Download Molecular Electronic Devices

Capacitance
µ1
s
µ2
1
1 /
H + U

 1
2 /

2
2



We will now spend time on the potential matrix U
1
Quantum effects on Capacitance
A metal sphere has a capacitance that describes its capacity
to hold charge. This is determined by electrostatics (Coulomb/
Gauss law), and thus depends only on the geometry (eg. radius)
CE = 4pe0R, with single electron charging energy U0 = q2/CE
What we will see in this chapter is that quantum mechanics can
further limit this ‘capacity’
We learned that small systems have discrete levels described
by a characteristic density of states. Even if electrostatics
allows charge addition (eg. large CE), it may be difficult to do
this if there are no levels near EF
This means there’s also a ‘quantum capacitance’ proportional
to the device density of states, CQ ~ DOS, which affects charging
2
Role of charging and DOS
U = UL + U0(N-N0), U0 = q2/CE
N = dED(E-U)f(E-)
N0 = dED(E-UN)f(E-)
N – N0 ≈ [CQ/q2](U-UN), CQ = q2D0
 U = (ULCE + UNCQ)/(CE + CQ)
3
Pictorially
U = (ULCE + UNCQ)/(CE + CQ)
U
UL
Applied
Laplace
potential
(e.g. Gate)
UN
CE
Electrostatic
capacitance
CQ
Neutrality
potential
quantum
capacitance
Next we’ll see how this affects the potential
4
Role of charging and DOS
V=0
V=0
V=1
What is channel potential?
U = (ULCE + UNCQ)/(CE + CQ)
Insulators, CE >> CQ, U = UL (levels slip with drain bias, 0.5)
Metals, CE << CQ, U = UN (levels tied with contact, 0)
5
S-D Current ID
Transistor
VC = VG[CE/(CE+CQ)]
S
D
Gate Voltage
OFF (Subthreshold)
CQ=0, U = -qVG
N = dED(E-U)f(E-) = dED(E)f(E-+U) = dED(E)f(E--qVG)
≈ N0exp(qVG/kT)  log10(N/N0) = (qVG/2.3kT)
6
Transistor
VC = VG[CE/(CE+CQ)]
Pdiss = 0.5 CVD2Nf
C, N, f fixed
VD decreased if we decrease subthreshold swing
N and thus I changes by one decade for
60 meV of applied gate voltage
7
The infamous SAMFET
20 meV/decade !
tox = 150 nm
L = 1nm !
Ref: Schon, et al., nature Vol. 413, pp. 713 2001
N and thus I changes by one decade for
60 meV of applied gate voltage
8
Charging  how to account for it?
e4
H=
e3
e2
e1
Device with
a few levels
9
How are these filled at equilibrium?

N=
f(e4-)
f(e3-)
f(e2-)
f(e1-)
Device with
a few levels
10
How are these filled at equilibrium?
0

N=
0
1
1
Device with
a few levels
At zero temperature
But this is in eigenspace where levels act independently
11
How does it look like in real space?
-t e -t

