak Ti-Ti interaction with W less than U gives upper and lower Hubbard bands split by a gap, a Mott-Hubbard insulator; strong interaction with W greater than U gives a metal.

Consider two extremes.

Titanium atoms far apart, or weak interaction: relatively narrow bands — band width smaller than the Hubbard U, and we effectively have a gap
A note: the Hubbard model is really a single-band system — the upper and lower Hubbard bands can each hold only one electron
Key point: when the on-site repulsion (Hubbard U) exceeds the band width, one filled band and one empty band — a gap
This yields a Mott-Hubbard insulator: insulating due to electron-electron repulsions
Conversely, strong overlap of the titanium d orbitals, W larger than U: partially filled bands — metallic behavior
In summary, the conductivity of transition metal compounds depends on two factors: the band width and the electron-electron repulsions estimated by the Hubbard U

Time to test this framework on real compounds.

Rock Salt Transition Metal Oxides

Properties of MO Rock Salt Compounds

Table of MO rock salt compounds from TiO to NiO with metal-metal distance, d-orbital radial maximum, electrical properties running from metallic through semimetallic to semiconducting, and magnetic properties.

Starting with titanium oxide (TiO), which has a d² configuration, we see metallic behavior.

The column shows the titanium-titanium distance and the maximum in the radial distribution function — a proxy for d orbital size
Vanadium oxide, V²⁺ with d³: semi-metallic, at the borderline between metal and semiconductor
MnO, d⁵: manganese ions now farther apart (Mn²⁺ larger than Ti²⁺), yet the d orbitals are smaller — clearly semiconducting, with antiferromagnetic ordering often associated with semiconductors
Semiconducting continues across the rest: the d orbitals keep contracting, while the distance between transition metal ions changes little — decreasing slightly, but not by the same percentage as the d orbital contraction
Contraction reduces overlap → bandwidth decreases; because the d orbitals are smaller, placing two electrons in the same orbital incurs a larger energy penalty U
Summary: bandwidth down, U up across the series — so only TiO is metallic, the later rock salt oxides are not

The rock-salt story has a perovskite counterpart.

Perovskite Transition Metal Oxides

Perovskites with M³⁺ and M⁴⁺ Ions

Table of LaM3+O3 perovskites from LaScO3 to LaNiO3 with electron configurations, electrical properties, and magnetic properties; a click adds the SrM4+O3 rows where metallic behavior becomes the rule.

Here are the properties of several first-row transition metal ions in perovskite structures.

The table: 3+ ions, from no d electrons in scandium to progressively filled d orbitals
Semiconducting everywhere across this series — the Hubbard U exceeds the bandwidth; only at LaNiO₃ do we see metallic behavior
Why: the 3+ oxidation state is quite high for nickel → substantial covalency with oxygen; and the partially filled eg band is wider than the t₂g
We might have expected metallic LaMnO₃, one electron in the eg orbitals — but the d⁴ count triggers a cooperative Jahn-Teller distortion, which narrows the bands and localizes the electrons
Ni³⁺ could also undergo a Jahn-Teller distortion — the compound relieves the instability differently, by delocalizing the electrons in the eg orbitals instead

[advance: add the 4+ perovskite table]

With 4+ ions, metallic behavior is the rule rather than the exception
Why: 4+ ions are smaller and more highly charged → larger covalency with oxygen and wider bands — this increased bandwidth is sufficient to maintain metallic conductivity for several of these compounds
Exceptions: SrTiO₃ semiconducting — no electrons in the d orbitals; SrMnO₃ semiconducting — d³ means half-filled t₂g orbitals
The U for SrMnO₃: moving an electron from one Mn⁴⁺ to another creates Mn³⁺ and Mn⁵⁺; Mn³⁺ is typically a high-spin d⁴ ion, so the electron must go into the eg orbitals — U especially large
In general, ions with half-filled configurations have especially large U — why SrMnO₃ is a semiconductor rather than a metal

One more pass through the series, now with the 4d metals.

4d Perovskites

Table of AMO3 perovskites with 4d metals from KNbO3 to LaPdO3, metallic almost across the board except the d0 and t2g-filled d6 compounds.

See Gradescope Quiz 3

1. Which of the following best explains the dramatic difference in conductivity between NiO and TiO, despite both having the rock salt structure and partially filled d-orbitals?

2. In the context of the Hubbard model, which condition is most likely to lead to a Mott-Hubbard insulator?

3. Considering transition metal oxides with the rock salt structure (TiO, VO, MnO), why does the conductivity trend from metallic (TiO) to semiconducting (MnO) as we move across the period?

4. Strontium manganate (SrMnO₃) is a semiconductor, even though other 4+ perovskites exhibit metallic behavior. This is attributed to a particularly large Hubbard U. What characteristic of the Mn⁴⁺ (d³) electron configuration contributes to this exceptionally large U?

Which of the following best explains the dramatic difference in conductivity between NiO and TiO , despite both having the rock salt structure and partially filled d-orbitals?

2. In the context of the Hubbard model, which condition is most likely to lead to a Mott-Hubbard insulator?
3. Considering transition metal oxides with the rock salt structure (e.g., TiO , VO, MnO ), why does the conductivity trend from metallic ( TiO ) to semiconducting ( MnO ) as we move across the period?
4. Strontium manganate (SrMnO3) is a semiconductor, even though other 4+ perovskites exhibit metallic behavior. This is attributed to a particularly large Hubbard U. What characteristic of the Mn4+ (d3) electron configuration contributes to this exceptionally large U? See Gradescope Quiz 3

When you are done, here is the homework.

Conductivity of Select Materials: Conducting Polymers

The conductivity table one final time, with doped polyacetylene at eight times ten to the four and Teflon at ten to the minus twenty-two highlighted in yellow.

Practice questions

Work the Lecture 20 practice questions before the next class. They cover intrinsic and doped semiconductors, the p-n junction and its devices, the Hubbard model, and conductivity trends in transition metal oxides.

Open the Lecture 20 practice questions

Every question carries a worked explanation, so you can check your reasoning as you go.