Week 2 Study Toolkit · Nanoelectronics

1This week's plan

Read
Datta Part A, Chapter 2 (pp. 21–30), Sections 2.1–2.5.

Run
Colab notebook week02_fermi_functions_C1.ipynb: Part 1 before the lecture, Part 2 in circular session C1.

Practise
Quiz (aim for 8/10), practice problems, and the C1 preparation list below.

By the end of the week you should be able to

Use the two key concepts, D(E) and µ₀, and the relation n = ∫D(E)f(E − µ)dE.
Sketch the Fermi function and the thermal broadening function, and give their key numbers.
Explain why current flows only in the energy window between µ₁ and µ₂.
Derive and apply linear response, f₁ − f₂ ≈ (−∂f₀/∂E)qV, and say when it fails.
Explain why conduction depends on −∂f₀/∂E rather than f₀, unlike the Drude picture.

Optional video: Datta's nanoHUB-U course, Unit 1: L1.2 "Two key concepts" (Sec. 2.1) and L1.9 "Drude formula" (Sec. 2.5.1).

2Key ideas in five sentences

Two key concepts. Draw the density of states D(E) (the container) and locate the electrochemical potential µ₀ (the water line); at equilibrium µ₀ is the same everywhere.
A voltage separates the contacts. The battery lowers all energies in the positive contact, so µ₁ − µ₂ = qV; each contact stays in equilibrium with its own Fermi function.
Current flows in a window. Between µ₂ and µ₁ the source fills states that the drain empties; above or below both, the contacts agree and nothing flows: I(E) ∼ f₁(E) − f₂(E).
Linear response. For qV ≪ kT, f₁ − f₂ ≈ (−∂f₀/∂E)qV: a pulse about 4kT wide centred on µ₀ selects the energies that conduct.
Conduction depends on −∂f₀/∂E, not on f₀. Materials with similar numbers of electrons (copper, glass) differ enormously because of the states available within a few kT of µ₀.

3Equation sheet

µ₁ − µ₂ = qVEq. 2.1

The battery separates the contacts' electrochemical potentials. The positive terminal (drain) has the lower µ, because the electron charge is −q.

f(E) = 1/(1 + e^(E−µ)/kT)Fermi function · Eq. 2.2

Probability that a state at energy E is occupied. f(µ) = ½; f goes from 1 to 0 over about µ ± 2kT. k = 8.617 × 10⁻⁵ eV/K.

F_T = kT(−∂f/∂E) = f(1 − f)Eqs. 2.3, 2.5

Thermal broadening function: symmetric about µ (Eq. 2.4), peak 1/4, area kT (Eq. 2.6), so about 4kT wide.

I(E) ∼ f₁(E) − f₂(E)Eq. 2.8

Current at energy E in a short conductor; f₁ = f(E − µ₁), f₂ = f(E − µ₂) (Eq. 2.7).

F(E) = (f₁ − f₂)/(qV/kT)Eq. 2.9

Normalized difference, with µ₁,₂ = µ₀ ± qV/2. Its area is always 1 (in units of kT); it approaches F_T as qV/kT → 0.

f₁ − f₂ ≈ (−∂f₀/∂E)qVLinear response · Eq. 2.10

Valid for qV ≪ kT (about 26 mV at room temperature).

v_d = −(qτ_m/m)F, σ = qnµ̃Drude · Eqs. 2.12–2.13

Field-driven picture with friction; µ̃ = qτ_m/m is the mobility. It suggests every electron conducts, which is misleading.

n = ∫D(E)f(E − µ)dEFilling the container

The number of electrons fixes µ for a given D(E); different D(E) shapes give different µ for the same n.

4Physics refresher: Fermi–Dirac statistics

Why a step?

Electrons are fermions: at most one per state (per spin). At T = 0 they fill the lowest states up to µ, so f is a sharp step. At T > 0, thermal energy kT lets electrons near µ move up a little, which rounds the step over a few kT.

Numbers to remember

f(µ)	0.5
f(µ ± kT)	0.27 / 0.73
f(µ ± 2kT)	0.12 / 0.88
f(µ + 3kT)	0.047
Peak of F_T	1/4
FWHM of −∂f/∂E	3.53kT

Taylor expansion behind linear response (Datta Appendix A)

For a small shift of µ: f(E − µ) ≈ f₀(E) + (−∂f₀/∂E)(µ − µ₀). Applying it to both contacts with µ₁,₂ = µ₀ ± qV/2 and subtracting gives f₁ − f₂ ≈ (−∂f₀/∂E)qV.