H=
-t e -t
-t e -t
In the real space basis, H is a full matrix with off-diagonal terms
N must be a full matrix too once levels get mixed up !
12
What does a full matrix N mean?
It means that in real space the electron occupancy at a given point is a
coherent mixture of many levels.
An incoming electron can jointly occupy multiple levels with definite
phase differences, like a laser source irradiating multiple slits, and
create quantum interference.
We can no longer count electronic charge by adding probabilities of
occupancy. Much like interference studies, we need to add probability
amplitudes and then square them. The cross-terms that arise are the
off-diagonal entries of your full matrix N, and represent interferences.
13
Does it matter?
Yes! There are many examples of quantum interference
Magnetoconductance
oscillations in an antidot
lattice (Nihey et al, PRL 51, ‘95)
Atom laser (Ketterle group)
Aharonov-Bohm interference
in a nanotube
(Dai group, PRL 93, 2004)
Fano interference between
a channel and a quantum dot
14
Quantum computation schemes
thrive on interference!
Scheme of a Si-based
quantum computer
Possibly the world’s first quantum
computer?
Questions remain….
(picture from D-wave website)
15
Small devices are better candidates for
observing quantum effects
We must thus learn to transform matrices, paying
special attention to their off-diagonal components
16
Handling quantum interference
The off-diagonal terms in the matrix carry information
about quantum interference
If we can calculate the matrix transformation rules
correctly, we should naturally get quantum interference
(see slide 34).
17
How do off-diagonal terms
describe interference?
Current: difference between left and right going states
eikx
e-ikx
In basis of free k waves,
currents are ħk/m, and
occupancies fk
Current = ∑kħk/m(fk – f-k)
18
How do off-diagonal terms
describe interference?
Now let’s choose different bases by
superposing eikx states
sinkx
coskx
Each state is a standing wave.
Current carried by each basis is zero!
But observable quantities should be basis
independent!
How would we recover the non-zero current?
Answer: we deal with a current matrix whose diagonal terms are now
zero. But the current is carried by the off-diagonal elements through
interference between sin(kx) and cos(kx) states !
See Slide 35!
19
So we need to learn how
to change representations
between basis sets
20
How do we change representations?
u1
u2
u3
u4
Say 4 atoms with periodic bcs
The real space bases are shown above
The H in this representation 
H=
2t
-t
-t
2t
0
-t
-t
0
0
-t
-t
0
2t
-t
-t
2t
21
Let’s diagonalize this and find its eigenspectrum
Since 1-D periodic solid, use
principle of bandstructure
Yk(n) = Y0einka
yk =
y0eika
y0e2ika
(site n=1)
y0e3ika
y0e4ika
(site n=3)
(site n=2)
(site n=4)
Each distinct eigenvector corresponds to a different k quantized by
the 4-site periodicity, k = (2p/4a)(0,1,2,3).
y0 is fixed by normalization, yk+yk = 1  y0 = 1/2
2t
The 4 corresponding eigenvalues are
-t
ek = 2t(1-coska), with the previous ks, k = 0, p/2a,
H= 0
-t
Thus
e1 = 0, e2 = 2t, e3 = 4t, e4 = 2t
-t
0
2t 3p/2a
-t
p/a,
-t
2t
0
-t
-t
0
-t
2t
22
Let’s write down these solutions
y4
y3
y2
y1
The eigenectors of H are
superpositions of basis sets.
Let us solve for this and the
corresponding eigenvalues
ya = jujfja
Note that we took our original
equi-energy y2, y4 and recombined
them as (y2y4) to get our new y2,y4
V=
y1
y2
y3
y4
1/2
1/2
1/√2
0
-1/2
1/2
0
1/√2
1/2
1/2
-1/√2 -1/2
0
1/2
0
-1/√2
We also calculated the coefficients for
the eigenvectors in the previous slide.
Let us store the coefficients a describing
y in terms of f as a matrix V)
Hl =
e4
0
0
e3
0
0
0
0
0
0
0
0
e2
0
0
e1
From the previous slide,
e1 = 0, e2 = 2t, e3 = 4t, e4 = 2t
23
How to change to the diagonal eigen-representation
?
ya(r) = juj(r)fja
[Hl]ab = y*a(x)Hopyb(x)dx
= ij f*iafjbu*i(x)Hopuj(x)dx
= ij f*iafjbHij
= ij f†aiHijfjb
= [V†HV]ab
Thus Hl = V†HV
--> Check that previous page’s V and H give Hl
24
How to go back to original representation?
Use orthonormality and completeness
V†V = VV† = I (Unitary Transformation)
Thus H = VHlV†
Now, in the eigen-basis,
the number operator:
So in original atomic basis
r = VrlV†
rl =
[r] is called the density matrix
f(e4-)
f(e3-)
f(e2-)
f(e1-)
25
How to go back to original representation?
Hl (the eigenvalue matrix) transfers to H using
H = VHlV†
How would Hl2 transfer?
VHl2V† = VHl(V†V)HlV†
