University of Kurdistan logoNanoelectronics: A New Perspective on TransportWeek 3 · The elastic resistor
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WEEK 3 · THE ELASTIC RESISTOR

Nanoelectronics:
A New Perspective on Transport

Flux = population ÷ transit time · G(E) = q²D/2t · degenerate and non-degenerate conductors
Prof. Mohammad RazaghiUniversity of Kurdistan · Graduate course in Electronics Engineering
Based on S. Datta, Lessons from Nanoelectronics, Part A: Basic Concepts, Chapter 3

RecapWhere we are

Week 2 gave us

  • Two contacts with Fermi functions f₁ and f₂; µ₁ − µ₂ = qV
  • Current flows only in the window: I(E) ∼ f₁ − f₂
  • Linear response: f₁ − f₂ ≈ (−∂f₀/∂E)qV

Today's question

"∼" is not an equation. How much current flows at each energy, and what decides it?

Week 3Today's goals

  1. Derive the conductance function G(E) = q²D(E)/2t(E) from "flux = population ÷ time in the channel".
  2. Write the elastic-resistor current I = (1/q)∫G(E)[f₁ − f₂]dE and its low-bias form.
  3. Explain where an elastic resistor releases its heat, and the sign convention for current.
  4. Explain why long resistors are linear over volts, and why R = RB(1 + NL/λ), not N·RB(1 + L/λ).
  5. Decide when G = G(µ₀) (degenerate) and when the Boltzmann tail is needed (non-degenerate).

Physics refresherRefresher: rate = population ÷ residence time

A programme enrols students who each stay a time t. In steady state:

flux = N / t
  • 100 students who each stay 2 years → 50 graduate per year.
  • Known as Little's law in queueing theory; it holds whatever happens inside.
Datta (Sec. 3.2): electron flux = electrons in the channel ÷ time each spends there.
Let the counter run: the measured graduation rate converges to N/t.

Datta Ch. 3 introductionAn elastic resistor: independent energy channels

  • Electrons travel along fixed-energy channels (Fig. 3.1): the current in E…E + dE does not mix with other energies.
  • So dI ∼ dE·G(E)·[f₁(E) − f₂(E)], and we just integrate.
  • Elastic ≠ ballistic: an electron may scatter and zigzag (diffusive) as long as it changes only its momentum, not its energy.

Datta Sec. 3.1How an elastic resistor dissipates heat

  • An electron crosses at ε and arrives as a hot electron: it releases ε − µ₂ in the drain.
  • It leaves a hole far below µ₁ in the source, filled by an electron that releases µ₁ − ε.
  • Total per electron: µ₁ − µ₂ = qV. With N electrons in a time t: I = qN/t and power = qVN/t = V·I = I²R.
We never need the details of the relaxation: we simply declare the contacts to stay in equilibrium at µ₁ and µ₂.

Datta Sec. 3.2Counting the flux at one energy

  • Take f₁ = 1, f₂ = 0 in the range E … E + dE, which has D(E)dE states.
  • The source fills, the drain empties: on average half the states are full, D(E)dE/2.
  • Each electron spends a time t in the channel, so the flux is
flux = D(E)dE / 2t

Datta: the half-filling holds in the ballistic regime, the diffusive regime and in between (Ch. 8).

Predict before the widget

With 24 states, f₁ = 1, f₂ = 0 and t = 1 time unit: how many electrons are in the channel on average? What is the flux?

(12 electrons; 12 per unit time)

InteractiveExplore: flux = electrons in channel ÷ time in channel

Flux = electrons in the channel ÷ time each one spends there  (Datta Sec. 3.2, Eq. 3.3)

Each box is one state of the channel at energy E. A contact tries to set every box to its own occupation f: an empty box is filled from a contact with probability f, a full box is emptied into it with probability 1 − f, each on a time scale t. Count what flows.

Electrons in channel (average)–
Expected D·(f₁+f₂)/2–
Net flux to drain, measured–
Theory D(f₁−f₂)/(2t)–

Try this

  1. With f₁ = 1, f₂ = 0: about half the boxes are full. Press Fast-forward and compare the flux with D/2t.
  2. Swap them (f₁ = 0, f₂ = 1): same flux, opposite direction.
  3. Set f₁ = f₂: boxes still flicker, net flux zero.
  4. Try f₁ = 0.7, f₂ = 0.3: occupancy (f₁+f₂)/2, flux D(f₁−f₂)/2t.
  5. Double t: what happens to the flux? To the occupancy?

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

The two streams superpose: this is Eq. 3.3 at a single energy.

Datta Sec. 3.2The current formula of an elastic resistor

Superpose the two streams (f₁ fills, f₂ fills from the other side) and multiply by q:

dI = q · D(E)dE/2t · [f₁(E) − f₂(E)]
I = (1/q) ∫ G(E)[f₁ − f₂] dEDatta Eq. 3.3
G(E) = q²D(E) / 2t(E)Eq. 3.2
Conductance = availability of states × ease of transport
D(E): how many states · 1/t(E): how fast they pass electrons

Exclusion principle? (Sec. 3.2.1)

Can the two streams block each other? For coherent transport they use orthogonal states and do not "Pauli block"; the relaxation-time BTE gives the same result.

Units check: D in states per joule, t in seconds, so q²D/t is in siemens.

