University of Kurdistan logoNanoelectronics: A New Perspective on TransportWeek 1 · Overview
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WEEK 1 · OVERVIEW

Nanoelectronics:
A New Perspective on Transport

Where is the resistance? Where is the heat?
Prof. Mohammad RazaghiUniversity of Kurdistan · Graduate course in Electronics Engineering
Based on S. Datta, Lessons from Nanoelectronics, Part A: Basic Concepts, Chapter 1
Welcome. This course re-thinks how current flows, starting from very small conductors, following Datta's bottom-up approach. Press N at any time to show these notes; M lists all slides.

MotivationElectronics has reached the atomic scale

chip, about 1 cmbillions of transistors gateSD one transistor channel L ≈ 14 nm≈ 50 atomic spacings
  • Commercial transistors already have channels L ~ 14 nm long.
  • Laboratories have measured the resistance of a single hydrogen molecule.
  • So we must ask again: what determines resistance, where is it, and where is the heat?
Datta's thesis: what we learned from these tiny conductors also explains big ones, "bottom-up".
Datta Sec. 1.1: modern transistors have L ~ 14 nm and people have measured the resistance of a hydrogen molecule. Ask the class: which formula do you trust for a conductor this small?

Course16-week roadmap

Weeks 1–4

Part A, Ch. 1–5

What determines resistance? Elastic resistor, ballistic vs diffusive, noise

Weeks 5–6

Part A, Ch. 6–7

Energy bands, density of states, modes, the nanotransistor

Weeks 7–11

Part A, Ch. 8–16

Where is the voltage drop? Hall effect, spin, thermoelectricity, entropy and information

Weeks 12–16

Part B, Ch. 17–25

Quantum transport: Schrödinger → NEGF, quantum Hall, Coulomb blockade, spin

One picture throughout: a channel between two contacts that disagree about how full the channel should be.
Part A builds the picture with simple physics; Part B shows the same picture from the Schrödinger equation (NEGF).

CourseHow this course works

Every week: fixed session

Physics refresher (10–15 min) → concept with live visuals → worked example → short concept check.

Every 2 weeks: circular session (C1–C8)

Colab lab, problem clinic and two-minute student explanations.

Your tools

  • Weekly Google Colab notebook (runs in the browser)
  • Interactive widgets, like the ones in these slides
  • Weekly study toolkit: summary, equations, quiz, practice problems

Suggested assessment

Concept checks 20% · Lab notebooks 20% · Midterm (Week 8) 25% · Final project (Week 16) 35%

Assessment weights are the suggested split from the teaching plan; adjust as needed.

Week 1Today's goals

  1. Explain why Ohm's law R = ρL/A fails for short conductors, and use the corrected law R = RB(1 + L/λ).
  2. Say where the length-independent resistance RB = h/(q²M) sits: at the contacts.
  3. Explain why an elastic resistor releases its heat I·V in the contacts, not in the channel.
  4. Place today's ideas on the map of transport theories: elastic resistor, Boltzmann equation (BTE), NEGF.

Physics refresherRefresher: Ohm's law, and a puzzle

R = ρ L / A    G = σ A / LOhm's law (Datta Eq. 1.1)
  • ρ: resistivity, σ = 1/ρ: conductivity (material properties)
  • Halve the length, halve the resistance
  • Puzzle: as L → 0, Ohm says R → 0. Is that true?
Experiments say no: even a perfect, very short conductor keeps a resistance RB.
Let the animation run: the resistor shrinks; grey dashed line = Ohm, red = experiment. Ask students to guess the intercept before revealing next slides.

Physics refresherRefresher: the numbers we will use all term

1 eVenergy gained by one electron across 1 V
= 1.6 × 10⁻¹⁹ J
25.9 meVthermal energy kT at 300 K
38.7 µSconductance quantum q²/h
25.8 kΩresistance quantum h/q²

Rules of thumb

  • Energies within a few kT of the electrochemical potential matter most.
  • A voltage V separates the two contacts' electrochemical potentials by qV.

Check yourself

What is kT at 600 K? What is h/q² divided by 10?

51.7 meV; 2.58 kΩ (the ballistic resistance of 10 modes).

Datta Sec. 1.2Ballistic conductance comes in units of q²/h

GB = (q²/h) MDatta Eq. 1.2b · ≈ 40 µS × M
RB = h/(q²M) ≈ 25 kΩ / MEq. 1.2a
  • M = number of modes: independent parallel "lanes" for conduction
  • Each mode adds q²/h: steps, not a smooth growth
  • Where M comes from (wave nature of electrons): Week 5, Ch. 6
Datta counts spin inside M. Experiments from the late 1980s on point contacts show steps of 2q²/h (two spins per spatial mode).

Datta Sec. 1.3What determines the resistance?

G = GB / (1 + L/λ)Datta Eq. 1.3 · λ = mean free path
R = RB (1 + L/λ)Eq. 1.8
  • L ≪ λ (ballistic): G ≈ GB, independent of length
  • L ≫ λ (diffusive): G ≈ GBλ/L: Ohm's law appears
  • Not two worlds: one continuum as L grows
Every line starts at R_B: the length-independent part. The slope R_B/λ is the familiar resistivity part.

