1This week's plan
Read
Datta Part A, Chapter 3 (pp. 31–42), Sections 3.1–3.4.
Datta Part A, Chapter 3 (pp. 31–42), Sections 3.1–3.4.
Run
Colab notebook week03_elastic_resistor.ipynb: both widgets, the I–V cell and Exercise 1.
Colab notebook week03_elastic_resistor.ipynb: both widgets, the I–V cell and Exercise 1.
Practise
Quiz (aim for 7/9) and the practice problems below.
Quiz (aim for 7/9) and the practice problems below.
By the end of the week you should be able to
- Derive G(E) = q²D(E)/2t(E) from "flux = population ÷ residence time".
- Write I = (1/q)∫G(E)[f₁ − f₂]dE and its low-bias form G = ∫(−∂f₀/∂E)G(E)dE.
- Explain where an elastic resistor releases its heat, and the electron-current sign convention.
- Explain why long resistors are linear, and why R = RB(1 + NL/λ) for N sections.
- Choose between the degenerate and non-degenerate forms of G.
Optional video: Datta's nanoHUB-U course, Unit 1: L1.3 "Why electrons flow" and L1.4 "Conductance formula" (Ch. 3).
2Key ideas in five sentences
- Energy channels are independent. In an elastic resistor electrons keep their energy, so each slice E…E + dE conducts on its own; it may still be diffusive.
- Flux = population ÷ residence time. With the source filling and the drain emptying, half of the D(E)dE states are full, so the flux is D(E)dE/2t.
- The conductance function. G(E) = q²D(E)/2t(E): availability of states times ease of transport; I = (1/q)∫G(E)[f₁ − f₂]dE.
- Measured G is an average. At low bias G = ∫(−∂f₀/∂E)G(E)dE: G(E) averaged over ±2kT around µ₀; degenerate → G(µ₀), non-degenerate → Boltzmann tail.
- Long resistors are elastic sections in series, each with a small voltage, which keeps them linear; count RB only for the real interfaces: R = RB(1 + NL/λ).
3Equation sheet
flux = D(E)dE / 2tSec. 3.2
D(E)dE states, half of them full (f₁ = 1, f₂ = 0), each electron spends time t in the channel.
G(E) = q²D(E) / 2t(E)Eq. 3.2
Conductance function. D per unit energy (1/J), t in seconds → siemens.
I = (1/q)∫G(E)[f₁ − f₂]dEEq. 3.3
Elastic-resistor current at any bias. Electron current, from source to drain.
G = ∫dE(−∂f₀/∂E)G(E)Eq. 3.1
Low-bias conductance: G(E) averaged over the thermal window (valid for V < kT/q per elastic section).
Power = V·I = I²RSec. 3.1
Per electron: µ₁ − ε released in the source and ε − µ₂ in the drain.
R = RB(1 + NL/λ)Eq. 3.5
Wire of length NL; not N·RB(1 + L/λ), which counts fake interfaces.
G ≈ G(µ₀)Degenerate · Eq. 3.6
G(E) nearly constant across ±2kT around µ₀.
G = ∫(dE/kT)G(E)e−(E−µ₀)/kTNon-degenerate
µ₀ many kT below the energies where G(E) ≠ 0 (Boltzmann limit).
4Physics refresher
Little's law
In any steady flow, throughput = population ÷ residence time. A programme of 100 students who each stay 2 years graduates 50 per year. It does not matter how they wander inside.
Units of G = q²D/2t
D in states per joule, t in seconds: q²·(1/J)/s = C²/(J·s) = A/V = S. With D in states per eV: G = q·D[1/eV]/(2t).
Electron current vs conventional current
Electrons carry −q, so the higher-voltage contact has the lower µ. Electrons flow from the source (higher µ, − terminal) to the drain; the conventional current points the other way. Datta always uses the electron current I.Why superpose the two streams? (Sec. 3.2.1)
For coherent transport the stream from the source and the stream from the drain occupy orthogonal states and do not Pauli-block each other; the relaxation-time Boltzmann equation gives the same result.5Glossary
| Term | Meaning |
|---|---|
| Elastic resistor | Electrons keep their energy in the channel; all energy exchange happens in the contacts. |
| Conductance function G(E) | q²D(E)/2t(E): the conductance of the energy slice at E, measured at low T with µ₀ at E. |
| Transit (residence) time t | Average time an electron spends in the channel between contacts. |
| Hot electron | An electron arriving in the drain with energy above µ₂; it relaxes there and releases heat. |
| Inelastic length Lin | Distance over which electrons exchange energy; the elastic model applies to sections shorter than this. |
| Conceptual interface | A boundary introduced when a wire is cut into model sections; its RB must be removed. |
| Degenerate conductor | µ₀ inside the band; G ≈ G(µ₀). Metals. |
| Non-degenerate conductor | µ₀ several kT below the band; Boltzmann statistics apply. Lightly doped semiconductors. |
6Calculators
One energy channel: flux and current
Electrons in channel
Net flux (per s)
Current q × flux
Average occupancy
Conductance function G = q²D/2t
G
In units of q²/h
Degenerate or non-degenerate? (G(E) = (q²/h)·M₁·(E − Ec), above Ec)
Exact G (Eq. 3.1)
Degenerate G(µ₀)
Boltzmann form
Regime
7Self-check quiz
Click an answer: you get instant feedback. Score: 0 / 0
8Practice problems
q = 1.602 × 10⁻¹⁹ C, q²/h = 38.74 µS, kT(300 K) = 25.85 meV.
