Lecture 7: Resistance: Ballistic to Diffusive
Transcript of Lecture 7: Resistance: Ballistic to Diffusive
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