Datta Secs. 1.3, 1.9Three formulas for the conductivity

New expression

σ = (q²/h)(M/A) λ

Eq. 1.4: modes per unit area × mean free path. Follows from G = GBλ/L.

Drude formula

σ = q² n τm / m

Eq. 1.5: free-electron density n, momentum relaxation time τm, effective mass m. Needs a simple positive m.

Degenerate Einstein relation

σ = q² (D/AL) Ddiff

Eq. 1.6: density of states per volume × diffusion coefficient. We derive it in Week 4.

Copper wire, 10 nm × 10 nm: σ = 5.96×10⁷ S/m and λ ≈ 40 nm give M/A ≈ 38 modes per nm², so M ≈ 3800 and RB ≈ 6.7 Ω. For L = 100 nm, Ohm's law gives 16.8 Ω but RB(1+L/λ) gives 23.5 Ω: the difference is exactly the contact part RB.
The copper numbers are computed in the Week 1 Colab notebook; λ for copper at room temperature is approximate.

Datta Sec. 1.4Where is the resistance?

R = RB + RBL/λ has two parts:

  • RBL/λ grows with length: it belongs to the channel
  • RB does not: it belongs to the interfaces between the narrow channel and the wide contacts
A wide contact offers very many modes; the channel accepts only M. The mismatch at the entrance is a resistance even with perfect contacts.

Datta: the minimum interface resistance is set by the channel, not by contact technology.

This was debated in the 1980s (Datta Sec. 1.4). The funnel is only a cartoon of mode mismatch; the precise statement comes with the quasi-Fermi levels of Ch. 8–10.

InteractiveExplore: the resistance of a channel

Where is the resistance?  (Datta Ch. 1 and 4)

Each dot is an electron injected from the source. It moves at constant speed and can bounce backwards (elastic scattering, mean free path λ). Count how many reach the drain.

● moving toward the drain    ● scattered back toward the source
Electrons that reached the drain0 / 0
Transmission λ/(λ+L)–
Conductance G–
Current I = G·V–

Try this

  1. Set L = λ = 200 nm: about half the electrons reach the drain, and R = 2RB.
  2. Shrink L to 10 nm: the resistance does not go to zero. Read off RB.
  3. Raise M from 1 to 4: everything drops by 4.
  4. Press send 2000 electrons and compare with λ/(λ+L).
  5. Ask: is every collision really a reversal? (next slide)

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

One-dimensional model: every collision reverses the electron, so λ here is the backscattering length that enters T = λ/(λ+L).

Preview of Ch. 4Scattering is random: most flights are short

  • The mean free path is an average. Each flight length is random.
  • Chance of flying x without a collision: e−x/vτ, so 63% of flights are shorter than the average.
  • In 3D a collision usually only deflects the electron. It takes several to turn it around.
Datta absorbs this into his λ (Eq. 4.10):
λ = vτ × 2 (1D), π/2 (2D), 4/3 (3D)
This answers the common question "why do electrons bounce back before one mean free path?". Week 4 derives the 2, π/2, 4/3 factors.

InteractiveExplore: electrons in a 3D nanowire

Electrons scattering in a 3D nanowire  (Datta Ch. 4 and App. B)

Both contacts inject electrons, each according to its own Fermi function: the voltage V only makes the source fill states (μ₁ = μ₀ + qV/2) that the drain leaves empty (μ₂ = μ₀ − qV/2). Electrons fly in straight lines. At random moments they collide and leave in a random 3D direction (orange flash). Bounces off the side walls are mirror reflections, not collisions. Drag the picture to rotate it.

● injected by the source    ● injected by the drain    ○ collision    (trails show where each electron has been)
Crossed the whole wire0 / 0
Datta's λ = (4/3) vτ–
Theory T = λ/(λ+L)–
Flights shorter than vτ–
Source → drain crossings0
Drain → source crossings0
Net = current0
I = (q²/h)·T·V, one mode–

Try this

  1. Drag the wire to rotate it. Follow one trail: it zigzags but keeps moving forward.
  2. Watch the histogram: the bars pile up near zero, with the average at vτ.
  3. Press quick measurement: compare with T = λ/(λ+L), λ = (4/3)vτ.
  4. Set V = 0: blue and green streams balance, net ≈ 0.
  5. Raise V: the source injects more inside the qV window.

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

Both contacts inject according to their own Fermi function; the voltage only unbalances the two streams.

Key ideaWhere is the voltage? Not in a force on the electrons

  • Electrons fly at constant speed between collisions. Nothing pushes them.
  • The battery separates the contacts' electrochemical potentials: µ₁ − µ₂ = qV.
  • Inside that window the source fills states the drain leaves empty, so the source injects more.
V sets how many electrons are injected per second (∝ qV).
Scattering sets what fraction get through (T).
Together: I = (q²/h) M T V.
Week 2 makes this precise with two Fermi functions. The field inside real channels is Ch. 8–9 (Week 7).

