Week 9 Study Toolkit · Nanoelectronics

1This week's plan

Read
Datta Part A, Chapter 12 (pp. 151–167): p-n junctions, spin valves, spin potentials.

Run
Colab notebook week09_smart_contacts.ipynb: both widgets, the two-level solar cell, the spin-valve network and the exercises.

Practise
Quiz (aim for 8/10), the practice problems and one mini-investigation. The models return in the C5 lab next week.

By the end of the week you should be able to

Explain why some devices need separate quasi-Fermi levels for different groups of states.
Derive I ∝ e^qV/kT − 1 from the two-level model, and I_sc and V_OC = ħω(1 − T/T₀)/q when the scatterers are hot.
Explain why asymmetric contacts are essential for a solar cell.
Compute R_P, R_AP and MR of a spin valve, and the spin potential in its channel.
Explain how a magnetic probe measures µ_S, and how spin-momentum locking creates µ_S from a current.

Optional video: Datta's nanoHUB-U course, Unit 3: L3.9 (spin potentials).

2Key ideas in five sentences

Smart contacts. A contact that talks to one group of states more than another drives the groups to different quasi-Fermi levels; scattering between the groups (R-G, spin flip, backscattering) pulls them back together.
The p-n junction is an inelastic device. No elastic path connects the contacts; the transfer A ↔ B limits the current, and detailed balance (Eq. 12.7) gives I ∝ e^qV/kT − 1.
A solar cell is a heat engine. Scatterers hotter than the device (sunlight, T₀ ≈ 6000 K) drive a current with no voltage; open-circuit qV_OC = ħω(1 − T/T₀) is the Carnot limit. It needs asymmetric contacts.
Spin valve. Majority and minority spins see interface resistances r < R; the circuit gives MR = P²/(1 − P²) when the interfaces dominate, and a spin potential µ_S = (µ_up − µ_dn)/2 in the channel.
Spin potentials can be measured (magnetic probe: 2Pµ_S) and even created by a plain current where spin is locked to momentum: µ_S = (q/2G_B)pI.

3Equation sheet

δµ = µ⁺ − µ⁻ = (µ₁ − µ₂)/(1 + L/λ)Eq. 12.1

Two groups (right/left movers) driven apart by contacts.

I_n/A = −(σ_0,n/q)dµ_n/dz, dI_n/dz = Σ(RG_m→n − RG_n→m)Eqs. 12.3–12.4

One µ per group, coupled by recombination-generation.

I = I₀(e^qV/νkT − 1)Eq. 12.5

Diode law; ν and I₀ set by the R-G processes.

I ∼ D_BAf_A(1 − f_B) − D_ABf_B(1 − f_A)Eq. 12.6

Two-level model; transition from the second index to the first.

D_AB/D_BA = e^{−ħω/kT₀}Eq. 12.7

Absorbing ħω from scatterers at T₀ is harder than emitting.

X = e^{ħω/kT₀ − ħω/kT}e^{(µ_A − µ_B)/kT}, I ∼ D_ABf_B(1 − f_A)(X − 1)Eqs. 12.8, 12.11

T₀ = T and µ_A − µ_B = qV give the ideal diode.

I_sc ∼ exp[(ħω/k)(1/T₀ − 1/T)] − 1, qV_OC = ħω(1 − T/T₀)Eq. 12.12, Sec. 12.1.1

Energy conversion; X = 1 at open circuit gives the Carnot efficiency (unnumbered in the book).

R_P = [1/(2r + R_ch) + 1/(2R + R_ch)]⁻¹, R_AP = (r + R + R_ch)/2Eqs. 12.13–12.14

Spin-valve circuit.

MR = (R − r)²/4rR = P²/(1 − P²), P = (R − r)/(R + r)Eqs. 12.15–12.16

R_ch = 0; tunnel junctions: 2P²/(1 − P²).

µ_P = (g^upµ_up + g^dnµ_dn)/(g^up + g^dn) = µ + Pµ_SEqs. 12.17–12.19

µ = (µ_up + µ_dn)/2, µ_S = (µ_up − µ_dn)/2. (Datta prints Pµ_S/2 in 12.19a; the algebra and Eq. 12.20 need Pµ_S.)

µ_P(+m) − µ_P(−m) = 2Pµ_SEq. 12.20

Measure µ_S by reversing the probe magnet.

µ_S = p(µ⁺ − µ⁻)/2 = (q/2G_B)pI, p ∼ (M − N)/(M + N)Eqs. 12.22–12.25

Spin-momentum locking: current-induced spin potential.

4Physics refresher: p-n junctions and magnets

p-n junction

Equilibrium: µ flat, bands bent by qV_bi = kT ln(N_AN_D/n_i²) over a depletion width W = [2ε_sV_bi(N_A + N_D)/(qN_AN_D)]^1/2. Forward bias V lowers the barrier to V_bi − V and splits µ into F_n and F_p by qV across the junction.

