Conductance Quantization

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© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips Conductance Quantization • One-dimensional ballistic/coherent transport • Landauer theory • The role of contacts • Quantum of electrical and thermal conductance • One-dimensional Wiedemann-Franz law 1

description

Conductance Quantization. One-dimensional ballistic/coherent transport Landauer theory The role of contacts Quantum of electrical and thermal conductance One-dimensional Wiedemann-Franz law. “Ideal” Electrical Resistance in 1-D. Ohm’s Law: R = V/I [ Ω ] - PowerPoint PPT Presentation

Transcript of Conductance Quantization

Page 1: Conductance Quantization

© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 1

Conductance Quantization

• One-dimensional ballistic/coherent transport

• Landauer theory

• The role of contacts

• Quantum of electrical and thermal conductance

• One-dimensional Wiedemann-Franz law

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© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 2

“Ideal” Electrical Resistance in 1-D

• Ohm’s Law: R = V/I [Ω]• Bulk materials, resistivity ρ: R = ρL/A• Nanoscale systems (coherent transport)

– R (G = 1/R) is a global quantity– R cannot be decomposed into subparts, or added up from

pieces

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© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 3

• Remember (net) current Jx ≈ x×n×v where x = q or E

• Let’s focus on charge current flow, for now• Convert to integral over energy, use Fermi distribution

q k k kk

J q g f dk v

Charge & Energy Current Flow in 1-D

q k k k k kk k

J q n q g f v v

E k k k k k k k E k k k kk k k

J E n E g f J E g f dk v v v

Net contribution

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© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 4

Conductance as Transmission

• Two terminals (S and D) with Fermi levels µ1 and µ2

• S and D are big, ideal electron reservoirs, MANY k-modes• Transmission channel has only ONE mode, M = 1

Sµ1

Dµ2

2 ( ) ( )qI f E T E dEh

q k k k kk

J q g f T dk v

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© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 5

Conductance of 1-D Quantum Wire

• Voltage applied is Fermi level separation: qV = µ1 - µ2

• Channel = 1D, ballistic, coherent, no scattering (T=1)

qV

1 dEvdk

01Dk-space

k

x

VI

2I qGV h

k k k kk

I q g f T v dk

1

21( ) ( )qI f E T E dE

h

quantum ofelectrical conductance(per spin per mode)

x2 spingk+ = 1/2π

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Quasi-1D Channel in 2D Structurevan Wees, Phys. Rev. Lett. (1988)

spin

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Quantum Conductance in Nanotubes

• 2x sub-bands in nanotubes, and 2x from spin• “Best” conductance of 4q2/h, or lowest R = 6,453 Ω• In practice we measure higher resistance, due to

scattering, defects, imperfect contacts (Schottky barriers)

S (Pd) D (Pd)SiO2

CNT

G (Si)

Javey et al., Phys. Rev. Lett. (2004)

L = 60 nmVDS = 1 mV

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Finite Temperatures

• Electrons in leads according to Fermi-Dirac distribution

• Conductance with n channels, at finite temperature T:

• At even higher T: “usual” incoherent transport (dephasing due to inelastic scattering, phonons, etc.)

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© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 9

Where Is the Resistance?S. Datta, “Electronic Transport in Mesoscopic Systems” (1995)

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© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 10

Multiple Barriers, Coherent Transport

• Perfect transmission through resonant, quasi-bound states:

• Coherent, resonant transport• L < LΦ (phase-breaking length);

electron is truly a wave

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© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 11

2 21 2

2 22 21 2

Resistance 1 12 2

r rh h Le et t

Multiple Barriers, Incoherent Transport

• Total transmission (no interference term):

• Resistance (scatterers in series):

• Ohmic addition of resistances from independent scatterers

• L > LΦ (phase-breaking length); electron phase gets randomized at, or between scattering sites

2 21 2

total 2 21 21

t tT

r r

average mean

free path; rememberMatthiessen’s rule!

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Where Is the Power (I2R) Dissipated?• Consider, e.g., a single nanotube• Case I: L << Λ

R ~ h/4e2 ~ 6.5 kΩ

Power I2R ?

• Case II: L >> Λ

R ~ h/4e2(1 + L/Λ)

Power I2R ?

• Remember. . . . . .

1 1 1 1 1 1

op phon ac phon imp def e e

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1D Wiedemann-Franz Law (WFL)• Does the WFL hold in 1D? YES• 1D ballistic electrons carry energy too, what is their

equivalent thermal conductance?

2 222 2

23 3B

th eB eG L T k kT

hhT

e

(x2 if electron spin included)

Greiner, Phys. Rev. Lett. (1997)

0.28thG nW/K at 300 K

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Phonon Quantum Thermal Conductance• Same thermal conductance quantum, irrespective of the

carrier statistics (Fermi-Dirac vs. Bose-Einstein)

>> syms x;>> int(x^2*exp(x)/(exp(x)+1)^2,0,Inf)ans =1/6*pi^2

Matlab tip:

Phonon Gth measurement inGaAs bridge at T < 1 KSchwab, Nature (2000)

SWNT

suspended

over trench

Pt

Pt 1 μm Single nanotube Gth=2.4 nW/K at T=300K

Pop, Nano Lett. (2006)

nW/K at 300 K2 2

0.283t

Bh

k TGh

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Electrical vs. Thermal Conductance G0

• Electrical experiments steps in the conductance (not observed in thermal experiments)

• In electrical experiments the chemical potential (Fermi level) and temperature can be independently varied– Consequently, at low-T the sharp edge of the Fermi-Dirac

function can be swept through 1-D modes– Electrical (electron) conductance quantum: G0 = (dIe/dV)|low dV

• In thermal (phonon) experiments only the temperature can be swept– The broader Bose-Einstein distribution smears out all features

except the lowest lying modes at low temperatures– Thermal (phonon) conductance quantum: G0 = (dQth/dT) |low dT

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• Single energy barrier – how do you get across?

• Double barrier: transmission through quasi-bound (QB) states

• Generally, need λ ~ L ≤ LΦ (phase-breaking length)

Back to the Quantum-Coherent Regime

EQBEQB

thermionic emission

tunneling or reflection

fFD(E)E

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Wentzel-Kramers-Brillouin (WKB)

• Assume smoothly varying potential barrier, no reflections

tunneling only

fFD(Ex)Ex

2.

20.

exp 2 ( )L

trans

incid

T k x dx

A BE||

#incident states ( ) 1A B A A x B BJ f g T E f g dE

k(x) depends onenergy dispersion

*

2 3 ( )2A B B A A B A BqmJ J J f f g g T E dE

E.g. in 3D, the net current is:

0 L

Fancier version ofLandauer formula!

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© 2010 Eric Pop, UIUC ECE 598EP: Hot Chips 18

Band-to-Band Tunneling

• Assuming parabolic energy dispersion E(k) = ħ2k2/2m*

• E.g. band-to-band (Zener) tunnelingin silicon diode

* 3/24 2( ) exp

3x x

x

m ET E

q F

F = electric field

3 * 3/2*

3 2

4 22 exp4 3

eff x GBB

G

q FV m EmJE q F

See, e.g. Kane, J. Appl. Phys. 32, 83 (1961)