University of Kurdistan logoNanoelectronics: A New Perspective on TransportWeek 2 · Why electrons flow
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WEEK 2 · WHY ELECTRONS FLOW

Nanoelectronics:
A New Perspective on Transport

Two Fermi functions, the energy window, linear response · and circular session C1
Prof. Mohammad RazaghiUniversity of Kurdistan · Graduate course in Electronics Engineering
Based on S. Datta, Lessons from Nanoelectronics, Part A: Basic Concepts, Chapter 2
Part 1 is the fixed lecture (Datta Ch. 2). Part 2, from the divider slide on, is circular session C1 covering Weeks 1–2.

RecapLast week in four lines

What we know

  • R = RB(1 + L/λ), with RB = h/(q²M) at the interfaces
  • Elastic resistor: resistance from the channel, heat in the contacts
  • µ measures how full the states are, not their energy
  • The voltage unbalances the injection from the two contacts

Today's question

If the electric field does not push the electrons, what exactly makes them flow, and at which energies?

Ask two students to state one Week 1 idea each before moving on.

Week 2Today's goals

  1. Use Datta's two key concepts: the density of states D(E) and the electrochemical potential µ₀.
  2. Sketch the Fermi function f(E) and say what µ and kT control.
  3. Explain why current flows only in the energy window between µ₁ and µ₂ (± a few kT).
  4. Derive linear response: f₁ − f₂ ≈ (−∂f₀/∂E)·qV, valid for qV ≪ kT.
  5. Explain why conduction depends on −∂f₀/∂E, not f₀, and how this differs from the Drude picture.

Datta Ch. 2What makes electrons flow? "The electric field" is not enough

  • Every atom has enormous fields around its nucleus, yet no current flows until a battery is connected.
  • A p-n junction has a large built-in field over micrometres, yet no current flows at equilibrium.
  • In conductors a few atoms long there is no clear line between "microscopic" and "macroscopic" fields.
Datta: a difference in electrochemical potentials makes electrons flow, just as a temperature difference drives heat and a level difference drives water.
Datta Ch. 2 introduction. Ask the class first, then reveal the two objections.

Datta Sec. 2.1Two key concepts: D(E) and µ₀

Density of states D(E)

How many states the channel offers per unit energy: the shape of the container. Set by the material and the dimensions (Week 5).

Electrochemical potential µ₀

The energy that separates filled from empty states: the water line. At equilibrium it is the same everywhere: source, channel, drain.

n = ∫ D(E) f(E − µ) dEelectrons = container shape × filling, summed over energy
"The first step in understanding the operation of any electronic device is to draw the available density of states D(E) and to locate µ₀." (Datta, Fig. 2.1)
Datta calls this a "point channel model": one D(E) for the whole channel. Spatial variation comes in Ch. 7.

InteractiveExplore: fill the states

Fill the states: density of states D(E) and the water line μ  (Datta Sec. 2.1, Ch. 6)

D(E) is the shape of the container: how many seats (states) each energy floor has. Electrons fill seats from the bottom, one per seat. The electrochemical potential μ is the water line, where seats are half full (f = ½).

Container shape:
Water line μ–
Seats at the water line D(μ)–
Fraction of seats filled–
Thermal blur kT–

μ of A–
μ of B–
μA − μB–
Net electrons moved A → B–

Try this

  1. Tab 1: fill the 3D band to 40%, then switch to 2D and 1D. Where does µ land each time?
  2. Choose Semiconductor: with no doping, where is the water line? Are there seats there?
  3. Add a doping of 10⁻⁶ of a band: how far does µ move?
  4. Tab 2: predict which way electrons flow, then press Connect.
  5. After connecting, raise the battery voltage: µA − µB = qV.

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

Key misconceptions to name: µ is not "where the electrons are"; D(E) counts seats, filled or empty.

Datta Secs. 2.1, 2.3A voltage separates µ₁ and µ₂

µ₁ − µ₂ = qVDatta Eq. 2.1
  • The battery lowers all energies in the positive contact (the drain) by qV.
  • Each contact stays in equilibrium with its own µ: they are too big to be driven out of equilibrium.
  • The small channel in between cannot satisfy both.
Electron charge is −q, so the positive terminal has the lower electrochemical potential (Sec. 3.2.2 convention).

Datta Sec. 2.1.1The energy window for current flow

  • Each contact tries to fill states below its µ and empty those above.
  • Between µ₂ and µ₁ they disagree: the source keeps filling, the drain keeps emptying → continuous flow.
  • Above both or below both they agree → no flow, however many electrons are there.
If the electric field drove the current, all electrons would flow. Only those in the window (± a few kT) do.
Datta Sec. 2.1.1. This is the central picture of the whole course.