Boltzmann limit

For E − µ ≫ kT, f ≈ e^{−(E−µ)/kT} and −∂f/∂E ≈ f/kT. This is the non-degenerate regime of semiconductors, where f₀ and −∂f₀/∂E look alike.

5Glossary

Term	Meaning
Density of states D(E)	Number of available states per unit energy in the channel (filled or empty).
Electrochemical potential µ	Energy at which states are half filled; at equilibrium the same everywhere. Often called the Fermi level.
Fermi function f(E)	Occupation probability of a state at energy E in equilibrium.
Thermal broadening function F_T	kT(−∂f/∂E) = f(1 − f); a pulse about 4kT wide centred on µ.
Energy window	The energy range between µ₂ and µ₁ (± a few kT) where the contacts disagree and current flows.
"Agenda" of a contact	Datta's word for the occupation a contact tries to impose on the channel, its Fermi function.
Linear response	The low-bias limit where current is proportional to voltage; requires qV ≪ kT.
Point channel model	Describing the channel by a single D(E), ignoring how it varies in space.
Degenerate / non-degenerate	µ inside the band (metal-like) / µ several kT below the band (Boltzmann tail); see Week 3.

6Calculators

Fermi function

E − µ (meV)Temperature T (K)

(E − µ)/kT

f(E): filled

F_T = f(1 − f)

Is my bias in linear response?

Voltage V (mV)Temperature T (K)

y = qV/kT

Peak of F (F_T: 0.25)

Largest |F − F_T|

Verdict

Wide conductor: I = (q²/h)·M·V

Modes MVoltage V (mV)

Current I

Conductance G

7Self-check quiz

Click an answer: you get instant feedback. Score: 0 / 0

8Practice problems

kT at 300 K = 25.85 meV; q²/h = 38.74 µS. Try first, then the hint, then the solution.

P1. Reading the Fermi functionEASY

Find f at E = µ + 2kT, µ − 2kT and µ + 3kT.

Hint

f = 1/(1 + eˣ) with x = (E − µ)/kT.

Solution

f(2) = 1/(1 + e²) = 0.119; f(−2) = 0.881 (= 1 − f(2)); f(3) = 0.047.

P2. A state above µEASY

At 300 K, what is the probability that a state 0.1 eV above µ is filled? What about at 600 K?

Hint

Convert to x = 0.1 eV/kT first.

Solution

300 K: x = 3.87, f = 0.020. 600 K: kT = 51.7 meV, x = 1.93, f = 0.126. Doubling T raises the occupation sixfold here.

P3. The thermal broadening functionEASY

Using F_T = f(1 − f), find F_T at E = µ and at E = µ + 2kT. What fraction of the peak is left at 2kT?

Solution

At µ: 0.5 × 0.5 = 0.25. At µ + 2kT: 0.119 × 0.881 = 0.105, about 42% of the peak.

P4. Linear response or not?MEDIUM

Find y = qV/kT for (a) 10 mV at 300 K, (b) 100 mV at 300 K, (c) 10 mV at 77 K. Which are safely in linear response?

Hint

kT at 77 K is 6.64 meV.

Solution

(a) y = 0.39: linear. (b) y = 3.9: not linear. (c) y = 1.5: borderline; cooling shrinks the linear range.

P5. A wide conductorMEDIUM

A channel conducts equally at all energies with one mode. Show that I = (q²/h)V at any temperature, and evaluate it for V = 1 mV.

Hint

Integrate f₁ − f₂ over all energies.

Solution

∫(f₁ − f₂)dE = µ₁ − µ₂ = qV exactly, so I = (1/q)(q²/h)·qV = (q²/h)V = 38.74 µS × 1 mV = 38.7 nA, independent of T.

P6. A narrow levelHARDER

A single level at ε = µ₀ is broadened by γ = 2 meV (Lorentzian transmission, peak 1). At 300 K, estimate the linear-response conductance G = (q²/h)∫T(E)(−∂f/∂E)dE.

Hint

γ ≪ kT, so −∂f/∂E ≈ 1/(4kT) across the level; the area under the Lorentzian is πγ/2.

Solution

G ≈ (q²/h)·(πγ/2)/(4kT) = 38.74 µS × 3.14/103.4 ≈ 1.18 µS. The exact integral gives 1.15 µS. The level is much narrower than the thermal window, so only 3% of the q²/h conductance survives.

