1This week's plan
Read
Datta Part A, Chapter 1 (pp. 1–18), Sections 1.1–1.9. Focus on 1.2–1.6.
Datta Part A, Chapter 1 (pp. 1–18), Sections 1.1–1.9. Focus on 1.2–1.6.
Run
Colab notebook week01_overview.ipynb: run all cells and do Exercises 1–2.
Colab notebook week01_overview.ipynb: run all cells and do Exercises 1–2.
Practise
The quiz and practice problems on this page. Aim for 7/9 on the quiz.
The quiz and practice problems on this page. Aim for 7/9 on the quiz.
By the end of the week you should be able to
- Explain why Ohm's law R = ρL/A fails for short conductors and use R = RB(1 + L/λ).
- Say where RB = h/(q²M) is located (at the contact interfaces) and why it cannot be removed.
- Explain why an elastic resistor releases its Joule heat in the contacts.
- Describe what the electrochemical potential measures (filling, not energy) and what the voltage does.
Optional video: Datta's course "Fundamentals of Nanoelectronics, Part A" on nanoHUB-U, Scientific Overview and Unit 1, L1.1.
2Key ideas in five sentences
- Even a perfect conductor has resistance. A ballistic channel with M modes has RB = h/(q²M) ≈ 25.8 kΩ/M.
- Short and long conductors are one family. R = RB(1 + L/λ): ballistic when L ≪ λ, Ohm's law when L ≫ λ.
- Conductivity = modes per area × mean free path. σ = (q²/h)(M/A)λ (Datta Eq. 1.4), an alternative to the Drude formula.
- Resistance and heat live in different places. Resistance comes from momentum loss (scattering, mode mismatch); the heat I·V is released where electrons lose energy, in the contacts of an elastic resistor.
- The voltage is not a force on each electron. It separates the contacts' electrochemical potentials (µ₁ − µ₂ = qV), so one contact injects more than the other; µ measures how full the states are.
3Equation sheet
R = ρL/A, G = σA/LOhm's law · Eq. 1.1
ρ resistivity (Ω·m), σ = 1/ρ conductivity (S/m), L length, A cross-section. Predicts R → 0 as L → 0, which is wrong for short conductors.
GB = (q²/h)MBallistic conductance · Eq. 1.2
q²/h = 38.7 µS, so RB = h/(q²M) = 25.8 kΩ/M. M = number of modes (parallel channels); Datta counts spin inside M.
G = GB/(1 + L/λ)Eq. 1.3
λ = mean free path (Datta's definition, a backscattering length). Equivalent form: R = RB(1 + L/λ) (Eq. 1.8).
σ = (q²/h)(M/A)λNew expression · Eq. 1.4
Modes per unit area × mean free path. Follows from G ≈ GBλ/L for L ≫ λ.
σ = q²nτm/mDrude · Eq. 1.5
n free-electron density, τm momentum relaxation time, m effective mass. Needs a simple positive effective mass.
σ = q²(D/AL)DdiffDegenerate Einstein relation · Eq. 1.6
D density of states, Ddiff diffusion coefficient. Derived in Week 4.
λ = vτ × {2, π/2, 4/3}1D, 2D, 3D · Eq. 4.10 (preview)
Converts the mean free flight vτ into Datta's λ, because a collision in 2D/3D usually only deflects the electron.
4Physics refresher
Units you will use every week
| 1 eV | 1.602 × 10⁻¹⁹ J (energy of one electron across 1 V) |
| kT at 300 K | 25.9 meV (k = 8.617 × 10⁻⁵ eV/K) |
| q²/h | 38.74 µS |
| h/q² | 25.81 kΩ |
| 1 nm | 10⁻⁹ m, about 4 silicon atomic spacings |
Ohm's law in two lines
Resistance grows with length and falls with area: R = ρL/A. Conductances in parallel add; resistances in series add.
Quick check: a 1 µm copper wire (ρ = 1.68 × 10⁻⁸ Ω·m) with a 10 nm × 10 nm cross-section: R = ρL/A = 168 Ω.
Why "modes"? An analogy
A highway's capacity depends on the number of lanes, not on how wide the city around it is. The contacts are a huge city; the channel offers only M lanes. Even with no potholes (ballistic), the lanes limit the traffic: that limit is RB. Potholes (scattering) add the length-dependent part RBL/λ.5Glossary
| Term | Meaning |
|---|---|
| Ballistic | Electrons cross the channel without scattering ("like a bullet"); L ≪ λ. |
| Diffusive | Electrons scatter many times and random-walk across; L ≫ λ. |
| Mean free path λ | Average distance between (back)scattering events; Datta includes a dimension-dependent factor (Eq. 4.10). |
| Mode (channel) M | An independent parallel path for conduction, like a waveguide mode; each adds q²/h to GB. |
| Interface resistance | The length-independent part RB, associated with the channel–contact interfaces. |
| Elastic resistor | A conductor in which electrons keep their energy in the channel; all energy exchange (heat) happens in the contacts. |
| Energy relaxation length LE | Distance over which hot electrons give up their excess energy; tens of nanometres or longer. |
| Electrochemical potential µ | The level that marks how full the states are (the half-filled point); different in the two contacts under bias: µ₁ − µ₂ = qV. |
| BTE / NEGF | Boltzmann transport equation (semiclassical) / non-equilibrium Green's functions (quantum): full theories that combine mechanics with heat generation. |
6Calculators
Resistance of a channel: R = RB(1 + L/λ)
RB = h/(q²M)
R = RB(1 + L/λ)
Transmission λ/(λ+L)
Contact share of R
From conductivity to modes: σ = (q²/h)(M/A)λ
Modes per nm²
Modes M in the wire
Ohm's law ρL/A
RB(1 + L/λ)
Defaults: copper (σ = 5.96 × 10⁷ S/m, λ ≈ 40 nm at room temperature). The difference between the last two numbers is exactly RB.
