Nanoelectronics: A New Perspective on TransportWeek 5 · Energy band model

WEEK 5 · ENERGY BAND MODEL

Nanoelectronics:
A New Perspective on Transport

Where the density of states and the number of modes come from

Prof. Mohammad RazaghiUniversity of Kurdistan · Graduate course in Electronics Engineering
Based on S. Datta, Lessons from Nanoelectronics, Part A: Basic Concepts, Chapter 6

Chapter 6 connects the elastic-resistor formulas of Weeks 3–4 to band structure and to the familiar Drude formula.

RecapWhere we are

Weeks 1–4 gave us

G(E) = q²D(E)/2t(E), G_B = (q²/h)M
G = G_Bλ/(L + λ), σ = G_Bλ × {1, 1/W, 1/A}
All averaged over the window −∂f₀/∂E around µ₀

Today's questions

Where do D(E) and M(E) come from? And how is σ = q²D𝒟/volume related to the Drude formula σ = q²nτ/m that everybody remembers?

Ask students to write down the Drude formula from memory: almost everyone can. Then ask what n means for a p-type semiconductor.

Week 5Today's goals

Count states in momentum space: p = integer × h/L, N(p) ∝ p^d, D = dN/dE.
Use the identity D·v·p = N·d to recover Drude σ = q²nτ/m with m = p/v.
Read the number of modes as wavelengths that fit across: M = floor(2W/λ).
Say what "n" means for n-type, p-type and graphene, and why conduction is a Fermi-surface property.
Explain G_B ∝ n^(d−1)/d and the √n of ballistic graphene.

Physics refresherRefresher: waves in a box and de Broglie

A guitar string only vibrates at wavelengths that fit; discrete wavelengths mean discrete frequencies.
de Broglie: an electron with momentum p is a wave, λ = h/p, p = ħk.
In a ring of length L: L = n·λ ⇒ p = n·h/L. States are evenly spaced in momentum.

Silicon-like band (0.2 m₀) at 25 meV: λ ≈ 17 nm. Copper at its Fermi energy: λ ≈ 0.46 nm.

Periodic boundary conditions (a ring) are what Eq. 6.15 really assumes; Datta comments on this in Sec. 6.4.1.

Datta Sec. 6.1Two formulas for the conductivity

	Drude (Eq. 6.1)	Datta (Eq. 6.5)
Formula	σ = q²nτ/m, µ̃ = qτ/m	σ(E) = q² (D/volume) 𝒟(E), averaged over −∂f₀/∂E
Counts	n: all electrons, every energy	only states within a few kT of µ₀
Needs	a mass m (unclear for graphene)	only D(E) and 𝒟(E)
Valid	one parabolic band	any conductor, even amorphous

Glass and copper have about the same number of electrons. Glass insulates because it has almost no states near µ₀.

Eq. 6.5 is the "degenerate Einstein relation". Thouless (1977) used essentially this to get G = q²D/2t.

Datta Sec. 6.2The E(p) relation: velocity is the slope, mass is p/v

E = E_c + p²/2mEq. 6.12 · parabolic (Si, GaAs, Cu)

E = E_c + v₀pEq. 6.13 · linear (graphene)

v = dE/dp, m(E) = p/vEqs. 6.14, 6.11

In a crystal the ions are absorbed into an effective mass: electrons move as if in vacuum with a different m.

Datta Sec. 6.3Counting states

p = integer × h/LEq. 6.15 · one state per h/L in each direction

N(p) = { 2L/(h/p), πLW/(h/p)², (4π/3)LA/(h/p)³ }Eq. 6.17 · 1D, 2D, 3D, per spin

D(E) = dN/dESec. 6.3.1

Picture it

N(p) = (volume of a sphere of radius p in momentum space) ÷ (volume per state). Count the dots inside a circle in 2D, inside a sphere in 3D.

N ∝ p^d for every E(p): this one fact gives the identity on the next slides.

InteractiveExplore: counting states in k-space

Where D(E) comes from: counting states in k-space (Datta Sec. 6.2–6.3)

In a box of size L the allowed wave vectors k form an evenly spaced grid (spacing 2π/L). The energy E = ℏ²k²/2m is not evenly spaced. Count the dots whose energy falls in a thin slice E to E + ΔE: that count divided by ΔE is the density of states.

Dimension:

Energy E Box size L

States with energy below E: N(E)–

States in the orange slice ΔN–

Counted D(E) = ΔN/ΔE–

Theory–

Try this

2D: press Sweep. The counted D(E) is flat: rings grow in circumference but get thinner.
1D: D ∝ 1/√E, with a peak at the band bottom.
3D: drag to rotate; D ∝ √E (shell area ∝ k²).
Double L: the dots get denser in every direction; D per unit volume stays the same.