Ferromagnets

The exchange interaction shifts the up- and down-spin bands. At E_F the majority spin has a different density of states (and interface conductance) from the minority spin. Polarization P = (D↑ − D↓)/(D↑ + D↓). Reversing the magnetization swaps up and down.

Detailed balance in one line

A system in contact with a bath at T₀ has occupation ratio e^{−ħω/kT₀} between states ħω apart. For that to be stationary, the ratio of upward to downward jump rates induced by the bath must be e^{−ħω/kT₀}: Eq. 12.7. Datta proves it in Ch. 15.

Why (1 − f)/f = e^{(ε − µ)/kT}

From f = 1/(1 + e^{(ε − µ)/kT}): 1/f − 1 = e^{(ε − µ)/kT}. This turns products of Fermi functions in Eq. 12.9 into the exponentials of Eq. 12.11.

5Glossary

Term	Meaning
Quasi-Fermi level (QFL)	The electrochemical potential of one group of states out of equilibrium with the others.
Smart contact	A contact that couples to some groups of states more strongly than to others.
R-G processes	Recombination-generation: inelastic transfer between bands (here A ↔ B).
Ideality factor ν	I ∝ e^qV/νkT; ν = 1 is ideal (59.5 mV/decade at 300 K).
Short-circuit current, open-circuit voltage	I at V = 0; V at I = 0.
Spin valve, GMR	Two magnets and a metallic channel; resistance depends on their relative orientation.
MTJ, TMR	Magnetic tunnel junction: an insulating barrier as channel; MR = 2P²/(1 − P²).
Spin potential µ_S	(µ_up − µ_dn)/2: spin accumulation.
Spin-momentum locking	Surface states whose spin is fixed by their direction of motion (topological insulators).
Rashba–Edelstein effect	A charge current creating a spin potential (and its inverse by reciprocity).

6Calculators

Two-level diode and solar cell

Level spacing ħω (eV)Device T (K)Scatterer T₀ (K)Dilution δ (1 = Datta)

Carnot qV_OC = ħω(1 − T/T₀)

with dilution: + kT lnδ

Dark slope kT ln10 / q

I_sc factor exp[(ħω/k)(1/T₀ − 1/T)] − 1

Spin valve (Eqs. 12.13–12.16)

Majority interface r (Ω)Minority interface R (Ω)Channel R_ch per spin (Ω)

R_P · R_AP

MR = R_AP/R_P − 1

P = (R − r)/(R + r) · P²/(1 − P²) · 2P²/(1 − P²)

µ_S/qV in the AP channel

Spin potentials: probe and spin-momentum locking

Probe polarization PSpin potential µ_S/q (µV)Channel polarization pModes M (spin-locked)Current I (µA)

µ_P(+m) − µ_P(−m)

G_B = (q²/h)M

µ_S/q = pI/2G_B

probe signal for that µ_S

7Self-check quiz

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

8Practice problems

kT = 25.85 meV and kT ln10 = 59.5 meV at 300 K; k = 8.617 × 10⁻⁵ eV/K; q²/h = 38.74 µS.

P1. Built-in potentialEASY

A silicon junction has N_A = 10¹⁷ cm⁻³, N_D = 10¹⁶ cm⁻³, n_i = 10¹⁰ cm⁻³ at 300 K. Find V_bi.

Solution

V_bi = 0.02585 × ln(10³³/10²⁰) = 0.02585 × 29.93 = 0.774 V.

P2. IdealityEASY

By what factor does the forward current grow when V increases by 0.1 V, for ν = 1 and ν = 2 at 300 K?

Solution

e^0.1/0.02585 = 47.9 for ν = 1; e^0.1/0.0517 = 6.9 for ν = 2.

P3. Carnot limitEASY

ħω = 1.4 eV, T = 300 K, sun at T₀ = 6000 K. Find qV_OC in Datta's model and with diluted sunlight δ = 2.16 × 10⁻⁵.

Solution

1.4 × (1 − 0.05) = 1.33 V; kT lnδ = 0.02585 × (−10.74) = −0.278 V, so 1.05 V.

P4. A ratchetMEDIUM

ħω = 0.1 eV, T = 300 K. Evaluate the short-circuit factor exp[(ħω/k)(1/T₀ − 1/T)] − 1 for T₀ = 400 K and T₀ = 250 K. What do the signs mean?

Hint

ħω/k = 1160 K.

Solution

T₀ = 400 K: e^−0.967 − 1 = −0.62 (electrons pumped B→A, out of contact 1). T₀ = 250 K: e^+0.774 − 1 = +1.17: the current reverses, like Feynman's ratchet.

P5. Spin valveMEDIUM

r = 2 Ω, R = 6 Ω. Find R_P, R_AP and MR for R_ch = 0 and R_ch = 4 Ω.

Solution

R_ch = 0: R_P = (1/4 + 1/12)⁻¹ = 3 Ω, R_AP = 4 Ω, MR = 33.3% = P²/(1 − P²) with P = 0.5. R_ch = 4 Ω: R_P = (1/8 + 1/16)⁻¹ = 5.33 Ω, R_AP = 6 Ω, MR = 12.5%.