7Self-check quiz
Click an answer: you get instant feedback. Score: 0 / 0
8Practice problems
Constants: q²/h = 38.74 µS, h/q² = 25.81 kΩ. Try each problem before opening the hint, and the hint before the solution.
P1. Ballistic conductanceEASY
A ballistic conductor has M = 5 modes. Find GB and RB.
Hint
GB = (q²/h)M.Solution
GB = 5 × 38.74 µS = 193.7 µS; RB = 25.81 kΩ/5 = 5.16 kΩ.P2. Short channelEASY
M = 4, λ = 100 nm, L = 50 nm. Find R and the fraction of injected electrons that reach the drain.
Hint
R = RB(1 + L/λ); T = λ/(λ + L).Solution
RB = 6.45 kΩ; R = 6.45 × 1.5 = 9.68 kΩ; T = 100/150 = 0.67. Note R = RB/T.P3. Measure λ like an experimentalistMEDIUM
Two nanowires of the same material and cross-section give R = 25.8 kΩ at L = 200 nm and R = 51.6 kΩ at L = 600 nm. Find RB, λ and M.
Hint
R is linear in L: slope RB/λ, intercept RB.Solution
Slope = 25.8 kΩ / 400 nm = 64.5 Ω/nm. Intercept RB = 25.8 − 200 × 0.0645 = 12.9 kΩ. λ = RB/slope = 200 nm. M = 25.81/12.9 = 2.P4. A copper nanowireMEDIUM
Copper: σ = 5.96 × 10⁷ S/m, λ ≈ 40 nm. For a 10 nm × 10 nm wire of length 100 nm, find M/A, M, RB, Ohm's-law R and RB(1 + L/λ).
Hint
From Eq. 1.4: M/A = σ/[(q²/h)λ].Solution
M/A = 5.96 × 10⁷ / (3.874 × 10⁻⁵ × 4 × 10⁻⁸) = 3.85 × 10¹⁹ m⁻² = 38.5 per nm². M = 38.5 × 100 ≈ 3850. RB = 25.81 kΩ/3850 = 6.7 Ω. Ohm: ρL/A = 16.8 Ω. RB(1 + 100/40) = 23.5 Ω. The difference is RB.P5. Heat budget of an elastic resistorMEDIUM
qV = 0.4 eV. Electrons cross at ε = µ₂ + 0.3 eV and the current is 1 µA. How much power is released in the source, the channel and the drain?
Hint
Per electron: µ₁ − ε in the source, ε − µ₂ in the drain. Electrons per second = I/q.Solution
Per electron: source 0.1 eV, channel 0, drain 0.3 eV (total 0.4 eV = qV). Power = (I/q) × energy: source 0.1 µW, channel 0, drain 0.3 µW; total I·V = 0.4 µW.P6. 3D mean free pathHARDER
Electrons in a bulk-like 3D wire scatter isotropically with mean free flight vτ = 30 nm. For L = 120 nm, estimate the transmission and R/RB. Why is the answer not λ = 30 nm?
Hint
Datta's λ in 3D is (4/3)vτ (Eq. 4.10).Solution
λ = (4/3) × 30 = 40 nm. T = 40/(40 + 120) = 0.25, R/RB = 1 + 120/40 = 4. In 3D a collision usually deflects rather than reverses the electron, so the backscattering length is longer than vτ. (The 3D widget confirms T to within a few percent.)9Common mistakes
MISTAKE"A ballistic conductor has zero resistance."
CORRECTIt has RB = h/(q²M): the contacts can feed only M modes.
MISTAKE"Better contact technology can remove all contact resistance."
CORRECTThe minimum interface resistance is set by the channel's M, not by the contact quality.
MISTAKE"Electrons travel one mean free path, then scatter."
CORRECTFlight lengths are random (exponential): 63% are shorter than the average.
MISTAKE"The heat is released where the resistance is."
CORRECTIn an elastic resistor the heat is released in the contacts, even though the resistance comes from the channel.
MISTAKE"The voltage pushes every electron with a force qE."
CORRECTThe voltage separates µ₁ and µ₂; the contact with more filled states injects more. The net imbalance is the current.
MISTAKE"A long wire of N sections has resistance N·RB(1 + L/λ)."
CORRECTRB(1 + NL/λ): only the two real interfaces contribute RB (Week 3, Eq. 3.5).
10Explore
These files are in the course folder (course_materials). Open the widgets in any browser; open the notebook in Google Colab (File → Upload notebook).
Week 1 slides
Widget: where is the resistance?
Widget: 3D nanowire
Widget: where is the heat?
Colab notebook
Mini-investigations (15 min each)
- In the 3D widget, set V = 0 and record the two crossing counts for 1 minute. Then set V = 100 mV. Explain the change in one sentence.
- In the resistance widget, find L such that half of R is contact resistance. Explain why the answer does not depend on M.
- In the heat widget, find the channel energy ε for which the drain receives three times more heat than the source.
11Can I…?
Tick what you can do without notes. Your ticks are saved in this browser.
12Looking ahead: Week 2
Why electrons flow (Datta Ch. 2): two Fermi functions, the energy window, and linear response. Think about these before class:
- What is the probability that a state at E = µ + 2kT is filled?
- If the electric field does not drive the current, what does?
- Why do copper and glass, with similar numbers of electrons, conduct so differently?