P1. Little's law for electronsEASY
A channel has 24 states at one energy; f₁ = 1, f₂ = 0, and each electron spends t = 1 ps in the channel. Find the average number of electrons in the channel, the flux and the current.
Hint
Half the states are full.Solution
N = 12; flux = 12/1 ps = 1.2 × 10¹³ s⁻¹; I = q × flux = 1.92 µA.P2. A conductance functionEASY
A channel offers D = 10 states per eV and electrons cross in t = 1 ps. Find G = q²D/2t.
Hint
With D per eV, G = q·D/(2t).Solution
G = 1.602 × 10⁻¹⁹ × 10 / (2 × 10⁻¹² s) = 0.80 µS, about 0.02 q²/h.P3. Do not count fake interfacesMEDIUM
A wire of length 3L (L = λ = 100 nm, M = 10) is modelled as three sections. Compare the correct resistance with the naive sum.
Hint
RB = h/(q²M) = 2.58 kΩ.Solution
Correct: RB(1 + 3) = 10.3 kΩ. Naive 3 × RB(1 + 1) = 15.5 kΩ. The extra 5.2 kΩ is two fake interfaces (2RB).P4. DegenerateMEDIUM
G(E) = (q²/h)·400·(E − Ec)/eV above Ec. With µ₀ = Ec + 0.1 eV at 300 K, compare the exact G with G(µ₀).
Hint
Use the calculator above, or ∫E(−∂f/∂E)dE = kT·ln(1 + eµ₀/kT).Solution
Exact 1558 µS; G(µ₀) = 1550 µS (0.5% low). The Boltzmann form would give 19 170 µS, far too large.P5. Non-degenerateMEDIUM
Same G(E), but µ₀ = Ec − 0.1 eV. Compare the exact G with the Boltzmann form.
Solution
Exact 8.29 µS; Boltzmann 8.37 µS (1% high); G(µ₀) = 0 is useless.P6. Heat budgetMEDIUM
qV = 0.2 eV, a one-level elastic resistor with ε = µ₁ − 0.05 eV carries I = 2 µA. Find the power released in the source, channel and drain.
Hint
Electrons per second = I/q.Solution
Per electron: source 0.05 eV, channel 0, drain 0.15 eV. Power: source 0.1 µW, drain 0.3 µW, total I·V = 0.4 µW.P7. Why a long resistor is linearHARDER
A 1 mm resistor with inelastic length Lin = 100 nm carries 5 V. Treat it as elastic sections of length Lin. What voltage drops across each, and is each section in linear response?
Solution
N = 1 mm/100 nm = 10⁴ sections; 5 V/10⁴ = 0.5 mV each, far below kT/q ≈ 26 mV, so each section (and the whole resistor) is linear.9Common mistakes
MISTAKE"Elastic means ballistic."
CORRECTElastic means no energy loss in the channel; the electron may still scatter and diffuse (change momentum).
MISTAKE"The heat is released where the resistance is."
CORRECTIn an elastic resistor the heat is released in the contacts: µ₁ − ε in the source, ε − µ₂ in the drain.
MISTAKE"All D(E)dE states are full when the source fills them."
CORRECTThe drain empties them as fast as the source fills them: on average half are full.
MISTAKE"The measured G is G(µ₀)."
CORRECTOnly for degenerate conductors; in general G is G(E) averaged over the ~4kT window.
MISTAKE"Use the Boltzmann formula for every semiconductor."
CORRECTIt fails when µ₀ is within a few kT of the band edge or inside the band.
MISTAKE"Current flows from the positive terminal through the channel."
CORRECTDatta's electron current flows from the source (higher µ, negative terminal) to the drain.
10Explore
Files are in the course folder (course_materials).
Week 3 slidesWidget: flux from state boxesWidget: degenerate vs non-degenerateColab notebookWeek 2 toolkit
Mini-investigations (15 min each)
- In the flux widget, set f₁ = 0.8, f₂ = 0.2 and double t. Predict, then measure, the change in occupancy and in flux.
- In the degenerate widget, find the µ₀ − Ec at which the degenerate and Boltzmann shortcuts are equally wrong (300 K).
- In the notebook I–V cell, move Ec to µ₀ + 0.2 eV. At what voltage does the current start to rise, and why?
11Can I…?
Tick what you can do without notes. Your ticks are saved in this browser.
12Looking ahead: Week 4
Ballistic and diffusive transport; conductance from fluctuation (Datta Ch. 4–5) and circular session C2.
- How does the time to cross a channel grow with L for a bullet? For a random walker?
- Why might Datta's mean free path differ from the average distance between collisions vτ?
- Can you measure a resistor's conductance without applying a voltage?