1This week's plan
Datta Part A, Chapter 7 (pp. 75–96), including the MATLAB codes of Section 7.6.
Colab notebook week06_nanotransistor_C3.ipynb: Part 1 before class, Part 2 (lab) in circular session C3.
Quiz (aim for 8/10), practice problems, and the C3 preparation list.
By the end of the week you should be able to
- Explain gate control with the point-channel model I = (1/q)∫G(E − U)(f₁ − f₂)dE.
- Explain why the current saturates and why α > 0 spoils the saturation.
- Set up and solve the self-consistent loop U = UL + U₀(N − N₀), and explain the ≈ 2.8× rise of Isat.
- Compute CQ = q²D₀, CE = ε/tox, their series combination and the gate efficiency.
- Explain the 60 mV/decade limit and the electrostatic rectifier.
Optional video: Datta's nanoHUB-U course, Unit 2: L2.8–L2.9 (quantum capacitance, the nanotransistor).
2Key ideas in five sentences
- A gate moves states, not electrons. VG slides G(E) and D(E) in energy (U = −qVG); current flows when states sit in the window between µ₁ and µ₂.
- Saturation. Once µ₂ drops below the band edge the drain only empties states, so I stops depending on VD, except through α (drain control) and charging.
- Charging is self-consistent. Missing electrons make the channel positive and lower its states, U = UL + U₀(N − N₀); with very strong charging Isat rises by 23/2 ≈ 2.8 in 2D, but the low-bias slope does not change.
- Quantum capacitance. CQ = q²D₀ acts in series with CE; only 1/(1 + CQ/CE) of a gate step moves the band above threshold.
- 60 mV per decade. Below threshold the band follows the gate and n is a Boltzmann tail: one decade per kT·ln10. Electrostatic asymmetry alone can make a symmetric device rectify.
3Equation sheet
4Physics refresher: capacitors and potentials
C = εA/t. Charge Q = CV. For the gate–oxide–channel stack: CE/area = εrε₀/tox; with εr = 4 and tox = 2 nm, 1.77 µF/cm². Adding one electron raises the channel energy by U₀ = q²/CE.
1/C = 1/C₁ + 1/C₂: the voltage divides in inverse proportion to C, and the smaller capacitor dominates. For the gate: CE (oxide) in series with CQ (the channel's states).
Units trick
F/m² × 100 = µF/cm². D per area in eV⁻¹cm⁻² × q (C) × 10⁶ = µF/cm² of quantum capacitance: e.g. 8.35 × 10¹³ eV⁻¹cm⁻² gives 13.4 µF/cm².Why U = −qV
Potential energy = charge × potential; the electron charge is −q. Datta draws energies, so a positive gate voltage moves levels down.5Glossary
| Term | Meaning |
|---|---|
| Point-channel model | The whole channel described by one potential U (Eq. 7.1); "top of the barrier" model. |
| Saturation current | Current once µ₂ is below the band edge; set by the gate, ideally independent of VD. |
| α (drain control, DIBL) | Fraction of qVD that lowers the channel; causes output conductance. |
| Self-consistent potential | U that satisfies the electrostatics and the electron count together (Fig. 7.7). |
| Charging energy U₀ | Potential-energy change per added electron, q²/CE. |
| Quantum capacitance | CQ = q²D₀: how easily the channel accepts electrons; in series with CE. |
| Subthreshold slope | Gate voltage per decade of current below threshold; ≥ 60 mV at 300 K for these devices. |
| Electrostatic rectifier | A symmetric device with asymmetric electrostatics (α = 0) and an asymmetric I–V. |
6Calculators
Gate capacitances and efficiency (2D channel)
Subthreshold switching
Low-temperature saturation current of a 2D channel (α = 0)
7Self-check quiz
Click an answer: you get instant feedback. Score: 0 / 0
8Practice problems
q²/h = 38.74 µS, kT = 25.9 meV at 300 K, ε₀ = 8.854 × 10⁻¹² F/m.
2D channel, m = 0.2 m₀, g = 2, W = L = 1 µm, tox = 2 nm, ε = 4ε₀. Find D₀ (states per eV), CE, U₀ and U₀D₀.
