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where
i
is the sum over the cells,
E
k
is the kink energy,
f
i
is the geometric factor that specifies the electronic falloff with distance between
cells,
x
i
is the polarisation of the
i
th neighbour cell,
γ
is the tunnelling energy between two logic states of a cell which is controlled
by the clock,
G
is the total kink energy caused by neighbouring polarized cells.
The energy of a QCA cell at each clock cycle is the expected value of the
Hamiltonian, which is given by [
22
,
30
]:
E
=
<H>
=
2
Γ
·
λ
,
(3)
where
is the reduced Planck constant,
λ
is the coherence vector and
Γ
is the
three dimensional energy vector:
1
Γ
=
[
−
2
γ,
0
,G
]
.
(4)
The equation for the instantaneous power can then be derived as follows:
d
d
.
P
total
=
d
E
d
t
=
2
d
d
t
)=
2
d
t
+
2
Γ
·
d
t
(
Γ
·
λ
·
λ
(5)
The first term represents the power in and out of the clock and the inter-cell
power flow. It is the second term, namely
P
diss
, that refers to the dissipated
power. Therefore, the power dissipation of a QCA cell can be calculated as:
d
.
P
diss
=
2
Γ
·
d
t
(6)
Energy dissipated in one clock cycle
T
c
=[
−D, D
] can be computed by [
30
]:
[
D
−D
Γ
·
D
−D
λ
·
E
diss
=
2
d
d
t
d
t
=
2
d
d
t
]
−D
−
Γ
·
λ
d
t
D
−D
λ
·
=
2
d
d
t
Γ
+
·
λ
+
−
Γ
−
·
λ
−
−
d
t
,
(7)
where
Γ
−
and
Γ
+
are used to denote
Γ
(
−D
)and
Γ
(
D
) and the same notation
is used for
is
a maximum under non-adiabatic switching. By modelling a step change with a
delta function, the upper bound of power dissipation for a QCA cell is derived
as follows [
30
]:
λ
. The maximum power will be dissipated when the change of
Γ
tanh
|
Γ
+
|
kT
+
Γ
−
|
Γ
−
|
tanh
|
Γ
−
|
kT
,
P
diss
=
E
diss
T
c
2
T
c
Γ
+
×
−
Γ
+
|
Γ
+
|
<
(8)
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