Datta Sec. 1.5The electrochemical potential measures filling, not energy

Landauer's 1957 problem: a ballistic channel with a hole (a scatterer) in the middle.

  • µ(x) drops sharply across the barrier and at the two interfaces (Fig. 1.6).
  • A hole has no atoms to jiggle, so it cannot be where the heat goes.
  • µ shows the degree of filling: a traffic jam before the bottleneck, an empty road after it.
The resistance is located at the bottleneck; the heat is released somewhere else.
Datta Sec. 1.5. The simulated road is a simple exclusion model: cars move only into empty cells, and the bottleneck lets them through slowly.

Datta Secs. 1.5–1.6But where is the heat?

  • Hot electrons relax over an energy-relaxation length LE, tens of nanometres or longer.
  • Elastic resistor (Sec. 1.6): a channel shorter than LE, where all energy exchange happens in the contacts.
  • Like a bullet: it heats the target it hits, not the air it flies through.
Per electron: (µ₁ − ε) released in the source + (ε − µ₂) in the drain = qV. Total power = I·V, none in the channel.
The bullet analogy is Datta quoting his colleague Ashraf (Sec. 1.6). Datta argues this model works even for diffusive conductors full of "potholes", as long as momentum relaxation is included.

InteractiveExplore: where the heat is released

Where is the heat? The elastic resistor  (Datta Ch. 1 and 3)

Energy runs up the vertical axis. In an elastic resistor an electron crosses the channel at constant energy, so the heat appears in the contacts, not in the channel.

Try this

  1. Put the channel energy ε in the middle: source and drain heat equally.
  2. Move ε up and down: the heat moves between contacts, the total stays qV.
  3. Turn on energy lost inside channel: now the channel heats, an inelastic resistor.
  4. Ask: if the channel stays cool, why does it still have a resistance?

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

Answer: resistance comes from momentum loss (scattering, mode mismatch), dissipation from energy loss. They can happen in different places.

Datta Sec. 1.6Elastic resistors: independent energy channels

  • In an elastic resistor each energy E₁, E₂, E₃ conducts on its own (Fig. 1.8a).
  • Inelastic processes add vertical resistors between energies (Fig. 1.8b).
  • If all channels look alike, no vertical current flows and the elastic model is accurate.
  • If not, vertical flow matters: Peltier effect (Ch. 13), p-n junctions (Ch. 12).
The animation alternates every 7 s between (a) elastic and (b) with vertical resistors where lower channels conduct poorly (Fig. 1.9a).

Datta Sec. 1.7Where this fits: transport theories

Classical dynamicsNewton's laws: force-driven Quantum dynamicsSchrödinger equation ++ Heat generationentropy-driven, irreversible == BTEBoltzmann · Part A NEGFGreen's functions · Part B Elastic resistormechanics in the channel,heat in the contacts:far simpler to teach

Datta Sec. 1.7: BTE and NEGF take semesters because they mix mechanics with distributed heat generation. The elastic resistor separates the two cleanly, yet its results are benchmarked against both.

Datta Secs. 1.8–1.9Is transport a many-body process? And why a new picture?

  • A bulb lights at once because a signal travels on an RLC line, not because electrons race from the switch (Sec. 1.8).
  • At low frequency only R matters, and R is about individual electrons in M parallel channels.
"A physical understanding is a completely unmathematical, imprecise and inexact thing, but absolutely necessary for a physicist." — R. Feynman, quoted by Datta (Sec. 1.9)
Drive current with the electrochemical potential, not the electric field: spin-up and spin-down electrons can even flow in opposite directions in the same field.
Spin example from Sec. 1.9 previews Ch. 12 (Week 9).

CheckConcept check

Discuss with a neighbour for one minute, then click to reveal.

1. "A ballistic wire has zero resistance because nothing scatters." What is wrong? (click)
RB = h/(q²M) remains: the interfaces can feed only M modes into the channel.
2. You double the number of modes M. What happens to RB and to RBL/λ? (click)
Both halve: more parallel channels. λ is unchanged.
3. In an elastic resistor, where is the Joule heat released, and why? (click)
In the contacts: electrons keep their energy in the channel and relax only after entering a contact.
4. With V = 0 electrons still cross the channel. Why is there no current? (click)
Both contacts inject equally; the two streams cancel. The voltage only unbalances them.

Wrap-upSummary and next week

Take-home

  • R = RB(1 + L/λ), with RB = h/(q²M) ≈ 25.8 kΩ/M at the interfaces.
  • σ = (q²/h)(M/A)λ: modes per area × mean free path.
  • Elastic resistor: resistance from the channel, heat in the contacts.
  • µ measures filling; the voltage unbalances injection.

Before next class

  • Run the Colab notebook week01_overview.ipynb
  • Work through the Week 1 study toolkit (quiz + problems)
  • Read Datta Part A, Ch. 2 (pp. 21–30)

Week 2: Why electrons flow

Two Fermi functions, the energy window, linear response, and our first circular session (C1).

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