Physics refresher · Sec. 2.2Physics refresher: the Fermi function

f(E) = 11 + e(E − µ)/kTDatta Eq. 2.2
  • Probability that a state at energy E is filled; f(µ) = ½.
  • Low energies full, high energies empty: "everything goes to its lowest energy" + the exclusion principle.
  • The change from 1 to 0 happens over about µ ± 2kT; k ≈ 86 µeV/K.
  • Datta plots it with energy vertical, so it sits beside D(E).
Quick check: f at µ + 2kT is 0.12; at µ − 2kT it is 0.88.

Datta Sec. 2.2.1The thermal broadening function

FT = kT(−∂f/∂E) = f(1 − f)Eqs. 2.3, 2.5
  • Symmetric about µ (Eq. 2.4); peak value 1/4 at E = µ.
  • Area = kT (Eq. 2.6): picture it as a pulse 4kT wide and 1/4 high.
  • −∂f/∂E itself has area 1: a weighting function that picks out the energies near µ.
We will meet this pulse every week: it decides which states matter.

Datta Sec. 2.3Out of equilibrium: two Fermi functions

f₁(E) = f(E − µ₁)
f₂(E) = f(E − µ₂)Eq. 2.7
I(E) ∼ f₁(E) − f₂(E)Eq. 2.8, current at energy E (short conductors)
  • Zero far above (both 0) and far below (both 1).
  • The difference in the "agenda" of the two contacts drives the flow.
Non-equilibrium statistical mechanics is hard in general; here it is easy because each contact stays in its own equilibrium.

InteractiveExplore: two Fermi functions and the window

Why electrons flow: two Fermi functions  (Datta Ch. 2)

Energy runs up the vertical axis, as in Datta's Fig. 2.4. Each contact tries to set the occupation of the channel states to its own Fermi function; current flows only where the two disagree, f₁(E) − f₂(E). The middle panel T(E) shows at which energies the channel can conduct: here one broadened level, standing in for Datta's D(E) (Chapter 3 makes this precise).

Channel:
Thermal energy kT–
Bias window μ₁−μ₂–
Current I (per mode)–
G = I/V–

Try this

  1. Press Equilibrium: f₁ = f₂ everywhere, no current, although electrons move.
  2. Press Linear response (V = 10 mV): the purple area is tiny and ∝ V.
  3. Press Large bias: the window spans many kT.
  4. Move the level ε outside the window. Does the current vanish? What does temperature change?
  5. Switch to Wide conductor: I = (q²/h)V exactly, Ohm's law.

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

The middle panel T(E) stands in for D(E): which energies the channel can conduct. Week 3 makes this G(E) = q²D/2t.

Datta Sec. 2.4Linear response

F(E) = (f₁ − f₂)/(qV/kT)Eq. 2.9, with µ₁,₂ = µ₀ ± qV/2
f₁ − f₂ ≈ (−∂f₀/∂E) qVEq. 2.10, valid for qV ≪ kT
  • As y = qV/kT → 0, F(E) collapses onto FT.
  • At 300 K, kT/q ≈ 26 mV: a 1 mV bias is deep in linear response, 100 mV is not.

Linear response: when does f₁ − f₂ become −∂f/∂E?  (Datta Fig. 2.5)

The purple curve is the difference of the two Fermi functions divided by y = qV/kT, so its area is always 1. The dashed curve is the thermal broadening function FT = kT(−∂f/∂E). Shrink y and watch them merge.

Datta's values:
Peak of F(E) (FT peak = 0.25)–
Area under F(E), in units of kT–
Largest gap |F − FT|–

Use the widget: click Datta values y = 7, 3, 1, 0.1 (his Fig. 2.5).

Datta Sec. 2.5Current depends on −∂f₀/∂E, not on f₀

  • f₀(E) counts occupied states; −∂f₀/∂E marks the states that conduct.
  • Far above µ₀ (E − µ₀ ≫ kT) they look alike: −∂f₀/∂E ≈ f₀/kT, the non-degenerate case (Sec. 3.4).
  • In general they are very different: deep below µ₀, states are full but carry nothing.
Conductance depends on −∂f₀/∂E: Datta's two contacts give this without advanced theory.

Datta Sec. 2.5.1The Drude picture and why it misleads

d(mv)/dt = −qF − mv/τmNewton's law + friction
vd = −(qτm/m)F,   σ = q n µ̃Eqs. 2.12–2.13 · µ̃ = mobility
  • Treats the field as the driving force on every electron.
  • Copper and glass have similar numbers of electrons, yet their conductivities differ by about 10²⁰.
  • What differs: the number of states within ~kT of µ₀.
The Drude formula needs "free electrons" and a simple positive effective mass; the window picture does not.

Datta Sec. 2.5.2The present approach in one line

f₁(E) − f₂(E) ≈ (−∂f₀/∂E) · qV

Driving term

The difference in the contacts' "agenda", not a force on each electron.

No mystery forces

No need to say "electrons in full bands cannot move": the window already excludes them.