P6. Tunnel junctionsMEDIUM

Which P gives MR = 100% in a metallic spin valve (R_ch = 0) and in a tunnel junction?

Solution

P²/(1 − P²) = 1 ⇒ P = 0.707; 2P²/(1 − P²) = 1 ⇒ P² = 1/3, P = 0.577.

P7. Measuring µ_SMEDIUM

In the AP configuration of P5 (R_ch = 4 Ω, no spin flips) a voltage V = 1 mV is applied. Find µ_S/q in the channel and the change in reading of a probe with P = 0.3 when its magnet is reversed.

Hint

In AP both spins carry the same current; µ_up − µ_dn = (R − r)qV/(r + R + R_ch).

Solution

µ_up − µ_dn = 4/12 × 1 mV = 0.333 mV, µ_S/q = 0.167 mV; signal 2Pµ_S = 0.100 mV.

P8. Spin-momentum lockingHARDER

A topological-insulator surface strip has k_F = 10⁹ m⁻¹ and width 5 µm (one spin-locked band, M = Wk_F/π). With p = 0.3 and I = 100 µA, estimate µ_S/q.

Solution

M = 1592, G_B = 38.74 µS × 1592 = 0.0617 S; µ_S/q = pI/2G_B = 0.3 × 10⁻⁴/0.123 = 243 µV.

P9. Derive Eq. 12.11HARDER

Starting from Eq. 12.9 and Eq. 12.10, derive X, and show that X = 1 at equilibrium (V = 0, T₀ = T).

Solution

f_A/(1 − f_A) = e^{−(ε_A − µ_A)/kT}, (1 − f_B)/f_B = e^{(ε_B − µ_B)/kT}, D_BA/D_AB = e^ħω/kT₀. The product is e^ħω/kT₀e^−ħω/kTe^{(µ_A − µ_B)/kT}. With T₀ = T and µ_A = µ_B, X = 1 and I = 0.

9Common mistakes

MISTAKE"A p-n junction is just a resistor whose conductivity depends on V."

CORRECTIts current is set by inelastic transfer between two groups of states; their conductivities hardly matter (Eq. 12.5).

MISTAKE"You need positive holes to understand the junction."

CORRECTDatta's model has only electrons jumping between levels A and B; holes are an optional bookkeeping device.

MISTAKE"Any absorbing material gives a photovoltage."

CORRECTWithout asymmetric contacts the pumped electrons leave equally to both sides: no current (Fig. 12.6b).

MISTAKE"Faster R-G is always better."

CORRECTWhen K ≫ γ the levels no longer follow the contacts; the transfer step must be the bottleneck for µ_A − µ_B = qV.

MISTAKE"MR depends only on the magnets."

CORRECTThe channel resistance and spin flips in series dilute it; MR = P²/(1 − P²) only for R_ch = 0.

MISTAKE"The probe polarization P and the channel polarization p are the same thing."

CORRECTp belongs to the channel and creates µ_S (Eq. 12.25); P belongs to the probe magnet and measures it (Eq. 12.20).

10Explore

Files are in the course folder (course_materials).

Week 9 slides Widget: p-n junction and solar cell Widget: spin valve Colab notebook Week 8 toolkit

Mini-investigations (15 min each)

In the p-n widget, record V_OC for T₀ = 600, 1000, 2000 and 6000 K and plot it against 1 − T/T₀. Slope?
Set K/γ = 10 with sunlight: does µ_A − µ_B still equal qV at the operating point? Why not?
In the spin-valve widget, with P = 0.6 and R_ch = 2r, find the spin-flip length at which MR has fallen to half of its no-flip value.

11Can I…?

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

explain why a device may need several quasi-Fermi levelswrite Eq. 12.6 and explain D_AB/D_BAderive the ideal diode law from X − 1derive qV_OC = ħω(1 − T/T₀) and explain the sign of I_scexplain why the contacts must be asymmetriccompute R_P, R_AP and MRfind µ_S from a probe-reversal measurementexplain spin-momentum locking and µ_S = (q/2G_B)pI

12Looking ahead: Week 10 and circular session C5

Thermoelectricity and phonons (Datta Ch. 13–14), then C5: a Colab lab on spin valves, p-n junctions and thermoelectric optimisation.

Two contacts at different temperatures and the same µ: which way do electrons flow if the channel only conducts above µ? Below µ?
Why is a good thermoelectric not simply a good conductor?
Heat is carried by electrons and by lattice vibrations. Which would you like to block, and how?

p-n junctions, solar cells, spin valves and spin potentials

1This week's plan

By the end of the week you should be able to

2Key ideas in five sentences

3Equation sheet

4Physics refresher: p-n junctions and magnets

5Glossary

6Calculators

Two-level diode and solar cell

Spin valve (Eqs. 12.13–12.16)

Spin potentials: probe and spin-momentum locking

7Self-check quiz

8Practice problems

9Common mistakes

10Explore

Mini-investigations (15 min each)

11Can I…?

12Looking ahead: Week 10 and circular session C5