ECE-656: Fall 2011

Lecture 7: Resistance:

Ballistic to Diffusive

Professor Mark Lundstrom Electrical and Computer Engineering

Purdue University, West Lafayette, IN USA

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resistors

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Landauer picture

1D, 2D, or 3D resistor

ideal contact ideal contact

Current is positive when it flows into contact 2 (i.e. positive current flows in the -x direction).

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driving “forces” for transport

Differences in occupation, f, produce current.

but differences in temperature also produce differences in f and can, therefore, drive current (thermoelectric effects).

Assumes TL1 = TL2 .

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this week

(next week)

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transport regimes

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outline

This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 United States License. http://creativecommons.org/licenses/by-nc-sa/3.0/us/

1) Review

2) 2D ballistic resistors

3) 2D diffusive resistors

4) Discussion

5) Summary

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an isothermal 2D resistor

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the ballistic conductance

(ballistic)

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T = 0 K ballistic conductance

Resistance is independent of length, L.

Rball =

1M EF( )

h2q2 =

12.8kΩM

“quantized conductance”

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If the number of modes is small, we can simply count them.

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quantized conductance

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B. J. van Wees, H. van Houten, C. W. J. Beenakker, J. G. Williamson, L. P. Kouwenhoven, D. van der Marel, and C. T. Foxon, ‘‘Quantized conductance of point contacts in a two-dimensional electron gas,’’ Phys. Rev. Lett. 60, 848–851,1988.

Conductance is quantized in units of 2q2/h - it increases with W, but not linearly.

The conductance is finite as L 0.

W -->

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for wider resistors, M(E) ~ W

M EF( ) = gVW

2m* EF − EC( )π

M EF( ) = gV

W kF

π

depends on bandstructure!

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2kF

2

2m* = EF − EC( ) How is M related to the sheet carrier density?

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sheet carrier density at TL = 0 K

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2kF

2

2m* = EF − EC( )

kx

ky

kF

nS =πkF

2

2π( )2A× 2 × gV ×

1A

nS = gV

kF2

2π

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M(E) and nS

M EF( ) = gV

W kF

π

M EF( ) =W

2gV nS

πdepends only on nS

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nS = gV

kF2

2π

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example: the channel of an n-MOSFET

nS = 1013 cm-2

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M EF( ) = ?

for a rough estimate, assume TL = 0 K

M EF( ) =W

2gV nS

π W = 2L = 100 nm

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quantum confinement in a Si inversion layer

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z

ener

gy --

>

W0

∞∞z

x

y

En =2n2π 2

2m*

m

* = 0.9m0

mt* = 0.19m0

m* = ?

Si conduction band

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bandstructure effects on quantum confinement

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z

ener

gy --

>

W0

∞∞z

x

ySi conduction band

unprimed ladder: m* = m

* gV = 2ε1

ε3

ε2

primed ladder: m* = mt* gV = 4

′ε1

′ε2

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quasi-2D carriers in the plane

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z

x

ySi conduction band

unprimed ladder: m* = m

* gV = 2

primed ladder: m* = mt* gV = 4

confinement masses:

kx

ky

mDOS* = mt

*

mDOS* = mt

*m

*

D2D = gVmDOS*

π2

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example: the channel of an n-MOSFET

nS = 1013 cm-2

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M EF( ) = ?

Assume TL = 0 K and that only the lowest subband is occupied.

M EF( ) =W

2gV nS

π

W = 2L = 100 nm

gV = 2

M EF( ) = 36

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T = 0 K ballistic conductance

Resistance is independent of length, L.

Rball =

1M EF( )

h2q2 =

12.8kΩM

“quantized conductance”

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If the number of modes is small, we can simply count them.

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conductance in 2D (T > 0K)

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Gball =

2q2

h+

∂∂EF

⎛

⎝⎜⎞

⎠⎟M E( ) f0 E( )dE∫( )

M E( ) =WgV

2m* E − EC( )π

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conductance in 2D (T > 0K)

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Gball =

2q2

h+

∂∂EF

⎛

⎝⎜⎞

⎠⎟M E( ) f0 E( )dE∫( )

M E( ) =WgV

2m* E − EC( )π

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conductance in 2D (T > 0K)

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conductance in 2D (T > 0K)

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M2 D =

2m*kBTL

ππ2

F −1/ 2 ηF( ) = M2 D E − EC = kBTL( ) π2

F −1/ 2 ηF( )

<M> is the no. of modes in the “Fermi window”

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example: the channel of an n-MOSFET

nS = 1013 cm-2

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M EF( ) = ?

Repeat the calculation for TL = 300 K and find <M>.

M EF( ) =W

2gV nS

π= 36 TL = 0K( )

W = 2L = 100 nm

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outline

1) Review

2) 2D ballistic resistors

3) 2D diffusive resistors

4) Discussion

5) Summary

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an isothermal 2D resistor

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(TL1 = TL2)

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the TL = 0 diffusive conductance

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ballistic to diffusive

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outline

1) Review

2) 2D ballistic resistors

3) 2D diffusive resistors

4) Discussion

5) Summary

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example: nanoscale MOSFETs

Question: Do we expect this device to be closer to ballistic or to diffusive?

(Courtesy, Shuji Ikeda, ATDF, Dec. 2007)

RCH ≈ 215Ω-µm

µn ≈ 260 cm2 V-s

nS ≈ 6.7 ×1012 cm-2

L ≈ 60 nm

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mfp in a Si MOSFET

RCH ≈ 215Ω-µm

µn ≈ 260 cm2 V-s

nS ≈ 6.7 ×1012 cm-2

L ≈ 60 nm

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Dn =

kBTL

qµn

(near-equilibrium, MB statistics)

Dn =

υTλ0

2(near-equilibrium, MB statistics, constant mfp)

λ0 =

2 kBTL q( )υT

µn

υT =

2kBTL

πm* = 1.2 ×107 cm/s λ0 ≈ 11 nm << L

diffusive

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40 nm channel length III-V FETs

D. H. Kim and J. A. del Alamo, "Lateral and Vertical Scaling of In0.7Ga0.3As HEMTs for Post-Si-CMOS Logic Applications," IEEE TED, 55, pp. 2546-2553, 2008. Lundstrom 2011

λ0 >> L ballistic

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outline

1) Review

2) 2D ballistic resistors

3) 2D diffusive resistors

4) Discussion

5) Summary

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key points

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1) Conductors display a finite conductance (resistance) - even in the absence of scattering.

2) The resistance is quantized, and the quantum of conductance is 2q2/h.

3) The ballistic resistance sets a lower limit to device resistance and is becoming important in practical, room temperature devices.

4) Transport from the ballistic to diffusive limit is easily treated by using the transmission.

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2D resistor summary

G =

2q2

hT M =

2q2

hλ0

λ0 + LM

M =

π2

WM2 D E − EC = kBTL( )F −1/ 2 ηF( )

ηF = EF − ε1( ) kBTL

(parabolic energy bands) (constant mfp)

M2 D E − EC = kBTL( ) = gV

2m*kBTL

π

TL = 0 K:

M = M EF( )

MB statistics:

M =W h

4υT

nS

kBTL

G =

λ0

λ0 + LGball

R = 1+ λ0 L( )Rball

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questions

1) Review

2) 2D ballistic resistors

3) 2D diffusive resistors

4) Discussion

5) Summary