Same idea for heat

Phonons follow the same logic with the Bose function instead of the Fermi function (Ch. 14).

CheckConcept check

One minute with a neighbour, then click to reveal.

1. At equilibrium electrons move constantly. Why is there no current? (click)
f₁ = f₂ at every energy: equal agendas, so every crossing is balanced by one the other way.
2. A level sits 200 meV above µ₀ at 300 K, small bias. Is its current exactly zero? (click)
No, but tiny: 200 meV ≈ 7.7kT, so kT(−∂f/∂E) ≈ e−7.7 ≈ 5×10⁻⁴, about 0.2% of its peak value 1/4.
3. When does linear response fail? (click)
When qV is not small compared with kT, or when the channel's conduction changes across the window.
4. Copper vs glass: why such different conductivities? (click)
Copper has many states within a few kT of µ₀; glass has a wide gap there and almost none.

Wrap-upFixed session summary

Take-home

  • D(E) = container, µ = water line; n = ∫D f dE.
  • µ₁ − µ₂ = qV; current flows only in the window, I(E) ∼ f₁ − f₂.
  • FT = kT(−∂f/∂E) = f(1−f): a pulse 4kT wide.
  • Linear response: f₁ − f₂ ≈ (−∂f₀/∂E)qV for qV ≪ kT.

Next: circular session C1

Refresher clinic → Colab lab → problem clinic → two-minute explanations. Open week02_fermi_functions_C1.ipynb (Part 2).

PART 2

Circular session C1

Covers Weeks 1–2 · about 90 minutes

A · 20 min
Refresher clinic
B · 30 min
Colab lab
C · 30 min
Problem clinic
D · 10 min
Two-minute explanations
The circular session rotates students through explaining, computing and problem solving.

Circular session C1Block A · Refresher clinic

Work in pairs: one explains, the other sketches. Swap after each item.

1 · Ohm's law

Why is R → 0 as L → 0 unphysical? Sketch R vs L with the intercept RB.

2 · Units

Show that (q²/h) × 1 V is a current, and that qV = 1 eV for V = 1 V.

3 · Fermi function

What are µ, kT and f(µ)? Sketch f(E) with energy vertical.

4 · A number

Probability that a state at E = µ + 3kT is filled? (0.047)

Circular session C1Block B · Colab lab: a conductance map

Run Part 2 of the Week 2 notebook. G(ε) = (q²/h)∫T(E)(−∂f/∂E)dE for one broadened level (γ = 20 meV).

  1. Why is the 50 K peak narrower and taller than at 600 K?
  2. At 300 K, at what ε does G fall to half its peak? Compare with 3.5kT.
  3. Repeat with γ = 2 meV. What happens at 50 K, where γ ≪ kT?
Hint: G(ε) is T(E) smeared by −∂f/∂E, which has unit area, so every curve has the same area, (q²/h)πγ/2.
The preview on the right is computed live in the slide with the same formula as the notebook.

Circular session C1Block C · Problem clinic

P1. M = 10, λ = 500 nm, L = 250 nm. Find R and the fraction of R at the contacts. (click)
RB = 2.58 kΩ; R = 2.58 × 1.5 = 3.87 kΩ; contact share = 1/(1 + L/λ) = 67%.
P2. qV = 100 meV at 300 K: linear response? What changes at 600 K? (click)
kT = 25.9 meV, so qV ≈ 3.9kT: not linear. At 600 K, kT = 51.7 meV and qV ≈ 1.9kT: closer, still beyond strictly linear.
P3. For a channel that conducts equally at all energies, show I = (q²/h)·V at any temperature. (click)
∫(f₁ − f₂)dE = µ₁ − µ₂ = qV exactly, for any T (the difference of two shifted steps). So I = (q²/h)V.

Circular session C1Block D · Two-minute explanations

Draw a card. Explain it for two minutes without equations, using one sketch.

Press the button.

Sketch check (everyone, 2 min)

Sketch D(E)·f(E) for a metal and for an n-type semiconductor at room temperature, and mark µ on each.

The cards

  1. Why current flows only inside the energy window
  2. Where the heat goes in an elastic resistor
  3. Why a ballistic wire still has resistance
  4. What µ measures, and what the voltage does
  5. Why copper conducts and glass does not
Sketch check answer: metal, µ inside the band with a sharp edge; semiconductor, µ in the gap below the conduction band with only a thin shaded tail.

Wrap-upNext week: the elastic resistor

Before next class

  • Finish the Week 2 notebook (both parts)
  • Work through the Week 2 study toolkit
  • Read Datta Part A, Ch. 3 (pp. 31–42)

Week 3 preview

How many electrons sit in the channel, and how long does each stay? That gives the conductance formula G(E) = q²D(E)/2t(E) and the full current I = (1/q)∫G(E)(f₁ − f₂)dE.

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