University of Kurdistan logoNanoelectronics: A New Perspective on TransportWeek 7 · Diffusion and Boltzmann equations
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WEEK 7 · WHAT AND WHERE IS THE VOLTAGE DROP?

Nanoelectronics:
A New Perspective on Transport

Diffusion equation for ballistic transport, Boltzmann equation, the two potentials
Prof. Mohammad RazaghiUniversity of Kurdistan · Graduate course in Electronics Engineering
Based on S. Datta, Lessons from Nanoelectronics, Part A: Basic Concepts, Chapters 8–9
Part 3 of Datta Part A begins: what and where is the voltage drop.

RecapWhere we are

Weeks 1–6 gave us

  • R = RB(1 + L/λ): an interface part and a channel part
  • Drift–diffusion: dI/dz = 0, I/A = −(σ₀/q)dµ/dz (end of Ch. 7)
  • So far almost no mention of the electric field

Today's questions

Where does the voltage actually drop? Can a diffusion equation describe a ballistic channel? And what is the difference between the electrostatic and the electrochemical potential?

Week 7Today's goals

  1. Get the new Ohm's law from the diffusion equation with modified boundary conditions (RB/2 at each end).
  2. Explain the two electrochemical potentials µ⁺ and µ⁻ and I = (q/h)M(µ⁺ − µ⁻).
  3. Write the Boltzmann equation and see it as Newton's law in phase space plus scattering.
  4. Separate µ (electrochemical) from U (electrostatic): equilibrium is constant µ.
  5. See why AC transport needs both: kinetic inductance and quantum capacitance.

Physics refresherRefresher: Newton's laws in phase space

  • State of a particle: a point (z, p).
  • dz/dt = v = ∂E/∂p,  dp/dt = F = −∂E/∂z.
  • Many particles: a distribution f(z, p, t) that flows along these trajectories.
The Boltzmann equation is this flow plus random jumps (scattering).

Physics refresherRefresher: charge bends the potential

  • Gauss / Poisson: d²U/dz² ∝ (n − n₀).
  • Junctions, heterostructures and contacts carry charge layers: built-in fields even at equilibrium.
  • Two potentials: U (where the states are) and µ (how high they are filled).
Predict: at equilibrium, which must be constant, U or µ?

Datta Sec. 8.1The diffusion equation and its boundary conditions

dI/dz = 0,   I/A = −(σ₀/q) dµ/dzEqs. 8.1–8.2
µ(0) = µ₁, µ(L) = µ₂ ⇒ I = σ₀AV/LEqs. 8.3–8.4 · standard
µ(0) = µ₁ − qIRB/2, µ(L) = µ₂ + qIRB/2Eq. 8.6 · RB = λ/σ₀A = h/q²M
I = σ₀AV/(L + λ)Eq. 8.5
Derivation: mu(0) − mu(L) = qV − qIR_B; I = (sigma0 A/qL)(qV − qIR_B); sigma0 A R_B = lambda.

Datta Sec. 8.1Why? Two electrochemical potentials

I = (q/h)M(µ⁺ − µ⁻)Eq. 8.8
I = −(σ₀A/q)dµ⁺/dz = −(σ₀A/q)dµ⁻/dzEq. 8.13
µ⁺(0) = µ₁,   µ⁻(L) = µ₂Eqs. 8.11–8.12

The average µ = (µ⁺ + µ⁻)/2 jumps by (µ⁺ − µ⁻)/2 = qIRB/2 at each end: that is Eq. 8.6.

Datta Sec. 8.2A ballistic channel: exactly two potentials

f⁺(z; E) = f₁(E),   f⁻(z; E) = f₂(E)Eqs. 8.15–8.16
  • Drainbound states are filled only from the source: in equilibrium with it.
  • f₁ = 1, f₂ = 0: drainbound lanes "bumper to bumper", sourcebound lanes empty.
  • Diffusive channel: guess Fermi functions with µ±(z) (Eq. 8.17), translate f into µ with Eq. 2.11.