Scroll inside the panel for all controls. Swipe or use the arrow buttons outside it to change slide.

Each dot holds two electrons once spin is counted.

Datta Sec. 6.3.1D(E) and M(E) in 1D, 2D, 3D

D·v·p = N·dEq. 6.9 · any single E(p) band

Why? (click)

N ∝ p^d ⇒ dN/dp = dN/p. Then D·v = (dN/dp)(dp/dE)(dE/dp) = dN/p.

So what? (click)

σ = q²(D/V)𝒟 with 𝒟 = v²τ/d becomes q²τvN/(pV) = q²nτ/m, using m = p/v and n = N/V: the Drude formula (Eq. 6.10).

It fails with several bands or no E(p) at all; Eq. 6.5 does not.

Datta Sec. 6.4Number of modes: wavelengths that fit across

M = floor{1, 2W/(h/p), πA/(h/p)²}Eqs. 6.19–6.20

N counts wavelengths that fit in the volume; M counts those that fit in the cross-section.
A partial wave is not a state, so M is an integer.
Steps are visible only when kT ≪ step spacing (small W, low T).

Datta: if M(E) runs from 201.59 to 311.67 over the window, nobody minds that it really runs from 201 to 311.

InteractiveExplore: the number of modes

Number of modes M(E): wavelengths that fit across the width (Datta Sec. 6.4, Eq. 6.20)

An electron of energy E has de Broglie wavelength λ = h/p. Across a strip of width W only standing waves whose transverse wavelength 2W/n is longer than λ fit, so the number of modes is M = floor(2W/λ). Each mode (per spin) adds q²/h to the ballistic conductance.

Dispersion:

Width W Energy above band edge, E − E_c Effective mass m/m₀ Temperature T

Wavelength λ = h/p–

2W/λ (continuous)–

Modes M (per spin)–

G_B = g(q²/h)M · R_B–

Try this

W = 40 nm, E − Ec = 60 meV, m = 0.2 m₀: predict λ and M before looking (11.2 nm, M = 7).
Switch to the linear band: steps become evenly spaced in energy.
Raise T to 300 K: the averaged M (red) loses its steps once 4kT exceeds the spacing.
Narrow W to 10 nm: steps far apart, quantization survives to higher T.

Scroll inside the panel for all controls. Swipe or use the arrow buttons outside it to change slide.

Link to Week 4: quantum point-contact steps of 2q²/h.

Datta Sec. 6.4.1The degeneracy factor g

25.8 kΩone mode, one spin: h/q²

12.9 kΩg = 2 (spin): GaAs, metals

6.45 kΩg = 4 (spin × valley): carbon nanotube

R_B = (h/q²)/(gM)

All g equivalent channels conduct in parallel (Sec. 6.4.1). Datta quotes ≈ 12.5 kΩ and ≈ 6.25 kΩ using h/q² ≈ 25 kΩ.

Ring or flat?

Eq. 6.17 assumes a ring (periodic boundary). A nanotube is a ring, graphene is flat: same physics when wide, different R_B when only a few nm across.

Datta Sec. 6.5What does "n" count?

∫(−∂f₀/∂E) N(E) dE = ∫ D f₀ dEn-type: window average of N = electrons

∫(−∂f₀/∂E) N(E) dE = ∫ D (1 − f₀) dEp-type, E = Ev − p²/2m: = holes

p-type: lower µ₀ (fewer electrons) ⇒ more conductance.
Graphene E = ±v₀p: a double-ended D(E) (Fig. 6.3).

Container analogy from Week 2: the container width at height E is M(E); the water line is µ0. Conductance follows the width at the water line, not the amount of water.

InteractiveExplore: who conducts?

Who conducts? Electrons, holes and the energy window (Datta Sec. 6.5–6.6, Figs. 6.1–6.3)

Each dot is a state (2D channel, 1 dot ≈ a fixed number of states per cm²). Blue = filled, grey ring = empty, red ring = empty state in a downward band (a “hole”). Only states inside the yellow window flicker: that is where the contacts can still add or remove electrons, and where conduction happens.

Material:

Electrochemical potential μ₀ Temperature T

Electrons in upward band n–

Holes in downward band p–

Window average of N(E) (Sec. 6.5)–

Ballistic G_B per µm width–

Try this

n-type: check that the window average of N(E) equals n.
p-type: lower µ₀. Fewer electrons, yet G_B rises.
Semiconductor: sweep µ₀ across the gap. The deep blue dots never change; G_B changes by orders of magnitude.
Graphene: G_B is smallest at the band-touching point; electrons above, holes below.