= (VHlV†)(VHlV†)
= H2
In general, any function F(Hl) transforms as
F(H) = VF(Hl)V†
26
Putting it another way
r = f(H-I) = [I + exp((H-I)/kT)]-1
Density matrix is the Fermi function of the Hamiltonian !
[V,D]=eig(H);
rho = 1./[1 + exp((diag(D)-mu)/kT)];
rho = V * diag(rho)* V’
27
What does it mean?
Let’s write out the real-space density matrix
r(r,r’) = ij u*i(r)rijuj(r’)
Thus electron density n(r) = r(r,r)
= ij u*i(r)rijuj(r)
But earlier in chapter three, we had assumed
n(r) = i|ui(r)|2fi
What did we miss?
28
We missed quantum interference
We’re missing cross terms in ui(r) and uj(r)
Can’t expect n to be diagonal in both real and
eigen-representations.
As long as wavefunction is a superposition
ya(r) = juj(r)fja
We expect cross terms in the coefficients !
(Called “Coherence” terms)
Then what’s the justification of dropping them in Ch. 3 ?
29
When can we ignore quantum interference?
If we’re dealing with classical states that vary fast
enough to wash out the interference terms
(eg. high temperature with phase-breaking scattering
that destroy coherence)
Only |uj(r)|2 terms survive the averaging since their
phases cancel, and rii = fi. We then get back our old result.
Cross terms u*i(r)uj(r) have a net rapidly varying phase,
so rij coherence terms stochastically average to zero
Random Phase Approximation (RPA)
30
When can we ignore quantum interference?
Semiclassical theories deal with Fermi function occupancies
alone, ie, only diagonal terms of the density matrix
(eg. Drift-Diffusion equation, Boltzmann transport equation)
 Worked for old devices
But we will deal with the full quantum transport by
retaining the full density matrix including its off-diagonal
coherent components
 Needed for present day and future devices where phasebreaking vibrations take longer than electron’s transit time
31
through device, and coherence becomes important
Total electron number
Individual electron numbers at various grid points require
us to go to real-space and observe the diagonal components
of the density matrix
But total electron number depends on the sum of these
diagonal components, ie, the trace of the density matrix.
Turns out the Trace is invariant under Unitary transformation
(like the determinant), we could have found the trace in
either representation!
N = Trace(r) = Trace(rl)
32
Example
Remove periodicity at ends of
1-D lattice [H].
Same process, except restore
periodicity and add large onsite
term at desired site
Then find trace(r), where
r = f(H-I)
Slide 27 shows how to implement this
Charge distribution of
well with walls
Charge distribution of
well with no boundaries
but with an impurity 33
Example
Friedel oscillations
Charge distribution of
well with walls
Charge distribution of
well with no boundaries
but with an impurity 34
Let’s implement it for the
earlier problem
eikx
V=
e-ikx
-i
1/√2 1
In this basis,
Jop=
ħk/m 0
0
-ħk/m
r =
fk
0
sinkx
Basis transformation
0
f-k
Charge N = Tr (r) = fk + f-k
Current = Tr(Jopr) = ∑kħk/m(fk – f-k)
i
1
coskx
Transform VrV+ etc to new basis
Jop=
r = 1/2
0
-iħk/m
iħk/m
0
fk+f-k
i(fk-f-k)
-i(f k-f-k)
f k+f-k
Same charge. Same current,
but in the off-diag terms! 35
What does this unitary transformation
physically mean?
36
We have a matrix in a given basis set (‘grid’). The unitary transformation
allows us to re-express it in terms of another basis set
Say we have a function of (x,y) whose coefficients are stored as a matrix [M]
A
f = Ax2 + By2 +2Cxy = [x y]
C
x
C
B
y
M
Basis transformation matrix
V=
y
y’
x’
q
q
{f} = (x,y)
{y} = (x ’,y ’)
cosq
sinq
-sinq
cosq
Check that
VV+ = V+V = I
x
y
= V+
x’
y’
x
Then new coefficients of the function
in terms of x ’, y ’ are M’ = V+MV
37
Now that we understand charge density
matrix and how to express it in various
basis sets (through unitary rotations),
let’s see how it influences the capacitance
and thus the potential
38
The infamous SAMFET
20 meV/decade !
tox = 150 nm
L = 1nm !
Ref: Schon, et al., nature Vol. 413, pp. 713 2001
N and thus I changes by one decade for
60 meV of applied gate voltage
39
How’s that possible?
40
Can we have a low
subthreshold swing?
Tunneling transistor
– Band filter like operation
Ghosh, Rakshit, Datta
(Nanoletters, 2004)
(Sconf)min=2.3(kBT/e).(etox/)
Hodgkin and Huxley, J. Physiol. 116, 449 (1952a)
J Appenzeller et al, PRL ‘04
Subthreshold slope = (60/Z) mV/decade
41
S-D Current ID
Above
threshold
ON
VC = VG[CE/(CE+CQ)]
S
D
Gate Voltage
CQ > 0
42