InteractiveExplore: µ⁺ and µ⁻ by counting electrons

Two electrochemical potentials: μ+ for right-movers, μ− for left-movers  (Datta Ch. 8, Figs. 8.2–8.5)

Linear response with μ₁ = 1 and μ₂ = 0: the source injects extra right-moving electrons (blue →), backscattering turns some into left-movers (red ←), and both contacts absorb whatever reaches them. The density of each stream, measured in bins, is its electrochemical potential (Eq. 2.11). Compare with Datta's solution of Eq. 8.13 (lines).

Transmission T measured · λ/(L+λ)–
δμ = μ+ − μ− (measured · Eq. 11.13)–
Drop at each contact = qIRB/2–
R / RB = 1 + L/λ–

Try this

  1. L = λ: predict δµ = µ⁺ − µ⁻ and the jump at each end (0.5 and 0.25).
  2. Slide to L ≪ λ: µ⁺ ≈ µ₁, µ⁻ ≈ µ₂ everywhere (Fig. 8.5).
  3. Slide to L ≫ λ: µ⁺ ≈ µ⁻ ≈ the standard linear profile.
  4. Check that the measured transmission equals λ/(L + λ).

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

The density of each stream, normalized to the injected density, is its electrochemical potential in linear response.

Datta Sec. 8.3Current from the quasi-Fermi levels

I(z) = I⁺(z) − I⁻(z)Eq. 8.18
I⁺ = q(D/2L)u f⁺ = q(M/h) f⁺Eq. 8.19 · half the states move right
I = (q/h)(µ⁺ − µ⁻)∫dE(−∂f₀/∂E)M(E)Eq. 8.23

Disclaimer (Sec. 8.1.1): in general each energy and angle can have its own µ±(z; E); the Boltzmann equation is the standard to check against.

Datta Sec. 9.1The two potentials

µ = (µ − U) + UEq. 9.1 · electrochemical = chemical + electrostatic
I/A = −σ₀ d[(µ−U)/q]/dz − σ₀ d(U/q)/dzEq. 9.4 · diffusion + drift, one σ₀
No current at constant µ (statistical mechanics), so drift and diffusion share one coefficient: the Einstein relation. That is why diffusion alone (U = 0) was enough to find σ₀.

Datta Sec. 9.1, 9.3Equilibrium: constant µ, not constant U

  • A built-in step: field and density gradient both present.
  • Drift pushes electrons back; diffusion pushes them forward; they cancel.
  • If current were driven by the field, every electron would feel it. In Datta's view only the window (f₁ − f₂) matters.

Datta Sec. 9.2The Boltzmann transport equation

∂f/∂t + vz∂f/∂z + Fz∂f/∂pz = SopfEq. 9.5
Sopf ≈ −(f − f̄)/τEq. 9.8 · relaxation-time approximation
Without scattering? (click)
f(z, p, t) = f(z − vΔt, p − FΔt, t − Δt): f is carried along Newton's trajectories (Eqs. 9.6–9.7).

"Semiclassical"

Classical particles, but with quantum input: the E(p) relation (e.g. graphene's E = v₀p) and the scattering operator.

The left side is mechanics; the right side (scattering) is what drives f towards equilibrium.

InteractiveExplore: the Boltzmann equation as particles

Boltzmann equation as particles: the two potentials  (Datta Ch. 9, Figs. 9.1–9.3)

Each dot is an electron moving by Newton's laws, dz/dt = v, dp/dt = −dU/dz (Eq. 9.7), with random elastic backscattering (relaxation time τ). Following all of them is the same as solving the Boltzmann equation (Eqs. 9.5–9.6). Both contacts inject a thermal (Boltzmann) distribution set by their μ. The lower panel shows the band edge U(z) and the electrochemical potential μ(z) = U(z) + kT·ln(n/nref) read from the local density.

