Charge quantization and
charging energy
Charging effects and
Coulomb blockade
¾ Charge quantization and charging energy
Isolated island: charge is quantized, Q=ne
charging energy
2
¾ Single-electron transistor
¾ Coulomb blockade
= addition energy E
¾ Master equations for sequential tunneling
¾ Tunnel rates
=
e 2
n = EC n 2
2C
Island close to a gate:
¾ Coulomb staircase
(the gate shifts potential – induces charge)
¾ Classical charge relaxation
¾ Tunnel rates with dissipation, P(E)
CV ⎞
⎛
E = EC ⎜ N − g g ⎟
e ⎠
⎝
¾ Coupling to one mode
¾ Ohmic environment
2
⎡C V 1 ⎤
N =⎢ g g + ⎥
2⎦
⎣ e
¾ General properties of current with dissipation
How many electrons?
Coulomb blockade
Tunnel junctions:
0.6
E
I
4
0.5
2
0.4
N
0.3
0.2
0
-2
0.1
-4
0
-4
-3
-2
-1
0
1
2
3
4
-4
-3
-2
-1
0
1
2
3
4
CgVg/e
CgVg/e
V
VL
Single-electron transistor:
1
VCΣ /e
I=0
Energy required to put an electron inside the island
Vg
VL
-2
Requirement on resistance
VR
Capacitances to source, drain, gate,
island
CL,
drain
∆t = RC typical time to charge/discharge of an island
∆E ∆t = (e2 / C ) RC > = Heisenberg uncertainty relation
=
R 2 = RQ ≅ 25kΩ
e
Requirement on temperature/voltage
k BT , eV ≤ EC
Qg /e
Single- electron transistor:
Equivalent capacitor circuit
Conditions for CB
source
1
-1
EC (n, Qg ) =
(ne − Qg )
2CΣ
CL
n
CR, Cg
CR
Cg
Vg
2
VL
VR
where CΣ = C L + C R + C g , Qg = C gVg + C RVR + C LVL
We ADD one electron: the change in charging energy is
EC (n + 1) − EC (n) = ∆EC (n, Qg ) = (n + 1/ 2 − Qg / e)
e2
CΣ
1
Single- electron transistor:
electron transfers
Coulomb diamonds
Where the transport is blocked? (For N=0 extra electrons)
VL
Given N: four possible
electron transfers
N
VR
assuming
CL = CR ; VL = −VR = V / 2
V
source
island drain
1: from the left : N → N + 1: ∆EFL ( N ) = E ( N + 1) − E ( N ) − eVL
2 : to the left : N → N − 1: ∆ETL ( N ) = E ( N − 1) − E ( N ) + eVL
3 : from the right : N → N + 1: ∆EFR ( N ) = E ( N + 1) − E ( N ) − eVR
FL : ∆EFL ( N ) = 2EC (1/ 2 + Qg / e) − eV / 2 > 0
TL : ∆ETL ( N ) = 2EC (−1/ 2 + Qg / e) + eV / 2 > 0
TL
FL
HERE
FR : ∆EFR ( N ) = 2EC (1/ 2 + Qg / e) + eV / 2 > 0
FR
Qg/e
TR
TR : ∆ETR ( N ) = 2EC (−1/ 2 + Qg / e) − eV / 2 > 0
4 : to the right : N → N − 1: ∆ETR ( N ) = E ( N − 1) − E ( N ) + eVR
Finite temperature : all transfers are allowed
T=0: a transfer can only happen if ∆E < 0
For N=1,2 and so on? The region is shifted by Qg=e,2e,..
Coulomb diamonds
⎧⎪
⎨
⎪⎩
1
-2
I = 0 for
Qg
e
−n−
VCΣ /e
1 V CΣ
>
2
2e
Master Equation
Charging STATE is given by n
T=0
Quantum coherence between different states is lost
pn(t) = probability to find the system in state n
I=0
Obeys Master Equation:
1
-1
d
pn (t ) = −(Γ F (n) + ΓT ( n)) pn (t ) +
dt
Γ F (n − 1) pn −1 (t ) + ΓT (n + 1) pn +1 (t )
Qg /e
Γ F = Γ FL + Γ FR ; ΓT = ΓTL + ΓTR
Master Equation: current
Goal: stationary current
If we know
probabilities
pn
I=IL=IR
Tunneling in metals (No CB)
• Tunnel barrier
we get currents
I L = e∑ [Γ FL (n) − ΓTL (n)]pn
• Tunnel current: from occupied to empty
transmitted
n
I R = e∑ [ΓTR (n) − Γ FR (n)]pn
n
Danger! It is not
the instant current
.
Current is transferred
in shots = tunneling
events
coming
I = GTV
• Rate: Γ =
I GT
=
eV
e e2
Width of the
available
energy strip
2
Tunneling and Coulomb
blockade
Tunneling rates
• Now the same with Coulomb blockade
• Electrostatic energy must be paid
Width is
reduced!