Solution
D₀ = gmWL/(2πħ²) ≈ 8.35 × 10⁵ eV⁻¹; CE = εWL/tox = 1.77 × 10⁻¹⁴ F; U₀ = q²/CE ≈ 9.05 µeV; U₀D₀ ≈ 7.6.Below threshold at 300 K with gate efficiency 0.8, how much gate voltage raises the current 1000×?
Solution
3 decades × 59.5 mV / 0.8 = 223 mV.For the channel of P1 (per area, CQ = 13.4 µF/cm²) with tox = 1 nm, find CE, C and the gate efficiency.
Solution
CE = 3.54 µF/cm²; C = 3.54 × 13.4/16.9 = 2.80 µF/cm²; efficiency 1/(1 + 3.78) = 0.21.At low T with α = 0 and U₀ = 0, µ₁ − Ec = 50 meV for the channel of P1. Compute Isat = (q/h)∫M(E)dE over 0 < E − Ec < 50 meV, and the low-bias conductance (q²/h)M(µ₁).
Hint
∫√(2mε)dε = (2/3)√(2m)ε3/2.Solution
Isat = (q/h)·g(2W/h)·(2/3)√(2m)(0.05 eV)3/2 ≈ 0.42 mA; M(µ₁) = 326, G = 12.6 mS. The notebook gives 0.43 mA and 12.5 mS at kT = 5 meV.Explain why very strong charging multiplies the saturation current of a 2D channel by 23/2.
Hint
Compare the electron numbers with U₀ = 0 and with U₀ → ∞.Solution
With U₀ = 0 the states between Ec and µ₁ are half filled (N = N₀/2). Strong charging forces N = N₀ with half filling, so the filled range µ₁ − Ec doubles (constant 2D D). Isat ∝ ∫√ε dε ∝ (µ₁ − Ec)3/2: ratio 23/2 ≈ 2.83 (2.80 in the notebook).GaAs 2D gas (m = 0.067 m₀, g = 2) under 3 nm of an oxide with εr = 10. Find CQ, CE, C and the gate efficiency.
Solution
CQ ≈ 4.48 µF/cm², CE = 2.95 µF/cm², C ≈ 1.78 µF/cm², efficiency = 1/(1 + 1.52) ≈ 0.40.What is the ideal subthreshold slope at 77 K? How much gate voltage would switch the current by 10⁴ at 300 K and at 77 K?
Solution
S = kT ln10/q = 15.3 mV/dec at 77 K (59.5 at 300 K). For 10⁴: 238 mV at 300 K, 61 mV at 77 K.9Common mistakes
10Prepare for circular session C3
- Why momenta are evenly spaced but energies are not
- Sketch D(E) and M(E) in 1D, 2D, 3D
- What "n" counts in a p-type band
- Capacitors in series: CE and CQ
B1: thin-film DOS staircase versus the 3D √E law. B2: count modes on a grid. B3: a nanotransistor with gate charging for tox = 1, 2, 5 nm. Write your prediction for B3 before running it.
Problems like P1, P2 and P6 above.
Saturation; charging and 2.8×; quantum capacitance; 60 mV/decade; the electrostatic rectifier; M = floor(2W/λ).
11Explore
Files are in the course folder (course_materials).
Mini-investigations (15 min each)
- In the nanotransistor widget, plot (by hand) Isat/Isat(U₀ = 0) against U₀D₀ = 0, 2, 5, 10, 20, 50. Where does it reach 2.5?
- In the quantum-capacitance widget, find the effective mass at which CQ = CE for tox = 2 nm. Check it with the calculator.
- In the notebook, run Datta's Fig. 7.11 cell at kT = 5 meV. Does the rectification become stronger or weaker? Why?
12Can I…?
Tick what you can do without notes. Your ticks are saved in this browser.
13Looking ahead: Week 7
What and where is the voltage drop? (Datta Ch. 8–9): the diffusion equation for ballistic transport, two electrochemical potentials µ⁺ and µ⁻, the Boltzmann equation and the two potentials.
- The diffusion equation with µ(0) = µ₁ and µ(L) = µ₂ gives Ohm's law G = σA/L. How must the boundary conditions change to give G = σA/(L + λ)?
- In a ballistic channel, which contact sets the occupation of the right-moving electrons?
- Can a current flow when the electrostatic potential U varies in space but µ is constant?