P7. Drude numbers for copperMEDIUM

Copper has n ≈ 8.5 × 10²⁸ m⁻³ and σ = 5.96 × 10⁷ S/m. Using σ = q²nτ_m/m with m = m₀, find τ_m and the mobility qτ_m/m.

Solution

τ_m = σm/(q²n) ≈ 2.5 × 10⁻¹⁴ s; mobility ≈ 44 cm²/V·s. These "free electron" numbers work for copper, but the formula gives no clue why glass, with a similar total electron count, is an insulator.

9Common mistakes

MISTAKE"The electric field makes the electrons flow."

CORRECTHuge atomic fields and p-n junction fields cause no current. The difference µ₁ − µ₂ does.

MISTAKE"The Fermi level is where the electrons are."

CORRECTMost electrons are far below µ; µ only marks the half-filled point.

MISTAKE"All electrons carry current when a voltage is applied."

CORRECTOnly those in the window µ₂ … µ₁ (± a few kT) do; states below both are full and carry nothing.

MISTAKE"Linear response holds for any small-looking voltage."

CORRECTIt needs qV ≪ kT: 1 mV is fine at 300 K, 100 mV is not, and the range shrinks at low temperature.

MISTAKE"Conductivity follows the number of electrons."

CORRECTIt follows the states within ~kT of µ₀ (−∂f₀/∂E), which is why copper and glass differ by about 10²⁰.

MISTAKE"µ must be at an energy where states exist."

CORRECTIn a semiconductor µ often sits in the gap, where D(E) = 0: a water line between floors.

10Prepare for circular session C1

C1 covers Weeks 1–2 and has four blocks. Come ready for each.

A · Refresher clinic (pairs)

Why R → 0 as L → 0 is unphysical
(q²/h) × 1 V is a current; qV = 1 eV for 1 V
µ, kT and f(µ); sketch f(E) with energy vertical
f at µ + 3kT

B · Colab lab

Run Part 2 of the Week 2 notebook (conductance map G vs ε at 50–600 K) and bring one-sentence answers to the three lab questions.

C · Problem clinic

Three problems similar to P2, P4 and P5 above.

D · Two-minute explanation

You may draw any of: why current flows only in the window; where the heat goes; why a ballistic wire has resistance; what µ measures and what V does; why copper conducts and glass does not. Practise one sketch for each.

11Explore

Files are in the course folder (course_materials). Open widgets in any browser; open the notebook in Google Colab (File → Upload notebook).

Week 2 slides Widget: fill the states Widget: two Fermi functions Widget: linear response (Fig. 2.5) Colab notebook Week 1 toolkit

Mini-investigations (15 min each)

In "fill the states", put 40% of the seats into the 1D, 2D and 3D bands. Rank the three µ values and explain the order using D(E).
In the semiconductor, find the doping that moves µ to about 0.85 eV. Check that n·p stays the same as without doping.
In the linear-response widget, find the largest y for which the gap |F − F_T| stays below 0.01.

12Can I…?

Tick what you can do without notes. Your ticks are saved in this browser.

explain D(E) and µ with the container and water-line picture give two reasons why "the electric field" does not explain current flow state µ₁ − µ₂ = qV and say which contact is lowered sketch f(E) and F_T and quote f(µ), the peak 1/4 and the area kT explain the energy window and why states outside it carry no current derive f₁ − f₂ ≈ (−∂f₀/∂E)qV and state when it holds explain why conduction depends on −∂f₀/∂E, using copper and glass run Part 2 of the notebook and explain the conductance map

13Looking ahead: Week 3

The elastic resistor (Datta Ch. 3). Think about these before class:

If a university enrols 100 students who each stay 2 years, how many graduate per year? What is the electron version of this?
With f₁ = 1 and f₂ = 0, what fraction of the channel states do you expect to be full on average?
Why might a long resistor stay linear over several volts, although a short one is linear only for V ≲ kT/q?

Why electrons flow: two Fermi functions and the energy window

1This week's plan

By the end of the week you should be able to

2Key ideas in five sentences

3Equation sheet

4Physics refresher: Fermi–Dirac statistics

5Glossary

6Calculators

Fermi function

Is my bias in linear response?

Wide conductor: I = (q²/h)·M·V

7Self-check quiz

8Practice problems

9Common mistakes

10Prepare for circular session C1

11Explore

Mini-investigations (15 min each)

12Can I…?

13Looking ahead: Week 3