Tunneling rate through resistance R
With el. energy change ∆E
Fermi Golden rule
∞
Γ ( n) =
∞
1
∫ d ε k −∞∫ d ε q f (ε k )[1 − f (ε q )] δ (ε q − ε k − ∆E )
e 2 R −∞
energy conservation
yielding
eV < ∆EC : Γ = 0
eV > ∆EC : Γ =
• blockade
GT
(eV − ∆EC )
e2
Γ=
1
e2 R
∆E
e
∆E
k BT
Γ
−1
∆E
Coulomb oscillations
good
oscillations
Coulomb oscillations: V small,
Conductance can be finite
∆V g
∆Vg =
w
Gate voltage
e
Cg
C
ew Σ k BT
Cg
e2 / CΣ >> k BT
T1>T2
Conductance
Conductance
Coulomb oscillations
T1
T2
Gate voltage
G∞ = (1 / GL + 1 / GR ) −1
peak maximum Gmax
independent of T
G
δ / k BT
=
G∞ 2sinh(δ / k BT )
RL= RR= RT
“High” Voltage conductance
δ = e(C g / CΣ ) Vg ,res − Vg Energy distance to the center
of the peak
Junction in the
environment
Coulomb staircase
Nonlinear transport:
upon increasing V more charging states
become available
I
R,C
Contact is characterized by
resistance and capacitance
Environment represented by impedance:
Ohm’s law: I=GV
−2 ≤ N ≤ 2
Vth ,n = (2n + 1)e / CΣ
−1 ≤ N ≤ 1
1
N =0
Qg =0
3
5
VCΣ
e
=
Z (ω )
…
Z (ω )
V
External voltage source
3
Classical charge relaxation
Z (ω ) = Rext Rext R
Purely resistive impedance:
R
Average charge of the junction:
Put the charge
C
Q = CV
Classical charge relaxation
Instantaneous charge:
R
Solution:
Q (t = 0) = Q0
C
and look at the charge dynamics:
Rext
Voltage drop at the junction:
V j = V − RextQ
Classical LC-circuit
Inductive impedance:
R
Z = iω L
Hamiltonian:
H=
C
L
V
Q2 Φ2
+
2C 2 L
Charge
Quantization of an LC-circuit
H=
Flux
ω0 = ( LC ) −1/ 2
wi→ f =
2π
=
2
f Vˆ i δ ( Ei − E f )
(probability of transition from i to f per unit time)
Total rate for tunneling from left to right:
Γ→ = ∑ f i (1 − f f ) wi→ f
i, f
=
1
dE dE ' f ( E )[1 − f ( E ')]δ ( E − E ')
e2 R ∫
i and f – states of the electron in the contact
ˆ2
Qˆ 2 Φ
+
2C 2 L
Flux and charge: conjugate operators
L
Commutation relation:
V
Equations of motion:
i=
ˆ ˆ⎤
⎡Φ
⎣ ,Q⎦ = − c
i
ˆ i ⎡ ˆ ˆ ⎤
Qˆ = ⎡⎣ Hˆ , Qˆ ⎤⎦ Φ
= ⎣H , Φ⎦
=
=
have the same form as for the classical LC-circuit
Tunnel rates in the
environment
Fermi golden rule:
CRext
Z t – effective impedance seen by the junction
C
Q = I = Φ / Lc
= − cV = − cQ / C
Φ
Harmonic oscillator with the frequency
CV with the typical time scale
Generalization to the impedance:
1
Q (t ) = CV + (Q0 − CV )CZ t (t ) Z t (ω ) = Z −1 (ω ) + iω C
R
Equations of motion:
Q (t ) = CV + (Q0 − CV )e − t / CRext
V
I = Q
Vext = Rext I
V j = V − RextQ
Charge relaxes to the “equilibrium” value
Rext
Voltage drop at the external resistance:
V
Q = CV j
Tunnel rates in the
environment
In environment: i and f – states of the
electron in the contact + environment
Total energy is conserved.
Energy of an electron is not conserved: Electrons can
exchange energy with the environment
Γ→ =
P( E )
1
dE dE ' f ( E )[1 − f ( E ')] P( E − E ')
e2 R ∫
-describes energy exchange between an electron
and the environment; related to the impedance Z.
4
Coupling to a single-mode
oscillator
One oscillator; frequency
ω0 =
Caldeira-Leggett model
Environment is represented as a set of harmonic oscillators
1
LC
P
Coupling to one mode:
An electron can lose or gain n oscillator quanta:
E
∞
P ( E ) = ∑ pnδ ( E − n=ω0 )
V
Quanta are emitted
independently
n =0
P
Two modes:
E
pn =
Poisson distribution for p’s
ρ
n
n!
e− ρ ; ρ =
Many modes:
EC
=ω0
P
Transition to continuous
frequency spectrum
Properties of P(E)
Current
∞
I = e ( Γ→ − Γ← )
∫ P( E )dE = 1
Probability normalization:
E
At zero temperature becomes
−∞
Detailed balance:
Low impedance:
P( E ) = δ ( E ) +
I=
P ( − E ) = e − E / k BT P ( E )
(it is easier to emit than to absorb)
2
1
GQ Re Z t ( E ), Z t ( E ) =
E
iEC / = + Z −1 ( E / =)
Zero quanta
High impedance:
One quantum
Examples:
Low impedance:
1
eR
eV
∫ dE (eV − E ) P( E )
0
I =V / R
I
High impedance:
P( E ) = δ ( E − EC )
RI = (V − e / 2C )Θ(V − e / 2C )
e / 2C
V
Ohmic environment
1
Z (ω ) = Rext- ideal Ohmic resistance Re Z t (ω ) = g (1 + ω 2 R 2 C 2 )
ext
Low energies:
P( E ) ∝ E
2 / g −1
; g = 1/ Rext GQ I ∝ V 2 / g +1
High energies:
P( E ) =
2 g EC2
π 2 E3
RI = V −
e
g e2 1
+ 2
2C π 4C 2 V
I
e / 2C
V
5