Scroll inside the panel for all controls. Swipe or use the arrow buttons outside it to change slide.

All with one formula: σ = (q²/h)Mλ averaged over the window.

Datta Sec. 6.6Conductivity versus n

σ = (q²/h) M λ {1, 1/W, 1/A}Eq. 6.22

G_B ∝ g^1/d n^(d−1)/dEqs. 6.24–6.25 · √n in 2D

Short ballistic graphene: G ∝ √n (seen experimentally).
Long graphene: σ ∝ n, because τ(E) ∝ √n for common scatterers: m(E) and τ(E) cancel accidentally.

ApplicationWorked example: n = 5×10¹² cm⁻² in a 1 µm wide channel

	Silicon-like 2D gas (m = 0.2 m₀, g = 2)	Graphene (v₀ = 10⁶ m/s, g = 4)
Fermi energy above edge	59.8 meV = n/D, D = gm/2πħ²	261 meV = v₀p_F
Fermi wavelength λ_F	11.2 nm	15.9 nm
Modes M = floor(2W/λ_F)	178 per spin	126 per spin and valley
G_B = g(q²/h)M	13.8 mS (R_B ≈ 72 Ω)	19.5 mS (R_B ≈ 51 Ω)
Datta mass p/v at E_F	0.2 m₀ (constant)	0.046 m₀ (grows with n)

All numbers are computed in the Week 5 notebook (Exercise 1) and the study toolkit calculator.

CheckConcept check

One minute with a neighbour, then click to reveal.

1. Same electrons per atom, conductivities 20 orders apart. How? (click)

Conduction uses states within a few kT of µ₀. An insulator has µ₀ in a gap: no modes in the window.

2. Why is D(E) constant for a 2D parabolic band? (click)

N ∝ p² ∝ E, so dN/dE is constant.

3. What does M count, and why an integer? (click)

The wavelengths (transverse standing waves) that fit across the cross-section; a partial wave is not a state.

4. Why is "the mass" of graphene ill-defined? (click)

E = v₀p gives m = p/v₀, which grows with energy and vanishes at the touching point. Eq. 6.22 needs no mass.

Wrap-upSummary and next week

Take-home

p = integer × h/L; N ∝ p^d; D = dN/dE
D·v·p = N·d ⇒ Drude is the single-band special case
M = floor(2W/λ): wavelengths across; G_B = g(q²/h)M
p-type counts holes; G_B ∝ n^(d−1)/d

Week 6: the nanotransistor + C3

A gate slides D(E) up and down. Why does the current saturate, what does charging do, and what is quantum capacitance? Read Datta Ch. 7 (pp. 75–96) and finish the Week 5 notebook and study toolkit.

Prof. Mohammad Razaghi · University of Kurdistan · after S. Datta, Lessons from Nanoelectronics

Nanoelectronics:A New Perspective on Transport

RecapWhere we are

Weeks 1–4 gave us

Today's questions

Week 5Today's goals

Physics refresherRefresher: waves in a box and de Broglie

Datta Sec. 6.1Two formulas for the conductivity

Datta Sec. 6.2The E(p) relation: velocity is the slope, mass is p/v

Datta Sec. 6.3Counting states

Picture it

InteractiveExplore: counting states in k-space

Where D(E) comes from: counting states in k-space (Datta Sec. 6.2–6.3)

Try this

Datta Sec. 6.3.1D(E) and M(E) in 1D, 2D, 3D

Datta Sec. 6.4Number of modes: wavelengths that fit across

InteractiveExplore: the number of modes

Number of modes M(E): wavelengths that fit across the width (Datta Sec. 6.4, Eq. 6.20)

Try this

Datta Sec. 6.4.1The degeneracy factor g

RB = (h/q²)/(gM)

Ring or flat?

Datta Sec. 6.5What does "n" count?

InteractiveExplore: who conducts?

Who conducts? Electrons, holes and the energy window (Datta Sec. 6.5–6.6, Figs. 6.1–6.3)

Try this

Datta Sec. 6.6Conductivity versus n

ApplicationWorked example: n = 5×10¹² cm⁻² in a 1 µm wide channel

CheckConcept check

Wrap-upSummary and next week

Take-home

Week 6: the nanotransistor + C3

Slides

Nanoelectronics:
A New Perspective on Transport

R_B = (h/q²)/(gM)