Net current, per unit time (as % of the one-way flux)–
Spread of μ(z) inside the channel–
Density right / left–
Electrons in the channel–

Try this

  1. Press Equilibrium with a field: density ratio ≈ e−ΔU/kT, net current ≈ 0, µ(z) flat.
  2. Watch phase space: below the step electrons turn back, above it they pass.
  3. Press Apply bias: µ(z) slopes and a current flows.
  4. Shorten τ: from ballistic to diffusive; the dots spread over phase space.

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

Each electron obeys Eq. 9.7; the ensemble solves Eq. 9.5 with an RTA-like elastic scattering term.

Datta Sec. 9.3The diffusion equation from the BTE

Step 1: equilibrium (click)
f₀ of E(z, pz) = ε(pz) + U(z) with constant µ₀ solves vz∂f/∂z + Fz∂f/∂pz = −(f − f̄)/τ, even with U(z) (Eqs. 9.9–9.12).
Step 2: two streams (click)
f± = Fermi functions with µ±(z) (Eq. 9.13); the field terms cancel and vz∂µ±/∂z = −(µ± − µ̄)/τ (Eq. 9.15).
Step 3: result (click)
∂µ⁺/∂z = ∂µ⁻/∂z = −(µ⁺ − µ⁻)/λ with λ = 2|vz|τ (Eq. 9.16): Eq. 8.13 with I = (σ₀A/qλ)(µ⁺ − µ⁻).

Equilibrium fields can matter (Sec. 9.3.1)

An applied field enters only at order V², so it drops out of the conductance. A built-in field (heterostructure, contact) is there at V⁰ and can change it. A magnetic field (Hall effect, Ch. 11) needs the F·∂µ/∂p term too.

Datta Sec. 9.4AC transport needs both potentials

∂(µ/q)/∂z = −(LK + LM)∂I/∂t − I/σAEq. 9.20a
∂(µ/q)/∂t = −(1/CQ + 1/CE)∂I/∂zEq. 9.20b
LK = (h/q²)/⟨2Mvz⟩,  CQ = (q²/h)⟨2M/vz⟩Eq. 9.21 · per unit length

1/√(LKCQ) = v exactly. With CE in series a nanotube carries waves at ≈ 3.1 × 10⁶ m/s, not c/√εr.

Numbers from the Week 7 notebook: M = 4, v = 8e5 m/s, diameter 1.5 nm, 1 µm above a ground plane, eps_r = 4.

CheckConcept check

One minute with a neighbour, then click to reveal.

1. Standard boundary conditions and L → 0: what goes wrong? (click)
I = σ₀AV/L → ∞. A zero-length conductor still has RB; Eq. 8.6 fixes it.
2. Why must µ⁺ ≠ µ⁻ when current flows? (click)
I = (q/h)M(µ⁺ − µ⁻): equal potentials, no net flow.
3. Ballistic channel: what are µ⁺ and µ⁻? (click)
µ⁺ = µ₁ and µ⁻ = µ₂ everywhere: each stream is in equilibrium with the contact it came from.
4. At equilibrium, which is constant: U or µ? (click)
µ. U can vary; drift and diffusion then cancel exactly.

Wrap-upSummary and next week

Take-home

  • Diffusion equation + RB/2 at each end = new Ohm's law
  • µ⁺, µ⁻: I = (q/h)M(µ⁺ − µ⁻), δµ = (µ₁ − µ₂)/(1 + L/λ)
  • BTE = Newton in phase space + scattering
  • µ = (µ − U) + U; equilibrium: constant µ

Week 8: quasi-Fermi levels, Hall effect + C4

The voltage drop across a single defect, what voltage probes measure (Landauer and Büttiker formulas), the Hall effect and edge states. Then circular session C4: mid-course review and midterm. Read Datta Ch. 10–11 (pp. 121–149).

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