Datta Sec. 3.2.2Which way does the current flow?

  • Electrons carry −q, so the higher voltage side has the lower µ.
  • Inside the channel electrons flow from higher µ (source) to lower µ (drain).
  • Datta uses the electron current I throughout; the conventional current points the other way.

Datta Sec. 3.3Low bias: G is G(E) averaged over the window

G = ∫ dE (−∂f₀/∂E) G(E)Datta Eq. 3.1
  • −∂f₀/∂E: a pulse of unit area, about 4kT wide, centred on µ₀.
  • The measured G is G(E) averaged over ±2kT around µ₀.
  • At low temperature, G ≈ G(µ₀): an operational definition of G(E).
Watch µ₀ move across the band edge: the measured G stays smooth while G(µ₀) has a kink.

Datta Sec. 3.3Why are everyday resistors linear over volts?

  • Eq. 3.1 needs V < kT/q ≈ 26 mV. Yet ordinary resistors stay linear over volts.
  • The elastic model applies only to a length < Lin, the inelastic scattering length.
  • Picture a long resistor as many short elastic sections in series, each with only a small part of V.
Example: 1 mm long, Lin = 100 nm, 5 V → 10⁴ sections, 0.5 mV each, far below kT/q.

Datta Sec. 3.3, Figs. 3.4–3.5But do not count the fake interfaces

R = RB(1 + 3L/λ)
and NOT 3RB(1 + L/λ)Datta Eq. 3.5, wire of length 3L
  • Cutting a wire into sections adds conceptual interfaces, each with its own RB.
  • The real wire has only two interfaces: subtract the fake ones.
  • Recipe: find the length-dependent part RBL/λ of one section, then build the bulk conductivity from it.

Datta Sec. 3.4, Fig. 3.6Degenerate and non-degenerate conductors

  • A · Degenerate: G(E) nearly constant across the window, so G ≈ G(E = µ₀) (Eq. 3.6). Metals.
  • B · Non-degenerate: µ₀ many kT below where G(E) ≠ 0. Use the Boltzmann tail:
G = ∫ (dE/kT) G(E) e−(E−µ₀)/kT
The semiconductor literature often uses B, but real devices are often in between. Datta keeps the general Eq. 3.1.

InteractiveExplore: which shortcut works?

Degenerate or non-degenerate? Averaging G(E) over the thermal window  (Datta Sec. 3.4, Fig. 3.6)

The measured conductance is G(E) averaged with the weight −∂f₀/∂E, a pulse about 4kT wide centred on μ₀ (Eq. 3.1). Where μ₀ sits relative to the band edge Ec decides which shortcut works.

Conductance function:
Exact G, Eq. 3.1–
Degenerate shortcut G(μ₀)–
Non-degenerate shortcut–
Regime–

Try this

  1. Set µ₀ − Ec = +150 meV: the exact G and G(µ₀) agree within 1%.
  2. Set µ₀ − Ec = −150 meV: G(µ₀) = 0 is useless; the Boltzmann form is exact.
  3. Near the band edge (±3kT): neither shortcut is reliable.
  4. Lower T: the in-between region shrinks.
  5. Switch to the constant-M shape: which shortcut changes most?

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

On the log plot, each kT that µ₀ drops below Ec costs a factor e: the straight-line Boltzmann tail.

Datta Secs. 3.3–3.4A short elastic resistor can be non-linear

  • If G(E) is flat, ∫(f₁ − f₂)dE = qV exactly: Ohm's law at any bias.
  • If G(E) changes inside the window (a band edge), I–V bends once V ≳ kT/q.
  • This is computed in the Week 3 notebook with Eq. 3.3.
Short devices are non-linear when their G(E) changes on the scale of qV: the starting point for transistors (Week 6).

CheckConcept check

One minute with a neighbour, then click to reveal.

1. Why are exactly half the states full when f₁ = 1 and f₂ = 0? (click)
Each state is filled by the source and emptied by the drain at the same rate, so it is full half the time.
2. Double the time electrons spend in the channel. What happens to G? (click)
G = q²D/2t halves: half the flux for the same population.
3. A wire of length 3L is three sections with R = RB(1 + L/λ) each. Is its resistance 3RB(1 + L/λ)? (click)
No: RB(1 + 3L/λ). The interface resistance appears once (Eq. 3.5).
4. An n-type semiconductor has µ₀ 0.2 eV below Ec at 300 K. Which formula for G? (click)
The non-degenerate (Boltzmann) form: µ₀ is about 8kT below the band edge.

Wrap-upSummary and next week

Take-home

  • Flux = population ÷ residence time ⇒ G(E) = q²D(E)/2t(E).
  • I = (1/q)∫G(E)(f₁ − f₂)dE; low bias: G = ∫(−∂f₀/∂E)G(E)dE.
  • Heat: µ₁ − ε in the source, ε − µ₂ in the drain.
  • Long wires: elastic sections in series, minus fake interfaces.
  • Degenerate → G(µ₀); non-degenerate → Boltzmann tail.

Before next class

  • Run week03_elastic_resistor.ipynb
  • Work through the Week 3 study toolkit
  • Read Datta Ch. 4–5 (pp. 43–56)

Week 4: what sets t?

Ballistic (t ∝ L) vs diffusive (t ∝ L²), the new Ohm's law, modes, and conductance from noise. Circular session C2.

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